Statistical Mechanics: A Set Of Lectures (Advanced Books Classics)

Frontiers in Physics

DAVID PINES, Editor Volumes in Preparation: E. R. Caianiello

COMBINATORICS AND RENORMALIZA TlON IN QUANTUM FIELD THEORY

STATISTICAL MECHANICS

R. P. Feynman

A SET OF LECTURES

PHOTON-HADRON INTERACTIONS G.

E. Pake and T. L. Estle

THE PHYSICAL PRINCIPLES OF ELECTRON PARAMAGNETIC RESONANCE. Second Edition, completely revised, reset, and enlarged

R.

P. FEYNMAN

California Institute of Technology

Notes taken by

R. Kikuchi and H. A. Feiveson Hughes Aircraft Corporation Edited by

Jacob Shaham University of Illinois, Urbana

THE BENJAMIN/CUMMINGS PUBLISHING COMPANY, INC.

ADVANCED BOOK PROGRAM Rea ding , Massachu setts

London · Amsterdam. Don Mills, Ontario · Sydney · Tokyo

CONTENTS

Statistical Mechanics : A Set of Lectures

First printing, 1972 Second printing, 1973 Third printing, 1974 Fourth printing, 1976 Fifth printing, 1979 Sixth printing, 1981 Seventh printing, 1982

International Standard Book Numbers Clothboun d : 0-8053-2508-5 Paperbound: 0-8053-2509-3 Library of Congress Catalog Card Number: 72-1769

Chapter

1.1 Chapter

Printed in the United States of America ISBN:

0-8053-2509-3

IJKLMNOP-MA-89876

2

Xl

Acknowledgments

xiii

Introduction to Statistical Mechanics

The Partition Function Density Matrices

2.1 Introduction to Density Matrices 2.2 Additional Properties of the Density Matrix . 2.3 Density Matrix in Statistical Mechanics 2.4 Density Matrix for a One-Dimensional Free Particle 2.5 Linear Harmonic Oscillator 2.6 Anharmonic Oscillator . 2.7 Wigner's Function 2.8 Symmetrized Density Matrix for N Particles 2.9 Density Submatrix 2.10 Perturbation Expansion of the Density Matrix 2.11 Proof that F :S Fo + (H Ho>o -

Chapter

3

3.1 3.2 3.3 3.4 3.5

Copyright © 1972 by W. A. Benjamin, Inc. Philippines copyright 1972 by W. A. Benjamin, Inc. Published simultaneously in Canada. All rights reserved. No part of this publication may be reproduc�d, stored i? a retrieval system, or transmitted, in any form or by any means, electromc, mechamcal, p�oto­ copying, recording, or otherwise, without the prior written permission of the publisher, W. A. Benjamin, Inc. , Advanced Book Program, Reading, Massachusetts 01867, U.S.A.

1

Editor's Foreword

Chapter

4 4.1 4.2 4.3 4.4

39 44 47 48 49 53 58 60 64 66 67

Path Integrals

Path Integral Formation of the Density Matrix Calculation of Path Integrals Path Integrals by Perturbation Expansion Variational Principle for the Path Integral An Application of the Variation Theorem

72 78 84 86 88

Classical System of N Particles

Introduction The Second Virial Coefficient Mayer Cluster Expansion Radial Distribution Function vii

97

100

105 111

viii

Contents

4.5 Thermodynamic Functions . 4.6 The Born-Green Equation for n2 4.7 One-Dimensional Gas 4.8 One-Dimensional Gas with Potential of the Form e - 1xl 4.9 Brief Discussion of Condensation Chapter

5

6

6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9 6.10 6.11 6.12 Chapter

7

8

127 130 131 136 149

Creation and Annihilation Operators

A Simple Mathematical Problem The Linear Harmonic Oscillator An Anharmonic Oscillator . Systems of Harmonic Oscillators Phonons Field Quantization Systems of Indistinguishable Particles . The Hamiltonian and Other Operators Ground State for a Fermion System Hamiltonian for a Phonon-Electron System Photon-Electron Interactions Feynman Diagrams

151 154 156 157 159 162 167 176 183 185 190 192

Spin Waves

7.1 Spin-Spin Interactions 7.2 The Pauli Spin Algebra . 7.3 Spin Wave in a Lattice . 7.4 Semiclassical Interpretation of Spin Wave 7.5 Two Spin Waves . 7.6 Two Spin Waves (Rigorous Treatment) 7.7 Scattering of Two Spin Waves . 7.8 Non-Orthogonality 7.9 Operator Method 7.10 Scattering of Spin Waves-Oscillator Analog Chapter

113 115 117 120 125

Chapter

198 200 202 206 207 209 212 215 217 218

Polaron Problem

8.1 Introduction . 8.2 Perturbation Treatment of the Polaron Problem. 8.3 Formulation for the Variational Treatment 8.4 The Variational Treatment . 8.5 Effective Mass

221 225 231 234 241

9 9.1 9.2 9.3 9.4 9.5 9. 6 9.7 9.8 9.9

Order-Disorder Theory

5.1 Introducti.on . 5.2 Order-Disorder in One-Dimension . 5.3 Approximate Methods for Two Dimensions . 5.4 The Onsager Problem 5.5 Miscellaneous Comments Chapter

Contents

Chapter

10

10.1 10.2 10.3 10.4 10.5 W.6 10.7 10.8 10.9 10.10 10.11 10.12 Chapter

11

11.1 11.2 11.3 11.4 11.5 11.6 11.7 11.8 11.9 11.10

ix

Electron Gas in a Metal

Introduction: The State Function rp Sound Waves . Calculation of P(R) Correlation Energy Plasma Oscillation Random Phase Approximation . Variational Approach . . . . Correlation Energy and Feynman Diagrams . H igher-Order Perturbation .

242 244 246 248 249 252 254 255 262

Superconductivity

Experimental Results and Early Theory Setting Up the Hamiltonian A Helpful Theorem . Ground State of a Superconductor Ground State of a Superconductor (continued) Excitations Finite Temperatures . ' Real Test of Existence of Pair Stat�s �nd En�rg; G�P Superconductor with Current . . . . . . Current Versus Field Current at a Finite Temperature Another Point of View .

265 269 273 274 277 279 281 285 290 293 298 303

Superfluidity

Introduction: Nature of Transition Superfiuidity�An Early Approach Intuitive Derivation of Wave Functio � s: Phonons and Rotons Rotons Critical Velocity . Irrotational Superfiuid Flow Rotational of the Superfiuid A Reasoning Leading to Vortex Lines The A Transition in Liquid Helium Index

G r�un d St�te :

312 319 321 326 330 334 335 337 339 343 351

EDITOR'S FOREWORD

The problem of communicating in a coherent fashion the recent developments in the most exciting and active fields of physics seems particularly pressing today. The enormous growth in the number of physicists has tended to make the familiar channels of communication considerably less effective. It has become increasingly difficult for experts in a given field to keep up with the current literature ; the novice can only be confused. What is needed is both a consistent account of a field and the presentation of a definite "point of view" concerning it. Formal monographs cannot meet such a need in a rapidly developing field, and, perhaps more important, the review article seems to have fallen into disfavor. Indeed, it would seem that the people most actively engaged in devel­ oping a given field are the people least likely to write at length about it. FRONTIERS IN PHYSICS has been conceived in an effort to improve the situation in several ways. First, to take advantage of the fact that the leading physicists today frequently give a series of lectures, a graduate seminar, or a graduate course in their special fields of interest. Such lectures serve to sum­ marize the present status of a rapidly developing field and may well constitute the only coherent account available at the time. Often, notes on lectures exist (prepared by the lecturer himself, by graduate students, or by postdoctoral fellows) and have been distributed in mimeographed form on a limited basis. One of the principal purposes of the FRONTIERS IN PHYSICS Series is to make such notes available to a wider audience of physicists. It should be emphasized that lecture notes are necessarily rough and informal, both in style and content, and those in the series will prove no excep­ tion. This is as it should be. The point of the series is to offer new, rapid, more informal, and it is hoped, more effective ways for physicists to teach one another. The point is lost if only elegant notes qualify. A second way to improve communication in very active fields of physics is by the publication of collections of reprints of recent articles. Such collections are themselves useful to people working in the field. The value of the reprints would, however, seem much enhanced if the collection would be accompanied by an introduction of moderate length , which would serve to tie the collection together and, necessarily, constitute a brief survey of the present status of the xi

xii

Editor's Foreword

field. Again, it is appropriate that such an introduction be informal, in keeping with the active character of the field. A third possibility for the series might be called an informal monograph, to connote the fact that it represents an intermediate step between lecture notes and formal monographs. It would offer the author an opportunity to present his views of a field that has developed to the point at which a sum­ mation might prove extraordinarily fruitful, but for which a formal monograph might not be feasible or desirable. Fourth, there are the contemporary classics-papers or lectures which constitute a particularly valuable approach to the teaching and learning of physics today. Here one thinks of fields that lie at the heart of much of present­ day research, but whose essentials are by now well understood, such as quantum electrodynamics or magnetic resonance. In such fields some of the best pedago­ gical material is not readily available, either because it consists of papers long out of print or lectures that have never been published. �. ..._-._-

�.

------

The above words, written i n August, 1961, seem equally applicable today (which may tell us something about developments in communication in physics during the past decade). Richard Feynman contributed two lecture note vol­ umes ("Quantum Electrodynamics" and "The Theory of Fundamental Pro­ cesses") to the first group of books published in this series, and, with the publica­ tion of the present volume and the forthcoming publication of "Photon-Hadron Interactions," it gives me special pleasure to welcome him back as a major contributor to FRONTIERS IN PHYSICS. "Statistical Mechanics : A Set of Lectures" will be of interest to everyone concerned with teaching and learning statistical mechanics. In addition to providing an elegant introduction to the basic concepts of statistical physics, the notes contain a description of some of the many original and profound contributions, (ranging from polaron theory to the theory of liquid helium) which Professor Feynman has made in this field.

Urbana, JIIinois June, 1972

DAVID PINES

ACKNOWLEDGMENTS

This volume is based on a series of lectures sponsored by Hughes Research Laboratories in 1961. The notes for the majority of lectures were taken by R. Kikuchi and H. A. Feiveson. Others who took notes for one or more of the lectures were F. L. Vernon, Jr. , W. R. Graham, Jr. , R. W. Hellwarth, D. P. Devor, J. R. Christman, R. N. Byrne, and J. L. Emmett. The notes were edited by Dr. Jacob Shaham who also prepared the Index.

CHAPTER 1

INTRODUCTION TO STATISTICAL MECHANICS 1.1 THE PARTITION FUNCTION

The key principle of statistical mechanics is as follows : If a system in equilibrium can be in one of N states, then the probability of the system having energy E" is ( l /Q) e E /kT, where -

n

k = Boltzmann's constant, and T = temperature. Q is called the partition function. If we take I i) as a state with energy Ei and A as a quantum-mechanical operator for a physical observable, then the expected value of the observable is

(A)

=

! L (iIA l i)e-EdkT• Ii )

Q

This fundamental law is the summit of statistical mechanics, and the entire subject is either the slide-down from this summit, as the principle is applied to various cases, or the climb-up to where the fundamental law is derived and the concepts of thermal equilibrium and temperature T clarified. We will begin by embarking on the climb. If a system is very weakly coupled to a heat bath at a given "temperature," if the coupling is indefinite or not known precisely, if the coupling has been on for a long time, and if all the "fast" things have happened and all the "slow" things not, the system is said to be in thermal equilibrium. For instance, an enclosed gas placed in a heat bath will eventually erode its enclosure; but this erosion is a comparatively slow process, and sometime before the enclosure is appreciably eroded, the gas will be in thermal equilibrium. Consider two different states of the system that have the same energy, Er = E.. The probabilities of the system being in states rand s are then equal. For if the system is in state r, any extremely small perturbation will cause the system to go into a different state of essentially the same energy, such as s. The

1

2

1.1

Introduction to statistical mechanics s

H

The partition function

Remember that Er «Eo. If it is true that (d/dE) In I1(E) = peE) is almost constant for E in the range under consideration, * then we can say

PeEr) PeEr')

so

------

------

Fig. 1.1

Er £,

Energy levels in a system S and a heat bath H.

same is true if the system is in state s. Since the system remains in contact with the heat bath for a long time, one would expect states of equal energy to be equally likely. Also, states of different energies would be expected to have different probabilities. , Because two states of the same energy are equally probable, the probability of a state having energy. E is a function only of the energy; P = P(E). Now consider a system, S, in equilibrium with a large heat bath, H (see Fig. 1 .1). Since experience shows that the behavior of a system in equilibrium is independent of the nature of the heat bath, the bath may be assumed extremely large and its total energy E very great. Also, the possible energy levels of the heat bath may be assumed quasi-continuous. Let the energy levels of the heat bath be denoted by Hi' These levels are distributed quasi-continuously. Let the energy levels of S be denoted by Ej• Then Hi » Ej for all i, j. The bath plus the system can be thought of as a new system, T, which is also in thermal equilibrium. T has a definite energy, but as that energy is not fixed exactly (the bath is in contact with the outside world), we can assume that the energy may be anywhere in the range Eo ± �. If � is sufficiently small, we can assume that the states of the heat bath are equally likely in the range Hi ± �. Let I1(Hr) be the number of states per unit energy range in the heat bath H around energy H r. The probability, P eEr), that S is in a state with energy Er is proportional to the number of ways S can have that energy. In other words, it is proportional the number of states of H that allow T to have energy in to I1(Eo - Er)2 � the range Eo ± �. Then =

PeEr) PeEr')

I1(Eo - Er) I1(Eo - Er,)

=

3

e1nq(Eo-Er)-lnq(Eo-Er').

e-P(Er-Er,).,

=

PeEr') oc e-PEr'. Normalization requires that P (Ei) = (l/Q) e-PE, where Q

=

Li e-pE,.

The fundamental law has just been shown to be quite plausible. But for those who doubt the constancy of peE), let us consider some examples; First, assume that the heat bath consists of N independent harmonic oscil­ lators. The energy of the bath is N F = L n/lwi' i= 1

where we assume that ni is very large, and we neglect the zero-point energy. 2, we have the How many states are there with energy less than F? If N situation. shown in Fig. 1.2. For N = 2, the number of states i&proportional to the area of the triangle. Clearly the number of states with energy less than F for large F is proportional to FN ; so the number of states per unit energy range is I1(F) oc (d/dF)FN oc FN-1• =

In I1(F) d In I1(F) dF

=

=

constant + (N

N- 1 F

--

1) In F,

- 1

N

�

-

�

since

E r « E o.

* The basic assumption made here is, that the system governing the probabilities has a quasi-continuous spectrum in the region considered and for which there is no partic­ u . energy is only defined up to an additive constant e, we should have SInce

/(e1) /(e2)

=

where/(e) is the probability. Defining we obtain

fee) = gee - e2),

g(e)g(e1 - e2)

which is (uniquely) solved by that is,

/(e1 + e) /(e2 + e) ,

=

g(O)g(el - e2 + e),

gee) = g(O)e-PE

(P constant),

4

The partition function

Introduction to statistical mechanics

5

\ \

\

\ .\

,. , .

·

·

. • .

.

·

F

� ,�.... . . . \- . .

.

\ . ,

, , ,

.

•

•

.

=

.

constant

.

.

--�----�--�nl Fig. 1.2

States with energy less than F for two independent harmonic oscillators.

For N large, p

=

1 energy per oscillator

=

peEr) peEr')

=

W'

F

_

e-Er/W e-Er'/W

as

-t -----

N

-t

=

2m

2m

+

*

=

2nfz(integer) L

Problem: Why is W independent of N?

=

•

P;N + P;N + P;N 2m

2nxnfz ' L

e -PA, L e-PAj

= -------,--,--

Now place SA and SB in loose contact with each other and consider the combined system S = SA + SB' with combined energy Tk = Ai + Bj•

T

PT( Tk)

=

=

Assuming periodic boundary conditions,

Px

2 2 2 2 ny, + nz, + nX2 + ... + nZN]

PA(A1.)

00.*

We then get the same value for p as by our previous method. As a second example, consider the case of a heat bath consisting of N particles in a box. F

+

(JF)3N. It follows that p is roughly constant, and equals 1/(tW), where W = energy per particle. Consider two independent systems SA and SB with energy levels Ai and Bj• The probability of system SA having energy Ai is

(Eo - Er)N-1 1](Eo - Er) (Eo - Er,)N-1 1](Eo - Er,)

P; + P;2 + P;2 .. . + + 2

2nfz [ -- nx2 , 2Lm

circle with radius JF(Lm/nh). For N particles, we must use a hypersphere in 3N dimensions, and the number of states with energy less than F is proportional to

and plug in N directly. We get

P;, + P;, + P;,

=

If there were just two 1]'S, we would calculate p with the help of Fig. 1 .3. The number of states with energy less than F is roughly equal to the area of the

1](Eo - Er) 1](Eo - Er,)

(1 - ErjNW)N-1 ( 1 - Er,jNW)N-1

Number of states with energy less than F is roughly equal to the area of the circle shown. where L is the length of the sides of the box. Then

1

where W = energy per oscillator. Alternatively, we can go back to the equation --

Fig. 1.3

The probability of system

e-PT(A,+Bj) __:_=_ ---, A -,---'" £...i e -PT , '" £...j e-PTBj e-PTA, e-PTBj A '" £...j e-PTBj' £...i e -PT , '" PT(A.1 + BJ.)

= -

ST being such that system SA has energy Ai is

6

Introduction to statistical mechanics

1.1

Similarly,

The partition function

7

where

(1.4) We see that if two systems are placed in loose contact with each other, i n equilibrium they have the same p. Temperature has a similar property, and i n fact, by the way temperature i s conventionally defined, p = l/kT where k is Boltzmann's constant. * From the basic principle of statistical mechanics, once

is known, all thermodynamic properties can be found. We define F, the Helm­ holtz free energy, so that n F = -kT In Q

S

=

=

-kT In

(� e-EnlkT) , L

entropy, is defined as -k

n

Pn In Pm

(1.1)

u

since

=

S

=

=

- + -=

C2

PI

P2

CI + C2 , P

which is, indeed, the form of nee) for microscopic bodies (see pp. 3, 5). Also, it can easily be seen (from 1.7 below) that 1_" ( en 2Q2 t,;.

_

em)2e-P<En+Em)

<

=

c

v

=

( )v au

aT

average energy

=

=

=

=

_1_

kT2

"E n

�

-T2

- Pn

n

� (�)

a

avo1

aF aT

-T

�

aT

T

aF

_

=

F - T

aF

oT

_

aF aT

:� _

=

aF aT

�

a(I T)

_

T

(1.5)

(1.6)

av

,

=

S

Q '; e-EnlkT

( ) G�)v

_

=

l � En

=

e-EnlkT

aE n

F + TS.

+

P

(1.7)

�,

(1.8)

a2F O T2

(1.9)

U2(P),

whence, since the U's decrease, PI < P < P2, and energy flows from the low-P body to the high-P body. This also fits into our intuitive perception of temperature.

=

I

-

L
Q i

Our alternative definition is equivalent to the first one if oH ov

0,

so that U is a decreasing function of p. Suppose we mix two systems having different P's, PI < P2' After mixing there will be a common p, according to UI(PI) + U2(P2) = UI(P)

=

=

In

POP

nee) = Ae"

__

pressure

n

To get a clearer idea of the nature of pressure, consider a possible alternative definition. The Hamiltonian operator for the system is dependent on the volume. Set H = Hamiltonian = H(V). We can take as a pressure operator = -oH/aV.

which is the well-known (experimental) formula. Equation ( *) has the unique solution

dU = dP

� :;

-aF/aT, U

C/P and we recognize in C the usual (total) heat capacity, so that CI

k

�

=

aT

U

[b e-EnlkT { ; - Q}] L ( )T

�

-k

=

aQ =

But

* Furthermore, suppose our spectrum has a density of states nee) which is relatively constant when e -> pe, that is n(pe) = n(e)q(p)

Then U (P) (see 1.7)

- (:�\ P

(1.2) (1.3)

(1.2) it can be seen that

From Eq.


oEi av

v

lim v'

....

�i.(V') - Ei(V) V' - V

=

=

oEi • av

v
lim v'

....

V' - V

where It) is the eigenvector of H' corresponding to I i), the ith eigenvector of H. H' = H(V').

1.2

Introduction to statistical mechanics

8

H' = H + (H' - H); so we can apply first-order perturbation theory to say that
oEi Ii (iIH' - Hli) ' oH ' -- m Il ) . - (l l oV oV y' .... y V' - V _

_

Linear harmonic oscillators

But each term i n the sum is negative, for the sign of Pn - Pm is opposite to the sign of (In Pm - In Pn ) . So dS/dt > O. Before we start to make calculations, note that if we have a system that is a combination of several independent subsystems, with Etotal = E = L E ; i n n Li energies of the subsystems,

Our two definitions are equivalent. Because H may be a function of the shape of the system, as well as its volume, our definition of pressure might depend on how the volume is changed. In general, for any parameter ex, there is a force that can be computed by

(Force)

OEi oex

a. = - ( - -) e-Ei(a.) /kT, 1 � 1-1 Q i

and we can write oEi oex

=

(il

oH oex

Ii).

=

- -(V, S). oU

oV

That this definition is equivalent to the other two may be verified without diffi­ culty from the equation U = F + TS. The equation S = - (oF/oT)v holds only at equilibrium, where F is defined. Away from equilibrium, S always increases with time. To see this, note that time-dependent perturbation theory gives dPm dt

=

L n

(IVmn l2P n - IV mnI2Pm) ,

where IVn ml2 is the probability per unit time of transition from state n to state m, and IVn ml2 = IVmn l2. Then dS dt

-k

since

L( L i

dPi dPi + In Pi dt dt

dPi

i dt

Then

dS dt

mn

)

=

�

L

dt i

Pi

=

=

dP dt

F

=

=

-kT In Q -kT In n t

-kT In

=

(L e-PEn;) "i

L e-PEn = n

=

-kT

L e-P En; � [ L e-PEn,] � -kT In

'

r.,

"1,"2, ...

In

"i

=

I

Fi•

The free energy of the whole system is the sum of the free energies of its non­ interacting parts.

1.2 LINEAR HARMONIC OSCILLATORS

A third definition of pressure is P

9

-k

=

dPi In Pi> i dt

Consider a system of harmonic oscillators i n thermal equilibrium. The partition function Q, free energy F, and average energy of the system of oscillators can be found as follows: The oscillators do not interact with each other, but only with the heat bath. Since each oscillator is independent, one can find Fi of the ith oscillator and then F

(M oscillators). Qi

Qi

=

L n

L1 Fi

i=

e-E�/kT

(LlO) from quantum mechanics (n

=

L e-A;w;(n+1/2) /kT

=

0, 1,2, . . . )

(Ll 1)

n

(Ll2)

L

O.

M

=

F i = - kT In Qi Ui

=

liw· --f + kT In (1

- e-Aw;/kT)

(Ll3)

average energy of a single oscillator in thermal equilibrium

= � Eni Qi n =

=

L e-E�/kT Iiw.e' --�Aw;/;k/TkT - + e w liw· ' 2

1 -

=

_0_ Fi 0 (1/T) T =

(Ll4)

liw· liwi -' + ---C-��-

2

e�w;/kT

- 1

(1.15 )

10

Introduction to statistical mechanics

1.3

F �i Fi Li [IiW2 i + kT - e-firodkT)] ' U i Ui i [IiW.2 ' + efiro,)kTIiW.] - • n;, Ui = + t)IiWi' , efirodkT U) Ui :::::; k T U. Iiw;/2 =

=

In ( 1

=

� '-'

=

� '-'

-

(1. 1 6)

Blackbody radiation

Because of the periodic boundary condition,

a

Ax nx

- =

1

(n,

n· =

1

1

and similarly

.

Returning to the example on p. 3 we see that for very high temperature we indeed have (and W:::::;

.

= an mteger.

ni, according to

It is customary to define an average

Thus

11

'?!s.

ckz 2n - ny, 2n nz• a dkx b dky c dkz dnx, dny, 2n 2n 2n dnz d3k (abc) dkx dky dkz d3n =dnx dny dnz abc (2n)3 (2n)3· k, k + dk k d3k 2 d3n =2-. (2n)3 (abc) W =kc c [-IiW(k)]) 2 d3k kbT ( fff kbT (2n)3 � hw/2 f [-IiW(k)]) 4nk2 dk . !. 2 kb T ( kbT (2n? U 2 f hW(k) [ -liw(k)/kbTJ 4nk2 dk [-liw(k)/kbTJ (2n)3 . =Iiw/kbT Iikc/kbT. _

-

=

= 1/13,

=

=

which is independent of » Note that the contribution to F of the ith oscillator is negligible if term). A t low temperatures the high-frequency modes are (except for the "frozen out" and do not contribute to the specific heat.

IiWi kT

=

=

=

For each there are two possible polarizations. Thus, the number of modes is per unit volume with wave number between and �-

1.3

BLACKBODY RADIATION

I n dealing with blackbody radiation, our point of view will be as follows: In a cavity (blackbody), there are a great number of modes of oscillation. The number of modes per unit volume per frequency bandwidth is given by classical considerations. Each mode, however, behaves as an independent quantum term is neglected. That is, En = harmonic oscillator, except that the because it leads to infinite energy when there We want to get rid of the is are an infinite number of modes. A Hamiltonian that eliminates the

hw/2 Iiw/2 i 1(p2i + Wi2qi2) - IiW2'i

H = "2

hw/2

nliw.

With the above assumptions, we can obtain an expression for the energy per unit volume per unit frequency. First we will find the number of modes per unit volume per frequency (or wave number). Assume a gigantic box of dimensions a, b, and c. The demand is made that the waves be periodic at the walls of the box.

Now where is the velocity of light. Also, the number of modes is so large that the sum over the modes can be replaced by an integral.

V

=

In

Let x

In

=

V

.

( 1 . 1 7)

term has been omitted.)

(Remember that the From symmetry,

V

1 - exp

1 - exp exp

=

1 - exp

=

( 1 . 1 8)

Then

! = number of waves/cm in x-direction

Ax Ax

� = number of waves in box (x-direction).

( 1 . 1 9)

12

1 .4

Introduction to statistical mechanics

The preceding results can be summarized and put in a more familiar form by replacing nw with hv. The number of modes/unit volume between k and k + dk is

2.

4nk2 dk (2n)3

k2 dk n2

8nv2 dv c3

since k =2nv/c. The average energy of an oscillator offrequency v is hv/ ehY/kT Thus, the energy per unit volume between v and v + dv is

_

dU U Y dv = V

=

8nv2 hv dv . c3 ehY/kT 1

l.

(1.20)

Uy dv

=

uT4

is the Stefan Boltzmann Law. For T very large (kT » hv), U dv =

Y

8nv2 hv dv c3 1 + (hv/kT)

_

1

=

8nv2 kT dv. c3

(1.21)

This is the Rayleigh-Jeans Law. Let F* =F/ V and U* U/ V. =

cv

=

a u* =4uT3 aT

is the heat capacity of a gas of photons in equilibrium with the container. If an oscillator (or mode) is excited to the Nth level, that is, if E =n W iN, one says that there are N photons with energy nwi . Since photons are defined as the degree of excitation of a mode, one cannot consider a permutation of photons as a new state. That is, photons are indistinguishable. This point is the basis of quantum statistics and will be dealt with in a later section.

1.4 VIBRATIONS IN

A

13

2. Find the normal modes of the system. There are as many normal modes as there are degrees of freedom; namely, 3(AN), where (AN) is the number of atoms in the crystal. The modes behave as independent quantum oscillators. 3. Given the modes, calculate F, the free energy. 4. From F, calculate Cv and any other thermodynamic quantities of interest. Once F is found, we can quickly calculate U, the total energy of the system. In the following derivation, we will calculate Cv directly from U without always writing down F first. Method of Labeling

-

This is the Planck Radiation Law, and is the same as Eq. ( 1 .18).

f

Vibrations in a solid

Consider a crystal with A atoms per unit cell. For convenience, assume that the unit cell is a rectangular solid of dimensions a, b, c, along three mutually per­ pendicular axes, x, y, z. Let the origin be at the "center" of a cell. This cell can be denoted by the triplet (0,0,0). The cell to its right along the x-axis is denoted by (1, 0,0),and so on. Thus, any cell can be denoted by a vector N =nxa + nyb + nzc where a =ai,b =bj,c =ck. If there are A atoms/cell, 3A additional coordinates must be given to locate each atom. Let IX denote one of these 3A coordinates. Call the displacement from equilibrium of the coordinate in the Nth cell Zrz,N' (We either consider m =mass equal to 1 or absorb it in Z) Zrz,N+M is the displacement of an atom in a cell close to N; if, for example, there are two atoms Al' A per cell, the displacement of A 1 in the x direction is 2 denoted by Zl,N, that of A (in same cell) by Z4,N' and that of Al in an adjacent 2 cell by Zl,N+1' 1 =(1,0,0) or (0,1,0), or (0,0,1). Normal Modes

T =(1/2) V = YeO) + L

",N

L

particles and directions

(�)

aZ",N z= o

Z;,N =(1/2) L Z;,N ",N Z",N

(1 .23)

SOLID

We want to find the specific heat of a solid. However, en route to the specific heats, we will derive important results that will prove useful in many other cases; chief among these results is the calculation of the normal modes of a crystal. The program will be as follows:

1. Consider the solid to be a crystal lattice of atoms, each atom behaving as an harmonic oscillator. These oscillators are, of course, coupled.

(1.22)

Assume that the electrons in the crystal always have time to adjust them­ selves to the configuration with lowest energy, even when the crystal is vibrating. In this configuration, there is no net force on the nuclei when the Z" N are zero, so

(�) OZ",N

z= o

=0 .

14

1.4

Introduction to statistical mechanics

The additive constant, YeO), will not affect our answers, so we might as well drop it. Let

a number that depends on the relative positions of the cells of the two atoms, and p . not on their absolute positions. Note that Neglecting higher orders, take (1.24)

For low temperatures it is not too unreasonable to neglect higher orders, because the separation between atoms is of the order of 1 A and at room temperature the vibrations have amplitude of the order of 0. 1 A. But we should not be too surprised if experiment shows our idealization to be false. In matters such as the one under consideration, the general approach is to make idealiza­ tions and then try to find corrections to our assumptions that will give better results. In order to motivate the procedure that we will use for finding the vibrations of a solid, let us consider the classical problem of vibrations of coupled oscil­ lators. Let the Hamiltonian be p,2 � - ��

i +

H

i 2Mi

Let the ath mode have frequency

Wa

15

so that

�a

for the motion of the ath mode, with a ) independent of time. Then

C �,

(OZa'NO;�'N+M)Z=O Ca� = C �M =

Vibrations in a solid

� 2IC'. . q.q. , , � ij 'J ' l'

2

wa,a) = j a J

The classical problem of vibrations of coupled oscillators has just been re­ duced to the problem of finding the eigenvalues and eigenvectors of the real' we must solve the equation symmetric matrix, IICijll. In order to get the Then the that

a�a)

= O.

( 1 . 26) (the eigenvectors) can be found. It is possible to choose the a�a) so '

L a�a)a�P) = i

bap.

The general solution for qi is qi

=

L Caq�a>,

a

Qa = Cae-iw�t, , = j a Qa = a ba).Qa = QJ" H = La Ha, Qj Ha tP; + tw;Q;. Qa Cae-iw�t.

where the Ca are arbitrary constants. If we take From this it follows that L

a a�a)Qa.

I.

P;

wa w2bijl

-

det IICij

� a(j)q. � ,

where the q; are the coordinates of the amount of displacement from equilibrium, = MJl; is the momentum, and C;j = Cji are constants. To eliminate the constants Mi, let

� C..a ( ) I.J l ) ·

a(

� a ( )a ! ) '-! I. l =

=

�

�

«,I

Making the change of variables,

we get qi

Li aF)qi' we get

where

=

This has the expected solutions

=

Now suppose we wish to solve the quantum-mechanical problem of coupled oscillators. Again we have p,2

H = L �i + L C!.q�q'. i i

Then we get

(1.25) The equations of motion are q.

=

oH =

p.

, OPi "

We now break the motion of the system into modes, each of which has its own frequency. The total motion of the system is a sum of the motions of the modes.

2Mi

where this time

j

'J

'

J'

�� P!= , . '" !

uqi,

Making the same change of variables as before, we get

Qa = Li

a�a)qi =

a

L a f )-/M i q; i

Introduction to statistical mechanics

16

Specific heat of a crystal

1.5

where

Then the general motion can be described by

Z�,N K,r C,(K) a�r(K)e-iw(r)(K), eiK' N =

It follows immediately that the eigenvalues of our original Hamiltonian are E The solution of a quantum-mechanical system of coupled + oscillators is trivial once we have solved the equation =

L� (N� !)l1w� . I Cij - w2bijl 1023 1023 !(1023) o

=

det

= det

Ilc/jJJMiMj - w2bijl .

If we have a solid with atoms we must apparently find the eigen­ values of a by matrix. But if the solid is a crystal, the problem is enormously simplified. The classical Hamiltonian for a crystal is

C:; Ci�M) Z�,N - M,p C�M,PZP,N+M'

and the classical equation of motion is (using ..

� £...

=

� £...

'Vr 11

Cr(K)JJ� IJJ� Q,(K) Cr(K)e-iw(r)(K)r. Qr(K)

are arbitrary constants, and

where the cells. The factor strictly necessary. = Let mode.

11

is the total number of unit

is inserted to make things look nicer later, but it is not describes the motion of a particular (1.28)

It follows that

Qr(K) �L,NZ�,Na:r(K)e-iK'N, oc

=

(1.29)

and the Hamiltonian for the system is

In a given mode, if one cell of the crystal is vibrating in a certain manner, it is reasonable to expect all cells to vibrate the same way, but with different phases. So we try

(1.30)

Z�,N aiK)e-iwteiK'N, =

K

eiK' N w w(K). w2a� eiK. N M,p (C�M,paP eiK' M)eiK'N Y�p(K) C�M,peiK·M. w2a�iK) � Lp Y�pap, (1. ) I Y�p - aib�pl O. w(r)(K) w�) ,

where expresses the relative phase between cells. The factor allows for wave motion. We now want to find the dispersion relations, or = =

Let

� £...

=

If we consider

Qr(K)

and its complex conjugate to be independent variables, we get the same equations of motion from

H

•

� £...

M

(Note that y is Hermitian. See end of p. 18). Then and we must solve the characteristic equation of a 3A-by-3A matrix: = det 27 =

The solutions of the characteristic equation are

=

KL,r [IQr(KW + w2(r>(K)IQ,(K)12].

(1.31)

1.5 SPECIFIC HEAT OF A CRYSTAL We now want to find F, so we must sum over all possible modes. For one mode, F = kT n (1 ) There is a quasi-continuum of values, and for each value there are 3A wit>'s. Assuming the crystal has volume V , and assuming periodic boundary conditions, we have in the range to + d approximately d 3 2n V modes with a given p. Thus,

K

K K

!. V

==

where r runs from 1 to 3A. The motion for a particular mode can be written where

17

=

e-hWK(P)/kT . I K KJ( )3 () ( hW ] d 3K T k / p) K ekT r LI1WK P + kTln[l JK p

+

2

d3K = dKx dKy dKz•

IK

is shorthand for

,,/a

"/b

,,/e

J J J -,,/a

-,,/b

-,,/e

(2n)3

K

(1.32)

18

Introduction to statistical mechanics

The reason for the limits -rrja, rrja is as follows: the factor K was introduced in nx, ny, nz are integers. But, Then

eiK·N•

eiK'N eiKxnxaeiKynybeiK.n.c;

19

Then, because f is an even function of K,

=

No new modes are introduced if K ranges beyond the prescribed limits. We now demonstrate that the specific heat of a crystal determined by our present method agrees with experiment; that is, at high T, Cv 3R. At low T, Cv --+ 0 as T3. We choose to look at U rather than at F. First we note that there is a max­ imum frequency, WM' Two adjacent atoms can be no more than 1800 out of phase, and thus the minimum wavelength must be of the order of twice the atomic spacing (call it 2a). Thus Kmax :::::; 2rrj2a nja, and WM W�)( KM )' First, we consider the high-temperature limit. For kT » hwM , all modes are excited to approximately the same energy; that is, if there are N atoms, =

=

U

=

=

nWi � � -

i 2

+

� t...

i

nWi

�

exp ( nwdkT) - 1

Uo + 3NkT

=

�

� Wi - t... 2 i

n

+

� kT t... i

Uo + 3RT,

--------

nw

exp( nw/kT) - 1

(1.33)

e:;

are real, y;aCK)

=

� t...

M eM eiK·M a, p

,

ML e�a e-iK'M LM ei.:'eiK'M

Also, for real (K), ytaCK) few , K)

=

=

=

YpaC -K), so Yap(K)

=

det IIYap(K) - w2c5apII

=

det IIYaP( -K) - w2c5apll

=

=

0

Since the characteristic equation must have a solution with w 1(0, 0) O. The characteristic equation becomes

=

=

0 for K

=

0,

=

o

=

w2

of 02f - � L KK· 2 ow 2! i,l I J OKi oKj

I

0

+ higher order terms,

so that, for low wand IK I we have

where cos lXi are the direction cosines of K. In other words, w IKI V, where V is the velocity of sound, which may depend on the direction of K when the eigenvalues of 02 f/oKidKjlo are not all equal. But, for sufficiently low frequency, V does not depend on the frequency. This is the approximation. For convenience, assume that the three sound velocities are equal; that is, w = VoK. =

Debye

:::::;0

when hw »kT. We also know that when K is zero, there are three zero-fre­ quency modes arising from translation of the entire crystal. For small K, there should be three very low-frequency modes. To see how those modes vary with K, consider YaP(K) for low K. Since Yap(K)

of (w2 0) oK '

=

if N is Avogadro' s number. Cv = 3R for large T. Note that we might as well neglect the zero-point energy, Uo. For very small values of T (kT « nwM), the behavior of Cv can also be approximately determined. Ignoring the zero-point energy, we see that the contribution to U of the high­ frequency modes is very small, because

where the

Specific heat of a crystal

1.5

K

=

w Vo

.

The factor 3 before the integral is there because for each K there are three modes with low w. U

=

U0 +

3n 22rr V 03

f"'M w3 e 0

hw/kT

-

1

dw,

where

(1.34)

Kmax can be found by setting the total number of modes equal to 3(AN), where =

Yap(K).

(AN) is the number of atoms in the crystal. 3(AN)

YpaC -K ), and

det lIypiK) - w2c5pa II

= f(w2, -K).

Then KM

=

=

rKrna•

J

0

3 d 3K V (2rr)3

=

rKrna• � 4rrK2

J

0

(2rr)3

dK

=

K� V 2rr2

.

(6rr2p)1/3 where p is the number of atoms per unit volume.

20

Introduction to statistical mechanics

1.5

liw

E>

kT

T'

W=

kTx

E> =

--'

Ii '

u

IiKV k

21

Cv

Let

X=

Specific heat of a crystal

3R

.

----------------

___________ _

(1.35)

V Here E> is the Debye tempe rature.

( 1.36) where A -+ n2k4jlOh3V� as TjE> -+ O. So Cv is proportional to T3 at low temperatures. The T3 dependence of the specific heat at low T can be made more plausible as follows : We assume that for liw ; > kT, the contribution to the energy of the W; mode is negligible and for liw ; < kT, the contribution is kT. The number of modes with wave number less than K is

3V

fK

4nk2

o (2n?

dk =

K 3V 2n2

.

At low T, K wjVo' Therefore the number of modes with frequency less than Wc is w� Vj2n2V�. But liwe = kT or We = kTjli and =

-t-"'-----_ Tie

Fig. 1.4 Specific Heat of a Solid. We wish to find the three natural frequencies associated with each K of the crystal. To do this, we must find C�p and then Ya,p' In complex coordinates

V=LVap a,p

where

(1.37) For example,

(1.37')

The energy varies as

u

V

If we express the displacement of atom N from its normal position as XN, then the potential energy from the distortion of the spring between atoms N andM is lK 2 M

And thus Cv is proportional to T3. The numerical factor is, of course, not correct. If the mode with frequency w;is excited to the nth level, Ei= liwi(n + 1-), we say that there are n phonons of frequency W; and energy hWi in the crystal. Figure 1 .4 shows the general form of the specific heat of a solid as a function of temperature. Example,' Assume the unit cell to be a cubic lattice with one atom per cell (Fig. 1.5). Each atom behaves as an harmonic oscillator, with spring constants kA (nearest neighbors), and kB (next-nearest neighbors). This case is fairly simple, and we can simplify the notation : a. = 1 , 2, 3.

Fig.

1.5

[

(XN

-XN+ M) .

J

M 2.

1M! '

Cubic lattice with one atom per cell.

22

Introduction to statistical mechanics

for N + a nearest neighbor to N, for N + a next­ = = nearest neighbor. In summing over N and to get the total potential energy we must divide by two, for we count each spring twice. If we use complex coordinates, however, we multiply V by two to get the correct equations of motion.

KM KA

M

KM KB

M

M

M (1 .38) N� KM I (XN - XN+M) . IMI I 2 Vl l = ! L KM I Mxl2 (X� - XN+M)( XN - XN+M) 2N,M IMI = ! L KM (Mx)2 [(X�XN + X�+MXN+M) - (X�XN+M + X�+MXN)J 2N,M IMI (1 .38') = L KM (Mx)2 [X�XN - X�XN+MJ . N,M IMI V

=

"21

The Mossbauer effect

1.6

suppose the temperature is absolute zero. Then the crystal must be in its vibra­ tional ground state before emission of a photon. We will find an expression for the probability that the crystal is in its ground state after emission. Let R be the position of the excited atom, and let P be the momentum of the photon. We will assume, without proof, that the amplitude for the itted em crystal being in a given final state is a

'

Comparing Eqs. (1 .37') and (1 .38'), we see that

KM (�y; c�tO - KM (�y. (1.39) Here (Mx IM I)2 = 1 for M ( ± 1 , 0, 0), 1- for M ( ± 1 , ± 1, 0) and M ( ± 1, 0, ± 1), and zero for the other nearest and next nearest neighbors. So C �l 2KA 4 KB, C t-P ) - KA' and so on. In this way, all the C�p can be found. We can then calculate y�,P(K) LM C�p[eiK'MJ. We wish to solve det I Ya,p - aiba,pl 0. w, wFor(p)(K).each K , there are three solutions for w. Thus we obtain 3N values of C� l

=

�

=

=

=

+

,O,O

=

=

=

=

1.6 THE MOSSBAUER EFFECT If a free, excited nucleus goes to its ground state by emission of a photon, the energy of the photon will be less than the excitation energy because the nucleus recoils. But if the excited nucleus is in a crystal, there is a finite probability that it will emit a photon with the full excitation energy. In other words, there is a finite probability that the state of the crystal after radiation will be the same �s the initial state. Similarly, there is a finite probability that the crystal state will be unchanged by absorption of a photon. This effect, called the Mossbauer effect, can be discussed in terms of modes of oscillation of a crystal. For example,

=

(finaI Ae P R h initial).

I i'll

Furthermore, we will consider only the vibrational states of the crystal, and will neglect any effect due to the fact that the nucleus changes state. Let the Mth atom be the one that emits the photon ; let Ro,M be its mean position, and let ZM be its displacement from that position. R

=

Ro ,M + ZM'

We can take the origin of our coordinate system at the mean position of the excited atom. Then RO ,M = o. A single one-dimensional harmonic oscillator in the ground state is described by the wave function

l/I (:;)1/4 e-w(Q2/2h),

=

/

23

=

where Q is the position measured in units such that the mass can be set equal to unity. The wave function of a crystal in its ground state is a product of the wave functions for each mode of vibration.

'l'crystal 0K,r (w(nnr)(K») 1/4 _w(r)(K)IQ�(K)j/2h We wish to calculate : C I fff l/I:rystaleiP'ZMlhl/lcrystaI12 ,I,

=

p

e

L

=

(1 .40)

where P is the probability that the crystal will remain in its ground state and C is a constant independent of P. As P --+ 0, F must tend to unity. So, if we can get a formula for F without the correct constant in front of it, the constant can be determined easily. We will ignore such factors as "A" and We take Z , = Z�.o as the displacement of the nucleus that emits the photon. For simplicity, assume the momentum of the photon is in the IX direc­ tion. Then we wish to compute a

"(w/nn)1/4."

M

pl/2 fff (0 exp [ _w(r)�Q;(K)J) (exp [iPZ�,o/hJ) (0 exp [ - W(r')(K')Qr2,(K . oc

x

211

Introduction to statistical mechanics

24

Quantum statistics for a many-particle system

1.7

25

But

so �

F Notice that

OC

=

OC

�

exp [-

(::)
exp [-

(::)
So we can write F Because F

= 1 at

oc

P = 0, we can write F

=

=

(1.41)

l
l

£.e-Ei/kT, Fie-Ei/kt n \
=

�

Q� ,

r

r ] 2(2n + 1) ar ] l a(K)f exp [=(2n�±..I)IiW( )(K) L n [1 - p r 2w( l(K)'11i 2kT [1 - p2(2n + 1) la�12] exp [-(2��+ 1)IiW] � 2w'1li 2kT K, n=O

n

[ - larl2 1 + e-hw/hW/kkTT] 2w'1li 1 _ eK,r [ p2 la�(KW 1 + e-hW/ kT e-hw/kT] r

2 n 1- p L

a

K,r 2w'1li

_.

F

and then F

i

k

· ··---

--

1.7

where In) denotes the nth state of a single one-dimens�onal harmonic oscillator with frequency w(rl(K). As f/ is very large, we can WrIte

=

exp

---�. ..-

1-

_ P2
(1.41)

QUANTUM STATISTICS FOR A MANY-PARTICLE SYSTEM

Consider a system of N identical particles, and assume that there is no interaction among them. Any two configurations of the system that differ only by an inter­ change of two or more identical particles are regarded as one and the same state. Thus the state of the system is determined by giving the number of particles na with energy ea' Our problem is to calculate the partition function Q with the condition L na = constant. The values admitted for every na may be (a) na (b) na

,

wr(K ]

r ] 2 2 1i n l - ! p (2 n + 1) ,ara(K),21 exp [-(2n + 1)liw( )(K) , r 2kT 2 '11i22w( l(K) K,

As before, we get for our answer

The value of F can also be found for nonzero temperature. If "i" is the number describing the state of the crystal, we must compute F;

Ii exp [ -en + 2) --;;y -) 1

X

=

=

0, 1,2, 3,4, . . . 0, 1

Bose-Einstein case Fermi-Dirac case

Bose-Einstein statistics must be used for particles with integral spin (for example, He4) and Fermi-Dirac statistics for particles with half-integral spin (such as, electrons). For Bose particles, any number of particles may occupy a given state. For Fermi particles, however, there can only be one particle at most in each state (Pauli exclusion principle).

26

1 .7

Introduction to statistical mechanics

A state of the system is described by the set of numbers na, which can take on any set of values allowed both by the statistics and by the condition La na N. =

(1 .42) If there were no restriction on the number of particles, we could write

Quantum statistics for a many-particle system

a function of J1, we can in principle select J1 so that any desired number of particles are in the box. For a given J1, the energy levels of a particle in the box become Ba - J1, rather an th Ba • By the key principle of statistical mechanics (as applied to the box), the ility of the gas having energy E La niBa J1) is proportional to e - pE• obab pr Q(Jl)

(1.43)

e- pg

a

Unfortunately, Eq. ( 1 .43) is incorrect, because we have an auxiliary condition that La na N constant. With this auxiliary condition, the problem of finding Q becomes much more difficult. It is possible, however, to get around the restriction by considering the system of particles to be a box connected to a large reservoir of particles (Fig. 1 . 6), and by assuming that it is possible for particles to pass to and from the reservoir. We further assume that the statistical mechanics of the total system acts as if it took energy J1 to remove a particle from the box to the reservoir. J1 can be adjusted by, say, a voltage regulator. In the case of electrons in a metal, for example, J1 is the work function of the metal. In general, increasing J1 will in­ crease the expected number,
=

exp - PEL niBa

L

-

J1)] .

(1 .44)

The summation can be made without restriction. We will now show how J1 is determined by N and how thermodynamic quantities depend on Q(Jl) or g == - liP In Q (Jl).

In the Fermi-Dirac case we would have

=

-

=

and we would have in the Bose-Einstein case

27

Q(Jl)

=

=

nt.n�

• • .

exp

( -13 [�

na(Ba

J)

- J1)

.

�:) 13 (� ) ( -13 [� - J) ( -13 [� = J)

o

=

=

na exp

L

But


=

=

J1)

na(Ba

L PN exp

niBa

-

J1)

.

( 1.46)

� L N exp ( -13 [� niBa - J1)J) .

(1.47)

Q Jl)

_1_ o Q(Jl) pQ(Jl) OJ1

( 1.45)

= ! � In Q(Jl) = - og . OJ1

f3 OJ1

This equation gives
• Problem: The function

Fig. 1.6 System of particles considered as a box connected to a large reservoir.

has

/(/1)

1

=

� e P<£v - Jl)

_

1

many plus-minus infinity points, and therefore the equation /(/1) N has many solutions. Why do we still talk about a unique chemical potential for bosons? =

28

1.8

Introduction to statistical mechanics

Evaluation of integrals

as computing the pressure, we do not consider p, to be part of the energy of the system. U = l

Q

L (L ) { -p [L - P,)J} a:pg (�av L ) { - p [:L l L .. naea exp

a

n l ,n2, ' "

Pressure = P

=

a

Q n l .n,. .

-

=

niea

naea exp

a

a

+

p,
niea

-

P,)J} .

p, can be treated as a variable or as a function of V and (N). If we treat it as an independent variable, so that 9 = g( V, p,), we get p =

\

- ag . av /l = const.

If we treat p, (and g) as functions of V and (N), that is, with g = g[ V, p,(V, (N) )], then

- ag av

is easily shown. Also,

If F = F(V,

(N»

= p+ ( N) = const.

ag aN

\

_

v

= g( V, p,( V.

aF p - -

av

--

where

\

\

-

_

(N » )



v

p, =

( N) = const.

Entropy = S = - k

aF

a
--

\

Q n l n2

--

,

�

= S -

oV

-

aT

-

• . . •

/l

(1 .53)

'

1.8

EVALUATION OF INTEGRALS

We will soon be dealing with integrals of the form J� 00 e - ax' dx and J� 00 x 2e - ax' dx. Let us pause for a second to calculate these integrals : e - x'

S

In summary :

_

L

_

ag

aT

P

=

niea

f�oo

e - Y'

dy

( 1 .55)

oo

e -ax2

- 1 dx Ja _

-

/l. V = const.

a

dx

.

f {- p [L

e - x'

V = const.

\

-l exp In Q(/l) = In g = -=-! n l .n,. . . . p

J f�oo

dx

- 00

It is easily shown that

( 1 .54)

'

/l, v

Soon we will be able to find 9 for an ideal Bose gas, and then for an ideal Fermi gas. But first we must evaluate some integrals.

[f�oo

·

+ (N)p,(V, ( N » , then

and

:L - ag \ - ag l

( 1 . 52)

1

P =

(1 .48)

-

P,)J} ,

29

( 1.49) ( 1 . 50) ( 1 .51)

f�oo foo x

_

X3 e - ax2

- 00

4

e -ax2

and so forth.

dx

f

foo

oo

( 1 . 56)

- 00

e _y2

dy

_

J� Ja

- -�

d e -ax 2 dx - - d da da � J� = !a - 3 / 2J�. da Ja -

- 00

= 0 (odd function)

d2 dx = da2

foo

- 00

(y =

fOO

- 00

J� x). e -ax

2

( 1 .57)

dx ( 1 .58) ( 1 . 59) (1 .60)

30

1.9

1.9

Introduction to statistical mechanics

2 where we have used J�", x2e - ax dx

THE IDEAL BOSE-EINSTEIN GAS

From Eq. ( 1 .45) we have e - Pg = L exp { - f3[ n l (e 1

-

11) + n 2 (e2 - 11) + . . . J }

= 0, 1 , 2, . . . 1 ' tem ' ), ( B ose EInS L e - Pn;(t; -Il) 1 - P(t; -Il) e n;

For a Bose-Einstein gas,

we obtain finally

na

_

��- - ----

(1 .62) (1.63) ( 1.64)

Consider a particle contained in a box but otherwise free. The number of modes in the box with momentum in the three-dimensional region d3p is s(d3k/(2n)3) V, where V = volume and s is the number of possible spin states (e.g., s = 3 for spin 1 when the rest mass does not vanish ; s = 2 for spin 1 when the rest mass vanishes, as for photons). The energy is 1i2k 2 p2 e = � = �2m 2m Here we have approximated the discrete set of modes by a continuum (which is the case for a completely free particle). The sum for 9 can then be replaced by an integration : d 3p In (1 - e - Pp2 /2 mePll) V ( 1 .65) = s (2nli) 3 f3 9 e - Pp2/2 mePIl d3p
!f

The ideal Bose-Einstein gas

f

_

_

( )

p

= s

=

31

(J;/2a)(l /J�), from Eq. ( 1 . 58). Letting

( )

m kT 3 / 2

( 3 / ia) .

2nh2

(1.68)

Thus, given p , we can find a = e PIl (in principle) by solving the equation 1 2nh2 3/2 p (3 / 2(a) = - --. 3 s

( ) mk

T /2

-

For the total energy of the system we have

( 1 . 69)

For very low p, high T, or both, p/T3/ 2 is very small ; then (3 2(a) is very small, / and thus a is very small. In this case (3 / 2(a) :::::: a, ( 5/ 2(a) :::::: a, so that ( 5/ia) (3 /ia)

::::::

1

and U/ V :::::: (3/2)kTp . This condition therefore represents the classical limit. Now let us look at the other limit. As T becomes lower and lower, with p fixed , ( / 2 (a) becomes larger and a approaches 1 . For a > 1 , ( diverges. 3 1 1 (3 /i 1) = 1 + __ + 3__ + . . . = 2 . 612 , 3 /2 2 3 /2 1 ( 5/ 2( 1 ) = 1 = -� + _5/_ + . . . 5/ 2 2 3 2

=

1 34 1 •

•

The temperature Tc at which a = 1 is called the critical temperature for Bose-Einstein condensation.
P ( )

( )

32

1.9

Introduction to statistical mechanics

Cv

The question arises : Why does our analysis break down at Te ? The answer is that for such low temperatures (T < Te), we cannot replace the sum, Eq. (1 . 64), by the integral, Eq. (1.65). Looking back at 9 and
=

� � In

(1 - e - Pe 'ePII)

=

� � In

The ideal Bose-Einstein gas

Cv

(1 - e - Pe'a) ;

1 1 e - p e'a - og - L N = -- L Li 1 - e - pe'a 1 = ll i e - P eP e, - 1 i a - eP e , - 1 OJl For small a, e - PII is large ; the terms with the lowest G i do not contribute much to the sum, and we can replace the sum with an integral. When e - P II is small, we cannot replace the sum with an integral because the first few discrete terms in the sum are important. Now og 1 ( na> = a;" = - ll �-=t. · -

-��

--

-r---+-----� T -Bose-conde

= ( no > ·

Without any loss of generality, we may take Go as the zero of the energy. Then

( no >

=

1

=>

Jl

=

-

kT

(

+

In 1

1_ _

)

::::::: - kT

A. Bose-Einstein gas around Te compared with liquid lambda pomt .

For T

<

Te

Nexc

-

=

=

1

(N

- no>

2.6 12

=

( )

1

00

.L1 p;.;-l e -

,=

mk T 3 / Z 2nli z

sV

=

fd ()

:::::::

( N)

sV

1 3p - L 3 -p2 mkT --l /Z (2 nrt ) e

T 3 /Z

Te

helium around the

the total energy is v

( )

mkT 3 / 2

=

fkT s

=

f kTNexc(O.51 34)

=

fkT(O. 5 1 34)

2nli2

V' s / z( l )

(�y

/2

=

( ) fk TNexc ' s / z I ( '3 / Z 1 )

(N) .

( 1 . 72)

Then Cy = a V/aT ex T3 /Z at sufficiently low temperature. For T > Te, we have
for large ( no >.

(no> (no> For low temperatures, Jl is very close to zero, so for the energy states above Go , we can neglect Jl. The sum for
Liquid helium

Fig. 1.7

Because
--+�-�-----· T

nsation

(e P ePe

(n 1 > =

33

.

We see, then, that our definition of the critical temperature is such that when T < Te, there must be a nonnegligible fraction of the particles in the ground state : <no) =
v

=

(a) fkT(N ) ' s / z . (0:) '3 /z

(1.73)

The behavior of the specific heat of a Bose-Einstein gas around T is rather . .lar to the behavior simI of liquid helium near the so-called "lambda point" (A point). See Fig. 1 .7. * For a given p, Te and A are very close to one another (the lambda point is a few degrees K). Because of the mutual forces between . parhcles liquid helium is certainly not an ideal Bose gas, but perhaps part of the � eXplanatIOn of the lambda transition involves Bose condensation.

� For a-calculation of the discontinuity in (oe ,fo T) at the transition point see 195�' Landau & E. M. Lifshitz, Statistical Physi�s, pr�blem of §59, Pergamon Press, ------

--

34

1.10

Introduction to statistical mechanics

The Fermi-Dirac gas

A more exact description of the behavior of liquid helium near the lambda point is given by the empirical formula : v

C

�

{aa' +

T= O

35

T>O

T,�.I ,

In IT + b In IT - TA l, b

The explanation of this behavior is left as an exercise for the reader. If successful, publish ! o

1.10 THE FERMI-DIRAC GAS

o

"O

Fig. 1.8 Occupation of states at T = 0 and T

For a Fermi gas, we proceed exactly as for the Bose case up to the point

>

o.

(1.74) Here

n 1 = 0, 1 ; n2 = 0, 1 ; and so on. Thus e -pg = n (1 + e- P(" - Il»)

(1.75)

i

(1.76) Once again, we can approximate the sum by an integral. This time there is no danger that a sizable fraction of the particles will be in the lowest state. Assume that the gas is of electrons, so that s 2.

=

2 d3p V 9 = (P!) f ln (1 + e -P(p2/2m-Il») (2nh)3 ' N) p = (­ V

(1.77)

_

(1.78) (1.79)

for = 0 and for any > 0 is shown in Fig. 1 .8. Now at = 0, = 1 if Ea fl ; and = 0 if > fl. Let fl = flo at = O. In other words, at = 0 all the states with energy less than flo are occupied, and all with energy greater than flo are empty. flo is called the Fermi energy and flo/k the Fermi temperature . O 4np2 ) dp = 2 (� 4np3 ) 1 (1.80) P = fP 2 ((2nh)3 (2nli)3 '

The form of T

T

T

T

<

Ea

0

T

Fig. 1.9 The density of states as a function of energy.

p�/2m.

where flo = For most "reasonable" values of flo and T, flo example, flo/k for copper is 82,ooo oK*). Since

2 2 4np dp (2nh)3

=

»

kT (for

4n(2m)3/2 rE V �' (2nh)3

the density of states with respect to energy is

f(E)

m) 3/2 ..;;. = 4n(2(2nh)3

apap + flN) = f( 2 ) +e-P(p2/2m-Il) 2 d 3p . u = -UV = -V1 (----1l -P(p2/2m-Il) 2m 1 e (2nh)3

This function is shown in Fig. 1 .9.

L

(1.81)

* Also, in states of very condensed matter, as in a neutron star, f.lo/k of the neutron, proton, and electron liquids are 101 0 - 101 1 OK whereas T 1080K. -

_

Introduction to statistical mechanics

36

At T = 0,

1.10

fPO (E�) ( dP) �� J_ (fPO dP) ---±� ( ) ! [� (fm) � P6J � ) 1J10 . .8 ) ( J [ "3 Po

4np 2 2 u(o) = o 2m (2nh)3 1 p5 2 _ - (2nh)3 2m 5 0, _

�(02 = p

=

(2nh)3 2m

p� 5 2m

p

- V -

_

-J

1 og _ - V OJ1

0, what is J1 and what is the

e -P(p2 /2 m)ePIl 2 d 3p 1 + e - Pp2/ 2 mePIl (2nh)3 '

I

� �

IoII g

2 (e) de + 2

f3

r oo . JO ;P

-J2me = p,

p

and

1

4n(2m)3 /2 -Je de - a (e - Il) + 1 (2nh)3

-

_

_

where

-foo gee)

foo 0

eP(e - ll) +

(1 .84)

1'

x dx

=

eX + 1 �

0

de +

0

n 2g ' (J1) 6f32

.

(1.87)

can be eval�ated by expanding xe X/(1 + e - X) in a power series, then integrating and summmg; the result is n2/ l2.) So, I

Remember that for 1 = p

But we have also

p

J: gee) e

= p,

d +

:2 k2 T 2g'(J1).

gee) = a-J�, and for I =

= a =

-J ; de

f oo ---- fll gee) r oo J x dx

dp = -Jm/2e de.

d3p becomes 4np2 dp de = 2n(2m)3/2.,fe de. Thus,

I

o eX + 1

For convenience, we wish to express these integrals in terms of the energy e: e = p2/2m,

g (J1)

(The integral

(1 .78) (1.81)

III o

u , gee) = ae3/2 .

Vre de + � (kT)2 �..c 6 2-JJ1 2

( ) J1 2 2a 3

(1.88)

3/ +

�� (kT )2 6

= (2a/3)J1 6 /2 from Eq.

a

2-JJ1 _

.

(1.89)

(1.80). Thus (1 .90)

4n(2m)3/2 -- · a = (2nh)3 In both cases, the integral that must be evaluated is I

-

-

de

eP(e -ll) + 1 ' 0

where gee) = a-J� for p, and gee) = ae3 / 2 for J1. Then I

=

J:

too eP(:��) d: J: gee) - J: f3(e - J1), J g(J1 Jilo gee) e fOO fL�� d

:��) : 1

eP(

=

In the first term let x = I

=

d +

(1.91)

��:� : d

de + ) e_ 1 1 and in the third term let x = f3(e - J1). PIl x/f3) dx xlf3) dx . ( 1 . 8 6) eX + l eX + l f3 f3 0 _

0

37

what we have called "reasonable" values of p and T not much For error W I·11 . . arIse I f we replace g(J1 ± x/f3) by g(J1) ± (X/f3)g'(p) and Sgll' by S O'. Then

(1 3

=

4n 2 3 (2nh)3 We will now tackle the case of finite T. For T # specific heat, etc. ? (N )

0

(1 .82)

=

_

2

p4

The Fermi-Dirac gas

u = u V = Uo + y V T 2 = Uo + y ' T 2 C y = au = 2y T , a

T

Cy of a metal =.,2y'T + elect.

I

rxT3 at low T

lattice vibrations

38

Introduction to statistical mechanics

At very low T the 1" T term dominates, and the contribution of the electron gas to the specific heat is noticeable. Cv of a metal = 2y' T + 3R at high T. Since the part of the specific heat due to lattice vibration is constant, the y'T term is again detectable. Note that at very high T, we have to include higher terms in the expansion of gee), and the part of the specific heat due to the electrons is no longer propor­ tional to T. We can demonstrate more easily than we have done above that the internal energy is proportional to T2. At any temperature less than the Fermi temperature almost all the states with energy less than the Fermi energy are filled, and almost all states with energy greater than the Fermi energy are empty. As the electron gas is raised to temperature T from T = 0, the average energy than can be im­ parted to an electron is about k T. Also, only those electrons within about k T of the Fermi level will be excited, because electrons with less energy have no place to go-the states are filled. The number of electrons with energies between Jlo and Jlo k T is

N

=

illO

1l0 - k T

-

ae 1 / 2 de.

For Jlo » k T, N is proportional to k T. Thus the internal energy added is pro­ portional to (kT)kT k 2 T2 .

CHAPTER 2

DENSITY MATRICES

2.1

INTRODUCTION TO DENSITY MATRICES

When we solve a quantum-mechanical problem, what we really do is divide the universe into two parts-the system in which we are interested and the rest of the universe. We then usually act as if the system in which we are interested com­ prised the entire universe. To motivate the use of density matrices, let us see what happens when we include the part of the universe outside the system. Let x describe the coordinates of the system, and let y describe the rest of the universe. Let cp ;(x) be a complete set of wave functions. The most general wave function can be written

t/J(x, y)

=

L Ci ( y)cp ;(x). i

=

(2.1)

At this point we will convert to Dirac notation. Let I CP i ) be a complete set of vectors in the vector space describing the system, and let I e ;) be a complete set for the rest of the universe. and The most general wave function can be written (2.2)

t/J(x , y)

=

< yl< xl t/J )

=

L Cij< xlcpi)< yl ej) ij

We can obtain Eq. (2. 1) by taking

C;( y)

=

L ci/ yl ej) . j

Now let A be an operator that acts only on the system ; that is to say, A does not act on the I e). When A acts on product states (for example, I t/J ») we really mean A la) l b) == (A la») l b). In such a case A does not equal

/A cp L ii' wl i>
40

, 1/2.1

Density matrices

but equals

Any system is described by a density matrix Li wil i) (i l and

L Aii'I
Pi'i

=

=

L C�Ci '/
=

L (
L C ij* C i 'j j

=

(tltIAltIt )

(2.3)

=

=

(A)

(2.4)

(2. 5)

L Wi = Tr P i If we let A be l i')( i' l we have Wi '

Therefore,

=

Tr p A

=

(A)

Wi

�

=

0

(2.6)

L w;/ i )( i l · i

=

(A)

(tltIAltIt)

and

=

(tit I tit)

=

=

Tr pA .

L ( i' l pAl i' ) i'

=

Li w;( i'l i) ( i IAl i' ) ii

L wi(i IAl i). i

(2. 10)

. system is in state i. If all but one of the W i are zero, we say that the system is in a otherwise it is in a It is easy to show that a necessary and sufficient conditi on for a pure state is p = p2 . If a system is in a pure state, t l ipure ) we can express the matrix as

.

pure state ;

i'

· Therefore it can be diagonalized From E q. (2 . 4) , it is obvious that P is hermitian. . . I ues W i ' ' ors with a complete orthonormal set of eIgenvect I I ) and reaI eigenva

If we let A be 1 , we obtain

Tr pA

=

.

L l
=

where p is of the form

Since < i I A l i) = the expectation value of A in the state Ii), it is obvious from : (b), (c) and Eq. (2. 10) that we can interpret the W as the probability that the i

Where we have used the result

P

=

(

l
=

(A )

Notice that

. . densIty rnat fiX.

We define the operator P to be such that Pi ' i the system described by x.

p,

41

a) the set I i ) is a complete orthonormal set of vectors. b) W i � O. c) L i Wi = 1 . d) Given an operator A, the expectation of A i s given by

Then

where

Introduction to density matrices

More generally,

mixed state.

p

=

P ij

= =

i

Iipure ) < pureI , <
(2. 1 1) (2. 1 1 a)

=

1.

=

� ( tltl i') IOj»)« Ojl ( i ' ltIt»)

=

L 1( i' I (OjDl tIt)12• j

(2.7)

If it is possible to discuss the system in the x-representation, we can write

p(x', x) d�. ( x' l p l x)

J

(2.8)

(2.9)

We now consider the concept of a density matrix indepe�dent of the pre­ ceding motivation. First let us reformulate quantum mechamcs :

=

L wi( x' l i) ( i l x ) i

L wi i(x') i *(x), i which, for a pure state I i ) , becomes p(x', x) .In the x-representation we write


=

Tr /A

=

=

i (x')i* (x)

f

dx<x l pA lx) .

(2. 12)

(2. 13)

42

%.1

Density matrices

But

<xlpAlx) = <xlp (f dX' I X ')<X' I ) Al x) = f dx'<xl p l x ')<x'I A l x ) = f dx' p(x, x')A(x', x).
Introduction to density matrices

In using Eq. (2.11), we took

14' 1 ) =

G)

(2.14)


The x-polarized state : Let a

The y-polarized state: Let

1 and b = O in Eq. (2.17). Equation (2.18) gives

(2.15)

The 45°-polarized state :

Wave function for the x-polarized state (2.16)

lal 2 I b l 2 = 1. +

When we use Eq. (2.11), the density matrix for the pure state Eq. (2.17) becomes (2.18)

(� �) .

(2.20)

a = 0 and b = 1. =

1/J2 gives

(t i).

(2.21)

a = - l/J"2 and b = l/J"2 gives

The 135°-polarized state:

P13S0.pol. =

(

-

t -!-

)

-t ' t

(2.22)

The p's in the last four equations are for pure states. Now consider the following two mixed states :

1.

Mixture of 50% x-polarized and 50% y-polarized states: the density matrix for this mixture is

2.

+

matrix for this mixture is

p = tP45°'pol. !b

From Eq. (2.11a)

p tpx.pol. -!-Py.pol. = (� �). Mixture of 50% 45°-polarized and 50% 135°-polarized states: =

(2.17)

(2.19)

P45'.pol. =

Wave function for the y-polarized state

Any pure state can be written as a linear combination of the two states in Eq. (2.16) :

G �) .

a = 1/.'/2 and b

p.

ab*) p = (aa* ba* bb* .

=

Py.pol. =

From our example in which we split the universe into two parts, we see that pure states are not general enough to describe a quantum mechanical system that does not include the whole universe. It is unknown whether or not the universe is in a pure state. To reformulate quantum mechanics in terms of the more general density matrices, it would be appropriate next to find the equation of But first, as a simple example of a density matrix, let us try to motion of describe polarized and unpolarized light. Consider a beam of light travelling in the z-direction. First define

where

=

Px.pol.

We see that in this case

(�) (�)

and

Now to examine four pure states and the density matrices corresponding to

them :

Considering again the problem of a system plus the rest of the universe, we can easily show

p(x, x') = f t/I(x, y)t/I *(x', y) dy.

43

+ tP13So.pol. =

(� �).

(2.23) The density (2.24)

us these two mixtures have the same density matrix and show the same phys­ Ical effect. Note that a given pure state (for example, x-polarization) determines a sta e vector ( ave fu ction) only IUP to a phase factor, whereas the density � . � � Illatnx IS determmed umquely.

