Geometry, Particles, and Fields (Graduate Texts in Contemporary Physics)

(1.20a) to make it look like (1.20b). This may at first sight not seem so impressing.

Although we have reduced the number of equations from

4 to 2, they are still complicated. Especially, they are still mixed -+ in

(1.6) represented by the gauge potentials
= -

V

aAo
then we can immediately find other gauge potentials representi:ng the same field strengths

Bo

and

trary scalar field.

Eo

To see this, let

x(t,i)

be an arbi-

Then:

will do the job:

Vx A by (1.12) and .. ait ;I: ;I: ax aAo a-+ -v

..

-V

-+ aAo -~

=

E0

Furthermore, the new gauge potentials

A ing

into (1.20a)

and

X drop out.

(l.20bJ, you will easily find that the terms contain-

By representing the electromagnetic field through

the gauge potentials

of motion are unchanged un-

der the transformations: (1.21)

.. A

A+

Vx

10

I

Such a transformation is called a gauge transformation and since physics is unchanged under this transformation, we speak of gauge symmetry. Hence, we see that a physical system, like the electromagnetic field, is described not only by a single gauge potential

(
but by a whole

family of
Con-

sequently, we may parametrize the solution in the following way:

(
(
lx at' A0

+

Vx)

By picking a special member of this family, we say that we have chosen a specific gauge.

This freedom of choosing a gauge is very important.

Let us now return to the equations of motion (l.20a) and (l.20b)

(
If

(
X

so that

satisfies:

o

(1.22)

To see this, we substitute

We then get:

o i.e.

~x - ~2 ~2 = - [V.A o + c at

1 Ha] c 2 at

This is a wave equation where we know the source term: ;!;

-

-+

[V'A o

1

a<po

+ c 2 at]

We can solve it and hence, we have obtained the appropriate gauge con-

dition

(1.22).

Observe

~hat

the solution is not unique:

We may al-

ways add a solution to the homogenous wave equation: (1.23)

o

Hence, we are still allowed to make gauge transformations:
A A + Vx -+

fies

(1.23) ~

tion, and we say that we work in the Lorenz gauge.

*)

-+


-

}t ,

X satisThe gauge condition (1.22) is called the Lorenz*~ondi­

without spoiling the condition (1.22) prcvided

In this gauge the

L.V. Lorenz (1829-91). Danish physicist who, independently of Maxwell, constructed a theory of light b~ing electromagnetic waves.

11

I

equations of motion simplify considerably, because the equations for ~

A

and

decouple:

(1. 24)

These equations are beautiful wave equations and they can be solved by standard techniques.

Passing by, we observe that the Lorenz condition:

(1. 22)

has exactly the same form as the equation of continuity: (1.17) In gauge theories it is a standard "trick" to let the gauge condition resemble the characteristic equation of the source.

1.3

MAGNETIC FLUX To get some experience with the

B

gauge potentials, we will deduce a formula

n/-y s

~

-+

B

/

a surface

r

_--------------------_

/

for the magnetic flux.

,

n \

S

Let us consider

bounded by the closed loop

We are interested in the magnetic

flux passing through the surface at a specific time

t

=

to .

By definition

the flux is given by the formula:

r

qI

=

J B.

r;

dA

S

Fig. 3

Substituting (1.18), we get: qI

~

J(VXA) . r;

dA

S

which, by Stokes' theorem (1.16), may be rewritten as:

(1.25)

Hence, we can express the magnetic flux in terms of

A

we have put no restriction On the electromagnetic field. be static or anything else.

Observe that It need not

This formula has important consequences.

12

I

Consider a closed surface S. We assume that the electromagnetic field is smooth in a neighbourhood of S but it may be singular at a finite number of pOints inside it.

(The electromagnetic field created

by an electron is singular at the actual position of the electron) . Let us now calculate the magnetic flux through

S

qI=JE.ndA=fA'dr s but since

S

gral vanishes and

+

qI

=

r r

is closed,

is empty.

Consequently, the last inte-

0

s

s

+ +:

SINGULARITIES

~

Fig.

4

You may feel I am cheating you! But let me try to convince you in another way: Choose a closed loop r lying on S, which divides S into two regions: Sl and S2 (See fig. 4). The corresponding fluxes through S1 and S2 will be denoted q\l and q\2' We then have: q\1

• n

n

dA =

because this time we integrate the other way around get as postulated:

r

dA

-+

dr

Adding the two fluxes, we

So we have shown: Theorem 1.

If the electromagnetic field is generated from a gauge potential

(~,A), then the magnetic fluz through any closed surface automatically vanishes!

To understand this result, let us consider the static Coulomb field created by a single electron.

Let

S

be the surface of a ball with

13

I

radius

r

and center at the position

of the electron.

Then the normal com-

....

ponent of the electric field

is

E

easily found:

. fi

....

1 r2 It is constant on the surface E

-'L 4'1TEO

S

and

the electric flux is then easy to calculate:

JE .

1

fi dA

~

r

s --.!L. J:.. 4 '1Tr 2

Fig. 5

4'1TEO r2

IdA S

= .3.. EO

All these computations were made just to remind you that the electric flux through a closed surface is pro-

portional to the charge contained inside it.

Therefore the above re-

sult about the magnetic flux simply excludes the possibility of magnetic charges, i.e. magnetic monopoles.

Of course, this is not sur-

prising because we started with the Maxwell equations etc. where we have included an electric charge density, but not a magnetic charge density. We are now prepared to consider a space region

n

field, which cannot be derived from a gauge potential

with a magnetic

A.

Consider:

(a)

The electromagnetic field created by a resting electron.

(b)

The electromagnetic field created by a hypothetical magnetic monopole.

0

In both cases the fields are singular at the origin clude that point.

The remaining part of space is denoted

n In

n

= R3 "

so we ex-

n:

(O}

the electromagnetic field solves the following Maxwell equa-

tions:

o

v•E

o

v

x

E

This can be shown by a direct computation, and we emphasize that they will hold for both case (a) and (b).

Observe that in

neither electric charges nor magnetic charges.

n

there is

Furthermore, the Cou-

lomb field and the monopole field do not differ by their field equations.