44

Density matrices

2.2 ADDITIONAL PROPERTIES OF THE DENSITY MATRIX

Recall that the density operator can be written

(2.25) where the system is in state Ii) with probability Wi ' As time changes, the possible states of the system also change ; so pet) = L wil i(t»
(2.26)

It is easy to find how li(t» changes in time, for we can expand l i(O» in eigen­ states of the Hamiltonian, H, and we know how those eigenstates change in time. Let HIEn) = EnIEn) .

(2.27)

n It follows that n

For convenience we use units in which Ii = 1. If we definef(H) for any function f by the equation f(H)IEn) = f(E.) I En ) , we can write

li(t»

=

e- iHtIEn)<En li(O» L n

=

e - iHtl i(O» .

(2.28)

Notice that, according to our definition off(H) , if f(En) can be expanded in a power series for any eigenvalue of H, then/(H) can be expanded in the same pOwer series. Substituting Eq. (2.28) into Eq. (2.26), we get pet)

Additional properties of the density matrix

1.1

=

L Wie - iHtl i(O» < i(O)l eiHt = e-iHtp(O)eiHt.

i

(2.29)

The signs of Eq. (2.30) and Eq. (2.31) are opposite. We could put Planck's constant back into the equations by replacing every H by Hili. From Eq. (2.29) we obtain Tr pOet) = Tr [e -iHt(p"(O)eiHt)] = Tr [(p"(O)eiHt)e -iHt] = Tr pOCO)

- i(Hp - pH).

p" = L wi l i )< i l , i

sp

f(p) = L f( wJ li )< i l .

. Y0.u may be suspicious about treating H like a number when taking the denvatIve of terms such as e - iHt, but such a procedure can be justified by expanding e - iHt in a power series. Alternatively, we can obtain Eq. (2.30) from Eq. (2.29) by taking the derivative of < En lp(t ) I Em ) . In any case, Eq. (2.30) is . true, and It plays the same role for density matrices that Schrodinger's equation plays for wave functions. It should be noticed that an observable A in the Heisenberg representation obeys A =

+ i(HA - AH).

(2.31)

(2.33)

i

Tr f(p) = :Ld(w), and the w; ' s are constant. [being the eigenvalues of p and thus unchanged by a unitary transformation such as Eq. (2.29)] ' It is possible to define "entropy" by the equation

(2.34) A pure state has "S" = 0, whereas very impure states have large positive "entropy." The quotation marks are used because, by our definition, the "en­ tropy" cannot increase in time but must remain constant. Next let us try to get a bit more experience in dealing with density matrices. In ordinary quantum mechanics it was easy to find such things as the expectation value of the position, the expectation value of momentum, and the probability that a system is in a given state. (a) Expectation value of position : According to Eq. (2. 14), if A is an operator,


(2.32)

where we use the fact that Tr AB = Tr BA . It follows that Tr p " is constant in time, and consequently Tr f(p) for any function f is constant. This result can also be obtained by noticing that

Taking the derivative of p with respect to time, we get p =

45

=

In this case A = x, and so

f

dx dx'p(x, x')A (x', x).

(2.35)

A (x', x) = <x'I A lx) = x<x'lx) = x c5(x' - x),

<x )

=

f

(2.36)

dx xp(x, x).

, " 0.) Expectation value of momentum : Now we must take A =

<xl pit{!) =

Ii -a <xlt{!),

-: I

ax

p.

Since

46

Density matrices

Density matrix in statistical mechanics

we have

A(x', x) = �i �ax f>(x' - x).

(2.37)

Integrate by parts ; for any function / we have

Thus the

density matrix is

-

Because

H I CPn)

I

(c) Probability that a system is in state

Ix):

x' � x

(2. 38)

Finding the probability that a system is found in the state I x ) is equivalent to finding the expectation value of the operator I x ) <x l . To see this for the case of a pure state, notice that if the system is in a state the probability of experimentally finding it in state I x ) is

I t/I), l <xlt/l) 1 2 = < t/llx) <xlt/l ) =
Therefore, the probability that the system is found in state I x ) can be written P

= probability = Tr plx><xl = Tr Li w;l i><xl = L wil 12 = <xl plx>·

(2.39)

P

so

= L wil12 i

Ixo), I t/I), p(xo, xo) = I t/I(xoW.

Xo

= En I CPn ), we can write Eq. (2.41) as 1 � e-PHI CPn>
The average energy of the system, U

�en we denved

(2.41)

p(xo, xo).

When I i ) is an eigenket (eigenfunction), and Ei is the corresponding eigenvalue of the Hamiltonian of the system, the probability that the system is in the state is

H

(2.40)

(2.42)

(2.43) (2 .44) (2.45)

U, can be written

= Tr pH = TrTr[H[ee--PHPH]] .

(2.46)

know the free energy and the energy of the system, the entropy is from

F = U - TS.

Ho is any other Hamiltonian,

(2.47) (2.48)

where

(2.49) and

So

Ho. F in Eq. (2.48) is the true free energy H, as written in Eq. (2.44). When we Ho as Fo =
and

subtract Eq. (2.50) from Eq. (2.48), we may write

is t �e entropy calculated from CO�espondmg to the true Hamiltonian Write the free energy corresponding to

DENSITY MATRIX IN STATISTICAL MECHANICS

CP ICPi)

p(x, x') = � � e - PE'cp.(x)cpi(x').

It can be shown that, if

is in accord with our interpretation of the W i as the probability that the system actually is in state Ii ) . The probability density at value for a coordinate is the probability that the system is in state which by Eq. (2.39) is For a pure state 2.3

where

i

The equation

where

n

Alternatively, the coordinate representation is

ff(X) �ax f>(x' - x) dx = f f>(x' - x) �ax f(x) dx = -f'(x') < p> = -f dX dx ' � f>(x ' - x) � p(x, x') ax -� f dx [� ax p(x, X')] I

47

F .:::;: Fo

+

Tr

[(H - Ho)e - PHO] . Tr [e - PHO]

(2.5 1)

48

Density matrices

Linear harmonic oscillator

This is a diffusion-type equation, and we can write down the solution readily :

This theorem will be proved later, in Section 2. 1 1 . Now, we regard the density matrix as a function of [3 :

p(x, x' ; [3)

(2.42)

J

=

p(XX' ; 0)

In place of P u we will hereafter write p. Next we will show that p([3) obeys the differential equation (2.53)

Hp.

The proof is as follows. In the energy representation, Eq. (2.52) can be written as (2.54)

Differentiating this equation, we have (2.55)

=

p(O)

e - PF

(2.56)

1.

=

HXp(XX' ; [3).

(2.57)

(2.61)*

=

f

p(X, x) dx

=

This is the partition function for the system.

Problem:

J

L

mkT . 2nli2

(2.63)

Show that the partition function for a three-dimensional system with is e - PF = V N(JmkT/2n Ii2)3N . (2.64) Another way of deriving Eq. (2.61) is as follows : =

L e - PEmt/lm(x)t/I!(x').

(2.65)

m If the particle is in a large box of volume V, we take

�

�

f Vd 3p

and

(2nli)3

Then Eq. (3.65) becomes

Here the subscript x on Hx indicates that Hx operates on x in p(XX' ; [3). The initial condition is (2. 58) p(XX' ; 0) = <5(x - x'). 2.4 DENSITY MATRIX FOR A ONE-DIMENSIONAL FREE PARTICLE

We want to solve the differential equation discussed in Section 2.3 for a one­ dimensional free particle. The Hamiltonian is H

=

and the differential equation (from Eq. (3.57» is - op(x , x', [3)

0[3

(2. 59)

p2/2m,

-

-

as

z.s

(2.66)

in Eq. (2.61). LINEAR HARMONIC OSCILLATOR

ib.e Hamiltonian for a linear harmonic oscillator is H

,

------

fz2 02 2 p(x x , [3). 2m ax

_

I

(2.60)

(2.62)

N particles

We can write Eq. (2.53) in the position representation as follows : - op(XX' ; [3)/0[3

=

p(x, x' ; [3)

from which we obtain Eq. (2.53). The initial condition for Eq. (2.53) is

J

<5(x - x'). For a linear system of length L, the trace of Eq. (2.61) leads to

(2.52)

=

[( )

m exp - � ( X - XI )2 . 2nfz 2 [3 2fz2[3

The numerical factor is chosen such that

The unnormalized p i s defined by

- op/o[3

49

=

p2j2m + m w2x2j2,

(2.67)

for constant, no expansio� i� powers of P exists near P O. This is, ��eeoted, tha� an Importan t feature of and \fIll mfluence the construction of perturbation 8eries. x,x

'

Pfr ee

=

50

Linear harmonic oscillator

Density matrices

so that the differential equation for p is 112 02 - op p 2 m 8x2 op __

=

-

_

_

Equation (2.76a) is integrated to give +

mw2 2 . xp 2

a

(2.68)

__

=

(t) coth 2(f

-

51

/0)'

wh�re the initial condition Eq. (2.73) requires that the integration constant .fc0 vanIsh : a = (t) coth 2/ (2.77)

Let us write (2.69)

Equations (2.76b and c) are integrated as

which simplifies Eq. (2.68) to - 8p df

_

=

The initial condition is p

=

p

=

or

�

<5(x - x')

J

02 p

b c

(2.70)

+ � 2 p.

for f

mw <5(� - 0 fz

=

(2.71)

0,

for f

=

p

+

=

-./

p

b(f)�

+

c(f)]}.

a'e

+

b'�

+

=

(I - 4a2)�2

-

4ab� + 2a - b2 ,

b'

-

=

- 4ab,

(2.76c) This is only true for '; -:f- ';'. When they are equal, p(';, ';', f) is equal to Pfree only for f o. See Section 3 . 1 0 for details. Obviously, when T is large and the classical , motion has large amplitude, the larger the amplitude the less will the oscillator feel a force between fixed x and x'. *

=

e coth 2f + 2

)J

A2 � coth 2f + smh 2f 2

]

[

.

B �2 + 2A� + A2 exp _ . 4f -./2f

-+

(2.79)

(2.80)

=

_

�

',

B

=

J':n� .

(2.81)

Using Eq. (2.81) in Eq. (2.79), we obtain the final form :

p(x, x' ; p)

=

J

mw 2nfz sinh 2f exp

x

(2.75)

(2.76b)

_

A

(2.74)

Eq. (2.75), where a prime (') means a derivative with respect to / From (2.76a) a' = 1 4a2 ,

[(

(2.78)

For this result to agree with Eq. (2.73) we see that

Use of Eq. (2.74) in Eq. (2.70) leads us to the equation

c'

B exp . smh 2f

-

ing form : Knowing the Gaussian dependence of Pfree on �, we try the follow exp { - [a(f)e

A/sinh 2/ (t) In (sinh 2f) + (A2/2) coth 2/ - In B,

When we let f -+ 0, Eq. (2.79) gives

(2.72)

0.

(2.73)*

p

=

where A and B are constants. Use Eqs. (2.77) and (2.78) to get

l, <5(x xo) = The factor comes from the scale change in the <5-function. In genera 1f'(x) I<5[J(x) - f(xo)] for f monotonic. t like a free For low f (high temperature) the particle should act almos that, for expect we particle, as its probable kinetic energy is so high. Therefore ly by ximate appro low f, the density matrix for a harmonic oscillator will be given

=

=

'or

•

;

p(x, x' ; P) .t

.

=

J

{

�i:l�2f

2fz

[

mw 2nfz sinh (fzw/kT)

x

exp

{

- mw

2fz sinh (fzw/kT)

The special case when x p(x, x ; p)

=

=

J}

(X2 + X'2) cosh 2f - 2XX'

[(

x 2 + x' 2) cosh

x' gives

J�- (.=!!!!!!. 2nn smh 2f

exp

n

- 2xx']} .

fzw kT

)

x2 tanhf .

(2.82)

(2.83)

(2.84)

52

Density matrices

2.6

This is the probability for finding the system at form. Equation (2.84) is used to calculate

2p(x, x) dx ( x2) = Jxp(x, x) dx J

=

x. Notice that it is of Gaussian

� coth f = _li_ coth liro . 2mro

2mro

)mro2 (x2 )

(Potential energy) = (t

(2.86)

We may compare this result with the average of total energy calculated by means of the partition function. We know (Energy) Equation

liro - - 2

Iiroe- liro/kT = liro 1 + e- 2f = liro coth f + li ro /kT e 1 2 1 e - 2f 2 -

(2.86) is exactly half of Eq. (2.87).

-

-

(Iiro/4) coth f

(2.88)

e- flF = f p(X, x) dx = J21tn sinh 2f Jmro 1t1itanh f ,�

=

1

2 sinh f

F = !p In (2 sinh !) = kT In (eliro/2kT e- liro/2kT) liro + kT In (1 e - liro/kT . = ) 2

(2.89)

_

This is the free energy, as already derived. We may examine the limit of high temperature, or small /in Eq. this limit �

exp

( m;;x2/2) -

=

exp

( :iX») -

,

(2.90) (2.84).

In

�

J:: (�k";) ( _ �:X2) ( -m:n2xl2) . exp

This limit is calculated as

p(x, x')

=

L i

exp

Consider the potential energy shown by the curve in Fig. 2. 1 , given by the follow­ ing function :

V (x) = mro2 2 x2 + kx3 �We will consider the area near x = 0 only, so that the region V (x) come into the calcula •

-

exp

e - PE'
(2.92) (2.93)

(2.95)

< 0 does not tion. Because of the anharmonicity, when the temperature .. increases the mean tposition of the oscillation moves out, as shown in Fig. 2.2. Bearing this in mind,

'., ,, '.

Vex)

\,\

" ,"

"

--�� ��

---

.x

-----

Fig. 2.1 Anharmonic oscillator (k

<

0).

(2.91)

except for the factor in front of the exponential. This result, Eq. (2.91), agrees with classical mechanics. The low-temperature limit is found by inspection of Eq. (2.83) :

p(x, x')

ANHARMONIC OSCILLATOR

\

_

p(x, x; P)

(2.94)

(2.84) :

mro

which leads to

(2.87)

Therefore we know that

(Kinetic energy) = (Potential energy) = The partition function is derived from Eq.


2.6

(Iiro/4) coth f

53

because when p � 00 , only the ground state has an effective contribution. Equation (2.92) agrees with Eq. (2.93) because we know that for the ground state

(2.85)

2kT

The average of the potential energy is thus

=

Anharmonic oscillator

\----1.- a __

---J Vex)

-_ _

I I I

I I

-------����---_ x

Fig. 2.2 Mean position of the oscillation is displaced.

54

Density matrices

2.6

we will treat the problem using the minimum principle discussed previously. The principle is written as (2.96)

F

which describes a harmonic oscillator with its center displaced by an amount a. The w in Eq. (2.97) is the same as that in Eq. (2.95). From Eq. (2.90) we find that

Fo = kT In (2 sinh (fzwj2kT)). o =

(

mwZ - xZ 2

+

)

mwZ kx3 - - (x - a) z . 0 2

If we make the substitution y = x - a, Eq. (2.99) becomes

o =

/\ mwZay

+

mwz 2 aZ

+

ky3 + 3kyZa

+

3kyaZ

+

ka3

+

Force = mwzx

< Force) = mwza + 3k
+

(mwZj2)aZ + ka3 + 3ka
a)

+

3kyZ

)

"

In order

mwza + 3k0 = mw2

3kh coth (fzwj2kT) 2mZw3

'

..

(2. 1 07) (2. 1 08)

pZ Ho = 2mo

'

o (2. 100)

,and vary a, mo, and wo. Show that m; i) mo

(2. 1 04)

(2.11 0)

(2. 1 1 1)

W o and a are determined from < V' (x) o = 0, <xV'(x) o =
(2. 1 02) (2. 1 03)

m w2 (x - a)2, 2

+ �

=

ii)

(2. 1 09)

to apply the variation principle in Eq. (2.96), try the following Ho :

Vex)

The best estimate of F from Eq. (2. 103) is obtained by minimizing the right­ hand side. In so doing we neglect the term ka3 because we are interested only in small oscillations, so that Vex) < 0 does not occur. Then the differentiation of Eq. (2. 1 03) with respect to a leads to that is,

+

H = pZj2m + Vex).

Substituting Eq. (2. 1 02) in Eq. (2. 1 00) yields


+ . . . .

+

(2.99)

From Eq. (2.85), we obtain

+

3kxz = mwZ(y

(2 . 98)

(2. 101)

F :s; Fo

(2. 1 06)

Therefore, when we neglect the higher powers of a, we arrive at Eq. (2. 104). !Problem: Consider a general bounding potential Vex) like that shown in Fig. 2.3. The Hamiltonian is

Because Ho = pZj2m + (mwZj2)yZ, we can easily calculate the expectation values of the various powers of y. By symmetry,


:s; Fo - (mwZj2)az.

I� Eq. (2: 105), th� displacement of the mean position, a, has a simple interpreta­ . tIon. It IS the pomt at WhICh the average force on the oscillator vanishes. This result is seen as follows : from Eq. (2.95),

(2.97)

Ho = pZj2m + (mwZj2)(x - a)Z,

55

In this case, Eq. (2. 103) becomes

Here F is the true free energy and H is the true Hamiltonian. When Ho is any Hamiltonian, 00 means the average taken in a system characterized by Ho. Fo is the free energy of the system of Ho. In the present problem we take for Ho

Now we have only to find

Anharmonic oscillator

---t-----------__ x

(2. 105) Fig.

2.3

A

general bounding potential.

(2. 1 12) (2. 1 13)

56

Density matrices

2.6

Here 0 0 is taken for Ho. Equation (2. 1 12) comes from the consideration of the absolute location of the system, and Eq. (2. 1 13) comes from the scale of this variable. The latter corresponds to the vi rial theorem, which states that (2. 1 14) < x V' (x) =
To show how Eq. (2. 1 14) is related to considerations of the scale of the variable, let us derive the virial theorem in a more general form. Consider a system inside a cubical box with sides of length L. The Hamil­ tonian is taken to be p'? (2. 1 15) + L V eri)' HL = L i 2mi ij i<j

(yVe assume that the motion of the system can be described by a potential depending only on the distances separating pairs of the particles making up the system.) r ij = Ir i - rj l . If the system is bound, then for a sufficiently large box the pressure is essentially zero. But in general, 1 aF aF P = Pressure = - -= -- - . (2. 1 1 6) 3I3 aL a (Vol)

=

Let FL F for a box of length L. When we later prove the theorem that F ::; Fo + < H - Ho)o, we will in fact prove that < H - Ho)o is the first-order correction in the perturbation expansion for F. We get an expression for P by writing FL( l +e) � FL + correction term or (2. 1 17) FL( l +e) � FL + (HL( l + e) - HL)Ih' Now HL( l + e) differs from HL only in that (ri)x can run from 0 to L(l + e) rather than 0 to L, and similarly for (ri)y and (rJz . Therefore let us make a change of variables, such that r[ = rd(l + e). As pf = - fz2 Vf, we obtain p'� 2 + L V [( l + e)r[J HL( l + e) = L ij i 2 m ;( 1 + e)2 i<j

[

]

(2. 1 1 8) L P? + L V(r ij) + e - 2 L E� + L r[X '(riJ i 2mi . ij . i 2m . ij . 1<) I <J Since the new coordinates (primed) run over the same range as the original (unprimed) ones in Eq. (2. 1 1 5), we can eliminate the primes over the r ij and Pi and write �

[

(HL( l +e) - HL) HL = e - 2 ( kinetic energy) +

(�<'jJ. rij V '(rij))J. ,

(2. 1 19)

Anharmonic oscillator

57

From Eq. (2. 1 1 6) and Eq. (2. 1 1 7), (2. 120)

From Eq. (2. 1 19), 3P V = 2 ( KE ) -

(

)

L rij V '(rij) . ij , < J.

(2. 121)

For the case of a harmonic oscillator, P is zero and we get <x V ' (x»

=


(2. 1 22)

This result should be compared with Eq. (2. 1 1 3). For the case of a particle moving in an inverse-square force field we get the ' classical result
(2. 1 23)

This method can be generalized. Let H be the Hamiltonian for the original system and let He be the Hamiltonian with the box changed in an infinitesimal manner. Suppose FH = FH and suppose that He - H = eS. Then the above 6 method gives <S)H = O. For example, in Eq. (2. 1 1 8) we have the case for which S =

[

-2

� {� + i�j rij V '(ri)

]

•

. As another example, suppose that the box undergoes an infinitesimal transla­ tIon, e, in the x direction. We get, by taking x ' = x - e, ,2 He � � J!L + L V e r i) + e L a v (1';) , (2. 124) 2m i ij ij i <j ;<j ax so that I

av (r ij ax i) i<j

(2. 125)

)

(2. 1 26)

S = L

and

(ri) /\ Lii � i <j ox Compare Eq. (2. 1 26) with Eq. (2. 1 1 2).

= o.

58

2.7

Density matrices

2.7 WIGNER'S FUNCTION The density matrix may be written in the coordinate representation and in the momentum representation as follows : p(x, x')

=

L e-PE'

p')

=

L e -PE'
pe p,

=

The diagonal element

(2.127)

i

f

p(x, x')e( - i/h)(p · x - p" X') dx dx' .

(2. 128)

p(x, x)

==

(2. 1 29)

P (x)

p ep, p)

==

(2. 1 30)

Pcp )

is proportional to the probability for finding the particle at P in momentum space. The expression Eq. (2. 1 30) is used in calculating the average value of the kinetic energy : =

Tr [(p2 j2m)p] Tr [pJ

=

f pe p , p)( p2 /2m)(dp/2nn) . f pcp , p)(dp/2nn)

(2. 1 31 )

In clissical mechanics, the density function f(p, x) in phase space has the following property : P c p)

P (x)

=

=

ff(p, x) dx,

(2. 1 32)

f f( P, x) � . 2nn

We ask ourselves if there is any fUI1ttion f(p, x) in quantum mechanics that satisfies Eq. (2. 1 32). One answer is Wigner's function fw(p, x) :

fw( p, x)

=

f(

p x

+

�, x

-

�)

e - i(p�/h )

d,., .

(2. 1 33)

This function is derived by regarding p(x, x') as a function of (x + x')/2 and (x - x')j2, and then writing (x + x')/2 as x and making a Fourier transform with respect to (x x')/2. -

59

Let us check that Eq. (2. 1 33) satisfies Eq. (2. 1 32). First,

dP ffw( P, x) 2nndp f p(x 2 f12Xf e- i(p�/h) 2nn ) d,., f p(x + � , x - �) dl] p(x, x) =

+ f1 ,

x

-

=

b(I])

=

=

P (x).

(2.134)

Thus the second part of Eq. (2. 1 32) is satisfied. Next, using Eq. (2. 1 28) and Eq. (2. 1 30), we have P c p)

represents the probability for finding the particle at x (neglecting the normaliza­ tion constant IjTr p) . Similarly

/�) \ 2m

Wigner's function

= =

f p(x, x')e - (i/h)p(X-x') dx dx' f p( y � , y - �) e- ip�/h dy d,., ffw( p, y) dy, pc p, p) =

=

+

(2.1 35)

where we have made the change of variables : x x'

=

=

y + ,.,/2 y - 1]/2.

(2. 136)

From Eq. (2. 1 32) we see that, if h(p, x) is either a function only ofp or a function only of x, then

(h( p, x»

=

f fw( P, x)h( p, x) 2nndp dx.

(2. 137)

However, Eq. (2. 1 37) is not true for a general function, h(p , x). Although Wigner's function fw(p, x) satisfies Eq. (2. 1 32), it cannot be regarded as the probability for finding the particle at the point x and with the momentum p, becausefw(p, x) can become negative for some values ofp and x.

Pro�lem : Calculate Wigner's function for a harmonic oscillator. This is a specIal case where fw(p, x) is always positive. In �q. (2. 1 33), we derived Wigner's function from the density matrix. We can denve the latter from the former :

ffw( p, x � x') ei[p/h](x -x') 2d:n

=

p(x , x') .

(2.138)

Thus fw(p, x) and p(x, x') contain the same information. Problem : Show that the momentum representation of the free-particle density matrix is pe p, p ') = 2nM( p - ')e- (p2/2 m)/kT: . (2. 1 39) p

2.8

Density matrices

60

2.8

SYMMETRIZED DENSITY MATRIX FOR N PARTICLES

We will now discuss the density matrix for a system of many identical Bose or Fermi particl�s. The symbols

will mean

D S A

Distinguishable Symmetric Antisymmetric

H = -2m-

� "'"

k

We can construct Ps starting from PD ' As an example consider the case of two particles : PD(X1X2 ; X�X2)

=

all states

L

e - fJE'l/J i(X1 , x2 ) l/JiCx � x;) .

PD(X1 X2 ; X2X� )

L

e - fJE'l/Ji(XI X2 )l/Ji(X2X� ) ,

L

e- fJE'l/J i(XIX2 H[l/J i(X� X2) + l/Ji(x2 x m

=

all states

Take the average of the two :

Vk2 •

t[PD(X1 X2 ; X�X2) + PD(X 1 X2 ; x2 x m

(2. 140)

all states

The density matrix PD satisfies the equation The solution is , PD(X l , X2" " XN ,. X l , X 2 , · · · XN , f3) "

.

( -) m

2n 1i 2f3

-�-

exp

[

-

m-2 1i 2 f3

� (xk - xk, ) 2

J

because

= all Lstates e - fJE'l/Ji(X1 ' "

(2. 142)

xN) l/Ji(x'l . . . x�) .

When the particles obey Bose statistics, then Ps(X 1 ' " ; x � . . . X� ; f3)

=

L

_

Symmetric states only

e - PE ' l/J(X l • • • xN)l/Ji(x'l . . . x�) .

(2. 1 43) (2. 144)

We introduce the notation P to indicate the permutation of particles. For example, for a permutation 1 2 3 4 5 P: L L L L L 5 4 3 1 2

1 2"

•

In general when particles are interacting with each other we can write PD(X 1 , X2 , · · · XN ; x � , Xl" . . x� ; f3)

L

symmetric states only

(2. 141) 3N/2

61

Now interchange x � and x; :

First consider the case of free particles. The Hamiltonian is _ 1i 2 -

Symmetrized density matrix for n particles

, 2' ) [,1,*( 'I' i X1X

+

,1,*( i X2, X 'I )] 'I'

=

{O

l/Ji(X�X2 )

anti symmetric symmetric l/Jr

,I,, 'I' :"

In general, a group-theoretical argument can show that

� � .I,(PX I . . . Pxk ' " 1:' 'I'

N!

PxN)

_ -

{

l/J(X I 0

. . . xk . . . xN)

symmetric for any othe r symmetry.

(2.145) Here Lp means that we sum over all perm utations. When we use Eq. (2. 1 45), the case of N particles can be written as

= N1' Lp PD(X I •

• • • Xk • • • ; Px� . . . Px� . . . Px�) .

(2. 1 46)

Thus the partition function beco mes

we have When the state l/J i is symmetric, (2. 1 47)

62

Use Problem : Derive the partition function for a noninteracting Bose gas. Eq. (2. 1 42) in Eq. (2. 147) and first write

N1 2 3 1 ( ) e -fJFs = N ! 2�h2/3 m

x

I p

Symmetrized density matrix for n particles

2.8

Density matrices

J { 2hn�/3. [(X l

h v = V (brh2/3v)3/2 . m

(2. 1 52)

We must now count the possible permutations. number of permutations that have

exp -

V

Using Eq. (2. 1 5 1) we obtain

C [ cycles of length 1

+

(XN - PXN)2]} dXl ' " dxN· (2. 148) In taking Lp , we start with the identity permutation that gives V N from the inte�­ . ral and next go to a pairwise interchange of particles, three-p� rtIcle cychc permutation, and so on. The combinatorial factors must be taken mto account.

C2 cycles of length 2

Cq cycles of length

Then

q

(2. 1 53)

Solution : Any permutation can be broken into cycles. For example,

P

=

(1 � � 1 f 1) 5 3 2 4 6 1

=

(1 56)(23)(4)

= (� f 1) (� �)(1). 5 6 1

3 2

4

=

Cv cycles of length v, where Lv vCv 3V/2 dXI dX2 . . . dxv ) hv = (2n_�_ h2/3 J (X2 - X3)2 . . . exp { �1_ [(Xl - X 2)2 2h2/3 + (Xv-l - Xv)2 (Xv - Xl)2]} h I = (2nh2/3)3/2 J dXl V (2nh2/3)3/2 .

Suppose a permutation contains > 1, let

v

_

x

(2. 149)

+

=

Then Eq. (2. 148) can be written

e -fJFs

=

It is not difficult to show that

J dy

exp [ -

m

As a preliminary to the determination of M(C[ , . . . , Cq), consider the case of N = 6, C[ = 3, = 1, = C4 = C5 = C6 = O. An example is the permutation P = (5)(3)(2)(461). Notice that the sequence Q = 5, 3, 2, 4, 6, I corresponds to P. In other words, from the sequence Q, we can uniquely con­ struct the permutation P with C[ = 3, = 1 . But from Q' = 2, 5, 3, 4, 6, I , or from Q" = 5, 3, , 1 , 4, 6, we can also construct in the same way the same P. 18 different sequences In fact, it can be easily seen that there are 3 ! times 3 that can be used in the way described above to construct the permutation P = (5)(3)(2)(461). Generalizing to any set of C ' s, we see that there are two ways we can get a new Q that leads to the same P :

C3

C2

C3

2

=

a) we can interchange cycles of the same length, b) we can make cyclic permutation within a given cycle.

TIv

Cv TIv Cv! TIvTIv vvCv

There are Cv ! ways of interchanging cycles of the same length, and cyclic permutations within cycles. Thus for each P there are different Q ' s that lead to the same Because there are N! possible Q ' s,

( v (h.)Cv) .

�I n N! p

P.

(2. 1 50) It follows that

a(x - y)2] exp [ -bey - Z)2] ( n )3/2 exp [ - �ab+ b (X 2J = a -b +

N. For

+

+

_

m

63

- z)

(2. 1 54) .

(2. 1 5 1)

where

Lv vCv

= N.

64

Density matrices

2.9

Instead of trying to do the sum in Eq. (2. 1 54), we will find the free energy of the grand canonical ensemble (that is, we will let N vary).

As usual, let

a

e-PF =

00

=

Then,

PF

=

/v)]Cv

n L

Cv = 0

I

�

00 .

C)

=

..-

2

1t

-

=

p 1 Tr P i Tr P 2m 1 - __ � Th p �

f �022I P(XI

'

. . . , xN ; x� , . . . , X�)

I

Xj = Xi'

dXI . . . dXN

(2. 158)

[hv(aV (hv aV) � v � a h �V : ( � y V � V��

-

\2m ) =

/Pi

e + I' P. Then,

00

=

65

Notice that we are not now specializing to the case of statistical mechanics. To compute the expectation value of Pi/2m,we proceed as follows :

(2. 155)

N= t

where now each Cq runs from 0 to e - PF

L e - PFNe + NI'P .

Density submatrix

( hv -aV) v

Interchanging f1 and L, we have

n exp v

2P

v

-

/2

=

exp L

2•

V

where we define

P I (x, X')

.

_ _

J p(x, x2 , . . . , xN ; x' , X 2 , . . . , xN) dX2 . . . dXN .

Tr [p]

•

(2. 159)

is the one-particle density submatrix. Using this matrix we can calculate, for example, PI

(2. 1 56)

This result is identical to the one we obtained in a much simpler way earlier. When the density of the gas is not large, only the identity permutation is important for high temperatures or for the limit of n -+ 0, since we have such factors as e - [(mkT/2A2)(x - x')21. This means that the quantum effect appears only for low temperatures. For the antisymmetric case,

\f J!)

=

-N

f[:; 0:�2 P l (X, X/)l =x dx

=

Tr

N

Tr [ VP l ] .

(assuming that the particles ate essentially identical). Similarly,

\� V(Xk») f V(X)Pl(X, =

N

x) dx

=

[J; P I]

N

(2. 160)

(2.1 6 1)

The two-particle density submatrix is defined as (2. 1 62)

where

( - I )p

=

{

I

for even permutation - 1 for odd permutation.

Problem : Modify the argument leading to Eq. (2. 1 56) so as to obtain the cor­ responding result for a noninteracting Fermi gas.

(2. 1 63) Special cases of density submatrices are the distribution functions . The single­ particle distribution function is defined as Pl(X ; x). P l (X) The two-particle distribution function is =

2.9 DENSITY SUBMATRIX For N particles, the density matrix can be written

(2. 1 64)

P 2 (X, y) P2(X, y; x, y) . (2. 1 65) . In a uniform substance, such as a gas or a liquid, P t (x) is independent of x, and P2(X, y) depends only on the difference (x y) . =

-

66

2.10 PERTURBATION EXPANSION OF THE DENSITY MATRIX

- =

8p 8fJ

There are very few Hamiltonians for which we can solve Eq. (2. 1 66) exactly, but it may be that H is close to one such Hamiltonian, Ho . (2.167) (2. 168) We would like to use Po to obtain an approximation to p, which we expect to be close to Po = e - flHo. Because p is close to e - flHo, we expect eHoflp to vary slowly with fJ. =

Ho eHoflp +

eHofl 8p 8fJ

=

eHoflHop

-

eHofl Hp

=

- eHoflH I P·

Integrating Eq. (2. l 69) from zero to fJ, and remembering that if fJ

(2.169) =

0, we find

=

pCP)

(2. 166)

- Hp.

Po(P) +

(fl

J

0

- I: dP, [po(P - P')HIPO(P')]

(2. 170)

( fl '

dP'

J

(fl

0

(fl'

dP''[po(P - P')H I PO(f3' - P")H I PO(P")] (fl"

- J dP' J dP" J dP"'[ ] + . . . (2.1 73) o o o We can easily rewrite Eq. (2.1 73) in coordinate representation. For example, p(x, x ' ; P) = < x l p(P) l x'> �

Po(x, x ' ; P) -

f: <xlpo(P - P') (Yx"><x"ldX ") H1Po(P')lx'> dP',

If HI

=

=

and Eq. (2. 174) becomes p(x, x' ; P) = Po(x, x ' , P)

- foo- 00 J

0

Po(x, x" ; P - P') V (x")p o(x", x' ; P') dP' dx"

=

Po(fJ)

- f: Po(fJ - fJ')HI P(f3') dfJ'·

(2.1 71 )

The last term in Eq. (2. l 7 1) is small if HI is small, and it is a correction term to the approximate equation p � Po . If the correction term is small, we can use an approximate p(fJ') to obtain a much more accurate approximation to p(fJ) . For example, if we make the approximation p(fJ') � Po(fJ'), we have p(fJ)

�

Po (fJ) -

J: Po(fJ - fJ')H1 Po(fJ') dfJ'·

(2. 1 72 )

+ ' "

(2.1 75)

2.11 PROOF THAT F :::; Fo + < H - Ho>o

Therefore,

(2.174)

5-0000 Ix"><x"l dx" 1 . Vex), then (x" I H1Po(P')lx') = V(x")Po(x", x' ; P'),

Iwhere we use

( fl

p(fJ)

67

In usin? Eq: (2.1 72) as a new approximation to p(fJ'), we can find a still better approxImatIOn to p. Continuing in this manner we get

Recall that the density matrix in statistical mechanics satisfies

8 (eHoflp) 8fJ

Proof that F :::; Fo + o

1.11

Density matrices

=

Now that .we ha�e a perturbation theory, we are in a position to prove as , m Sec�lOn 2.3, that F � F + (H - H . Let p�omIse� o V H - Ho . o)o Smce F IS determmed by the equation e flF = Tr e fl (Ho + V) , the mos t 0bVIOUS ' , IOn , to approach IS ' to find an approxImat F by using the perturbation expansion of e- fl (Ho + V) = p.

-

-

(2. 176)

68

2.11

Density matrices

e-fJF = Tr e-fJ(Ho+V) = e- fJFo f: Tr [e -(fJ-u)Ho Ve - UHO] du fJ UI dUl dU 2 Tr [e -(fJ-UI )HO Ve -(UI-U2)Ho Ve -UzHO] - . ff

Taking the trace of Eq.

1m)

(2. 176) we obtain

o

0

Using the fact that Tr AB

f: J: II

e -fJF e -fJFo =

+

=

It is possible to simplify the second-order term. Let w x = Ul ' Then the integral becomes

L

Tr

f

Tr [e -fJ H oeWHoV e -wHoV ] .

A'

e -fJF = e- fJFo

=

..

13 Tr

69

Em and En be their respective

Im)<ml 1 . Equation (2. 181) becomes e- fJF = e -fJFo Ln f3e - fJEn Vnn + 132 Lmn e-fJEEm -- Ee- fJEn I Vmn1 2 . . . . (2.184) n m Note that if m = n , e - fJEm e - fJEn e -fJEn efJ(En-Em) - 1 = f3e -fJEn. En - Em En - Em In order to get an expansion for F, we write F = F2 + . . . , where Fl Fl is a first-order perturbation and F2 is a second-order perturbation. Then e-fJF = e-fJFoe - fJ(FI + Fz + " ' ) 2 = e - fJF0( 1 - f3Fl - f3F 2 132 (Fl + F2 + . . . ) 2 + . . . ) 2 = e-fJFo [ 1 - f3Fl + � Fi f3F ) (2. 185) ( 2 + . J. up to and including the second order. We have a perturbation expansion for F we set - e - fJFof3F1 = Ln f3e - fJEn Vnn and (2. 186) 2 m - e -fJEn E fJ e 13 13 e-fJFo ( 2 F 21 f3F2) = 2 mn� E E I Vmn 12' n m =

(2. 178)

= Ul - U2

and let

=

_

_____

(2.179) 13

-

x.

(2.180)

+

-

_

• • •

=

Fo

[e -fJHo V ] + � f: dw Tr [e- fJHoewHo Ve -WHo V ] + . . .

.

(2.181)

x

+

+

+

-

(2. 179) and (2. 180), we find, from Eq. (2. 178), -

+

(2.183)

nm

where we have used the fact that Lm

13 - w and let x'

:$;

[e-fJHoV] = Ln ( nl e - fJHo V l n ) = Ln e-fJEnVnn' (2.182) Tr [e-fJHoewHo Ve - wHo V ] = L ( n le -fJHoewHo V L Im) ( ml e -WHO V l n ) n m

(2.177)

[e -fJHOeWHOVe - WHo V ],

where A is the shaded region in Fig. 2.4. Now let w' Then the integral becomes

Averaging Eqs.

.

du Tr

.

F Fo o

Tr

Tr BA, we find that

e -fJHoV z dU l dU 2 Tr [e -fJHoe (ul-uz)HoVe -(UI-u )HoV ] -

-

In)

and Let be eigenstates of Ho, and let genvalues. Then ei

-

+

Proof that

.

.

if

-

_

_

_

�

_

Solving these equations we obtain

and -4�

__

Fig.

Fl = TrTrVee -fJ-fJHHoo = < V )Ho

(2. 187)

J---------� w

__ __ __ __

ted. 2.4 The shaded area A is the region over which Eq. (2. 1 79) is integra

(2. 188)

70

Proof that F

Density matrices

With the Cauchy-Schwarz inequality,

:::; Fo

+

o

71

(2.190) we apply the perturbation analysis to F(rx) for each rx; F(rx + y) Fo(rx) + yFJ(rx2 ) y2F2(rx) + 0(y3), where Fo(rx) F(rx), and yFJ(rx) and y F2 (rx) are as in Eqs. (2.187) through �.189), with Ho replaced by H(rx) and H by H(rx y) H(rx) yv. The sult of Eq. (2.189) is then To prove Eq. is, we write

that

Wn, 2 2 [LnLnWnWna�J LnLnwnlWnanl Applying this inequality to Eq. (2.188), we have (2.189) F2 -21 ePFo Lmn e - PEEmn -- Ee -mPEn I Vmnl 2 O. m*n Now, let H(rx) Ho + rx V, and let F(rx) be the free energy calculated from H(rx). Then F(O) Fo and F(I) From the preceding perturbation2analysis, F(rx) Fo + rxF1 + rx F2 + 0(rx3). So F' (0) F If we can prove that (2.190) F"(rx) 0 for all rx, then F(rx) will be concave downward (see Fig. 2. 5) and thus it lies below the line + rxF'(O) Fo + rxF1• From this we can conclude that F(O) F FQ. + Fl Fo + < V)Ho' or (2.191)

it is easy to show that for any set of positive real numbers, �

�

-

-----�

�

=

=

=

F.

=

=

l'

�

=

�

=

which was to be proved.

F(O) + aF1

-+----------------------�------. a

__

o

F·Ig 2 5 The situation when F(cx) is concave downward. •

•

=

=

+

=

+

re

so that

+

F"(rx) d��y F(rx + y)/ y = o 2F2(rx) O. This proves Eq. (2.190) and thus completes the proof of Eq. (2.191). Applying this same theorem to Fo, we obtain Fo + < - V)H' (2.192) Combining Eqs. (2.191) and (2.192), we see that Fo + < V)H F Fo + < V)Ho' (2.193) more accurate approximation F is obtained by including F2 : F Fo + < V )o + �2 < V ) 6 - 2! \IfP ewHo Ve-wHo V dW) . (2.194) =

=

�

�

F

�

�

A

�

0

0

3.1

Path integral formulation of the density matrix

CHAPTER 3

u

2. E

3.1

and has the formal solution

ap(u) -Hp(u). au

Fig. 3.1

(3. 1)

=

_.

where it will be noted that has been redefined to be {Hi, which has the dimen­ sions of time. Furthermore, the "time" can be broken up into divisions of length u) so that can be developed incrementally as follows :

a(na

u p p(u) = e -He/he - He/h . . . e -He/h = PePe . Peen factors).

=

where

(3.3)*

"

_ __ _ _ _ _ _ __

-

p(x, x' ; u) the solution is p(X, X'; U) r · · fP(X, Xn-l; a)p(Xn-I, Xn- 2 ; a) " , p(X I , XI; a)dx I " ' dXn-I '

In the coordinate representation for the density matrix represented as

- - - - - - __ _

The representation in Fig. 3. 1 is useful in the interpretation of the above expres­ sion. The particle travels from to through a series of intermediate steps, which define a "path" . The total amplitude u) for the particle to begin at and end up at is given by a sum over all possible paths, that is, for all possible values of the intermediate positions As the time in­ crement is allowed to approach zero, the number of integrations on the inter­ mediate variables become infinite and Eq. (3.4) can be written symbolically

X l' X2 . . . , Xn - x' b

x' x X

a

p(x, x' ; U ) .IT [x(u)] E&x(u), =

p(x, x', Xi'

(3.5)

* The reason we want to reduce Ii as much as possible is, of course, that the asymptotic behavior of p at high temperatures can be easily found (see below).

72

----

A particle

�------------

--__

travelling from

___

x

'

to along a path Xl' X2 etc. x

[x(u)] = lim P(X, Xn-l; a)p(Xn-l, Xn_ 2 ; a) " 'p(xl, x'; a)

'

(3.6)

m; = u

and

E&x(u) = nlim-+ oo dXl dX2 ' " dxn-l .

�.1though all possible paths are to be considered in Eq. (3.5), the main contribu­ tion comes from the limited set of paths for which none of the e) is too small. For example, we know that for Iow ,

a

m e - (m/ 2 he)(x - x')2 p(x, x' ; a) J2nfza '

p(x, x' ;

�

=

(3.4)

x2

£�o

(3.2)

u

__ _

---

Earlier it has been shown that the density matrix satisfies the equation h

-------

o�

PATH INTEGRAL FORMULATION OF THE DENSITY MATRIX

x

_ _ __ _ _ _ _ _ _ _

3.

PATH INTEGRALS

73

p(x, x' ; a)

which is the densit)"- matrix for a free particle. The term is small if x - > ,J2h/m,J so the main contribution comes reasonably smooth path�. Now. let us find more explicitly, first for a free from partic le and then for one mOVIng subject to a constraint.

X'

Free Particle :

a;

The Hamiltonian is H =

From previous results we know

p(x, x' ; a) = J-� e -(m/2he)(x -x')2 2nne

(3.7)

(3.8)

74

and

,

p(x, X ; U )

=

Path integral formulation of the density matrix

3.1

Path integrals

. f . f exp { me21i [(x-e�) (x � Y . . (Xl � x'YJJ

!��

- Xn _ l 2

-

. .

+

Xn - 2

n- 1

where in our case, Po = Pfree . For low e, because Pfree is a very localized Gaussian, most of the contribution in the integral over x" occurs near x" = xo, where it can be shown that Xo

.

dXn - l �2nlie/ m

(3.9)

So we can, for small e, write Eq. (3. 14) as ' dU V(xo) oo dx" Po(x, x" ; e - u)Po(x", x' ; u) ()p(x , x'; e) � - o Ii

<J>[x(u)]

=

dx(u) --

=

_

du u=ke

x(u .

)\

()p(x, x'; e) (3. 1 1 )

�

e

- - V (X)Pfree(X, x '; e) fz

under the assumption that Ix - x' i and e are small. Of course, as Ix - x'i increases, our conclusion that Io V(xo) du � e Vex) becomes less accurate, and it might be thought preferable to write something like

Particle Moving in One Dimension in a Potential Vex) : =

H

The equation for P is explicitly op = - Ii ou

or

1i2 02 - - - + V (x).

in place of e Vex). But for our purposes, Eq. (3.13) is adequate. So

2m ox2

_

pee)

� � p + V(x)p .

(3.12)

2m ox2

In an infinitesimal "time" interval e the particle may be considered as moving freely with a small correction due to the presence of the potential Vex). To a very good approximation Vex) epfree· (3. 13) {)p = - -Ii

- f oo f

This result is demonstrated as follows : From perturbation theory we have d 4) dx" < Po(x, x" ; e - u ) V(x")P o(x", x' ; u ) ' (3. 1 ()p(x, x' ; e) � - 00

0

(3.1 5)

°

Now, if x � x', xo is also close to x and V(xo) is constant over the range of integration. Equation (3. 1 5) goes over to Eq. (3. 13) :

u=k, I exp {-� f: ; [X(U)]2 dU} .

-'"-_�-' --+

f f ' dU V ) -f (XO Pfre.(X, x' ; e) . Ii

=

(3.1 0)

xk - Xk - l e and Eq. (3. 10) becomes

ux + (e - u)x' e

=

- 00

Thus, for a free particle,

As the interval e is made smaller,

75

:

�

[

Pfr (e) 1 ••

Thus

-

�

J

V x) e

J2nliem exp [

�

Pfree(e)e - V(X)
(3.16)

]

m(x - X')2 eV(x) . (3.17) Ii 2lie To find p(x, x' ; u) it is only necessary to apply the correction e- V(x)
�

_

IT exp { - IiI f U [mx(u)2 -- + V(x(u» du } .@x(u ) 2

--+

'.

p(x, x , U )

=

==

.

.IT <J>[x(u)] .@x(u), 0

J

(3.18)

76

where x(O)

==

x' and x( U)

[x(u)]

=

=

exp

{ -� f: [mx;u)2 x. Thus

] }

+ V(X(U») dU .

The free energy is derived from e -flF

= =

Path integral formulation of the density matrix

3.1

Path integrals

(3.18a)

In the classical limit, the temperature is high and U is small. We want the ssic cla al partition function, and we would like to see how it is related to the quantum-mechanical partition function. From Eq. (3. 1 8) U mx(u)2 rT exp p(x, x ; U) = -- + V(x(u») du f0x(u). 2 - Ii lJ x(O ) = x(U) = x (3.20) Ifwe assume that the most important paths are those for which V(x(u») � Vex), Eq. (3. 1 9) can be approximated by U mx(u)2 f0x(u) p(x, x ; U) � exp exp 1 v ex) Ii 2

{ If [

] }

[ -¥ ] If

[ - - i --]

0

f p(x, x) dx + V(X(U»)] dU } f0X(U) } f { ff exp { -�Ii f U [mx(u)2 2 0

x(O)= x(U) = x

dx

(3. 19)

where U = pli. Having derived the general expression for the question now is, which "paths" are the most important? If the potential is not very dependent upon position, then the more direct paths contribute the most. This is because longer paths, with higher mean square "velocities," [X(U)]2, make the exponent of the weighting functional [x(u)] more negative. On the other hand, suppose the potential does depend on position. For simplicity, let the potential increase for larger values of Then a direct path such as 1 in Fig. 3.2 does not necessarily give the largest contribution, because the integrated value of the potential is higher than over another path such as 2 or 3. Consider a path such as 2 that deviates widely from the direct path. Then Vex) decreases but x increases over the path. In this case we would expect that the increased velocity x would more than compensate for the decreased potential over the path. Thus, the most im­ portant path would be one for which any smaller integrated value of potential energy is more than compensated for by an increase in kinetic energy (sym­ bolically represented by 3 in Fig. 3.2). X.

x

[ - � vex)] Pfree(X, x ; U ) x(O ) = x(U) = x

= =

exp

J2n�U exp [�_m!�;_ X')2

0

]

U V(X) Ii

(3.21)

Equation (3.21) is just a specialization of Eq. (3. 1 7). But from the derivation of Eq. (3.21 ) it is easy to see the direction of the error in the classical limit. We assumed that the most important paths had x(u) � x, whereas in fact the most important paths are those for which Sg [mx(u)2j2 + V(x(u»)] du is a minimum which occurs when (x(u») goes over paths that tend to let V(x(u») be lower tha� Vex) . It follows, then, that Pelassical(X, x ; U) < p(X, x ; U) . (3.22) The classical partition function is e - flFc\

=

f oo

- 00

Pcl(X , x ; U) dx =

From Eq. (3.22) we obtain

m- f oo e- UV(x)/n . J2nliU dx _ 00

(3.23)

Fel :<: F. (3.24) . To find the condition for Eq. (3.21 ) to be a good approximation to the path mtegral, note that the paths for which 1 { U -mx2 - du 1 Ii J o 2 do not contribute much to the path integral. Using the Cauchy-Schwarz in­ equality, find »

we

Fig. 3.2 Three paths between x' and

77

X.

78

Path integrals

With ! =

X

and 9

Calculation of path integrals =

1,

we have for 0

::;

UI

::;

79

Consider the element

U: (3.25)

(3.28)

So for the important paths, (m/2IW) l x(ul) - x(OW < l . If d is the maximum value of Ix(ul) - x(O) I for 0 < U I < U, then the paths that contribute signifi­ cantly satisfy

where x(O) = X l and x(U) = x2 . This is the density matrix element for a free particle. Let us expand each path about the straight-line path in the xu-plane. Call the straight line path x(u). Then

mkTd 2

--.� � �

2h2

< 1. -

(3.26)

For the hydrogen atom at 1 0K Eq. (3.26) yields d ::; 9 A. For oxygen at ::; 2.5 A; for oxygen at lOooK, d ::; 0.25 A. If the potential varies so slowly that it is approximately unchanged even when the path varies by more than d, then Eq. (3.21) is good. Thus, for heavy particles at normal temperatures, systems may be treated by classical statistical mechanics. However, the classical approximation does not apply to liquid helium, nor does it apply to electrons in a metal, nor even to solids if the tem­ perature is too low. A few things should be noticed about this formulation of statistical mechanics : 1 . We have discussed only the case of one coordinate. However, the extension to N degrees of freedom is obvious; if x( l !, x( 2 l, . . . , X(N ) are the coordinates, 9Cx(u) becomes (3.27) 9Cx(u ) = lim dXll ) dx12 ) . . . dX1N ) dx�l ) · . . dX�N ) . . . dX�N ) .

X(u)

and

1 0K, d

2. If "u" is replaced by "iu" in Eq. (3. 1 ), we have Schrodinger's equation. In a similar manner to that of statistical mechanics, quantum mechanics can be formulated in terms of path integrals. For mathematicians, statistical mechanics is easier to deal with because the exponentials in the path integrals are exponen­ tials of real quantities. 3. We have not dealt with relativity (in which particles appear and disappear) and spin (which is a discrete, rather than a continuous, variable) in the path­ integral formulation. If such effects exist, we can only get approximation s . Actually, spin can be put into the formalism, but our interpretation of E0x(u) as a set of paths differing by an infinitesimal amount no longer holds. 3.2

CALCULATION OF PATH INTEGRALS

We will discuss two techniques for the calculation of path integrals : 1 . Integrals whose exponents are quadratic in X and x will be calculated exactl y . 2. Perturbation expansions.

f(x2 , Xl'

U)

=

Now

x(u)

-

X2 - X l

= -

.- .. �

U

=

y(u)

+

.IT exp [ � � LU

dx(u) du so

=

(.i{u)

v

=

(3.29) +

y(u W

dU

J

9Cy(u).

. velOCIty

(3.30)

(3. 3 1)

that

LU [X

+

yJ2 du =

But y( U)

=

LU (v 2 + 2vy + y2) du = yeO)

=

v2U +

2 v [ y(U ) - yeO)]

+

LU y2 duo (3. 32)

O. Therefore

(3.33)

Thus we have split �he path integral into a factor depend ent the endpoints and U, and a path mtegral dependent only upon U. Writinupon g

We get

.IT

exp

y(O) = y( U ) = 0

[ -� LU y2 uJ 9Cy d

=

F( U ),

P(X 2 ' Xl , U ) = F( U) exp [ - m(x2 - X I)2/2hU].

(3.34)

We can obtain an equation for F ( U) by using p(x, y ; U t

+

u2) =

f

dx' p(x, x'; u )p(x ', y ; u l ). 2

(3.35)

80

Path integrals

The reader should derive that equation for F(u) and show that its most general continuous solution is (3. 36 ) F (U) = .j mJ2nliU e"u, where 0( is an arbitrary number. Physically, the value of 0( is immaterial, because a change in the zero of the potential does not change the physics; it does, however, change the value of 0(. As a matter of fact, since the expectation value of any quantity, A, is given by
=

F(U) need not be found at all. It cancels out in the division. We know that 0( is zero, however, from other methods of calculating p (see Eq. (3.8»). Now consider the path integral for the harmonic oscillator :

!(X2, Xl' U) = jJ exp {- � ioU [m;2 + m;2 X2] dU}

For any path, we write

o.

X=x+y

yeO) = 0, y( U) = 0 and the integral over u becomes ioU [� (x + y)2 + 1'11;2 (x + y)2] du = Jruo [X2 + + y2 + W2X2 + w2 xy w2y2] Jruo (x2 W2 2) du Jruo + w2xy) du so

11'1 2

=

� 2

2xy

+

X

+

2

+

� 2

(2xy

2

J �use y(0)

=

0 and

y(

U)

du

=

=

2YX

l: Iou

81

2xy duo

-

0, we have

� Ptce x satisfies the Euler equation.

�'"

Thus, by expanding about the classical path, .;we have eliminated the integrand linear in the deviation. Thus we may write

f(X2 , Xl ;

with yeO)

=

=

U)

=

exp

[ -�1 Jrou ("2m x2 + m-2w-2 x2) du] F(U),

.IT [ - � IoU (; y2 + exp

0 and y( U)

=

;

m 2

) ]

y2 dU 9y(u),

Xl integral and X . 2 integral dePendent upon alone. Thus the important dependence on Xl and X2 can be W,und by simply solving the minimizing differential equations (Eq. 3.38) subject the end conditions Xl and = X2 and calculating the integral IoU (; x2 + m;2 X2) duo 0, is a factor independent of the endpoints

. Thus expansion about the classical path allows us to separate the path into a factor dependent upon U and the endpoints, and another path

U

to

x(O)

In fact

=

x(U)

even this integral can be simplified. We have um m m . du - X2 0 2 2 2 0 Therefore

r u (x Jo

fo

= - xx· I U - - f U xx. duo

= m2 xxl U + Jrou m2 ( w2x) du = m2 xxI U .* J haAs a new problem, suppose we have two interacting systems, one of which is .f. rmonic oscillator. Let X describe the position of the oscillator and that of other system. A possible Hamiltonian is mx2 M·2 mw2 . + H = - + --- X2 + 2 2 2 f

du

I: xy 2

F(U)

ioU G(u) du = ioU [m;2 + m;2 X2] du = minimum, where x(O) = Xl' x(U) = X . Thus, our integrand must satisfy the Euler 2 equation, � oG (u) oG (3. 38) = 0 or mx - maix = ou ox ox

floW

�x.

Let us expand each path about the path that makes the largest contribution to P. For this path,

_

�t

Calculation of path integrals

m 2

2

+

W2X2) du

x

0

-

x +

q

�e

k

_ _ _ _

0

--.!L

, You can compare your result with Eq. (2.83).

V(q) - yxq

82

Path integrals

IT [ I ( [ I (Mi/

The partition function is

Q=

exp

x

-

exp

-

Eventually one will arrive at the result :

) J

U mx 2 m oix 2 Ii1 -- + -- - yq (u)x(u) du 2 2 0

Ii1

U

) J

-- + V(q ) du fifix fifiq . 2

0

(t[/] = exp

We cannot, in general, evaluate this path integral, but we can integrate out the x variable. To do so we must find

F [q (u)] =

IT [ - � t ( ;

U m 2

exp

+

;

mw x 2

) J

exp

>:( 0 )=>: >:(U) =>:'

-

I(u) (3.39)



1 =-

where

II

-

U U mw e-WIU-UV(u)f(u' ) du du' + -. 2 smh wU 4mw 0 0 x

exp

[(x 2 + X, 2) cosh wU

-

I I

2xx' + 2A(xeW U - x') + 2B(x'eWu - x) + (A 2 + B 2)eWU - 2AB], (3.40)

. U e-WUf(u) du, A = -' 2mw 0 . U e-w( U - u)f(u) duo B = -' 2mw 0 -

-

Now let

(t[J] =

/ [ -� f \ exp

U

f(u)x(u) dU

h 0 J FlU; x, x) dx J Fl(O ; x, x) dx

J)

(3.41 )

(3.42)

J

t U ) f(u)f(u du dU ' . ')

(3.43)

f oo

�

u

�

U by letting

I

be periodic with

u

J

r -w1 u u'lj 1_ e - (u)f(u') du dU ' . 4mwh - 00 J o

_ _

=

( -( --

--) -)

� a cos 2nnu + sm 2nnu bn . n

1.J

n= l

(3.43a)

P

P

,

00 . 2nnu . fo + L fn cos 2nnu + gn sm n= l P P

The path integral

can then be written as

Fl[j] =

fff . . . f

dao dal . . . db l db 2 . . · x

and can easily be evaluated. • The equation

harm. osc.

[

x(u) = ao +

The result is where

=

-

An alternative way of finding Fl(f; x, x) is to Fourier analyze/(u) and x(u) :

U mx 2 mw 2x 2 + ij(u)x(u) du fifix . + -20

Ii

I

U r u cosh w( / u - u' l . 4mwh 0 J o smh tw u

1

__

outside the interval 0 we may od U, write peri

(t[j]

) J

IT [ 1 I (2

[-

I

If we define

- yqx du fifix .

This integral can be evaluated by the same method as that used for a har­ monic oscillator. More generally, we can evaluate

Fl[j; x, x'] =

83

Calculation of path integrals

1 2a2

+

exp

*

�

[ � tU (m;2 m;2 x2 -

+

-

cos n(} = !!.- cosh (n 1(}l)a 2a sinh n a + a2

n = l n2

)J

+ ij(U)X(U»

du,

is useful. FI (f ] as found by the above method has an unknown multiplicative constant a consequence of the Jacobian for the change of variables implicit in fifix(u) -+ dal dbi db2 ; but Eq. (3.43) can be found unambiguously by the above method. as

• • •

• • •

84

3.3

Path integrals (.0

Path integrals by perturbation expansion

0 [I] .IT [ - � f: [!mx(u)2 + if(u)x(u)] dU] £»x(u) 21t m b[f: feu) dU] { ! (f: f: Iu - u'lf(u)f(u') du du ' J� + � [f: uf(u) duJ)} . either by letting

For a free particle, we can find FI calculation. The result is

--+

or by direct v

------ -------------

exp

exp

=

85

X2

4 n

(3.44)

We will make use of this last result later.

3.3

Fig. 3.3 Integrand of first-order term PATH INTEGRALS BY PERTURBATION EXPANSION

PO(X2, Xl' Yo, (X ' Xl' P 2 Vo + V', V' Yo ' P(X2' Xl ' .IT {- � f: [m;2 + Vo(X) + V'(X)] dU] £»X} .IT [ {- � f: [m;2 + Vo(X)] dU} {- � f: V'(x) dU}] £»x {- � f: [m;2 + Vo(X)] dU

In order to calculate the path integral, we first integrate over all paths for which x(U) = x2) and then integrate over y. = y (with x(O) =

Suppose that we have found the density matrix element U) for some system of potential and we wish to know U) for another system with potential where « Then U) =

' x(u) P(X2 , Xl

!

exp

x

jJ

exp

} [1 - � f: V'[x(u)] du + 2�2 (J: V ' [x(u)] duY + ] Po - � lJ f: {- � f: [m;2 + Vo(X)] dU'} { - � IU [m;2 + Vo(X)] dU'} V'[x(u)] du £»x + . . . Po - � f: du JJ {- � f: [m;2 + Vo(X)] dU'} V'[x(u)] { - � tU [mt + Vo(X)] dU'} + . . . e xp

=

exp

x

=

exp

exp

x

exp

Po,

u + � f U du f " dv f oo dy f oo dz Po(x2 , n

. . . £»x

x

0

!

Expanding the second exponential, we have =

- u)V(y)po(Y, X l' u) dy du + . . .

U all y i Pictorially, the integrand may be represented as shown in Fig. 3.3. i We can think of the zeroth-order term, as being the contribution to the i density matrix that results from the particle going from to in "time" U without being affected by the potential. Then the first-order term, which has its cintegrand represented in the Fig. 3.3, may be thought of as the contribution : resulting from the particle being essentially free, but being "scattered" once at some "time" and some position y. Similarly, the second-order term can be considered as the contribution from two "scatterings", and can be written as i

exp

=

U) =

Xl' Po - � f U f Po(X2, y,

0

0

- 00

- 00

X

.

Z,

U

Xl X2

- u)V(z )po(z, y, u - v )V(y)po(Y, X l ' v )

Pictorially, this integrand may be represented as shown in Fig. 3.4. Notice that we have not found any new results. We have already derived this perturbation expansion without the use of path integrals.

Problem : Find to what order the nth-order integral expansion of

solves

knowing that

- :: =

- 2: ::z + [VoCx) + V'(x)]p,

p(Xz , Xl'

U)

86

3.4

Path integrals

Fig. 3.4

Variational principle for the path integral

87

Integrand of second-order term. Fig. 3.5

Geometrical interpretation of (e-f ) ;;:: e-(f>.

3.4 VARIATIONAL PRINCIPLE FOR THE PATH INTEGRAL

(3.45a)

The geometrical interpretation of this relation is as shown in Fig. 3.5. In Fig. 3. 1 , Note that Eq. (3.50) does not depend on is always above the curve how thej's are distributed. When we apply Eq. (3.50) to Eq. (3.49) we can write

(3.45b)

where

For the free energy we know the relation

F :::; Fo + (H - Ho>o,

or

e - f.

(e - f>

(3 . 5 1 )

Ho.

This is written in terms of the Hamiltonians H and We ask ourselves if we can write a corresponding minimum relation using the path integral. In the path-integral form, the partition function is written as

e -PF .IT e -S[x(u)].@x(u). So, "S", (3.46) e -pF IT e -(ITSe--Ssoo)ePfix- so ;;z;x e-PFo,

(3.46)

=

x(O ) = x(U)

that is easier to work with. (Here may be written as

Suppose we have another mean entropy). Then Eq.

rM

e -PFO .IT e-so fi)x.

(3.48)

[ -(S - So)J (3.47)

=

Now we suppose that

and

as

(3.49)

are real and use the inequality �

e - u>.

When

We

(3.53)

=

show that

The first factor of Eq. (3.47) has the form of an average of exp as the weight for a certain path x(u) . We may write Eq. with

e -PF <e-(S -So» so e -pFo. S So (e - f>

Exercise:

F :::; Fo + p1 <S - So>so' S f: [mx�(u) + V(X(U»)] du, 2(u) + Vo(x(u»)] du, mx2So fp [-P1 <S - So>o o.

(3.47).

(3.54)

=

==

e -so

does not

Usually Eq. (3.52) is simpler to calculate than the first factor in Eq. thus have a theorem :

(3.47)

=

where

So

(3.52)

(3.50)

0

=

Thus Eq.

(3.55)

(3.53) contains Eq. (3.45) as a special case. (Hint : from Eq. (3.54»), S

- So f: (V - yo) duo =

The numerator of the right-hand side of Eq. the integrand of which is independent of u .

(3.52) becomes an

(3.56) integral over

u,

88

Path integrals

An application of the variation theorem

3.5 AN APPLICATION OF THE VARIATION THEOREM

Finally in this chapter let us explore an application of our variation theorem, Eq. (3.53). In a one-dimensional, one-particle case, we have

e - PFO

=

exp

�

f { - L [mx�u)2 V (X(U»)] dU} �x(u) dx(O). x(O)x(O).

x(O),

Here, we first fix and x(U) = and then do the path integral over all paths ; then we vary The zero in § indicates that we integrate over In the classical limit of high temperature, or small Ii, U is small, so that the path does not deviate much from the initial point Thus the first approxima­ by the initial value tion in Eq. (3.59) is to replace to obtain e - pF -

x(O). V(x(O») V(x(u») m21thk T2 f e - x. -J V (x)JkT

x(O).

d

(3 . 60)

This is the well-known classical result. How can we improve on this approximation to take the quantum effect into account? First we observe that, because and U) are equal, it appears more natural to use an average 1 U (3.61) x == -

x(O) x( x(u) du UI in place of x(O). Second, the deviation of the path from the classical straight line may be taken into account by some average of Vex) over the path rather than by a constant V(x(O») . These considerations lead to the following trial So : 2 du + Uw(x). (3.62) So = Iou mx(u) 2 Here w(x) is a still undetermined function, which is to be varied later to minimize Eq. (3.53). In the first approximation one would expect to choose w(x) Vex) but we can do much better with our variational principle. 0

--

=

] dU] 2)x

UWeX)

"" clO ", " with fixed average x

x - x(O). Since y(u) = x(u), we obtain e - PFo = f dx e - Uw(x) .IT exp [ - LU m;] �y.

(3.58)

To avoid unnecessary clutter, we will set Ii equal to unity in most of the following work. Every once in a while we will restore the Ii's, so that the reader can see where they belong. The partition function may be written as U e - PF = exp (3.59) +

+ UW(X)

all closed paths

Now let y =

U == {Hi.

U

exp

(3.57)

where

.IT �x [ - L [m;2 ] dU] f dx [ [ - nmt du +

89

all closed paths y with y(O) 0

In the �ath integral above, there is no restriction on y because y = x and we mtegrate over The reasons for integrating over the paths chosen can be made clearer by drawing diagrams ; the reader should do this if he needs clarification. e - PFo = e - Uw(x)pfree(0, , U) e - Uw(y) (3.63) 1t Ii U - OCJ It should be noted here that in the classical case Eq. (3.60), w(y) is replaced by V(y). ��xt, we calculate Using Eqs. (3.57) and 3 . 6 , we see by defimtIOn 1 (S =

x(O).

O · - J2 m f OCJ

f dx

<S - So>.

- x(O),

dy.

( 2)

So> = .IT l/P(S - So) exp [ J mx 2/2 du] exp [ - Uw(x)] �x .IT exp [ - J mx 2 /2 du] exp [ - Uw(x)] �x .IT { l / U J� V(x(u'») du ' � w(x) } exp [ - J mx 2 /2 du] exp [ - Uw(x)] --- {0x . .IT exp [ S mx 2 /2 du] exp [ - Uw(x)] {0x (3.64) Here x is defined in Eq. (3.61) and is functional of x(u). The first term in the nUmerator is simplified as follows : IT h LU V(x(u'») du' exp [ - LU mx�u)2 dU] exp [ - Uw(x)] �x LU �� .IT V(x(u '») exp [ loU mx�u)2 dUJ exp [ - Uw(x)] {0x s: d;' f dx' Vex').JL exp [ r mx;(u) du r mx;(u) dU J exp [ - Uw(x)] �x(u). P

-

--

-

=

=

-

-

�

x(u') = x'

x

90

b.s

Path integrals

An application of the

, "

r The first term in the numerator of Eq. (3.64) becomes

u = v ---,r-----r---�--

: d;' f J

.IT

..

dx' V (x')

.IT

[L � [L ; J V

exp V

x(O) = x( V)

91

]

m x (V) dV exp [ - Uw(x)]

x(O ) = x( V ) = x'

V(x(O)) exp -

variation theorem

mx (u) du exp [ - Uw(x)] �x(u).

(3.65)

Thus Eq. (3.64) may be written as

! <S - So)so

u = o --�L-------��--�-Fig. 3.6a

f3

Paths of integration.

= ( V(x(O))) - ( w(x) .

(3.66)

To transform further we use the following result (proved previously) :

'il [ L � J [ - iLv ] J ; (L ) { � [L L exp -

V

=

mx u) 2 du exp

2 m

V

c5

f(u)x(u) du �x(u) V

feu ) dU exp

V

4

l u - u' lf(u)f(u') du du' (3.67)

v = u - u' --��----_1�--�-

In evaluating Eq. (3.65) we first do the path integral keeping x fixed ; then , Eq. (3.65) is written

Fig. 3.6b

We are integrating over paths like those in Fig. 3.6a. By making the change of variables o ::::;; u ::::;; u' u + U - u'

V =

{

u' ::::;; u ::::;; u

u - u'

-

�

we integrate over paths like those in Fig. 3 ..6b. �ec�us� of the condition th� x(u = 0) = x(u = U), we do not get a dlscontmUlty m x at V = U u , therefore

Also,

fov

m x2 (u) du = 2 1

x=U

f

V

0

f

V

0

mx 2(V) dV. 2 1

x(u) du = U

f

V

0

xCV ) dV .