But they differ when we try to represent them by gauge potentials!

14

I

The Coulomb field may be represented by the gauge potential

"'-~ 471E:

'i'

o

1.

A = -0

r

because this immediately leads to:

B = V x A = -0

E = -VIj>

and

But the monopole field cannot be represented by a gauge potential (1j>,A): This is because the magnetic flux through the unit sphere

S

is dif-

ferent from zero: q\

=

JB

...9:.. • 471

fi dA

471

=

g

*

0

S

We immediately see that this magnetic field contradicts theorem 1. Thus, we have shown that even if

.... 'V

in a space region

Q

....

B = 0

, we cannot in general conclude that

written on the form: (1.18)

x

B

may be

....

A

We may summarize the preceding discussion in the following way: The concept of magnetic monopoles and the concept of gauge potentials sean to be in conflict with each other!

Later on we shall spend much time

constructing a theory where magnetic monopoles and gauge potentials live in peaceful co-existence.

I

15

1.4

ILLUSTRATIVE EXAMPLE: THE GAUGE POTENTIAL OF ASOLENOID Up to this point we have discussed gauge potentials in very general

terms only, so let us pause to compute a gauge potential in a special situation. Consider a long coil of wire wound in a tight helix.

A current

through the wire will produce a very strong magnetic field inside the solenoid, and if the solenoid is very long compared to its diameter, the magnetic field outside the solenoid will be negligible.

....

....

B

B

Let us consider

the ideal case of an infinitely long solenoid.

B= 0

Then we can safely put

outside it.

Inside the sole-

B

noid the magnetic field

will

everywhere be pointing in the same direction. Now, let us try to compute the gauge potential

A

which is the only gauge

potential we need to be concerned about.

Inside the solenoid we have a constant magnetic field:

This may be represented by -+

A =

Bn """2

(-y,x,O)

1

:2

-+

B

-+

x r

as you can easily verify. This choice of gauge has nice properties: (1)

The magnitude of the gauge field is proportional to the distance

(2)

p

from the z-axis.

It is always perpendicular to the z-axis and the radial vector

-+

p.

Outside the solenoid the magnetic field vanishes:

A=0 ,

it would be tempting to put a closed loop through

r

r

but this is wrong!

B = o.

Hence,

If we consider

as shown in fig. 8, then the magnetic flux passing

is

JB

s

A

ndA

16

I

but it can also be expressed as the line integral: (1.25)

...

fA

=



dr

and this is certainly inconsistent

A

with the choice: we do then?

0

! What can

First, you observe that

the gauge potential inside the solenoid is circulating around the z-axis, but then it is tempting to let the gauge potential outside the solenoid do the same:

A

(-y,x,O)

a:

Second, we know that the line integral should have the constant value B o na 2 But if we choose r as a circle with the radius Po then it is clear that the line integral is proportional to the magnitude of and proportional to the radius ly proportional to

IAI

Hence,

must vary inverse-

Po:

IAl

a:

1

j

x 2 +y2

Combining these two things, we look for

A=

A

on the form:

k(- ~ , _ _ x_ ,0)

x 2 +y2

x 2+y2

The first thing we should check is the following one: really vanish?

This is easily verified.

Does its curl

The second thing we should

check is whether it produces the correct magnetic flux or not.

A

tangential component of k •

The is

and so we get:

1 Po

...

k •

dr

J:.. • Po

211 P

2nk

O

Hence, we can determine the constant k , because

...

A



<1-...OJj)

Fig. 9

IIII1

2nk

=

=~

Bona2i.e. k

B oa 2 .

Thus, we have completely solved our problem: ... A inside

Bo

= ""2

... ·A outside =

Bo

(-y,x, 0) ;

""2.

a

2

x 2 +y2

(-y,x,O)

17

Observe that Ainside ary of the solenoid.

and

I

Aoutside

match continuously on the bound-

You might find it puzzling that we can have a non-vanishing gauge potential A in a space region where the magnetic field vanishes!

At

first you might think that it occurred only because we worked in a "bad" gauge, so maybe, if we used a gauge transformation, we could "gauge away"

A?

have seen that

A

But we have already killed this hope because we

=0

through the solenoid!

is incompatible with the magnetic flux passing Hence, we cannot gauge away

A

throughout the

exterior of the solenoid. However, it is possible to gauge away the gauge potential a.lm:Jst everywhere!

To see this, we consider the function: Arctan 'I.. x


y

Observe that

(0,1) (x,y)

7r
r
\



simply produces the

polar angle and that

is smooth


except at the negative x-axis, where it makes a jump from

+11

+7r

to

-11

X

a 'I.. ax Arctan x

~ ax


- '1..

2 __ x_

1+('1..)2

x

7r
-11

First we compute the derivatives of

Fig. 10 a Arctan 'I.. x ay

E.1 ay

1 x

x

1+('1..) 2

x

They look very much like

Ax

and

~

x 2 +y2

~

X +y

Ay

Now consider the gauge transformation with X(t,x,y,z)

B

a2

o -2
= -

This produces the equivalent gauge potential: -~

+

A'

+

A + 'ilX

B a2 2

__ x_

~

x 2 +y2

B0 a 2 -2-

o Hence, we have managed to gauge away y

=

0, x < 0 , where

X

-~

o

x -2--2 x +y

o

o

Q

x 2+y2

x 2 +y2

is singular.

A

except for the half plane:

Since there is symmetry about

I

18

the z-axis, we may ignore the z-coordinate for a moment.

The problem is then really

two-dimensional, and we see that everything is all right except on the negative x-axis. Since

X

is

discontin~ous

along the nega-

tive x-axis, we conclude that singular - like the negative x-axis.

Vx

becomes

a-function - on the

If we insist on perform-

ing this singular gauge transformation, i.e.

x

insist on using a singular gauge, we see

Fig. 11

that we have concentrated a singularity in the gauge potential along the negative x-axis.

SUch a string on which the gauge potential is singular is called a DiraC! string. We will discuss them in greater detail later on. Now what about the fact that

~ =

fA.

dr

=

BoTIa

Z

Does it hold in the singular gauge, too?

Dirac string

Yes!