, .IT

[f � J

-

2 V(x(O)) exp - mx u) dU exp [ Uw(x)] �x

where

K (y) =

=

.IT

V(x(O)) exp -

.IT

V(x(O)) exp -

[L

V

f �

[L

V

For this we use the Fourier transform :

Vex)

==

f

J

mx u) 2 dU �x

Such that x = y. Equation (3.69) may be written

K (y) =

-

K(y) exp [ Uw(y)] dy

�

J

(3.68)

(3.69)

mx u) 2 du c5(x - y) �x(u) .

(3.70)

v(q )eiqx dq

(3.71)

92

(3.70) as K (y) = J v (q) dq J dk IT exp [iqx(O)] exp [ik(x - y)] [ ru 2 mx2 du] �x. (3.72) exp J I?ed, and we are ignoring Here b(x y) of Eq. (3. 70) was also Fourier transfor WrIte to factors such as 2n. Use Eq. (3. 6 1 ) in Eq. (3. 7 2) K (y) f v (q) dq f dk exp [ -iky] .IT exp [iqx(O)] exp [i � f: x(u) duJ [ Ju 2 (3.73) mx2 du] �x. exp

to write Eq.

x

-

0

-

�

x

-

0

ssion can be brought Now we see that the path-integral part of this expre if we define form of Eq.

(3.67)

feu) qb(u - 0) + k -

-

=

because

u

J: f(u)x(u) du = qx(;Q) + � J: x(u)

du o

into the

(3.74) (3.75)

(3.74) in Eq. (3.67) we see that (3.76) J: feu) du + k, (3.77) J: uf(u) du = k� , J: J: lu - u'lf(u)f(u') du du' = 2J: du f (u - u')f(u') du'f(u) = 2J: du [J: (u - u ' ) � dU' + uq]f(U) ru k 2 ] k2 U = 2 du [ � + uq feu) = -3- + kqU . Jo 2 (3.78)

When we use Eq.

An application of the variation theorem

3.5

Path integrals

=

q

93

(3.78)asin Eq. (3.67) and also in the path-integral part (3.77),theandlatter (3.(3.73)76)to, write K (y) = f dqv(q ) f dk exp [-iky]b(q + k) exp [4� (Y + kq + �2)J f dqv(q) f dk exp [ - iky]b(q + k) exp [ - �;�J f dkv( -k) exp [ -iky] exp [ -�;�J . (3.79) Using the inverse Fourier transform, we find that Eq. (3.79) becomes K(y) = f dk exp [-iky] f V(z) exp [ikz] dz exp [ �:�J f dzV(z) exp [ - 6� (y - Z)2J . (3.80)

Use Eqs. of Eq.

=

=

-

=

When the correct factors are taken into account, the final form is

K (y) J6mnn2kT fdZV(Z) exp [ _ 6mkT(nY2 - Z)2J . (3 .8 1) It should be noticed that K(y) is V(z) averaged over a Gaussian. The root mean square of the Gaussian spread is (3.82) �12 mk T ' This spread is about A for a helium atom at 2°K , and is a very narrow spread at room temperatures. At the limit of infinite temperature, the Gaussian becomes a delta function, and K(y) V(y). Summarizing the results obtained so far, we have (3. 83a) (3.83b) (3.83c) =

1

-+

94

Path integrals

An application of the variation theorem

Equation (3.83b) is Eq. (3.63). Equation (3.83c) follows from Eqs. (3.64), (3.65 ) , and (3.68). Further, K(y) in Eq. (3.83c) is given in Eq. (3.81). Now we ask what is the best choice of w(y) ? Changing

w(y)

--+

w(y) + l1(y)

(3. 84)

l1(Y) is small, we see from Eqs. (3.83b) and (c) J n e - Pw dy (3.85 a) J e - Pw dy e P (K - w) dy] + J e - PW(K(- w)- dy J f3'1e- Pw . {J1 <S - So > - J - W[ - f311e-Pw dy J e Pw dy) 2 J

where

l:F0

u

�

_

.

,

Using this result in Eq. (3.90), we obtain

e-PFcl K = J2rcmli2f3 f exp [ -13 m2w2 (y2 + 1i2f3 )] dy 1 2m

- 11

� Fo +

leads to

w(y) = K(y).

so that, from Eq. (3.83a),

-

(3.86)

(3.94c)

(3.8 7)

K

So>so = 0,

(3.88)

Fel K

(3.89)

�

When we write

e- PFC1 J-2rcmli U f e-PK(Y) dy (3.9 0) compare Eq. (3.60) . K(y) is defined in Eq. (3.8 1). K is a better approxima' tion than the ordinary classical form in Eq. (3.60). Example: To find out how good K is, let us consider the harmonic oscillato r . In this case the potential is

Gtrue Gel

=

Fel

Fel

Fel

K( y)

=

J

6 kT m w2 . �

rcli 2

�

2

f

[

]

Z 2 exp - 6 m kT (y - Z ) 2 dz 1i2 --

=

mw 2 2

�

(

-

)

.

(3.95)

(3.95a )

K = f1 In (2f) + j-62 ,

(3.95b)

-

Gel

=

1

J

In (2f).

(3.95c )

K

Table 3 . 1 is a numerical comparison of Gtrue, Gel and Gel ' This table shows the remarkable improvement GCl has over Gel ' Remember that f = ! cor­ resPonds to the temperature k T = liw where the quantum effect is large.

K

Table 3.1 1/2

/= 1

/= 2

0.08263 0.08333 0

0.85 44 0.85 98 0.693 1

0.9908 1 .0264 0.693 1

/

1i2f3 y2 + 1 2m (3.9 2)

Iiwf3 . 2

= � In (2 sinh f), f

(3. 9 1 )

�

f ==

Eq. (3.94) becomes

where Fel means the classical free energy with the potential V(y) replaced by K(y). Alternatively,

From Eq. (3.81)

(3.94a)

'

(3.94b)

and

F

K

[ ( )]

. IiWf3 2 = {J1 In 2 smh

(3.85b)

This is the best choice of w(y). In this case Eq. (3.83c) yields <S

Ftrue

dy

i <S - So >so) = 0

(

Thus

(3.93)

Thus, we have the following results :

-

_

95

Gtrue GelK Gel

=

96

Path integrals

�

better than the The expression Fe! K i � Eq. (3.90) comb in;d wit Eq. (3. 8 1 ) is : . . . , y . . series expansion of K(y) 10 terms of V(y), d V/d . . FIrSt, �t Nevertheless, Fe! K is not as useful as Table 3 . 1 might 1Od�cate. s eXIst. . Se�ond, .It cannot be used when quantum-mechanical exchange effect derivative as III large very a fails in its present form when the potential V(y) has the case of hard-sphere interatomic potential.

CHAPTER 4

CLASSICAL SYSTEM OF N PARTICLES

4.1

INTRODUCTION

The classical partition function of N interacting particles is written as (4 . )

1

where ZN

=

f

e - PV(R, R2 · · · RN) d 3R l d 3R2 • • • d 3RN•

(4.2)

These equations can be derived from the quantum-mechanical results we have obtained previously. From Chapter 2, Section 8, we have e-

PF

s

=

1 L _

N!

P

f

PD(Xl , . . . , XN ; PX l , . . . , PXN) dXl . . . dXN

for Bose statistics. For Fermi-Dirac statistics, e-

PFA

=

_1 L

N!

P

( - ltpD(Xl , . . . , XN ; PXl , . . . , PXN) dXl · · · dXN •

In either .case, only the identity permutation is important, because at 2 sufficiently high temperatures, factors such as e - mkT/2h2( X - PX) kill the other term s. It follows that : e-

PF

�

.�

N!

f

PD(Xl , . . . , XN ; Xl , · · · , XN) dXl . . . dXN•

We have already estimated this integral for the case N = 1 by the use of path integrals (see Chapter 3, Eq. (3.60)). Generalizing the method used to the case of N particles in three dimensions, we arrive at Eqs. (4. 1) and (4.2). When a gas particle is a polyatomic molecule, the internal motion of a molecule and the motion of the center of gravity can be separated ; then the specific heat is a sum of the two contributions. In Eq. (4.2), Ri is regarded as the center of gravity of the ith particle ; the partition function for the internal motion is not included here. The latter can be calculated using information 97

98

Classical system of N particles

Introduction V(r)

about the energy levels obtained from the infrared spectra, or calculated using quantum mechanics. . . For a dense system such as a liquid the internal motion of polyatomIC molecules and the motion of center of gravity are entangled with each other, and the two are hard to separate. The following discussion excludes the internal motion of a particle, and the system is regarded as a gas or a liquid of inert gas particles. The free energy is written from Eq. (4. 1 ) as F

=

-

3N 2P

In p

+

-

3N 2P

2n h 2 1 In - - - In ZN P

m

+

N N - In - , P

(4.3)

e

The e in the last term of Eq. (4.3) is the base of natural logarithms, rather than the electronic charge. This e appears because we used the Stirling formula : N!

�

(�r �

2nN

In N !

�

�

U =

_

T2

a(FjT ) aT

where

R

tRT

=

+

( �).

N In

(4.4)

R T 2 aZN , NZN aT

(4.5)

From Eq. (4.3), we have for the internal energy U

Fig. 4.1

The equation of state is found to be aZN V PV - = -_ . _ , RT NZN a v

from Eq. (1 .6).

Example : For an ideal gas, V(R1, Eq. (4.7) becomes

•

•

•

, RN)

=

(4.7) 0 ; therefore ZN

=

V

N and

(4.8)

P V = RT.

When we assume that the potential energy can be written as a sum of pair­ wise potentials : (4.9) V(R1R · · · RN) = 'L V (ri), 2

ij

Eq. (4.2) can be written ZN =

f [

exp - p

An unrealistic potential ; Vo is the volume of a cluster of N particles.

! Note that this expression contains all the information about the phase transitions :A:,fthe system. In other words, the nature of the phase transitions can, in principle, be discussed by examining the behavior of the integral in Eq. (4. 10) purely inathematically, without physical knowledge. Equation (4. 1 0) suggests that when the number of particles N is finite there is no discontinuity in physical qualities. Only when N --+ 00 do we expect discontinuities, and this is the case we are interested in. The assumption of a central force potential V(rij) in Eq. (4.9) is not exactly justified. The necessity of a noncentral force is illustrated by the following observation : In the solid state, if we assume the nearest-neighbor interaction only, the face-centered cubic and the hexagonal close-packed structures are not distinguishable. When the second nearest neighbors are taken into account the central-force assumption leads to the conclusion that hexagonal close packing has the lower energy, whereas face-centered cubic is actually more stable for solid argon for example, which is inert in its gaseous phase. The free energy F of Eq. (4.3) is not necessarily a function of VIN only. If we take an unrealistic potential like the one in Fig. 4. 1, the energy of the �ystem is roughly of the form CN2, where C is a constant and N is the number of particles forming a cluster of volume Do . Thus, we can think of two extreme forms of ZN :

(4.6)

Nk.

=

�

paIr

]

pair

v(ri) d 3R 1 d 3R2 · · · d 3RN·

99

(4. 10)

Z<J" )

�\vhen the temperature is low, for large

i

=

N V .

(4.1 1)

N, the contribution of Z�1 ) can be larger an ) Z12 !th ; then F is not a function of V IN only. r-

100

Classical system of N particles

4.2

The second virial coefficient

101

4.2 THE SECOND VIRIAL COEFFICIENT When the gas is not dense, the data of the equation of state are summarized iJ practice as PV RT

B C 1 + - + -- + · · · V

V2

(4. 12

where B is called the second virial coefficient, and C the third virial coefficien1 In this section we will derive an expression for B. We assume a pairwise potenti� as in Section 4. 1 . Consider a system of N + 1 particles. In this case, Eq. (4. 10) of Section 4. becomes

Weight : 1 (a)

(b)

Fig. 4.2 Weighting factors in the derivation of the ideal gas equation of state. Thus, when we neglect the three-body and higher-order colIisions ' we can approximate Eq. (4. 1 5) as

When we integrate over X first, all the R/s are fixed. If the gas is not dense the volume for which VeX - Ri) is appreciably different from zero is small ; s we may approximate (4. 1': Equation (4. 1 4) will lead to the ideal-gas equation of state. To obtain a bette estimate of ZN + l , we write Eq. (4. 1 3) as ZN + l -

_

N S exp [ - f3 L; VeX - R ;)] exp [ - f3 Lpair Veri)] d 3X d 3 R N S exp [ - f3 " �pair V (rij )]d3 R x

f exp [ - f3 �

paIr

]

N Veri) d 3 R .

(4. 1 5)

The ratio part of this equation has the form of the weighted average of

We will evaluate this average by neglecting the three-body and higher-order collisions. When X is close to R;, this pair is assumed to be far away from any of the rest of the Rj • This assumption is equivalent to replacing the weight factor

[

exp - f3

�

pair

]

Veri)

by unity (see Fig. 4.2). When X approaches R; in Fig. 4.2(b), we have a three­ body collision among X, Ri, and Rj • Therefore, we neglect the configuration (b).

S exp [ - f3 Li V eX - RJ] d 3 X d 3 NR ZN + l =:= ZN S d 3 NR =

S d3 X U exp [ - f3 V(X - R)] d 3RY V

Now, we make the transformation :

f

e -PV(X-R ) d 3R = V -

When we write the second term as a,

a

I'"

==

f

N

ZN ·

[ 1 - e-PV(X - R )] d 3R .

[1 - e-PV(r)] 4n r2 dr.

(4. 1 6a) (4. 16b)

(4.17)

(4. 1 8)

Using Eqs. (4. 1 7) and (4. 1 8), we write Eq. (4. 16b) as _ V(V - a) N ZN + l ZN · VN

(4. 1 9)

This is a recurrence relation for ZN . We can solve for ZN from Eq. (4. 19) as follows : ZN = V ZN - l

=

V

(1 ;Y (1 �y

-l

-

_

-2

ZN - l , ZN - 2 ,

. .. ...................... .

.

(4.20) .

102

Classical system of N particles

4.2

(4.20), we have ( )N(N -1) 2 ZN V N (1 V)(N-1)+(N- 2)+ . . ·+1 V N 1 - -aV / . (4. 2 1) When N is very large, it can be approximated as (4.22) This exponential part is interpreted physically to mean that each pair contributes ) a factor - (a/V The equation of state is derived from Eqs. (4.22), and (4.7): PV .!:'. In ZN V � (N In V N2V2a) , (4.23) N RT N

Multiplying all the equations in Eq. =

e

_

�

When the temperature is high we may approximate Eq.

=

•

=

The second virial coefficient

a

=

av

av

or

_

(4.24) PVIRT 1 Nal2V Comparing this with Eq. (4.12), we obtain the expression for the second vi rial coefficient : (4.25) B Na12. =

+

=

This is an exact result, in the sense that the three-body and higher-order col­ lisions that we have neglected have effect only for the third virial coefficient and higher-order terms of Eq.

(4.12).

Example : Let us consider a potential as shown in Fig. evaluated as

VCr)

4. 3 . Then Eq. (4.18) is (4.27)

a

=

4nb3

3-

- kT Jroob 1

(4.27) as


(4.28)

which we may write in the abbreviated form :

a Vb - -DT . When we use this in Eq. (4.24) , PV 1 NVb ND . R T 2V 2VT

(4.29)

=

=

+

(4.30)

_

We can find out if this expression fits experimental data well by comparing

it with the Van der Waals equation of state, which is a good fit to the data. The

Van der Waals equation :

(4. 31) (4.32)

can be expanded as

PV 1 bw - awlR T b!2 RT V V we set NVbl2 b w and ND/2 awlR, we arrive at Eq. (4.30) up to the second virial coefficient. If the Van der Waals equation is valid, it gives a second virial coefficient of (4.33) A special feature of this expression is that, because aw is positive, . Bw becomes negative for low temperatures. Experimental data for helium are drawn in Fig. 4.4. The slight decrease for high temperatures does not follow =

=

If

+

+

+

. . .

•

=

B

o T

o

Fig. 4.3 A hypothetical potential.

103

Fig. 4.4 Experimental data for helium gas.

104

Classical system of N particles

Mayer cluster expansion if,3 I'

from Eqs. (4.33) or (4.30). It must be remembered that Eq. (4.30) is based on the hard-core assumption of Fig. 4.3. In actuality, the potential in Eq. (4. 1 8), which is accurate, is slightly soft inside r = b, so that for high temperatures, when particles are hitting each other hard, they can come slightly coser than b . Thus there is a decrease i n the "hard-core" radius, and hence a decrease i n B .

MAYER CLUSTER EXPANSION 'From �q. (4.2),. we know that the classical partition function for an N-particle interactmg gas IS

VCr)

Numerical results:

- C2/r6,

(4.34)

= A'/r" - B'/r6,

(4.35)

r = e- c,r

or the Lennard-Jones potential

VCr)

in Eq. (4. 1 8) we can calculate a . Incidentally, in Eqs. (4.34) and (4.35), the term comes from the Van der Waals attraction. Table 4. 1 lists the numerical results for helium gas (in units of cc/mol).

r-6

Table 4.1 TOK 350 250 1 00 35 20 15 *

*BCalc (CL.)

1 0.80 1 1 .34 1 0.75 -6.95 - 20.0

BObS

1 1 . 60 1 1 .95 10.95 4.80 - 4.00 - 14.0

The Slater potential i s used. J . G . Kirkwood, Phys.

t J. G. Kirkwood, Phys. Zs. 33, 39 (1932).

tBCalc (Q.M.)

1 0.82 1 1.16

ZN f

Rev. 37, 832 (193 \ ).

For a mixture of two gases with atomic fractions Xl and

show that the second virial coefficient is

=

X2, such that

e -PF = e - P V(Rt,R2,

�

(mkT)2 3N/2 ZN N! 2 d 3Rl d 3R2 . . . d 3RN. • . •

nn

, RN )

(4.36) (4.37)

The equation of state can be written, as in Eq. (4. 1 2), as B

PV 1 + - + -2 + . . . RT V V C

(4.38)

This is the virial expansion ; as before, B is the second virial coeffic ient, C the "third, and so on. l This formal expansion can, in principle, give all the vi rial coefficients. It . . was ongmally hoped that this expansion would yield the triple point, and related phenomena. This was not to be. For reasons to be discus sed later' the formal program, which we now review, has not proved practical. . In the followmg work, it is implicitly recognized that the lth virial coefficient ;( IS due to I-fold clusters. ,�

.

4.44 - 5. 1 4 - 15.1

The quantum-exchange effect on the second virial coefficient was neglected in this section. This effect becomes important only for temperatures belo w about 1 0K. The general references on the subject of the second virial coefficient are : J. de Boer, Report on Progress in Physics, Vol. 2, p. 305 (1 948-1949). Mayer­ M ayer, Statistical Mechanics (John Wiley & Sons, Inc. New York, 1940).

Problem:

"

When we use the Slater potential

VC )

105

VeRb R2 , . . . , RN) can be written (we assume) as L Veri). ZN f exp [ - P L VCri)] dNR . WN(R 1 , R2, . . . RN) exp [ - P Veri)]. =

Let If there

=

is one particle, then

For two particles, �t

W2(Rb R2)

L

(4.39)

(4.40) (4.41)

= e - PV(r12 )

106

Classical system of N particles

Mayer cluster expansion

Wz (R1 ' R 2) � U1(R1)U1(R 2 ). If the points . �re close to ea�h other, this approx­ imate equation must be corrected by the addition of a quantity, called U2 (R1, R 2 ) , that is large only when R1 and R 2 are close. Thus W2 (R1, R2 ) can be written as

!W(G ) is a product of its disconnected parts.

U3(R1, R2 , R3) =

U2 (R1, R 2 ) (4.45)

U3 (R 1 > R 2 , R 3 ) is the contribution due to all three particles" close to one another ; U3(R1, R 2 , R3) � 0 unless the particles are "clustered. . WN(R1, R 2 , . . . , RN) = U(R1) . . . U(RN) + . . + UN(R1, R 2 , · . . , RN) · (4.46) There is another way of expressing the above equations. If we set U2 (Ri, R) = 1 + J1"i) and, from Eq. (4.40), J1"tJ' then e -P V(rij) =

• .

For example, if N = 3 ,

W3(R1, R 2 , R3) = 1 + 11 2 + 11 3 + 12 3 + 11 2 /2 3 + 11 3/2 3 + 11 2 /1 3 + 11 2 /2 d1 3 · We can express each individual term in this sum as a graph consisting of points and lines. If N = 3, then 3 3 3 3 � � 1� � 11 2 /2 3/1 3 L · 11 2 � --=-. 11 2 /2 3 1 2 1 2 1 2 1 2 If G represents a graph, and WCG ) is the integral over all space of its cor­ responding function, then (4.48) ZN = L W(G), •

•

•

G

3

�

1 2

+

3

A

1 2

+

3

L

1 2

+

3

Do ·

1 2

'The UN can also be expressed in terms of the W' s. From Eq. (4.43) U1 (R1) = W1(R1). From Eq. (4.43) and (4.44),

Similarly, WiR1, R2 , R3 ) can be written as

. •

A cluster of N points is represented For example :

�Y a sum of all possible graphs in which those points are connected.

(4.44)

W3(R1, R2 , R3) = U(R1) U(R2 ) U(R3) + [U2 (R1, R2 ) U1(R3) + U2 (R2 , R3) U1(R1) + Uz (R1 ' R3) U1(R2 )]

107

where the sum is taken over all possible graphs with N points. For disconnected graphs, such as 4 5� -3 1 2

=

W2 (R1, R2 ) - W1(R1 )W1(R2 ) ·

(4.49)

From Eqs. (4.45) and (4.49),

U3(R1, R 2 , R3 ) = W3(R1, R2 , R3) - L W2 (R1, R2 )W1 (R 3 ) + 2 W1(R1 ) W2 (R2 )W3(R3) '

where

L W2 (R1, R 2 )W1(R3) In

general,

==

(4.50)

Wz (Rh R2 ) W1(R3 ) + W2 (R2 , R3) W1(R1) + W2 (R1, R3)W1(R2 ) ·

U,(R') = U,(R1, · . . , R,)W,(R1, · . . , R,) - L W2 (R;, R)W'- 2 (Rk' . . . , R,) + . . . + ( _ 1), - 1 (1 - I ) ! W1(R1) · · · W1(R ,), (4.51 ) 1 Here the coefficient for a term with k groups is ( _ I)k - (k - I ) ! .

Thus, from Eq. (4.40), all the WN are known. From Eq. (4.51), U(R') i n terms of the Wi i s given ; thus the U,(R') can be found. (4. 52) In any particular partition of the N particles, there will be m, clusters of size I; and of course LI Im, = N. For example (as is evident from Eq. (4.45»), a three-particle gas can be partitioned into 1.

1 cluster of 3 , m 3 = 1 , m 2 = m 1 = O . 2. 3 clusters of 1 , m 1 = 3 , m 2 = m 3 = O. 3. 1 cluster of 2 and 1 cluster of 1 (this can happen 3 ways), m 1 = m 2 m3 O.

=

1,

=

From the above discussion, with Eqs. (4.45) and (4.46), L im, = N.

(4.53)

S( m r) means the sum over all possible partitions of N, that is, all divisions of particles into groups.

108

Classical system of N particles

From Eqs. (4.37), (4.39), and (4.40), XN Now construct the expression

bl =

==

ZN = N!

_1_ f

VI!

f

WN d NR. N!

(4. 54)

I UzCR ) dRI . . . dRI•

(4.55 )

This equation is known as Mayer's cluster integral. From Eqs. (4.5 1 ), (4.53), and the statement labeled (4.52), XN =

�! N

S(m , ) n

f

I Uz(R )

R calling a similar summation encountered earlier (for the quantum . . � statIstICS of a many-partIcle sys�em), we note that if the restriction L mil = N could be removed, XN can be easIly evaluated. Using the same methods as before we defin e

and where

(4.57)

( )

32 a = � / ellP. 2nli 2f3

But if N = 5, for example, U(RIR2 ) U(R3, R4) U(Rs) gives the same contri­ bution as U(RIR3) U(R2 R S) U(R4), which is a different partition with the same set of m i ' For each possible set of ml there are several terms, all of which are equal. To see how many such terms there are, note that there are N! permuta­ tions of the N coordinates, and each such permutation corresponds to a term in S(m , ) with the same set of mi' But we are counting each term more than once. We should not count different permutations of the coordinates within a given U, for example, U(RI' R2) == U(R2' RI). We also do not count changes of order of the U ' s as separate contributions. For example, U(RI' R 2) U(R3 ' R�) U(R5) U(R3 ' R4) U(RI, R 2) U(Rs) are not counted separately. Thus for a given set of mZ, there are N!/DI (/!)m 'ml ! terms. So we can write =

XN - ZN - � L possible sets N ! - N ! ofallm,: "Elm, N

But first ta �ing the sum L ( I ) with L mil = N and then summing over all N is � clearly eqUIvalent to summmg over all ml with no restriction. ( Vb )m 'alm , e _ pg = L n ' , (no restrictions) m I mi ' _ __ _ _ _

m l,m 2, " ' .

[

= exp V

.

•

•

•

(4.58)

This is known as Mayer's cluster expansion. From Eq. ( l .5 l )

(4.56) It is important to note that (for large volume), bl is independent of volume. That this is so can be seen as follows : UI(RI) vanishes unless the particles are clustered. Holding dRI fixed and varying dRz , . . . , dRI then gives some number completely independent of volume, because unless R 2 , ,RI are near R l UI(R ) = The integral over dRI then gives a quantity proportional to the volume, V. and as bl was defined with a V, in the denominator, bl is independent of V. We must now evaluate XN = S(m , ) TIl ( Vb,) ml/ml ! subject to L m , l = N.

J

� albz

l L a bl · g = - kTV - I


=

O.

109

Mayer cluster expansion

Therefore the equation

=

_

og all

+ kT V L la ' - 1 �a bl 011 I p

=

V

= < N)/V = L lal bl

determines ':I. if we know the dens ity. Fro m Eq. ( 1 .53) the pressure is given by

og p = - --

I

oV p.

=

I

k T L a'b I ' I

� [alb . 7" I

(4.59)

110

Classical system of N particles

Radial distribution function ...4

Therefore we get the equation of state PV RT

PV = Ll b,(X1 NkT Ll lbp,l .

(4.60)

1 « For small density, (X is small-as can be seen from Eq. (4.59). Then (X l + (X l ; so that expansion CEq. (4.60)) need only be carried out a few terms . For small I, h, can be calculated easily. Inverting Eq. (4.59) to get (X as a powe r series in I / V, and writing PV RT

RADIAL DISTRIBUTION FUNCTION

consider a system of N particles interacting with pair-wise forces. The partition (unction ZN is (4.63) The probability

density for particle I being at R1, particle 2 at R2, etc., is

C B 1 + - + -2 + ' " ' V V

we have

where the factor of I /ZN provides normalization. The probability that a given particle, say particle 1, is at RI and the rest of the particles are anywhere is

+

(;J f exp

)

e- P[V<'12) +V<'23) +V<' ,3)l dR dR · 2 3

(4.62)

When the density becomes large, the Mayer cluster expansion is of little use. This is because at very high density (liquid), the most important term in the expansion becomes the Nth term. That is, almost every particle becomes a member of a very large cluster- and in a liquid all the particles are in an N-fold cluster. Hard Sphere

Gas

The question may be asked : What is the partition function of a hard-sphere gas-that is, a gas composed of small inpenetrable spheres that exert no forces on one another when not touching? Unfortunately, this question is still un­ answered. The question may be stated in a different way.

i-

ZN =

But

111

1 d NR.

no two closer than 2a. a = radius of sphere VN

=

J.

d NR.

no restrIctIOn N N Thus i- d R/S d R is essentially the probability that if N points are "dropped"

into a volume V, no two particles will be closer than 2a.

[ - 13V (RI " ' " RN)] d 3R2 d 3 R 3 , · · · , d 3RN·

Because there are N particles, the probability density for one of these particles being at R is n l (R)

=

N

f exp = N f exp [-13 � Veri)] ZN

[ - 13V(RI ' R2,

palr

ZN

nl

•

•

•

, RN)] d 3R2 ' " d 3RN

d 3R2 ' " d 3RN ;

(4.64)

is the one-particle density, or distribution function. The two-particle distribution function n2(R l > R2) is defined as

- f [-13 � ]

N(N 1) n 2 (R1 , R2) = ZN

exp

palr

V d 3R 3 ' " d 3 R N ;

(4.65)

n2 is the probability density for finding a particle at RI and another particle at R2• For a liquid or a gas, n zCR1 , R2) is a function of the distance between the two

particles.

n 2 (R1 , R2) = n 2 (r 1 2 ) ·

The .Beneral shape of n 2 (r) is as shown in Fig. 4.5. When r -+ 00, n2 (r) a�proaches ni where n l is the one-particle density ; this is because when the �stance becomes large, the existence of one atom does not influence the distribu­ tion of the other atom. The radial distribution function n2(r) is related to x-ray or neutron diffrac­ ti.on as shown in Fig. 4.6. The x-ray is coming in along the z direction and is SCattered by atoms. The scattered x-ray is observed at P somewhere far from the

112

Classical system of N particles

4.5

Thermodynamic functions

113

So Eq. (4.66) becomes - - - - - - - - nr

or

eikR Cp

Amplitude = a(8) -- . G,

with Fig.

4.5

Two-particle distribution function.

z ---...,. ..

Incoming x-ray wave vector K in

\

Rj

C, reference center

C

�ip,

[i(Kin . Rj

- Kout .

R)] =

L exp (iq . R), j

(4.68) To compute the intensity, we need to know the expectation value of I G 1 2 . Now

unit vector Outgoing wave wave vector Kout = k Ip

IGI 2

( I GI2)

a(8)(eikRjR).

Here R is the distance from the scattering atom and a(8) indicates the �os�ible angular dependence of scattering, which is not important here. Takmg lll t o account the phase of the incoming wave, eik ' z, we write Amplitude = a(8) L (eikRjpeikZjjRjp), j

P is the point of observation. Then, approximately,

i

=

N I eiq · ( Ri - Rj). i,j = 1

For a liquid, neglecting the = j terms in the double sum, we have

X-ray diffractions.

liquid. The amplitude observed at P is the sum of amplitudes scattered by each atom, the l�er being of the form

and we may write

j

Intensity = IAmpiitude l 2 = (la(8WjR�p) IG I 2 .

p

4.6

= L exp

The intensity of the x rays is

somewhere in the liquid. We can define Z = 0 at C.

Fig.

G

where

@

(4.67)

R cp

=

N(N2- 1) f

eiq ' (R , - R2 )

[

[ - f3

exp

�:air

V(Rij)]

]

N d3 R

(4.69)

because the term in brackets, [ ] , is the probability for the configuration R 1 , Now, using Eq. (4.65), we write

R2, • • • , RN•

( IGI 2 )

= 1-

f

eiq ' (R, - R2 )niR 1 ' R2) d 3R1 d 3R2 �

(4.70)

When n2 (R 1 , R2) depends on I Rl - R2 1 only, V = the volume of the liquid, and q = I q l ,

(4 . 66)

(4.71) 4 .5

THERMODYNAMIC FUNCTIONS In the classical case, if we know n (r ) , we can derive all the thermodynamic 2 functions. The internal energy is

U

1

aZ 3 = zRT - - --N , ZN ap

(4.72)

114

,

\,",

Classical system of N particles

ZN fJp

_

=

_

f f

l Equation (4.74b) follows. Equation (4.74a) is nothing but the obvious h eraliz �n �tion to_the case of potentials that are not spherically symmetric. The ::p� roduct R · V VCR) in the integrand of Eq. (4.74a) is called the virial.

_

[ -13 L V] d 3NR

V(r12) exp

When the force is homogenuous of degree (n + 1 ),

ZN

[ -13 L V] d 3NR · · ·

V(r2 3) exp

- f

ZN

N(N

1)

V(r1 2)

2

exp

';50 i

[ -13 L V ] d 3NR. ZN

or

ZN fJp

=

U

=

-

-! 2

f

-f

2 V(r)n2(r)4nr dr.

or

pV

=

RT

Vol

2 rV '(r)nir)4nr dr.

(4.73)

(4.74b)

These are classical formulae, which do not hold for liquid helium. Equation (4.74b) is derived as follows. From Eq. (2. 1 2 1) 3pV

= 2(KE)

As in Eq. (4.73), (KE) written

=

- (�

' rij V'(rij» j I < J.

)

'

=

n Crn

=

n V(r)

=

(3 + in)RT.

(4.76)

This relation is called the virial theorem.

This equation could have been expected. The first term of Eq. (4.73) is the kinetic energy, and the second term is the potential energy. The latter comes from the pair potential VCr), and the factor 1/2 appears because each pair is counted twice. The pressure is expressed as follows : Vol (4.74a) pV = R T - 6 R · VV(R)niR) d 3R

f -6f

r V'(r)

3p V + n U

V(r1 2)n2(r1 2) d 3 R l d 3R2

3 Vol RT + 2 2

that

�e can use these in Eqs. (4.73) and (4.74b) to see that

Here, we use Eq. (4.65) to arrive at

1.- fJZN

115

f

where ZN is the partition function defined in Eq. (4.2), now

1.- fJZN

The Born-Green equation for n2

i4-6 la

(4.75)

(3/2)RT. The second term of Eq. (4.75) can be

Problem:

Show that at low densities n i r)

=1

-

e -P V(r).

Using this result in Eq. (4.74), derive the virial coefficients in the virial expansion.

�4.6 THE BORN-GREEN EQUATION FOR

n2

�In Section 4.5 we pointed out that one can find all thermodynamic functions

t�ven n2(r). In Section 4. 1 of this chapter we showed how we could find (in rinciple) e - PF given the potential between pairs of partic1es, and from F we can .also get all thermodynamic functions. But to get e - PF we must integrate over i:something like 102 3 variables. Computers cannot do such an integration ; so we ·have got to find some way of getting the answer approximately. The Mayer ' cluster expansion was one way, and it was satisfactory for a gas. So far, for a liquid, we have replaced F by n 2 (r), which requires six fewer than the 102 3 integrations for F. Clearly, we still have quite a way to go. One approach toward approximately evaluating our monster integral is by means of the Born­ Green equation.

�

niR1 ' R 2 ), defined in Eq. (4.65) can be transformed as follows : niR1 ' R2 )

=

=

N(N

- f [-13 � ] - 1) f [ -13

ZN

N(N ZN

1)

exp

exp

palf

3 3 V d R3 ' " d RN

L V(R l i)

i* l

-

13

]

L V(Rij)

i*l i* l

116

Classical system of N particles

4.7

4.7

Differentiate the above equation to obtain V 1 nzCR1, R2) =

_

f3N(N ZN

When we define n 3 (R l ' R2 ' R 3) = -�3) ! ZN (N _

1) f [

V1V (Ru)

f exp [

+ ;;1L

V 1V (R li)

]

- f3 L v d 3 R4 ' " d 3 RN, pair

] (4.78)

we can write Eq. (4.77) as V1 n zCR 1 , R2) = - f3V 1 V(R 1 2) n zCR1 ' R2) - f3

This is the Born-Green equation. *

f

V 1 V(R1 3)niR1 , R2 , R3 ) d 3 R 3 ·

-+

Our attempts to evaluate the integrals involved in the thermodynamics of liquids and gases have not met with complete success. The Kirkwood approximation • might not be bad, but we do not know exactly how to generalize it, or whether a generalization would give better results . What we want is a method that would give progressively more accurate results as we increased the computational labor. No such method is available today. We are faced with a problem that is too hard for us. We may be able to gain some insight by picking a soluble problem that has some of the features of the insoluble one. For example, consider gas particles constrained to move on a line (Fig. 4 . 7). First we assume only nearest-neighbor interactio n. The partition function is

f

PN(y, L)

f

=

Y < Xl < X2 <

which is called the Kirkwood approximation,t or the superposition approximation. When we use Eq. (4.80) in Eq. (4.79) we can solve for n 2 '

Born M. and Green H. S., Proc. Roy. Soc. (London) A18.8, 10 . ( 1 946). See J Yvon Actualities Scientifique et Industrielles (Herman and Cle, Pans 1 935), No. t' Kirk':ood J. G. and Boggs E. M ., J. Chern. Phys., 10, 394 ( 1 942). *

.

.

•


e - P[V(Y-X d+ V(X I - X2 ) + " ' ] dx 1 dX 2 . , . dxN·

���

.

(4.82)

In Eq. (4.82) Y is fixed. Now, consider the case shown in Fig. 4.9. From Eq. (4. 82) we can write

PN+ l(Z, L)

f

=

Z < Y < Xl < . . . < L

e- P[V(z - y)+ V(y - xd + " ' ] dy dX l . . , dxN·

(4.8 3)

o ,

n2(R1 Rz)nzCR2R 3 )nzCR 3 R 1 ) n 31

(4.8 1 )

The length L is fixed. We would be making a good start toward solving our problem if we could get a recursion relation with ZN + 1 expressed in terms of ZN ' Since we cannot even do this, we will modify the problem in such a way that the macroscopic properties of the system are not changed but a recursion relation for the partition function can be found. We put one extra atom before Xl and fix its position to be at y (see Fig. 4.8. ) The resulting partition function we call PN( y, L).

Considerations such as this led Kirkwood to the approximation niR1 ' R2 , R3 ) =

e - P [V( XI - X2) +V( X2 - X3 ) + " ' ] dx l dxz " ' dXN'

O < X l < X2 < " ' < xN < L

(4.79 )

n zCR1 , R2)n 1 ·

117

ONE-DIMENSIONAL GAS

'ZN(L) =

The special feature of this equation is that both n2(R1, R2) and n3 (R1, R2, R 3 ) are involved. When w e g o further and write the equation for n 3 w e need n4 · The . equation for n4 includes ns, and so on. Thus, we have a hi�rarchy of equatlOn s. . This kind of hierarchy appears in other branches of physIcs-for exampl�, In meson field theory-but no general, accurate way of treating the set of e�uatlOns has been established. When we make a guess for ns, say, and cut off the hIerarchy there and calculate back n2 , we might get a good estimate of n 2 ; but that an error in the guess of ns becomes unimportant as we approach n 2 has not been proved. . . . An ingenious way of closing the hierarchy IS due to KIrkwood. When R 3 i , very far from R1 and R2, n 3 (R1, R2 , R 3)

One-dimensional gas

L ,

Fig. o I

4.7 Gas particles constrained to move in one dimension only. y

Fig.

4.8

L I

One atom fixed at y.

1 18

Classical system of N particles o I

4.7

Using

Fig. 4.9 One atom fixed at z. In the integral,

z

PN + l (Z, L)

f e -PV(z- y) dy z
PN( y,

g

=

=

=

is fixed, but y varies. From Eq. (4.83) =

=

= F

dXN

=

o.

y,

(4.84)

ePp(L -y)

Using PN (y, L), w e can construct the grand partition function

I/N!

N

/1N/1,,

F

or

e - /ll'

V F Vf(V/N).

(4.85b)

N to obtain

=

I'"

e - p/lxe -/lV(x) dx.

(4.87 )

The integral is insensitive to the upper limit, so we have replaced that limit with CXl . This is the desired result from which we now can find all the thermodynamic quantities. For example

N

F

N

(4 . 86)

( 4.85a)

As in Section 1 . 1 , g for a combination of several independent subsystems. each with the same is the sum of the g i for each subsystem. In Section 1 .6 \\ e defined = g + where is the expected number of particles in the syste m. Clearly, for a combination of independent subsystems is also a sum of the f ' s for each system. If there is a sufficiently large number of particles, we can consider the line t o be a combination of line segments that interact with each other very little. With such considerations, it is easy to see that F/ is a function only of the densit; (at a fixed temperature), and can be written = It is not hard t o show that of the above form is equivalent to the equation g = - p V. The proof is as follows : From Section 1 . 6 :

FF

L ePI'NpN( y , L) .

which leads to

/1.

N L ePI' pN(y, L) .

=

Multiply Eq. (4.84) by e(N + l )I' P and sum over

In our case, ZN/N! = PN(y, L) (the factor comes from the fact that we ha\ e in the integral for PN restricted the range to Y < X l < . . . < L instead of allowing each particle to be anywhere in [0, L]). We can absorb the factor N Then (m/2n n 2 {3) / 2 in the e/lIlN by adjusting =

N

v

z
e - Pg

119

F Vf(V/N) we then have + P V F - /1N + P V F - N oOFN I V oOFV I F - N [ - ;: 1' (�)J - V [f (�) + � 1' (�)J - Vf (�) Because the "volume" of the system is L - we can now write Eq. (4.85a) as

L I

y

z

One-dimensional gas


= -

(/1 0/1o op ,0/1)!) Vol g

=

=

Vol

Equation (4.87) can be written 'as

.

"rom 0

o = {3/1 +

=

df/d/1,

roo

log J e -P/lxe- P V(x) dx o it follows that

Op a/1

_

=

0/1op L 0/1op =

f[/1, p(/1, {3,

af/al:! = So�P/lXXe -PflVV(XX)) ' aflap So xe-P/l e- (

{3)] .

120

Classical system of N particles

One-dimensional gas with potential of the form e - I

4.8

Problem: When the nearest- and next-nearest-neighbor interactions are in­ cluded, derive the chemical potential. (Hold two atoms fixed near the left end of the line and derive the recursion formula. An integral equation results.) We see now that we can solve the problem of particles fixed on a line if the potential acts only between the nearest few neighbors. But what if all the atoms in the system interact? For a certain special form of the potential we can still solve the problem. Section 4.8 discusses the case of an e- ixi potential. 4.8 ONE-DIMENSIONAL GAS WITH POTENTIAL OF THE FORM

e- Ixl

We will now discuss the problem of a one-dimensional gas, all particles of which interact repulsively with all others with a potential Vex) = e - i x i . When there is attraction and no repulsion, the preferred configuration occurs when the particles all sit on top of one another. The potential is pro­ portional to the number of pairs, N(N - 1 )/2. The more pairs there are, the lower the potential can get, and ultimately all particles collapse to a point. This result does not tell us much, so we assume a repulsive force. Assume N particles are distributed in a line of length L (Fig. 4. 1 0). We wish to evaluate

x

I

121

Suppose, as before, we mod ify the problem sligh tly by placing an extra atom before X l ' and fix the p s!tion of that atom at y. Let a = eY• Later we will � set y equal to zero. The partItIon function for this syste m is

IN(a, L)

f

=

O
exp • • .


{- [

/3 a(e-XI + e - X2 + . . . + e-XN) +

Then

]} dx . . . dx .

L e(Xi -Xj)

j>i� l

I

2

(4.88)

V

L

L

�

L e(Xi - Xj)

j>i� 2

]} dXI

1+----- L --------+1

Fig. 4.10 One-dimensional gas.

exp ( - /3ae-XO)

O
+

f {- [

=

exp

/3 (a + e-XO)(e - Xl +

... +

exp ( - /3ae-XO)IN(a + eXO, L

-

+ e - XN)

]} dxo . . . dx

. L e(Xi -Xi)

J > i� l

y) dxo .

N

(4.89)

in Section 4.7, we can again modify the problem in a manner that does not a ect �he macroscopic propertie s. We allow N to vary' put in a chemical potentIal, and define •

•

.

dXN

·

(4.87)

=

L IN(a, L)eP/lN. N ca�e IN(a' 9 is the artition function for a system with one particle fixed � y - l g a, Its lengt h IS L - log a, and g = -p eL - log a). � MultIply �q. (4. 89) by e(N + 1 )/lP, sum, set y = 0 and replace L by 00 . The result after a lIttle algebra is e- pg

�

;�:

(4.90)

m Eq. (4.90) we can obtain all the thermodyn amic properties, and we can , ablY assu e that the ato fixed at zero does not change those propertie s. � But e us see w at we can do wIthout fixing an atom at zero. The force on the

��

:

122

Classical system of N particles

: 4.8

first atom, the one located at X l ' is due to the rest of the atoms in the gas, and the effect of the other atoms can be described with a potential involving o ne extra parameter. Thus the first atom moves in a potential eeX ! - 1 , where e = Li e - Xi. Instead of specifying the position of one of the atoms as an extra parameter in the problem, we will specify e, which determines the potential felt by the atom at Xl ' We now calculate GN(e), which is the partition integral :

f exp [

]

L

-

fJ

We

o

Xl . .

.

<

XN

<

f

L. Mathematically,

[

� - rx l i j e (X i

exp - fJ

O < X l < . . . < xN < L

Note that GN(e) = 0 if e L

<

e - PFN

}(

, e -

� e-x

)

dXl . . . dxN.

0, and

foo = (--) N1 2

m

2nn 2fJ

0

(4.91)

=

f exp [ x

where

(

- fJ

�

GN(e) de . L

' e - (xi - xd - fJ

b e - e - X! -

i

L'

==

e

- ,-" - ,-" � ' =

)

- ,'

�

e"'O

(" , ,

- 1

-

f:'

,

- ,'

(Jla) b(x). Equation (4.92b) can be rewritten

(x , ) exp

dx , exp

exp

,

[

-

fJ

[ - p(ee'"

- 1)]

]

L'j e - (Xi - Xli dx� ' " dx'N

i<

r- " [ 0

L ( dXl exp (X l) exp [ - fJ(eeX! - I)JGN l (eeX! L --x !

Jo

ee'" - 1

-

123

)

-

1).

I X ;

�'

,

-"

]

(4.93 )

�n general, the solution comes from Eq. (4.93), but it is very difficult to find. For subsequent

use, let us define (for N particles)

(4.94)

To get a recursion relation we will hold Xl constant until the rest of the integra­ tions are done. Equation (4.91) can be rewritten G�(e)

r x

L e - Xi = e <

(

because b(ax)

with the range of integration subject to the restrictions that

and 0

know that

6{-e)

- ( Xi - Xil dXl . . . dXN je i�

i

One-dimensional gas with potential of the form e -

�' e -x

)

]

� ( r i) i ; e- X X

dXl . . . dxN,

:and

p =

Combining Eqs. (4.93) and (4.94), we find that gee)

(4.92a)

iL* · 1

Because of the action of the b-function, the L; e-Xi in the exponent of the inte­ grand can be replaced by e - e - X ! . To make the argument clearer substitute x; = X - X l ' Thus

i

(4.92b)

pressure.

= rx

100

dx exp (x) exp [ - fJ(eeX - l )J e - PPXg(eeX - 1),

(4.95)

= Xl and extens ion of the limit L -+ 00 have been made. In obtaining the eigenvalues of this integral equat ion, a relationship will :be f�und between rx and p ; gee) itself is not of intere boundary value on '(I(e) IS gee) = 0 If. e < O. To put Eq. (4.95) in termsst.ofThe differentials, substitute

' where the substitutions X

100

e - P Cg(e)

h (e)

=

gee)

= rx

g'(e)

= rx

foOO

dxeXe - /iPXh (eeX

-

1)

dxe2Xe - /ipxh ' (eeX - 1)

Classical system of N particles

124

- Jofoo - 1)(1 - 1) -

Integrate by parts : g'(e)

=

� h(eeX

e rx(f3p

=

- 1) - 1) 1)

e rx(f3p

=

gee) gee)

e

�

l)eXe - pPxl�

_

_

_

h(eeX

e

If instead

o/(W)

e

� e-P (c - l )g(e

e

g'(e)

=

=

. we get (4.96)

1).

e-Pc2/ 2f(e) _ f3ee -Pc2/ 2f(e) + e- Pc2/ 2!'(e).

Then Eq. (4.96) becomes ef'(e) - f3e 2f(e) - rx(f3p - l)f(e)

=

- rxeP/ 2f(e

We can use the Laplace transformation o/(q)

=

too

-

e - qCef'(e) de

=

dq

(:;

eiWCf(e) de

+ rxeP/2 eiw + rx(f3p

- 1) t)) +

t/I(w).

The boundary conditions are not known definitely in this transform notation. , However, it may be that the only restriction necessary is that the function I/I(q ) or o/(w) remain finite everywhere. We can now, in principle, get a functional relationship between rx and p. BRIEF

DISCUSSION OF CONDENSATION

As the temperature is increased there is a continuous transition from the "solid" to an aggregate of groups of atoms : Cf:XX) CJ:XXX) Cf:XX) . The energy required for breaking up one group of atoms into two groups is roughly the energy required to break two atoms apart and we call it 8. The probability of such a break �s e - e/kT, and when the temperature is raised it happens more or

e - qcf'(e) de

0

less uniformly. In three dimensions one might have a group of N atoms

and Therefore

v

!£ [q o/(q)]

+ f3

�2 o/(q) dq

+ rx(f3p - l)o/(q)

=

rxeP/ 2 e - qo/( q).

Substituting we eliminate the first derivative term : - IN//(q )

=

[- :;

- rxeP/ 2 e-q + rx(f3p

(4. 98)

� ....

d - - [qo/(q )] dq

dq

=

f

For a one-dimensional gas such as the one we have just considered with repulsive forces there is no condensation. Let us consider instead the case where there are both repulsion and attraction, such as the one depicted by the potential of Fig. 4.1 1 . At T = 0° all atoms are adjacent to each other in positions corresponding to Vmin in Fig. 4. 1 1 :

1).

e - qcf(e) de

- � foo

f3t/1//(w)

4.9

to obtain an equation in o/(q). Thus :

too

=

and substitute

� h(e

125

we use Fourier transforms

f3p)eX - PPX dx

is not, Equation (4.96) appears at first glance to be a differential equation, but enter into it. because gee) and g(e For another form, substitute gee)

Brief discussion of condensation

4.9

- 1-] 1) +

t/I .

(4.9 7)

Fig. 4.11 A potential with attraction and repulsion.

�.

126

Classical system of N particles

The energy required to cut across a group is e = N 2 / 3 A, where A is roughly the value of the energy required for taking one atom away from the group. Thus, the probability for splitting up the group is e- N2 /3!::./ KT. The larger N is, the higher T is required to be for a finite probability of breaking up a large group of atoms. Thus, in the one-dimensional case, a finite energy is required for breaking up a group of atoms, a finite number of groups exist, and the fraction of atoms (essentially infinite in number) in the condensed state (that is, in the largest group) is zero. In the three-dimensional case, because of the probability factor e- N2 /3!::./ KT, a finite fraction of the total gas can exist in the condensed state at a finite temperature. For solids the difference is more subtle. In one dimension, as the temper­ ature is raised, let the mean square error in the position of one atom relative to the ones on each side of it be 15 2 • Then, over a group of N atoms, the mean square error in positions of the first and last atoms is NJ 2 from the perfect lattice configuration. The error "propagates" (in a sense), and when temperature is increased, the solid expands monotonically. There is only one path along which information can be transmitted. In three dimensions, the position of an atom in one place influences that of a distant atom much more strongly than in one dimension-there are many paths connecting the two along which forces are being transmitted. Thus to melt a solid T must become high enough that the thermal energy can overcome these "multipath" forces.

CHAPTER 5

ORDER-DISORDER THEORY

5.1

INTRODUCTION

We will now consider a �ew class of statistical-mechanical problems, problems . tha� do not mvolv� motIOns of atoms but instead involve their positions in a lattI �e. More preclsely, we will consider alloys made up of two types of atoms. . For mstance, m the cubic lattice shown in Fig. 5. 1 , there will be some preferred arrangement for the manner in which the lattice sites are occupied by type A and type B �toms. Let us take the case of zero temperature. Essentially, two types of lattice arrangements are possible, one of which is depicted in the figure, and �hey ca� be unders�ood from a physical point of view as follows. Let VAB be the mteractI?n potential between unlike atoms and VAA, VBB the potential between the lIk� ato�s. If VAB > (VAA + VBB)/2, then A atoms prefer to be near A a�oms and likeWise for the B atoms. The solid would tend to exist in two . parts-one of A atoms only, and one of B atoms only. If VAB < dlstmct (VAA + VBB)/2, the A and B atoms would tend to alternate in position throughout . . If the solid is heated, then gradually A and B atoms will exchange the s?lId . . p�SltI ?ns m a random way until, at the Curie temperature, the order "melts" ; at s�Ill hlghe� temperatures, disorder sets in. It is of interest to compute the parti­ tion functIOn for such a system :

(-

L

exp

I I

I I

I I

I

I

I

I

I

I

arrangements of atoms

L

(i,j,bonds)

)

P Vij .

--�--�--�--

- -� - - �--�-� � � I I I

--

Fig.

5.1

I

--

I

--

I

--

Cubic lattice with two types of atoms, A and B.

127

128

Introduction

Order-disorder theory

Define :

J;But w�en we sum quantities over all possible arrang�ment� of the atoms, t�e

�trictlOn that NA - NB be constant may cause dIfficultIes. To relax thIS

�' teStriction, we put in the usual "chemical potential." In other words, we change

Il i = + 1 for A atom - 1 for B atom NA, NB = number of A, B atoms, respectively.

Thus,

Ilillj

V.

'j

=

Then,

( )

2a

�

'--

+

NB

as

+ j all,.llj. + b � Il + c. 2

VAA = a + b + VBB = a - b + VAB = - a + c

and

the problem (but not the results of the solution of the problem) by saying that NA - NB can vary. Associated with variation of NA - NB is a proportionate amount of chemical potential, which has the effect of making a certain ratio NA/NB the most probable. In such a case, we cannot ignore

+1 +1 -1

if AA pair if BB pair if AB pair. Considering nearest-neighbor interactions only and the total NA constant, we will write the interaction energy in the form =

-

_

VAA + VBB 2

C C

VAB'

)

) (

Ili + Ilj exp ( - f3c). exp - f3a � 1l;llj exp - f3b � ',j 2 ',j

The last exponential, e - Pc, is a constant and does not enter into calculations .. In the sum Lbonds (Il ; + 1l)/2, each atom contributes to the sum a certalll the � number of times. The number of times an atom contributes is equal to For number of bonds it makes, that is, the number of nearest neighbors it has. ur example, in the case of a two-dimensional square lattice, each atom has fo nearest neighbors ; so i + III = 2(NA - NB)· L Il bonds 2

In general, if NA and NB are kept constant, we can ignore L (Il ; + 1l)/2 bec� use it does not depend on the arrangement of the atoms. The problem then IS to calculate for the partition function the quantity

(

)

(

bonds

Ili + Ilj 2

=

,...., Ili · '--

atoms

Our problem reduces to the calculation of

(

L exp - H L lliBj -

Thus, the exponential in the partition-function equation above is

(

129

)

L Ilillj ' L exP - f3a iLj lliBj = L exp - H bonds ,

bonds

J

)

L Ili ,

atoms

where, in summing over possible arrangements, we still take NA + NB = N fixed but no longer fix NA and NB, separately. If J = 0, we are obviously solving the problem of (NA) = (NB) ' For practice, we will evaluate the above sum for the one-dimensional case. Then we will consider an approximate method that works for two and three dimensions, and finally we will find the Sum exactly for two dimensions. The exact solution for three dimensions has not yet been found. It should be noted that the above model can be taken as a model of ferro­ magnetism. We interpret II I = + I as spin up and III = - 1 as spin down, where the atoms are taken to have half-integer spin. A better model for ferromagnetism is to take the Hamiltonian

=

H L Si ' Sj bonds

where Si is the spin operator for the ith atom. This model is not equivalent to the model with Billj terms. For example, a two-atom system has SI . S2 = H(SI + S2)2 - tl12 ] ,

Which has the value tn; with a priori probability three times as large as that for Which the value of SI . S2 has value - in2 • That is, there are three ways in Which two atoms of spin -t can combine to give a system with total spin I , but only one way in which they can combine to give spin 0. On the other hand, Ilillj' takes on its two possible values with equal a priori probability. Although the two models are different, the qualitative results might be similar. For the rest of this chapter we will discuss the model in which the energy has terms Ilillj. This is the Ising model.

Approximate methods for two dimensions

5.3

Order-disorder theory

130

If we think of (XN( l ), XN( - 1 )) as an eigenvector and e{JJl as an eigenvalue of the above equation, then it is easy to see what to do next. To find e {JJl it is necessary to solve

5.2 ORDER-DISORDER IN ONE-DIMENSION

For practice, let us evaluate L exp ( - H L BiBi + l - J L B;) ±1 ±1

e -{JF =

o =

'1 = '2 =

in one dimension. First, suppose J new variable :

=

O.

131

Then an easy way of doing the problem is to define a if Bi = B i + 1 if Bi = - Bi + l

Then

(

J (J det e - H - -- e Jl eH - J

The result is

cosh J ± eH�l + e-4H sinh 2 J. There are two solutions here; which one do we want? If we look back at the case of J = 0, it is obvious that efJJl becomes 2 cosh H = e - H + eH, so the general result is T efJJl = e - H cosh J + eH�1+e- 4H sinh j . ef1Jl = e-H

Problem : Try to get the free energy as a function of the ratio between the number of A's and the number of B's. =

So

(2 cosh H)N.

and we have completed the calculation. . cosh FIf J -"# Nj0,f3theIn (2above method does not work. We have to use what =

H),

IS,

5.3 APPROXIMATE METHODS FOR TWO DIMENSIONS b'y

large N, now, a familiar trick. That is, we make up a new problem that,n for found . can relatio almost the same as the old one, but for which a recursion in which thebeparticle We will try to evaluate the partition function for a system on the end has a fixed B . Let IS

Then

You may have noticed that in Section 5.2, none of the equations were considered important enough to number. This is because we were not especially interested . in the answers, but instead wanted to examine some possible ways of looking at . problems. Perhaps you will be able to improve upon and use some of the methods described here, or maybe you will find these methods useful for solving a completely different problem. In this section we will nibble at the edges of another problem : that of order-disorder for a two-dimensional square lattice. For simplicity, let us take J = O. One approach is to find a recursion relation. Tq, do this, find the partition function for the case where all the B'S in the last column are fixed. Suppose t�ere are M columns of N particles each, and let XM(B1 , B2 , . . . , BN) be the parti­ bon function. As in the one-dimensional case, the free energy is proportional to the number of atoms. It follows that N ef1Jl XM(B1,

The free energy for a large number of atoms should be proportional to the num ­ ber of atoms ; therefore XN + 1(BN+ 2)

=

e{JJlXN(BN + 2)'

where /l is independent of N for large N. The free energy is then F = recursion relation then becomes two simultaneous linear equations where

y

= ± l.

-

•

•

•

, BN) = X M+ 1(B1 , . . . , BN) L exp { - H[8101 N x

/IN. Th e

+ 8202 + . . - + BNON]}

61, . . . ,6

exp { - H[8182

+ . . . + BN 1BN]}XM(OI , " " ON)' -

(5.1) �he analogous equation in Section 5.2 required for its solution the diagonaliza­ bon of a two-by-two matrix. The above equation requires the diagonalization of a 2N_by_2N matrix.

132

Order-disorder theory

Approximate methods for two dimensions

133

Given the matrix equation M A = JeA with M hermitian, a well-known way for finding the solution with the minimum or maximum Je is to try to find the extremum of A . MAlA ' A . This method might be applicable to Eq. (5. 1). Let XN(B1 , · · · , BN)

=

exp { - -tH[BIB 2 + . . . + BN - IBN]}Z(B1 , . · · , BN)'

Then Eq. (5. 1 ) becomes L 01,

x

x

• • •

,ON

exp { - tH[BIB2 + . . . + BN - IBN]}

exp { - tH[8182 + . . . + 8N- 18N]}

exp {H[BI81 + . . . + BN8N]}Z(81 , . · . , 8N). We made this substitution in order to get an eigenvalue equation with a sym­ metric real matrix, that is, a hermitian matrix. Then, to solve this equation, we must find the extremum of liD, where I = L Z(81, . · · , 8N)Z(Bl, · . . , BN) exp { - tH[BIB2 + . . + BN- IBN]} .

8i,ei

x

exp { - tH [81 82 + ' " + 8N _ 1 8N]} exp { - H[B1 8 1 "+ . . . + BN8N]} . D

=

2 L IZ(Bl , . . . , BN) 1 . E;

Of course, it is not likely that you will be able to get an exact extremum of IID. But it should be possible to get an approximate one by cleverly choosing a n appropriate function Z(B l ' . . . , BN) ' To be more systematic, one might choose a class of functions ZiBI " . . , BN) and maximize liD with respect to IX. eP�.� should be equal to the maximum possible value of liD.

Find as large a value of liD as you can. In evaluating certain sums. setting try 8N = 81 and BN = B1 ; it might simplify matters if the two-dimensiona l wrapped around a cylinder. were lattice been considering interactions such that when the temperature j , have We atoms of type A and of type B alternate in the lattice. For thd zero, absolute case of two or three dimensions, if the temperature is low enough, the crystal has long-range order. If we know what type of atoms are found at each lattice point in one region of the crystal, we know what type of atoms are likely to be found fairly far off. Note that in one dimension one wrong atom at a latt ice point reverses the expected order of atoms on the other side of the defect . Thus, in two or more dimensions, we can speak of IX sites and f3 sites, whe re to be found. In on e IX sites are those lattice points at which an A is more likely Bethe used the dimensions, two of case the For this. dimension we cannot do ly . '" approximate phenomena er order-disord describe to sites f3 concept of IX and

Problem :

* Bethe, H. "Statistical Theory of Superlattices," Proc. Roy. Soc., 150, 552 (1 935).

Fig. 5.2 Two-dimensional lattice with IX sites and P sites. Following Bethe, we will work with only a few atoms of the lattice and will We will look closely at only the atoms pictured within the dotted line (region I) of Fig. 5 .2. .• The relative probability of a given distribution of atoms in the entire crystal is exp ( Lbonds B/;) (we are assuming there are just as many A atoms as B atoms). If we consider only the five atoms in region I we have to allow for the fact that the four outer f3 sites are affected by region II. We do this approximately by saying that the relative probability of a given distribution of the five atoms is t!' exp ( - L B iBj) where n is the number of f3 sites in region I that have the wrong type of atom (A) on them. The sum is only over the four bonds in region I. B can be thought of as try to describe the effect of the rest of the atoms by one parameter.

-H H

B

_

-

a

a

priori probability that a f3 site in region I has an A atom . priori probability that a f3 site in region I has a B atom _ _ _ _ _ _ _ _ -=.. _

By "a priori probability" I mean the probability that would be assumed before the t�ermodynamic effects of the interaction are taken into account. That is, even If there were no interactions between the atoms of region I, the interaction of the four external atoms of region I with region II would still tend to cause the fJ sites to be occupied by B atoms. Notice that n is not the total number of sites of region I that have the wrong atom. If the IX site in the center has a B atom we �o not mUltiply the relative probability by another B because region II does 'not Interact directly with that atom. Note also that the atoms inside region I can �ffect each other by way of region II, so B is only an approximate way of describ­ Ing the effect of region II. . �or a .given value of B we can compute w, the relative probability that the IX .SIte III �e�lOn I IS' o�cupied by a B atom. We can also compute w ' , the relative . . pr obabIhty that a gIven f3 SIte of regIOn I is occupied by an A atom. We require �hat w = w' , because the probability of a wrong atom occupying a given site is Independent of the site. The equation w = w ' will allow us to fix e.

134

Order-disorder theory

Approximate methods for two dimensions

Some of the results of Bethe's method will be derived below.

4 1 =0

L (relative probability that the rt. site in region I is occupied by a B ato rn and there are P sites occupied by A atoms) There are

W

=

n

4! (4n ) = n! (4--=n)-!

Then W

=

SiS 4 - 2n. j =

4) s"e - H(4-2n) ( 1 =0 n

'

+

H H. +

-

(eH se - H)4.

(5.3)

r) ; so the actual probability that the site has

H :t seH)4 �(_e - H_ � � seH)4 (eH se - H)4 . +

_

+

_

___

+

__

t (expected number of P sites of region I with A atoms)

4 11=0

+n

= t L n[(relative probability that P sites are occupied by A atoms and the rt. site has a B atom) (relative probability that n P site s a r e occupied by A atoms and the rt. site has an A atom)] t

4 ± n ( ) s" [e - H (4 - 2 1 ) + eH ( 4-2 1 )] 1 =0

1 n ± ( 3 ) sn [e - H(4 - 2n) eH(4 -2 1 )] 1= n 1 eeH(e-H + seH)3 se - H(eH se - H)3. Equating w and w' from Eqs. (5. 2) and (5. 4), we find (e - H + eeH)4 eeH(e - H + eeH)3 ee - H(eH ee- H)3 . =

I

=

e2 H

(5.6)

The normalization factor is I /(w a B atom is

W

e

Notice that if satisfies Eq. (5.5), so does l is. This fact holds because nowhere mathematics of our derivation did we specify that rt. sites, rather thart' P the locations for which A atoms are more probable. are sites, Solving for in terms of s, we find two solutions : either

±

r =

w'

_

in the

w is a relative, not an actual, probability. The actual probability that the 'J. site has a B atom is w multiplied by the appropriate normalization factor. Let r be the relative probability that the rt. site has an A atom. The equation for r is the same as that for w, except <-4

ex

(e - H + seH)3(e - H + eeH eeH) = se - H(eH + ee - H)3. Simplifying our equation for s, we get (5.5)

ways in which np sites may be occupied by A atoms. It is easy to see that L bonds

+

+

=

+

+

or else = 1 and H is arbitrary (the latter solution is most easily seen from Eq. (5.5) . The minimu m value of is so if H is too small (high temper­ ature) we must take = 1. From the defin ition of e, s = 1 corresponds to complete disorder. We can expect discontinuities in certain thermodynam ic qua�tities or their derivatives at the temperature for which complete disorde r sets Ill . As an example of a thermodynamic calculation using the above model for order-disorder, let us calculate the energy. S�ppose a bond is an A-B bond with probability p. Then with probability p a given � ond contrib utes energy HIP, and with probability (1 p) the bond contnbutes HIP. The expected value of the energy is

s

E

-I e SI/3 + S /3 2 ;

+

=

(number of bonds)

-

-

[p ( -pH) + p) ( +pH)] 2�H (1 (1

-

=

_

2p) (5.7)

Where N is the number of atoms.

+ [seH(e - H + eeH)3] 1

-� -��

w

+

-

135

( 5 . 4)

r

thet theprob�bility that a given f3 site of region I is occupied by an A atom given L p.te of

rt. SIte has a B atom. Excep t for the normalizing factor l (w + r) ' this is the terms in w'.

136

Order-disorder theory

The Onsager problem

is the probability that a given {J site of region I is occupied by a B atom , g i \ c n that the a site has an A atom. The probability that a given bond between a n J. and a {J site is an A-B bond is therefore - 3 p = BeH(e - H + Bell? + 1 _ ����H + Be H) . ( 5 . 1\ )

(

w + r

w + r

)

HI{J is a constant independent of temperature. Using Eqs. (5.7) and (5.8) we can get Cy = aElaT. We get the entropy from the equation Cy dT' ? SeT) SeT) S(O) = ( 5 . ,! I 0 At T = 00 , we can compute the entropy exactly from its definition. Recall that

-

=

I

T

S = - k Ln Pn In Pn, where Pn is the probability of state n. There are 2N state,

of our N-atom system, and each has equal probability at infinite temperature.

S = - k(2N)

(� )

In � = Nk In 2. 2N

2N

So

(5. 1 0)

Our approximate calculation, using Eq. (5.9) gives S = Nk(O.697) instead Nk(O.693).

of

5.4 THE ONSAGER PROBLEM Consider a two-dimensional square lattice of N atoms. Each atom is designated by + 1 or - 1 . That is, the ith atom carries a number Bi' which is + 1 or - I . The energy of interaction of nearest neighbors is + HBiBj in units of kT. A state of t h e system is denoted by giving the Bi for each atom. The problem is to calculate the free energy of the system. The partition function Q = L e -PEK all states

Q =

L

all states

e - r.*llE,Ej =

L

all states

�"js obviously true.

So

eHE,Ej = cosh H + BiBj sinh H = cosh H( 1 + BiBjT),

(5. 1 2) T = tanh e H. (Note that T does not represent temperature in this section.) \Vher From Eqs. (5. 1 1) and (5. 1 2) we find Q = L D* [cosh H(1 + BiBjT)] ' Remembering that for an N atom lattice there are 2N bonds, we have Q = cosh 2N H L rr*(l + BiBjT) == 2 NQ ' cosh 2 N H . all states

and take H positive. Since BiBj = ± 1 ,

e- H + e - H 2

+

L

all states

B ·B · I

J

rr*eHE'Ej

--- eH

_

2

(5. 1 3) (5. 1 4)

We must now find the modified partition function Q' =

� L rr*(l

2 N all states

BiBJ) .

+

(5. 1 5)

Expanding the product we use the notation Li to indicate a sum over all possible sets of 1 different bonds. Then

L { 1 + T r.t BiBj + T2 r.i (BiB)(Bi'Bj ') + . . . } . (5. 1 6) EN = ± 1 Note that in each term, the B'S occur in pairs corresponding to nearest neighbors, and no such pair occurs twice in the same product. Associate with each pair BiBj a bond connecting the ith andjth atoms (i and j are nearest neighbors). With each term o r product of 21 B'S, we associate a graph or set of 1 bonds. Because no pair BiBj appears twice in a given term, no bond appears twice in a given graph. For example, Fig. 5.3 gives the graph associated with the term (B I B2 )(B3B4)(B5B6)(BIB5)(B2 B6)(B3B 7 )(B4B8)(B 7 B8)(8 1 oB I 1 )(B 1 0 B I 4) ·

rr*e- HE,Ej,

where L* and D* mean the sum and product on pairs of points i and j that a re adjacent on the lattice. It is easy to see that if the sign of H is changed. Q remains the same. For convenience we will write Q =

137

5

(5 . 1 1 )

13 '

Fig.