When we integrate, we

necessarily pass the negative x-axis

A'

where

\

has a

a-like singularity

and this gives the correct contribution to the line integral.

x

We may clarify these remarks by recalling the definition of a a-function. It is defined by the property that

A' is

singular!

+

Fig .. 12

~

J

a(x)f(x)dx = f(O)

for any smooth function which vanishes at infinity. Now consider the Heaviside function S(x) defined by: for

e (x)

x
0

for x < 0 This function makes a jump of height 1 at x = 0 and we are going to show that: o(x)

=

de dx

In fact. we immediately find: +

+

~

-J

de f (x)dx dx

[e(x}f(x)]: : -

~

Jf' (x)dx

- [f(x)]~

f(O)

~

J~

e,(x}f' (x)dx

19

I

where we have integrated by parts and extensively used that f vanishes at infinity. Hence, we have demonstrated in all details that if a function makes a· jump, then the derivatives get a-like singularities.

Worked exercise: 1.4.1 Problem: Prove the following formula: +

where

1.5

g

~

J f (x) 0 (g

(x) )dx

o

=

is a monotonous function!

RELATIVISTIC FORMULATION OF THE THEORY OF ELECTROMAGNETISM Up to this point we have carefully separated space and tLme, and

we have not used Lorentz invariance.

However, it will several times

be more convenient to translate the results into a Lorentz invariant form.

We assume that you are familiar with the basic principles of

special relativity, but to fix notation we have collected some useful formulas:

In relativistic formulas we put

Indices:

Space-time indices:

a,

c = 11 = EO = 1

(Natural units)

a,y, ...

=

0, 1, 2, 3

Space indices:

i,j,k,l

=

Other indices:

a,b,c,d

=

1, 2, 3 1, 2, ... , n

,)l,V

SEace-tLme diagram Coordinates: P(t,x,y,z)

+

t

xa : xO=t

Cl

= 0, 1, 2, 3 xl=x

x 2=y

x 3=z

Parametrization of world line: x

Cl

=

Xa(T)

T : ProEer time x

1

)

y

dt = ydT Y

~=i (t)

=

1

7i7

20

I

Metric:

Four vector: ka ka

= =

w: k :

o

(k ,k i ) naS

kS

=

(w,R)

= (-w,k)

Scalar part Vector part

ka k

a

=

-w 2 +

k2

Tensor:

s 00 =

[ ~:O

II

-:-

oj ;i~

S OJ

Four-velocity:

Velocity: +

v

=

~

(dX

dz)

dt, dt, dt

va

is a time-like unit vector, pointing towards future

Energy, momentum: E

pa

= my

..

+

p =

Four-momentum: dxct

= m dT

-p pa = rn 2

myV

ct

= (E,

=

p)

square of rest mass

Charge density, current: -t

Four current: +

p, J = pv

Gauge potentials:
Gauge potential

A

Derivatives:

A

ct

= =

Maxwell field:

(-
Derivatives:

a = __3_ = a

axa

(!t' 0

V) d'Alembertian

21

Field strengths:

I

Field tensor:

-+ -+ E,B

Fas

=

0

-E x

Ex

0

E

Y

-B

EZ

-E

Y

Bz

-B

z Y

0

B

-B x

0

z

By

-E

x

Lorentz force:

F=

d-+ ~ dt

= q(E+~"B)

Fa

=

dpa a de = qF SuS

(1.26)

Maxwell eguations: -+

->-

VoB

-+

=

VOE =

0

dB + at

...Q..

dE dt

EO

VXE

- c 2V x

= -0 -+ B

daFSY + dsFya + dyFas

=

E~ation

of continuity:

-+ ~+ V

-T

]

at

Gauge -+ B =

d FCl.S S

=

~

d JCI. = 0

0

=

"" - ~ EO

=

CI.

0

(1. 27)

(1.28)

.

(1.29)

Eotentisl~:

Vx

-+ A ;

E = -V

dit -at

F

Cl.S

=

dCl.AS

-

dSACI.

(1.30)

Gauge transformation:
-+


_2.x

-+ -+ A-+ A +

at

A -+ A + dCl. X CI. CI

(1.31)

d ACI. CI.

(1.32)

Vx

Lorenz gauge:

V

0

-+ A +

...l..ll = 2 at c

0

=

0

22

Eq. of motion

I

for

the gauge potentials:

ll -

.1:..n 2 c 2 at

-E..._.1... (v·it at

+

EO

1

c

2

2.1)

(1.33)

at

2-+-

llA -

.1:..2 ~2 c

at

1 -+- -1-2 -t-J+ V( V°A + 2 2.2.) at E c c 0

Observe that the gauge potentials and A have been collected into a ACt = (,A).

single four-vector field

This four dimensionalgauge potential

will be referred to as the Maxwell field.

Worked exercise: 1.5.1 Problem: Use the relativistic formulation to re-examine the introduction of potentials, the equation of continuity and the equation of motion for the gauge potential.

1.6

THE ENERGY-MOMENTUM TENSOR As a more exciting thing we will discuss the energy-momentum ten-

sor of the Maxwell field. particles with position The charge density

As a preparation we consider a system of N

~n(t) and charges p is defined as:

qn0

3 -+- -+-

j

and the current

p(t,~) = L qno (x-x n (t) ) n is defined by: 3 -+- -+-

We may collect

p

and

3(t,it) = L qno (x-xn(t» n -tinto the four-vector: J

d-+xn °

at

(For Ct = 0 we remember that xn = t). This is not manifestly a Lorentz invariant but we may rearrange it.

I

23

To prove the Lorentz invariance of the proper time. JCL

we use the invariant paraCL

particle-trajector~es

metrization of the

xn

(T)

where

T

is

Using exercise (1.4.1), we rewrite the expression for

:

(1.34 )

dt dT

I t=x~ (T)

The last expression is obviously Lorentz invariant.

Observe that we

may deduce the equation of continuity directly from the definition:

Hence, we have obtained the conservation of charge without using the Maxwell equations.

But of course, it is not surprising in this very

simple model where we have a finite, but fixed, number of particles with fixed charges. This was entertainment.

Now we must go to business.

Each of the P~(t) = (En(t),Pn(t». Hence, the density of energy and momentum is given by: particles carries a four-momentum given by

T~~CH (t,~)

=I

n

p~(t)03(~-~n(t»

and the current of energy-momentum is given by:

We may unite them into a single formula:

(1.35)

CLf! ... T (t,x) MECH

24

I

The important thing is to recognize that this is a tensor. we rewrite the expression for

CLS -.. T (t,x) MECH

To see this

T~~CH

L n

n

i.e. (1. 36)

which immediately shows that it is a tensor.

The components of this

so-called energy-momentum tensor have the following meaning:

[

(1.37)

~~E_R~Y_ ~E~~I~~ ~ ~ ~~E~G_Y C~~R~~T 5-1 :

MOMENTUM DENSITY

g

Now we want the total energy to be conserved.

_

_

I

MOMENTUM CURRENT

:

"STRESS-TENSOR"

Hence, we expect the

energy density and energy current to satisfy an equation of continuity:

~~ +

V.

5 =

0

~ doToO + djToj = 0 ~ aSTo S

In the same way the total momentum should be conserved.

=0 Hence, the

density of the x-component and the x-component of the current should satisfy an equation of corttinuity:

etc!

Consequently, the conservation of the total energy and momentum

can be expressed by an equation of continuity, satisfied by the energymomentum tensor:

(1.38)

o

25

Returning to the system of

o[ IpCl(t)o

't o

n

n

3

I

N particles we get:

(~-~ (t»

o{

n

+

I

oxl.

d i

pcl

n

i

dx n03 (x-xn(t» ->-> + - . LPCl -dt

]

n

(t)~ dt

n

]

n

[_0_. 0 (xLxjn oxl. 3

(t»

1 J

The last two terms kill each other and we have shown:

(1. 3 9 )

If the particles are free,

i.e. they are not charged and they do not

experience any forces, etc., then

P~(t)

are constants and we get

immediately: (for a system of free particles!) However, if they are charged, they will create an electromagnetic field and then experience forces.

In this case we get:

'" 0 and, of course, this is not surprising because we have only counted the mechanicaZ part of energy and momentum and not the part of it stored in

the electromagnetic field!

Nothing can prevent mechanical energy from

being converted to field energy and vice versa! Now, introducing the energy-momentum tensor of the electromagnetic field

T~~, the total energy-momentum tensor may be decomposed:

and the conservation of energy and momentum ,requires:

26

I

CLS dSTMECH ' we may use this to determine

Since we know

dpCL

=- L

n

~

n

3

~

T

CLS EL

-+

0 (X-Xn (t) )

CL

dPn dT 3 -+- -+dt 0 (x-xn(t»

- I -aT n

~

- L qn n

FCL

dx S n dT 3(-+- -+- ( S-aT dt 0 x-x n t»

(due to the Lorentz force)

i.e. we have shown (1.40)

We have got to rearrange the term on the right hand side.

Worked exeraise: 1.6.1 Problem: Show that Maxwell's equations imply the formula FCL JS = 3 [FCL FYS + !nCLS FYOF } S B Y 4 YO From this exercise you immediately read off: (1.41)

This was a very long computation!

Let us summarize the results: We CLS CLS have introduced two energy-momentum tensors: T MECH and TEL' They obey the equations (1.39l and (1.40)

n

For a system of free particles this immediately implies: CLS dSTMECH because

pCL (t)

are constants.

n

field obeys: because

JS

In a similar way a free electromagnetic CLS dSTEL

=

0

= 0

for a free field.

=

0

Finally, if we have a system of

charged particles interacting with the electromagnetic field, we get:

27

I

T~S

because the total energy-momentum tensor d T~S

=

S

0

Exercise 1.6.2 Problem:

Show that

are symmetric tensors, i.e.

and

When we know the energy-momentum tensor of the electromagnetic field, we can especially find the energy density and the momentum density: TOO =-FO F YO -

E(t,X)

cY

~ (FYOF

E2 + 1 4 (2S2_ 2E2) =

yo 1

E

x

1 (E 2 + 2

a2 )

S

i.e. (1.42)

where we have re-introduced

EO

and

c.

We will finish with some remarks about energy-momentum tensors in general: Let us consider the conservation laws. implies conservation of charge and that tion of energy and momentum.

We have seen that

dST~S

=

0

a;~ = ~

0

implies conserva-

Actually, they imply something more:

Theorem 2 {Abraham's theorem} (a)

If

Ja(x)

is a field of four-vectors,

J

that: Q

3

JOd x

then

d~Ja

=0

implies

is a Lorentz-scalar, i.e. it is independent

xO=tO of the observer. (b)

If

T~S (x)

,. p"~es

th a: t

is a field of Lorentz tensors, then

p~

--

JT~Od3X

trans f orm

aST~S = 0

im-

as t h e components of an

ordinary four-vector. This will be proved in chapter 11, section 1.

28

I

T~6

Observe that in the cases we have been discussing

is symme-

tria (See exercise 1.6.2)

This has an important consequence. M~6Y(x)

M~ey

(Observe that

= x~T6Y

Consider: - x6T~Y

x~

is not a tensor because

is the coordinate of

point, and not the component of a vector!) Taking the divergence; we find:

=

M~6Y Y

a

o~ T 6Y + X~(d T 6Y ) - 06T~Y - x 6 (a T~Y) Y Y Y Y

T~6

Hence, the symmetry of conserved! instance:

J~6

Now

=

...g

f

- T~6

M~6od3~

is the four-dimensional angular momentum!

f 12 where

J~6(tO) =

implies that

= T6~

E

is For

( j ko xkT jO ld3~ ijk x T -

f

is the momentum density. Since we expect this to be conserved for a closed system, we must demand that the energy-momentum tensor is symmetria.

SOLUTIONS OF WORKED EXERCISES. No. 1.4.1

Suppose

g(x)

is a monotonous function and consider: +<0

f fix)

f

o(g(x»dx

f(x) o(g(x»

dg(x)

x=- g' (x)

x=-oo

where g(x o ) = 0 ] and g is increasing f (g_1 (u» u=-oo

If

g

O(u)du

=

g' (g-l (u»

is increasing, then

g' (x

) > O

f(g-l(O» g' (g-l(O»

0, and if

g

=

~;[ -g' (x o)

is decreasing, then

Hence, we may collect the above in the following formula:

-J

fix) o(g(xl)dx

x=-

f(xo) = ----"-

Ig' (x o) I

where g(x o) = 0 ] and g is decreasing g' (x )

o

<

O.