3

6

7

4

D D

9'

e-H

2

1O

r---- 1 1

I)

'15

' 12 •

16

5.3 A graph or set of I bonds.

138

The Onsager problem

Order-disorder theory

139

a

nondosed graph, which contributes zero. Each term in Eq. (5. 1 6) correspond­ ing to a closed graph has each e i occuring to an even power (0, 2, or 4), and therefore equals simply TL, where L is the number of bonds. Summing over all 8; gi ves a factor 2 N , which is cancelled by the 2N on the left side of Eq. (5. 1 6). Each closed graph of length L thus contributes a term TL to Q' ; so

o

(5. 18) where geL) is the number of closed graphs of length L that can be drawn on the lattice. Notice that geL) = ° unless L is even, so the sign of T (and therefore the sign of H) is immaterial, as we expected. Method of Finding Transition Point

Fig. 5.4 Example of a closed graph

Now,

L e = E=

± 1

°

and

L e 2 = 2. E=

(5. 1 7 )

±1

s whe�e each F. , From Eqs. (5. 1 6) and (5. 1 7), it is clear that only those te�� This I S eqUl�alent . to appears a n even number o f times will contribute to Q'. Q') are those III wh� cl: saying that the only graphs that count (or contribute to 4) of bonds en:�natlll g each atom or lattice point has an even number (0, 2, or be superposItions from it. In other word s, the contributing g.raphs must graphs may be calle simple closed polygons having no common Sides. Such . "closed graph s." the graph of Fig , Figure 5.4 gives an example of a closed graph. Note that is another exampl e o j 5.3 is not closed and does not contribute to Q'. Figure 5.5

o�

We now digress for a moment to show a method (due to K ramers and Wannier) by which the transition point of the Onsager problem can be found. This method is not powerful enough to yield an analytic expression for the partition function. Consider a closed graph on the lattice. Label the midpoints of the lattice squares Ili = + 1 if inside the graph and Il i = - 1 if outside. (See Fig. 5.6.) Consider a bond ( ) between midpoints of unlike 11 values. Each of these bonds corresponds to a bond of the closed graph (because each -- bond cuts a - bond). We now have two lattices. The new lattice is formed by putting "atoms" described by Ili at the midpoint of the old lattice. For the new lattice we can discuss Q at a temperature T' and such that the interaction energy is h in units at kT'. Then ---

Q(h)

0

C).--

OJ Fig. 5.5 Example of an "Unclosed" Graph.

0

--+

? I

+ I

=

L

all states

rr*eh/l;/li.

0

0

0

0

+

0

+

+

0

0

0

0

- - '<)

0

0

+

0

+

0

0

0

+

+

0

0

0

0

0

0

0

0

0

+

+--

l1'ig. 5.6 Midpoints of lattice squares labelled + inside the closed graph and - outside it.

140

5.4

Order-disorder theory

" ·" . = + 1 , and there is a contribution e + h• If P i = - f"-J /I ' If r' " . = + t"""J. ' then t"'f.rJ then the contribution is e-h• Now, if in a given state there are L unlike bonds and 2N - L like bonds. then its contribution to Q(h) is so that Q(h) = e 2 Nh L m(L)e- 2 Lh L

where meL) is the number of ways by which unlike pairs can be arranged on thl' lattice. But from the preceding argument (concerning --- and - bonds), meL) geL), where geL) is defined by Eq. (5. 1 8). Thus Q( h) = e2Nh L g(L)e- 2 L\ L Q(H) = 2N cosh 2 N H L g(L)TL . L 2 Let us now define h by e - h = T = tgh H. Then : Q(h) = tanh- N H L g(l)T' I

and we get the identity

Q(H)

H tanhN H Q(h). Now, when H is large, h is small ; if Q(H) has a singular point, so doe. Q(h). That is, if there is a transition at Hen there is also a transition at he r. If assume that there is only one transition or Curie point, He r = her and He r - t In (tanh He,) . This can be solved to give He r = t log (1 - /2). = 2N cosh2 N

We' c=­

You might hope that we could use the identity Q[x] = 2N cosh 2 N x tanhN xQ[ - t In tanh x]

The Onsager problem

Continuation of Onsager Problem

Call h(l) the number of ways that, starting from the atom (origin), we can , proceed in 1 steps (l bonds) through the lattice and return to the origin without using the same bond twice. In h(l) we do not count separately the two or more different paths we can use to form the same polygon. For carrying out the calculation we shall need the sum over all different 1 :polygons of the crystal. Our h(l) polygons for atom A in Fig . 5.7, say, will be [ counted again in the h(/) polygons of other atoms, say B in Fig. 5.7. Each ! polygon will occur 1 times. Therefore, the number of polygons per atom is h(l)/l. Let

(5 . 20) ' We have Q' = L g(L) TL

L

Unfortunately, x"

=

-

t In tanh x '.

x, so this method is not useful.

L

closed diagrams

( )( )

(5 . 1 9 I

B

x' = - t In tanh x.

x" =

=

but need an expression for Q' that we can hope to evaluate. We will work up to such an expression by finding an approximation to Q', which we will then correct , by means of a trick. A term from a closed diagram can be thought of as a product of terms made up of graphs formed by polygons. The contributi on to Q' from no polygons is simply 1 . The contribution from graphs consisting of one polygon is Nq, for the number of graphs consisting of one polygon of length 1 is Nh(l)/I. The contribu­ tion from pairs of polygons, one of length 1 and the other of length l' is (Nh(l)/ I) (Nh(l')/I')T / + l' so the contribution from pairs of polygons of any length is about ! L Nh( l) Nh{l ') T '+ l ' , I' I 2 1,1'

to find Q(x) for many values of x given Q(x) for one value of x. Just plug in x 0 1 1 the left side of Eq. (5. 1 9) and get Q evaluated at another value, We can then get Q evaluated at

141

Fi.g.

:

Origin

)

5.7 A. polygon formed by p �oceeding through the lattice back to the starting point, Without usmg the same bond tWice.

142

where the factor of 1 /2 is to account for the fact that we count each pair twice . we have So the contribution from pairs is about !N2q2 . I say "about " because counted such graphs as

CD

and

is a pair of squares and is also a single polygon of length 8, ;so we have counted it twice. If we continue to ignore such errors, we get

143

in both second and third order. In second order, there are two terms that form a diagram of type 5.8c, and both have a negative sign :

even though that pair of squares is not "closed," since one bond appears twice . Furthermore

om ITJ O

Fig. 5.8c

that is, there are two ways of writing 5.8c as a product of 5.8a and a 5.8b type term. In third order, one of the two second-order terms is canceled by t he positive product of three type (a) terms. To cancel the other second order term of type (c), we must insert (c) with a positive sign into h(l 2). In Eq. ( 5.21), we counted polygon 5.8d twice ; therefore when we get a rule for what terms belong in our corrected h(l) and what their signs should be, we will have to have an excuse for not counting (d) in the corrected h(l).

. (5. 2 1 ) 1 + Nq + !(Nq)2 + t !(Nq? + . . = e" q. We Our next step logically should b e t o correct q i n order t o get Q ' right. terms. tion correc the of few can correct q by correcting h(l). Let us write down a Because of the presence of polygon 5.8a Q'

The Onsager problem

: 5.4

Order-disorder theory

�

Fig. 5.8a

in the first-order term in Q, we obtained the forbidden graph of 5.8b

m

Fig. 5.8b

5.8b, we in second order (i.e., as a product of two polygons). To cancel graph to get seem we insert graph 5.8b into h(8) with a negative sign. But then in Q' terms such as

\ \ \

Fig. 5.8d

Now, let us look at where we stand. In evaluating h(/) we expected trouble because of the restriction to certain allowed polygons. In correcting h(/) we f�und that there must be other polygons with not necessarily positive contribu­ tion, and some of the allowed polygons should not be counted. The natural thing to �o is assign each polygon an appropriate weight, and not ignore any polygon III the sum. Recall that formerly we did not count separately the two or more different paths we can follow when starting our from an atom and forming a polygon. Now we will count all the different paths, but we will choose . the weight of each path so that our corrected "q " will lead to the right value of Q' when we use Eq. (5.21). A Topological Theorem

Give a weight (X = eiTt/4 for a left turn and (X - I = e - iTt/4 for a right turn, and keep track of the left and right turns as a closed path (returning to origin) is traversed in the lattice. The topological argument of Kac and Ward is that the

144

�e forbidden closed graphs (or ones that we want to count) will be counted, and t mg them are travers of ways graphs (e.g., Fig. 5.5) will cancel out when different strated demon be now will considered. That this theorem is true for simple cases is rather and an* Sherm by by a few examples. The complete proof was given involved. Consider the simple closed polygon in Fig. 5.8a. that ..\ Starting at A and return ing to A in a clockwise direction we note right turns were made giving a total weight of - 1. (e - ilt/4)4 a-4 Proceeding in a counterclockwise way gives a weight of (at

=

(eilt/4)4

=

-

1

.

d. To get The total weight is thus 2, and the simple polygon is indeed counte the correct weight of 1 , we multiply everything by - 1/2. Now, consid er the polygo ns in Fig. 5.8e and e' : -

---,= : =±:o:tA=::::!tLn L.:r=' D� jD D (e)

Fig.

(e')

5.8

=

=

Method of Calculating partition Function

Let q denote our corrected value of q. Then

and =

eNq

_

2T

T if one continues forward : Ta if one turns counterclockwise : a = eilt / 4 ; e - ilt /4 ; Ta- 1 if one turns clockwise : a - 1 =

o

if one turns 1 80°.

{corrected 11(1)} T ' 1 1= 1

�

dq dT

=

-2

�

1=1

{corrected h(l)}T'

; 4 , 1 2 1 3 (1 963). * Sherman, S., J. Math. Phys., 1, 202 (1 960)

In our rule for determining the amplitude for each lattice point passed, we have neglected to say how to deal with the first step, whose amplitude depends on the last step. In what follows, we will calculate the amplitude to start at the origin and arrive back at the origin from below, that is, moving upward. Thus, we wiII choose as our rule for the amplitude of the first step : T if one starts upward ;

o

=

=

which is the sum over all paths leaving and returning to the origin of a complex "amplitude" or weight for each path. This amplitude is the product of ampli­ tudes for each lattice point passed :

Ta if one starts to the left ; Ta - 1 if one starts to the right ;

A) that there In Fig. 5.8(e), the path is traced out in such a way (starting from 2 (a4) = 1 . are 6 left turns and 2 right turns for a total weight of a6(a- ) turns and In Fig. 5.8(e') the path is traced out in such a way that there are 4 left 1 . The total weight then o� the two 4 right turns for a total weight of a4a 4 O. It is clear that the two counterclockwIse paths clockwise paths is - 1 + 1 have weight 0 also. a d all Similarly you should be able to show that 5.8(b), 5.8(c), and 5.8(d) ? I S mg simple polygons give the correct contribution (remember that everyth multiplied by - 1/2).

q

145

=

=

Q'

The Onsager problem

5.4

Order-disorder theory

-

if one starts downward.

M ore generally, we will use the above rule for the first step and work with amplitudes for starting at the origin and arriving at any lattice point (x, y). We shall have to describe the amplitude by giving four components : U(x, y), the amplitude to arrive at x, y moving upward ; D(x, y), the amplitude to arrive moving downward ; R(x, y), the amplitude to arrive from the right; and L(x, y), the amplitude to arrive from the left. This, as mentioned before, is because the amplitude to make the next step depends on the arrival direction, and we shall have to keep track of it. Roughly, what we will do is calculate the amplitudes for arriving at (x, y) from various directions in exactly n steps, and then sum over n. UnCx, y) is the . am�htu de to arrive at (x, y) in exactly n steps with the last step upward and . SlI�l1�arly for Dm Rm and Ln. We assume that in the zeroth step we arrive at the bx.oby.o, Do = Ro Ongm moving upward. Thus Uo(x, y) Lo = O. In other words, the amplitude to arrive in zero steps is one if we arrive upward at the origin and zero for any other point or any other direction of . aI'Ilval . With these amplitudes for the zeroth step, we can get the amplitudes for any number of steps by recursion. We can write for the amplitude to arrive �pward at (x, y) in n + 1 steps in terms of amplitudes to arrive from (x, y 1) III n steps. =

=

-

146

Order-disorder theory

The Onsager problem

If we came to x, y - I going up, we have an additional amplitude factor T. If we came to x, y - I from the left, we need to turn counterclockwise and ad d a factor Trx, and so on. We cannot get to x, y going up if we previously entered x, y - I going down, for we do not allow paths that reverse themselve� Similar expressions for Rn(x, y), Ln(x, y), and Dn(x, y) are easily written. Then we see that our assumption for the amplitude of the zeroth step w i l l give, for the first step, the correct amplitudes (such as T if one starts upward s). When summing over n, we will eventually have to subtract out the n = 0 term. because we do not consider a figure with zero side to be a polygon, even though it starts and ends at the origin. To simplify the summation over n, we define the transform of Vn(x, y) 00

L

x= - oo y = - oo

Then

V.(x, y)

=

f J: 1t

Vn(x, y)e-i�Xe-iqy .

1t ei�xeiqY V (�' Yf) n

For example, the transform of Vn(x, y - I) is e-iqun(� ' Yf), and the transform of Vo (x, y) is 1 . The transformed equation then reads

Vn+ l (� ' Yf)

=

Te- iqV. (�, Yf) + o · e -iqDn(�' Yf) + Trx- l e-i�Rn(� ' Yf).

+

Trxe-i�Ln(�' Yf)

(5.23)

and similarly for Dn+ l (�' Yf), Ln+ l (�' Yf), and Rn + l (�' Yf) . If we let "'n be a four-component symbol (column vector) with components (Vn, Dn, Ln, Rn), we can write the matrix equation (5. 24)

where M (�, Yf) is the matrix

M(�, Yf)

rxe-i� rx- l e-i� eiq rx- l ei� rxeiq 0 0 rx- l e-i� rxe-i� e-i� ei� rxei� rx - l ei� 0 e - iq

=

0

"'o (�, Yf) = ( 1 , 0, 0, 0). "' l (� ' Yf) = TM",o "'2 = TM"' l = (TM) 2",0

(5.2 5)

= nL (TM)"",o = =O 00

1

is the transform of the ampli tude to arrive in any number of steps. The ampli ­ Jude to arrive at the origin movin g upward is

t

n l V.(O, O)

=

-1

+

=

�

n o Vn(O, 0) =

1

-1

+ "'0 .

"'0 d� dYf -M (2n)2 1 ---T 2 1t 2 1t "'0 . --- "'0 o 0 1 - TM

x

(5.22 1

gn�; .

1TM "'0

!Then '"

147

ii(

1

So21t So21t ei�Oei�O

1) d�(2n)2dYf . __

We could have arrived at the origin movin g in any direction, so we must sum over all four values of "'0, that is "'0 = ( 1 , 0, 0, 0), "'0 = (0, 1 , 0, 0), "'0 = (0, 0, 1 , 0), "'0 = (0, 0, 0, 1 ). The result is a trace. Thus,

i i [ 1 1] d): dYf 1 (2n)2 ii [iT 1 1 ] dt 2t ii [iT ] 1 (1 ii (1

- 2 T dq = 2 1t 2 1t Tr dT 0 0 - TM 2 q = 1t 2 1t d� dYf Tr o 0 (2n)2 0 2t(1 - tM) 2 1t 2 1t d� dYf -M = -- Tr dt 2 o 0 (2n) 0 2( 1 - tM) r 21t r 21t d� d Tr [log - TM)] ; � "2 Jo Jo (2n) 1 2 1t 2 1t d� d = -� 2 0 0 (2n) log det - TM),

'"

-

--

+

-

We have used the following theorem :

Tr log A = log det A . 1'0 prove this identity, note that

det eB = lim (det eB/Nt = lim

N -+oo

=

lim

N �oo

(1

[ ( BN)JN det 1

N -+oo N + . � Tr B = eTc B . N

)

+

(5 .2 6)

(5.27)

148

Order-disorder theory

Miscellaneous comments

Therefore, log det eB = Tr B. Let B = log A . Then Tr log A M is given by Eq. (S.2S), and one obtains det [ 1 - TM]

=

log det A .

(T2 + 1)2 - 2T(1 - T2)(COS � + cos '1).

=

'continuing as before, we get in place of Eq. (S.29), f3F N

(S.28)

- In 2 -

Summarizing, we see that Q

=

q

= ff _

T

-

So

5.5

- In 2 -

f f

__

MISCELLANEOUS COMMENTS

In the above we sometimes assumed that the lattice was finite (with N points) and we sometimes assumed it was infinite (when we neglected the boundary for example). Because N was assumed to be very large, it is plausible to suppose that our inconsistency made little difference. However, let us modify the above work to make it consistent. Assume the lattice has the topology of a torus. If we pass through L points in one direction (or L' points in the perpendicular direction) we finish back where we started. Formerly, to compute the amplitude to arrive back at the origin moving upward, we took the sum

Now, we must take the sum w

00

00

L U L . p� - oo q� - oo n� l n(p , q I:) L

L

2 l[ 2 l[ d� dlJ L eipL�eiqL'� 2 0 0 (2n)2 p,q

In [cosh2 2H - sinh 2H(cos � + cos '1)] .

[

(

7

* A slightly different treatment for the Onsager problem was given by N. V . Vdovichenko, Soviet Phys. JETP 20, 2, 477 ( 1 965) ; also see Soviet Phys. JETP 21, 2, 350 ( 1 965).

; )] ,

In COSh2 2H - sinh 2H cos 2 n + cos 2 n

As L and L' tend to infinity, Eq. (S.30) approaches Eq. (S.29).

1 21[ 21[ d� d'1 In [cosh 2 2H - sinh 2H(cos � + cos lJ)J. 2 0 0 (2n)2 (S.29)

-

x

tanh H.

=

Putting all this together we obtain* BF N

f3F N

TM ),

( T 2 + 1 ) 2 - 2T(1 - T2)(COS � + cos '1)

=

!f f

But

=

1 21[ 21[ _ d� d --'1 log det ( 1 2 0 0 (2n)2

det ( 1 - TM )

x

- In Q/N, 2 2N(cosh N H)Q', BF/N

149

(S.30)

A few comments about the form of Eq. (S.29) are in order. If we use the expression f3F = - log L e- PE states

we can verify that Eq. (S.29) is correct for high temperatures (H � 0) and low temperatures (H � 00). Another check of Eq. (S.29) is by way of Eq. (S. 1 9). If we consider Q as a function of y = sinh 2H instead of as a function of H, Eq. (S. 1 9) can be written in an especially simple form. Because sinh 2H = 2 cosh H sinh H 2 cosh2 H tanh H, =

Q(y)

=

2 N cosh 2N H tanhN HQ

Equation (S.29) can be written f3F N

-

=

- In 2

-

G)

=

sinhN 2HQ

G)

[ -

=

yN Q

G) ·

(S. 1 9a)

]

l[ d� dll i In y - -1 2 2rr --'2 In y + 1 - (cos � + COS IJ) 2 0 0 (2n) y

f f

•

(S.29a) from which it can be seen that Eq. (S. 1 9a) holds. Kac made the conjecture that the solution for the three-dimensional Onsager problem (cubic lattice), is of the same form as Eq. (S.29) except that cos � + cos '1 is replaced by cos � + cos '1 + cos ( and d� dlJ/(2n) 2 by d� dlJ d(/(2n)3 . However, this conjecture is false. Instead let us conjecture that the specific linkage of the Ising lattice will not greatly affect the character of the transition point in the specific heat curve. Also , an approximation starting from the transition point should prove very

150

Order-disorder theory

valuable. If a magnetic field is added to the Onsager problem, the problem o f evaluating the partition function can again be converted to a path problem. which has not been solved. The problem here is to keep track of the area en­ circled as well as the length of the path. We have done a lot of work just to evaluate the expression Q

=

(

' CHAPTER 6

)

CREATION AND ANNIHILATION OPERATORS

L exp - H L Bi Bj .

E; =

±1

bonds

One reason for going to all this trouble is that maybe you will be able to genera l­ ize the methods used, or apply them somewhere else. For example, we considereJ a statistical-mechanical problem for which we eventually had to evaluate a S u m over all paths leaving and returning to the origin, with an amplitude depending on the preceding step. This is essentially a Markovian walk, with each step depending only on the one previous step. A Markovian-walk problem, however. can be handled when each step depends on a specific finite number of preceding steps. In a way analogous to the method used above, Qn can be found for the case where the amplitude of a step depends on the n steps preceding it. The question is : Can one find an interesting problem whose partition function is Qn ?

6.1 A SIMPLE MATHEMATICAL PROBLEM In this chapter we shall describe an operator formalism that has widespread applications in quantum mechanics, notably in dealing with harmonic oscillators and in describing many-particle systems. We begin by formulating and solving the following simple problem : Suppose an operator a satisfies (6 . 1) The problem is to find the eigenvalues of the Hermitian operator a + a, and to relate the eigenvectors. (Note : a+ denotes the Hermitian conjugate of a, and [A, B] is, of course, the commutator AB - BA.) We first note that, if /IX) is a normalized eigenvector with then

IX

=

a + a/IX)

=

IX / IX),

(6. 2)


=

! Ia / IX) 11 2 2: O.

(6. 3)

Using the identity

That is, the eigenvalues are all real and nonnegative. [AB, 'C] = A [B, C] + [A, C]B, we observe that [a+a, a]

=

[a+ , a] a

=

[a+a, a + ] + a + [a, a + ]

or, equivalently,

(a+ a)a (a+ a)a +

=

=

- a, =

a+ ;

=

a(a+ a - I ) /IX)

(6.5)

a(a+ a - I), a+(a+ a

+

=

a(1X

-

(6. 4')

I).

From Eq. (6 .4') we have, for an eigenvector /IX), (a+ a)a/IX)

(6.4)

1 ) 11X)

(6. 5') =

-

(IX - l)a llX).

(6.6)

The�efore allX) is an eigenvector with eigenvalue IX 1 , unless a llX) = O. . SImIlarly a + IIX) is an eigenvector with eigenvalue a + 1 , unless a+ l a) = O. The nor m of allX) is found from I la llX) 112

=

< cx la+a la) 151

=

a < lX la)

=

a,

152

Creation and annihilation operators

6.1

or

A simple mathematical problem

and in general

( 6 . 7)

(6. 1 2)

Similarly, ( 6.8) Now, suppose that d'la) i= 0 for all n. Then by repeated application of Eq. (6.6), d'la) is an eigenvector of a+ a with eigenvalue a - n. This contradict:, Eq. (6.3), because a - n < 0 for sufficiently large n. Therefore we must have

d'la)

i=

.ja - n

=

Iia i a - n) 11

=

0,

and therefore a = n. This shows that the eigenvalues of a+ a must be nonnega ­ tive integers, and that there is a "ground state" 10) such that

alO)

=

(6. 1 0)

o.

By repeatedly applying a+ to the ground state we see that (a+)" I O) has the eigenvalue n and, because of Eq. (6.8), it is never zero. Thus the eigenvalues of a+a are 0, 1, 2, 3, . . . . If In) is a normalized eigenvector with eigenvalue n, then, from Eq. (6.8),

I n - I)

=

:(Note that we could have included arbitrary phase factors in the definition of ,In ) ; our convention here is to make them unity.) With this definition, the In) ;are orthonormal* and satisfy

a+ In)

=

aln)

=

a + aln)

=

(6 .9 )

but

0

for some nonnegative integer n. Let l a - n) = d'ln)/ I d' ln) II , so that l a - n) is a normalized eigenvector with eigenvalue a - n. Then from Eqs. (6.7) and (6.9),

I

- 1. Also

•

a+ ln I ) = (l/.j�)a+aln) = .j�ln). So applying a+ to In - 1 ) gives us back I n) (within a factor), rather than some other state with eigenvalue n. We may then construct the eigenstates of a + a as follows : First we find

.j;;--+ l in + I )

(6. 13)

.j� I n nln).

(6. 14)

I)

(6. 15)

Equations (6. 1 1 ) through (6. 15) form the answer to the problem posed at the .:beginning of this section. The operators a + and a are called "raising" and "lowering" operators, respectively, because they raise and lower the eigenvalue of a+ a. In later applications a+ a will be interpreted as the observable representing the number of particles of a certain kind, in which case a+ and a are called "creation" and "annihilation" (destruction operators, or "emission" and "absorption" operators. Equations (6. 13) and (6. 14) may be alternatively expressed in terms of matrix elements :

( 1 /.j�)aln)

is a normalized eigenvector with eigenvalue n

For,

(ml a+ ln)

=

.j� bm ,n + l '

(6. 1 3')

(m laln)

=

.j� bm,n - l '

(6. 14')

by (6. 1 2) we have

-

a

state 10) such that

153

a lO)

= o.

(6. 1 1 )

( 10) may be unique ; if not, we find 'other operators commuting with a and a - . and classify the 10) ' s according to their eigenvalues.) Then we define

From Eq. (6. 1 ) we easily obtain so that

(O l an(a + nO)

=

(Olan - 1 (a+)maIO ) + (O l nan- 1 (a + )m-1 IO ) n (Olan - 1 (a+)m- 1 I O )

=

n(n - 1 ) " ' (/1

=

and the orthonormality follows .

-

m

+ 1 ) (O l an -m IO)

154

Creation and annihilation operators

The linear harmonic oscillator

6.2 THE LINEAR HARMONIC OSCILLATOR Our first application of the results of Section 6. 1 will be to the one-dimension al harmonic oscillator, which has a Hamiltonian of the form 1 H = - p2

2m

mw 2 2

+ _··· x 2 ,

(6. 1 6)

where x and p are the position and momentum operators for the particle and satisfy (6. 1 7) [x, p] = in. Our task is to find the eigenvalues and eigenstates of H. Note that J(mw/n)x and (l/Jmwn)p are dimensionless. Let us define (6. 1 8) Because x and p are Hermitian it follows that a+ = From Eq. (6. 1 7) we obtain

J2 (J

� )

mw p . x - i J wn n

J: �

a+ , 2

(6.21)

a - a+ p = J-mwn '

(6.22)

x

=

a

w

I

J2

We get, for the Hamiltonian,

H

=

nw 2

(6.23)

Thus, the eigenstates of H are those of a + a. Now we can ap�ly the results o f Section 6. 1 , obtaining the eigenstates 1 0 ), I I ), 12), . . . that satIsfy + t)nw ln) .

The energy levels are thus En = (n + t)hw.

J (x mw 2n

)

� i p 1 0) . mw Applying < x / and noticing that < x/p/

= a / O)

o

=

=

J (

+

)

mw n d <x/O) x + mw dx 2n

�

�

(6.25)

(6.26)

-

(where x is now a number, rather than an operator.) Equation (6.26) is merely Eq. (6. 1 1) in coordinate representation, in which it takes the form of a differ­ ential equation. Solving it, we get

<x/ O) = Ae- (mw/ 2I1)x2 ,

1 = (010) =

= / A/ 2 so

f�oc;

Jmw nn


= /A /2

,

f�oo e- (mw/�)x2

dx

( )

mw 1 /4 . A = e'B. � nh

The phase 8 of A is arbitrary, and we set it equal to zero. Then

So

(a +a + aa+) = hw(a +a + t).

Hln) = (n

and (6. 1 1 ) ,

(6. 19)

Expressing x and p in terms of a and a+, we have

.

The eigenstates themselves are given by Eqs. (6. 1 1) and (6. 1 2). We can easily obtain the wave functions
where A is a constant. Normalization requires that

(6.20)

155

<x/O)

=

(::) 1 /4 e - (mw/2h)x2•

(6.27)

We have thus found the wave function for the ground state. For the other states we apply a+ according to Eq. (1 . 1 2) :

(6.24) (6.28)

156

Systems of harmonic oscillators

Creation and annihilation operators

Since

) Jmea 21i ( meai p ( meaIi dxd ) Jmea 21i (mea)n/2 ( � �)n mea 21i n e-(mro/2h)x2• �) (meali ) 1/4 (mea211 )n/2 ( � mea dx n =

we have

< xl n)

= =

1 .I n ! 1

.J n !

x

<xl

=

-

x

us 1 6 terms, but, thanks to the raising and lowering ' Expand'.mg ( a � a + )4 gives es of a and a, erh the only terms giving a nonzero expectation value are prop . , s a two with and two a + ' s : those

_

_


x - - - <xl,

_

<x I O)

dx

x

_

(6.29)

p2m2 + mea22 + AX4. Assume that A is small enough (<< liea) that we can use first-order perturbation Then theory, treating h4 as a perturbation of the Hamiltonian (Eq. ed energy levels are -

--

the perturb

(6. 3 1 )

where

(6.32)

From Eq. (6.21) we have

=

* Problem .' Prove that [x>

=

( )

mw 1/ 4 --;;rz

exp

( )

< n l (a

2n x2 exp mw

L anCa+)n,

n(n - 1) + n2 + n(n + 1)

+

aaa + a+) [n)

+ n(n + 1) + (n + 1 )2 + (n + 1 )(n + 6n2 + 6n + 3,

�n

=

3).

(�2mea)2 2 2 (2n

__

2)

+

n

+

1 ).

6.4 SYSTEMS OF HARMONIC OSCILLATORS Suppose a system has the Hamiltonian

where Qi and P i are canonical coordinates and momenta :

+

(6 . 33 )

a + )4 I n ) .

[( - :21)( a+ =

[Q i ' QJ

=

[P i ' PJ

-

x

0;

J )] 2mw 2 rz-

(6.36) V ij

=

2

1- -= \/ m i m j -

[q i' pJ

V;j '

iM ij ,

- l '<:' i2 + 1 '<:' Vi H "2 � 2 �P jqiqj ' I

[0),

x . From this formula , find =

(6. 35)

and

and in terms of them the Hamilto nian is

n f'(a+).] Ln a[a,f(a+)] n

=

and Vij = Vji' To simplify the presentation a little let us make a change of scale, defining

Then q i and Pi are also canonical :

when /ex) + is interpreted as Prove first that : (Hint . ) n [ (x for n functio ing a generat

(a )

where /

=

+ aa + a+ a + aa + aa +

where we have used Eqs. (6. 1 3) and (6. 1 4) repeatedly. Therefore

(6.30)

x2

(6. 1 6)).

�n A (_2mea1i_)2


(6.34)

6.3 AN ANHARMONIC OSCILLATOR Suppose a system has the Hamiltonian =

=

=

The matrix elements of observables between harmonic oscillator states ca n be found without having to express the states in coordinate representation and integrating over x. We simply express the observable in terms of the raising and lowering operators. An example of this procedure is given in the following section. *

H

157

i,j

(6.37)

(6. 38)

(6.39)

We shall expres s H I� ' terms 0 f ralsmg ' . and lowering operators as we did for the . ofine-d'ImenSlOnal OSCillator. The procedure involves two steps '. the fi rs t IS . ndm g a set 0 f no r mal coord mates ' ii, with respect to which the potential is in .

158

Phonons

Creation and annihilation operators

diagonal form, and the second is expressing the coordinates and momenta in terms of raising and lowering operators. Let the coordinates q i and iirt be related by

q" =

(6.40)

Prt =

.

Because ( Uij) is assumed to be real and symmetric, the transformation matrix Crtl that diagonalizes it is orthogona l : � CrtiCrtj rt

=

�ij'

(6.42)

=

(6.43)

Finally, we define Prt in such a way as to preserve the canonical commutation relations : (6.44) (6. 45)

[iirt' PPJ = iMrtp •

H = t � (p; rt

+

(6.46)

w;ii;) ,

which means we have a system of decoupled harmonic oscillators (one for each value of oc) . Using the methods of Section 6.2, we form lowering and raising operato rs for each mode : i _ , _ 1 (6.47) art = .J n V Wrt qrt + .J Prt ,

) ( 2 + = .J21i ( /- - .J ) , wrt

art

1

_ V w" qrt

i

w"

(6.50)

[a rt, apJ = [a; , an = 0,

(6.51)

_ p"

(6 .48)

(6.52)

+ t)·

The eigenstates of H are described by giving, for each a: art' Thus,

(6.53) oc,

the eigenvalue n rt of (7.54)

I nl n2n3 • • • ) =

[IJ (a +y«] 1000 " Jnrt !

> is defined by 0 for all art 1000 . . . >

where the ground state 1000 · . .

W;�rtP '

and thus

The result of our efforts is that

w" + 2 (art - art ) '

.

- I

H = L liwia; a " rt

We further assume that the eigenvalues of (Ui) are all positive, that is, that the matrix is positive definite (this ensures that qi = 0 is a point of stable equi­ librium). Denoting these eigenvalues by w�(Wrt > 0), we have � CrtiCpj Uij i,j

(6.49)

[a", an = �rt(J'

(6.41 )

The inverse transformation of Eq. (6.40) is then

Ii

(art + art+ ),

_

Then,

J�2wrt n J�

159

=

' ),

oc.

(6.55)

(6.56)

Note that the energy of the ground state is Lrt -!liw". For a system with infinitely many degrees of freedom (which we will consider shortly), this quantity will generally be infinite. Because the zero point of energy is a matter of defini­ tion (only the difference between levels being of physical importance), it is convenient to redefine the Hamiltonian of such a system so that the ground­ state energy is zero. Thus, if we let H = t L ( p; rt

+ w;q;

- -!liwrt)

(6.57)

with a corresponding (but more complicated) expression in terms of the original coordinates q i ' then H = L liw"a; art, (6.58) rt and H l n1 n2n3 • • • ) = L n"liwrt l n 1 n2n3 • • • ). (6.59) "

6.S PHONONS The states of the system considered in the preceding section can be given a simple interpretation in terms of "noninteracting phonons". Assume that the lIamiltonian is given by Eqs. (6.57) and (6.58), so that the energy of the ground

160

Phonons

Creation and annihilation operators

represents state is zero. The ground state is then called the "vacuum state" and is in the system the If ns. phono no are there which the state of the system in = 1 , 2, 3, . . . ) . a(a type of ns phono n" are there that state In l n 2n3 ' ) we say state is The n" are called "occupation numbers." Note that the energy of this is nw,. a type of phonon . . single a of energy the that nl nWI + n 2 nw 2 + " so ns. In phono ual individ the of s energie the of and the total energy is the sum g. eractin nonint other words , the phono ns are Since I . . . ), a : l nl . . . n"_ l n,,n,,+ l . . . ) = . n,, + l i n, . . . , n,,- l ' (n" + 1 ) , na + l (6 . 60) •

•

values I , 2, 3, . . . . Then the state with n l , of the a ' s equal to I , n 2 cc ' s equal to 2, and so on, is 1 1, . . . , 1 , 2 , . . . , 2, . . . ) = ( a tY'(a iY2 . . . 1 0) ------"1

------"2

(6.66)

that its norm is .J� l ! n 2 ,-. When we deal with only a few phonons at a time, it is usually more con­ venient to use the notation I cc l , . . . , an) rather than the occupation-number description I n l n2 ' . . ) , especially when cc can take on a continuum of values. The effects of creation and destruction operators on lal, . . . , an) are sO

• .

•

(6.61) we may call a: and a" creation and annihilation operators for phonons of type a. The operator for the number of phonons of type a is a: a", and the operator for the total number of phonons is

(6 . 62) Let the vacuum state be denoted by 10), and let l a)

=

a: 10)

=

10, . . . 0 1 0 ,

,

,

,

.

.

.

),

(6.63)

system be the state with n" = 1 and nfl' = ° for a' '# a . A phono n, then, is the l hanica m-mec quantu whose system that is decribed by the states la) (i.e., the states, n phono (one-) are ) 113 and Hilbert space is spanned by the la»). If la) then < al P) =

+

b"p)I O )

(6.64)

on of the so that the states la) are orthonormal. To each normal mode of vibrati (since original system of harmonic oscillators corresponds a one-phonon state they are both indexed by a). ns. We can also use a similar notation for states of two or more phono defining

161

(6 . 67) a"l a l . · · · , all )

n

=

L b"akl a l. · · · , ak - l , ak + l , · · · , an) k= l

(6.68)

Equation (6.68) comes from Eq. (6.65) and the relation

Note that phonons act like Bose particles (insofar as we can call them particles*), � s an arbitrary number of them may be in any given state (i.e., la, . . . , a) eXists for any number of a ' s). Their Bose nature is also reflected in the symmetry of the states (e.g., la, 13) = 113, a») . In Section 6.7 we shall show how the �rdinary rules for quantum-mechanically describing systems of many Bose particles lead to a set of states and operators with the same form as those obtained here, so that the interpretation of the oscillator as a system of many Bose particles is correct. �e conclude this section by considering the qualitative effect of an anhar­ mOlliC perturbation on the oscillator system . Suppose the perturbation has terms of the form � f...

i,i,k

rij·kq i qj·q k

and

� f...

i,i,k,1

riiklq iqi qkq /.

In terms of creation and annihilation operators, the cubic terms are of the form a"+ ap+ ay+ , a"+ a{J+ ay, . . . , a"a{Jay,

(6.65)

Which always changes the number of phonons (e.g., the first term creates three

t ; otherwise, These states are normalized as they stand if a i ' . . . , an are distinc takes on the a that eness definit they have a norm larger than 1. Assum e for

:The phono�s will l ook �ore like particles (e.g., carrying momentum and energy) hen the OSCillator system

IS

a field as described in the next section.

162

Creation and annihilation operators

Field quantization

new phonons). Thus, if we start with a definite number of phonons and let the Hamiltonian drive the system forward in time,

11/I(t»

=

=

Note how Eqs. (6: 69) and (6.70) resemble the corresponding equations for . the system of harmOnIc oscIllators described in Section 6.4 :

e - iHtI1/l(0» 11/1 (0» - itHI1/I(0» + 0(t 2),

o

we will soon start finding different numbers of particles. The quartic term s similarly change the number of particles, except for terms like

which conserve the number of particles but act as a mutual interaction between them ; that is, the particles are no longer independent. The description of a mutual interaction will be considered in more detail later. Exercise : Verify that the number-of-phonons operator N, defined by Eq. (6.62), commutes with a product of creation and destruction operators if and only if the number of a + 's equals the number of a ' s in the product.

L(fP , cp)

=

1-

f

=

K(x' - x). o

=

=

*

f f d 3x

d 3x' K(x - x')fP(x)fP(x'),

(6.69)

The classical equations of motion, found by

.£ �

o t bfP(x) cp(x)

iii

+

L Uijqj ' j

K(x - x') = - C 2 V' 2<5 \X - x') .

+

_

L

=

t

and Eq. (6.70) becomes

I

d 3x[cp(x)cp(x) - c 2 V'q;(x) ' V'q;(x)] ,

f

d 3x' K(x - x')fP(x ') .

( 6.70)

The following procedure can be generalized for a multicomponent field by putt ing indices on everything. t Classically q/(x) depends on t, but (as with q i in Section 7.4) we will not show it explicitly. Besides, in the Schrodinger picture, the operator q/(x) is time-independent. t We assume that the system is invariant under translations, so that K is only a function of x - x'.

'

c2

cp(x) =

0,

(6.72)

(6.73)

which is the usual wave equation. If we assume that q; (x) is a coordinate of the system for each x , the conjugate momentum to fP(x) is D(x)

=

The Hamiltonian is then H

= =

bL _ bfP(X)

(6.7 1 )

Then Eq. (6.69) becomes, after a few integrations by parts,

V' 2 q;(X) -

d 3xcp(x)cp(x) - 1-

where K(x - x') varying fP(x), are

=

Th us, we are justified in treating the field as a system of harmonic oscillator s (at least formally) : q; corresponds to the symbol " q " and x correspo nds to i. fP(x) can be thought of as a separate coordina te of the system for each . x As an example, suppose that

6.6 FIELD QUANTIZATION A notable example of a system with infinitely many degrees of freedom is a field. Examples are the amplitude of sound waves, drumhead vibrations, light, and so on. Consider a real scalar* field fP(x) whose motiont is described by the Lagrangiant

163

I

t

d 3xD(x)cp(x) - L

I

d 3xD(x)n(x) + t

-

<5L

- =

<5 cp(x)

I I d 3x

cp(x) .

(6.74)

d 3x'K(x - x')q;(x)q;(x').

(6.75)

To quantize the system we let q; (x) and n(x) be Hermitian operators satisfying [q; (x) , q;( x')] = [n(x), D (x')] = 0,

(6.76)

[q;(x), D (x')] = ili<53(x - x'),

(6.77)

;md assume that the Hamiltonian is given by Eq. (6.75), except for a scalar term

o make the ground-state energy zero. We next express everyth ing in terms of "norma l modes." The situatio n tur�s out to be slightly different from that of Section 6.4 because it is con­ VenIent here to use "compl ex" (that is, non-her mitian) normal coordin ates.

164

Field quantization

Creation and annihilation operators

Because the system is translationally invariant, we expect that it might he lp to express the fields i n "momentum" representation. Therefore we define ip(k) = fi(k) = The inverse transformation is* qJ (X)

=

f f

Next we wil l define annihilation and creation operators (compare Eqs. (6.47) through (6.50») :

d 3xqJ(x)e - ik ' X

(6. 78)

a(k)

d 3xTI(x) e- ik ' X•

(6.79)

a + (k) =

f g:�

3

ip (k) eik ' x

=

Now let w2(k) =

f

(6.83)

d 3 xK(x)e- ik ' x .

(6.84 )

(6.85)

w2(k) = w2(k)* = w2( -k).

f

2

= 1

d3 � [fie - k)fi(k) + w2(k)ip( - k)ip(k)J

( 2n)3

f [ti +(k)ti(k) 2 (2n)3 d3 ,,-

+

w2(k) ip +(k) ip(k)] .

(6.86)

e. [Tl1li' We assum e that w2(k) > 0, so that the Hamil tonian is positiv e definit w(k) is real, and we take w(k) > O.J ( e lk l · In the examp le described in Eqs. (6.7 1 ), 6.72), and (6.73), w(k) = *

useful : Throughout all of this the follow ing integrals will be

f

d 3 xe- '' k · x

3 3 = (271:) ,) (k ) ,

f

d 3k

( 2 71:) 3 .�

ik ' x

e

J

]

i - - k) - - ,._--= - - k) , v w(k) qJ( TI( V w(k)

--n = [a(k) 2w(k)

J

+

(6.88)

a + ( - k)J,

(6.89)

;

n w k1 [a(k) - a + ( - k)].

(6.90)

[a(k), a(k')J = [a+ (k), a + (k')J = 0,

(6.9 1 )

[a(k), a+ (k')J If we

=

(6 . 92)

(2n)3b3(k - k').

write H in terms of a and a + , making the change of variable k H =

1f

-2

d 3k nW(k)[a +(k)a(k) (2n)

- 3

+

-+

-k

a(k)a +(k)J

plus a correction term to make the vacuum energy zero. Hamiltonian is evidently

The corrected

(6.93)

Rewri ting the Hamil tonian of Eq. (6.75) , we get

1

=

J2 n

; when necessary, we obtain

From K(x) = K( - x) = K*(x) it follows that

H =

.

(6.87)

,ifhe commutation relations are, from Eqs. (6.82) and (6.83),

(6.82)

[ip(k), ti (k')J = in(2n)3;j3(k + k').

]

fi(k) ,

Jw(k)

J2 n

fi(k) = - i

From Eqs. (6. 76) and (6. 77) we obtain [ip(k), ip(k')J = [fi ( k ), fi(k')J = 0,

[ 1 [ r-

�_ J;(k) ip(k) + - ",-i

ip(k) =

(6.80)

we h a\e with a similar expres sion for n(x) . Since qJ(x) and n(x) are Hermitian, (6.8 1 ) fi + (k) = ti ( - k). ip + (k) = ip( - k) ;

165

-

< 3 (x'J .

u

[� ote that the correction term is the infinite quantity --!- S d3k/1w(k)b3(O)] . Fmally, we express the original field variables in terms of the creation and destruction operators, using Eqs. (6.80), (6.89), and (6.90) : qJ(x)

=

Il (x)

=

1 3� J-fz. f !(2n)3 [a(k) ik 2w(k)

f J � 3k_ (2n)3

e

X

+

nW(k) [ - i a(k) eik ' x 2

a + (k)e - ik ' xJ +

i a + (k )e- ik ' x] .

(6.94) (6.95)

Equations (6.9 1 ) through (6.95) are the important results of the quantization procedure. The commutation relation of Eg. ( 6.92) may appear strange, in that

166

Systems of indistinguishable particles

Creation and annihilation operators

[a(k), a+(k)] is infinite (instead of unity), so that the analysis of Sect�on 6 . 1 does not apply directly to a(k) . Suppose, however, that we use a less s mgu lar representation. Choose a complete orthonormal set of functions t/I ik), where

'l.

is a discrete index :

f (2rr)3 -��� t/I * (k) t/IP(k) a

L t/lik)t/I:(k') a

=

=

[)afJ '

(6.96)

(2rr)3[)3(k - k')

(6.97)

and define

�ch as the Hamiltonian, can be written in terms of the creation and destruction �perators using any basis of states. Note : In relativistic quantum mechanics, when one quantizes a free-particle field with w(k) = ,jkZ + mZ (11 = c = 1), a different normalization and sum mation convention is commonly used for momentum states. Everything is written in terms of S d3kj(2rr?2w(k) and (2rr)32w(k)[)3(k - k'), which happen to be relativistically invariant ; to accomplish this change of normalization one ,/ uses a "relativistic" a(k) equal to 2 w(k) times our a(k) . Equations ( 6 . 92) , (6.93), and (6.94) then become [a(k) , a + (k ' )]

(6.98 )

Then

167

=

H =

c,o(x)

(6.99 )

=

(2rr)32w(k)[)3(k - k' ) ; d3� w(k)a +(k)a(k) ; (2rr)32w(k) . d 3,,�. [a(k)eik ' X a + (k)e- ik ' x ] , (2rr)32w(k)

ff

+

.) so that we can apply previous results and construct In! . . . na ' : . But these states may not be eigenstates of H. The states

In some texts the (2rr)3 is also treated differently. We will not use the relativistic normalization here, but it is mentioned in case the reader fi n ds it elsewhere and wants to reconcile it with our notation.

and so forth, though unnormalizable, are eigenstates of H. What kind of phonons do that unnormalizable states represent? The state Ik) is a phonon of energy I1w(k), and we may also say that it has momentum 11k. To discover the reason for this, consider the operator

6.7

(6. 1 00)

which satisfies so that

- l1kaCk)

(6. 1 0 1 )

+ . . . ) l k 1 , k z, · . . , ) .

(6 . 1 0 2 )

[P, a(k)] Plk ! , kz, " . )

=

(11k + I1kz

=

Now, from Eqs. (6.94) and (6. 1 0 1 ) we obtain [P, c,o(x)] = iI1Vc,o(x). One can then show that ea ' P li� c,o (x)e - a ' Pli�

=

c,o(x

+

(6. 1 03 )

a) ,

(6. 1 04)

so that P generates translat ions and is therefore the momen tum operato r. t he In Section s 6.7 and 6.8 we will describe further the relation between rs . o operators and the states they create and destroy, as well as how other operat

SYSTEMS OF INDISTINGUISHABLE PARTICLES In the preceding sections we considered the quantum states of an oscillator system as being states of various numbers of a "particle" called a phonon. We identified certain states as one-phonon states, and others as states containing more than one phonon. In this section we shall follow a different line of reasoning. We will start with a space of states describing a single particle, either Bose or Fermi, and construct the multiparticle states according to standard methods. For the Bose case we will arrive at a system of states and operators that is mathematically the same as that found previously for a system of oscillators, thereby showing that the interpretation of oscillator states as many-phonon states is consistent with the usual description of many-particle systems. In the meantime we will have also developed a formalism for dealing with Fermi particles, for which the states do not resemble those of a harmonic-oscillator system. We will treat the Bose and Fermi cases simultaneously, distinguishing them by the number ( ; ( 1 i f the particles are Bose = (6. 1 05) 1 if the particles are Fermi. We will use the symbol (P (where P is a permutation) to denote 1 for the Bose �ase and ( - 1 t for the Fermi case.

{+

-

168

Creation and annihilation operators

Systems of indistinguishable particles

Consider first the case of distinguishable particles. If 1 1/1 I ) ' . . . , 1 1/1n ) one-particle states, then describes the n-particle state with the ith particle I (Jl I ) I (Jl 2 ) ' . . l (Jln ), then we write «Jl I I/I )

III

state 1 1/1 i ) '

=

« (Jl I I «Jl 2 1 ' . . «JlII I}( I I/I I ) 1 1/1 2 ) . . . II/In» )

=

« Jl I I I/I I ) «Jl 2 1 1/1 2 ) ' . . «Jln l l/ln ),

If 1 (Jl )

are

Example :

particles) ,

"=

If (

f

d3xl ' . .

f

=

£53(X1 - Y I ) ' . . £53(Xn - Yn) ,

d3xIIClxl ) ' . . Ixn »)« x 1 1 ' . . (xn l)

=

-I

=

I.

1

x

I b) = la, b)

la )

x

l a ) = l a , a ) = vl2 1a ) I a ) .

( Ferm i particles) la, b) =

1

;

,/ 2

=

1

'\/ 2

( l a) l b ) - I b)l a »),

l a , a ) = O.

Thus, we have the expected result that two Fermi particles cannot be in the same state. What is the inner product of two of these n-particle states ? The answer is given by the following theorem* «JlI l l/ln ) «Jlnll/ln )

(6. 108) (6. 109)

Using this basis we can express the n-particle states in coordinate representation :

If ( = + 1 (Bose

( l a ) l b ) + I b ) I a »),

l a)

(6. 1 07)

which defines the inner product «Jl I I/l ) . The Hilbert space describing t h e n-particle system is that spanned by all nth-rank tensors with the form of Eq. (6. 1 06). The state in which the ith particle is localized at the point Xi is Ix I ) Ix 2 ) ' . . I Xn ) . As each Xi runs over all space, the resulting states form a complete orthonormal set for the n-particle space (ignoring spin and other variables) : « x I I ' . . (xn l}( I Y I ) ' . . IYII»)

Let la ) and I b ) be two single-particle states.

169

where, for any n

x

n matrix A = (A i), I A I(

==

/

(

(6. 1 1 3) '

L ( PA 1 P( I ) ' " A IIP(n) ' P

A,

(6. l 1 4)

(6. 1 10)

That is, I A 1 - is the determinant of permanent of A .

(6. 1 1 1 )

«({Jh " " (Jln ll/l l " ' " I/In ) = 1 ; L L e(Q« (Jlp( I ) 1 . . . «((JP(n) I )( II/IQ( 1 » ' I " I/IQ(n» ) n. P Q

For the particle 1 1/1 ) of Eq. (6. 1 06), we have, using Eq. (6. 1 07),

and I A I + is what is often called the

Proof:

where I/I Jx J = (Xd l/li ) ' Next, let us consider indistinguishable particles. We assume that the particles obey Bose or Fermi statistics, which means that we must symmetrize or anti­ symmetrize, respectively, the states obtained in Eq. (6. 1 06). We therefore define

-

_

I

OP(Q � , L. � L. (, � ( (Jl I I I/IQP - ' ( 1 »

11 .

P Q

' " «Jln I I/lQP - ' (n»

(Where we have permuted the factors by P) where P runs through all permutations of n objects. It will often be conven ien t to write 1 1/1 1 , 1/1 2 , ' . . , I/In ) for 1 1/1 1 ) x 1 1/1 2 ) x . . . x II/In ) . The space of n-particle states is that spanned by all "products" of the form of Eq. (6. 1 1 2). Note that 1 1/1 1 ) x . . . x I I/In ) is totally symmetric in the Bose case and totally anti symmetric in the Fermi case, as it should be.

•

. Compare thOIS t heorem wIth the well-kno wn formula for vectors in 3-space,

( a

x

h) '

(c

x

d)

=

I

c a. a'd h·ch·d

'I .

170

Creation and annihilation operators

Systems of indistinguishable particles

171

II .

We have constructed, for each n, a Hilbert space that describes a system of n In many processes the ,.umber of particles is not constant : particles can be created and destroyed. To cfescribe such processes we need a Hilbert space that contains states of varying JI:llmbers of particles. To get such a space we simply combine all the n-particle spaces into one big space that we may call the multiparticle space. A general slate in the multiparticle space is of the form

reticles ; thus we have an infinite sequence of spaces. (letting R

=

QP- l )

,:,

where

which is the desired result. Now let { I I ), 1 2), . . . } be a complete orthornormal set of states :

L IIX)< IX I

=

I.

(6. 1 1 5 )

A complete set of n-particle states consists of I IX I ' 1X 2 ' . . . , IXn ) , where IX I :::; . . . :::; IXn in the Bose case, and IX I < . . . < IXn in the Fermi case. These states are orthogonal to one another, but not always normalized. The reader can show. using Eq. (6. 1 1 3), that for a complete orthonormal set of states we may take I IX I ' .

"_"_,!;d

J nl ! n2 ! ' "

We define states of different numbers of particles to be orthogonal to each is another multi particle state and is expressed in the manner

other, so that if l


<


,

==

<
+

<
+

. . '.

(6 . 1 1 8)

If { IIX)} is a complete orthonormal set of states, so that Eq. (6. 1 l 5) holds, then using Eqs. (6. 1 1 3), (6. 1 1 5), and (6. 1 1 8) we may summarize orthogonality by

< IX

(IX I :::; . . . :::; IXn) for bosons

I IX I , . . , , lXn) (1X 1

(6. 1 1 7)

II/I(n» is an n-particle state.

I

, . . . , lXnlf3l , . . . , f3m )

=

t5nm

l�

a lpl anP l

(6 . 1 19)

From Eq. (6. 1 1 6) we also have the completeness relation

< . . . < IXn) for fermions,

where na is the number of times that IX occurs in the sequence lX I ' . . . , IXn ( for Fermi particles, na = 0 or I). For either case the completeness relation can be written in the following convenient form :

(6 . 1 1 6 ) Here the range of each lX i is unrestricted, duplication of states being taken care of by the l in ! and the normalization. In the Fermi case, the terms with non­ distinct lXi are, of course, zero. The I on the right-hand side of Eq. (6. 1 1 6) means the unit operator on the space of (properly symmetrized) n-particle state s. Equation (6 . 1 1 6) can be verified by applying the left side to a state 113, . . . , /3,) and using Eqs. (6 . 1 1 3) and (6 . 1 1 4). The case n = 0 may require some explanation. The zero-particle states are tensors of rank zero, that is, scalars (complex numbers). They form a onedimensional space, all of whose elements are proportional to the number I . The "state" I will be denoted by Ivac) (or sometimes by 10» and called th e "vacuum state." The zero-particle states are thus spanned by the state Ivac).

(6. 1 20) this equation " 1 " means the unit operator on the whole multiparticle space. As an example, suppose we describe the states in coordinate representation. The (unnormalizable) state Ix I ' . . . , xn) describes the situation in which there is one particle each at points x I , . . . , Xn' Then Eqs. (6. 1 19) and (6. 1 20) become In

(6. 121) (6. 1 22) We may expand an arbitrary multipar ticle state 11/1) as follows, using Eq. (6 . 1 22) :

11/1) = lIere

f L, n.

n=O

f

d 3X I

•

•

•

f

d 3XnIX b " " xn)I/I(II)(x 1 , · · · , xn)'

(6. 123) (6 . 1 24)

172

173

Systems of indistinguishable particles

Creation and annihilation operators ,

'

1 1/1) X i' 1 1/1 1 " " , I/In),

1 1/1)

is normalized) the amplitude for the state to have n particles, o ne is (if is symmetric or antisymmetric accordi ng Note that at each is an n-particle state and is of the forIll to the statistics. Note also that if is a one-particle state, then where each

l/I(n )(x l , . . . , xn) 1 1/1) I I/I i)

(6. 1 2

5)

= Equation (6. 1 25) follows from Eqs. (6. 1 24) and where (6. 1 1 3). In the Fermi case the determinant is called the "Slater determinant . " We are now ready to define creation and destruction operators. The,e operators are fundamental for two reasons ; first, we constructed the m ulti­ particle states so that we could describe changing numbers of particles, and we need some operators that can effect this change, and second, other operators, such as the total energy, will turn out to be simply expressible in term.s of the creation and destruction operators. Let be any one-particle state. We define to be that linear operator which satisfies

I/I /x) < xj l l/l j) .

Icp)

a+ (cp)

(6 . 1 26) =

for any n-particle state 1 1/1 , " I/In ) For n 0 this is understood to mean a + (cp) l vac) = I cp) . We call1 "a+(cp) the. creation operator for the state I cp), and its adjoint a(cp) the destruction operator.

A creation operator clearly converts an n-particle state into an (n + 1 )­ particle state. It is easily seen that a destruction operator turns an n-parti cle state into an (n - I )-particle state and annihilates the vacuum state. To find the effect of on an n-particle state 1 1/1 1 ' we multiply on the left by a n arbitrary (n - I )-particle state

a(cp)

. ' I/In) <X I . . . Xn - I I ·

<X I ' " Xn - Il a(cp) 1 1/I1 ' " I/In ) < 1/1 1 " ' I/Inl a + (CP) l xl . . . Xn - l)* < 1/11 . . ' I/InI CP , X I . . , Xn - I>* 1 <l ) .. .. ..
(an expansion by minors)

f (k - I
In

Equations (6. 1 26) and (6. 1 27) describe the action of creation and destruction operators on many-particle states. From Eq. (6. 1 26) it follows that

a+(CPI )a + (CP2) = 'a+(CP2)a+(CP 1 ), [a+(cp I )' a+(cpz)] - t; = 0, (6. 1 28) where [A, B]_I; AB - ,BA ; that is, [AB]_(, = + 1 ) the commutator and [A , B] «( - 1 ) is the anticommutator. + Taking the adjoint of Eq. (6. 1 28) we obtain the further result (6. 1 29) or

=

IS

=

=

Thus, the creation operators commute for Bose particles and anticommute for Fermi particles, and similarly for destruction operators. Now, what is Does it (or any similar expression) reduce to anything simple ? We first calculate

[a(CP I )' a+(cpz )] - t;?

a(cp I)a+ (cp z) 11/I 1 . . ' I/In > a(CP I )I (cp z , 1/1 . . ' I/In> n L (k < CPIll/lk ) I CPz , 1/1 1 . . . (no I/I k) " · I/In ) . + k=1 =

=

=

=

=

I

=

(6. 1 30)

and then

a+ (CP2 )a(CP I ) 11/I 1 " ' I/In> = k=tI (k - I ICP2' 1/1 . . . (no I/Ik) . . . I/In ) ·

t;

=

-

1

(6. 1 3 1 )

174

Creation and annihilation operators

Multiplying Eq. (6. 1 3 1 ) by ( and subtracting it from Eq. (6. 1 30) we see that

(6. 1 32) Equations (6. 128), (6. 1 29), and (6. 1 32) are the fundamental "commutati on" relations for creation and destruction operators. The relations we have derived are usually stated in terms of an orthonormal basis, and we shall now do that. Let { I a ) } = { I I ), 12), . . . } be a complete orthonormal set of one-particle states. It is usual to let a� = a( a). Then, since ( a l P) = ij�p, we have [a�, an - � = ij�fl ' We consider the Bose and Fermi cases separately. Bose Case

Let

1 1 , . . . , 1 , 2, . . . , 2'-"-"--"-L , I n 1 n2 . . . ) .. VI n l '' n 2 ', . ��

(6. 1 33 )

� ��

where n� is the number of times a appears in the ket on the right. Then the In 1 n 2 ' ) (each n� = 0, 1 , 2, 3, . . . ) form an orthonormal basis for the whole multiparticle space. From Eqs. (6. 1 33), (6. 1 26), and (6. 1 27) we find •

•

a+a I n 1 n2 . . . na . . . )

=

Systems of indistinguishable particles

6.7

I n 1 n 2 . . . n� + I · . . V/� a

),

(6. 1 34) (6. 1 35)

175

Fermi Case

Using the notation 1 r:J. ] , . . . , Ci,. ) , we have

(6. 137)

and aa I Ci ] ,

k ) = L ( - I ) - lij.,.! CX ] , . . . , Cik _ ] , Clk + 1 , . . ] = k n

. . . , cx,.

. , Cin

)

(6. 1 38)

We could also use the occupation-number notation where cx ] < CX 2 < " ' , and na is the number of times cx occurs (n, = 0 or I) in this sequence. If n� = 0, then a; changes it to I , whereas aa annihilates the state. If na = I , then a� changes it to 0, whereas a; annihilates the state. There are also factors of ± I involved, depending on what other states are occupied. It is easiest merely to remember Eqs. (6. 1 37) and (6. 1 38). Note that a(cp) 2 = a + (cp) 2 = 0 for any one-particle state Icp). This state­ ment follows from Eqs. (6. 1 28) and (6. 1 29) (with ( = - I and CP l = CP2 = cp), and it is also equivalent to the fact that two fermions cannot be in the same state, that is, Icp, cp) = o. One could also derive Eqs. (6. 1 37) and (6. 1 38) directly from the anti­ commutation relations

(6. 1 39)

The commutation relations are

(6. 1 36) Equations (6. 1 34), (6. 1 35), and (6. 1 36) are identical to Eqs. (6.60), (6.6 1 ). (6. 5 1 ), and (6.52) for raising and lowering operators for a system of harmon ic oscillators. The operator for the number of particles in the state la ) is

The notation of Eq. (6. 1 33) (in terms of "occupation numbers") is generally not the most convenient. It is more natural, in fact, to continue using the nota ­ tion we have been using all along in this section, that is, the notation la1 , a 2 , . . . . an ) . This notation was discussed in Section 6.5 in connection with phono ns. Note that Eqs. (6. 1 26) and (6. 1 27) of this section, when applied to states of the form la 1 , . . . , r:J.n ) , become identical to Eqs. (6.67) and (6.68). Thus, creation and destruction operators for a system of Bose particles look just like those for what we called a "phonon" system.

as was done in previous sections for the a;s of the harmonic-oscillator system. But for the Fermi case there does not appear to be any a priori reason for pos­ tulating Eq. (6. 1 39). (Remember, for oscillators the corresponding commutation rules followed from the canonical quantization procedure.) One may rather consider Eq. (6. 1 39) as derived from the antisymmetrization postulate for fermions. Let us return to the general case where Eqs. (6. 1 26) through (6. 1 32) apply. One advantage of deriving them in such a general form is that we are not tied down to a particular basis of states. Suppose we use a basis of momentum eigenstates, Ip). Because (p ip') = (2rr)3ij3(p - p') we have

[a(p), a + (p')] _ , [a(p), a(p' )] _ ,

= =

(2rr)3ij3(p - p'), [a + (p), a+(p ')] _ ;; =

O.

(6. 1 40)

Pro m the vacuum state we can construct the other states by

(6. 1 4 1 )

The Hamiltonian and other operators

Creation and annihilation operators

176

I x), then since <x I x ' ) = b3(x [a(x), a+(x')] _1; = b3(x - x'), (6. 1 42 ) (6. 1 4 3 ) I x l , . . . , xn) = a+(x l ) ' " a + (xn ) l vac ) . If we use a basis of hydrogen-atom energy eigenstates Inlm), then [a(nll/1j. a+(n'l'm')]_\ = bnn, bu' bmm" and so on. x'J.

If we use a basis of position eigenstates

How do the creation and destruction operators change when we ma ke change of basis ? This question is easily answered by noting that if

a

(6, 1 44) Ix) = I 1/1 ) + fJl cp ), a+(x) = exa+ (I/1) + fJa+(cp) , a(x) = rx*a( l/I) + fJ*a( cp), (6. 1 45 ) This means that creation operators "transform" like kets, whereas destruction operators "transform" like bras (because (x l rx * <1/1 I + fJ*< cp I), Equation ex

then

=

(6. 1 45) are readily generalized to infinite series and integrals. Now if we change. for example, from position to momentum representation, so that =

1

X

1

=

(6, 1 46)

P/

=

a+ (x)

=

f d3xa + (x)eiP 'x, f (d32 p) 3 a +(p)e-ip . x•

(6, 1 47)

11:

a(p)

a(x),

and To relate the destruction operators simply take the Hermitian adjoint of Eq, (6. 147), One proceeds in a similar way for any other change of basis.

I rx ) ,

Suppose we have a complete orthonormal set of states and we let the "wave functions" of these states be Write down the in terms of and and vice versa. formulas for and Exercise :

a+(x)

6.8

a(x)

<x l rx ) = uix). a: aa'

ii.

We will now show that these operators have other uses creating and destroying states. erely m tJtan is an operator that acts only on one-particle states. We Suppose that represents the "sum of over all of the wish to find an operator That is, for any n-particle state s." icle pIlrt

A( l )

1

THE HAMILTONIAN AND OTHER OPERATORS

In the last section we developed a method for describing systems containi ng many Bose or Fermi particles, and we defined creation and destruction operato rs

A( l )

A

(6. 1 48)

we

A 11/1 ) to satisfy I I/In) AII/I) A(I)I I/I l ) 1 1/1 2) . . . 1 1/1,) + 11/11 ) A(I) 11/I 2) . . . + 11/1 1 ) (6. 149) 11/1 2) . . . A(I)I I/In) ' I To see what this means, suppose each 11/1 ; ) is an eigenstate of A( ) with eigen­ value a i ' Then Eq. (6. 1 49) implies that want

=

x

+

x

x

x

x

x

x

. . .

x

x

(6. 1 50)

For example, if is the single-particle Hamiltonian, then is the total energy (ignoring mutual interactions, which we shall consider later i n this section). If is the momentum operator for a single particle, then is the total mo­ 1 mentum. = 1 ( ) (the unit operator on one-particle states), then = N, the "number-of-particles operator." The desired operator is easy to find. We first fi n d it for the special case fJ l . I n this case Eq. (6. 1 49) becomes

A( I)

A

If A( 1 )

A A II/I) (fJl l/I l ) l rx , 1/1 2, · · · , I/In) (fJ l I/I2)1t/l1 '

A(l ) = I rx)<

the creation operators are related by

a+ (p)

tc)c the particle states.

A(l)

f d3xlx)<xlp) = f d3xl x)eip,X, d3p - i ' p I ) = f (d3 2;)3 p )< pI x ) f (2�) 3 e p x ,

Ip)

177

=

+

Now look back at Eq. (6. 1 3 1 ) and notice that when have a +(

ex,

CP2

A

A

. . . , t/I,)

=

'l.

and

(6. 1 5 1 )

CPI = fJ we

rx)a(fJ) 1 1/1 ) k�1 ,k - \fJ/l/lk ) l rx) l/Il" ' " (no I/Ik)" ' " I/In) · (6. 1 52) But (k - I l ex , 1/1 1" " , (no I/Ik)" ' " I/In) = 1 1/1 1, " ' , I/Ik- I' ex, I/Ik + l , " " I/In ) due to the symmetry property of the n-particle state. Using this equation in Eq. ( 6. 1 52) and comparing it with Eq. (6. 1 5 1 ) we find when (6. 1 53) A(I) = lex)
>.P

>.P

178

The Hamiltonian and other operators

Creation and annihilation operators

Then, from Eq. (6. 1 53) and linearity,

A

=

L A�1)a +(�)a(f3).

a,(J

As a first example we consider A ( !) = 1 ( 1 ), so that A the number of particles. Using various bases we have =

1 (1)

=

179

where x is the position operator. In coordinate representation,

(6. 1 55)

(6. 1 6 1)

N, the operato r for

so

that

H

L I � )(� I a

= =

f d 3x f d 3x' [ - 2� V2c53(x - x') f d 3xa + (x) [ - ;: vex)] a(x).

+

( p I H ( I )/p' )

from which we can immediately write

- XI)]

a +(x)a(x') (6. 1 62)

+

In momentum representation*

(6. 1 56)

V(x)c5 3 (x

=

=

where

p2 (2n)3c53(p 2m

_

2

p' ) +

f,,;; (2n)3c53(p - p')

+

f

' d 3 xe - iP ' X V(x)eiP . XQ

V(p - p '),

(6. 163) (6. 164)

Therefore

(6. 1 57) Next, consider the momentum operator. Because

p (l )

= =

f �g p l p )( p l (2n)3 f d 3xlx) � V(xl

=

(6. 1 58)

= =

d 3P f (2n) pa +(p)a(p) 3 f d 3xa + (x) � Va(x).

=

p 2 + Vex ), 2m

q)a(p).

(6. 165)

(6. 167)

(6. 1 59)

(6. 1 60)

+

(6. 1 66)

then

(Compare the first of these expressions for P with Eq. (6. 1 00), another expres sio n for P.) Finally, suppose the Hamiltonian for a single particle is H( l )

+

The term V(q)a + (p + q)a(p) annihilates a particle of momentum p and recreates it with momentum p + q, the amplitude for this process being V(q). If we use a basis of eigenstate I � ) of H( l ), so that

we have for the total momentum

p

d 3p p2 f (2n)3 a +(p)a(p) f (2nd3� f (2nd 3�3 V(q)a +(p 2m

which is what we had for phonons with E, = ftw, . J

..

The second term is obtained by noting that V(Xp)

=

J

d3xV(x) lx ) (xl

an d putting this between states
180

�

jl.8

Creation and annihilation operators

The Hamiltonian and other operators

181

I

,f

�his is in fact .correct, as can be verified by applying V to l XI ' . . . , xn) .

r

All of these expressions for H can be derived from each other using the formulas relating a+ (x) and a + (p), and so on, described at the end of Section 6. 7 . But it is often simplest to obtain such expressions directly from the single­ particle Hamiltonians, as we have done here. The density of particles (that is, number per unit volume) at the point x is given by the operator

p(x)

=

a+ (x)a(x)

(6. 1 6 8 )

(which corresponds to the one-particle operator Ix)<xl). Thus the number-of­ particles operator in Eq. (6. 1 57) may be written as N

=

=

=

1-

f d 3x f d 3ylx, y) V(Z)(x, y)<x, yl , =

V

=

-!

d 3x

d 3 ya + (x)a + ( y) V(Z)(x, y)a(y)a(x).

if j < if j >

k k'

&

; � ultiplying by -! V( Z)(x, y) and integrating over x and y, we find that V as ,given by Eq. (6. 1 7 1 ) indeed satisfies Eq. (6. 170). { In view of Eq . (6. 1 68) and the remarks following it, we might expect that . . the mutual mteractIOn could also be described in terms of the particle density by the operator (6. 172) However, V' is not quite the same as

L V (Z)(x;, xj) l x l , · · . , xn)

i<j

(6. 170)

Looking at Eq. (6. 1 69) and noticing that a + (x)a + (y) creates the state l x, y), whereas a(y)a(x) destroys the same state, we might guess that

f f

!where

(6. 1 69)

as can be verified by applying it to a two-particle state l XI , Xz ) . We now want an operator V on the whole multiparticle space such that

Vlx l , · · . , xn)

�.

,Then :.a + (x)a + ( y)a(y)a(x) l x b " " xn)

f d 3x V(x)p(x).

(6. 1 7 1 )

Using

r

t

This last equation represents the integral of the potential energy weighted by the density. So far we have described a system of independent particles, each particle being (possibly) acted on by an external potential but no two particles influencing each other. Suppose, however, that there is an additional potential V ( 2 ) (Xi' x) between any two particles at Xi and Xj (giving rise to a "two-body" force). We assume that V ( Z) (Xi' Xj) = V( Z \xj , xJ On two-particle states the operator is then

V(Z)

(6. 1 27) tWIce, we have

! k

f d 3xp(x),

and the potential-e nergy term i n Eq. (6. 1 62) as v

fEq.

So

that

V.

To see the difference we write

p(x)p(y) = a +(x)a(x)a + ( y)a(y) = (a + (x) a + ( y)a(x)a(y) + J 3 (X - y)a+ (x)a(y) = a + (x)a + ( y)a(y)a(x) + b3(x - y )a + (x)a(x), (6. 1 73)

. Thus V' contams an extra term, which may be interpreted as a self-energy ' .

'

It contn'butes even when there I' S only one particle present. The true mutual . Interaction V is zero unless there are two or more particles. We want only the

Creation and annihilation operators

182

Ground state for a Fermion system

183

Then, ignoring mutual interactions, we have for the multiparticle Hamiltonian : (6. 1 77) Every state may be built up from the vacuum state by applying creation operators : (6. 1 78) Fig. 6.1 Operator V adds momentum q to one particle and subtracts it from the other. mutual interaction, because any self-energy (ifit exists in nature) can be included in the Hamiltonian of Eq. (6. 1 62) . Besides, for many potentials (for example, the Coulomb potential), V' is infinite and is not what we would consider to be the true energy. If we express V in momentum representation, using Eqs. (6. 1 71) and (6. 1 47), and assume (6. 1 74) V( 2)(x, y) = Vex - y) = V(y - x), we arrive at

V

=

d 3P d 3p� V(q)a + (p + q)a +(p' - q)a(p')a(p) . f f 2 (2n) 3 f (2n)

!

d 3q ( 2n)3

(6. 175)

Here V(q) is as defined in Eq. (6. 1 64) (in proving Eq. (6. 1 75) we use V(q) V( - q), because Vex) = V( - x»). Note that

V lp l ' p z)

=

f

d 3q V(q) l p l (2n)3

=

+ q, pz - q) ,

which says that V adds momentum q to one particle and subtracts it from the other with amplitude V(q). This process is denoted by the diagram in Fig. 6. 1 . The total momentum is conserved, as we would expect, because Eq. (6. 1 74) implies that the mutual interaction is invariant under translations. The Hamiltonian and mutual interactions described here all conserve the total number of particles (see the exercise at the end of Section 6.5). 6.9

GROUND STATE FOR A FERMION SYSTEM

In this section we will consider fermions, so that the results of Sections 6.7 and 6.8 apply with , = - 1 . Suppose we know the eigenstates and eigenvalues o f the single-particle Hamiltonians, H( 1 ) : (6.1 76) (ex = 1 , 2, 3, . . . ) .

(Both sides vanish unless the exi are all distinct.) It is often inconvenient to refer everything to the vacuum state, as in Eq. (6. 1 78). In practice we may be considering states that differ from some "ground" state only by the presence or absence of a few particles. Suppose there are G particles present. Assume that the energy levels are ordered such that El :s; Ez :s; E :s; . . . . 3 Then the state of lowest energy is I gnd)

=

1 1, . . . , G)

=

at . . . a ci l vac),

(6.1 79)

which we call the ground state ; its energy is (6. 1 80) Any other G-particle state will have some of the levels 1, . . . , G unoccupied and so �e higher levels occupied. It is convenient to use Ignd) as a reference, . descnbmg the removal of a particle from 1 , . . . , G as the creation of a "hole." Particles in the levels G + 1 , G + 2, . . . are still called "particles." If a particl� is excited from the level ex to the level [3(ex :s; G :s; [3), then we say that a hole WIth energy - Ea has been created as well as a particle with energy E{J' The concept of holes may be formulated mathematically as follows : Define for ex :S; G. From the anticommutation relations in Eq. (6. 1 39) we have' if ex, [3 [3' < G, [aa' aa ,J + = [aa ' bpJ = [bp, bp ,J = 0, + + [aa, a:J + = baa' ; [bp, b;J + = bpp', [aa, b;J + = o. ,

(6.181) (x '

> G and

_

(6. 182)

T hus the operators aa+ (ex > G) and ba+ (ex :s; G) behave like creation operators. We can now express the states in the form

G, Pi :<:::; G). We call this a state with ground state acts like a vacuum state in that

(lXi >

aa l gnd )

=

l

bp gnd )

m

=

particles and n holes. The O.

(6. 184)

The Hamiltonian of Eq. (6. 177) may be written in the form H

=

=

cx :s G

'X > G

L Eaa; aa + L Ei - b; ba + 1 )

a>G

cx :s G

(6. 185 ) so that the energy of a hole in state IX is - Ea . In other words, it takes an energy - Ea to create the hole. The number-of-particles operator is

L a; aa; + L ( - b; biZ + 1)

a>G

a:5 G

(6. 1 86) thus a hole counts as - I . The number of particles and holes outside the ground state (that is, the number m + n in Eq. (6 . 1 83)) is given by the operator N'

=

L a; aa + L b; ba,

a>G

a:5G

(6. 187)

which counts a hole as + 1 and the ground state as nothing. Suppose now that a perturbation is applied to the system i n the form of an external potential (6. 188) '" Uap ( 1 )aa+ ap , U = � iZ.P

where we assume that the single-particle potential U�J ) may be nondiago na l. Then U

=

'" ( l )aa+ bp+ � Uap � Uap( l )aa+ ap + '" a>G a>G P :5 G P>G + L U�J ) baap - L U�J ) b; ba + L U�; ). Cl :s G a::;; G IX�G P :5 G P> G

185

Suppose further that there is a mutual interaction, a two-body potential th f o e form described in Section 6.8 : (6. 190) As in Eq. (6. 1 89), we can express V i n terms of aa; (IX > G) and ba (IX :<:::; G), and then use the anticommutation rules of Eq. (6. 1 82) to write in a form such that in every term all of the creation operators are to the left of all of the de­ struction operators. Such terms are called "normal products." The result will be a number of terms involving four operators, plus other terms that can be lumped into U or added to Egnd•

V

L Eaa; aa + L Eabab;

=

Hamiltonian for a phonon-electron system

6.10

Creation and annihilation operators

184

6.10 HAMILTONIAN FOR A PHONON-ELECTRON SYSTEM Now we will consider, as an application of the creation and destruction operator formalism, the interaction between the electrons and lattice vibrations in a crystal. It is this interaction (as well as the presence of crystal impurities and imperfections) that accounts for the finite conductivity of metals under most conditions. (One might otherwise expect from the band theory of metals that, once an electron got into an unfilled band, it would move unhindered, resulting in infinite conductivity.) We will show how the Hamiltonian is derived and written in terms of creation and destruction operators. We start with the electron in a lattice with no vibrations. Let N = n1a1 + n2a2 + n3a3 (where n 1 , n 2 ' n3 are integers) describe the positions of the nuclei. The potential Vex) felt by an electron in the lattice is periodic and has the form V1(x)

L N

Vo(x - N).

(6. 191)

The Hamiltonian for a system of independent electrons in the lattice is, from Section 6.8 ,

(6. 192) Now, suppose we have solved this part of the problem and know the one­ particle eigenstates I IX) and eigenvalues Ea;. Denoting <X l IX) by CPa;(x) we have

[ 2� V2 -

(6. 1 89)

The first and fourth terms of this equation modify the energies of the particles and holes, respectively, and the fifth term modifies the ground-state energy· The second term creates particle-hole pairs, and the third term destroys them . Note that N, as given by Eq. (6. 1 86), is conserved, whereas N', as defined i n Eq. (6. 187), is not.

=

+ V1(X)

] CPa(x)

=

EaCPa(x) .

(6. 193)

Then we can express the electron Hamiltonian as

(6. I 94) Because

Ix) = L Ict)
a

186

Creation and annihilation operators

6.10

we have

a +(x) = L aa+ cp:(x) a

(6. 195 )

and similarly

a: =

f

d 3 xa+ (x)cpaCx).

Hamiltonian for a phonon-electron system

k

(k,

c5 form an orthonormal basis of 3-space for each (in that et,a ' ek,b = ab) , a) is a normal mode. chosen so that the mode You should check the consistency of Eqs. (6.200) as an exercise. It is convenient to use the relation

(6. 196)

To obtain relations between the destruction operators, simply take the Hermiti an adjoints of Eqs. (6. 195) and (6. 196). If, as in Section 6.9, we refer all states to a ground state (called the "electron sea"), which is normally filled, then in the notation of that section,

187

v

I

K

d3 k (2 n) 3

eiK ' (N-N')

= c5N,N , ,

and the easiest way to verify it is to make a change of variables 'i N' can be written as

N-

=

K · a i'

(6. 197) In what follows we shall ignore the ground-state energy, so that the electron Hamiltonian becomes H;I :

where m i are integers, so that K · (N - N') transformation is

l,il aak, /

(6. 198) Note that, in terms of electron and hole operators, Eq. (6.195) and its adjoint become a +(x)

=

L a: cp:(x) a> G

+ L bacp:(x), a ,s G

(6. 1 99)

Consider next the lattice vibrations. For each N let ZN be the displacement of the corresponding nucleus from its equilibrium position N. The ZN form a set of coordinates for a system of harmonic oscillators. The procedure for finding the normal modes and quantizing this system is similar to the field quantizatio n of Section 6.6, except that Fourier transforms are replaced by Fourier series, with the "momentum" of the phonons running over a limited region. The n or­ mal coordinates qk a are related to the ZN by equations of the following form (assuming for simpiicity that "a" runs from 1 to 3 as in the case of one atom pe r unit cell of the crystal) :

k

qk,a

=

ZN

-

.JV

'"

It

I

V

K

e- ik ' Nek,a * ZN, •

d3 k � '-(2n) 3 a = l

-. ek ae ik· N qk,a ' .J v

{kl

k

(6.200)

. ai :$; n ; i = 1 , 2, 3 } ; Here the region of integration is K = - n :$; V = lal x az ' a 3 1 is the volume of the unit cell of the crystal lattice (recall that the ai are vectors describing the periodicity of the lattice), and ek ,a (a = 1, 2, 3)

=

I. aarlk,[ 1

.

=

=

m ' r,

and the Jacobian of the

1

iT '

'

Equations 6.200 are essentially a special case of Eqs. (1 .28) and (1.29). The change of notation is Chapter 1

Chapter 6

QrCk) a�(k)

The IXth component of ek,r The IXth component of ZN

v

v

n

The normalization conventions used in Eq. (6.200) are convenient because, in the limit that the lattice approaches a continuum (i.e., ai � 0), K becomes all of momentum space and, letting x = N, we have and

(6.20 1 )

Thus we have obtained the normalization of Section 6.6 for a field cp(x), where -iMjV ZN

�

cp (x) .

Creation and annihilation operators

188

6.10

Following the procedure of Section 6 . 6 we have an expression for ZN in terms of creation and destruction operators :

ZN

=

f

d 3k K (2n) 3

�-

�

�

-J

Ii---V - [A(k a)eik . N ek , a 2Mw(k, a) '

----

[A(k, a), A + (k ', a ')] = (2n) 3 c5 \k - k )c5aa . '

,

f

d3 k + � L liw(k, a)A (k, a)A(k, a). 3 a K (2n)

(6 .204)

(We have used A for the phonon operators* to avoid confusion with the electron creation and destruction operators.) Having written down Hamiltonians for the electrons and the phonons (lattice-vibration states), we now turn to the interaction between them. The potential energy of an electron in an undisturbed lattice was given by Eq. (6 . 1 9 1). If the nucleus at N is displaced by an amount ZN' then the potential energy changes to

V1(x) + A V1(x)

=

L Vo(x - N - ZN), N

where

VaP =

(6 . 203)

M is the mass of the vibrating atom, and w(k, a) is the frequency of the cor­ responding mode. The Hamiltonian for the oscillator system (apart from a constant term) is =

Hint =

+

(6.202)

(6.205)

where we have assumed that each atom in the crystal acts like a rigid body when it is displaced, so that the potential Vo arising from it is simply displaced by ZN ' (In practice, not all the electron shells around the nucleus move by the same amount, so that the potential changes its shape as well as being displaced : however, we shall ignore this fact.) The potential V1(x) was already included in Hel . Writing

=

J

-

J

d3 X AV1(x)a + (x)a (x) (6.207)

d 3 xq;;(x) AV1(x)q;p(x)

� ZN .

J

d3 xq;:(x)VVo(x - N)q;p(x).

AV1(x)

=

-

ZN . V Vo(x - N). L N

(6.206)

* Note that A + (k, a) (as well as the state it creates) has the dimensions of (length) 3 / 2 , as do the corresponding operators for other particles in momentum representation.

(6.208)

Now Yap is also an operator on phonon states, because ZN is an operator. . Usmg Eq. (6 . 202) we get

(6.209) where

CaP(k, a)

=

-

J:::;, :

ek,a

•

� eik ' N

J

d3 xq;:(x)V Vo(x - N)q;p(x) . (6.210)

Therefore, Hint

=

Ix (�:�3 � b

[ CaP(k, a)A + (k, a)

+ ClaCk, a)A(k, a)]a; ap.

(6.21 1)

The total Hamiltonian of the system is H

Hel + Hose

+ Hint. Note: The states of our system are of the form =

or, if we use the hole notation, we have for the interaction energy (that is, the extra energy of the system due to displacement of the lattice) :

189

For the many electron system the interaction Hamiltonian is

+ A (k , a)e - ik ' Nek*,a] ,

where

Hose

Hamiltonian for a phonon-electron system

l ocb . . . ,

C(m :

(6.212)

(n electrons and m phono ns),

f31 " . . , f3n : k l ' a I ' . . " kp, ap)

(n electrons,

m

holes, and p phon ons). The effect of a creation operator A + (k, a) on such a state may be defined as

A + (k , a) lelectrons, holes, k l o a l , . . . )

Which is in turn defined to be

=

Ik, a, electrons, holes, a l , . . . ),

l e lect rons , h oies , k, a , k l ' a l , . . . ) .

190

6.1 1

Creation and annihilation operators

The result of this definition is that all phonon operators commute with all electron (and hole) operators. (In general, it is conventional to say that creation and destruction operators for different particles always commute unless the particles are both fermions. In the latter case it is convenient to define the operators to anticommute ; then everything is consistent if we decide to call the particles different states of the same particle, as with the proton and neutron.) If we write Eq. (6.21 1) in terms of electron and hole operators, we get an expression four times as long, involving terms of the form A +a+a,

A +a+b + ,

A a+a,

A +ba,

A +bb+ ,

A ba,

A bb + .

These terms represent the transition of an electron o r hole from one state to another, or the creation or annihilation of an electron-hole pair. In each process a phonon is emitted or absorbed. The foregoing derivation assumed that there is only one atom per unit cell of the crystal. However, the results are similar if there are more atoms (say A) per unit cell, the only difference in the final result being that there are more phonon modes for a given k (a running from I to 3A). As a simplification it may turn out that some of the modes do not "couple" to the electrons, that is, they do not influence the potential felt by the electrons ; these modes are independent of the others and may be ignored. Such is the case in the Polaron Problem (Chapter 8), where only one phonon mode contributes for each k. 6.11 PHOTON-ELECTRON INTERACTIONS Suppose we shine light on the crystal of Section 6. 10. What is the Hamiltonian now? To H we must add a term Hy for the free electromagnetic field (y refers to phonons) and a term Hey for the interaction between electrons and phonons. For the free electromagnetic field, the classical Lagrangian density is .!f = !eE2 - c2B2) =

t["f2

- c2(V

X

Photon-electron interactions

The equation of motion resulting from the Lagrangian is o

C 2[V2A - V(V ' A )] - A'

=

=

(6.217)

,The Hamiltonian is Hy

=

f

d 3k 2 L Iic C + (k ' r)C(k" r) (2n)3 r = 1

k

- �-

(6.218)

To find the photon-electron interaction we replace the operator* P by (the charge of the electron being - e) in the single-particle :,Hamiltonian 'F ,

,P + eA(x)

so that

H�/ )

A(X)

(6.214)

.. We assume also that the scalar potential is zero. We are using "rationalized units," but with eo = 1 ; thus e 2 /4rcnc ;::; 1 / 1 37 and V . E = p, V x B (l/c2)(j + oE/o!), etc.

(6.215)

ipolarization e" ,r :

We have

o.

�J

= 1 , 2) are two unit vectors, perpendicular to each other and to k, ,and C +(k, r) is the creation operator for a photon with momen tum lik and

(6.213)

=

[

c2 V2 - � A. c2 at2

' Here e" ,r (r

Writing

where A(x) satisfies the subsidiary condition*)

=

Carrying out the quantization procedure of Section 6.6, we find w(k) elk l = ck, and

=

A)2]

V·A

=

191

=

+

f

�

2m

(p . A

+

A.

d 3xA(x)lx><xl ,

P)

+

2 2m

!- A 2 .

(6.219)

(6.219 ')

(6.220)

� .. P and X are the electron momentum and position operators, and A(x) operates on photon states, so that A(X) (see Eq. (6.21 9'» operates on both photons and electrons. -----

192

Creation and annihilation operators

where P )(x)

=

-

Feynman diagrams

-:- (Plx)(xl

2m

+

of each term. (The diagrams assume their full power in the relativistic case,

which we are not concerned here.) To show how the method works we consider the Hamiltonian for a single, otherwise free, electron interacting .will with the electromagnetic field : with

I x ) ( x I P)

- � i[(Vlx» ) ( x l - Ix)(V(x D] '

(6.2 2 1 ) 2m Note that j < l )(x) is the charge - e multiplied by the probability current-densit! operator. The expectation value in any one-electron state 1 1/1 ) is 1 (I/IU( )(x) I I/I) = -

�

2m

i [I/I(x)VI/I *(x) - I/I *(x) VI/I(x)].

By inspection of Eq . (6.220) we have, for the interaction between photons and systems of arbitrary numbers of electrons,

Hey

=

-

where j(x)

+

f d3 xA(x) . j(x) i� f d 3 xA(x)2a + (x)�(x), =

e - li i [Va + (x)a (x) - a +(x)Va(x)] 2m

(6.222)

C(k, r)a: ap,

C(k, r)C(k' , r ')a:ap,

-e

C(k, r)C +(k', r')a: ap,

+

d3p p' p + ( ' a( e- i(P' -p) · x/h . f (2nd3p'li)3 Jr (2nli)3 a p ) p) 2m

(6.225)

;and

Hint l'

( 6. 224)

=

-

+

f d3 xj(x) . A(x) f d3

X

� a + (x)a(x)A(x)2.

A(x)

=

f (2n ��)3 f"�l J_l2ck [A(k' r)eik . x r

+

.

2m

�ere we denote the photon-creation operator by

(6.223)

and so forth, which have interpretations similar to those of the phononelectron interaction. The electromagnetic current defined in Eq. (6.223) may be expressed i n momentum representation as follows : ' J (x) =

f

!where

�C+(k, r) :

is the electromagnetic current-density operator In Eq. (6.222) , A(x) is itself an operator for each x, given by Eq. (6.216). If we use Eq. (6.21 6) in Eq. (6.222), and also express a+(x) in terms of a: using Eq. (6. 1 95), we get terms involving C +(k, r)a: ap,

193

A + (k, r )

instead of by

A + (k' r) e -ik ' X] ek.r>

andj(x) is given by Eqs. (6.223) and (6.224). In this section, Ii The states under consideration are of the form

(6.226)

=

(6.227)

1.

(one electron and n photons).

(6.228)

According to standard quantum-mechanical techniques, the transition amplitude Mf i for a transition from an initial state Ii) to a final state If) can be expressed in terms of the Hamiltonian (to second order) as follows : (2n)3 c5 3(Pr

- pJMfi

=

(fIH l i )

+

IHln) ( nl!f� L (f + n E; - En

Ie

+

0(H 3 ) .

(6.229)

Cross sections and transition rates are proportional to the absolute square of the amplitude :

This expression will be useful in Section 6. 1 2. 6.12 FEYNMAN DIAGRAMS

In

of

A graphical method employing what have come to be called "Feynman diagrams" has proven to be very convenient in dealing with perturbation solutio ns e complicated Hamiltonians. These diagrams serve as a "bookkeeping" devIc to keep track of all the perturbation terms and a guide in writing down the value

practice, the states Ii) and If) are usually states of the form given in Eq.

(6.228), that is, states of particles of definite momentum . It is therefore con­ venient to express Hint in terms of the A (k, r), substituting Eq. (6.227) in Eq. (6.226) , and to use Eq. (6.224) for j(x). (6.230)

194

where

HI =

e p' + p . (2n)3 � �

f d 3p' f d3p f d3 k (2n)3

x

and

H2 =

Feynman diagrams

Creation and annihilation operators

(2n)3

p'

[(2n)3b3(p' - p - k)a +(p')a(p)A(k, r) + (2n)3 b3(p' + k - p)a +(p')a(p)A +(k, r)],

f d 3p' f d3p f d3k' f d3 k (2n)3

x

1

.J 2ck

ek r ,

(2n)3

(2n)3

e2

195

p

(6.23 1 )

(p +k + k' = p )

1

(2n)3 r� 2 m .J2ck '.J2ck

ek

r ', "

(p +k = p' + k' )

(a )

ek , r

[(2n)3b3(p' - p - k' - k) a +(p')a(p)A(k', r' )A(k, r) + (2n)3b3(p' + k' - p - k) a+(p')a(p)A+ (k', r' )A(k, r)

(b)

p'

+ (2n)3b3(p' + k - p - k')a +(p')a(p)A(k', r')A + (k, r) + (2n)3b3(p' + k' + k - p) a +(p')a(p)A +(k', r')A+ (k, r)]. (6. 232)

p

Note that the total momentum is conserved. Each of the terms in Eqs. (6.23 1) and (6 232) is represented by a diagram in which a straight line denotes an electron and a wavy line a photon. For HI the diagrams are given in Fig. 6.2a and 6.2b, and the amplitude for each is

(p H' = p + k)

.

+P . e p'�

1 ek , r

.J2ck '

For H2 the diagrams are those shown in Fig. 6.3a through 6.3d, and the amplitude for each is 1

e2

-

(e)

,

2 m .J 2ck '.J 2ck ek 'r

"

ek r · .

(p + k = p' )

(p = , . + k)

(a)

(b)

Fig. 6.2 Feynman diagrams for

HI '

Fig. 6.3 Feynman diagrams for

(d)

H2 •

Note that parts b and c of Fig. 6.3 represent essentially the same process. In fact, after the integrations are performed, the second and third terms of Eq. (6.232) are equal ;* so in any process involving such terms we need only calculate for the case shown in Fig. 6.3b, say, and multiply by 2. Now suppose we have a definite amplitude to calculate, for example, that for Compton scattering. The whole process is denoted by Fig. 6.4. To find the

Fig. 6 4 Feynman diagrams for Compton scattering. .

"w A

'

e Ignore the infinite self-energy of the electron that comes from rearranging the and A + of the third term in Eq. (6.232).

Creation and annihilation operators

196

Feynman diagrams

The total amplitude for Compton scattering is M = M(a) + M(b) + M(c ) + O(e4)

In equations

Fig. 6.5

(e)

(b)

Feynman diagrams with the same ingoing and outgoing lines as Fig . 6.4.

amplitude we draw all diagrams having the same ingoing and outgoing lines as that of Fig. 6.4, using those of Figs. 6.2 and 6.3. Thus we get the three cases shown in Fig. 6.5a, b, and c. Parts 6.5a and b correspond to the second-order term of Eq. (6.229), in which the intermediate state In) consists of one electron and H = Hl + O(el). (The fact that the photon lines cross in Fig. 6.5b is of no significance.) Using Eqs. (6.229) and (6.23 1) and removing the factor (2n)3c53(pz + kl - Pl - kl ) , we get for these contributions to the amplitude M (a)

and

(-.J

e

Pl

2c kl

(�

+

Pl

2m

+

kl .

pi 2m �l

�{�

-

+

kl .

pi 2m

+

) (.J

ek z • rz

e

2ckl

P l + kl + Pl . ek l •r l

l ck l _ (P l + kl) 2m

ek 1 •r

ck l

) (� �L

__

(Pl - k l) l 2m

2m

)

(6.233)

�� . ek2•r2)

(6.234)

We may associate the two factors in the numerator with each vertex, and th e denominator with the "propagation" of the intermediate electron. Figure 6.5c corresponds to the first-order terms of Eq. (6.229) in which only Hl can contribute. Remembering the factor of 2 due to the two ways in which this process occurs in Eq. (6.232), we get

(6 .23 5)

(6.236)

(6.233) and (6.234)

fact, if the initial electron is at rest (i.e., P l = 0), then M( a) and M(b) both vanish. The results we have derived are satisfactory for low photon energies (Ckl « m el). For high energies we would have to use a relativistic theory and also take into account the spin of the electron. In

Ca)

197

�.1

Spin-spin interactions

199

t

to direct magnetic interactions, a term of the form of Eq. (7. 1) ,rean In addition appear indirectly as a result of spin-independent interactions combined ion principle. As an example, consider the hydrogen molecule ; �ith the exclus . . nuclei

CHAPTER 7

to be fixed and Ignore their spins. The two electrons may be by their positions X I and X2 and some component of their spins SI d ibe scr ,de ' d S2 ' If !/Ja and !/J b are energy eigenfunctions for a single electron with energies an E,. and Eb, then (7.4) ::take the

SPIN WAV ES

7.1 SPIN-SPIN INTERACTIONS In this section we will consider interactions that give rise to a coupling between the spins of two objects of the form

has energy Ea + Eb, neglecting mutual interactions between the electrons. The state !/JbaCX 1 ' X2) = !/Jb(X I )!/Jb(X2) has the same energy. If we add a mutual interaction V(X I ' x2) = V(X2 , X l ) which is spin­ independent between the electrons, then the two-electron eigenfunctions become

(7. 1 ) In later sections we will find out what happens when such a coupling appears in a lattice full of spins. An interaction of the form of Eq. (7.1) can be due, first of all, to the magnetic interaction between two magnetic dipoles. The interaction between two magnetic dipoles III and 112 with separation r is

H=[

11 1;/2

3 (11 1 . r;�2 ' r ) -

]

_

831t II I . 112 c53 (r )

= H ' + H ",


-

.

(7. 3) for a particle with mass defines g i» '

m i'

charge qi and g-factor gi ( Eq. (7.3) in fac t 198

1 = .J2 (!/Jab +

til A

= /i (!/Jab

with their respective energies

(7.2)

where H' is the term in brackets. (The term HI! with c5 3 (r) arises from a corres­ ponding term in the magnetic field of a dipole, which appears when one evaluates V x (II x r )/r 3) . Suppose the two particles are orbiting around each other (as in an atom) and we consider the matrix element of H between the two states of orbita l of r, H ' is a angular momentum [ and ['. As a function of the direction (e, element i s matrix the combination of spherical harmonics of order 2 ; therefore zero unless I[ 2 1 ::;; [' ::;; [ + 2 . I n particular, the first-order splitting o f the S-state (l = [' = 0) is zero. (This argument is just another way of saying that S states are spherically symmetric and H' averaged over all directions is zero). However, HI! contributes only for S states, because only those have nonze ro wave functions at r = 0 (a state with angular momentum I varies as r l near the origin). Thus, for S states we have an interaction proportional to II I 11 2 ' which is proportional to SI . S2 because

!/Js

Here

and

J=

-

f

Es

=

EA

=

1

Ea

+

Eb

Ea + Eb

!/J:b V !/Jba d 3X l d 3X2

!/Jba) ,

- !/Jba) ,

+I-J + I + J.

=

-

f

(7.5)

(7.6)

( 7.7) !/J:a V !/Jab d 3X l dx2 ·

As implied by the notation, !/Js is symmetric and !/J A is antisymmetric under exchange of X I and X 2 ' We have not yet reckoned with the electron spins or the exclusion principle. We must mUltiply our spatial wave functions by spinors that describe the Possi ble spins of the electrons, and the total wave function must be antisym­ . metnc under the simultaneous exchanges of coordinates and spins. Now for two spin-t particles, the symmetric spin states have total spin I , and the anti­ sym metric states have spin 0 (see Section 7.2). The total wave functions are then of the form l (7.8) !/Js = !/JsX(O), !/J A = !/J A i ) w?ere X (s ) is any two-electron spin state with total spin S. (Note that S and A stIlI refer to only the spatial symmetry). In other words, the energy eigenstates

200

�.z

Spin waves

are also eigenstates of the total electron spin. We can then write the energy in terms of the spin as follows : If S is the total spin, S · S has eigenval ue:, s(s + 1) 0 and 2. The energy of a state with total spin s can be written in t h e form =

(7 .9) because it equals Es when s 0 and EA when s = 1 . Therefore we can write the mutual-interaction Hamiltonian for the system in the form

The Pauli spin algebra

'where

(7 . 1 6)

and where we take Ii = 1 (otherwise S = hO'I2). Equation (7. 14) implies that using a basis of states in which the z-component of the spin is diagonal­ ized ; the states

we are

and

IP)

=

=

(7. 1 0)

are states of spin up (Sz J3q. (7. 14), we have ux l ex )

Now,

=

!) and spin down (Sz

=

IP),

Es + (EA - Es)(i- + Sl . S2 ) t(Es + 3EA) + (EA - ES)Sl . S2 Ea + Eb + I + -V + 2JS1 S2 ' In other words, the splitting of the energy levels is described by H

and cyclically in x,

(7. 1 7)

- !), respectively. From

uz lex)

=

lex),

uz lP)

=

- IP) .

y,

(7. 1 8)

(7. 19)

z;

(7.20)

=

=

=

•

AH

=

2JS1 S2 , •

(7. 1 2) (7 . 1 3 )

which is of the form of Eq. (7.1). We have thus taken a spin-independent inter­ action (e. g., a Coulomb potential) between the electrons and made it look like a spin-spin interaction. Note that the splitting is proportional to the "exchange integral" J defined in Eq. (7.7), which depends on how much the wave function s ds !/IaCx) and !/Ib(X) overlap. Note also that the derivation of Eq. (7.12) depen a get would we spins crucially on the fact that electrons have spin ! ; for higher . polynomial in Sl S2 if we went through a similar procedure. The interaction of Eq. (7. 1 3), due to the exchange integral, turns out to be very strong compared to the usual magnetic-dipole interactions. 7.2 THE PAULI SPIN ALGEBRA It is convenient and conventional to use the Pauli spin matrices when deal in g with a particle of spin ! :

Alternatively one may use the following elegant form of Eq. (7.20) : (a · a)(h . a)

l exex) ,

=

!O'

(7. I 5)

a ' h + ia

l exP),

x

h · 0',

(7.21)

IPex),

I PP),

(7.22)

Where, for example, l exP ) is the state in which the first particle has its spin up and the second has its spin down. We then define a 1 to act on the first spin and tl2 on the second. That is, for example, u 1 y l exex)

=

u2 y lexex)

=

iI Pex), i l exP ) .

Not� that Ul iU2 j = u2 P l i' which is to say that operators pertaining to different particles commute. We have previously considered (and will consider further in the section to fOllow) the operator SI . S2 = ta l ' a2 ' By explicit calculation, we find 0' 1 . 0'2 1 exex) a 1 a2 l exp) 0'1 . 0'2 1/h) 0'1 . 0'2 11313) •

S

=

where the a i and hi are numbers (or other operators, provided that they commute with a). Consider next a system of two spino! particles (ignoring spatial and other degrees of freedom). As a basis of states we may take

(7. 1 4) The spin is then

=

(�)

The algebra of the u-matrices is as follows :

because Therefore

201

- ---'

=

l exex), 2 1 Pa ) 2 l ap)

=

11313) .

=

=

l exP), IPa ),

(7. 2 3)

202

Spin wave in a lattice

Spin waves

lexex) and similarly for 1 [3[3 ), we immediately come uPo n

Writing lexex ) = 2 lexex) the useful relation

(7.24)

les ; that is, where pi ,2 is the operator that exchanges the spins of the partic pl,2 Iexex) = lexex ), p1. 2 I ex[3) = l [3ex), and so on. The symmetric states

11, 1 )

=

lexex ), 1 1 , 0)

=

lex[3)

)

)

+

J2

---

l [3ex ) , 1 1 , - 1 ) --

all satisfy pl,z l l , m = 1 1 , m and therefore I, 0, I ). The antisymmetric state -

=

(11 . (12 11, m )

1 [3[3 ) =

1 1 , m)

(7.25) (m

'=

lex�) ; 1 [3ex � (7 2 6) 2 0) Therefore satisfies pl,z IO, 0) = - 1 0, 0) and thus (11 . (12 1 0, 0) = - 3 10, . is spin total the eigenvalues of (1 1 . (1 z are 1 and - 3. Now the 1 0, 0)

=

.

(7.2 7)

and its square is Therefore

(7 .2

S · Sl l , m)

=

)

S ' SIO, 0)

2 1 1, m ,

=

8)

)

which is derivable from Eqs. (7.27) and (7.28) . 7.3 SPIN WAVE IN A LATTICE ction betwe en Consider a lattice, the ith member of which has spin (1 i' The intera given by the spins is due to the exchange effect, and the Hami ltonia n is (7. 3 0) H = � A · .(1 . . (1 " i..J i ,j

J

. . be denoted Consider the lattice to be cubic , and let each pomt on the lattIce spactng· the lattice by the integer vector N = a(nxex + nyey + nzez , where a is Then (7.3 [) I,)

1

)

=

It is clear that the interaction depends only on the distance N - N' M, d .an if i n addition it is assumed that only nearest-neighbor interactions are important, (7.32) Occasionally next-nearest-neighbor interactions are important. They may be even more important than the nearest-neighbor ones, but such a situation can be analyzed in the same way that we use here. We now return to Eq. (7.32) and consider the one-dimensional case. The extension to three dimensions will be immediate. We also are interested in the ferromagnetic case ( A 1 < 0) , and we can neglect boundary effects by con- U� l ' sidering the system to be a closed loop. Let there be N lattice points ,' (JN + l = Thus, (7.33) where A = - A I ' The minus sign in front of A is included to emphasize the fact that in the case of ferromagnets, A < 0 . That is, the energy is most negative when the spins are parallel. The (rare) instances when A > ° are due to a double-exchange effect. Recall that we used ex to represent spin up and p spin down, and we defined the spin exchange operator p l , 2 = t( 1 + (11 . (1 ), 2 satisfying -

-

p i , 2 I exex ) p i , 2 1 [3[3 ) p i ,2 Iex[3) p i ,2 1 [3ex)

0,

have spin 1 and so that (as we anticipated in the notati on) the states 1 1 , m 1 and the anti' 1 0, 0) spin 0. The fact that the symmetric states have spin relatio n symmetric ones have spin ° may also be seen directly from the (7.2 9) S . S = pl,Z + 1 ,

203

=

I wx), 1 [3[3 ) , l[3ex ) , 1C([3).

(7. 34)

Then H = -A

N

L (2p"," + 1 n=1

�

-

1 ) = NA - 2A

N

L p" , " + 1 .

n= 1

(7.35)

o solve Schrodinger's equation with the above Hamiltonian, we must find the elgenstates and eigenvalues of H. To do so, notice first that Eq. ( 7.34) implies that H operating on a state with a fixed number of [3's gives a state with the same number of [3's. Since all states can be represented as linear combinations of tho se with a fixed number of {J's, we have a method of partially diagonalizing H. . Rather than trying to completely diagonalize H, let us merely find eigenstates With low energy. One . suspects that the lowest state (most negative energy) occurs when all the SPInS are parallel. For example IC(C( ' . . C() is such a state. More generally ' . �en we say all SpInS are parallel what we really mean is that the total spin is as l arge In " magmtude as possI. ble, and It. can be shown that the states that satisfy . are compI ete I y symmetnc . states. The completely symmetric thl.s con d'Itlon

204

Spin waves

Spin wave in a lattice

so p. '. +

under pn .n +

1; 1 can be replaced by I in computing the states are invariant energy of those states . Thus, we expect that Egnd

=

NA - 2 A L 1 = NA .= 1 is the lowest-energy state. That this is the lowest energy is obvious. Th e eigenvalues of p. ' . + 1 are ± I ; thus the minimum conceivable energy can OCC ur only when for all n we get the maximum possible expectation value for p. '. + I . . when we can repIace p• n + 1 WI' th + 1 that IS,

8A

N

-

,

Problem :

E



=

205

In order to take the ground state as the zero of the energy spectrum, defin e =

+ NA

=

are equivalent and that none of them can be an eigenfunction. We look for a superposition of the 1 1/1.) . Let (7.37) 11/1) = L a. 11/1.>. To solve the equation H ' II/I)

L ( Ea.) II/I.> •

=

E II/I>

7.1 Energy as a function of o.

=

Neglecting boundary conditions as we have done is tantamount to assuming that the N lattice points are arranged in a circle, or that the N + I point is the first point. In other words,

N

- 2A L ( p•.• + l - 1 ) . .= 1 One may suspect that the next-to-Iowest states has all but one spin parallel. But which one ? It is clear that 11/1 1 ) = IPrxrx · · · rx), 1 1/12) = IrxPrx " . rx) , (7.36) (P in nth place) H

Fig.

•

If - A > 0, what is the lowest state and what is the lowest energy')

H'

- 7r

E II/I) ,

=

H 'II/I)

N

N

- 2A L a. L ( pm.m + l - 1 ) 11/1.) n= 1 m = 1 N - 2 A L a. [( pn .n+ l - 1 ) + ( pn - I .• - 1)] 1 1/1.) .=1

- 2A L a .[II/I. + l ) + 11/1. - 1 ) - 2 11/1.)J . .= 1 Because the 1 1/1) are orthogonal, we can equate coefficients of 1 1/1) to get (7.3 8) Ea. = - 2A(an + 1 + a. - 1 - 20.) . To solve this set of equations, let an = eio• and substitute into Eq. ( 7 . 3 8). We thus obtain 28 (7. 39) E = - 2 A(eiO - 2 + e- iO) = 4A ( 1 - cos 8) = 8A sm 2 ' N

•

or This means that 8

=

or

(2njN)I; I an integer ; - Nj2

�

We have, from Eqs. (7.39) and (7.40) :

I �

Nj2.

(7.40)

8A sin 2 � = 8A sin 2 2 n i . (7 . 41) 2 N 2 This function is plotted in Fig. 7. 1 . There is, in other words, a band of energie s, the minimu m energy occurring at 8 = O. If 8 = 0 we have a completely sym­ metric state, which we have already concluded has the lowest possible energy, E = 0. Another way of seeing that the state with 8 = 0 has the same energy as Icxrx · . . rx) is as follows . Let each spin be tipped from its "up" position by an angle e, where e is very small. Such a state is given by E

I X)

=

I(

cos

=

� rx + sin � P)(cos � rx + sin � p) . . . ) .

Because ej2 is small, cos ej2

�

�

I] « I (I] = 8j2 ). Then I x ) = I (rx + I]P)(CI. + I]p . . . » . Neglecting terms in 1]2 or higher powers of 1], we have I x > = I rxCl. . . · ) + 1]( I Prxrx ' " rx) + I Cl.Prx . . rx) + ' ) . �Ubtracting out the ground-state term, we obtain Ix') = L. 1] 1 1/1.) where 11/1.) 18 give n by Eq. (7.36 ). Since I x) must be an eigenvector of H with the same eigenvalue (energy) as I(X, (X, . . . , rx ) ,* so must I x '), and (1/1]) 1 x ') = L. I I/In). ------I and sin 8j2

·

'

'

• Since in the absence of external interac tion there could be no change in energy when spins are rotated in a certain angle.

011 of the

206

Two spin waves

Spin waves

For J #- 0, we see that the spins are out of phase. This suggests the concept of a wave, and we now rewrite e i�n to make the nature of the wave someWhat clearer. The one-dimensional integer "vector" is na, where a is the lattice spacin g . We can write an as an = e i�n = eik (an), which describes a plane wave of wa\ e number k . Such waves are known as "spin waves." Thus, E

=

4A(1 - cos ka) ,

k

�

=

a

For small ka, which means long waves, cos ka � 1

E

�

2Aazkz

=

2nl Na

-

(7.42)

.

k1aZ /2,

=

SA sin Z K

For long waves

{

t

+

+

(e - iKxa - 1 )

t

sin z K

+

+

(eiKya - 1 )

t}

sin 2 K

+

so

that (D'Z . D' 1 )O'l x - O' l x(D'Z . 0" 1 )

1 /2J.l

=

(7.44)

The Hamiltonian is N

=

N

- A L (O"n ' O"n + l - 1). n= 1

+

O"ZYO"IY

+

O'ZzO"I .)O' l x - O" lx( -

. .) (7.49)

Using Eq. (7.49) in Eq. (7.47), we can write

(7. 50) •

{

Ii�n,x hO'n,y

: �A(20'n,y - O'n + l ,y - O'n - l )

(7.5 1 ) 2A(20"n,x - O"n + l ,x - O"n - l ,J . We expect the vector O"n t o rotate around the z-axis, and we also expect the solution to be of the form n O'n,x � c sin w te i ak, (7.52) n O"n,y � C cos w te i a\ �

li w

(7 . 46)

= 4A(l

-

cos ak) .

(7.53)

TWO SPIN WAVES When the spins at positions n and nz point downward we define 1 the wave fUnction (()n n . 't' l, 2 l({Jn , .n ) = l a · · · afJa . . . !Y.fJa . . . a ), (7.54) . . . Where the B' s are In posItion s n 1 and nz. The Hamiltonian is the same as that of J:� ..... . 7.46, and the Schr6dinger equation is •

Where

- 2A L ( pn,n + l - 1) n=1

(O'z xO" lx

=

7.5

2 Aa2•

7.4 SEMICLASSICAL INTERPRETATION OF SPIN WAVES =

=

Where a is the lattice constant. When we use Eq. (7.52) we see that

To find the next band of energies, we must consider the case when all t he spins but two are parallel. But first we will digress to discuss the semiclassical interpretation of spin waves. Problems dealing only with spins have often an immediate semiclassical interpretation. Let us look at the spin wave case.

H

(7.47 )

= 2 i(0"IYO'ZY - O"l zO'Zy) 2i(0"1 X O"z )x '

(7.4 5) where

(HO"n - O"nH ).

(7.4S)

.

...}

.

i

Ii

We transform this using the commutation relations :

is a nortlinear equation,

and, in a manner similar to that used in the one-dimensional case, E = - L A M( e iK ' M - 1). and M For nearest-neighbor interactions (in a cubic crystal),

_ 2A{(e iKxa - 1)

= -

We regard this as the classical equation of motion for the vector D'n This ' but it can be linearized when we consider 0"nz to be close to 1 , In WhICh case we can approximate thus :

H ' = - L A M(O'N ' O'N + M - 1 ) N,M

=

.

O"n

( 7. 43)

The extension to three dimensions gives us no trouble : The lowest state again has all spins parallel. Subtracting out the ground energy, we obtain

E

!lfhe Heisenberg equation of motion for the nth spin is

and

(long wavelength).

207

Energy

=

4A e( Z )

(7.55) (7.56)

208

Spin waves

Two spin waves (rigorous treatment)

209

When Eq. (7.56) is used in Eq. (7.55) and the coefficients of ({Jn t ,n2 are compared we can distinguish two cases. When n2 #- n l + 1 , in other words when the two down spins at n l and n2 are nonadjacent, we have 2eant,n 2 = ( ant,n 2 - ant - I ,n) + (a.t,.2 - a.t + 1 ,.2) + (a.t,.2 - an lon2 - 1 ) + (an t,.2 - a. Io•2 + 1 ) ·

(7. 5 7) Fig. 7.2 Two spin waves, or the scattering of two quasi-particles.

On the other hand, when n 2 = n l + 1 the two down spins are next to each other and Eq. (7. 57) does not hold. We then have 2ea.,. + 1 = ( an,. + 1 - a. _ I ,. I ) +

+

(a.,. + 1 - a.,n + 2) '

(7. 58)

Before solving Eqs. (7.57) and (7.58) rigorously let us derive some appro x­ imate results. Because the case n 2 = n l + I is only one out of N (the number of lattice points), for most of the time Eq. (7. 57) holds. Therefore, the first crude approximation should correspond to two noninteracting waves :

and ( 7.62)

The energy is (7.63)

(7.5 9) This energy depends on T to the 5/2 power :

When we use Eq. (7.59) in Eq. (7.57), we see that 2 e( ) = (1 - cos k l a) = e(kl ) + e(k2 ) ,

+

(I -

(7 . 60)

where 4Ae(k l ) is the energy for one spin wave of wave number k 1 . At low temperatures, only the lowest energy modes are excited ; thus for one spin wave, the energy might be,

and for two spin waves E =

ki

2J1

+

f In

k� . 2J1

( 1 - e - PE(k »

Therefore the specific heat is proportional to the 3/2 power of T: C

ex

T3/2 .

This result for spin waves was derived by Bloch. Altho�gh the system is made up of ideal Bose particles, Bose-Einstein condensatIOn does not occur because the number of particles is not fixed.

Pr�blem: When there is an external magnetic field, derive the energy of the spm waves. Find the magnetic susceptibility.

We can form wave packets with group velocity ow/ok, where E = I1w and the phase velocity is w/k. The statements above bring out strikingly the similarity of spin waves to crystal vibrations or phonons. By analogy to the concepts of "phonon," "photon," and so on, we call a spin-wave excitation a "magnon." Since the approximate system discussed above is made up of independent Bo se particles, we can use the expressions that we discussed before. The free energy F is F = kT

U ex T 5 /2 .

cos k2 a)

d 3k . (2n) 3

( 7 .6 1 )

For low temperatures only the E(k) near the bottom are significant, so that we may approximate

7.6 TWO SPIN WAVES (RIGOROUS TREATMENT) Bethe treated the two spin waves in a linear system rigorously. When there are two spin waves in a system, the problem can be looked upon * as a scattering of two quasiparticles, as in Fig. 7.2. For the scattering process the energy and !he m?mentum are conserved, and the eigenstate is a superposition of the InCommg and outgoing waves : (7.64)

. From this point onward we will let a = 1 . That is, for convenience we are elth�r assuming unit lattice spacing or absorbing the factor a in k. As mentioned earlIer, n 1 < n 2 ' The ratio rx/fJ of the coefficients is related to the phase shift

-------

• See A. Sommerfeld and H. Bethe, Handbuch der Physik, Vol. XXIVj2, p. 604 ( 1 933) .

210

7.6

Spin waves

resulting from the scattering, and it is to be determined from the scatterin g equation. The left-hand side, anl ,n 2 ' of Eq. (7.64), must satisfy Eqs. (7.57) and (7.5 8) First , when an 1 ,"2 is substituted in Eq. (7.57), we see that 2 e( ) =

(l - cos k l ) + (1 - cos k2)'

(7.65)

This equation has the same form as Eq. (7.60), although we expect kl now satisfies a different relation from that found in Section 7. 1 . Next we can use Eq. (7.64) in Eq. (7.58). Rather than doing that immediately, notice that Eq. (7.58) is a special case of Eq. (7.57) if we define an ,n (which was left unspecified before) by 2an,n + I = an,n + an + I ,n + I (7.66) Then, substituting Eq. (7.64) in Eq. (7.66) is equivalent to substituting Eq. (7.64) in Eq. (7.58). From Eqs. (7.66) and (7.64) we have 2(a eik2

+

peik l ) = a

It follows that 1 1

a P

-

+

+

+

P

+

ei(k l + k2) ei(k l + k2)

cos (k J cos (k1

+

+

a ei(kl + k2)

_

_

+

2 eik, 2eik2

(7.67)

If k ) and k 2 are real, a P

z

z*

'

cp 2

--

(7.7 1 ) When we impose the periodicity requirement with the period of N, we see that (Because n l + N > n2, we do not write an I + N .n2 satisfied when, from Eq. (7.7 1 ),

or 2 cot - = 2 cp

Nk l = 2nm l

(7.68 )

!( 2

+

=

(7 . 72)

anl ,n2' ) Equation (7.72) is

cp

(7.73)

Nk 2 = 2nm2 - cp,

where ml and m 2 are integers. It must be noticed here that Nkl is not equal to 2rrm l , as was the case for one spin wave, but is changed slightly by the amount of cp. Using Eqs. (7.69) and (7.73) we can eliminate k l and k 2 ' and determine cp as a function of m l and m2' It can be shown that the real values of k do not exhaust the possibilities. We can examine the case of complex wave number by putting +

iv, k2 = U - iv.

Re z = -1m z cos (k) =

(7.70)

From the substitution of Eq. (7.70) in Eq. (7.64), this becomes

kl = U

where cp/2 is the phase of z : cot - =

2 a = eicp/ 2 P = e - icp/

where

Therefore la/P I = 1 , and in fact

2

k)

cot - - cot 2

k2 2

(7.74)

Equations (7.70) through (7. 73) still hold (although cp IS complex ). Eqs. (7.74) we can rewrite Eq. (7.73) in terms of u and v :

With

(7.75)

sin (k ) - k2)/2 + k2)/2 - cos (kJ - k2)/2

cot � - cot

211

Equation (7.69) holds, for such cases, as an algebraic identity or by analytic continuation from the case of real cp. For scattering problems we assume k l and k 2 to be real . Then l a/P I = I , expected, and cp, which is real in this case, is the phase shift after the is s a scattering. One special solution is

{

pei(kl + k2 ) .

2 k 2 )/2 - ei(k l - k2)/ 2 k2)/2 - e- i(kl - k2) / .

Two spin waves (rigorous treatment)

cp = n(/11z - m l ) + iNv. (7.76) From Eq. (7. 75) it follows that u is real. Using Eqs. (7. 74) in Eq. (7.67), and (7 .76) in Eq. (7.68), we obtain a relation between 11 and v :

)

k2 , 2

(7. 69 )

We will also consider cases in which k ) and k2 are complex. Then l a/P I is no longer 1, but cp may still be defined by Eq. (7.68) (cp will now be complex) .

i

e ,,(m2 - m i l e - N v

Now, if we hold v fixed and let N large N

--+

= 00 ,

cos u - e - v cos u - eV

the left hand side

--+

0 ; so we have for

(7.77)

212

7.7

Spin waves

so that v is approximately real. (We have assumed that Re v > 0, but t he following result will be the same if Re v < 0.) The energy is derived from Eqs. (7. 65) and (7.74) ; use of Eq. (7.77) leads to 2 B( ) = 2 - cos (u + iv) - cos (u - iv)

2 ( 1 - cos u cosh v) � sin2 u = t(1 - cos k),

=

where we have defined

k

=

2u

=

kl

+

(7.78 ) (7.79)

k2 •

To get an [ ,n2 we use Eqs. (7.74) and (7.76) in Eq. (7.71). The result is (with a change in the normalization) or

an"n2

=

ei(n [ + n2)kj 2

cosh v

(l'i + 2

n I - n2

)

(7.80)

according to whether m l + m 2 is even or odd. an[ ,n 2 has a sharp maximum when nl = n2 (or nl = n2 - N) and decreases as n2 - nl increases (remember we assumed nl � n2 ) ' Our solution represents a bound state. What is the binding energy for this quasiparticle? When there are two independent spin waves both having wave number K, the energy is (7. 8 1 ) = K 2 - K4/12 + . . . B = 2(1 - cos K)

For a quasiparticle o f wave number 2K, we find from Eq. (7.78) the energy 4 B = t( 1 - cos 2K) = K 2 - K /3 + . . . . (7.82)

Equation 7.82 gives a lower energy than Eq. (7.8 1 ) does, but the difference is very small, namely, K4/4. In other words the binding is very weak, and it depends on K in such a way that the long-wavelength spin waves do not interact much. This is the reason why the approximation of non-interacting spin waves is a good approximation for low temperatures. 7.7 SCATTERING OF TWO SPIN WAVES In this section and the one that follows we will approach the two-spin-wave system in a slightly different way. In Section 7.6 we found the eigenstates of the Hamiltonian containing two spin waves. Here we will consider the simple waves found in Section 7.5, which are not eigenstates of H but describe pairs o f independent spin waves. Then we will apply perturbation theory to calculate the amplitude for scattering of these waves.

Scattering of two spin waves

213

The general Hamiltonian for spin waves in three dimensions is written as H = - L 2 A M( pN ,N - M - 1 ) ; (7.8 3) M,N where � M is a constant. For the one-dimensional, nearest-neighbor exchange interactIOn case, we may write H =

- 2A L ( pn ,n + 1 - 1). N

(7.84)

The scattering of two spin waves will be described in this section for the one­ dimensional case, although a similar analysis holds for the case of three dimen­ sions. We know that a ground state of Eq. (7.84) is I IP ( O»

laa · · · a),

(7.85)

L eikll l lPll)'

(7.86)

==

for which all spins point up. When one of the spins points down, 1 1/Ik )

=

j-iV

is an eigenfunction of Eq. (7 . 84) with the eigenvalue Ek = 4A ( I

- cos k). When there are two spins pointing down, 1 1/1kl , k2 )

=

� I 2 "'-' V N " ' "2

1

eik [ Il ' eik21l2 l m n ) Y t"2

(7.87)

(7. 88)

d�scribes � state in which two spin waves exist. However, Eq. (7.88) is not an eigenfunctIOn of Eq. (7.84), as we saw above, and the two spin waves scatter each other. Let us examine how the two spin waves interact. By definition I �nln, ) = I IP1l 1l , ) is the state where spins are down at n l and n , and I IPll , ) = 0 : 2 ln 2 We want to evaluate � N 2 H I '" "rktk2 )

= � i...

"1"2

eik ' "keik212 H l mnl ) • 't' "2

Here HIIPllln, ) is calculated for four cases : i) nl and n 2 are separated, that I' S nz .... ..... -r. n l , n z -r n l + 1 , n 2 -:f. nl - 1 . ii) n 2 = n l + 1 . iii) n 2 = n l - 1 . iv) n 2 = n l . In this case I IPIl , Il , ) = O. F or case (i), B I IPll , ) = - 2A[(I IPIl , + I ,Il) - 1 <'PII ,n » + ( I <'PII , - I ,n ) - l <'Pn , » ln n ) ) » + ( 1 <'PIl , ,1I2 + I - 1 <'PII ,Il + ( 1 <'PIl , ,1l2 - 1 - l <'Pn, "»)] , (7,89)

214

Spin waves

! 7.8

For case (ii),

H!({)n ,n + l ) = - 2A [( I({)n - I , n + l ) - l({)n,n + I » + ( I ({)n,n + z ) - l({)n,n+ I » ]. In order to conform with Eq. (7.89) we will write the equation above in the form - 2A [ - 2 1({)n,n+ l ) + ( l({)n- l ,n+ l ) - I({)n,n + l » + ( I({)n,n + z ) - l({)n,n + I » ] - 4A I({)n, n+ I ) ' (7.90) where the term in square brackets is Eq. (7.89) with n z = n l + 1 . Similarly, for case (iii), HI({)n,n+ l )

HI({)n,n - l )

=

- 2 A [ - 2 1 ({)n,n - l ) + ( I ({)n,n - z ) - I ({)n,n - I » + ( l({)n+ l ,n - l ) - I({)n,n - I » ] - 4A I({)n,n - l ) '

=

Ix) with

r

( Ek ,

+

Ek ,}l t/lk 1k ) + Ix),

� ei( kl +k 2) n !' l m _� �2 � ":t 'f'n,n + l ) ,

JN n

(7 .9 1 )

ei(k 1 n+k2(n - l ) I ({)n ,n _ l ) m + m _ ei ( k1 n +k2n) ( 't'n,n 1 + 'rn," 1 )] - cos kz )] J N 2 t/lk l k 2 + JN 2 Ix) . (7.93) +

_

4A [(l - cos k 1 ) +

(l

(7.96) (7.97)

I

-

N

(7.95)

'ri.8 NON-ORTHOGONALITY

0 = - 2A [I({)n + l ,n) + I ({)n - I ,n) + l({)n,n+ l ) + I({)n,n - l )] (7.92) + 4A [ I({)n,n + l ) + I({)n,n- I )] . Here the first term is Eq. (7.89) with n l = n z· When we use Eqs. (7.89) through (7. 9 2) to evaluate JN 2 Hlt/lk , k, ) ' the first terms of Eqs. (7.90), (7.9 1 ), and (7.92) combined with Eq. (7.89) lead to the sum of Eq. (7.92) over nl and n z without restriction ; thus ei( k ln l +k 2n2) I({)n l n, ) [( eik l - 1 ) + ( e- ikl - 1 ) J N 2 H 'I'.1.klk2 = - 2 A � £...

=

H l t/lk,k,)

,where

fYhe general theory of scattering tells us that the cross section

�

The "remainder" Ix) is written as (7.94 ) 1 - ei(kl + k 2 » I({)n,n + l ) J N z Ix) = - 4A L ei(kl +k2)n(eik 2 + eikl n The first term of Eq. (7.93) is, as seen in Eq. (7.88), a double sum over I I , whereas Eq. (7.94) is a single sum over n . There:ore, the second term o . Eq. (7.93) can be regarded as expressing the scattermg of the two spm wave that appear in the first term of Eq. (7.93). N otice that when H operates on I t/lk t k, ) a new state emerges, but the neW state is still a combination of states with two flipped spins. In other words, when two spin waves encounter, two spin waves come out.

-

!

(1

satisfies

(7.98)

=

- 4A L [ei(k 1 n + k 2(n+ 1 ) ) I({)n,n+ 1 )

215

With the help of Eq. (7.87) we may write Eq. (7.93) as

=

For case (iv),

HI({)n ,n)

Non-orthogonality

where

,

v

is the velocity of the colliding particles. From perturbation theory,

H' H' Mf i = H �i + L ��

.

+ . . (7.99) n * i Ei - En �. �fi is the matrix element of the perturbation Hamiltonian between the initial �nd final states. In this section the discussion will be limited to the first term ff Eq. (7.99). � We are interested in the scattering from the initial state I t/l kl ,k ) to the final fstate It/lL 1 ,L, ) where f (7. 100) :'Then we cannot use Eq. (7.99) as such, because the states It/lklk,) are not ,orthogonal to each other. This is shown as follows. We compute the inner tproduct, using Eq. (7.88) :

\

(7 . 101) From the definition of I ({)n t n ), we see that

«({)nin,,I({)n, n)

= =

Thus, we can write

1 when n � = n 1 =1= n z = n; or n� = n 2 =1= n 1 = n;, 0 otherwise, including the case n� = n l = n z = n� . (7. 1 02)

(7. 103)

216

Operator method

7.9

Spin waves

In the first sum of Eq. (7. 103), n 1 and n2 are summed without constraints, so that we see

217

When we expand Mf i for small kj we see that IMI 2 (k1 . k 2 ) .

2

�

Therefore, unless kj is large, the scattering cross section is small. =

(7.104)

0 otherwise.

If the second term of Eq. (7. 1 03) were absent, that equation would show the orthogonality of I/Ik l k2 ' If k 1 ' k2 are not equal to L 1 , L2 but

(7. 105)

where "0" denotes any integral multiple of 2n. Then, although the first sum of Eq. (7. 1 03) vanishes, the last term contributes - 2IN. Thus, I I/Iklk, ) are not strictly orthogonal to each other. To calculate the scattering amplitude we consider the first term of Eq. (7.99), using for the perturbation Hamiltonian

(7. 106) Therefore,

Mf i

=

< I/ILIL21(H - (Ekl + Ek2)) ll/Iklk, )

=


(7. 107)

=

n + 1 or

n1

= n

Remembering that

ax

=

+ =

we have

a+

=

ax

-

+

iay 2--- ;

( �) (� �) 0 1

+ 1 , n2

=

a_

and

ax - iay 2

=

ay

-�

=

a_

and

( - �) G �) . 0

(7. 1 1 1)

•

I

,

=

=

(7. 1 12)

(7. 1 1 4) The state l
II

l
=

(fN �

)

eik ' NaN _
=

at l
Further, in Eq. (7. 1 08)

where we define the creation operator of a magnon to be

n

a k+ =

(7. 1 10)

---

(7. 1 1 3) ! + !az• When l
(7. 108) - n, n2 - 1 when n 1 't'n.n + 1 ) < In't'nln2 l in = 0 otherwise.

a

It is easy to show that

where we used Eq. (7.95). Using Eqs. (7.96) and (7.88), we see that

because

7.9 OPERATOR METHOD It is possible to rewrite the Hamiltonian for spin waves in terms of creation and annihilation operators. Using the spin operators ax and ay, we define a + and a_ by

0 otherwise.

=

The destruction operator is then

Thus, using Eq. (7.97), we have

(7. 109) when k 1 + k2 = L1 + L2 + "0". Because of this relation between k and L, and also the relation in Eq. (7. 1 00), Mfi is symmetric in k and L.

ak

_1_ � eik ' Na N-'

� N t:

- )N t:

_

1 � e - ik ' Na N+ ·

(7. 1 1 5)

(7. 1 16)

(7. 1 1 7)

Further, I I/Ikl .k,) of Eq. (7.88) is then

It/lk1k2)

=

a ,;tA;21
(7. 1 1 8)

218

Scattering of spin waves-oscillator analog

7.10

Spin waves

219

To see that this is correct, remember that (J _ (J _ = O.

Equation (7. 1 1 7) may be solved for (IN + as (IN+ =

1

� ik . N

.IN t-

L I = KI + Q L2= K 2 - Q

( 7 . 1 1 9)

ak•

N N+ The Hamiltonian of Eq. (7.83) includes p , M, which contains (IN ' O'N + M o r terms of the form

in which the first two terms can be rewritten in terms of products of two a's. The last term corresponds to products of four a ' s because of the result shown in Eq. (7. 1 1 3). Now let us find out if the a' s satisfy commutation relations such as (7. 1 20 )

The answer is no, and is related to the fact that the I/Ik l k 2 are not orthogonal to each other. We can disprove Eq. (7. 1 20) using Eqs. (7. 1 1 6) and (7. 1 1 7) together with the fact that [(J + , (J _ ] = (Jz : + [ak, aL ]

-

1 � 2 i.. e - ik ' NI eiL ' N [(JN I - ' (JN 2 + ] N N I N2

_

-

1 - L e - i(k - L) ' N(INZ N N

1 - i(k-L) .N ((J � Nz - Uk,L + - L e N N

-

(7. 1 2 1 )

- 1 ),

Fig. 7.3 Scattering of two spin waves. For small Kl, K2 , Mf i approximately by

L 1 N N

-

�

Uk

' L-

L M

i eiQ ' M( 1 - e - M ' K2)(1

(7. 1 23) �

_

e iM ' K I

).

(7. 1 22 )

.....

2

.....

(7. 1 24)

It is plausible to suppose that in a virial expansion we must first consider independent spin waves, for the next term, two-particle collisions, then three, .and so on. Such plausibility arguments can be made rigorous by the use of an "harmonic oscillator analog." In this analog, corresponding to the spin state where the ith spin is down we . wIll take the oscillator state where the ith oscillator is excited to the first level and the others are not excited at all. Hence, 001 00 corresponds to rxrxf3M, and so on. Let rxi; be the creation operator for the oscillators. That is, rxi; creates a phonon (say) in the Nth oscillator. � N + e - iK N v..N .IN N 1

__ _ --

i..

'

Now the real H (for spin waves) acts as follows : H(rxrxf3rxrx)

--+

•

(rxf3cxrxrx) . (Maf3a)

(7 . 1 2 5)

(7. 1 26)

We want now to construct an analog H for the oscillators so that H(001 00)

Figure 7.3 shows two waves of momenta Kl and K interacting and "emerging" 2 with momenta Ll and L • The scattering matrix Mfi is given by

The scattering cross section (J is given

�

7.10 SCATTERING OF SPIN WAVES-OSCILLATOR ANALOG

Mfi = - 2A

..... K2.

+ aK =

On account of the second term in Eq. (7. 1 2 1 ), Eq. (7. 1 20) does not hold exactly. although the second term is numerically of the order of l iN when there are only a few particles with spin down.

2

2

iAt a temperature T, IiK2/2f1 = kT and K 2 ::::; T. From Eqs. (7. 1 22) and '(7. 1 23), (J T2. For ordinary particles, Mf i 1 ; hence (J 1 , and the virial ;coefficient b ..... T1/2. In the case of spin waves, however, (J T2 and thus

because e - i(k -L) 'N -

Kl K

Now

H =

� [ - 2A (ai+;' 1 ai

--+

(01 000) (00010) .

- at et:J - 2A(ai-+:' 1 al - aj+ a;)]

I

A M ( a� + N - et:; )aM - L N

=

L eK a ; a K , K

(7. 1 27)

220

Spin waves

where

BK

=

- L, AM (e iK . M - 1)

CHAPTER 8

M

will do the trick as long as only one p is present. However, consider the following (rxprxprxrx) (7.128) H(rxrxpPrxrx) -. (rxrxprxprx) '

POLARON PROBLEM

However, the H defined by Eq. (7. 1 27) acts differently :

H(OOllOO) -. (010100) (001010) J 2. (000200) J2 (002000)

�

(7. 129)

�

Note that the arrowed functions in Eq. (7. 1 29) have no analog in Eq. (7 . 1 28). Therefore we must add terms to H so that these arrowed terms will disappear : that is, we want to define a new H' = H + L1 such that

H (00100) I

(010100) -. (001010) .

Hence the term L1 that is added to H behaves like

(000200) L1(001l 00) -. J � . -J2 (002000) Then

(7.130)

is the new term we need. For example, rxi+ lrxi annihilates two adjacent phono ns at i and i + 1, and rxt+ 1 rxt+ 1 creates two at i + 1. H ' = H + L1 defined as above is not Hermitian, and terms such as + only rx., + 1 rx.,+ rx ".rx . must be added. H is an exact analog for the real H if states with . ' red : p conslde two are or s one with states if exact one p are considered ; H' is and so forth. . Note that the state (00200) may contribute an energy E/ such that e pEJ IS not negligible, in which case the e - P EI term must be subtracted out when forming the partition function, e - PEn • Problem : Using rx ' S and rx� ' s which satisfy anticommutation relation� will s� l\ � N the problem of states with more than one "particle." What are the difficulties .

8.1 INTRODUCTION An electron in an ionic crystal polarizes the lattice in its neighborhood. The interaction changes the energy of the electron, and furthermore, when the electron moves the polarization state must move with it. An electron moving with its accompanying distortion of the lattice has sometimes been called a polaron . It has an effective mass higher than that of the electron. We wish to compute the energy and effective mass of such an electron. Even without any vibration, there is a (periodic) potential acting on the electron. This potential will be approximated by assuming that its presence causes the electron to behave as a free particle but with an effective mass, M. Hence th� energy of the electron when there are no lattice vibrations is P2j2M, where P IS the electron's momentum. It wil l be recalle d that the dispersion in a crystal is as shown in Fig. 8. 1 . We will be interested in the region described by the almost constant part of the optical branch, that is, the branch for which the positive ions move in a direction opposite to that of the neighboring negative ions. If we consid er the permittivity as a function of frequency (Fig. 8 2), the portion of the curve that will interest us is that part where the lattice . vibrati ons are important. w

-

��----� K

Fig.

8.1 Dispersion in a crystal. 221

222

Polaron problem

8.1

Introduction

of the form

P(x)

=

a

' f d3K 3

�

(2rr)3 a�1 -

223

.

[a(K' a)e, K ' :reK,a +

where a' is a real constant. The charge density is then p(x) = - V , P(x) =

- Lattice vibrations

L-----��----_. w

Fig. 8.2 Permittivity as a function of frequency. Let us take the crystal lattice as being a continuum with a unit cell at each point x. The lattice vibrations are quantized according to Section 6. 1 0, where the transition from a discrete lattice to a continuous one is made according t o Eq. (6.201 ) of that section. Before writing down the Hamiltonian let us consider which vibrational modes actually interact with the electron ; the others may be ignored in this problem. U sing the notation of Section 6 . 1 0, we assume that the electron ' s potential due to an undisturbed lattice is either zero or else is taken into account by reducing the mass as already described . If the lattice is displaced, the potential A V1 (x) felt by the electron is due to the charge density p(x) resulting from the displacement, and this in turn is due to the polarization P(x) : v2 A V1(x) = ep(x) = - eV '

P(x)

(8. 1 )

(where the electron has charge - e). Assume that the crystal lattice has a positive ion and a negative ion in each unit cell. There are then six modes of vibration for each wave number K. In three of the modes both ions move the same distance in the same direction ; as K -+ 0 these modes approach rigid translations of the whole crystal and w (K) -+ O. Such modes do not contribute much to the polarization P(x). In the other three modes the positive and negati\ e ions move in opposite directions in each unit cell, producing a polarization th at is proportional to the amplitude of the mode. For these modes the freque ncy approaches a nonzero value w as K -+ 0, and we shaH make the approximati on that w(K) = w for all K. That is, we assume that the frequency is indepe nde nt of the wave number. Let a +(K, a) be the creation operator for phonons. By analogy wit h Eqs. (6 . 200) and (6.202) (in which A + (k, a) is the creation operator for pho no ns) we see that the polarization (which must be proportional to the displacemen t) is

" f d3K- - L3 [a(K, a)eiK ' :rK ' eK 2 )3 a=

la

(

n

1

,a

- a + (K a)e - iK · :rK · eK* ,a] . ,

To a very good approximation we have one e which is in the K direction Ka (the longitudina� optical mode) and therefore two which are perpendicular to it (transversal optIcal modes). Actually these differ in frequency' and have a fi�equen�y rafIO 0f (8 0 I 8ee ) 1 / 2 , where 80 and 8 00 are the static and high frequency dIelectrIC constants of the crystal s. Only the longitu dinal mode the one with (eK ,a I l K) contributes to p(x) ; denote the creation operator for th;t mode simply by a� and ignore the other modes. Thus

p(x) = ia'

f £li K [aKeiK ' :r - aK+ e - iK ' :r] , (2rr)

3

(8.2)

�o that, from Eq. (8 . 1 ), the electron's potential energy due to the lattice vibrations

IS

=

-

i(J2rrCl. )1 / 2

(1i5W3) 1/4 f -d3K . - -- 1 [a + e - ,K ' :r - a eiK . :r] -

M

(2rr)3 K

K

K

,

(8.3)

where a is a dimensionless constant. It can be shown that

(8.4) Where 80 and 800 are again the static and high-frequency dielectric constants of the crystal. In a typical case such as sodium chloride, a is about 5, and in general it runs from about I to 20. The Hamiltonian for the free electron and lattice is

(8 . 5) Where we are consl'denng ' on Iy one electron with momentum operator P and . ' We are IgnOrI . ' ng the other lattice ' m odes (which do not interact with the electron acCording to our assumptions).