29

I

No. 1.5.1 Some time ago we solved the Maxwell equations:

~ • B=

0

dB + , dt

~ x E = -0

with the ansatz:

E=

aA

?4J _

-

dt Now we have reformulated these Maxwell equations as

o=

daFay

+

daFya

+

dyFaa

and we solve them with the ansatz: Faa = daAa - dSAa We can easily check this by explicit computation: daFBY + 0aFya + dyFaa da(daAy-dyAa) + da(dyAa-daAy) + Oy(daAa-daAa) (dada-dada)Ay +

{ayda-dady)Aa + (dady-dyda)Aa = 0

due to the fact that partial derivatives commute

(dada = dada)

From the two remaining Maxwell equations ~

• E =L. £0

we deduced the equation of continuity:

~

+

~

7-

o

J dt Now we have comprised these Maxwell equations into a single equation:

Taking the divergence, we get: but in

d~dBFaa = 0 because d d is symmetric in aa while Hence, we recover th~ equation of continuity:

as.

d

~

P

FaS

is antisymmetric

= 0

Finally, we may deduce the equations of motion for the Maxwell field: Aa. Substituting Faa = daAa - dSAa in the Maxwell equation, dSFaB = J~, we get: d (daAB_dSAa )

S

= J~ ~

If we choose the Lorentz gauge where

o~(o AS) (ODoS)A~ 6 ~

daAa = 0

= J~

~

these equations simplify to:

30

I

No. 1.6.1 a [Fa F8Y ] Y B

Fa Ja = Fa a FaY 8Y 8

l

(1.28)

J

[a Fa ]F8Y Y 8

The last term is a mess and we have got to rearrange it separately: INDEX - GYMNASTIC:

a Y a8 BY [ayF 8]F = [a F ]FBy

~(aYFa8)F 2

1 Y -(a Fa8)F

2

By

+ ~(aYFa8)FQ 2 ... 1 __ fJy_ _ __

8

~

Y

Y

~

8

1" ~ -(a 8Fay)F 8Y 2 1..._Y_I!_ _ _ _ Fay = _Fya . , r - - - - - - - Fy I! = -F 8Y

~(aYFa8)F8Y

+

~(a8FYa)F8Y

l[aYFal! + a8Fya]Foy = - l[aaF8Y]F

2

2

y

L-

"

(1.27)---..!

This was the complicated part of the computation. aoTal! = - FaoJ8 =-a [Fa F8Y] fJ

EL

fJ

=~ay[Fa8FBY

Y

+

~

8

Now we are almost through:

~ a [naY'F8oF u.]

4

Y

8

n'JYF8oFao]

In the last expression we interchange the dummy indices

8

and

y

and get:

31

I

ch"pter Z INTERACTION OF FIELDS AND PARTICLES 2,1

INTRODUCTION It is as we have seen possible to introduce a gauge potential

A~ ~ (~o,Ao)

describing the electromagnetic field, and this potential

has the peculiar property that the gauge transformed potential (1.

31)

describes the same electromagnetic field and satisfies the same equation of motion. Now a classical particle interacts with the electromagnetic field in the following way:

The field produces a force and this force de-

pends only on the value of the field strengths at the momentary pOSltion of the particle.

What is going on at other places is completely

irrelevant. Hence if there is a space-time region where

...E

and

...B

...B

...B

Bok

vanish, a

classical, charged particle will experience nothing.

But we have

seen that even if there are no field strengths in a space-time region

Q

there may be a non-

trivial gauge potential region.

A in

, I

...=...:::.lL~B a 2 A x 2 +y2

this

Here non-triviality

means tha t the gauge potential cannot be completely gauged away.

(L.!JJ..?

[-y]x IQIT..!) ~

0 ~

1111/

For

Fig. 13

instance, we have seen that outside a sOlenoid there is no magnetic field but there is a non-trivial gauge potential:

A=~~[-~] 2 x 2 +y2 0 Thus

you might ask:

Does there exist physically measurable effects

associated with the motion of a charged particle completely outside the solenoid?

In other words:

Can we distinguish between the situa-

tion where there is no current in the solenoid and the situation where there is one, just by letting a charged particle go around the solenoid?

32

I

Clearly, if there is such an effect it cannot be a classical one, because a classical, charged particle will experience no forces in either of the situations.

However, there could be a quantum mechan-

ical effect: To study this possibility we are faced with the problem of how to quantize the physical systems above.

If the fields are strong so

that the particle can be regarded as a test particle, then we are allowed to keep the fields on a classical level. quantized.

Hence instead of characterizing the particle by its clas~=~(t)

sical trajectory ~(r,t)

We treat them as

However, the motion of the particle should be

external fields.

we introduce a Schr5dinger wave function

with the well-known interpretation:

solute square

I~(r,t)

12

particle at the position

At the time

t

the ab-

is the probability density for finding the -+

r.

Instead of the usual Newtonian equa-

tion of motion: :;.

where the dots represent] [ time derivatives the dynamical evolution of the quantum-mechanical system is governed m x

(2.1)

by the Schrodinger equation:

a~

.

~ 11 at

(2.2)

= (-

(fi

112

2m

= 2~

i

h = Planck's constant)

~ + V(r)) W(~,t)

Although you probably already know the Schrodinger equation when there are electromagnetic fields present, it will be rewarding to examine the quantization procedure carefully.

2,2

LAGRANGIAN FORMALISM FOR PARTICLES: THE

NON~RELATIVISTIC

CASE

The first step wheq you want to quantize a system consists of casting the problem into the Lagrangian form. ClaSSically it is a compact way of deriving the equations of motion. Consider for simplicity a one-dimensional motion of a particle, say along the x-axis. We want to consider those motions of the particle where it moves from the space-time point The question is:

A to

A: (t1,x 1 ) to the space-time point B: (t 2 ,x 2 ) How can we find the classical path leading from

B?