224

Polaron problem

Perturbation treatment of the polaron problem

'8.2

225

To simplify slightly the calculations of the following sections we shall assume the units to be such that Ii = M = w = 1 . Also we shall sometimes use the notation P-K

With the assumptions made above (due to Frohlich*), the problem reduced to that of finding the properties of the following Hamiltonian :

l�

(8.6)

Here, X is the vector position of the electron, P its conj ugate momentum, and at, aK are the creation and annihilation operators of a phonon of momentum K. The polaron problem is not of special importance, but we have now reduce d it to mathematics. The method we shall use to solve this mathematical problem will be applicable to different problems of a similar nature. Our method will be valid for arbitrary coupling, but we will first assume small rJ. and will use con­ ventional perturbation theory to get, first, an answer that we can compare with our results for arbitrary coupling, and second, a description of what happens if the electron goes too fast. 8.2 PERTURBATION TREATMENT OF THE POLARON PROBLEM We wish to find !lEo, the perturbation energy when there are originally no lattice vibrations. The Hamiltonian is H = Ho + H' where

Ho

=

p2 - + L ataK 2 K

H �nH �o + . . . (8.7) H'00 + � ' '-' n E OO E nO where H�n (m I H' ln), 1 m) and In) are eigenstates of Ho and E� is the eigenvalue of Ho corresponding to In). Because H' acting on a state changes the number of phonons, we have H �o O. We are interested in the energy of an electron traveling with the momentum P. Therefore, we consider the diagram in Fig. 8.3. Here the initi al !lEo

=

_

=

=

*

H. Frohlich, Adv. in Physics,

3, 325 ( 1954).

P

Fig. 8.3 Electron with momentum (P - K) and phonon with momentum K. state has an electron of momentum P and no phonons. In the intermediate state the electron' s momentum is (P - K) and a phonon of momentum K exists . Thus the energies Eg and E� are

Eg

=

E�

=

�� ,

(!'- - K)� + 2

1.

(8. 8)

Here " I " in E� is the energy of the phonon (in our present units).

HnO

=

1 /i(v 2 rrrJ.) 1 / 2 (n l L K (at e - iK . X K

- a KeiK ' X)IP; no phonons),

which is 0 unless In) represents a state with one phonon and an electron of momentum P' . If the phonon has momentum K,

Hn o = i(12 rrrJ.)1/ 2 (p', no phonons 1(l/K)e - iK ' X IP, no phonons). Since e - iK ,xIP) I P - K) , P , must be P - K, as expected, so Hn o

and

=

(8.9)

i(12 rrrJ.)1/ 2 (l/K)bp',P_ K '

�

- J2 rrrJ. � 2 (P _-I() 2 1+ _ p ( 8 . 10) /2 1 2 /2 2. The extra factor of 2 is a consequence of the spin of the electron . There are two Intermediate states ":1' th momentum P - K. Generaliz ing Eq. (8 . 10) to three . . !lEo

=

x

dimenslO ns and wrItmg the sum as an integral, we find !l Eo

=

- 2,/2 rrrJ.

f-

_ 2d 3K/(2rr)3__ - 2P ' K + 2) '

K 2 (K 2

(8. 1 1)

226

�

Polaron problem

I

In evaluating this integral, we may make use of the identity

I f (�:� I - I f I f - I I 1 ab

dx . b(�)] 2

(8. 1 2) t�-x + 0 Regarding b = K 2 and a = K 2 - 2p · K + 2, we can write Eq. (8. 1 1 ) as x L'iE; = - 4.J2 rJ.n 2 2P ' K 2) + ( 1 - X)K 2 ] 2 [x( K - 4n.J2 rJ.

dx

0

:

_

[(K

_

Now we use the integral :

E =

- p2 2

- rJ.

p2

--

12

L'i Ei

=

- 4n.J2 rJ. -

Thus, finally,

---

r

"If;: '

m eff

In Eq. (8. 1 3), if P >

m

L'i Ei When P

=

2

(8. 1 3 )

If the perturbation expansion is carried to higher order,* Eq. (8. 1 4) becomes L'iE = - rJ. + 1 .26(rJ.( IO) 2 . This is the perturbation energy when the electron is at rest. When P is sma iL we may expand Eq. (8 . 1 3) in the form

- � [52 � (5 Y 2

- rJ. -

+

p2 12

rJ.

+

2

-" ,

2(1 + a/6)

-

rJ.

+

..'.

+

�

6

.

(8. 16)

+

a/6)

(8. 17)

This implies that the

-

K) 2

p2

+ 1

,

(8. 18)

2

(8. 19)

= P.

the smallest P that can satisfy Eq. (8. 1 9) corresponds to the minimum value 9f the left-hand side, which is .J2, so that

.

(8. 1 4)

rJ.

1

- + -I K 2

0, Eq. (8. 1 3) gives

L'iE ; =

=

K

.J . - I p - rJ. p sm · .J2

=

-

p2

-

.J2, L'iE; becomes imaginary.

2 ,�quation 8. 1 8 can be rewritten as

IoI 8nv/2xdx- x2p 2 ---=

=

tnergy of the electron dissipates by creating a phonon in the same sense as in terenkov radiation. Let us evaluate the smallest value of P sufficient for this Jecay effect. Obviously, the process that could give the smallest possible P is te one in which P is in the direction of K. Consider the case shown in Fig. 8.4. I>' conservation of energy, we know that

d 3K/(2 n)3 . XP) 2 + (2x _ X 2 p 2 Y

4n.J2 rJ. 2 . - I P - sin 8n P .J2

+...

rJ.

�uation (8 . 1 6) shows that the mass of the electron is increased ( l rmes by the interaction with the phonons. That is,

(P

to find

227

.,hen we combine this perturbation with the kinetic energy, the total energy if

t,

d 3K( 2n)3 (K 2 _ 2xP ' K + 2X)2

dx

0

- 4n.J2 rJ.

=

Perturbation treatment of the polaron problem

+

....

* E . Haga Progr. Theoret. Physics (Japan) 11, 449 ( 1954).

...

]

(8.20)

�, .

is the condition for the " Cerenkov" effect to occur, as we might have expected from Eq. (8 . 1 6). Let us next calculate the rate of this Cerenkov effect. This rate is given as (8.21) Rate = L 2n I H roI 2 <5(Er - EJ.

Final states

The initial state i consists of an electron of momentum P, and in the final state (P - K) and a phonon of momentum K. From Eq. (8 .6), we have

,.-e have an electron of momentum

H f i = ( P , , l KI H int l P ) =

P

(8. 15)

-----_.

Fig.

i

K

-

(.J2 na) 1 / 2<5p' ,P _ K' K

� ----• •

P- K

8.4 Electron dissipates energy by creating a phonon.

228

Polaron problem

Perturbation treatment of the polaron problem

where 1 K means one phonon of momentum K. Thus Rate

�1

=

=

)

P Z d 3K - K)Z nrt. b 2n J(2 --2 -- + 1 - 2 (2n) 3 2 .

f "l2 ( P

/

(8.22)

The b-function insures conservation of energy. We may write

d 3K

so that Rate

=

�------�----------. E�

471Kz dK(do.j471)

=

2../ 2 rt. do. dK b - PK cos 4n

f f (

where () is the angle between P and When K9 is a solution of - PK9 cos

KZ 2

() +

+

K _Z9 2

Fig. 8.5 The summand in Eq. (9.7) +

1

=

(8.24)

0,

we may transform

fdK b ( - PK cos + �z + 1) fdK b [t( K - P cos ()

=

Rate

2 -I K9- -----

() 2

- (K9 - P cos W)]

P cos el

Thus Eq. (8.23) becomes =

- f

4../2 rt.

dO.

4� I K� _ P cos

() I

.

(8.24)

When the momentum of the incoming electron P is given, K9 of the created phonon is a function of the angle (), and is given by Eq. (8.24). Solving Eq. (8.24) we obtain 2 . �_ . (8.25) sin () de - 2rt. i cosh Rate = 2 V/2 rt. / ../p z Z P () 0 COS 2 v2

- fPCOS 9=v'2

SIn - 1 •

-

_

_

It should be noticed that since

Pj../2 >

P

../ 2

=

�/

(8.23)

K. () +

71

2+

-1

I , we may write I

•

229

cosh - 1 � .

../ 2

Thus cosh - 1 Pj ../2 in Eq. (8.25) is the imaginary part of sin Pj../2, a factor we obtained in Eq. (8. 1 3). The connection between Eqs. (8. 1 3) and (8.25! ca � be seen as fol lows. When the rate of transition is y, the amplitude of remaJll� n� . wave func tIO in the original state has a factor e - yt/ Z ' and the tIme-dependent has a factor

thus - yj2 is the imaginary part of the energy. Equation (8.25) agrees with Eq. (8. 1 3) because both are for the limit of small rt. and both are based on a perturbation treatment. Although we have found Eq. (8.25) to be consistent with Eq. (8. 1 3), you The situation is may object ; Eq. (8. 1 3) should be valid only when < l E � = E g, or when hold not does (8.7) Eq. speaking, Strictly follows. as d e arifi c when the sum over E � diverges near E g . Schematically drawn, the summand in Eq. (8.7) has the shape shown in Fig. 8.5. The correct way to take care of the region near E g is to write the perturbation as

P ../2.

/).Eo

and Wtake the limit as s e see

-->

=

�

f... n .. E 0O

__

HonHnQ.. . E ,,O + 18 _

(8.7')

_

0.

-

- --- - -- -is -X Z + S2 XZ + SZ •

1

x

x + is

The first term has the shape shown in Fig. 8.6, and to use this term is equivalent to taki ng the principal value when integrating. The imaginary part approaches n times a b-function as s --> 0, because

-1

f OC

- Of:

Thus W e may write lim £� o

--

X

+

.

IS

=

�

B

.:;:- S2

princi pal

dx

= n.

value

(1) X

i71

b(x).

(8.26)

230

Formulation for the variational treatment

Polaron problem

a" FORMULATION FOR THE VARIATIONAL TREATMENT ]pe partition function may be written as

f(x)

e-{JF

Jben /3 ---

�------------� x

__ __ __ __ __ __ __

00 ,

-x -2 X2 + e

Tr [e - {JH]

=

L e-{JE,.

(8.28)

i

the leading term is e-{JEo. Therefore, we claim lim

{J� 00

Fig. 8.6 The function /(x) =

=

231

[-

� In (Tr e -

/3

PH)]

=

Emin.

ius, we first calculate the partition function to evaluate Emin . iarlY interested in the case of large /3.

(8.29)

We are partic­

Using the path-integral representation, the partition function is written as

o

'

(8.30)

'Ie path-integral

representation is used when the Hamiltonian is written in .s of the coordinates and momenta. jl Since H is written in terms of the creation and annihilation operators we .st first undo it and write it using coordinates and momenta. If we quantize tie motion of a crystal, we must choose the creation and annihilation operators SO that h

.. x

(8.31)

..... ---<

Fig. 8.7

The integral

fOO

e 2 _ oo x +

e

2

dx = n.

i

Qpt then the interaction between the electron and the phonons is of the form 1

Thus Eq. (8.7') may be written as �

H onHno

": E O principal part

Hint

0

_

( 8.27)

E On

Equation (8.27) shows the consistency of Els. (8.7) and (8.2 1). Although l Eq. (8. 1 3) was originally calculated for P < J2, the analytic continuation w as tacitly made by raising the pole slightly. Problem :

For low-frequency sound waves OJ =

=

KCs'

Cs being the velocity of sound. Calculate the �ritical value of P for radiat! on

sound. Calculate the direction of the emergmg sound waves as a functIOn energy. Evaluate the rate of emission.

��

j(/l. 1tIX) 1 /2 it / -

j(J2 naY /2 =

�/ [a� e- iK ' X - aKeiK ' X] 1

� /K/ (a � K - aK)eiK ' X

J2 (J2 na) 1 /2 it

I� PKeiK ' X.

We do not find the above form for Hint suitable, because in path integrals we 'lant potentials that are functions of position, rather than of momentum. 'e can exchange the roles of qK and PK either before quantization of the crystal 'lotio n (by means of a canonical transformation) or afterward, as follows. Let a� = - ia_ K . Then if we define q� and p� as in Eq. (8 . 3 1) we find

232

Polaron problem

Formulation for the variational treatment

233

The resulting Hamiltonian (dropping the primes) is H

=

+

iP 2

L (1p� + -tq �) K

+ J"2 (J2

cor) 1 / 2

L K

q Ke

iK ' X

K

( 8 .32)

Thus the Hamiltonian is written in terms of the electron coordinate mom�ntum P, phonon coordinates qK and momenta PK ' We write Tr (e - PH )

f

=

X.

S

(8 . 3 3 )

e - s �x(u) �q l (U) �q z(u) ' "

X( O ) = X(P ) q,(O) = q,( P )

+

=

+

} exp { + � f: f: 4

w 2 q 2)

q (u)y(u)] dU

=

=

f f exp ( -i J:

X

2

i

-

( U

,,'S

e-s

9 (path)

�q (u)

}

y(t)y(s)e - w l t - sl d t dS .

(8.35)

x

• Equation (3.39.) t The complete form

exp (iK '

) exp (2J"2 �n � J: J: 2�2)

(X(t ) - X(s»)e- 1 t - s l dt ds) �X(u)

to

But remember that P

is very large.

0

IX( t)

_

X(.0 1

d t ds.

(8.36)

=

- So) e - So �(path ) S---e - (-S--- --S e - so �(path) ( e - ( S - So »

f

e - so

f

e - so

� (path)

9(path).

(8.37)

write Eq. (8.37) as

where

f

e-s

� (path)

(S

_

So )

�

e - ( S - so >

�(path),

(8. 38)

=

US - So) e - S �(path) . S e - so 9(path)

(8.39)

f

e - so

O

Because for the approximate choice of the Hamiltonian the free energy Fo is

liven by

e - P FO

We can write Eq. (8.38) as

=

f

e - so

�(path),

(8.40)

is written by replacing e - w l t - sl in Eq. (8.35) by ew l r - s l e- wP e- w l r - s l - --. - + 1 - e - wp 1 - e - wp

0

e - 1t - s l

P

where So is an appropriately chosen approximation of S. We use the inequality

e - s �x(u)

X 2(U) d U

P

:J.

=

Since y and q are complex here, we modify Eq. (8.35) to read y*(t)y(s) instead of y(t)y(s). We will need this modification after Eq. (8.36). Also, Eq. (8.35) IS approximate, for e - Pw is neglected.t When we use Eq. (8.35), Eq. (8.33) IS written as Tr (e - P H )

f

f

The advantage of this method is that the path integral over the phonon coordinates can be performed because qK and 4K both appear quadratically (and linearly) in Eq. ( 8.34). We use the result proved previously*, and find [1(4 2

:2 1

=

Actually, we are in error when we treat all the qK ' S as independent real variables as we did in Eq. ( 8 .35) above. The qK ' s are complex, with q k = q - K ' B ut if we went to the trouble of doing it properly, Eq. (8.36) would still come out as it did above. Our aim in using this path-integral approach is to combine it with a variational theorem described before. We write

where the action integral S is

fexp { - J:

ff

'So

B efore choosing S , let us examine the meaning of S in Eq. (8.36). The irst term is analogous oto a kinetic energy, and the second term represents the �tential energy. The special feature of this potential is that the potential at t, 'j

� j

<

234

Polaron problem

where t is thought of as a "time," depends on the past, with �eight �- It-Sl: This is a type of retarded potential, and the significance of the InteractlO� ,:Ith} h e . past is that the perturbation caused by the mOVIng electron takes , time to propagate in the crystal. When rJ. is small, we may take only the first term of Eq. (8.36) as So : So

=

-t

f2 X.

, 18.4 I

Electron

..

:s;

Eo

+

Fig. 8.8

lim 1 <S - So) .

P -+ 00

f3

This gives the old perturbation answer E - Eo

:s;

-

'Y. .

8.4 THE VARIATIONAL TREATMENT We have seen that the use of So for the kinetic energy -t Sg X 2dt gives the per­ turbation result !J.E � rJ. (second-order perturbation) and furthermore this result is shown to be an upper bound (a result proved only by much greater effort with the usual methods). However, it is possible to imitate the physical situation m �ch better with a more judicious choice of So and thus obtain a much better estImate of E. !he next most obvious thing to try for So is the action for an electron bound In a classical potential VeX) . This choice can be shown to be equivalent to the use of . la r, some trial wave function in the ordinary (Ritz) variational method. In particu if one chooses VeX) to be a Coulomb potential, one obtains for E (at large :x ) the same result as that which follows from a trial wave function of the fo rm . e - Kr• If one chooses VeX) to be a harmonic potential, then one obtaIns an . improved estimate for E that could also be obtained with a trial wave funct IOn of the form e - KX2 (at least at large rJ.). However it can be shown that for rJ. less than approximately 6, no VeX) ca n improve on the result V = O ! This indicates that a classical potential is not a very good representation of the physical situation, except possibly at very larg e binding energies. There are two major reasons for this. First , the electron is not constrained to any particular part of the crystal b ut is free to wander. Any potential VeX) obviously tends to keep the eIectron n ear its minimum. . Second ' we see from the form of the exact action that the potentiaI. the . .. . ' With a electron feels at any "time" depends on Its pOSitIOn at prevIOus t Imes, weighting factor e - 1t-sl• That is, the effec � the electron has on the crY tal propagates at a finite velocity and can make Itself felt on the electron at a I ter

-

:

An electron

..

(� Force - K f

Mass = I

235

Particle

f

0

(u) duo

Then, from, Eqs. (8.40) and (8.29), we obtain E

The variational treatment

i

0 M

=

coupled by a "spring" to another particle of mass M.

time. This is less so for tighter binding, where the reaction of the crystal occurs much faster; hence a classical potential might well be expected to be a good approximation only for tight binding. A model that would not suffer from either of the above objections would be one where, instead of the electro n's being coupled to the lattice, it is coupled by some "spring" to another particle and the pair of particles are free to w ander. See, for example, Fig. 8.8. The trial action for the system after the coordinates of the mass M have been eliminated is (8.41) where W

==

J KIM and C'

=

M W3/4.

Of course, we could have seen mathematically that So would overcome the objections raised above without deriving So from any physical model, because we are free to choose any form for the trial action that we please (if it is not Complex). However, the recognition of the physical nature of So greatly helps in avoiding mathematical difficulties when questions of boundary conditions, asymptotic behavior and so on, arise, and also may facilitate computation (see below). We write according to our variational principle (8.42) Where Eo is the binding energy of our model and < e-so " .

Let

I(K, t, s)

= =

) means "average with weight

( eiK ' ( X(t) -X(S» ) S _�iK ' (X(t) � X(S))e � �X _ S e -so gX

236

The variational treatment

8.4

Polaron problem

t, s), then it is easy to find (S - So> : I P P e - 1t - s-=-l-dt ds ) (S - So> = - -S \ f f 0 0 I X(t )- X(s) 1 J - � (J: J: {X(t ) - X(sWe - Wl t - sl dt dS) .

If we can evaluate I(K,

a

For example,

(J: J: IX(t) - X(sWe - Wlt - sl dt dS) J: J: dt dse - Wl t - sl (IX(t) - X(s)n J: J: dt dse - Wl t - sl [ -Vi/ (K,t, S)]K�() ' ( f(I X(t) IX(t ) rrei��:(t) -X(S)l (�:�) f (�� �� I(K, t, s). ) (L� ( IX(t ) � X(S)I

and reduced mass M/(l + M). We have already done the path integrals for a driven free particle and a harmonic oscillator, and hence we may write the desired result immediately. c) Alternatively, we may proceed to perform the integral by a technique* (used in previous lectures for solving the simple harmonic oscillator) that which makes the driven action an extremum consists of finding the path, and noticing that (within a normalizing factor) the desired result is

X'(t),

I

=

=

X(s) l , we can evaluate

In general, for any function f of Xes) !»� by Fourier analyzing !

-

=

=

(K t, s) we will work with the more general I(g, t, s) = (exp [ i f: g(u) . X(u) duJ) , and then will specialize to the case g(u) = K[c5(u - t ) - c5(u - s)] . proceed in several ways to do this general path integral :

may

=

exp

D J: g(t) · X'(t) dtJ . nrrt X(O) 1

We will use a fourth method. Let

X(t )

=

+

00

L an

n=

sm •

13

.

Note that = X(f3). Integrating over all satisfies the requirement that paths is equivalent to integrating over all and all an" Working in one dimension (for simplicity), we first must find So in terms of and

X(t)

f p 2 fP X

To evaluate I ,

We

2

_

0

00

nL =1

13

2

fP fP [X(t) - X(s)Ye- Wlt- sl dt ds o

.�

- - -1 2 2

X(O) X(O) X(O) an " n2rr2 . (8.45) n2rr2- COS2 -nrrt. dt = -1 L an2 -an2 -

0

13-

00

4 n= 1

13

0

a) If we divide "time" into small intervals, we have a problem of the form

f f f · · · dX dX2 dX3 ' "

( - 11 AijXiXj + � BiX;) . (8.43) This integral has been developed* for Aij a positive definite and adjoint matrix. The result is that the expression 8.43 becomes : (rryl2 (Det A) - 1 /2 exp [ - t Lij BiB/A - 1)iJ . (8.44) 1

n

exp

i

s elf­

Here, is the number of time intervals in 13, and it may tend to infinity a fter dividing by the proper normalization to make the quantity approach 1 as B i -+ With this method we must insert the end-point conditions X = Xo 0 by any one of several artifices (such as an added term in the driving force B). b) Alternatively we may proceed by recognizing from the physical m ode l that So can be broken up into "normal modes," that is, a free particle with th e + -4 0 JI . total mass M + I plus a harmonic oscillator of frequency v = .J

O.

n

W2

* B. Friedman, "Principles alld Techniques o/ Applied Math," Wiley, N.Y., 1 964.

237

(8.46) Equation (8.46) holds only approximately because we have made certain approximations that are valid in the limit of high 13. Then

(8.47) where

=

And

* R.

bn

=

I(K, t, s)

i J: g(u)

=

sin

n;u du = iK (sin n;t - sin n;s) .

-

Hg)�e_xp J 2::'= 1 (2:::'= 1 SH � oo exp

-

�

( Ana� anbn21_c!�da 2 Ana�) da 1 da 2 ' "

P. Feynman, Phys. Rn·. 97, 660, ( 1 955).

.

.

.

__

(8.48)

238

Polaron problem

8.4

.

where

J

I (K,

t, s)

=

exp

(

00

n=

As f3 approaches

b2

00 ,

n -

the sum in Eq. (8.49) becomes an integral with -> This integral can be evaluated to give

nnlf3 x, nlf3 (nlf3) dn dx. 2 W 2 It _ sl + 4C [1 e - 1t - s! V I (K, t, s) exp [_ � { WV 3 2 V2 + e -(t+s)V - te - 2 Vt - te - 2 VS] , }] =

->

=

We can now we wanted in Eq. (8.42) as follows : e - IIFo(C) e - So(C) t})X impliesfindthat,the forEo large f3, E �(C) F �(C) -f3e -1 IIFo(C) d!£C e - IIFo(C) - � ( -� f f dt dsI X(t) - X(s) 1 2 e - Wl t - sl ) 3 3 V W WJW 2 + 4CjW .

(8. 50)

=

t

=

•

From Eq. (8. 5 1 ) we get for one dimension

(f: f: (X(t) - X (sWe- Wlt -sl dt dS) 2 _ fll fll dt dse- Wlt-sl ddK I (K, t, s) I o

-2

0

Eo(C)

ll IX(t) - X(sWe - Wl t - SI ) f If 2 \ 11

0

0

=

=

- W).

=

!( V

0, we have

=

K=O

=

3f3� .

vw

2f3

V W as f3

�

-

->

co .

3 in three dimen­ (8.52)

I(K,

t, s)e - 1t - sl g(lt - sl ) .

t, s)e -1t - sl dt ds.

= Here we are assuming Eq. (8.51 ) to hold Let J(K, because f3 is large. Then it is easy to show that

f: J: g(lt - sl) dt ds f:

g(u)

=

2

(f3

- u)g(u) duo

u, f3g(u) ug(u) whenever there is a contribution to the ug(u). We can also let the upper limit of the integral

!�ince � 0 for large mtegral ; so we can neglect be 00. Then

»

l1 e -1t - sl dt ds f Jo ( IX(t) _ X(s)1) f (�:� ��n f: du [ - {�T::u + :�3 e -u du _- V f oo J2n JW2u [( V 2 _ W2)IV]( 1 e - Uv) [II

0

2

=

)

0

�

(8.53)

f: J: (IX(te)-�-:(S)I ) dt ds f (�:� �� f: J:

exp

4f3

For three dimensions we must triple the result (since V . K sions). We find that �

=

= .

To get <S - So > we must evaluate

V2 W2 + 4Cj W. J(K, t, s) is used only in integrals over s and t with large f3. In such integrals there is a contribution only when K 2 W 2 /2 V 21 t - sl is not much greater than unity, that is, in an area of order 2f3 V21K2 W2 (in the plane formed by the s and axes). In essentially all of the regions of integration, the fact that f3 is large implies that e -(t+ s) V, e - 2 Vt, and e - 2 Vs are very small, and we can write K2 W2 4C I (K, t, s) exp [ - { -- It - sl + ---- [ 1 - e- 1 t - s ! V] }] (8.51) - 2 V2 WV3

=

=

Integrating, and using the fact that Eo(O)

_

where

239

=

1 4An)

L

The variational treatment

+

(1

- e - UV)] + u}] (8.54)

_

Using Eqs. (8.52), (8.53), and (8.54), we arrive at the final result : E $ Eo

+

! <S - So > f3

� - W)2 - �� Sooo [ Wi� --j--(eV 2 _ d:2�;V)( 1 e-UV)J 1/2 .

11-:00 4

(V

_

(8.55)

rl .

The best W

P

==

From this, we obtain

=

3 and V

� [(1

W

-

-

=

3(1 + 2aO - P)/3 W), where

,Taking e - P(F - Fo)

W)1/2 - I J . (8.56)

The correct result from perturbation theory to this order is E =

-a

l .26(a/ I O)2.

The best W = 1 and V = (4a2/9n) - (4(In 2 + C/2) - I ), Euler Mascheroni constant = 0.5772, . . . , and a2/2 n - -t(2 In 2 + C) i + 0(1 /a2) . E :s; (8.5 7)

For large

C

=

rl.

which, up to a otherwise.

=

-

-

•

•

1 0 3 , is lower than any other estimate, upper bound or

For intermediate rl. A very great advantage of the method here is that it gives a smooth result for E from weak through intermediate through strong coupling, and it gives the only known reliable results for this problem in the intermediate coupling region. Unfortunately, a little machine calculation is required here to perform the integration in 1 /f3(S). In case you would like to compare some other calculation, we give in Table 8.1 a few results obtained by Schultz* using Eq. (8.55) and a computer.

3

Upper bound of E:

- 3 . 1 333

-

5 5 .440 1

7

9

11

- 8. 1 1 27

- 1 1 .986

- 1 5.710

When using an approximate method, we like to have some idea of the size of the error. Our approximation was to assume that the inequality (e- f ) � e - U > =

=

( 1 - f + �� + . . -)

=

1

- (1) + Hf2)

•

•

.

e - <J> + t(f2) - (f/ ) - . . . .

* T. D. Schultz, "Electron-Lattice Interactions in Polar Crystals," Thesis, MIT, 1 956.

(e - (S - So» o , we get

'If S - So is not small, the correction term,

-

-

So 1 2 ) - (S - So )2 J , 2f3 ;nay not be an improvement, but it should be an order-of-magnitude estimate bf the size of the error in the uncorrected estimate, F :::::; Fo + O /f3)(S - So). !Such an order-of-magnitude estimate of the error in the polaron energy has 'been carried out, and it turns out to be small.

'8.5

� [ (IS

EFFECTIVE MASS

A variational principle along the above lines has not been found for the calcula­

!tion of the effective mass of the polaron because we need paths e-s' where S' ris complex. In fact, for the lowest energy of an electron with velocity U we ( need to path integrate using I. �

=

S(Xt

+

i Ut),

as we would expect in "imaginary-time" space. However, if U were imaginary, then we would again have a real S' and could use the variational method. Now one can show that E(I UI), considered as function of a complex variable, is analytic near l U I = O. Hence E in the neighborhood of l U I = 0 can be obtained by knowing E for imaginary l U I . Therefore, one can deduce that for small velocities the effective mass is given by m

represents an equality. To check the error involved i n this assumption, proceed as follows : (e-f)

=

+ higher-order terms in (S - So).

Table 8.1 (X =

241

For small f, a good approximation is to neglect all but the e - < I > term on the right. But a better approximation (for small 1) is to take ( e - f ) = exp [ - (1) + t( 12) - (1)2)J + higher-order terms.

We may now vary V and W to obtain the lowest upper bound. (These resul ts are the lowest ever obtained as upper bounds for this problem.) For small

Effective mass

8.5

Polaron problem

240

=

1 + tn- 1/2a f oo [ W2r-+«-VZ dWrr2e2)/-rv)0 V3

0

_

_

;-':'-Vr)J 3 /2 '

(8.58)

where V and W are the best parameters found for E( U = 0). It is interesting to note that Eq. (8.58) gives a value for m that is always within a few percent of the total mass, 1 + M = V2/ W2, of the trial model. * *

For more details, see R. P. Feynman, Phys. Rev. 97, 660 ( \ 955).

Introduction : the state function

243

Suppose that VCR) = O. Then (reviewing results obtained in Chapter 1), we have

CHAPTER 9

h =

and

The energy is* ELECTRON GAS IN A METAL

ft2K 2

E = L Si = L -

K

i

9.1 INTRODUCTION : THE STATE FUNCTION

cp

(9. 1)

and

L "i = i

N.

(9.2)

From the Pauli exclusion principle, nl = 0 or I , and the wave function of the entire electron gas must be antisymmetrical. At moderate temperatures, the electron gas is almost in its ground state. That is, the N electrons occupy the lowest N states-almost. For a gas with all spins in one direction (for example, all up or all down), the ground-state wave function is cp =

1

-.IN !

L ( - ltu I (PRI)u (PR ) P

2

2

•

•

•

(9 .3)

t 1 means a sum over all permutations, an even permutation having weigh . and an odd permutation, weight - 1 . PRI , which might better have been wntte n RpI , means a permutation of the electrons. The normalization factor 1 1"; N ! will be derived later. With these definitions of Lp and PRI , it is seen that Eq. (9.3) can b e written as � �P

1 -=

-/N !

x

(9 . 4)

Det [u ;(R)]

N

V

o

ft2K 2

2

--

--

2

=

o

eiK · R ==

----=-

-.Iv

ft2K 2 d 3K ---

-

2m (2n)3

(9 .5)

.

VF(K).

(9.6)

dK =

1 ft2 1 dK = V - - -- Ko5 ' 5 2m 2n 2 3

(9 . 7)

3�.

0

(2n)3

Thus, if P o = N/ V and S = E/ V, 3

_ S - - Po

5

ft2 K �

--

2m

- -1 ft2 (6n2)5/3 P o5/3 = ap o5/3 . 5 4mn2

_

(9.8)

--

It is sometimes convenient to define ro and rs so that I /p o = volume per electron = 4n/3 r5. rs = ro/ao

: where

ao = ft2/Zme2

=

Bohr Radius.

(9.9)

is of course dimensionless, and for most metals, rs is between 2 and 6. The energy is often expressed in Rydbergs where I Rydberg = me4/2ft2 . Th.:: results of Eqs. (9.7) and (9.8), plus, of course, many more, can be ob­ tained from


' r.

Also,

f

2

Ur(Rl )U�(R ) ' " u I(R )uiRI) ' " d3 Rl ,d3R 2 2 2

•

•

•

= o.

'" We assume periodic boundary conditions. t The factor liN ! did not appear in Eq. (6. 1 25) because th enormalization there was l iN ! frp*rp d3Rl . . . d3Rn 1. =

242

f

fKO 2m 4nK (2n)3 K 4nK ±nK Vf O (2n)3

E =

and the energies given by E = L "iSi, i

2m

-+

u(R)

V = volume and F(K) = I if IKI ::; Ko, F(K) = 0 if IKI > Ko , where ftKo is ; the maximum momentum, that is, Eo = ft2K � /2m = Fermi level.

The actual behavior of a metal is very complicated : The metal electrons interact with the lattice, with the lattice vibrations, and with one another. As a first approximation, we will assume the lattice to be rigid and will neglect the mutual electrostatic and magnetic interactions, except insofar as each electron is affected by some average potential due to the other electrons and the periodic lattice. Hence, the Hamiltonian is given by

cp =

'P

244

Electron gas in a metal

9.2

In other words,

f qJ*u 1 (Rz)uiRl) · · · d 3R l d 3Rz . . .

=

1 , etc.

(9. 1 0) Consider a single term of qJ, which is now assumed to be normalized. Such term is 1 /) N ! u 1 ( Rz)uz(Rt) · . . and (dropping the I /)N!),

a

z z - li Vi/2m u1 (Rz)uz(R 1) · · · = u 1 (Rz)u3 (R 3) · · · ( - li Vi/2m uz(R1)).

Since V = 0, * the kinetic energy T equals the total energy E and the plane waves. Hence AK = Ii z K 2 /2m and V

=

li z K 2

t� =

f� liZ K z d 3K F(K) rr = (2 )3

Equation (9. 1 1) is the same as Eq. (9.8). The energy per electron = �(1i 2/2m)K1; has been used.

3

sP a

The velocity of an excitation

v

9

liZ K � -2m -

= !Ii � aKa

IiKo m

=

Po

(9. 1 3)

11'1

where Po is the Fermi momentum (not the pressure). Equation 9. 1 2 is a general result ; if the velocity ofa gas particle is a constant, C, the velocity of sound is c;li For, suppose P is the pressure of the gas, 8 its internal energy density and p its internal momentum density. Then, to first order in the sound perturbation, we have, using standard hydrodynamics, e

A2 U2 (R1),

where A 2 is some eigenvalue, this must be so since electron number one has nowhere except Uz to go. That is, vi operates only on u2 (R 1 ) and does not bother other electrons. The argument used to evaluate the normalization integral now may be used, and since

E

(9. 1 2)

p = - vp

Although it is not a priori evident that

=

Ui

are

=

- C2V ' p

(energy conservation).

But for constant velocity we also have P = t8. Combining these equations we obtain

which represents a wave propagating with velocity c;li The perhaps startling conclusion may be drawn that sound propagation in . the collisionless electron gas is impossible. In other words, the energy of a den­ sity disturbance will be dissipated very quickly. This may be seen as follows : Consider a Fermi sphere (Fig. 9. 1 ).

(9. 1 J )

A

B

2.22/r; rydbergs, where Eq. (9.9)

9.2 SOUND WAVES It is interesting to ask what the velocity of sound Co i s in an electron gas. I t can be found by introducing a perturbation of density P = Pavg + (<'5p) cos kx and determining how it propagates, or more directly by calculating c6 = ap/cp *

Try not to confuse V

=

volume with V

=

potential.

245

where P is given in Eq. ( 1 .53) and P = the mass density (not the number density Po). Then

There are N! terms of the type u1 ( R t )uz(Rz) · . . and thus S qJ * qJ d3NR = N ! for an unnormalized qJ . When ui(R) i s normalized, the normalization factor i s therefore 1 /) N ! . The kinetic energy T can be calculated in the same manner as the normalization integral.

( - liz Vi/2m)u2 (R 1 )

Sound waves

Fermi surface

Fig.

9.1

A Fermi sphere.

246

Electron gas in a metal

Calculation of P(R)

A given excitation of momentum K can increase the total energy of the gas by as small an amount as we please (for instance when an electron is excited from A to B tangentially to the Fermi surface). Thus a great many energy leve ls of excitation exist for a given momentum increase. These energy levels a r e below the energy of a sound wave and hence the sound-wave energy will be quickly dissipated.

247

P(R)/R2

9.3 CALCULATION OF peR) We now remove the restriction V

=

=

V

R)q»

=

--�O�------� R

ij

f

� [U i(Ri )UJ(R2 ) I)

J)

Let

L V eRi ' R).

The reader can easily verify that

(q>, � VeRi'

O.

'Fig. 9.2 Probability peR) of two electrons (of same spin) being some distance R apart.

ui(R2) uJ (R i )]

X V(Ri ' R2 )[ui(Ri)u/R2 )] d 3 Ri d 3 R2 .

�oulomb

interactions are now understandably not important-although the long-range ones are. It is also clear that Eq. (9. 14) could have been written as

What is the probability, peR) , of two electrons (of same spin) being a given distance R apart? =

pea, b)

(q>, �

J(Ri - a) J(Rj - b)q»

is the probability that there is a particular at a and one at b. This is of the same form as Eq. (9. 1 4), so pea, b)

Take V

=

O.

=

L {lu;(aWlu/b) 1 2 - [ui(a)u/a)] [uJ (b)u;(b)]} . ij

Then UK pea b) = '

=

peR)

==

=

f

eiK . a/Jv, and L i

ff �

d 3K d 3 L 1 [ �nr

J v2

If

_

�

K R3

(sin KoR - KoR cos KoR)

(9. 17)

(potentIal energy between two particles at a and b and the probability of two lparticles being at a and h. , Suppose that one wants to know the number n of electrons that are in some region R. n

where

=

� R(R;) = �

f

J(Ri - a)R(a) d 3a

pea)

(9. 1 5 )

Y ea, b)P(a, b) d 3a d 3b.

p'he m,ean potential energy is the integral over a and b of the product of the

and

N2

eiK . (a -bl F(K) d 3K 2 N (2n) 3

; {I - [ ;

=

R)

R(a)

ei(K -Ll · (a -bl] F(K)F(L)

d 3a d 3bP(a, b) J(a - b - R)

4n 2

(9. 14)

J d3KV/(2n) 3, so that

I

(q> � VeRi, q>) f

(9. 1 4)

= =

{I

is the density of electrons at point a.

o

=

f

p(a)R(a) d 3 a

inside R outside R,

L J(Ri - a) i

Fluctuations can also be handled. For example, using the definition of

J}

pea, b), we have .

(9. 1 6)

Equation 9. 1 6 is an important result. For example, it tells us that the hard-sphe re model of a one-spin gas should be easy to formulate, since under most conditio n s the electrons are not very close anyway (P(R) = 0 at R = 0). The short-range

(q>, n2 q»

= =

(q>, � f f

J(Ri - a) J(Rj - b) R(a)R(b) d 3a d 3 bq»

R(a)R(b)P(a, b) d 3a d 3b . PK =

f p a)e + iK . a d 3a (

(9. 1 8)

248

or

PK =

Clearly < PK >

=



0 for K =

(cp,

Li eiK · Ri

=F

=

I PKI 2 cp)

=

Recalling that

V2

N

2

(9. 1 9 )

L c5(R i - a)eiK . a d 3a . i

O. But by squaring Eq. (9. 19), we derive =

J

i where S(K) = J e K ' RP(R , 0) d3R Eq. (9. 1 5) we see

S(K)

Plasma oscillation

9.5

Electron gas in a metal

J

F(K - L)

eiK ' (a -b)P(a, b) d 3a d 3b =

{�

if IK - L I if I K - LI

From

� >

=F

O.

(9. 20)

Ko, Ko,

we can show that S(K) is the shaded volume in Figure 9.3. Calculating this volume, we obtain

( )]

[

K 3 3(2n) 3 K =F 0, for K � 2Ko, _ � + .L !£ S(K) _ - _ 4 K� 1 Ko 6 1 Ko 4 n (9.21) K > 2Ko· S(K) = 0, 2 . Are the fluctuations here Problem : Calculate < I P 1 > for sound waves K greater or less than the fluctuations of the real electron gas ? 9.4 CORRELATION ENERGY

e charge. !he Consider an electron gas with the background of uniform positiv the exclUs lOn­ determinantal wave function for electrons (Eq. (9.4)) leads to Coulom b The volum e radial distribution function peR) plotted in Fig. 9.2. energy is then calculated as 4 e S(K) (9.22) , = peR) d 3 R = 2�

cp

(cp, 11 ;: cp) J �

f�

(cp, � ) - --

0.91 6 e2 rydbergs. (9.23) cp = r rs 'J ij Here rs is defined in Eq. (9.9) as rs = ro/ao with ao = the Bohr radius and (4n/3)rg = the volume per electron. The total energy of this system is usually written as E

K

Using the form of S(K)

�

VS(K) ,

Fourier transform of P(R, O).

d3L F (L)F(K - L) (2 ) ' n3 =

=

where S(K) is the Fourier transform of peR). in Eq. (9.2 1), we calculate Eq. (9.22) as

249

(�:�

=

2.22/r; - 0.916/rs

+

Ce

rydbergs.

(9.24)

The first term is the kinetic energy calculated in Section 9. 1 . ce is called the correlation energy, but actually it is the difference between the true energy E and the first two terms on the right-hand side of Eq. (9.24). Ce is also sometimes called the stupidity energy Es. There are attempts to compute ce. When the electron density is low and rs is large, we see from Eq. (9.24) that the kinetic energy is negligible compared to the potential-energy part. When we distribute the electrons in a regular array, the body-centered cubic structure gives the lowest potential energy, which is Epo! =

-

L2 rydbergs for large rs. 's

Comparing Eqs. (9.24) and (9.25) we see - 0.88 Ce = for large 's'

--l's

(9.25)

(9. 26)

Wigner examined the case of small rs also, and his best guess is* - 0.88 Ce = rydbergs. (9 . 27) 1's + 7.8 When Eq. (9.26) was calculated, electrons were put on a body-centered kubic array with the wave function taken to be a c5-function. We may get a better estimate of energy if we use a different wave function. A calculation was done assigning a Gaussian distribution for an electron around a point of the body-centered cubic lattice and varying the shape of the Gaussian distribution. A reasonable numerical result was obtained. 9.5 PLASMA OSCILLATION In a conductor, electrical resistance increases when impurity ions are introduced, the extra resistivity being due to the scattering of electrons by the extra positive charge. However, the resistivity calculated from the scattering is much larger

Fig. 9. 3

distance K apart. Intersection of two spheres, radius Ko, centers a

'" See Wigner, E. P., Phys. Rev. 46, 1 002 ( 1 934) ; Trans. Far. Soc. 34, 678 ( 1 938). Also see, for correction, footnote i n Pines, D., Solid State Physics, 1 , 375 ( 1 955).

250

Electron gas in a metal

,.5

than the observed value. This indicates that the extra positive charge is part ly neutralized by the higher density of electrons gathered around the positiv� charge. In the Fermi-Thomas model, for example, the variation of the electron density is related to the (slowly varying) potential


C5 n n where EF is the Fermi energy. Thus div D = div E

-

4ne C5 n

so that the dielectric constant is

=

[

6n n e2 V2 - � F

]

Plasma oscillation

whose solution is

v

The potential energy arising from the change of local density is written as

c;

1 2

-Z (C5p) Z cos z kx dx, op

(9. 29)

where is the energy of the Fermi sphere, which depends on the density. Fro m the theory of sound we know that the velocity of a sound wave Co is written as

(9. 30)

where m is the mass of an electron. For the plasma of charged particles, we have another contribution to the potential energy, namely the electrostatic potential . It is calculated from the Poisson equation

_ V2 V

=

- 4ne(C5p) cos kx,

(9 . 32)

( - ep) devol) 4ne 2 k2 (C5p)l cos1 kx dx .

(9.33)

iAdding Eqs. (9.29) and (9.33), we find the total potential energy to be

Let us consider a one-dimensional sinusoidal fluctuation in the (number) density of electrons : P = Pay + (C5p) cos kx. (9. 28 )

- f iPc;

1

=

Obviously, this means that the field of a point charge decays to zero over a distance of ( EF/6nne 2) 1/ 2 (the Fermi-Thomas wavelength), and so does C5n in the neighborhood of the point charge. The point charge is therefor� screened (if positive) by the electrons, which form a small cluster around it. The above applies also to each one of the individual electrons in the con­ ductor, C5n being now negative, so that each electron creates to himself a "hole" in which it moves. This is beside the effect of Fig. 9.2. The same kind of fluctuation of electron density causes the plasma oscillation. It can be calculated as follows :

=

f 2f

cos kx . k2

'The electrostatic potential energy is then

E1

E1

-4ne(C5p)

=

E2 = t
251

(9. 3 1 )

+

E1

=

!2

f(�

c5

Pay

)

+ 4n�1 (C5p)l cos1 kx dx. k

(9.34)

This quantity is equal to the kinetic energy of the wave in Eq. (9.28) : K =

- f - -1

1 2

m OJ1 (C5p)l cos1 kx dx, Pay k

(9.35)

where OJ is the angular frequency in the wave : Equating Eqs. (9.34) and (9.35) we may write

(9.36) where we have put Ck

=

OJk /x ,

I/.nd no is used for Pay ' For the second term of Eq. (9.36) we introduce (plasma frequency, as 1 OJp = (4ne1no/m) / 2 . Then Eq. (9.36) may be written as k

OJ

=

.JOJ;'

+ c6k1 .

OJp ,

the

(9.37)

The minimum energy required for exciting the plasma oscillation is nOJ which is about 10 to 15 eV. It has been observed by sending a beam of electro ;� through thin films of aluminum or beryllium . On� mig�t wonder how this semi-classical treatment fits into the quantum �echallical picture of the degenerate electron gas. To see this we note that the �Implest way of creating an excitation of wave vector q from the Fermi sea IS by the operator ak+ qa" with a certain allowed k, such that k + q is outside

252

Electron gas in a metal

9.6

the Fermi sphere whereas k is inside. Since we see no reason to restrict ourselves to a special k, we form o = A

� f(k)akt + ak • 1... q

=

Lk

(eo + ()(k, q))lf(k) 1 2 +

(()(k,

q)

=

are PK

L (V (q) - V (k - k ' »)f *(k)f(k')

k,k'

. -k --q m

2

)

P. K

and let us try to minimize it with the normalization condition

L If(k)1 2

allowed k

=

1.

Obviously, this amounts to a diagonalization of a matrix A (allowed k's only), Ak,k '

=

(eo - ), +

()(k, q)) bk,k ' + V(q) - V(k - k').

For large q, ()(k, q) seems to control the result so that the favorable excitations are electron-hole pairs, f(k) = (jk,ko ' But when q is small enough, the long range of the Coulomb potential suggests that V(q) controls the result, so that the min­ imizing f(k) is f(k) const. Then the favorable 6 is �

which is exactly the operator for a density fluctuation with w�umber q . We therefore conclude, by this qualitative argument, that for small density fluctuations should play an important role. Let us now see how we can make the microscopic picture more exact.

q,

9.6 RANDOM PHASE APPROXIMATION Bohm and Pines discussed the plasma oscillation in the following way. Hamiltonian is

The

L (j(Ri - x)

=

(9.40)

i

Li eiK · Ri.

(9.41)

The potential energy is Fourier transformed and written as

=

j.)

K ( PI -+ K-/2).1... e iK ' R, .. · --_ I'(HPK - PKH ) - I. � m

(9 . 42)

where PI is the momentum operator of the lth electron. In deriving Eq. (9.42), we used the fact that PK of Eq. (9.41 ) commutes with the potential energy of Eq. (9.39), and also we used the relation

(9.43)

(remember that in our units Ii = n. In evaluating PK ' we note that Eq. (9.42) no longer commutes with the potential part of Eq. (9. 39), so that we obtain 1 � 4ne2 iK ' R, - iK" R, iK " R 'K • ' 2 2 K '• e 2e e PK = - L e iK ' R, [� _(�±_ �L )] - - 1... J i m V K' mK ' I,j (9.44) Now we make a crude approximation :

-

+ 2 K L eiK ' R, [ ' (Pt m _K/ )] 2 I

;:;:;

K 2 pJ

LI eiK . R,

3m

(9.45)

where pJ is the momentum at the Fermi surface. The "3" in the denominator comes from averaging over three directions. For the second term of Eq. (9.44), we separate the K' = K term from K' i= K. The former gives �7re 2 N 4n e2� ' (9.46) PK L eiK · R j = m

(9 .38 )

(Subtracting N eliminates the self-energy term with i

=

Let us examine the Heisenberg equation of motion of PK ' The equation of motion of PK is calculated as follows

q + 2m

253

We will proceed to write the equation of motion of density fluctuations. The Fourier components of the density operator

p(x)

The average energy of the state 6 1 0 > is E

Random phase approximation

V j

m

where n = N/ V is the number of electrons per unit volume. For K i= K', we may write K" RJ' � i(K - K') ' R, , 4ne � ' � 1... . - K K' ei 1... e (9.47) 2 K',j mK , K' * K

L

Bohm and Pines showed that this term is small for the relatively high-density case. When we neglect this term we call the result the random phase approxima­ tion. This approximation is interpreted physically as the fact that the sum Ll is

254

R �

small if are distributed over a wide variety of positions, so that the various compone ts making up the term tend to cancel. It should be emphasized that Bohm and Pines proved this approximation rather than simply assuming it. After these considerations, Eq. (9.44) simplifies to (9 .48 ) with w

i

= p}K 2 /3m 2

The last term gives the plasma frequency

+ 4ne2 n/m.

; = 4ne 2 n /m.

(9.49)

9.7 VARIATIONAL APPROACH The plasma state can be investigated using a variational method. It can be regarded as the state of free electrons modified by plasma modes. Thus, we may use as a trial function CPTrial = exp

[

-

it ClKI PKI 2

J

CPFree'

(9.5 1)

where PK is the Fourier component of the density operator i · PK = Li e K R,. In Eq. (9. 51) ClK is the variational parameter. When we minimize the energy we find that Bohm and Pines 's treatment corresponds to the fol lowing choice of IY.K : 2ne 2 for K < Kcritical � 2 (9.52) K hW p = ClK for K > Kcritical o

{

Thus, the ground state for the plasma oscillation is 2ne2 (9.53 ) I pKI2 CPFree' exp L 1-_ cP I k l < Kcr TlWpK 2 n Variational calculations done on functions ofthe above type* give correlatio s energies which were slightly larger for small rs, but coincide with Wigner' es values for rs ;<: 4. These correlation energy values are also larger than v�lu. obtained by modifications of the RPA theory for larger (and more realIstiC) densities. t

[

-

]

>I< T. Gaskell, Proc. Phys. Soc. (London), 72, 685 ( 1 958). t J. Hubbard, Proc. Roy. Soc. (London), A243, 336 ( 1 957) ; P. Nozieres and D. Pines Phys. Rev. 1 1 1, 442 ( 1 958).

255

9.8 CORRELATION ENERGY AND FEYNMAN DIAGRAMS We will be concerned next with finding the correlation energy of an electron gas. Although there is an easier way to do this (to be shown later) the present method, based on a paper by Brueckner and Gell-Mann* has the advantage of exempli­ fying the Feynman-diagram procedure. The diagram method has proved very valuable and has wide application in quantum electrodynamics. Indeed it was for this latter field that the diagrams were invented (see Chapter 6, Section 12). The problem can be stated as follows : In a metal electron gas, the electrons interact via the Coulomb potential. The Hamiltonian is given by

(9.50)

w

=

Correlation energy and Feynman diagrams

9.8

Electron gas in a metal

H =

h2

-

L vt 2m i

+

e2 1 . L 2 i*i rij -

(9.54)

We wish to find the correlation energy !J.E due to the Coulomb interaction. Equation (9.54) can be converted into creation and annihilation operator language. Let at be an electron-creation operator, that is, a; creates an electron of momentum = hK with wave function e iK R. Then ap annihilates an elec­ tron, and as usual ap,a; + a; ap' = (jp,P" With the methods of Chapter 6, Section 8, Eq. (9.54) can now be written as

P

'

p2 2ne2 h 2 1 a; ap + LQ 2 L L a;, - Qa;2 + Qap2ap, H = L p 2m Q P, P2 -

= Ho

+ HI,

V

---

(9.55)

where the Q-dependence may be obtained formally by Fourier analyzing Eq. (9.54). However, Q is nothing more than the momentum transferred by a Coulomb interaction ; 4ne2 /Q 2 is the amplitude of such a momentum transfer. Several simplifications are possible. Chief among these is the introduction of b and b + , the annihilation and creation operators of holes. At T = 0, all states less than the Fermi level eo = Pl;/2m are filled. We then define b + and b as follows : ap = ap if IPI > Po ap = b:p ' if IPI < Po (9.56) ap+ = at if IPI > Po a; = b _ p if I P I < Po · In other words, if an electron below the Fermi level jumps above the Fermi level we say that a hole and an electron have been created, and so on. We also have *

Some references related to the correlation energy are : D. Pines, Solid State Physics

I, 367 ( 1 955). D. Bohm and D. Pines, Phys. Rev. 92, 609 ( 1953). M. Gell-Mann and

K. A. Brueckner, Phys. Rev. 106, 2, 364 (1 957). D. Pines and R Nozieres, The Theory of Quantum Liquids, W. A. Benjamin, Inc., Menlo Park, California, 1 966.

9.8

Electron gas in a metal

256 =

hj'Y.ro , where V

=

N(4n/3)r6 defines ro (N is the number of electrons, V is volume of the gas) and 'Y. = (4/9n) 1 /3. The momentum, P, will be expressed in Fermi momentum units (Po = 1) and the energy in rydbergs. Po

From ordinary perturbation theory

AE

=

(OIH 1 10)

+

+

J L (H l )o H l )n" n * O Eo - En

Equation (9.57) may be written as +

AE "equals" (OIH 1 10)

(O I H

1

( ) L L Ll! l )"", HJ mn(H I� n * O m * O (Eo - Em)(Eo - En )

1

+

. . ..

- H 1 10)

1 1 ( 9.58) H1 10) + . . . . (OIHI ---- H I Eo - Ho Eo - Ho The quotation marks around "equals" in Eq. (9.58) are to indicate that the +

equation is valid only if a comment is added to it. This comment will tur? out to be surprisingly simple when stated in Feynman-diagram language. We will come to this shortly. The notation of Eqs. (9.57) and (9.58) is familiar. That is, (0I H1 10) = S CP�HICP O dx and

(H l )mn

f cp!H CPn

=

1

dx,

etc., where H O CPn = Encpn"

(9.59)

Before continuing, let us try to see the "sense" of Eq. (9.58). First, it is ea� y to demonstrate the (near) equivalence of Eqs. (9.57) and (9.58). By the ordmary rules of matrix multiplication,

(OIH 1

1

- ----- --

Eo - H o

H 1 1 0)

==

(HI Eo -1 Ho H I ) --

_

_

_

00

(9.60) But, excluding pathological cases,

Hence

f cp!f(Ho)CPn

dx =

f CP!CPn

dx =

DmJ(En) ·

(Eo -- H )mn - Dmn Eo - En ' 1

_

0

and Eq. (9.60) equals

f (En)

1

__ _

_

257

The comment to be added to Eq. (9.58) is that when we mUltiply out all matrices we exclude terms with n = O. One could express this more precisely by writing (1 - 10)(01 )/ (Eo - Ho) instead of I /(Eo - Ho) · Second, it is possible to understand more directly why Eq. (9. 58) represents a perturbation expansion. This will not be demonstrated here except to say that .the key to the demonstration is the matrix identity convergence* 1

(9.57)

Eo - H o

Correlation energy and Feynman diagrams

---

A + B

1 1 1 = - - - B -A

A

A

+

1 B -1 1 B - - . . ..

A

-

A

A

(9.6 1 )

We return now to the mainstream of our problem. Using Eqs. (9.55) and (9.56), we can illustrate the possible fundamental processes allowed by H I by the set of diagrams in Fig. 9.4. A downward arrow represents a hole and an

upward arrow an electron. The space below a dotted line stands for "before the interaction" and above, "after the interaction." Some comments on Fig. 9.4 may prove helpful. Process (a) is the only one ithat can start from vacuum or the ground state. The interpretation of (a) is !that an electron and hole are created, the difference in momentum Q is carried along the dotted line, and another electron-hole pair created, the momenta being such that the total momentum is conserved. In general, each diagram junction represents a two-particle interaction with conservation of momentum. As another example, (c) describes the annihilation of one electron and the creation of two electrons (of different momenta) and a hole. To demonstrate how Fig. 9.4 can be used to calculate the desired matrix elements in Eq. (9.58), we first focus on the second-order term,

The first operator, H I' operates on the ground-state wave function and thus the first operation must be (a) in Fig. 9.4. The next H I operation must get us back to the ground state, because (O I A 10) = 0 unless A cpo = acpo . (The (Eo - Ho) - 1 operation does not change the state.) All the possible operations are given in Fig. 9.4, and it is quickly '" Equation (9.61 ) can be proved as follows. Write

S = I_ - ! B ! + � B l- B -� _ · · · . A A A A A A Multiply

this with A from the left :

AS =

1 -

B I + B1 BI

A

A

1 -

_

A (A + B)S

=

1.

BS,

258

---- -----D

Electron gas in a metal

Correlation energy and Feynman diagrams

1\--7\ >------7\

V=V

(b)

(a)

v < _____

(d)

(e)

V--7\

v-----< >-----7\ (f)

>--------< (e)

(a)

0----------0 ---- ---0----------0 --- --- ---

259

(b)

Fig. 9.6 Two third-order diagrams.

---

-

Fig. 9.7

A

-

- -

fourth-order diagram.

(hl

(g)

>--------�

>--------< (j)

(jl Fig. 9.4 The fundamental processes allowed by Hi '

+

Fig. 9.5 The operations of figure 9.4a and b added together. seen that only (b) added onto (a) can return us to the ground state or to vacuum conditions. See Fig. 9.5. Before discussing the amplitudes or numbers associated with each diagram of Fig. 9.5, let us pause to look at some third and fourth-order diagrams­ Figs. 9.6 and 9.7. Figures 9.6(a) and (b) are both composed of the fundament al diagrams shown in Figs. 9.4(a), (b), and (e).

Q=q

Fig. 9.8 The first diagram of figure 9.5. When we reach fourth order a new ingredient appears. The diagram in Fig. 9.7 illustrates an evidently valid process. Note, however, that the diagram in this case consists of two independent parts-or unlinked clusters. It can be shown that unlinked clusters must be forgotten and not counted (because they involve intermediate states In) = 1 0) in the expansion of Eq. (9.58») . In other tWords, the process illustrated in Fig. 9.7 is not valid and will not occur. The ,statement that unlinked cluster diagrams must not be counted is the afore­ mentioned comment that must be added to Eq. (9.58). With the addition of this statement the quotation marks can be removed from that equation. Returning to the calculation of 8(2) we ask, what is the number associated with each diagram of Fig. 9 . 5 ? First, all fundamental diagrams (Fig. 9.4) have amplitude 2rre2 11 2 / VQ 2 where Q is the Coulomb interaction momentum (see Eq. (9.55») . Now consider the first diagram of Fig. 9.5, shown again in Fig. 9.8 . q , P I ' and P2 are arbitrary but then all other quantities are fixed.

260

Electron gas in a metal

261

Correlation energy and Feynman diagrams

,.8

An examination of Fig. 9.9 . enables us to write down the amplitude im mediately : from Eg. (9.58). From Eg. (9.62), we see that the amplitude for a given q mus t he

( )

2neZ n Z Z } 1 1 . _ q Z q Z Eo - Hn ' vol

where In) is the intermediate state in Fig. 9.8. Now z + q) pt + (P2 - q) z En _ Eo = (P I 2m 2m 2m

AT

= (- 1) x

pZZ 2m

fff e :� ) C � ) Z

n

n nZ

l qi

f(P) =

A

( )

2neZnZ Z 1 1 = --V--qZ qZ q2

+

-m (PI - - P2 ) . q .

(9.65)

To find the total amplitude A T (associated with the diagram of Fig. 9.8) \\ C must sum over P I ' Pz , and q, thus : z �� n = T A q . q q Z + (P P2 ) . q ( n 3 V (9.6(1 )

fff e : y \ \

�:

: �� (i��\ (���23 .

The integral is such that if R stands for the Fermi sphere, then Pz is in R . P I in R, and PI + q and Pz + q are outside R. Summing over electron spiJh gives us a factor of 4, two for each loop. · Consider next the second diagram of Fig . 9.5, shown again as Fig. 9.9.

+

q - Pz )Z qZ

3 3 3 V 3 d q d P 0! pZ f( p I )f(P )[1 - f(PI Z (2nn)9

Here

Note that the energy associated with a hole is negative since we are measurin l! energy from the Fermi level. 1 /(£0 - Ho) is called the propagator, an J _ m/[q Z + (PI - P2 ) · q] is called the propagator Jactor. (2ne zh 2/vol ) : ( I /qZ)( I /q z ) is the coupling. Thus for a given PI > P2 , q the amplitude associated with Fig. 9.8 is

(P I

{�

+

+

-(i:� Pz) . q

q)] [ l - f (Pz - q)] .

if I PI < Po if I P I > Po ·

The introduction of the last factor (involving the J(P») enables us to dispense with the comments after Eg. (9.66) explaining the limits of integration. Now the integrations can go from - 00 to 00 . The sign of each amplitude is determined the sign is - I for as follows. For Fermi-D irac statistics (the present case), every closed "matter" loop. For Bose-Einstein statistics the sign is + I for every closed matter loop. Figures 9 . 1 O(a) and (b) are each considered as one closed matter loop. Figure 9. 1 1 is considered+as two closed matter loops. Hence the sign of the first term in Fig. 9.5 is ( ) and the sign of the second term is ( -). To sum over spins for the second dIagram of Fig. 9.5 gives another factor of 2 in Eq. (9.67).

o (a)

(b)

Fig. 9.10 Each of these is considered as one closed matter loop.

Q�q

Fig. 9.9 The second diagram of figure 9.5.

(9.67)

0 0 Fig. 9. 1 1 Two closed matter loops.

262

Higher-order perturbation

9.9

Electron gas in a metal

263

9.9 HIGHER-ORDER PERTURBATION

9.12.

Consider the diagram of Fig. It should be noticed that the interaction i � always wave vector The perturbing energy contributed from this diagram is calculated (including the sum over spins) as

q.

x

-m -m � 2 2 2 P � ) aq + q . PI + aq + ' 2) (!q +-i P ) +- (�q 2 + q ' P3) ' q

(9.68 )

where the integration is over the appropriate range. Because of the existence of the powers of q in the denominator, diagrams o f this type diverge when integrated over q, even with the restriction on the range of integration. However, the final quantity of interest is the sum of contributions from these diagrams, and the final result is expected not to diverge. In order to arrive at the final result without getting involved in the divergence, is using a small e, the integrations are carried out first modified into including and at the last stage e is brought to zero. The worst divergences comes from diagrams such as those of Fig. 9. 1 3 . These are called the sausage diagrams, and the q's are the same for all inter­ actions shown by dashed lines. Consider the integrals

d3p F(p)(l - F(p + q))e -l tl (q2/2+q .p) F (t) = f (2rr)3 --

q

and

An = -n f W . . . f ro dtl dtn b(tl + t3 + . . . + tn)

(9.69)

.

The simplest case of

•

.

- 00

(9.70 )

A n is

A 2 = ! f�CXJ f�'l) dt2 f b(tl + t2)e- l td(q2/2+q'p de -l t2 1 (q2/2 +q 'P2) 3pJ �-d3p---2 el(2rr) 3 (2rr)3 dtl

X

---

D

�o D

0---0--0

-�O

_ _ __ _ _ _ _

Fig. 9.13 Sausage diagrams.

1

- 00

Fig. 9.12 Diagrams of this type diverge when integrated over q.

l/q 2

1/(q 2 + e2)

e,

q

....

where all integrals are over the appropriate ranges. Thus we see that the contribution from an n-Ioop diagram is

(9. 2) n = f (2rrd3q) 3 (2rreq2h22 )n An(q)( _I)n. In order to simplify A n(q), use the integral expression of the b-function o(t ) = f ro e ·2rrd-u so that Eq. (9.70) becomes A n = !n f elt . . . eltnei(t, - t2+' " q(t )F (t2 q(tn) 2drru c

7

�

iut

- 00

1

(9.7 1 )

V

+ tn)UF

1

q

)'" F

(9.73 )

Electron gas in a metal

264

Use Eq. (9.72) in Eq. (9.71) to write the sum from all sausage diagrams as

f !- 3 (q) (2ne 1£) ( 1 '; " :- (2n)3 n n e �� = f d3 q- f L ( - 1): [2n n f d teiIUFi t)] (2n)3 2n n �q- f dU In [ 1 f dteiIUFit)] . (9.74) f (2n)3 2n The integral over S dt . . . has been worked out for small q and is q ' p)- F(p)[1 - F(p q)] d3p Q (u) = f d t eilUF (t) = f q. (2n)3 �

C =

�

2

tj A

- )"

"

Vq 2

CHAPTER 1 0

2 2

Vq 2

n

+

=

2(-tq 2

q

q

(

� 4n 1 For small q,

Vq 2

+

p) 2

(-tq 2 +

: tan - l �) .

+

+

u2

(9.75 )

Thus Q,/u) is finite for small q . The final answer for the contributions from all the sausage diagrams is

where

=

4IXrs � (l - In 2) In n n2

[

R

=

I -

y

+

S (In R)R 2 S R2

tan - 1

IX = (4j9n) I / 3 . rs

10. 1 EXPERIMENTAL RESULTS AND EARLY THEORY

Just about any material can be brought into a superconducting state, in which there is no measurable electrical resistance, by cooling it below a certain critical temperature. That temperature depends on the material ; it gets as high as 2 l oK for Nb 1 2 AI Ge. Although the vanishing of the resistance is perhaps 3 the most spectacular effect in superconductors, we can perhaps find clues to the ;cause of superconductivity by examining other effects. One such effect is the discontinuity in the specific heat at the critical temperature. The specific heat less the a T3 contribution of the lattice gives the specific heat of the electrons, Ceo Comparison of normal elements with superconducting ones gives the two curves shown in Fig. J O . 1 . For the normal element Ce varies as T, because the thermal energy is k T and only those electrons within an energy range kT near the Fermi surface can be excited. Thus, the thermal energy varies as T2 . Integrating the specific-heat curve shows that the superconductor is lower in energy than the normal element. �

(9.76)

e

SUPERCONDUCTIVITY

�n;e 2 n2

y,

dy

dy !] 2 -

(9.7 7 ) (9.7 8)

is the average spacing of electrons measured in units of the Bohr radius. It should be noticed that a In rs term appears in Eq. (9.76). The final result for the total energy is 2.22 0.9 1 6 + (9.7 9) 0.06 22 In rs - 0.096 + O(rs) ' E = 2 rs 1's Here, the last constant term - 0.096 is the exchange term calculated apart from the diagram sum. It should be remembered that the mathematics in this section summed only the sausage diagrams as shown in Fig. 9.13. Although it is true that each of these diagrams causes the worst divergence, the theory would not be complete if the rest of the diagrams like the one in Fig. 9.9 were neglected. Gell- Man n and Brueckner examined the contribution from these diagrams also and - 0.09 6 is the result; as was expected, it is smaller (for small r.) than the contributi on from the sausage diagrams, but it should not be neglected.

Ce

_

Superconductor

--��-----L--� T Tc

Fig. 10.1 Specific heat of electrons as a function of temperature for a superconductor and a normal element. 265

266

Superconductivity

Meissner Effect. A changing magnetic field produces an electric field, S o something with zero resistance cannot contain a changing magnetic field. I f a superconducting sphere is placed in a magnetic field, the field lines will be force d away from the sphere. If a ring is cooled in a magnetic field until it is a super­ conductor, the flux through the ring will remain after the external source of the field is turned off. Apparently the resistance is exactly zero, for experimental!) the flux remains constant indefinitely (provided the ring is not allowed t o warm up). More surprising is the Meissner effect. If a solid (simply connected) piece of superconducting material is placed in a magnetic field and then cooled belo\\ the critical temperature, the magnetic field is pushed out of the superconductor . Technically, some lines might be trapped i n the object, because some parts reach the superconducting state before others. Furthermore, if the magnetic field is strong enough, it might not be pushed out at all. In such a case, the material does not become superconducting. Its resistance and specific heat are normal. Because of its magnetic domains, iron cannot be cooled into super­ conductivity. The Gibbs function is defined as G = F + PV where P is the pressure and V is the volume. Here the pressure can be taken to be the energy per unit volume required to push the field out. From classical thermodynamics, the Gibbs function has the property that in a reversible change of phase at constant temperature and pressure, G does not change. So the critical field is characterized by

Fsupereond

Experimental results and early theory

10 .1

+

pH;, V

8n

Gsupereond = Gn ormal

Fnormal '

(l0. 1 )

At the critical temperature, Her = 0, and as the temperature decreases Fnormal i ncreases, so that Her increases.

267

London observed, on the Meissner effect, that if n = density of electrons, mass = m, charge = - e and electric field = E, then acceleration is given by - eE = mi,

and the current density j is - nex = j.

In a magnetic field of vector potential A (if we use cgs units), 1 aA

E = - - -c

at

gives a rate of change of current density of dj dt

Hence,

aA -A - · at

.

- A(A - A eonst) = j,

where A = ne2 /me = constant. A eon st is constant i n time, but may vary with position. It is fixed by the Meissner effect, so that no arbitrary magnetic fields can be put into the superconducting region. London proposed that A eon st be taken equal to zero for superconductors. We then must satisfy the boundary condition that jn ormal = 0 at the surface of the superconductor, so that A n ormal = O. We do this by an appropriate choice of gauge, called "transverse gauge." We conclude then that j = - AA . (10.2) Equation ( 1 0.2) implies a modification of the statement that there is no magnetic field in a superconductor. Since v 2A

= _ 4n j c

=

4n AA e

,

(10.3)

there is no sudden drop in the magnetic field to zero when we enter the super­ conductor. In one dimension, for example, A

---+----�--,. H

Fig.

10.2

Variation of critical field with temperature.

ex

exp ( ± )4�A/� x),

where the sign in the exponential is chosen so that A decreases as the distance into the solid increases. The magnetic field penetrates to a depth of order 700 A () c/4-;A). The existence of a finite penetration depth can be determined experimentally by measuring the diamagnetic susceptibility of small drops of superconducting material, or by working with thin films. The above theory suggests where the A might come from and approximately how big it should be. But remember that electrons in a metal are not free, so A

268

10.2

Superconductivity

Setting up the Hamiltonian

269

is not exactly ne 2 1me. The theory can be made to corresp o?� with. reality bet ter where is averaged over the pOSItIOn .uSlllg an appro­ by taking j = priate functio n. For example, in one dimens ion the average might be taken as follows :

-AA

A' A

I,

A ' = normalizatIOn constant fA( x

y)e - I x - y l /� dy .

For "hard" or "type II" superconductors, � is much smaller than the p��etration depth. For "soft" or "type I" superconductors we have large �. ImpurIties make a superconductor hard. . . A further contrib ution by Londo n came from consideratIOn of the quantummechanical electric current, which is e times the probability current : -

ne = - (I/J *VI/J

j

2im

= =

With

;; { 0 I/J *

- ne [I/J *VI/J 2im

A = 0 we get

j When

A

#-

A

V +

=

_

:)

e

(VI/J)* I/JJ

=

_

[(� A e2

me

V +

_

e2A -I/J*I/J me

=

�

-

:) I/JJ I/J}

e

rl/J .

I/J � V I/J o - (Vl/J o) * l/J oJ

0, I/J o changes into I/J . If I/J

j

I/J +

-. ne [ 2zm

=

(lOA)

O.

I/Jo , then Eq. ( 1 0.4) gives ne2

- �

me

=

A -AA .

( 1 0.2)

change We see then that if the wave functio n is "rigid" (that is, if it does not th iS cause could What 0.2). 1 ( Eq. when is introduced) then Eq. ( 1 0.4) implies rigidity? From perturbation theor y, we have

A

I/J

=

� n I HpertIO) I n) , I/J o + L * O En - Eo n

state.. If where Eo is the groun d-state energy and En is the energy of an excited eXCited st r fi there is a gap between the ground-state energy and the energy of the state, then Ell Eo is large and I/J � I/J o· -

Fig. 10.3 Variation of energy gap L1 with temperature.

- (VI/J)* I/J) .

In a magnetic field the momentum operator becomes p + e/e , so that the current is then j

--�-------L--� T

The assumption of an energy gap can also explain the anomaly in the specific }heat. Instead of the energy varying at T2 , a gap would cause i t to vary as i e-A/k T, where L1 is the size of the gap. More direct confirmation of the existence of a gap has been afforded by experiments involving microwaves . The energy needed to excite the material across the gap could even be measured as a function of temperature, and L1 decreases as the temperature increases. It took almost fifty years for the problem of superconductivity to be reduced to that of explaining the gap. In what follows we will explain the gap following the theory of Bardeen, Cooper, and Schreiffer. This theory is essenti­ ally correct, but I believe it needs to be made more obviously correct. As it stands now there are a few seemingly loose ends to be cleared up. 10.2 SETTING UP THE HAMILTONIAN 3 The energy gap is L1 � kT� � 1 0 - eV/electron, which is a small quantitative effect to produce a large qualitative effect. Thus the Coulomb correlation energy is too big, and can therefore be neglected. If this reason for neglecting Coulomb effects seems odd, remember that we are not trying to explain and predict everything about the solid. We are just trying to understand superconductivity. If some effect is associated with an energy that is too large, then we know that effect is not the cause of super­ conductivity. Similarly, as we know that the energy associated with super­ conductivity is small, we can predict that certain phenomena (such as e + e ­ annihilation) are not going t o b e affected much b y superconductivity. If we change the isotope out of which the superconductor is made, the critical temperature changes. Spin-spin and spin-orbit interactions do not change with changes in the isotope, so they should be neglected.

270

Superconductivity

1 0.2

The speed of sound is definitely dependent on the mass of the atoms of t h e superconductor. It is plausible then to guess that superconductivity has so m e_ thing to do with interactions between the electrons and the phonons. Let U s try, then, a Hamiltonian of the form H = L eKat aK + L WKC;CK + L MKK,CK' - Ka t,aK K,K ' K K ( 1 0. 5 ) + L M:K , Cit - Ka � aK " K ,K' a t creates an electron, ct creates a phonon. eK is the energy of an independen t electron of momentum K, and WK (Ii = 1) is the energy of an independent phonon. We measure the energy of the electrons from the Fermi surface, an d holes are treated as electrons with negative e . The last two terms of Eq. ( 10 . 5) represent the interaction between phonons and electrons. Usually this inter­ action is taken to be responsible for resistance. High resistance at normal temperatures implies high M , which in turn implies a special propensity towards superconductivity. There are many effects that cause large energy changes, but which are easily understood and have nothing to do with superconductivity. For example, the effect represented by Fig. 10.4 changes the energy of the electrons by an amount larger than the gap, but does not produce a gap. The diagram in Fig. 1 0.5 just corrects the properties of the phonon.

Setting up the Hamiltonian

271

H

Fig. 10.6 A two-electron process. K 2' K 1'

Fig.

10.7

One electron distorts the lattice, the other is affected by the distortion.

We see, finally, that the superconductivity must involve more than only one electron or one phonon. We must consider diagrams such as that in Fig. 1 0.6. Physical ly, this diagram can be interpreted as a consequence of one electron distorting the lattice, which in turn affects another electron. Figure 1 0.7 shows that two electrons K1 and K2 interact via a phonon, causing K{ and K). to come out. For the momenta the relations are K{ - K1 Fig. 10.4 An effect that causes large energy changes. The initial energy IS Ein itial virtual energy is

Q,

K2 - K'2 Q. eK I + eK 2 in the intermediate region, and the

eK; + eK l + IiwQ • The perturbation energy due to the mechanism of Fig. 1 0.7 is Evirtual

Fig. 10.5 An effect that corrects the properties of the phonon.

=

( 10.6)

272

Superconductivity

10.3 K ; ....t------

K2'

K ,'

A helpful theorem -----�

273

K2

.1------ K 2 ...

-"''-

Fig. 10.9 A case where IKI - K{ I is large.

,-�-...Q

- - ---------

���

------------

Fig. 10.8 Another two-electron process, of similar type to Fig. 1 0.7. Figure 10.8 shows another mechanism of a similar kind. The perturbatio n energy coming from this mechanism is

Vz = MKtK;

1

M:,K2 ' (CKt + CK,) - (CKt + CK 2 + hWQ)

Looking at Eqs. ( 10.8) and ( 1 0.9) we see that for overall attraction when � cK " larger values of IK, - K{ I are more favorable. For example, look at the case shown in Fig. 1 0.9. We can summarize these results by writing a new Hamiltonian for the electrons alone. The phonons have the effect of modifying the CK for the electrons and of modifying the interaction between the electrons so that the interaction is under certain circumstances, attractive. We will be working with th� Hamiltonian IlKt

( 1 0.7)

For the calculation of the perturbation energy, the initial state and the final state have the same energy : or The perturbation energy is the sum VK ; K2 ;K t K2 ' We are mainly concerned with electrons near the Fermi surface, so we can take all c's to be approximately equal. Then 1 ( (10. 8) VK\ K ,;K t K2 � - -- MK2K2MK*I K t + MKtK;MK*1K2) . ,lWQ

If the M's are approximately equal, the perturbation energy is negative ; so the electrons near the Fermi surface attract each other. In metals, the Fermi surface is curved. If we disregard the curvature of t he surface, the Hamiltonian can be treated exactly. However, when we do that V.e find no superconductivity. Therefore the curvature of the Fermi surface is essential for superconductivity. The Coulomb interaction among electrons gives the interaction energy : 4nez ( 1 0.9) - ------::----VcouIomb K{ I Z + (const)2 I K, _

which is always positive, so the interaction is repulsive. The "(const)2" is a consequence of shielding. In order to achieve overall attractive interaction between electrons we need v + Vcoulomb < O.

(10. 10) Our problem will be to discover how Eq. ( 10. 1 0) can lead to a ground state with especially low energy. 10.3 A HELPFUL THEOREM

Consider a Hamiltonian

H = Ho + U. The eigenvalues and the eigenfunctions of Ho are written as Ei and CPi' Uij

= (CPi, UCPj) '

If all the E;'s are nearly equal, and also all the Uij's are nearly equal, then a large amount of lowering of energy can be achieved. This is shown as follows : Try a wave function and evaluate the energy expectation value for this state : � = L Eil a ;l z + L Uijaiaj. i ij Following the assumption let us put

Ei The normalization of the a /s is

�

Eo ,

Then (10. 1 1)

274

Superconductivity

Ground state of a superconductor

10.4

Suppose there are m states of the required nature ; what is the best choice the a ;'s that makes ( of Eq. ( 1 0. 1 1 ) a minimum ? It is* for

ai = when Eq. ( 1 0. 1 1 ) becomes

of

I /J�l,

275

D D D D k I'

k 2'

k"I

k 2"

Fig. 10.10 k-states arranged with pairing states side-by-side.

( = Eo - m V.

Thus, when V is positive, or Uij is negative, the stabilization is m-fold intensi fied . 10.4 GROUND STATE OF A SUPERCONDUCTOR With the theorem of the previous section as a guide, let us proceed to find a set of such
both occupied or both empty, and the configuration i n which one of them is occupied and the other empty is not allowed. When we make this requirement on the configurations we can show that the sign in Eq. ( 1 0. 1 2) is always positive. Suppose all the k-states are arranged in such an order that the pairing states sit side by side as shown in Fig. 10. 1 0. Using the rule ( - I t + N' and the fact that N = N' = 0, we see that the sign is always positive. Because Vk;k2,k I k2 should not vanish, we require for every pair k;

+

k; = k l + kz = p = canst.

This constant p can be chosen as any vector, but for the ground state we choose

where

p = We can consider this sum to be restricted to, say, k; > k; and k 1 > k 2 . Remember that ak I ak2 = - ak2ak, and at; at:i = - at,at; . The two states con­ sidered differ only in k{, k� and k 1 ' kz. This matrix element is (10. 1 2) The ± sign comes from the ordering of k's, and the rule is easy to obtain from Fermi statistics. The sign is ( _ I ) N + N' where N = number of states between k l and k z in the function describing the initial configuration, and N' = number of states between k; and k; in the function describing the final configuration . For example,

O.

Thus, the pairs we consider are

How about spins ? In computing the potential, we included the diagram i n Fig. I 0. 1 2a but neglected Fig. 1 0. 1 2b, plus and minus. The sign i s + for states that are symmetric in space, that is, for states that are anti symmetric in spin . To make the potential a s large and negative a s possible, w e take the states with antisymmetric total spin, that is with total spin equal to zero. We choose

< k" , k;, kp , k; W lk l , k" , kp , kz > = ( - I ) I + Z Vk;k 2; k , k2 '

x I

Bardeen, Cooper and Schrieffer proceed to show that if a particular set of configurations is chosen, we can make Vk;k2,k I k2 in Eq. ( 1 0. 1 2) all real and nega­ tive, and the sign in front of it always positive. The idea is to take the states in k-space always in pairs. This means that the pair of states k l and kz are either * To prove this, use the Schwarz inequality :

with equality if (a i ' a 2 , chosen real.)

. . .

)

ex

( 1 , 1 . . . ), that is, if the

' ai S

are equal (and may be

x

I

Fig. 10.11 Pairs in the k-space.

Superconductivity

276

Ground state of superconductor (continued)

10.5

277

Let and

(10. 1 5)

Then (10. 16) (a)

If

(b)

Fig. 10.12 Included and excluded states in the potential calculation.

then

To simplify the notation, the pair (k'i ; - kl) is denoted by "k" in this section. Consider all the pairs in the k-space and arrange them in a certain order The Uk 'S for the unperturbed state are

A pair ki is either occupied or unoccupied. Let us define :

for k

=

the wave function for kj and - kl both unoccupied, and I C)JK( l » the wave function for k j and - kl both occupied. l C)Jk(O»

Then a wave function that is a possible candidate for the ground state is ( 1 0. 1 3) where and The actual ground state is some linear combination of states of the form given in Eq. ( 1 0. 1 3), but we will make the simplifying assumption that with appropriate choice of Uk and Vk , Eq. ( 1 0. 1 3) is the ground state. By adjusting the phase of l C)Jk( l » and l C)Jk(O» , we can take Uk and Vk real . Our program wil l first be to find the ground-state energy by varying Uk and Vk = )I=- Uf. Then we will find the energy of the excited states, which wil l turn out to be a finite, macroscopic amount above the ground state. The energy of a candidate for the ground state is E

=


=

(10. 17)

L Vk 'k L
where all sums are taken over half the k's and where V� 'k = Vk ' , - k ' ; k, - k . Then E = L ek
for k

<

>

kFermi. kFermi.

The best choice of the ground state wave function is obtained by minimizing Eq. ( 1 0. 1 7) with respect to the Uk 'S, Vk being a function of Uk . In the minimiza­ tion it is not necessary to fix the total number of electrons explicitly. By varying Uk and Vk , we vary the number of electrons. If we choose an appropriate zero of the energy ek ' we wind up with the correct number density of electrons after minimizing the total energy. We wil l take the zero of the energy to be at the Fermi surface, so that eFermi = o. 10.5 GROUND STATE OF SUPERCONDUCTOR (CONTINUED)

Following our program we must next minimize Eq. ( 10. 1 7) with respect to the Uk 'S. A set of equations results :

(

)

U� = 4ek Uk + 2 L Vkk, Uk,vk ' Vk k' v� Because Vw is real and negative, we can define and we write Eq. ( 1 0. 1 8) as

Llk = - L Vkk, Uk, Vk, k'

>

0

O.

(10.1 8)

(10. 19)

( 1 0. 20) Introduce x as

u i, = -t(1 + x)

( 1 0.21)

278

1 0.6

Superconductivity

so that

1 - ui

and

1 - 2ui

=

t( l

= - x.

x)

-

Excitations

// /

I / I f I \ \

( 10.22)

Square Eq. ( 10.20) and use Eqs. ( 10.2 1 ) and ( 10.22) to solve for x2 :

\ I I I

\

\

"

..... ....... -----_/

where we define Je; + Ll; . Taking the right sign we obtain from Eq. (10.23) : Ek

=x=

- (1 - 2Ui)

>

so that for ek < 0 (below Fermi level) x Putting this back into Eq. ( 1 0.2 1 ) we have

( n = ! (1

U 2k

( 1 0.24)

==

=

�. 1 2 2

_

+

_

ek Ek

)

0 and Uk

) ek ) .

ek_ Ek '

U�

-+

(to.25)

Vk , as it should be.

>

Fig. 10.14 The range of nonzero Vkk', as assumed by Bardeen, Cooper, and Schrieffer.

Ek

-+ 0 far outside the Fermi sphere (ek

(10. 1 9), we shall find Uk Vk . Because Eq. ( 10.26) gives Uk V.k

Eq. ( 10. 19) can be written as ilA k

( 10. 26)

1 deep inside the Fermi sphere (ek

< >

0) 0).

These limits suggest the distribution shown in Fig. 1 0. 1 3 . U�

( 10.27)

=

=!

J1 2

_

� V.kk '

k - £...

ei Ei

=

� 2Ek '

(10.28)

Llk ,

/

(10.29)

2...; ek2 ' + Llk2,

This is the equation to be solved for Llk • In order to obtain some insight, Bardeen, Cooper, and Schrieffer assumed Vkk' is a constant for k and k' within the range liwer above and below the Fermi surface, and zero otherwise. See Fig. 1 0. 1 4. When we make this assumption, Llk of Eq. (10.28) becomes a constant independent of k, and the equation for it is

1 = - V Ik

1

2Jei + Ll2

Changing Lk into an integral over e, we have 1

=

fliwer M (e) de I 2 I -liwer Je2 + Ll V

2

�

I V I M (O)

Iliwer 0

de Je2 + Ll2

.

M(e) is the density of states for energy e ; it is close to a constant M(O) near the Ferm i surface. Solving for Ll, we obtain

�----�- - ------- l

I I+I I I I I

//

I / / /

To complete the calculation we have to solve for Llk• If we are to use Eq.

U� is the probability that the pair k is occupied. The form of U� in Eq. ( 10.26) shows qualitatively that the distribution U� is a rounded Fermi distribution. Assuming Llk is small, we see U�

./ hwCT "< \ \

\ \

( 10. 2 3 )

and

.......------ ......."

279

Ll Unperturbed distribution

U: in

Eq. 10.26

�O+-----------------.LFe-rm-i�==�======--� ·k

Fig. 10.13 The rounded Fermi distribution.

=

liwer

sinh ( I /I V IM(O))

(10.30)

10.6 EXCITATIONS

Now we are ready to find the excitation energy. To describe excited states we will have to consider the possibility of a given pair of momenta being half­ occupied. Define (10.31)

280

)

Superconductivity

Then It/I ) is some linear combination of the four following orthonormal states : II

It/lk(O)) = Uk
If t/lk , =
=

ek , ( 1 - 2Ui) - 2

(�

=

( )

=

To find the energy of a system at a finite temperature, we will use :

e� , + Il� , Ek ,

( 1 0.33)

If t/lk, =
E

= L (Probability of state i) i

x

(energy of state i).

Describe a state by the product Ilj I t/lk/n) , or by the sequence n 1 > n 2 , , where nj = 0, 1 , 2, or 3. I � Pk(n ) is the probability that I t/lk) = I t/lk(n» , then the probability of a state IS Ilj Pk/nj). Of course, LII Pk(n) = 1 . Then .

E

=

+

But

•

•

[

� • • • 1} Pkin) t ek,sk,(n/) + f; Vk,kmtk,( n/)tZjnm)]

1I 1 '1I

=L /

= Ek, .

=

Expectation value of the energy

)

( )=

_

10.7 FINITE TEMPERATURES

Vk,k' Using Eqs. ( 10. 1 9), (10. 25),

k, ek , � - 2( 11k , ) Il E Ek , 2 k,

281

Finite temperatures

Vk , kUkVk Uk Yk , .

We have used the fact that Vkk, = Vk�k and (10.28) we see that

E ' - Eo

10.7

( L. . n ) L )� � (II ' � " Jt II ,

.

not n,

/

.

] i' /

Pkin)

II,

Pk,(n/)ek,Sk,(ll/)

Pk in)

not n" nm

lI m

Pk,(n/)Pkjnm) Vk,kmtk,( n/) tZjnm) '

1, = =

and similarly,

( 1 0.3 4)

Note that Ek , > Ilk,. Suppose t/lk l =
=

Thus,
=

Ek l + Ek2 ·

=

t � Pk,(n)ek,Sk,(n) + /� t;. Pk,(n)Pkjn' ) Vk,kmtk,(n) tZjn').

(10.36)

Suppose we call fk the probability that a k state is excited. Then the prob­ ability that a given pair of k states is unexcited is

because Vk,k 2 Uk , Vk , Uk2 Vk l is an infinitesimal quantity. Thus, Etwo excitations - EO

E

( 1 0.35)

If t/lk , = Vk l
Pk( O)

= (1

- J,y.

( 10.37a)

The probability that one of the states (say k) is excited, and the other is not, is Pk(2)

=

Pk(3)

= j,,( J

- j,,) .

( l 0.37b)

The probability that both states of a pair are excited is ( 1O.37c)

10.7

Superconductivity

282

t,,(I ) = - V"V", t,,(2) = t,,(3) = o .

S = entropy = -k L probability of state i i

Thus, from Eqs. ( 1 0.36) and ( 1 0.37), E

= L 28,, [ V � (1 - 2j,,) + I,,] + L V"" , V" , V", V" V,,( l - 2/",)(1 - 2j,,). " "'" (10.38 )

=

For a given set of orthonormal states V" and /", define a set of Pi = prob­ ability of state i. Then

o

= L EiPi i

S = -k L Pi In Pi (k = Boltzmann constant) i

F(Pi)

= E - TS.

(10.39 )

If the Pi are proportional to e - E;jk T, then F(Pi) = free energy. It is easy to prove that if F(P;) is minimized with respect to Pi (subject to the condition that L Pi = 1), then the Pi have the correct values for temperature T. Using the method of Lagrange multipliers, we find that o

=

8F 8Pi

_

Pi A L = Ei + kT (In Pi 8

8Pi

+

1)

_

" 1 ,n2, . . .

-k L L P,,(n) In P,,(n). "

n

Set

=

{)F {)j"

=

{)E {)j"

_

{) TS {)j"

= 28,,( 1 - 2vD - 4 L V""' V,,, V" ' V"V,,(l - 2j",) + 2p- l In � . '" 1 - j" (10.40) (1 0 .4 1 )

Then

( 10.42) If we now take

,.

we have (n J + n 2 + n3 + . . . )P(n J ' n 2 "

In [probabilty of state i]

)

Then Pi = e().jkT - l )e - E;jkT. A must be chosen so that L Pi = 1 , in which case the Pi are correctly given for equilibrium. For the case of superconductors at a finite temperature, minimizing F with respect to Pi is equivalent to minimizing with respect to V" and /". Before we do this, note that we are really considering an F that has an undetermined number of electrons. We then minimize subject to the condition that the expectation value of the number of particles is a given number. So, for example, Pi should be written P (n l , n 2 " . . ) where n l , n 2 , . . . are the number of particles in state 1 , 2, . . . respectively. In addition to the condition that

L

x

It follows that

Here we assume that vi + V � = 1 , but we do not assume that V" is such that E is a minimum for zero temperature. V" and /" are functions of temperature.

E

283

The method of Lagrange multipliers then gives what amounts to the chemical p otential of Chapter 1 , Section 6. When working with superconductivity, we do not explicitly write the chemical potential because it is taken care of by an appropriate choice of the zero of the energy.

From Eqs. ( 1 0.37) and ( 1 0. 1 5), S,,(O) = 2 V � , S,,(I) = 2V � , S,,(2) = S,,(3) = 1 ,

Finite temperatures

. . ) = fixed number.

we have (in principle) equations that can be solved for both V" and /". Now, define

(10.44) Then Eq. ( 10.43) becomes identical to Eq. ( 10.20). All our results from Eq. (10.20) through Eq. ( 1 0.28) are true at finite temperatures. Equation ( 1 0.41) becomes
8,,(1 - 2vD + 2V" V"�,, = 8"

(�) E"

+

�" 2 - �" = 2E"

E",

( 1 0 . 45)

where we have used Eg. ( 1 0.44), then Eqs. ( 1 0.25) and ( 10.28), and finally Eq. ( 10.24).

284

10.8

Superconductivity

Real test of existence of pair states and energy gap

285

The final result is Ek , , � Vkk, Ilk ' tanh Ilk = - .i.J k' 2kT 2Ek, -

Ei

=

ei

+

To simplify this equation, assume <

Vkk' = - V

(10.46)

Il�.

0 for l ek l

fliwcr

<

---=0+----'-- TITc 1 .0

nwcr.

Then Ilk becomes independent of k, and Eq. ( 10.46) reduces to

1 M(O) V

=

0

Fig. 10.15 1l(T) as a function of T I Te.

E de tanh ' E 2kT

(10.47)

SiWcr

S<:' hWcr

where M(O) is the number of states per unit energy at the Fermi surface. The "2" in the denominator of Eq. ( 10.46) drops out because de and de are equal. When T -+ 0, Eq. ( 1 0.47) gives 1 __

M (O) V

�

. smh 1 nwc r - 2 _

__

Il

fliwcr 0

de exp [ - v'e 2 + 1l 2 jkT ] . v'e 2 + 11 2

2.52

(10.48)

�o���-----+--. ��

Assuming M(O) V is small, we find 1_ __

M(O) V

�

2 In nwcr .,

1l(0)

1l(0)

= Il

at T = O.

It is easy to see that as T increases, Il decreases, There cannot be a value for Il at all temperatures, because in the limit of high temperatures we get an equa­ tion independent of Il :

and this equation may simply not hold. In Eq, ( l0.47), Il appears only as a square (in E = v'e 2 + 1l 2 ). The effect of decreasing Il is to decrease E(e). But E(e) cannot be lowered below the valu e it has for Il = 0, The critical temperature, then, is at Il = 0, above which there is no solution for Il . Putting Il = 0 in Eq. (10.47) we have

1 M(O)V

�-

=

fliwcr 0

( ) (

de e tanh e 2kTe

�

--

�

Fig. 10.16 Specific heat of a superconductor, Ce... as a function of T/Te.

(10.48')

)

nw In 1 . 14 _cr . kTe

( 10.49)

Equating Eqs. (10.48') and ( 10.49), we get,

(10.50) 21l(0) = 3.52kTe • We write "21l(0)" because in experiments in which a superconductor is excited, we create both a hole and an electron. Equation ( 1 0.50) is experimentally

fairly good. Let T = Te r. As r increases from zero it is clear that if Eq. (10.47) is to remain true, E must change from e by an amount proportional to r . But E = v'e 2 + 112 � e + 1l 2 /2e, so 11 2 is proportional to r . That is, if T � Tc' 11 2 = A(Tc - T) -

for some positive constant, A. We conclude that the general shape of 1l(T) is as in Fig. 1 0. 1 5. Another check with experiment is the specific heat whose electronic part is of concern to us. The normal specific heat Ce•• is proportional to T: Y

Ce,� =

yT,

being a constant. For the superconductor, Ce•s is plotted in units of y Tc ' It is as shown in Fig. 1 0. 1 6, which is in agreement with experiment. 10.8 REAL TEST OF EXISTENCE OF PAIR STATES AND ENERGY GAP

Any phenomenon in which scattering of electrons is involved will serve as a test for the existence of the pair states. Attenuation of phonons and para­ magnetic relaxation are examples.

Real test of existence of pair states and energy gap

10.8

Superconductivity

286

287

I.fJk, (2) and t/lk wave function is t/lk' element for the transition is

Fig. 10.17 Electron scattering.

In such phenomena the interaction part of the Hamiltonian includes thl form (10.51 where a is the spin coordinate. Equation 10.51 represents the amplitude for a J electron with ka being scattered into k' a', as shown in Fig. (10. 1 7). When thl pair states proposed in the BCS (Bardeen, Cooper, Schrieffer) theory exist, ; scattering of an electron kj induces an interference with the paired electron a - k!.

The coefficient B usually has a symmetry like that shown in Fig. 10. 1 8 There are two cases to be distinguished :

Bk'u';kU = B_k;-u , - k',-u' Bk'u' ;ku = - B_ k, -u;-k',-u'

(Case I), (Case II).

(10.52

In the ordinary scattering case we write a table :

Initial final

I

kj

k' j

occupied empty

empty occupied

That is, we consider the scattering of an electron from kj to k'j. For the BCS case, the table should include kj and - k! together. When we speak of the scattering of a kj electron into k' f, we mean that we take the initial wave function with t/lk = I.fJk(2) and t/lk ' = Uk'l.fJk ,(l) + Vk'l.fJk' (O). Then final

/ -k', -,,'

�

� -k' - "

Fig. 10.18 Symmetry of the coefficient B.

(t/I 2 IH' I t/ll ) = < l.fJd2) 1 < Ukl.fJk( I ) + Vkl.fJk(O) I ( Bk' i k i at' i ak i + B- k � -k ,� a � k � a _ k' � ) x Il.fJk(2) l Uk ' l.fJd I ) + Vk'l.fJk'(O) = Bk' i k i Vk Vk , + B-k� -k' � Uk Uk ' = Bk' i k i ( Vk Vk , ± Uk Uk')' (10,53)

The ± sign gives the two alternative cases (1) and (II) in Eq. (10.52). Changing �he sign from plus to minus causes a big difference in the scattering. Qualitatively !we can understand the change when we consider the case k' � k. Then we see

:vi

+

ui

=

1;

:When the sign is positive, the scattering is enhanced, whereas when it is negative the effect is reduced. Phenomena such as attentuation of sound, thermal conductivity, and paramagnetic loss have been examined and the theory can explain the experiments well. Let us now discuss some gap experiments. Suppose we form a junction by \placing a superconductor with a thin insulating layer (such as an oxide of the �uperconducting material) against another metal, which can be either a normal :metal or a superconductor. If a voltage is placed across the junction, the amount of current will be affected by the presence of the gap. Semi quantitatively, we can set up the problem as follows : Label the metals 1 and 2 . Electrons in a state described by "k" in metal 1 will have an amplitude to go to state "q" of metal 2. Let that amplitude be a constant, M, in the energy range that interests us (near the Fermi levels). The probability of transition from state k of metal 1 to some state of metal 2 is

�1t Lq 1 M 1 2 (probability of state k being occupied) x

(1

-

probability of state q being occupied)

x

c5(Ek - Eq),

Where Ek and Eq are the energies of the respective states. The probability of state k being occupied is

and similarly for state q. If voltage - V is applied to metal 2, the energy of state q becomes E� = Eq - V, and the probability of q being occupied is the same as before, namely f(Eq) = f(E� + V),

10.8

Superconductivity

288

Real test of existence of pair states and energy gap

289

The current from metal I to metal 2 is

2 ne L IM I 2f(Ek)[1 - f(E� + V )] c5(Ek - E �). ,I kq

If

I

=

the total current = current from I to 2 - current from 2 to 1 ,

we find that --o�------�--� 61 =

2 ne I M I 2 L [J(Ek) - f(E� k, q h

+

V

Fig. 10.19 Current across a junction with one superconducting metal at zero temper­

V )] c5(Ek - E �).

ature.

Now Lk,q can be replaced by S dEk dE�PI(Ek)P 2 (Eq), where PI(Ek) and P 2 (Eq) are the density of state functions in the respective metals. Then

I(T)

=

:

2 e

At absolute zero,

1(0)

=

f

-

1M 1 2

2ne ft

I

dE [f(E) - feE + V)]PI(E)P 2 (E + V). IM I 2

fO

dEpl(E)pz(E + V ).

(10.55)

-v

Thus the current depends on the density of states near the Fermi level, EF, for the two metals. For normal metals we can consider pee) to be roughly constant, but for superconductors the interactions between electrons cause the spectrum to be distorted. The state that would have had energy ek without the interaction actually has energy IEk l = Jei + Ai. For simplicity, take Ak = � independent of k. Thus there is a gap around Ek = O. While dN/de (which is the density of states if there were no interactions) is roughly constant,

Fig. 10.20 Current in the same junction at finite temperatures.

�

Unoccu pied Iml

rises to infinity as the gap is approached, then drops to zero within the gap. If metals I and 2 are normal, PI(E) and piE + V) are constants, and I i s proportional to V. But if metal I is a superconductor, PI (E) is zero for all E in the integral of Eq. (10.55), unless V > A. Thus for V < A there is no cur��t. For V > A, it is easy to show with our simplifications that I oc. JJ72-=- A i· and looks like the function in Fig. 10. 19. At finite temperatures, the curve looks like the function in Fig. 10.20. This situation, with metal I a superconductor and metal 2 normal, can be visualized as shown in Fig. 10.21 . At V = 0, EFI is the same as EF2. Whe n v lowers EF2 more than Al below EF" current can flow from I to 2.

p

Filled levels

{E',

{

(

----

I

61

----

---

1

V

---

(superconductor)

2 (normal)

�ig. 10.21 Levels in a junction of two metals, separated by an insulating layer. Metal l IS

a superconductor and metal 2 is normal.

290

�.9

Superconductivity

�b

T> O

The number density of electrons is

�Qjm. Then

�.

excess energy

Vte c urrent is

�.

,AI

=

= S 2 ui, and we can define v =

,AI(m

v 2 j2) .

(10.56)

INhere p is the momentum operator. It follows that

o

,

J

Fig. 10.22 Current in a junction of the superconductors, at zero and finite temper­ atures.

Suppose both metals 1 and 2 are superconductors. The current as a function of voltage looks like the curves in Fig. 1 0.22. From the second curve in Fig. 1 0.22, both � l and � 2 can be obtained. For more details concerning the energy gap. see Progress in Low Temperature Physics, Vol. IV, page 97 and on (especially. page 140 on).

In Section 10.4, we assumed that the electron pairs had zero total momentum, so that k l + k2 = O. The more general situation is k l + k2 = 2Q = constant for all pairs, in which case there exists a current. As before, we can describe a pair by a single momentum vector, k, but in this case,

k l = k + Q, k2

= -k

+ Q.

Formerly, the energy of a pair was Ck + L k fz 2

� (k + Q) 2 + � ( k 2m

fz 2 2m

Ck l + ck2

-

=

+ Q) 2

= �efz L [(k + Q) + ( - k + Q)] u t m k

-

fzQ � 2 U k2 e � 1... m

=

2Ck' But

=

fz 2 Q 2 fz 2 k2 2 - - + 2 -- - , 2m 2m

2Ck + 2 ( fj 2 Q 2 j2m)

.

In analogy to what we did before, we define V ��\ as Vk+ Q, - k + Q ;k ' +Q , - k ' +Q' Notice, however, that V ��\ � Vkk , + terms of the order Q2 or less. The change in Vkk, of order Q2 is similar to the doppler frequency shift in sound emitted from moving objects. Neglecting "Doppler shift'" we modify Eq. (10. 1 7) only by the replacement of 2Ck by 2Ck + 2fi2Q2j2m. The energy of the ground state is increased by ( fz 2 Q2j2m) Lk 2 ui·

- ,;V'ev.

k

(10.57)

, How is the wave function changed by replacing pairs ± k with pairs \j:k + Q? The equations for Uk can be brought up to date by replacing III: with Ck + fi 2 Q 2 j2m . As well as fz 2 Q 2 j2m , we must add another constant Jo make the zero of the energy appropriate for the correct number density of electrons. The net result is that the zero of energy is again at the Fermi surface, and the Uk come out the same as before. But now the Uk describe the amplitude for ± k + Q to be occupied. S uppose we describe the ground-state wave function by 1/Io(x1, X2 , . . . ) . . We want to modify this wave function to become 1/IQ(x 1 , X2 , ). This modifica­ ition is effected by replacing each component of momentum p by a component of :�he same amplitude, but of momentum p + fz Q. That is, eiPe xe/h is multiplied 'by e iQ ' x e . Then v. 1/I Q = exp i Q . Xe 1/10 = exp xe 1/10' (10.58) •

10.9 SUPERCONDUCTOR WITH CURRENT

so

291

Superconductor with current

(

�

•

•

'

)

[i �

�

J

Equations (1 0.56) and (10.58), it should be noted, represent simply the Consequences of a Galilean transformation of Schr6dinger's equation. According 'to our approximations, a superconductor with a current is the same as a super­ : conductor in its ground state as seen by a moving observer. In a real metal the current can vary from place to place. Because k is large near the Fermi surface, the lowest-energy wave functions are not affected m Uch in energy if the variation in current is slow. To see how to deal with variation of current with respect to position, l assume a metal is divided into regions. Each electron is constrained to move within a given region, and for each region there is a different current. Then the '. wave function becomes 1/1

=

(

exp i '!!. V l ' fi

L electrons in region 1

) (

Xe

exp i

m

Ii

V2 '

�

region

2

)

Xe · · ·

1/10 (10.59)

292

Superconductivity

Current versus field

10.10

We can now let v(xe) vary gradually with xe ' But this is not quite right. To s( why this modification of t/! is inaccurate, remember that the equation for c UneI density involves derivatives. The fact that v(xe) is not constant modifies n current. To improve our description of a current in a superconductor, let us ne: try

E

=

=

=

t/!*Vt/!

The number density of electrons is given by

t/!"t, Vt/!o ·

Equation ( 10.62) becomes (if Eo is the energy for t/!

If we remember that the current for t/! =

=

=

=

t/! 0) :

t/!o is zero, we find

Ii - .!Ve - VO. m

(10.60)

For example, if 0 = mv ' xjfi, j(R) is the same as that for Eq. (10.57). Note that we have only described currents for which V x j = O. For t he steady state, V . j = O. In a ring, the above conditions can be satisfied with nonzero current. I n this case, f j · ds = - JVe(lijm) AO, where (AO) is the difference in 0 as we go around the ring. A() need not be zero, but it must be a multiple of 2n, for we require that the wave function be single valued. If we define v = -jj A' e == (lijm)VO, then

t v ' ds

( 10.62)

Also,

and the current density is given by

j(R)

=

f t/!*Ht/! f t/!* [� ��2 V; + V ] t/! ;� � f Vet/!* . Vet/! + f t/! * Vt/! ;� � f d3R f (Vet/!* . Vet/!) (j(Xe - R) + f t/!*Vt/!

293

= 2nn

�,

( 1 0.61 )

where n is an integer. If the ring's central hole shrinks to a point, the only \\ aJ n can be kept greater than zero is to have lines along which the superconducto r breaks down. By calling v = -jjJVe, we have implied that this quantity can be thought of as a velocity. This interpretation is confirmed by computation of the expected value of the energy.

= Eo +

But jo(R)

=

� d 3 RVO(R) . jo(R) e

f

current for ground state

=

+

� d 3R IVO(RW .!V(R). 2m

f

( 1 0.63)

O. Then ( 10.64)

as we would expect from the interpretation of vCR) as a velocity. 10.10 CURRENT VERSUS FIELD In the equation j(r ) = - AA (r ) A' is averaged over some region about the point r. In other words, we must use a sort of "spread-out" A (r ) . We can interpret this spreading as an effect of the size of the electron pairs, and we can calculate the spreading by means of the BCS theory. Empirically we have '

A'( r ')

=

C

f

,

A r_e ) - l r - r' l � o . r - r ')J( ') d 3 ----"-�_ - r_ ( ( r' Ir - r ' 14 [ _ _

_____

(10.65)

294

�O.10

Superconductivity

where C A' = A).

is an appropriate normalization factor (so that constant A gi ves Note that Eq. (1 0.65) has the form of a convolution. If we write j(q)

K(q)

=

=

f f

Current versus field

�fter some manipulation we find

. j(r)e- iq r d 3r, ( 1 0. 66)

summing over all particles, we get

K(r)e- iq ' r d3r,

then j(q)

=

A(q) . K(q).

( 1 0 . 67 )

In what follows we will find the component of j(q) that is in the directio n of A(q). Suppose that A(r) is in the z direction and has magnitude A oe - iq ' X, where q is perpendicular to A (we are using a gauge in which V . A = 0). Then

By, say, L I e - iq . �, PI, .we a�e representing the operator, acting on all the particles, that on one partIcle IS e - 1q ' rp (r and P are operators), Using

and

P =

it is easy to show that where everything is normalized within a unit volume. In Section 1 0. 1 , we showed that if the wave function is "rigid," .iD = - AA. In so far as the wave function is not rigid, we get a contribution from the paramagnetic current of j

295

=

- ne * . [1jJ VIjJ - (VIjJ)* IjJ] ' jtotal = i + iD' 2 1m

Set n = 1 . By considering IjJ as an operator (as in second quantization) we could get the operator for i(r). But we will use a slightly different approach. For a single particle, the operator for i l(r) is

f

d 3p PIP)< P I (2n)3

e - iq ' rp = L klk - q )
In �hapter 6, Section 8, we learned how to find the corresponding operator on arbitrary numbers of particles. The result is Similarly,

L/ P/e - iq ' r,

=

L (k - q )a + (k - q )a(k). k

Equation ( 1 O.68a) becomes (10.68b) Because q is perpendicular to the

where rl is the operator for position and r is a vector (a set of three numbers) . Then /

il (q) =

f

( j t (r) e- iq ' r =

f


z

axis (i.e., to A), ( 10.69)

. lhe wave fun�tlOn ' I IjJ ) can be easIly found by means of perturbation theory. . The Hamlitoman for the superconductor with no field has a term r. �t . I

2m

296

1 0. 10

Superconductivity

kzattak t

0

Using the fact that V . A = and neglecting terms of second order in A, see that the Hamiltonian is modified by

\\ e

( 1 0.70)

The wave function is 11/1 )

=

1 1/1 0) + L An l l/ln )

n

where

( 10. 7 1 ) n Eo
A =

::::;

,,-

-

o

-e

We will calculate the third terms of the above result, and it should then be easy to see that the first two give zero.

<01 � kzat qak n) =
From

(t kz kr t =

at- q a

and setting k

+

q

=

l

kzat ak _q

-

=

k', we find

tz k

(because kz +

k at+ q a

0) 1 2 z} (q ) = m 2e n I
I/Ik(O)l/IdO) gives kzUkVk,IO, - kl ) lk'i , 0) .

kzat rak r operating on takes the state words,

This is modified to become

j

Current versus field

e2 Ao

� 1...

o

The ground-state wave function is a product of terms such as

ki, - kt occupied

i nto the state

kJ

In other

k'j, - k' t empty

kj, - k' t ( - k)t, (k' i) empty occupied We can get a similar final state by using the operator with amplitude This operator takes

-kza= k ;a_q. kzUk Vk,.

ki , - kt empty

k' i , - k' t occupied into the same state as that found by using Then -

kzattak t' but with amplitude � L I
k z Uk ' Vk.

e 2 ()

_

e AO

-

� ------ - 1... -

- ----

•

The algebra in the above derivation is not rigorous. For example, the signs have not been unequivocally proved. The reader should do the algebra himself to check such things as the correctness of the signs. It is important to be con­ sistent ; decide, for example, whether Iki, -kl) is or The following steps are straightforward :

ak�a= kt 10,0) a= kt ak+; 10,0) . ( UkVk, - VkUk,)2 UZVZ' + VZUZ' - 2 UkVkUk,vk" From Eqs. (10.26) and (10. 2 8), the right-hand side becomes ! ( 1 AkAk GkGk ) 2 EkEk, and Eq. (10. 73) becomes (because .iz(q) = Kzz(q)A z(q)) k; AkAk'GkGk') KzzCq) = � L _ _ ( 1 k 2m e Ek + Ek, EkEk, with = .JGZ + Ai (Eq. (10.24)). Bardeen, Cooper, and Schrieffer find what amounts to K(q) for arbitrary components and nonzero temperature. They also show that at zero temperature, the approximates the empirical formula Eq. (10. 6 5), with 0.15 result to 0. 2 7 in place of the factor 0.18. Considering the approximations we =

_

qz =

297

'

'

'

_

EK

0

( l . 7 2)

�o =

(and Bardeen, Cooper, and Schrieffer) made, the theoretical results are very good .

298

10.1 1

Superconductivity

10.11 CURRENT AT A FINITE TEMPERATURE To compute the energy of a superconductor with a current and at a n o n zero temperature, we must modify Eqs. ( 1 0.36) and (10.37) as shown in Table 1 0. I . Table 10.1 What was formerly

Becomes

8kSk(O) = 28k Ui

(8k + Q + LUQ) ui

8ksk(1)

=

28k V i

8kSk(3) Pk(O)

=

=

So far, we know neither what fk is nor what Uk is. As before, we must minimize F = E - TS, where

TS

=

r r 1 L L Pk(n) log Pk(n) k k l - p - L [fUQ 10g fu Q + f-u Q 10g f-u Q k -

-

k

If we vary with respect to Uk and let

L k +Q

(1 - Ik)(l - I-k)

Pk(l ) = Ii

=

(1 - Ik)2

( l - IUQ)( l - l-k + Q)

that neither (k excited.

+

Probability

Q) nor ( - k + Q) are =

Ik+ QI-uQ

Pk(2) = Ik( l - Ik)

IUQ(l - l- u Q) (k + Q) is ( - k + Q) is

-

-

Llk =

8k

299

+ (1 - fU Q) log ( 1 - fUQ) + ( 1 - f- k + Q) log ( 1 f- k + Q)] - 2p - 1 L Uk + Q 10g fu Q + ( 1 fU Q) log (1 fUQ)]. (10.75)

(8UQ + LUQ) Vi

8kSk(2) = 8k

Current at a finite temperature

Probability occupied (excited) not. =

that

and

l- uQ( l - IUQ)

The energy is

-

(10.76)

L n�l Uk, Vdl - 2fk' + Q)' k'

then we get an equation similar to Eq. (10.20). In fact, if we neglect errors of order Q 2 , we get Eq. ( 10.20) exactly, and everything from Eq. ( 10.20) to Eq. (10.28) is again true. More generally, we can define eCk) = -t(ek + Q + "-u Q)' and Eqs. ( 10.20) to ( 10.28) hold with eCk) in place of ek · Next, vary with respect to fk + Q . The result is o =

- 2e(k) U i + .ek + Q + 2Ll.l,uxVx + p - 1 log

=

ek + Q - e(k) + Ek + [3 - 1 log

�.+JL

=

1i2 k · Q + Ek + [3 - 1 log fk + Q , 1 - fUQ m

1

_

fUQ

1 - fk+ Q

Jfk + Q

(10.77)

from which we find fU Q

x

[( 1 - fk' + Q)(1 - f-k' + Q) - fk' + Qf-k' + Q]

= L [(euQ k

+ " k+Q) U i( 1 - 2fuQ) + 2eu <J"uQ] -

(1 0.74)

=

1 2 (k + exp {[Ek 1i · Q( m)] [3} + 1

( 10.78)

Equation (10.78) is what we would expect from a Galilean transformation of velocity v = Ii Q(m. We might also expect that the current should be the velocity times the total density. In fact, this is not the case. The actual current is less than this, for the energy of excitations causes them to prefer to move in such a way as to oppose the flow. The effect described above should not be confused with "back flow," which is instead an effect that comes about as follows. Consider the problem of a wave packet of excited electrons in a superconductor. Such a wave packet can be described by an approximate wave function, which, in turn, can be used to find energies and localized currents. But it turns out that the approximate wave function first used did not conserve current. An improved trial wave

300

Superconductivity

function (which does not improve the energy much) does conserve current. "Back flow" is the additional current that arises from the correction to the wave function. Now, let us see how much current we get from the wave function we have chosen for a uniform current in a superconductor . Electric current

x

m( - e) = momentum

= L Probability of state i i = L L Pk(n)

k

n

= L (1

k

-

x

Current at a finite temperature

10.1 1

x

momentum of i

contribution of the ± k + Q states to the expectatio n value of the momentum

fk+ Q)( l - f- k+ Q) U � [(k + Q) + ( - k + Q)]

+ L fk+ Q(1 - f- k+ Q) (k + Q) + L f- k+ Q( 1 - fk + Q)( - k + Q) k k + I fk + Qf-k+Q Vi [(k + Q) + ( - k + Q)] k (10.79)

The term multiplying Q can be written as

s(k) Q L 2 U � + Q L (l - 2 U i) (2fk + Q) = ffQ + 2 Q L fk + Q Ekk k k (see Eq. ( 1 0.25)). Lk fk+ Qs(k)/Ek is roughly zero. To see why, consider fk + Q and B(k)/Ek as functions of energy, s. (See Fig. 10.23.) fk+ Q is roughly symmetric about the Fermi energy, and B(k)/Ek is roughly anti symmetric. Their product is roughly antisymmetric ; so it integrates out to zero.

--+------+.--_ Energy

F�.23

The functions /k + 2 and s(k)(Ek•

The term in Q is therefore,

.At Q.

301

The other term in the current is

Up to first order in Q , the integrand of this term is __ __ � __ _ _ _ _ __ _ _ E exp {[ k + (k · Q/m)] /1 } + 1

�

k eEklJ + 1

k(k · Q/m)f1eEkP - (eEkP

+1) -£

The only contribution after integration comes from the second term in the integrand, which by symmetry gives

-2

· Q /m)f1eEkP = (eEkP + 1 ) 2

f (2nd3k)3 k(k

_

2/1Q 3m

f (2nd 3k)3

�2_eE�P _ _ . (eEkP + 1 ) 2

Summarizing, we see that Eq. (10.79) leads to : j = - .Atsev, . where .Ats = effective density of superconducting electrons = .At ff

ffn

v

= 2 =

f (2n�3k) 3 f d 3k

2/1 3m

Q. = m

-

.At ; n

U � = numerical density of electrons,

k 2 eEkP (2-n)3 (eEkP +-1)2

, (1 0 .80)

As the temperature tends to zero, /1 approaches infinity, Jv" approaches zero, and ffs tends toward .AI. ffn comes from the excitations, and can be considered as a number density of normal (nonsuperconducting) electrons. To help us see how ffn behaves, let us make some appr�ximation s. . Suppose that the integrand of ffn is appreciable only near the Fermi surface III . k space. This assumption is a consequence of the assumptIOn that fk + Q con­ tributes only near the Fermi surface, as in Fig. 1 0.23. Then we can replace k 2 with ki-, and

where M(O) is the density of states at B = o. We have M (s) de

_

-

4n:k 2 dk - --(2n:) 3

4n:km k dk _ �n:k l!2 _ --- - -_ (2n:)3

m

(2n:)3

I (1I� .

302

10.12

Superconductivity

Also

4nkFm . (2n) 3

.

f d 3k

k

where V = Q/m . We have already worked with the integral in this equation, and it is easy to see that

3 (2n)

The factor of 2 comes from spin. Putting all this together yields

j = - .Ale V + .AI"e( V - Vn) = - .Alse V - .Alne VIl"

(10.82)

Another result obtained using the same ideas is

ffy(P/:,.). (10.8 1 )

We already know that at absolute zero, y = 0. At the critical temperature, e, and y = 1 . In other words, %. = ° at the critical temperature, and there is no superconducting current. %. is given qualitatively as a function of temperature by Fig. 10.24. Without much effort we can generalize this derivation. Suppose that instead of simply minimizing F, we minimize F subject to the condition that there is a fixed expectation value of the momentum. We then do not get a stable state, but instead a state that will decay into a stable one. An outline of the procedure follows. Where before we minimized F, we now minimize F + Vn '
2e

] = - ffe V � (2n) 3 exp{[-Ek + -k-'---V]P--:-}-+-1 (V ,,)=

4 nkJ, . ff = 2 X 3

Here y is a function defined as

303

We fix the value of < P) at the beginning of the problem, and at the end of the problem we adjust V" so that
Therefore M (0) =

Another point of view

= 0, Ek =

Energy (at fixed entropy) = t.Alsm V2 + t.Alnm V;.

(10.83)

It can be shown (with some difficulty) that V" is the velocity of wave packets in the superconductor. Equations (10.82) and ( 1 0.83) can be summarized by the following slightly nonrigorous statement :

In a superconductor there are two types of electrons-normal electrons, which carry currents that decay due to resistance, and superconducting electrons, which "short circuit" the current carried by normal electrons. This statement is misleading because %. and %" do not really count electrons. %" V is defined in terms of a certain integral, and %. is given by ff - .JY�. In a similar manner, liquid helium can also be thought of in terms of a two-fluid model, and here, too, the model is satisfactory if used carefully, although it should not be taken too seriously. 10.12 ANOTHER POINT OF VIEW Schrodinger's equation can be found from a variational principle of the form (j

f 2 d3x dt = 0,

(10. 84)

where the Lagrangian density is 2(1/1 , 1/1 *, VI/I, VI/I*, ifJ, ifJ*)

= --�------------��--� T Tc

Fig. 10.24 Number density of superconducting electrons N., as a function of temperature.

� [G V - q:)l/Ir ' [G v - q:)I/1] + q VI/I*1/I + � 1/I*ifJ.

2

(10.85)

Finding 1/1 and 1/1 * such that S 2 d 3x dt is an extremum is equivalent to solving the equations b2/bl/l = (10.86) bfi'/bl/l * = °

o}

Superconductivity

304

where, for example, i5!l' i5 1/!

Another point of view

10.12

a!l' aI/!

a!l'-d --;;- L3., ·

d ()!l' � ,�j = l dXj u(vl/!/ox)

dt ol/!

( 1 0 . 87)

rx does not come from electrostatics, for the electrostatics specifies only that %s + %" = constant, where ffs = p. We will not discuss rx any further.

Equation ( 10.89) can be replaced by the equations

Using Eq. ( 1 0.87) with Eq. ( 1 0.86), we get Schrodinger's equation and its complex conjugate. If we write I/! = J P eiO, we can consider !l' as a function of p and f) an d their derivatives. Then i5!l' i5!l' vI/! i5!l' al/!* . = . = 0 - + i5 1/! * vp i5p i5 1/! ap

_

and

.

-

.

- -

( 1 0.88 )

It follows that

(

--

)

-

-

'

!mv2 + qV + U = - 116,

where v = (l /m)(nV8 - qA/c)

V · pv + p = 0,

and

[

]

av + (v · V)v at

( 1 0 . 89) (10 . 90)

= q (E + v

= - VV -

1 aA , c at

rx 2 (p - Ps) 2 ,

V U,

B) +

B = v

- -

(10.91)

x

A.

Equations ( 10.9 1) are merely the equations of hydrodynamics. The first equation is the equation of conservation of the fluid, and the second is F = ma, with U interpreted as a pressure. Because of the rx/2(p - p.)2 term in !l', U is no longer given by Eq. (10.90). Just as Maxwell's equations describe the motion of many photons in the same state, Eq. (10.9 1) are macroscopic equations for many superconducting "particles" in the same state. From Eq. (10.90), mv + qA/c = nV8 is a gradient ; therefore its curl is zero. If B = V x A is zero, . V x v = 0 and the fluid is irrotational. Fro m Eq. ( 10.90), h(8 - 8 1) = 2

If point 1 = point 2, we get is the probability current, not the electric current. Now, how does all this apply to superconductivity? In superconductors. the electrons in the ground state come in pairs with opposite spin. The linear combination of such pairs with the lowest energy can be considered a particle. Since the spin is zero, the "particle" obeys Bose, statistics, and the ground state of a superconductor has huge numbers of "particles" in the same state. If p is such that I/! is normalized, Eq. (10.89) describes the motion of a single "particle, " but because the "particles" are bosons, any number can be in the same state and p can be the number density. Note that q = 2e and m = twice the effective mass of the electrons. Note also that if p is the number density, we are obliged to modify !l' in some way so that p does not deviate too much from some value characteristic of the material, say Ps' We must add to !l' a term

x

where

E

. n I . h2 qA 2 + . p hV8 ( vJp) ( V'J P) + q Vp + np () + - p . 2i 2m c 2m 1

m

i5!l'/i58 = O.

In terms of p and 8, !l' =

... _-

305

n(2nn) = =

r(

f ( q:) . � f (A} mv +

mv +

ds =

:) . ds.

q

� f (m;v + A ) ' ds

+ A) . ds,

where n is an integer and A' = mc/qps = - 2mec/2eps.

! (A'j :r

+ A) = -

2nnc n 2e

;::::;

- (2.09

x

1 0 7 gauss cm 2) n . -

(10 . 92)

Hereafter, replace n by - n o For a simply connected superconductor (i.e., one without holes) V and A are continuous everywhere ; so n = O. It follows that A = - A}. * But suppose

* This equation is the same as Eq. ( 1 0.2). Remember that the j of Eq. ( 1 0.2) is the charge current, not the probability current as above.

Superconductivity

306

10.12

the superconductor is in the form of a ring. The field does not penetrate verv deeply into the ring (see Section 10. I ), so there is no current deep in the rin�. For a path of integration near the middle of a reasonably thick ring,

2.09

x

1O- 7n

=

f(A]

+ A) . ds =

f

A ' ds =

ff

B · dS.

Another point of view

We will take a real, and leave the more general case for the reader to work out. If we write 1/1 = .jp e i8, Eq. ( 1 0.94) yields

( 10.95)

(10.93)

Equation (10.93) predicts that the flux through a ring is quantized, and this prediction is confirmed by experiment. Furthermore, if q were e instead of 2e, the coefficient of n would be twice as large as experiment gives. There are some other, very interesting, applications of the point of vie" introduced in this section. Consider an insulator between two identical pieces of superconductor (Fig. 10.25). If i\x is sufficiently large, the two superconductors do not affect each other. and

307

and a similar equation with " I" and Eq. (1 0.95) we find that Ii'\ =

a

Jp-; PI

"2"

exchanged . Equating real parts of

cos (02

- ( 1 ) + £1

and exchanging " I " and "2" gives

Similarly,

h .

- - 1/12 = E2 1/1 2 ' i

E1 can differ from E2 if there is a voltage across the insulator. If i\x is very small, the phenomenon of barrier penetration allows some of 1/1 2 to leak into the region of 1/1 1 , and vice versa. Since Schrodinger's equation is linear and homo­ geneous, the rate of leakage into region 1 is proportional to 1/12 and vice versa. It can be shown that

ft

- -;- 1/12 = E2 1/1 2 1

where

.

'It I

-

P I = the current that flows through the j unction . Define

.

If P I

�

10

==

2a ft

-

/-V P I P2 � -

P 2 , we get

2aps

--

Ii

.

j = jo sin D,

(10.96)

(10.94)

+

a is real if there is no vector potential.

Su perconductor

j=

a *1/1 1 If there is a vector potential,

Su perconductor

'lt2

where V is the voltage across the junction. Now, suppose we construct the device shown in Fig.

11 II Junctions

IE: )j

Fig. 10.25 An insulator, width i\x, between two identical superconductors.

Fig. 10.26 A two-junction device.

1 0.26.

If b = O2 - 01

308

10.12

Superconductivity

and /j' = e� - e;, then

( 1 0.97)

}tot = current that flows freely through the device = io(sin /j + sin /j').

+ (qA/e)) , so

Inside the superconducting part of the loop ve = ( 1 fI)(mv 2n:n = change i n e as we pass around the loop = /j' - /j +

But if the wires and junctions are thick enough, 2n:n = /j' - /j

+

lA ne J'

!L

v

f� (

mv

+

Another point of view

;)

q

io/J � o' �-- -

. dol'.

{

is small, and

. ds = /j' - /j + q c£ ,

ne

( 1 0.9�)

where cp = S B · dS i s the flux through the loop. Equation ( 1 0.97) can be graphically represented if we notice that

-------- - -

--

Fig. 10.27 Current jo in the complex plane.

itot = the imaginary part ofJo(eid + eid'). I n the complex plane, i behaves as shown in Fig. 1 0.27. A small voltage can cause /j and /j' to change with time, but /j' - /j is a function only of the flux through the loop. With a given flux through the loop, the maximum value itot can have is by simple geometry . ' cos W lrnax = 2Jo

-

2

/j) = 2Jo' cos - . ecp

fIe

( 1 0.99)

We predict Jrnax to vary periodically with cp as shown in Fig. 1 0.28. The distance 7 between successive maxima represents a vary small flux (of the order of 1 0 gauss cm). Something like the graph of Fig. 1 0.29 is actually observed. Figure 1 0.29 differs from Fig. 1 0.28 in three respects : 1.

the scale is different ;

Fig. 10.28 Theoretical periodic variation of maximum current jrnax .

2. as cp increases, the oscillations die out ;

3. a nonzero lower bound arises for the oscillations as cp increases.

The above experiment is analogous to the double-slit experiment in optics In optics, if the slits have finite width, the intensity must be multiplied by the diffraction pattern of a single slit, and for large angles the oscillations die out. In our experiment, there is some magnetic field passing through the j unction material, so /j and /j ' are not uniform along the junctions. The spreading of ,) along the junction corresponds to the "width" of the slit in the optical doub\e­ slit experiment, and it has a similar effect. The nonzero lower bound that appears as the oscillations die out is a consequence of the fact that v, the velocity of the electron pairs, is nonze r o . The physical apparatus a s pictured in Fig. 1 0.30 does not really have its left side identical to its right side, so t v ds need not be zero.

Fi g.

1 0.29 Experimental variation of jrnax .

309

310

Superconductivity

10. 12

If the magnetic flux through the loop is produced by a solenoid through t he loop, the oscillations do not die out because no magnetic field passes thro ugh the path of the current. The fact that varying


W fv

.

ds =

�

� . b0.-/ Superconductor .· . · .

II

311

t

Fig. 10.30 The "double-slit" apparatus.

� fj . ds = h�� f qj ' ds

to the phase, 6 - 6'. The integration path is along the bottom film of superconductor. By measuring the electric current needed to produce a given shift in the position of the maxima of the oscillations, we can find m/lipsq. Recall that A = psq/m. Another experiment we can perform is to rotate the device pictured In Fig. 10.30. If b is a vector,

()

db measured in a system d t as rotating with angular velocity w

db - - w X b. dt

()

So, for example, if the position of a particle is called r,

(v )rot = ��

( )

Another point of view

dt rot

= v

- w x

r.

d(v) arot = cl�rgt - w X (v )rot dt dt rot dv dr - w x - - w x ( v ) rot dt dt = a - w x [(v ) rot + w x r ] - w x (v )rot = a + 2(v )rot x w - w x (w x r).

Neglecting second-order terms in w, we see that in the rotating frame there is an added effective force of v) 2mwc 2m( v )rot x w = q ( rot x c q that is, an effective magnetic field B = 2mwc/q.

-

-

Fig. 10.31 The same, with the bottom film doubled on itself. Going back to quantum mechanics, we see that this effective magnetic ;field causes an effective A to appear in the Hamiltonian. In our experiment Iwith the rotating loop, the maximum current varies as the angular velocity is �varied. By means of this experiment we can measure q/m. Devices related to the one described above may have many uses. They react so quickly that there are rumors that one can detect radio signals by the :vibrations of the magnetic fields. More likely, such devices might be useful for :measuring small magnetic fields simply by counting the maxima of the current oscillations. Perhaps they could be used to maintain fields (or rather fluxes) fconstant to a high degree of precision. Maybe they would be useful for informa­ :tion storage in computers. There may be circumstances in which they could be used as a strain gauge, by keeping B fixed and detecting changes of area.

1 1.1

Introduction : nature of transition

313

"A" and as a result the transition temperature 2. l so K is often called the "),­ 'point." The specific-heat curve on both sides of the A-point is given empirically by (see 1 .9.)

CHAPTER 1 1

SUPER FLU IDITY

11.1 INTRODUCTION : NATURE OF TRANSITION Liquid helium4 undergoes a transition at 2. I SDK, which may be demonstrated in a spectacular manner. When normal liquid helium, known as Hel, at the boiling point is cooled to 2. l soK, the boiling abruptly stops and the quiescent low-temperature modification, liquid helium II, (Hell) appears. Hell does not boil. This abrupt transition also can be demonstrated by specific-heat measurements (see Fig. I 1 . 1). The shape of the specific heat curve resembles a c

u

c

4

2

Fig. 11.1 Specific heat of helium near the transition temperature. Pressure

where a > is the constant relevant for T > TA (HeI) and a < for T < Ti- (Hel l). 'This result has not been obtained theoretically and remains one of the unsolved problems of superfluidity. At very low temperatures the specific heat is propor­ tional to T 3 , a relation that is understood. Liquid helium below the A-point, that is, Hel l, or "superfluid" helium, exhibits several remarkable properties.

,Properties of Liquid Helium :

1 . As has already been noted, Hel l does not boil (although there is evapora­ tion). This can be explained by assuming the heat conductivity of Hel l to be essentially infinite.

2. More remarkable than the apparent infinite heat conductivity is the zero viscosity (under certain conditions) or superfluidity of Hell. It has been shown that below a critical velocity Vc' HeI l flows through a thin capillary or "super­ leak" with zero resistance. Furthermore, Vc increases as the capillary diameter ,decreases. Observing flow through a capillary is not the only way to measure iviscosity, however. If a cylinder is placed in a liquid helium I I bath and rotated, ithere is a momentum transfer from the rotating cylinder to the helium, indicating :that under the conditions of this experiment, the viscosity is not zero ! These :viscosity experiments can be crudely explained if it is assumed that Hell is ,composed of an intimate mixture of two fluids-one fluid with zero viscosity and density PS ' and the other with normal viscosity and "density" Pn "* Thus it is the zero-viscosity component that flows through capillaries and. the normal component that interacts with the rotating cylinder. To explain the observed data, we must assume that Pn lPs is a function of the temperature. , 3. A third remarkable property of Hell is the "thermomechanical" or [ "fountain effect," together with the related "mechanocaloric effect."

Consider two containers of HeH connected by a superleak (Fig. 1 1 .3). Let the density, p, and temperature T be kept constant on each side. Then there will be superfluid flow through the superleak until the pressure head AP is equal to AP = ps A T. s is the specific entropy. This temperature difference giving rise to a pressure difference is known as the thermomechanical effect. If o ne container is a thin tube the pressure head will cause a fountain of liquid heli um (Fig. 1 1 .4) . Fig. 11.2 Phase Diagram of helium near the transition temperature. 31 2

... The reason for the quotation marks about "density" will become evident later.

314

1 1.1

Superfiuidity

PI '

Tl

PI + Llp ,

H e II

Tl + Ll T

We have seen that both the viscosity experiments and the thermomechanical effect can be explained (crudely) by a two-fluid model for liquid helium II. A phenomenological theory using the two-fluid concept was introduced in 1940 by Tisza* and in another form in 1 941 by Landau.t Since Landau's view is the deeper one, we shall concentrate on his formulation. Consider Hell to consist of a perfect background fluid (which has zero entropy and viscosity), and some type of excitations, which for the moment may be regarded as phonons. This simple hypothesis explains a great many of the superfluid properties. (The model is somewhat analogous to a solid con­ sisting of a background lattice plus phonon excitations.) The excitations accord­ ing to the Landau view are the normal component. First, the specific heat of a "phonon gas" is proportional to T3 at low temperatures in agreement with Fig. 1 1 . 1 . Second, as the perfect background fluid flows through a superleak the phonons are inhibited because they collide with the walls. Thus the perfect fluid emerges with no excitation and hence zero entropy. The two-fluid model is further strengthened by the followin� ingenious experiment due to Andronikashvili.j A pile of closely packed discs i� rotated in a HeIJ bath. The superfluid component is unaffected, but the phonom (and any other excitations) are dragged around with the discs and have an inertial effect, which can be measured. In this way the "density" Pn of the inertial or normal component can be measured. (Fig. 1 1 .5). The two-fluid concept suggests that the two components can oscillate ou1 of phase in such a way that the total density of Hell at a point is essentiall) constant but the difference or ratio of the superfluid and normal components i� not. The density of excitations, however, is a function of temperature, and hence as PnlP varies so must the temperature. This leads to a new type of wave propag�. tion known as "second sound." Second sound is a temperature wave and wIi be excited by heat rather than pressure pulses. According to the Landau view, one can think of second sound as a densit) wave in a phonon gas. If the phonon velocity near T = 0 is c (c is, of course, the Landau Two-Fluid Theory :

He I I

Fig. 11.3 The thermomechanical effect.

Heat -f------.

315

will be either rapid evaporation from the wall layer or the formation of droplets that will fall back into the liquid. In Hell, however, the abnormal heat con­ ductivity prevents the establishment of any temperature differences, and as a result the layer creeps up the side and over the top.

B

A

Introduction : nature of transition

He Il

\

Fig. 11.4 The fountain effect. If the containers of Fig. 1 1 .3 are kept at a constant pressure and there is mass flow from A to B, container B will cool. This is the mechanocaloric effect. If the two-fluid concept is extended, so that now it is assumed that the superfluid component not only has zero viscosity but also car�ies zero entropY, the thermomechanical and mechanocaloric effects can be explailled both qualIta­ tively and quantitatively. For example, the cooling of container B in the mechanocaloric effect is exactly what one would expect if the mass transferred had zero entropy and hence zero temperature. The superfluid flow of a zero-entropy, zero-viscosity fluid also explains the abnormal heat conductivity of Hell. Most of the other "superfluid" properties of Hell can be explained in terms of the properties described above. For example, liquid helium II placed in a beaker will creep up the surface of the beaker and spill over the sides. This wIl. l continue until the beaker is empty (it being assumed throughout that the tem­ perature of the helium is always below the A-point). This rather peculiar behavi or can be explained by the ordinary physics of evaporation and the infinite thermal conductivity and zero viscosity of Hell. If any normal liquid is placed in a beaker we would expect that, due to Van der Waals attraction between liquid and beaker molecules, a layer of liquid' whose thickness decreases with height, will be formed on the beaker walls. In all liquids but Hell, however, the layer for�ation is inhibited by smal but finite temperature, and hence vapor pressure, dIfferences between the wal and the liquid. Thus, depending on whether the wall is warmer or colder, there

:

Some ofTisza's ideas can be found in his following papers : Nature, 171, 9 1 3 ( 1 938) Phys. et Radium, (8) 1, 1 64, 350 ( 1 940) ; Phys. Rev. 72, 838 ( 1 947). t L. D. Landau, J. Phys. USSR 5, 71 ( 1 94 1 ). See also : L. D. Landau and E. M Lifshitz, Fluid Mechanics, Addison-Wesley, Reading, Mass., 1 959, ch. XVI ; ane L. D. Landau and E. M. Lifshitz, Statistical Physics, Addison-Wesley, Reading, Mass. 1 959, ch. VI, §§66-67. 1: E. L. Andronikashvili, J. Phys. USSR 10, 201 ( 1 946) ; Zh. Exsperim. i Teor. Fiz. 1 8

J. *

424 (1 948).

316

Superfluidity

11.1

Introduction : nature of transition

P,,/P

317

E(P)

! --4''-----'---------' --1---"------1 ----L__ P (momentum) Po

Fig. 11.7 Excitation curve for phonons and rotons.