First we will introduce a function

L(x,x)

of two variables.

It is called the Lagrangian, and the explicit form will be derived later on. With each

number:

path leading from

A

to

B

we now associate the

33

I

S

(2.3)

called the action associated with t-axis

that path.

SPACE-TIME-DIlIGRAM B (t ,X )

We want to choose the

path x = x(t) in such a way that the action has an extremum.

2

Suppose

x

=

xo(t)

produces such

an extremum and consider another path:

t

x x-axis

Fig. 14

= xo(t)

+ Ey(t) ; y(t l )

We will allow

8

= y(t 2 ) = 0

to vary but

keep y fixed for a moment. know that the action

We

t2 S (8)

=

J

L (x

o

(t) + E'y(t) '*0 +

y)dt

tl which may be considered a function of E

=

E, has an extremum for

0 . Consequently

t2

o = s'

(0)

=

J

[~~

yet) +

a~~)

y(t)] dt

tl

aL [ ax

d

aL

y(t) - dt ~

y(t) ] dt

t

J2 [~Lx o

_ ddt

aL]

0(*)

yet) dt

where we have used the boundary conditions: 0 = y(tl) = y(t 2 ) . However yet) was an arbitrarily chosen function. Therefore the above identity is only consistent if

x = xo(t)

satisfies the fol-

lowing differential equation known as the EuZer-Lagrange equation

(2.4)

34

I

Illustrative example: One-dimensional motion in a potential V(x) , Let us try to determine the Lagrangian in such a way that the Euler-Lagrange equation aL d aL (2.4) reproduces Newton's equation of motion:

av

d (m:ic) dt You innnediately read off the following relationships aL av L(x,x) (TERMS INVOLVING x) - Vex) ax = - dX ( 2. 1)

- ax

aL = mx ax i.e.

=

..

mx

=

L(x,x)

!m(x)2 + (TERMS INVOLVING

L(x,x)

!m(x)2

x)

Vex) + CONSTANT

Consequently we have determined the following suitable Lagrangian (2.5) L(x,x) = !m(x)2 - Vex) The above scheme is easily generalized to more complicated Pl\RI'ICLE 2

systems.

Suppose we have a

system characterized by the following set of generalized coordinates: ql, ••• , qn. If we

CENTER OF MASS: (x, y, z) = (ql rq2,q

have two particles for instance,

Z

then PARTICLE 1

ql,q2,q3

could be the

Cartesian coordinates of the center of mass, and

q 4 ,q 5 ,q 6

could

be the polar coordinates of the

y

vector connecting particle x

1 2. These generalized coordinates take their value in a subset ~ of the and particle

Fig. 15

n-dimensional space

R n . We call

Q

the configuration space.

The

dynamical evolution of the system is described by a curve qi

=

qi(t)

Lagrangian

in the configuration space. L(qi,qi,t)

We now introduce a

which may depend explicitly on time too.

Next we consider paths connecting the point with the point

A: (tl,q~,

, qn)

B: (t2,q~, " ' , q~). With each of these paths we as-

sociate the action: S

t2

J L(q

i

oi

,q ,t)dt

and we want to determine a path Again we suppose that

qi

sider another nearby path:

q~(t)

q

i

=

i

qo(t)

extremizing the action.

actually extremizes it and con-

35

q

We will allow

Ei

I

i

to vary but keep

y(t)

fixed for a moment.

Con-

sider the action, t2

i i • i i. L(qo +E y, qo +E y,t)dt

J tl E1,""",E n

as a fUnction of

J [~y aql

as

o

-r aE

It assumes an extremal value for

Therefore t z

(0,'"",0)

I (0,'·",0) tl

+

aL •i aq

Y]

dt

t

-~ ~] J zr~ La dt a"l q

t

1

yet) dt

q

1

But this is consistent only if

o ,

(2.6)

Thus

ial ,

•.. , n

each of the components satisfies the appropriate Euler-Lagrange

equation. Let us return to the case where a single particle moves in the three-dimensional space. tential

veil

If it moves under the influence of a po-

then we can immediately extend the above analysis.

It corresponds to a Lagrangian: 1

which reproduces the (2.8)

-+-2

'2 mv

(2.7)

m

...

- Vex)

well-~

Newtonian equation of motion

dZi dt

Z

But we are really interested in the motion of a charged particle in an electromagnetic field.

This corresponds to the equation of mo-

tion (2.9)

m

dZ~

dt

Z

and this equation cannot immediately be reproduced by a Lagrangian

36

I

of the form (2.7) for two reasons: 1.

The force may contain an explicit time dependence because the field strengths may depend explicitly on time. sider for instance the electric field

-+

E

(Con-

in a capacitor

joined to a circuit with an oscillating current.) 2.

The force is velocity-dependent due to the

-+

-+

v • B-term.

Hence we can try to replace the simple-minded potential by a generalized potential

U(x,v,t)

V(~)

which may depend explicitly

both on pOSition, velocity and time.

So we look for a Langrangian

on the form: (2.10)

-+

-+

L(x,v,t)

=

21

-+2

mv

-+ -+

- U(x,v,t)

Substituting this Lagrangian into the Euler-Lagrange equations (2.6) we get aL

0

d aL dt --::;: av

--::;:

ax

-

au -+

ax

d -+ d dt (mv) + dt

au -+

av

Le.

au -+

ax

d

+ dt

au -+

av

To determine the generalized potential we must rearrange the expression for the Lorentz force (2.9)

d2~

m --2

dt First we introduce the gauge pctentials

through the well-known

(<j>,A)

relations (1.19)

-+

B

-+

...