Fig. 11.5 Andronikashvili's experiment.

lowest temperatures, the mean free path of the phonons becomes large ; so i t becomes difficult t o measure the speed o f second sound. At low temperatures, E(P) = cP and the excitations are phonons. In the region around Po,

Cs

where II is a constant and J1 is some effective mass. Landau called these excita­ tions "rotons." The two-fluid model of Landau can be put on a more quantitative level by considering the statistical mechanics of a phonon-roton gas. The free energy is

velocity of first sound), the ordinary theory of sound propagation predicts that

the velocity of second sound Cs = crlj as T -+ O. (See also Section 9.2.) Tisza associated second sound with the superfluid component rather than the normal (phonons), and as a result he predicted that c -+ 0 as T -+ O. For a while there was a controversy on this point, but experiment soon proved Landau's supremel] confident prediction that Cs -+ cjJ) to be essentially correct. (Actually ' Cs > cjJ 3 at very low temperatures, less than O.5°K.) The actual curve of second sound vs. temperature is given in Fig. 1 1 .6. To explain this curve between the A-point and 1 0K, it must be assumed that there are other excitations besides phonons. Landau derived the excitation curve of Fig. 1 1 . 7 empirically ; it has since been developed theoretically. At the

=

F

Fig. 11.6 Velocity of second sound.

k T ln (1

- e - E(P)/k T)

d 3L V . (2n n)3

( I Ll)

It i s clear that the low-energy excitations (phonons) and the excitations around

Po (rotons) contribute the most to the integral. Well below 1 0K , however, the contribution due to the phonons dominates. Above 1 0K, the rotons dominate. The average energy and specific heat due to the phonons (E = Pc) is found as follows : The expectation value of the number of phonons of energy E is

L:� o ne - B( nE) L� o e - B( nE) �---

Then

E ph

too

Pc

e Pc/kT

_

1

4nP2 dP (2ITh)3

e BE

- --------

ePc/kT

_

4n sk4 T4 15h3c 3

,

Cph

_

1 6 n seT3 15h 3 c3

( \ 1 .2 )

318

Superfluidity

11 .2

The specific heat due to the rotons (E = C

rot

-

A (P - Po)2/2J1) 2J/1 1 /2pl/ �A2/ [ 1 kT �(kT)2J e -!J.lkT. (21t)3 2k 2 T3 2 h3 A 4 A +

+

is

+

Pn'

(11.3)

To understand the Andronikashvil i experiment and the meaning of we must consider the statistical mechanics of a moving phonon-roton gas. Perhaps the simplest way to proceed is to assume that the superfluid background component is moving with velocity Then the energy of an excitation, phonon or roton, is given by

Vs'

E

=

E(

P) p . VS'

(11.4)

+

A similar equation was derived for superconductors. An easy way to see this relation is as follows. Suppose a gremlin floating in the fluid gives it a kick and produces an excitation of momentum Then the velocity of the rest of the fluid is decreased by where is the mass of the background fluid. If the fluid winds up with velocity plus the excitation, it must originally have had velocity along with the energy necessary to create the excitations. + The total energy must therefore be + + + E(P) � + E(P). The energy the fluid has if it moves at velocity without an excitation is Therefore the energy necessary to excite the fluid without changing its velocity is E( Now, applying ordinary statistical + mechanics, we find that the expectation value of the number of phonons of momentum is

P. M Vs tM(Vs P/M)2 Vs tMV; p . Vs p. V. (11.5) e(E(P)+P'Ys)1 lkT 1 P Vs < P) d3p (P ) f e(E+P·y.)PlkT -1 (21th)3 d3p - fP(P(eElkVsT )/kT1)2eE1k_T (21th)3 Vs (11.6) - Pn Vs· e lkT � (1 1 .7) Pn - f(p2/3kT) e( ElkT 1)2E (21th)3P ' P/M, Vs P/M, tMV;. P) P Np

=

_

The total momentum density is evidently given by

+

where

=

+ higher order in

•

_

By definition

_

_

In other words, from the point of view just adopted, Pn is a derived con­ cept and is not the density of anything. It is clear from Eq. ( 1 1 .6.) that (p = where is the density of superfluid. As in Section 9 of Chapter we can consider the case of an unstable state by minimizing E - TS, subject to the condition that there is a fixed

Pn) Vs Ps Vs

Ps 10,

Superftuidity-An early approach

319

expectation value of the momentum. In other words, now the excitations can be forced to drift. The result is

and

(P)

=

- Pn( Vs -

Np

Vn) .

=

1

[ (P) -':-P ' Yn + p · y,] eP E

_

1

11.2 SUPERFLUIDITY-AN EARLY APPROACH We saw in the previous section that the Landau two-fluid theory and the empirically derived excitation curve ( Fig. 1 1 .7) explain a great many of the superfluid properties of liquid helium. Thus, a deep understanding of super­ superfluidity can be obtained by a theoretical deduction of the excitation curve and of course, the transition. In particular, we must explain : 1.

Why there are so few excitations at low energy-which is the central feature of superfluidity.

2. Why the excitation curve has its particular form. 3. Why there is a transition, and

4. The quantitative nature of the transition.

Proof that superfluidity is not explained by quantum-hydro­ dynamics. Although the answer to why there is a transition has been known

The Transition :

for some time, * this problem has a rather curious history. The Einstein-Bose condensation ( Section 9 of Chapter I) predicts that an ideal Bose gas at the density of liquid helium will undergo a sharp transition at which is remarkably close to the A-point, Liquid helium4 obeys Bose statistics, but is of course not an ideal gas. Nevertheless, it had long been felt that the A-point transition is some sort of Bose-Einstein condensation, and as wil l be shown later, this is indeed the case. However, a seemingly alternative explanation of the transition was supplied by Landau in Landau, using a procedure termed the "quantization of hydrodynamics," developed a set of commutation relations that seemed to suggest that there were an energy gap, phonons, rotons, and so forth. These were never deduced by Landau. If, however, it is assumed that the quantization of hydro­ dynamics gives rise to the excitation curve, it is not unlikely that all of super­ fluidity including the transition can be explained by "quantum hydrodynamics. " Furthermore, since quantum hydrodynamics has nothing t o do with statistics (the way Landau developed it), the validity of Landau's approach would imply that 3 He as well as 4 He undergoes a transition, and the Bose condensation is not the cause of the transition.

2.l8°K.

3.2°K,

1941.

* Feynman, R. P., "Application of Quantum Mechanics to Liquid Helium" in Progress in Low Temperature Physics, Vol. I , edited by C. J. Gorter ( 1 955). Dingle, R. B., "Theories of Liquid Helium I I," Advances in Physics 1, No. 2, p. I I I ( 1 952).

320

11.3

Superf\uidity

Thus, for a while there were two alternatives purporting to explain the liquid-helium transition . The refusal of 3 He to undergo a transition as lower and lower temperatures were attained perhaps threw doubt on the Landau hypothesis. However, I think I have proved that quantum hydrodynamic s does not predict the excitation curve as had earlier been hoped by Landau and others. An outline of this proof is as follows : We consider a fluid and adapt the Lagrangian point of view. That is, we describe the state of the fluid by giving the displacement R(yo, t) of each particle. Yo is the position of the particle at time t = ° and has the effect of labeling or distinguishing it. Each particle obeys quantum mechanics, and since the particles are distinguishable, quantum statistics need not be considered. It is then possible to find a set of commutation relations, and to show rather easily that there are a tremendous number of low-energy excitations (i.e., there is no energy gap, etc.). Now it is possible to convert the commutation relations from the Lagrangian to the Eulerian point of view. In the Eulerian point of view, the density, velocity, and so on, are functions of y and t where y is a fixed point i n space. The result is that the Eulerian commutation relations so obtained are precisely those obtained by Landau ! Thus, according to Landau' s quantum hydrodynamics there are a multitude of low-energy excitations, and hence this theory does not explain superfluidity. Landau' s error is that by tacitly assuming the particles to be distinguishable, he neglected the effects of quantum statistics. Theory of Superfiuidity :

We now want to explain superfluidity from first principles, given a collection of N He4 atoms at density Po. The two-body interatomic potential is given in Fig. 1 1 .8 . It is clear from this figure that the helium atoms resemble hard spheres of diameter 2.7 A. Since at normal density there is 45 A 3 per atom the atoms are not tightly squeezed. The first question one may ask is, why is He4 still a liquid at T = 0, at zero external pressure ? If we calculate the potential energy of a helium lattice

Intuitive derivation of wave functions : ground state

and add the zero-point kinetic energy, we obtain the total zero-point energy, which is approximately the correct value. However, the kinetic or oscillatory energy is so large that the oscillatory amplitude is of the order of a lattice spacing. Thus the lattice "melts" due to the lattice vibration. With the exception of H e3 , no other substance has such a high ratio of zero-point kinetic energy to zero-point energy. This is because helium is very light (and hence has high natural frequency) and has a saturated outer shell, causing very weak interatomic forces.

11.3 INTUITIVE DERIVATION OF WAVE FUNCTIONS : GROUND STATE We now turn to the central feature of Hell, which is the scarcity of available low-energy excited states. Rather than look at the Hamiltonian we shall "wave our hands," use analogies with simpler systems, draw pictures, and make plausible guesses based on physical intuition to obtain a qualitative picture of the solutions (wave functions). This qualitative approach will prove singularly successful. Consider the ground state Rz, . . . , RN), where R i is the radius vector of the ith atom. Each set of 3N numbers (N radius vectors) will be called a "configuration" and can be represented by N points in 3-space, the N points representing the positions of the N helium atoms (Fig. 1 1 .9). (For ease of illustration, the figure is, of course, two dimensional). For each configuration there is an amplitude or number
10 Q � � c: 2 0 �

5

0

-5

- 10

Atom separation

(0

4

2

c:B

(A)

Fig. 11.8 Two-body interatomic potential.

321

Fig.

(0 (0

2.7

A

0

1 1.9 A configuration of helium atoms in 3-space.

322

To see that

qJ

can be taken to be real, consider Schrodinger's equation for (p :

_h2 L Vi2 + i j Veri)] qJ EqJ [i < iqJ qJl qJ � 1...

2m

qJ*

=

.

qJ qJ*

+ is a Clearly, satisfies the same equation ; so either is real or nonzero real solution to Schrodinger's equation energy E. If there are several degenerate eigenfunctions, first pick one solution, use it to construct a real eigenfunction, I , then consider the space of degenerate eigenfunctions orthogonal to and repeat the procedure. In this way we can obtain a set of real eigenfunctions with energy E. To show that does not pass through zero if its energy is as low as possible, we will show how to find a function of lower average energy when a wave function with nodes is given. Since is the lowest possible in the ground state, it will show that is not the ground state, and therefore that the ground state has no nodes. Suppose we fix all the arguments of except �, and let . . . ' �, . . . ) = <1>(�) look like Fig. 1 1 . I Oa. Then construct a new wave function that looks like Fig. 1 1 . l Ob. This wave function has the same value of

qJ

qJ

Ix),

Ix)

qJ

qJ

qJ(xl , X2 '

(OqJ� )2 (OXqJ)2 ' m2 o Oi J qJHqJ dN R lI!/�m Li (ViqJ) 2 + L VqJ 2 ]d N� J qJ2 d NR J 0/2 d NR ,

(VqJ)2

C l qJ l C2qJ 2

that has nodes, and therefore does not have the lowest possible energy. Thus, we can take the state of lowest energy to be real and everywhere nonnegative. Figure 1 1 .8 indicates that the helium atoms are almost hard spheres. Thus any configuration (e.g., Fig. 1 1 .9) that includes two overlapping atoms gives a very smal l amplitude--call it zero. From Eq. ( 1 1 . 8) we see that the energy is high where there are large gradients in Thus we want the ground-state wave function to vary as slowly as possible. Now consider in Fig. 1 l . l 2, the atom ® to "move" about, while the other atoms remain fixed . for configuration (b) must be very smal l ; otherwise there would be a high gradient as ® moves the small distance from (b) to (c). In other words, configurations where two atoms are very close are unlikely, and in the ground state the most likely configurations are those where the atoms are approximately evenly distributed (Fig. 1 1 . 1 2a). Thus we have deduced that is symmetric, real, always positive, largest when the atoms are evenly spaced, and lowest when there is clumping. A n expression that satisfies these requirements is :

qJ.

qJ

qJ

,

and since the energy may be expressed as

(E ) =

=

( 1 1 .8)

our new wave function has the same energy as the old. We can now lower the energy by smoothing out the wave function (see Fig. I Ll I ), that is, we can

(b)

Fig. 1 1.10. To show that a solution with lower energy can always be found, given a wave function with nodes.

qJ2 C1 C2 '

drastically decrease near the node without increasing the very much. It fol lows, by the way, that the state of l owest energy is nondegenerate. If it were degenerate, we could, by appropriate choice of make a wave and function + IjJ =

qJ

and

't'

323

Intuitive derivation of wave functions : ground state

11.3

Superfluidity

qJ

= exp

[ - � f(ri)J 'J

=

0 F(ri)

( 1 1 .9)

lJ

where rij is the distance between the i and j atoms, and F is given approximately in Fig. 1 1 . 1 3 . F is occasionally approximated by some simple function that fits Fig. 1 l . l 3, such as F = I a/rij. Since is expressed in terms of the one func­ tion F, it is natural to ask why a variational principle to determine F is not used. The answer is that the integral over all the 3N variables cannot be expressed simply in terms of F.

qJ

-

0 o

o 0

o o

(a)

C0

o

o

0 0

(b)

o

o

o

o (e)

Fig. 11.12 a, b, and c. A configuration in which one atom "moves" about while all Fig. 11.11 A smoothed version of Fig. I t . 1 Ob.

the others remain fixed.

324

11.3

Superfluidity

Intuitive derivation of wave functions : ground state

00 0 0 0 0 0 0 0 0",('." : 0 0 0 0 0. 0 00 0 0 00 0 0 o 0 0 0 O. 0 0 0 )-) o 0 0 00 0 0 . 0

F

a

Fig. 11.13 The function F of Eq. ( 1 2 .9). N ote that the nondegeneracy of the ground state implies that it is a sym­ metric state for any many-particle system, when the particles are identical. For, since the Hamiltonian is invariant under a permutation P of the identical particles, then cp(PR) will also be a ground state when cp(R) is. But since there is no degeneracy, we must have

and since cp(PR) is always positive and real (all cp values are) we have a = I . Thus cp is totally symmetric. (Remember that this function might yet not be allowed by statistics.) Again, in the analogy to simpler Bose systems, the first excited state wave function, 1jJ, must have one node and be symmetric. Since cp is always positive and J cpljJ dNR = 0, IjJ must be positive for half the configurations and negative for the other half. We now come to the key argument of superfluidity : The only low-energy excitations are phonons. Phonons are excited states of compression, or states involving small displace­ ments of each atom with a resultant change in density. For the phonons to have low energy, the density fluctuations must be over large distances ; in other words, the phonons must have long wavelengths. The central problem is to see why no states other than these phonons can have such low energies. Now 1jJ, which we assume is not a phonon state, must be orthogonal to each phonon state as well as to cp. This means that IjJ must vary from plus to minus for configuration changes that do not alter the large-scale density. If IjJ is to have low energy, this change from plus to minus must be very slow, that is, it �ust involve large-scale "movement s" of the atoms. But the fact that the particle s are indistinguishable and obey Bose statistics makes such a movement impossible !

Excited States : Explanation of Superfluidity :

325

Fig. 11.14 Two configurations (solid and dotted) that result from large displacements (long arrows) of the atoms, can actually be accomplished by much smaller adjustments (short arrows) because of the identity of the atoms. (From Reference I). Now let us quote from the first reference cited in Section

I 1 .2 * :

"The function IjJ takes on its maximum positive value for some configuration of the atoms. Let us call this configuration A, and the particular locations of the atoms a-positions. We said that the a-positions must be well spaced, so that the atoms do not overlap, and further that they are, on a large scale, at roughly uniform density. Similarly, let us call configuration B, with atomic positions /3, that for which IjJ has its largest negative value. Now we want B to be different as possible from A. We want it to require as much readj ustment over a long dis­ tance as possible to change A to B. Otherwise IjJ changes too rapidly and easily from plus to minus, our wave function has a high gradient, and the energy of the state is not as low as possible. "Try to arrange things so that A requires a large displacement to be turned into B. At first you might suppose it is easy. For example (see Fig. I l . I 4), in A take some atoms in the left side of the box containing the liquid and move it across to the other side of the vessel, and call the resulting configuration B. One objection to this is that an atom is moved from one side to another, so a hole remains at the left and an extra atom is at the right. This represents a density variation, To avoid this we may imagine that another atom has been removed at the same time from right to left, and the various holes and tight squeezes have been ironed out by some minor adjustments of several of the neighboring atoms. This movement of two atoms each through a distance equal to the size of the vessel, one from left to right and the other from right to left, is certainly a long displacement, so B and A seem very different. But they are not. "The atoms must be considered as identical ; the amplitude must not depend on which atom is which. One cannot allow IjJ to change if one simply permutes atoms. The long displacements can be accomplished in two steps. In the first step permute the atoms you wish to move to those a-positions closest to the * Feynman, R. P., Ibid.

326

Phonons and rotons

11.4

Superfluidity

ultimate position they are to occupy in the final configuration B. This step does not change t/I because all the atoms are still in the same configuration of C(­ positions. Then the change to the B configuration is made by small readjust­ ments, no atom moving more than half the atomic separation. In this mino r motion t/I must change quickly from plus to minus and the energy cannot be low. Because the wave function is unchanged by permutation of the atoms it is impossible to get a B configuration very far from the A configuration. No very-low-energy excitations can appear (other than phonons) at all. "In the phonon case we consider configurations in which, as t/I changes sign, the density distribution changes. A change in density cannot be accomplished by permuting atoms. That is why the Bose statistics does not affect phonon states. But it leaves them isolated as the lowest states of the system, so that the specific heat approaches zero as T approaches zero according to Debye T3 law. This is the key argument for the understanding of the properties of liquid helium."*

We have been assuming that large-scale density variations are long-wavelength phonons. That this is so can be seen as follows : Let peR) be the density of the helium liquid. Then, Fourier analysis of peR) gives peR) = LK qK e- iR ' K and =

L !(4 i + C 2 K 2 q i),

(1 1 . 1 0)

K

which is the Hamiltonian for a collection of 3N independent harmonic oscillators (normal modes) each of frequency w = KC. Each mode is quantized and, by common usage, if the Kth mode is excited to energy EK = fzKC(n + !), we say that there are n phonons of frequency KC. The ground-state wave function for a set of independent oscillators is

[

exp - ! i gs = n K 2 KC

x

t/lK = qKCP '·

q

J

= exp

[

J KC

q -! � i

2 1....

.

(1 1 . 1 1)

The low excited state of one phonon of momentum fzK is given by ( 1 1 . 1 2) However, Xgs is not the true ground-state wave function in the case of liquid helium, since it does not contain enough short distance detail . For example, it doesn't tell us that two atoms cannot overlap, and so forth. Let cp be the true

* Feynman, R. P., Phys. Rev. 91, 1 29 1 , 1 301 ( 1 953) ; 94, 262 ( 1 954). Feynman, R. P. and Coh en, M., Phys. Rev. 102, 1 1 89 ( 1 956). Feynman, R. P., Progress in Low Temper­ ature Physics (edited by C. J. Gorter), ch. II, ( 1 955).

Then in ( 1 1 . 1 3)

We are talking about long-range density fluctuations, and K is small. Since the operator peR) = L i b(R - RJ, we obtain

f p(R)eiK ' R d 3R

qK =

(

and from Eqs. ( 1 1 . 1 3) and ( 1 1 . 1 4) ./,

=

� eiK . Rj,

( 1 1 .14)

)

( 1 1 .15)

� eiK ' R , cp . o 'l'Phonon - 1... Small K i _

Actually, we should write in place of Eq. ( 1 1 . 1 0)

peR)

1 1 .4 PHONONS AND ROTONS

H

ground state, which takes the short distance factors into account. analogy to Eq. ( 1 1 . 1 2), the true low excited state t/lK is given by

327

=

Then

H

=

L K

L (qK

sin

4i

4 ��

K

(

2

+

2

K · R + q � cos K · R).

)

+ C 2 K 2 (q i + q i ,).

The singly excited states are t/lK = qK CP and t/I� = q �cp. Then the state t/I� + it/lK also represents a one-phonon state, and is equal to ( L i e iK ' R,)cp. Equation ( 1 1 . 1 5) helps us to understand why it is impossible to have as a low energy non-phonon excitation a single atom moving through the helium fluid. For the wave function of such an excitation is e iK . R,cp, which is not symmetric. Blindly symmetrizing this function gives L i e iK ' R,cp, which represents a phonon excitation. It is possible, however, to have a (symmetrized) wave function L i e iK ' R,cp if I jK is of the order of the atomic spacing, a. The energy associated with such an excitation is E = fz 2 K 2 j2m fz 2 j2ma 2, �

which is very close to the excitation energy of a roton. Rotons : Higher Energy Excitations : We have seen that low-energy non-phonon excitations are impossible. In other words, there are no possible long-distance movements of the atoms that do not change the density. So now let us try to construct a wave function t/I that goes from plus to minus, without a density change, as slowly as possible, but which of necessity involves only short movements. Again let A be the configuration for which t/I takes on its maximum positive value, and B the configuration for which t/I takes on its maximum negative values. The positions of the atoms in configuration A will be called C(-positions and, in configuration B, /1-positions. The change from A to B must involve only a small displacement of each atom and yet involve as large a total movement as possible.

328

11.4

Supertluidity

.. (,'�)

O" 8

CD

( 1 1 . 20) where

(1 1 .21)

Fig. 11.15 In this configuration the fi-sites are between the IX-sites, We know that the IX-sites in configuration A must be evenly spaced, and the same for the Ii-sites in B, These conditions imply that the Ii-sites are between the IX-sites as i n Fig, 1 Ll 5, Suppose for a moment that each atom is either on an IX- or a Ii-site, and let Na and Np be the number of atoms in each type of position, Then l/J = (Na - Np)cp is maximum positive for configuration A , maximum negative for configuration B, and passes smoothly from A to B, l/J can be expressed mathematically by defining a function f(R;) to be + 1 if R i is at an IX-site and - 1 if R i is at a Ii-site,

( 1 1 . 1 6) For an intermediate position (that is, one for which R i is between an IX and Ii-site), l/J will vary smoothly if f(RJ is taken to vary in some smooth way between its extreme values of + 1 and 1. We now want to use the variational principle to find the best possiblef(R), if

-

I ..!(Ri)cp i

=

Fcp :

F

=

'[ .f (R;), i

PN is evidently the probability o f a given configuration. s is the energy of an excitation, and Eo is the energy of the ground state, Returning now to the specifics of the liquid helium problem, we have F = "LJ(RJ and we wish to minimize Eq. ( 1 1 .20) to find the best value for f The denominator of Eq. (1 1 .20) becomes

f lF l ZPN d3NR

=

i� l i ( h 2 /2m)IVil/J 1 2 + V I l/J1 2J d 3NR S I l/J1 2 d 3 NR

P og (R I - R2 )

<E )

=

{� ;� [f I F I2(Vjcp)(ViCP) d 3NR f (F *VjCP)CP(ViF) d 3NR f (FViCP)CP(ViF *) d 3NR f cp2 1VjF I2 d 3NRJ

L Equation

�, � J

f

(j(Ri

k * ',J

- R 1 ) (j(Rj - RZ )PN d 3NR '.

=

Po !i� 2m

f IVf(RW d 3R.

( 1 1 . 23)

( 1 1 . 24) By setting the variation (js = equation

0 to minimize

� VZf(R) 2m

= S

f

The solution to Eq. (1 1 .25) is feR) =

( 1 1 . 1 9)

[f PN n . d 3RkJ d 3Ri d 3Rj

( 1 1 .20) can now be rewritten as

-

+

=

f !!.2m� IViF I 2PN d 3NR

+

+

, J

Here cp is normalized so that S P d3NR = N = number of atoms. P o is the N liquid number density in the ground state and g(R1 - Rz) is the probability of finding an atom at Rz per unit volume if one is known to be R 1 . The numerator of Eq. (1 1 .20) can be written as i

When l/J = Fcp , this integral can be written in an interesting way, which is quite general (though we take cp real and also assume cp vanishes at infinity). Replacing l/J by Fcp in Eq. ( I l . l S) gives us

f

� � f * (RJf(Rj)

where

(11.17)

( 1 1 .1 S)

=

(1 1 . 22)

The energy integral to be minimized is

<E )

329

Integrating the first integral on the right of Eq. ( 1 1 . 1 9) by parts and making the substitution Hcp = Eocp gives finally

8 CD

l/J =

Phonons and rotons

l/J

=

(� eiK ' R) cp,

9(R

s

we obtain the integrodifferential

- R')f(R') d3R' .

eiK ' R ' .

.

( 1 1 2 5)

( 1 1 . 26)

330

Superftuidity

11.5

and £

=

2

112K 2

where

m S K) (

331

S(K)

( 1 l .27 )

S(K) is the form factor for scattering of neutrons from the liquid and can be found experimentally (Fig. 1 1 . 1 6). From Eq. ( 1 l .27) it is the Fourier transform of g(R), which is the probability per unit volume of finding an atom at a distance R from a given atom in the liquid in the ground state. Since states of different momenta are orthogonal, the energies of Eq. ( 1 1 . 27) are significant not only for K near the minimum but also in the neighborhood of this value. Although in the derivation of Eq. ( 1 1 .27) it has been assumed that K 2n/a (i.e., K is large) where a is the atomic spacing, the analysis is valid for all K as long as IjJ = ('I J(R i))CfJ is a reasonably good wave function. But we see from Eq. ( I 1 . 1 5) that for small K, IjJ = ("L; !(R;))CfJ represents phonons, and is a very good wave function. Hence the results of Eqs ( 1 1 .26) and ( 1 1 .27) are valid over the entire range of K. From Eq. ( I 1 .27) and Fig. 1 1 . 1 6 we can plot £(K). The minimum occurs near K = I /a, and the excitations in this region are the rotons. It is interesting to note that the excitation curve (Fig. 1 1 . 1 7) can now be determined directly, by measuring the energy loss suffered by a scattered neutron of incident momentum 11K. For a given angle of scattering, there will be a minimum energy loss, which is the energy £(K) of a single excitation.

Rotons

1 .0

1---+-

""I"""-+_--I"""7"'-�_+�-+_-

_____

0.5

6

....

K (A - 1 )

o

�

Fig. 11.16 Experimental result for the form factors for scattering neutrons from the liquid.

fi OJ > ·S

cr'

'"

� c:

kJ >..

e'

�

'"

c:

1 1.5 ROTONS A roton can be visualized as a group of children going down a slide. The slide is the AB part in Fig. 1 1 . 1 8. After sliding down, the children go around and come back to the s lide again. The returning-flow part is not included in Eq. ( 1 1 .26) of Section 1 1 .4 for the roton wave function ./, 'I'

- (�-7 iK . R,) _

e

CfJground

40

(From Feynman's theory)

30

,, J , , ,,

20

Ex perimental

'... _ ..... '

IO 0

'- Rotons

1 .0

2.0

Momentum (wave number in

3.0

A-I)

Fig. 1 1.17 Experimental and theoretical excitation curves. [Feynman, Progress in Low Temperature Physics ( I 955) ] .

( 1 1 .28)

but is derived as follows. When we construct a wave packet out of the rotons of Eq. ( 1 1 .28), we have a localized excitation. This localized wave packet does not move, because the group velocity of the rotan state is zero as is seen in Fig. 1 1 . 1 7 (where d£/dk = 0). This localized excitation is associated with the momentum hK. When we calculate the quantum mechanical current density j for a rotan, we find it is not zero :

j =P o.

( 1 1 .29)

Fig. 11.18 Schematic of the movement in a roton.

332

Superfiuidity

Rotons

11.5

333

Fig. 11.19 A wave packet of rotons.

Fig. 11.21 A roton has the same shape as a perfect smoke ring.

Air flow ----..

Fig. 11.22 The attraction between two opposite vortices of a smoke ring is balanced by its motion.

Fig. 11.20 Conservation of current. Then we have the picture shown in Fig. 1 1 . 1 9. We see j i= 0 in a localized region and j = 0 outside. This contradicts the Schrodinger equation, since the current must be conserved. This difficulty is avoided by considering the slow return of current to the packet (Fig. 1 1 .20). Mathematically this is achieved by improving Eq. ( 1 1 .28) by writing

(� [ . Ri

-

( 1 1 .30) �J(Ri R)J) CPground and finding the best form of f(R i - R). After the variation we find the best

t/J

=

exp

iK

form is

,I,

'I' =

i · H·

1" e K '"'

'

(

1 '+

+

'"'

rJ. 'J

- )

Ri Rj

Ii; --- -Rj I 2

CPground'

(1 1 .3 I )

This form conserves the current. As i s suggested by Figs. 1 1 . I 8 and 1 1 .20, the roton is like a classical smoke ring, as shown in Fig. I 1 .2 1 . There is one difference, though. The smoke ring

moves, whereas the roton does not, as has already been mentioned. Because two vortices of opposite sense attract each other (Fig. 1 1 .22), there is another important difference : In the case of the smoke ring, this attraction is balanced by the flow of air through it, so that the ring does not shrink to a point. For the roton, since it does not move, the diameter of the ring shrinks down to the atomic scale. When the atom i n position A in Fig. I 1 . 1 8 goes to position B, the state returns to itself. This leads to a condition on K in K is of the order of I /a where a is the average distance between atoms. This agrees with the curve in Fig. I 1 .20. Thus the roton is associated with a momentum and velocity of approximately

eiK ' Hi.

p

= h/a ;

v =

h/ma.

( 1 1 . 32)

When there is more than one roton, they interact. The interaction becomes important for high temperatures. Landau and Khalatnikov * calculated the attenuation of the first and second sounds from kinetic theory taking into account the cross sections of phonon-phonon, phonon-roton and roton-roton scattering. * For detailed discussion and references, see I. M. Khalatnikov, An Introduction to the Theory of Sllperj!llidity, W. H . Benjamin, 1 965.

334

Superfluidity

1 1 .7

11.6 CRITICAL VELOCITY

Let us find the energy necessary to create an excitation of momentum p. Suppose a body of velocity V and mass M creates the excitation and winds up moving with velocity V'. We see by conservation of momentum MV

=

M V' + p,

=

2 lM 2 V

-

V 'p + p

2

2M

( 1 1 . 33)

The energy of the excitation is written as e(p). The excitation cannot be created unless

1 1 .7 IRROTATIONAL SUPERFLUID FLOW When liquid helium is undergoing a mass motion of velocity V the wave function

<

2

V 'p - L

2M

�

V 'p

(i i � V ' )

IS

�

=

.

P Rc.G.

R

exp


( 1 1 . 35)

( 1 1 . 34)

for large M. The critical velocity necessary to create an excitation is then derived by drawing a tangent to the e(p) vs. p curve as shown in Fig. 1 1 .23 . The critical velocity calculated from this condition i s about 60 m/sec. We might say that this is the reason why helium is a superfluid ; it is not excited for small perturbations. A lthough this explanation seems to be qualitatively satisfactory, excitations occur for velocities of about I cm/sec. Now we have to reverse our position and look for the reason why liquid He can be excited with such a small velocity. ,(p)

=

=

exp

Rd N

[i � ] e(RJ

= m

V . � Rj,

( 1 1 . 36)


( 1 1 .37 )

and the velocity field VCR) at R is found from OCR) by using

C

V R)

=

We see from Eq. ( 1 1 . 38) that

V

x

h

m

VCR)

VeeR). =

( 1 1 .38)

0,

so that when the system is singly connected there is no rotation. When the system is doubly connected, we can have a circulating V in the system as pictured in Fig. 1 1 .24. We take e = ka. For this choice of e, VCR) is perpendicular to the radius and has magnitude

Il k

!VI

mr

h

--��----� p

Fig. 1 1 .23 The critical velocity is at tangent to the e(p) versus p curve.

(� )

Nm V .

where P is the momentum of the whole system, Rc.G. the center-of-mass coordinate, and N the number of particles in the system. As in the case of superconductivity, for a nonuniform velocity we assume

�

so

e(p)

335

Note the difference between Eq. ( 1 1 . 35) and the roton or the phonon wave function of Eq . ( I I . I 5). The exponential part in Eq. ( 1 1 . 35) may be interpreted as follows.

so that the new kinetic energy of the body is 2 lM 2 V'

Irrotational superfluid flow

Fig. 1 1 .24 Circulating

V

in a doubly connected system.

336

Superfluidity

11.8

k is restricted to the integers by the requirement that the wave function must be single valued.

e

= WI.

WI =

Ii

mr

n

V ' dS = 2rr

m

and

f

The unit of quantization is 2rr

Ii

m

=

2rr

x

1 .5

x

*( 1 1 .39)

.. Ii

n.

1 0 - 4 cm 2 /sec.

What i s the energy for this flow ? If Po is the fluid density in atoms per cc, the kinetic energy is ( for n = I ) =

KE

t r (;/:Y pom

where a and b are the radii as shown in Fig. liquid. Thus : Line energy per unit length

-

=

porr

�

In

2ITr drL, I

1 . 24, and

(�)

=

( 1 1 .40) L

(! )

ergs/cm, ( 1 1 .4 1 )

where w e used Po = (3.6) 3 A 3 = -is- A 3 . If we neglect atomic structure and assume a classical continuous liquid with surface tension, a unit line makes a hole opposed by surface tension that we can calculate to be only 0.4 A in radius. That means that there is no real hole in the liquid. The energy associated with a vortex line can be estimated from Eq. ( 1 1 .4 1 ) . The lower limit a of the integral in Eq. ( 1 1 .40) is considered as a length of the order of atomic spacing, because within about the atomic spacing the velocity formula is meaningless ( and also the density is low near the center of the vortex). The exact determination of a would require us to solve the difficult quantum­ mechanical problem. In almost all applications the ratio b/a will be very large. and the logarithm large enough to be insensitive to the exact value of a . For this reason we will not attempt a detailed evaluation, but will simply choose a to be close to the atomic spacing. We may arbitrarily take a = 4.0 A .

* The quantization o f a vortex line was first suggested by L . Onsager, NIiOVO Cimenta 6, Supp\. 2, 246 ( 1 949).

337

11.8 ROTATION OF THE SUPERFLUID* Let us consider the state of rotating liquid helium in a can (see Fig. 1 1 .25). Operationally this state is defined by starting from solid helium in the can ( under pressure > 25 atm at OaK). First rotate the sol id helium and then release the pressure to melt it ; the liquid helium is then rotating with the angular momentum initially given to the solid. What is the final state of helium ? We ask then for the lowest state of this rotating helium. For a system of given angular momentum, the kinetic energy is the l east if the angular velocity m is a constant throughout the liquid. However, this motion is not rotation-free, because V x V = 2m, and a high energy (see Eq. 1 1 .4 1 ) is required in order to set any small part of l iquid helium into a rotational state. If we consider the helium to be a rigid body rotating, we get a reasonable energy for the rotation, but we cannot consider helium to be rigid. The energy to clamp the atoms into a rigid relationship with one another is very high, and must be taken into account if one wishes to calculate the energy for uniform rotation, with V x Vs #- 0 everywhere. A simple way to check the validity of the above reasoning is to consider two atoms of mass m in a harmonic potential. The energy of excitation of one of the atoms is hm = h

is the depth of the

1 0 - 8 In

Rotation of the superfluid

(�y

/2

.

But if the atoms could be rigidly held together, that energy would become h(k/2m) 1 I 2 , which apparently would mean that the presence of a second atom allows a lower energy of excitation. This paradox is resolved by noting that i t takes energy t o hold the atoms together. If only a limited energy is available, nearly all the parts of the fluid must be frozen out i n their ground states. That is, nearly everywhere the local angular momentum is zero, so that, V x V = 0 except at certain discontinuities. Suppose we assume that the angular momentum is carried by excitations. That is, the liquid is not free of excitations such as rotons

Fig. 11.25 Rotating can of helium. * R. P. Feynman, chapter II in vol. I by C. J. Gorter ( 1 955).

Progress in Low Temperature Physics, edited

338

11.9

Superfluidity

and phonons even when the temperature is oa K . The number and type of excitations is to be found by minimizing the energy for the given value of angular momentum. This energy, for vessels of centimeter dimensions turning at about one radian per second, turns out to be nearly 1 04 times the energy of a rigid body rotating at the same angular velocity. We must reject such a model. Thus, in order to achieve a state of lower energy for the given angular momentum, we examine velocity fields that are not everywhere continuous. As we saw in Section 1 1 . 7, we know that if there is a hole at the center of the liquid, circulation can exist. Thus a solution is suggested : the liquid forms a vortex around a hole with constant circulation as analyzed in Section 1 1 . 7 . The velocity varies inversely a s the radius, rising to such heights near the center as to be able to maintain the hole free of liquid by centrifugal force. The energy for this state is stil l quite a bit higher than the kinetic energy for the rigid body case, because the velocity instead of being distributed proportionally to the radius (the solid case), actually falls as the radius increases ( see Eq. 1 1 . 39). Nevertheless it is orders of magnitude below the continuous V-field with excita­ tions as suggested above. Once we admit the possibility that V is not necessarily continuous we can think of a state of lower energy. Suppose that the liquid has not only one vortex at the center, but several vortices. For example, suppose beside the central one there were a number distributed about the circle of radius R/2, half that of the vessel R, and all turning the same way as shown in Fig. 1 1 . 26. If the number of vortices at R/2 is large, this state is like a vortex sheet so the tangential velocity can jump as we pass from inside R/2 to outside as shown in Fig. 1 1 .27. The gain in energy resulting from this improved distribution may more than com­ pensate for the energy needed to make the additional holes (and, further, the central vortex need not now be so large and energetic) . Continuing in this way with ever more vortices we soon notice that the energy can always be reduced if more vortices form. However, there is a limit. Due to the quantization of the vortex strength, the smallest vortex has cir­ culation 2rrh/m. The lowest energy results if a large number of minimum-strength vortex lines (which we will call "unit lines") form throughout the fluid at nearly

o o

0

A reasoning leading to vortex lines

339

Tangential velocity

�:------------

'-----------

R/2

Radial distance

Fig. 11.27 Discontinuity in tangential velocity at R/2. uniform density. The lines are all parallel to the axis of rotation. Since the curl of the velocity is the circulation per unit area, and the curl is 2w, there will be (by Stokes's theorem) 2w �- = 2 . 1 X 103w lines/cm 2 2rrh

( 1 1 .42)

with w in radians/second. For w = I radians/second the lines are about 0.2 mm apart, so that the velocity distribution is practically uniform. The energy associated with one vortex line is derived from Eq. ( 1 1 .4 1 ) . Taking a = 4.0 A and b = 0. 2 mm, and using Eq. ( 1 1 .42) w e find Total line energy per unit volume = Powh In

(!)

= 1 4Powh.

( 1 1 .43)

Thus the total line energy in the rotating l iquid of radius R and a unit thickness is 1 4powfmR2. The kinetic energy for a rigid rotating body of radius R and unit thickness is rrmpoR4w2/4. The ratio of the two is of order 4

�/m

R2w

x

14 � 1 0 - 2

( 1 1 .44)

( the number 1 0 - 2 is calculated for R = I cm and w = 1 rad/sec). For macro­ ' scopic laboratory dimensions the excess energy to form the lines is small . They would form i f rotating solid helium i s melted by releasing the pressure, the angular velocity distribution would differ imperceptibly from uniformity, and the surface should appear parabolic.

o o

o

Fig. 1 1.26 The vessel as seen from above, with vortices all turning the same way.

11.9 A REASONING LEADING TO VORTEX LINES Suppose liquid helium is separated into an upper and a lower part by a fr icti on­ less surface. and the upper part is moving with velocity V (see Fig. 1 1 .28).

340

1 1 .9

Superfluidity (V pper part)

V

• •

_ _ _

exp

(i '!l V ' 2.: R) i h

i

A reasoning leading to vortex lines

341

X if>G II

(Lower part)

11.28 A frictionless surface separating moving liquid helium from stationary liquid helium.

Fig.

\ , '..... -

Fig.

' I -//

\

" ...._---/

Fig.

I /

No surface tension here

11.29 Discontinuities at the dividing surface.

n G G , , \ \

....

.....

_ _ .....

I I / /

\ \

'"

Fig.

.....

_ _ .....

/

I I I

I \,

11.31 Irrotational flow from an orifice.

•

'"

.... _ _ .....

I I

I /

11.30 Vortex lines.

~--

d

,

j�

I �

r------.--- ----,

j�

-

�J ,j

I I L--- ______ ...L _____ � - - - -__ J

- v

I n the upper part the wave function contains the factor Fig.

When the dividing surface is removed there appears the surface free energy or the surface tension. Because at periodic points eXPUm/li)vR = I , there is no surface tension at these points. The dividing surface is pictured in Fig. 1 1 .29 : around a discontinuity there is a circulation. Now you may ask, what is the best size of the slot for discontinuity ? A calculation shows that the discontinuity is not the shape of a slot, but a line, so that the dividing surface would look l ike Fig. 1 1 . 30. Thus, this reasoning may lead to the assumption of the existence of vortex lines. Consider what would happen if liquid is flowing out of an orifice into a reservoir of fluid at rest. If the flow is irrotational, it looks like Fig. 1 1 . 3 1 . A very high velocity develops near the corners and there are large accelerations there. A n ordinary fluid, such as water, flows in a more complicated manner. For a rough estimate, let us suppose that the tube is a long narrow slot, and suppose the fluid tries to go out in a jet, at first of the same width and velocity as in the tube. Then outside this tube we have a situation similar to that shown in Fig. 1 1 .28 and following the same reasoning as before, we expect to see the

11.32 Vortex lines outside an orifice.

vortex lines shown in Fig. 1 1 .30 or in Fig. 1 1 .32. Integrating over the dotted line we find VI =

f

V dr •

=

II

C,:Ii) ,

where II i s the number of vortex lines in length 1 . The number of vortices per unit length along the line of flow is V

mV

(2rrli/m)

( 1 1 .45)

2rrli

These vortex lines have an associated energy. From Eq . ( 1 1 .4 1 ) the energy per unit length of the vortex line may be written as �

rrp

1i 2

m

In

() '

(

a

•

( J 1 .46 )

342

1 1 .10

Superftuidity

where p is the density of liquid helium, d is the width of the jet in Fig. 1 1 . 32, and a is the spacing between atoms, roughly 4 A as was stated at the end of Section 1 1 . 7. The vortices must move out at velocity V/2, the velocity at the location of the vortices. Thus the energy of vortices created per unit time per unit length of slot is V 2 X - X ( 1 ) X (2) : 2

( 1 1 .47)

In order to form the vortex lines, the kinetic energy of the jet per unit slot length must be greater than Eq. ( 1 1 .47). The kinetic energy of flow coming out of the orifice per unit time per unit slot length is Eflow = m

pV2 -- V d. 2

( 1 1 .48)

The critical velocity is obtained by equating Eq. ( 1 1 .47) and Eq. ( 1 1 .48) : Do

=

!!- I n ( ) .

dm

d

( 1 1 . 49)

a

For velocity greater than Do, there is enough energy in the flow to create vortices. To get a rough idea, if we use d = 1 0 - 5 cm, so that In (d/a) = 6, we see ( 1 1 . 50) Do = 1 00 cm/sec. The experimental value is about 20 cm/sec. The difference suggests that the model we have used is oversimplified. There is a connection between the concepts of vortex lines and rotons. Starting from the vortex lines as shown in Fig. 1 1 . 32 we see that they will gradually change into rotons as shown in Fig. 1 1 .33. A vortex ring can be broken into two smaller rings. A roton may be regarded as a small vortex ring, as discussed in Section 1 1 . 5 . There are experimental verifications of vortex lines. See, for instance, W . F. Vinen and H . E. Hall, Proc. Roy. Soc. 238, 204 ( 1 956) ; 238, 2 1 5 ( 1 956) : H . E. Hall, Proc. Roy. Soc. 245, 546 ( 1 958) : W . F. Vinen, Nature 181, 1 524 ( 1 958) .

�

I

000 Fig.

1 1 .33

o o

Vortex lines changing into rotons.

o o o o

1 1.10

The ). transition in liquid helium

343

THE ;" TRANSITION IN LIQUID HELIUM

When we know the curve for E( p), the energy versus momentum function, then the partition function and the specific heat can be calculated if we assume each excitation is independent. This calculation leads to the curve of the specific heat, which agrees with the experiment for the part drawn in solid in Fig. 1 1 . 34. The above approach, however, does not explain the existence of the transi­ tion shown by the dotted curves in Fig. 1 1 . 34, for the interaction between excitations is not taken into account. For the region near the transition point the E(p) curve in Fig. 1 1 . 34 is not of much help. We start afresh and use a different method. In terms of path integrals we have e - fJF

=

1 L

N!

p

fD

x,(O) = x, x,( U ) = px,

exp (

_1

(U

�� mx]

Ii J 0 l

I

2

+

.L Ve x; .

I

<j

-

XJlf dU) ( L 1. 5 1 )

As we concluded in Section 4. 1 for high enough temperatures only the identity permutation counts, and we get approximately the classical partition function of helium. In this case, "high enough temperatures" means ( I /Ii) J� (mi2/2) is large for x(O) = position of one atom, x( U) = position of another atom. If d is the average interatomic distance, i iS 'on the average dju and ( l jli) n (mi2j2) is of order mk Td2/2li2. For T » 2°K, exp [mkTd 2 ] » 2li2

I,

and permutations cannot be important in Eq. ( 1 1 . 5 1 ). For low T (high U) permutations are i mportant. Furthermore, the approximation that V[x;(u) - xj(u)] is independent of u for important paths is true only if U is small, so that x(u) ;:;:; x(O) to have an appreciable contribution to the path c

I I I I " I \ I \ \ I ' I ... - - -- I I I

--::o¥---- T Fig. 1 1 .34 Specific heat of liquid helium ; t heory and experiment agree for the part shown as a solid line.

344

1 1.10

Superfiuidity

i ntegral. For low temperatures, x(u) can vary greatly without giving rise to a large value of x2(u), and thus V(xi - x) can vary over the pa�h . T o calculate the A transition, we must concern ourselves with what hap pens at low temperatures. We can visualize each path of the integral as a motIOn of every particle of the liquid from one position to another. "u" is thoug�t of �s time, and xi(u) is the position of the ith particle of "time" u. If, �urmg this motion, x/u) becomes too large, or if V becomes too large, the p�th gl:,es o � ly a small contribution. V[x;(u) - x/u)] is large if at "time" u particles l and ! �re close to each other. Of course we are not claiming that x;(u) is really the pOSitIOn of some atom at an actual time. But thinking in the manner described above allows one to apply physical intuition effectively. Consider a permutation in which a particle at position A mo� es to the position B, while the particle at B moves to a third position. (See Fig. 1 1 . 3 5 . ) W e will consider the contribution t o (l lli) L S (mx2j2) caused b y A . . As A moves to position B, it must somehow avoid passing thro�gh particles such as C. To get to position B in time U, particle A must move qUickly enough to cover the distance. Since x is large for particle A, it is increased a large amount by increments of velocity so A does not have time to avoid C by trying to move around it. Instead, C is jostled out of the way. Suppose, now, that the distance from A to B is r, and suppose that there are n particles that have t� be pushed out of the way. n is of the order of rid ,where d is the inte�atomic �pacl �g. Each particle along the path has to move a dista�ce of order � to aVOid bemg too close to particle A . Particle A is close to particle C for a time o � orde� �/n, so that particle C must move with velocity d'/Culn) during that time, gIVIng contribution

( )

m d' 2 u 21i uln n

to

1

Ii

JU 0

The A transition in liquid helium

345

m��:n 2 = iiz (�') r; .

Thus the n particles give contribution

Particle A itself contributes mr2/21i2 [3. Thus the effect of the potential is to change e - mr2j 2 h2 to e - ( - ymr 2j 2 h2p ) where y = I + (d'l d)2 . We are going to proceed as if y were independent of r and [3, and our reasoning above shows that this assumption is not too unreasonable . But remember that for displacements of order d we may not need as much adj ust­ ment, so y may be less if r is small. Also if the velocity is especially high, it may be preferable to pass through relatively high potentials, rather than have rapid adj ustments. Thus again y would be lowered. Let us ignore such details. It is convenient to define the effective mass m' as

my = m' so that we may write e - m'/2 h2Pr 2 . Thus in the liquid state, so far as the path integral is concerned, a particle behaves like an ideal-gas particle* of mass m'. We have arrived at the important conclusion that the interaction among particles is "kinetic" rather than "potential." Kinetic interaction varies as r 2 1[3, while potential varies as r[3 (as can be seen by estimating the potential contribution for paths that do not avoid high potentials). Therefore the partition function for low temperatures has the form

mxf 2

L-' i

0 0 0 00 0 0 0 0) 0 0 0 000 0 00 0 0 0 0 0 0 0 0 0 o 0 0

. 1 1 ;35 Particle at A moves to B while particle at B moves elsewhere. Fig.

The last factor comes from the potential-energy contribution of the initial configuration. The reason for the factor (m'/2nIi2[3)3NI2 is not obvious, for it i s not obvious how a change i n the effective mass i n the path integral will change the normalization of the i ntegral . To be complete, we should instead write a factor Kp(m'/2nIi2[3)3N!2. For simplicity, we will pay no attention to Kp. The factor p is clearly small for i nitial configurations (equal to the final configurations) such that atoms overlap. We can, for example, expect results that are qualitatively correct by choosing p 0 if any two R's are closer than b (with "b" roughly 2.6 A) and p = I otherwise. A permutation among particles can be visualized as a polygon with arrows. The example shown in Fig. 1 1 . 36 consists of several cycles of length 1 (particles

=

*

The above results apply only to Bose particles. Fermi particles behave differently.

346

Superfluidity

11 .10

o o

o

o

0

o

0

g

O

°

O O 0 0 0 0 0

e PF = -

0

347

constant

x

( ) �_ 2rrh2 p

3 N I2

X

L'

p

e - m ' d 211(p) I Z h 2 P

.

In this equation, the difficult problem is the calculation of the sum. So ignore the first two factors on the right and set y = e - m'd 212h2 P . We must compu te e - PF

Fig. 1 1 .36 Permutation among particles ; here there is one cycle of length 2 and two of length 3.

that are not permutated), one cycle of length 2, and two cycles of length 3. Thus, the sum Lp in Eq. ( l 1 .45) is a sum over all possible polygon p��terns. A way to estimate Eq. ( I 1 .45) for temperatures near the transition tem­ perature will be explained here.* For such temperatures, a side (of a polygon) longer than the average interatomic distance is not important. 'Thus we may assume that the sides of all polygons that contribute have length of order Given a particular configuration (R Rz, . . . , RN) we restrict our permuta­ tions to be such that Ri and PRi are close. In order to get a contribution to the integral in Eq. ( 1 1 .45), P forces the atoms to be roughly uniformly distributed. Consider the sum

t,

transition in liquid helium

n(p)

[;:h

d

A

where L; is the sum over all permutations of a fixed lattice such that the resul­ ting polygons are made up of lines joining near neighbors, and is the total number of sides in the set of polygons defined by P. We can now take the above sum out of the integral in Eq. ( l 1 .45) and write

0

0 0 0 o 0 o

The

d.

=

L' P

y"(P).

So we have the following problem : Given a lattice of points, lines can form between adjacent points to form polygo ns. With each line is associa ted an energy We wish to calculate the free energy of this system. Let be the numbe r of polygons with s sides and let . . . ) be the number of permutations with polygons of s sides. Then

(m'd z/2h2P). ns

ns

C(nt, nz,

e - PF =

nl'�" "

C(nt, n 2 , · · . ) {Jz snsJ

•

This sum is restricted by the equation

s = sns cYJ

L

1

=

N

=

number of atoms.

The restriction on the sum can be eliminated by the usual method of letting N vary while assigning to each atom a weighting factor, (in Chapte r I , "r" was eP Il) .

r

Large shifts in the Ri that preserve the uniform distribution effectively permute the atoms and thus leave the above sum unchanged, except for the change that would result from small shifts in the Ri. The effect of smaller shifts in the R i that do not permute the atoms can be included by replacing Ri - PRi with an average interatomic distance, For all uniform distributions of atoms, we can therefore replace

d.

Q = ( 1 1 . 53)

From Eq. ( 1 1 .46) it is easy to see that the expectation value of N is

(N )

by �,

-7

exp

(_I11'2dh22n(p p)) ,

* For a more complete discussion, see R. P. Feynman, Phys. Rev. 91, 1 29 1 ( 1 953).

=

r

��o�Q =

dr

- rp

cig .

dr

To evaluate Eq. (\ 1 .46), we must find a way of approximating C(n t , We will approximate by writing

C

( 1 1 . 54) 11 2 "

. .

).

348

1 1.10

Superftuidity

where R is the number of ways of drawing a polygon with s vertices on a lattice of
The number of ways of drawing a polygon with s sides is roughly independent of the polygons that have already been drawn. The effect of the competition between polygons for available atoms is taken into account by the factor

"r. "

b.

c.

If we allow overlapping of polygons there will not be much error.

ns small enough so that given ns random polygons with s sides, there prob­

The

Equation ( 1 1 .46) becomes

Q

or g =

s r"t Rt sfI �� ( yr ) ns . . nl ns ! ; "t,1I2, . ! 2 L

[

- kT R l r +

r is chosen so that

=

n L�� �� � fI

lit

.to R.(yry}
=

f

n l ! ,; 2 ns ; O

[Rs(yr)S] ns n, !

( 1 1 . 55a)

00

Rlr + sL sRs( yr)'. (1 1 . 56a) ;2 We now must find Rs' If s = I , we wish to find R the number of ways of picking one point out of
349

dJs

(d = length of each step), the volume in which a typical walk ends is s proportional to S3/ 2 . Thus hs is proportional to I /s 3 / 2 for large s, where the proportionality constant is independent of s. Equations 0 1 .55a) and ( 1 1 . 56a) become g =

- NKTr

+

C

(ryl)S sL ; 2 S5/2 ' 00

--

( 1 1 . 55b) ( 1 1 .56b)

where c is a proportionality constant. For extremely large s, our result above for hs is incorrect. If one chain covers a volume comparable to the size of the whole system, then hs = IS/N, where I /N is the probability that a random walk will end at the original point (or any other particular point). Then a more realistic hs might be

hs

�

(� l) I S. S 3 /2 N +

which has the correct behavior for medium and large s. In any case, the sums in Eqs. ( 1 1 . 55a) and o 1. 56a) converge for ryl < 1 and diverge for ryl > 1. At reasonably high temperatures y is small and the sum converges. Also, at such temperatures there are not too many polygons, so assumptions (a), (b), and (c) are not too unrealistic. Thus we expect our results to be qualitatively correct above the A point. Equations ( l 1 . 55b) and (I 1 . 56b) resemble the results for a Bose gas (see Section 1 .8) and lead to an increase in specific heat with decreasing temperature. The shape of the specific heat curve is shown in Fig. 1 1 .37.

I,

Specific heat

==

drawing a polygon with s vertices if we started at any atom ; but we have counted each polygon s times. Thus Rs = Nhs/s. To avoid worrying about what happens when we get near the boundary, we simply consider the system to have periodic boundary conditions. That is, in two dimensions the system would have the topology of a torus. We will now approximate hs for fairly large s assuming that the temperatures are such that small s does not contribute much to the sums in Eq . ( 1 1 . 55a) and Eq. ( 1 1 . 56a). Suppose I is the number of nearest neighbors of each lattice point. Consider the problem of random wal k on the

transition in liquid helium

lattice. There are IS equally probable random walks that start on a given atom and make s steps. In hs' we wish to consider only those ways in which the walks terminate at the origin. The fraction of walks that end up at the origin is in­ versely proportional to the volume of space in which the walk is likely to end. Sin:e the distance from the origin of the typical random walk is of the order

ably will not be any that are on top of each other.

Then we can choose ns polygons of length s in about R:s/ns ! ways. By assumption (a), the total number of ways of choosing sets of polygons with ns polygons with s sides is

l

---; ;--!£-----L---_ T T,

Fig.

1 1.37 Specific heat near the lambda point.

350

Superftuidity

This analysis (which is supposed to be good for T > T;) thus explains why C increases as T approaches T;. from above. This treatment does not lead to the discontinuity of the kind shown in Fig. 1 1 . 34, probably because of the approx­ imations made between Eqs. ( 1 1 . 52) and ( 1 1 . 5 5a). In other words, we expect that the volume of the helium atoms described by p in Eq. ( 1 1 . 52) is the cause of the discontinuity in the specific heat. Kikuchi and others have written papers in which it is indicated that more accurate computations starting from Eq. ( 1 1 .52) would lead to a discontinuity. * How do these polygon calculations differ from the ones we did in Chapter 5 for the Onsager problem?

INDEX

Problem :

Annihilation operators, see creation and annihilation operators Bardeen-Cooper-Schrieffer (BCS) model, 269 BCS model, see Bardeen-Cooper-Schrieffer model Blackbody radiation, 1 0- 1 2 Boltzmann's constant, 1 , 6 Bose-Einstein condensation, 3 1 -33 Bose-Einstein statistics, 30-33 classical limit of, 3 1 density matrix for, 60-64 field operator for, 1 73 , 1 74- 1 76 free energy for, 64

* Kikuchi , R., Phys. Rev. 96, 563 ( 1 954). Kikuchi, R . , Denman, H. H . , and Schrei­ ber, C. L., Phys. Rev. 1 19, 1 823 ( 1 960).

Chemical potential, /1, 26-27 in a Bose-Einstein system, 26 in a classical system, 1 09 determination from number of particles, 27, 28, 30, 34, 1 09 in a Fermi-Dirac system, 26 for the lambda transition problem, 347 for a one-dimensional classical gas, 1 19, 121 for a one-dimensional order-disorder problem, 1 30 for a two-dimensional order-disorder problem, 1 3 1 Classical systems, 97- 1 26 condensation of, 1 25-1 26 equation of state of, 98, 1 00, 1 02, 1 03, 1 10, 1 1 4 Mayer cluster expansion for, 1 05-1 1 0 of one-dimensional gas, 1 1 7- 1 25 partition function for, 97 radial distribution function for, 1 1 1 - 1 1 3 second virial coefficient for, 1 00- 1 04 thermal energy of, 98, 1 1 4

Cooper pairs, 274 Correlation energy of electrons, 249, 255-264 Creation and annihilation operators, 1 53 for Bose-Einstein systems, 1 73, 1 74-1 76 expression of other operators in terms of, 1 76-1 82, 1 88, 1 89, 1 9 1 , 1 92 for Fermi-Dirac systems, 1 73, 1 74- 1 76 and the harmonic oscillator, 1 54 for phonons, 1 59, 1 62 for spin waves, 2 1 7 for a system of harmonic oscillators, 1 57- 1 59 and an unharmonic oscillator, 1 56 Critical temperature for Bose-Einstein condensation, 3 1 for liquid helium, 3 1 3 for order-disorder systems, 1 40 for a superconductor, 284 Debye approximation, 1 9 Debye temperature, 20 Density matrix, 39-7 1 , 40, 47 in the classical limit, 77 differential equation for, 48 for free particles, 48-49, 59, 73-74 for free Bose-Einstein particles, 60-64 for free Fermi-Dirac particles, 64 functions of, 45 for an harmonic oscillator, 49-53, 80-84 for an interacting harmonic oscillator, 8 1 -84 for one-dimensional motion, 74-76 path integral formulation of, 72-96, see also path integrals perturbation expansion for, 66-67 for polarized light, 42-43 for a pure state, 41 in the x-representation, 41 351

352

Index

time derivative of, 44 of an unharmonic oscillator, 53-55 Wigner's function and, 58-59 see also density submatrix Density of states, 2 connection with temperature, 3 independence of energy, for a heat bath, 3-5, 6 Density operator, 248, 253, 326 equation of motion of, 253 Density submatrix, 64-65 see also distribution function Distribution function, 65 Kirkwood approximation for three particles, 1 1 6 for a single particle, 65, I I I for two particles, 65, 1 1 1-1 1 3, see also two particle distribution function Effective mass, 227, 241 , 3 1 7, 345 Electrons in a metal, see electrons in metals and phonons, 1 85-1 90, 221 , 224, 23 1 , 270-273 and photons, 1 90- 1 92 superconducting, 301 Electrons in metals, 37, 242-264 correlation energy of, 249, 255-264 energy of, 243, 244, 249, 264 Fermi energy for, 243 Fermi-Thomas model for, 250 Fermi-Thomas wavelength for, 250 phonons in, 244-246 plasma oscillations in, see plasma oscillations specific heat of, 37-38 structure factor for, 248, 330 two particle distribution function for, 246 Energy, thermal, see thermal energy Entropy, 6, 45 of Bose-Einstein particles, 29 of Fermi-Dirac particles, 29 law of increase of, 8-9 of liquid helium, 3 1 3-3 14 for order-disorder problems, 1 36 for a superconductor, 283, 299 Equation of state, 98, 100, 1 02, 1 03, 1 10 Equilibrium, thermal, see Thermal equilib­ rium

Index Exchange integral, 1 99, 200 Expectation value, of an observable, 1 , 42, 59 of energy, 7, see also thermal energy of kinetic energy, 52, 57, 58, 65 of momentum, 45-46 of position, 45 of potential energy, 52, 57, 65 Fermi-Dirac statistics, 34-38 density matrix for, 64 field operator for, 1 73, 1 74-1 76 free energy for, 64 Fermi energy, 34, 243 Fermi temperature, 34 Fermi-Thomas model, 250 Fermi-Thomas wavelength, 250 Ferromagnetism, 1 29 Feynman's diagrams, 1 92-196 correlation energy of electron gas and, 257-264 Field, 1 62 Field operator, 1 63-1 67 for Bose-Einstein systems, 1 73, 1 74-1 76 for Fermi-Dirac systems, 1 73 , 1 74-1 76 see also, creation operators, annihilation operators Free energy, 6 of blackbody radiation, 1 1 of Bose-Einstein particles, 30, 64 in the c1assical limit, 77, 94, 95, 97, 98, 1 09 of combined systems, 9 of a crystal, 1 7 for Fermi-Dirac particles, 64 of free particles, 49 of an harmonic oscillator, 9, 52 inequality for, 47, 54-55, 67-71 , 87, 93, 233 for liquid helium, 3 1 7, 345, 347 for one-dimensional motion, 76, 1 1 9, 1 20 for one-dimensional order-disorder problems, 1 30 perturbation expansion for, 56, 67-71 of spin waves, 209 for a superconductor, 266, 282, 283, 299, 302 for two-dimensional order-disorder systems, 1 48, 149

Gibbs function, 27-28 for Bose-Einstein systems, 30, 32 for Fermi-Dirac systems, 34 for a superconductor, 266 Graph representations for cluster expansions, 1 06 for the correlation energy, 257 Feynman's diagrams, 1 92-196, 257-264 for the lambda transition, 347 for the Onsager problem, 1 37 sausage diagrams, 262 in superconductivity, 270-273 Ground state for a fermion system, 1 82-1 85 for liquid helium, 321-324 for many particle systems, 1 52 for a superconductor, 274-279

353

viscosity of, 3 1 3 vortex lines in, 339-342 Mayer cluster expansion, 105-1 10 Meissner effect, 266-269 Mossbauer effect, 22-25 Normal modes, of vibrations in a solid ' 1 3-1 7 Onsager problem, 1 36-1 50, 350 Order-disorder problems, 1 27-1 50 critical temperature for, 1 40 in one dimension, 1 30- 1 3 1 i n two dimensions, 1 3 1 -1 50

Partition function, 1 -9 of Bose-Einstein particles, 25-27 for classical systems, 97 Heat bath, 1 , 2 for Fermi-Dirac particles, 25-27 of harmonic oscillators, 3-4 of an harmonic oscillator, 9 of free particles in a box, 4-5 for order-disorder systems, 1 27 Heat capacity of a macroscopic system, 6 for the polaron problem, 231 Hydrodynamics, 305, 3 1 9 for two-dimensional order-disorder systems, 1 36-14 8 Junctions, with superconductors, 287-290, Path integrals, 72-96 306 calculation by expansion about the classical path, 79-8 1 Kirkwood approximation, 1 1 6 calculation by perturbation expansion, 84-85 Lambda point of liquid helium, 33, 3 1 3, definition of, 72-73 343 and the polaron problem, 232 Liquid helium, 3 1 2-350 variational principle for, 86-87 critical velocity, 3 1 3 , 334, 342 Pauli matrices, 200 effective mass, 345 spin exchange operator and, 202 energy, 3 1 7 Permutations, of identical particles, 62 entropy of, 3 1 3-3 1 4 Phonons, 1 2-22, 1 59-1 62 excitations in, 324 in a cubic lattice, 20-22 free energy in, 3.1 7, 345, 347 dispersion relations of, 1 6, 221 ground state of, 321-324 and electrons, 1 8 5- 1 90, 22 1 , 224, 23 1 , lambda point of, 3 1 3, 343 270-273 mechanocaloric effect in, 3 1 3-3 14 in an electron gas, 244-246 phonons in, 3 1 5, 326-327 free energy of, 1 7 quantization of vorticity, 336 i n liquid helium, 3 1 5, 326-327 rotons in, 3 1 7, 327-333 thermal energy of, 1 8 , 1 9-20 second sound in, 3 1 5 Planck radiation law, 1 2 specific heat of, 34, 3 1 3, 3 1 5, 3 1 8, Plasma oscillations, 249-254 349-350 dispersion relation for, 2 5 1 , 254 structure factor of, 330 random phase approximation for, 253 thermomechanical effect in, 3 1 3-3 14 variational approach to, 254 two fluid theory for, 3 1 5 wave function for, 254

354

Index

Polaron, 221 dissipation of energy of, 227-228 effective ",ass of, 227, 241 energy of, 226, 227, 235, 239, 240 partition function for, 2 3 1 path integral for, 232 variational treatment for, 234-241 Pressure, 7-8 of Bose-Einstein particles, 28 of classical particles, 98, 1 14 of Fermi-Dirac particles, 28 Probabilities, of a quantum mechanical state, 1 , 2-3 of combined systems, 5-6 Propagator, 260 Propagator factor for electron gas, 260 Rayleigh-Jeans law, 1 2 Random phase approximation (RPA), 252-254, 254 Rotons, 3 1 7, 333 RPA, see random phase approximation Second sound, 3 1 5 Specific heat, 7 of blackbody radiation, 1 2 of Bose-Einstein particles, 33 of Fermi-Dirac particles, 37-38 of liquid helium, 34, 3 1 3, 3 1 5, 3 1 8, 349-350 of spin waves, 209 of a superconductor, 265, 285 of vibrations in a solid, 1 8, 20 Spin exchange operator, 202 Spin-spin interactions, 1 98-200 Spin waves, 204, 206 combination of two, 207-2 1 7 energy of, 205, 206, 209 free energy of, 209 scattering of, 2 1 2-220 semiclassical interpretation of, 206-207 specific heat of, 209 Stefan-Boltzmann law, 1 2 Structure factor for an electron gas, 248 for liquid helium, 330 SupercC'nductors, 265-3 1 1 BCS model for, 269

critical temperature of, 284 with current, 290-3 1 1 energy of, 277, 28 1 , 293, 298, 303 energy gap in, 269, 279, 284, 285 entropy for, 283, 299 excitations in, 279-280, 283-284 free energy of, 266, 282, 283, 299, 302 ground state of, 274-279 Hamiltonian for, 273 Hard, 268 and hydrodynamics, 305 junction of, 287-290, 306 Lagrangian formalism for, 303 and magnetic field, 266, 267-268, 293-295, 3 1 1 in rotation, 3 1 0-3 1 1 soft, 268 specific heat of, 265, 285 type 1, 268 type II, 268 Superfiuids, 3 1 2-350 Temperature, 1 , 6 of harmonic oscillators, 4, 1 0 increase o f energy with, 6 Thermal energy, 7 of blackbody radiation, I I of Bose-Einstein particles, 3 1 , 33 for classical systems, 98, 1 1 4 of a crystal, 1 8, 1 9-20 of Fermi-Dirac particles, 35-37 of an harmonic oscillator, 9, 52 of spin waves, 209 of a two-dimensional order-disorder problem, 1 35 Thermal equilibrium, 1 Two particle distribution function, 65, 1 1 1 -1 1 3 Born-Green equation for, 1 1 5-1 1 6 for an electron gas, 246 at low densities, 1 1 5

: , '

ALSO BY R I CHARD P. F E YNMAN TH E THEORY OF F U N DAMENTA L PROCESSES

1 72

pp., i l i u s., paperbou nd

( Frontiers i n Physics Series)

32507 ( 1 962).

.

,

"The entire treatm ent i s of a refresh i ngly aristo cratic s i m p l icity . . . the book is ideal and most desira ble readin g to fol l ow al most any conve ntiona l i ntrod uction to field theor y. Not the least of its advan tages is that it afford s a l i ttle i nsight into the ma n ner in which a creati ve m i n d works i n th is f i e l d . "

Mathematical Reviews

QUANTUM E LE CTR ODYNAMICS

1 98

pp., i l lus., hardbound

32500; paperbound 32501 ( 1 962) .

( F rontiers i n Phsyics Series)

The ai m of th is book is to presen t the mai n resu l ts and ca lculat ional proced ures of q � antum electr odyna m ics i n as si mple and straigh tforward a way as possib le. . I t IS I n tende d for the studen t worki ng for a degree i n exper iment al physic s who does not i ntend to take more advan ced graduate course s i n theore tical physic s. The materi al consists of notes on the th ird of a three-s emeste r course given at the Ca l i forn ia I nstitute of Tech nology . "The author ach ieves exactly what he promis es, and the manne r of h i s present a­ tion is well enoug h k n own by now not to need the recom menda tion of a reviewe r . "

Mathematical Reviews

Vibrations in solids, see phonons Virial coefficients, 1 00-1 04, 1 09 Virial theorem, 55-57 Vortex lines, 336, 338-339 Wigner's function,

see

density matrix

7

'J