-+ -+

'V x A

E

-

4-

'V<j>

dA

--at

giving: d2~

m --2 dt

-+

-+ -+ -+ -+ aA q(- 'V<j> - at + v x ('V x A»

Clearly, this is a mess and we will have to rearrange the -+ x A)-term. v x USing the standard rules of vector analysis we get

-+

(V

37

I

v-+ x -+

dA

-

V·-+

ax

substituting this into the equation of motion we now obtain

x

-+

d2 m-2 dt

q(-

q(-

q(-

dA lP._ at: -+

ax a

-+

-+

=

d

-

-+

-+

a -+

a -+

ax

-+

(v'A)

ax -+

v'A]

-

ax

and we are almost through. A

[
+

_ aA -+ dx -+) ax dt

aA _ aA.d;:) ax dt

at

-+ -+ dA\ [

Performing the trivial substitution:

d

-+-+

-+-+

-(A'v-
-:;:(A'V) av

which is permitted because

-+

does not depend on


v

,

we get

-+ -+ ) v'A]

and from this formula we can immediately read off the generalized potential U(~,~,t)

(2.11)

-+

-+

q[

Thus we have found the total Lagrangian:

L(x,~,t)

(2.12)

21

-+2

mv

-+

-+

-+

-+

- q
We may think of the above result in the following way:

If we have

a system without electromagnetic fields, then to any path joining A: (t l ,Xl)

and

B: (t

2

,X 2 ) t

2

1 J (2

there corresponds an action -+2

mv

If we switch on an external electromagnetic field

(
have to add another piece of action,the interaction term:

then we

3&

I

t

The total action now becomes:

\

S

=

So + Sr' We may rewrite the in-

teraction term in a more elegant manner. Let us denote the path join-

\ i

"'-A

ing the space-time points A:(tl,Xl) and B: (tz,xz) by

)-------~----~y

f. We then obtain

Fig. 16 t2 Sr

f ~dt

-q

tz + q

tz q[ f-~dt +

~

f A.~~

dt B

tz

f

A·d;]

q

f

[Aodx" + A.dx i ] 1

A

i.e. (2.13)

In the Lagrangian formalism this formula replaces the Lorentz force! Observe that in this formalism the particle interacts directly with the gauge pctential and not with the field strengths. Now remember the

gauge symmetry: We are allowed to replace Au by Au + dUX without changing any physics. This is obvious if we look at the equation of motion:

because the field strengths are gauge invariant! But does there exist a quick way to see it in the Lagrangian formalism? What we want to show is that the form of the Lagrangian immediately implies that the motion of a particle is usaffected by a gauge transformation. rt should be possible to see this without explicitly computing the equation of motion. The particle is following a path fo which extremizes the full action S

= So

+ Sr. Performing a gauge transformation we get

a new interaction term: (2.14)

S'

r

qfA'udxU f

qfA dx u + u

+ q[ X

(B)

But the last term does not at all depend on the path f !

-

X (A)]

For any path

the action will therefore be shifted by the same amount. But then fo is still an extremal path and we have explicitly shown the gauge invariance.

39

I

2.3 BASIC PRTNCIPLES OF QUANTUM MECHANICS We are now prepared for the quantization procedure! There is actually two alternative ways of quantizing a system: 1. The path integral technique (Feynman's procedure), 2. the canonical quantization. The first one is the procedure which serves our purpose best, so we will start with that following Feynman [1964] . The fundamental concept in quantum mechanics is the probabitity amplitude. Let us assume that we have an electron gun A emitting electrons, a screen B with two holes, and a screen C with a detector that can measure the arrival of electrons

~ --~.

~ ::..-:?'>-------~---------~,-)....~

PIx)

~,,-

-------'--

~.

Fig. 17

C

An event is the registration of an electron in the detector at a specific position x. With each event tude

~

we associate a probability ampli-

(x), which is a complex number, and the probability P(x) for

the corresponding event is found by squaring the amplitude P(x) = I~ (x)

(2.15)

,2

Furthermore, the electron may have passed through either of the holes 1 and 2. Hence we have two alternative ways in which the event can occur. Each of these is characterized by a probability amplitude ~,

(x) and

(2.16)

~2

(x). The total amplitude is found by adding these two ~

(x) =

~,

(x) +

~2

(x)

When you apply this principle it is important that we have no way of veryfying experimentally, which of the alternatives actually occured. If we make the experiment in such a way that we can decide which of the alternative trajectories the electron followed, then interference is lost, and we see no wave pattern!

(See fig. 18).

40

~-----Fig. 18

I

~\V _----

----r.=== ~.

PIx)

Now consider an experiment with polarized light. We have a beam of photons, which has passed through a polarizer O. In the direction of the beam we have put two more polarizers

=s

1

1 and 2.

0

(

DErEX:'IDR

o

2

Fig. 19

For a given photon there is a probability amplitude

¢lO

that it

will pass through polarizer 1, and once it has passed through that there is a probability amplitude

¢21

that it will pass through pola-

rizer 2. We are interested in the situation where a given photon actually hits the detector after passing polarizer ¢20

2. This amplitude

is found by multiplying the amplitudes corresponding to the

in~

dividuaZ steps. (2.17) Observe that this time there are no alternative ways to arrive at the detector. There is

only one way, which consists

of two steps. First

the photon has to pass through polarizer 1, and then it has to pass through polarizer

2.

We have now stated all the fundamental laws of probability amplitudes. They are basic laws, i.e. you cannot derive them from some underlying principles. We simply have to accept them as a starting point for the quantum mechanical description of a system. For later references we have collected them in the following scheme:

41

I

THE BASIC PRINCIPLES OF QUANTUM MECHANICS: ~.

1.

Each event is associated a probability amplitude event is found by squaring the amplitude:

The probability of the

2.

If an event may classically occur in several alternative ways i = l, ••. ,n each of which is characterized by the amplitude $i' then the total amplitude of the event is found by addition:

(2.15)

n

(2.16) 3.

$ = l: $.

i=l

1

If an event occurs in a way that can be decomposed into several individual steps j = l, ••• ,m, each of which is characterized by the amplitude $., then the total amplitude of the event is found by multiplication: J m $ = IT $.

(2.17)

j=l J

Now we are prepared for the discussion of a system consisting of a single particle. Consider two fixed times t , and t2' At the time -+ ~(r,tl)

tl we have a probability amplitude

of finding the particle

at ;. At the time t2 we have another probability amplitude ~(;, t2) -+

of finding the particle at r. The amplitude

-+ ~(r,

t) is of course

nothing but the Schrodinger wave function. The dynamical problem ,

we must solve is the following one: How does

-+ ~(r,

t)

develop in time?

We will solve it by finding a formula which connects ~(;, t) at times t

=

t,

and t

=

t2'

Consider ~(;2,t2)' It is the amplitude of finding the particle -+

-+

-+

at r2 at the time t 2 . NoO! let K (!'2; t2 I !'1; t,) denote the p!'obabi lity -+ amplitude fo!' the event, that a pa!'ticle emitted at !', at the time

t, O!i II be 'bb se !'ved at ;2 at the time t2. Then we can argue in the -+

following way. A particle which arrive at r2 at time t2 must have been somewhere at time t,: Hence there are several alternative paths the particle could have followed, each characterized by the position at time t,. Each of these possibilities is composed of two individual steps: 1. 2.

The particle was at the point ;, at time t, -+

The particle was emitted from r, r2 at time t2

( amplitude

=

(amplitude = ljJ(~l,t) ).

at time t, , and observed at

K(r2;t2Irl;tl)

).

According to (2.17) each possibility is then characterized by the amplitude: -+

-+

K(r2;t z lr, it,

-+

)~(r"t,).

To find the total amplitude ~(;2,t2) for arriving at;2 at time t

2 ,

we must sum op all these amplitudes (cf.

(2.16», i.e. we get!

42

I

(2.18)

This is the basic dynamical equation of the theory. Although it is an integral equation, we shall see later on that it is completely equivalent to the Sahpodingep equation (2.2).

2.4

PATH INTEGRALS - THE FEYNMAN PROPAGATOR t-axis

Consider now the amplitude -+

,

7f

r

I

-,

I

Feynman ppopagator and if we can determine that, we will controll

I

the dynamical evolution of the

I \

Schrodinger wave function! Now

K(B,A) I

-+

-+

K(r, ;tllrl ;tl) is the amplitude for

\

\

A(r, It,) Fig. 20

-+

K(r,;t,lrl ;tl). It is called the

the event that a particle which was ... . r-axu;

-+

emitted at rl

at time tlwill be observed at r, at time t , . TO come from -+

(rl,tl) to (r2,t2) a classical particle must have followed some path r. Consequently the classical particle could have rroved in several ways, each corresponding to a path leading from A: (;1 ,tl) to B: (;"t,). Let ~r(BIA) denote the amplitude that a particle, emitted at A and observed at B, actually moved along the path r. Then we get: (2.19 )

K(BIA)

=

where we sum over all paths r leading from

A

to

B.

Obviously this sum is very complicated, since there is an infinity of paths! It is called a path integpaZ. The precise definition of such a sum is very complicated, and as we are interested only in the basic physical ideas we will continue writing the path integral in the above naive form! We have reduced our problem to that of finding

~r(BIA).

Now this

cannot be deduced from basic principles, so we must simply postulate the value of it. This is where we make contact with the Lagrangian formulation of classical mechanics. With each path r we hava associated a classical action:

43

S(f)

t2

I

d;

-+

f L(x, a:F' t)dt tl

Following a suggestion from Dirac, Feynman now used the folloving expression (compare Dirac [1935], §33, p. 125-26)

exp[~

(2.20)

Since each

path~epresents

S( r)]

a possible history and each history

contributes with a phase factor, a complex number of modulus 1, this is often referred to as Feynman's principle of the democratic equali-

ty of all histories! This prinCiple leads to the following formula for the propagator:

Now, this is a very complicated formula and you might wonder what happened to classical physics? In classical physics you are used to considering only one single path, the path which extremizes the action. In the above formula all paths are on the same footing. There is no special reference to the classical path. But how can the classical path then play such a major role in our everyday experiences? To understand this we must first try to characterize what we expect to be a typical classical problem, and what we expect to be a typical quantum mechanical problem. First you might be tempted to say that a typical classical problem is one where the involved actions are very great compared to h, and a typical quantum mechanical problem is one where the involved actions are comparable to h. However, this is not a good formulation because the action is not at definite number. It is only determined up to a constant and hence the action for at single path has no absolute meaning. But if we take two paths ence

~S

to say:

= S(r2)

rl

and r 2 , then the differ-

- S(rl) has a unique meaning. Therefore we are allowed

44

t-axis

CUlSSICAL

I

P~1

A

Fig. 21 x-axis

x-axis

A typiaal alassiaal ppoblem is a problem whepe small ohanges in the path oan produae a great ohange l:.s aompared to

Ii, and a typiaal

quantum meohaniaal problem is one where eVen great ahanges in the path only produae a change l:.S aomparable to

Ii.

Suppose we consider a classical situation and pick an arbitrary path r,. It contributes to the propagator with a phase factor

exp[~ S(r,)]. But very close to the path r, there will be another path

r2

,

which differs in the action by rrli, but then the correspon-

ding amplitudes will cancel each other, because i

exp[ Ii S ( r 2

)]

= - exp[

i

Ii S ( r, ) ]

Thus they do not contribute to the general sum due to the destructive interference. Although this is the general situation, i t is not always true. Let rc I be the classical path, then S (r C I ) is extremal and all the nearby paths will have almose the same action. Hence we will have constructive interference from all the paths close to the classical one! Since

~he

main contribution to the propagator co-

mes from paths near the classical path, and since they all have approximately the same action as the classical path, we can put as a first approximation:

~

(2.22)

This is known as the alassical approximation.

L(~C I , dxc I ,t) dt]

crr-

45 t-axis

I

As we have seen the main contribu-

B

tion to the propagator comes from a "strip" around the classical path

REl3ION WITH CONS'l'RUCI'IVE

where the action varies slowly, and

INl'ERFERENCE

where the changes in the action are smaller than rrh. For a typical clas-

CUlSSlCAL

sical problem this strip is very

PATH

"thin", but for a typical quantum

A

mechanical problem the strip is veFig. 22

ry "broad". ConsequenUy the classical

x-axis

path looses its significance in typical quantum mechanical situations, sayan electron orbiting around a nucleus. The path of the electron is "smeared" out.

Worked exercise: 2.4.1 Problem:

Consider the quadratic lagrangian:

Show starting from (2.21) that the is given by the formula:

Feynman propagator

. t2

(2.23)

(2.24)

K(X2,t2; XI,ttl =A(t 2 ,ttl exp[~

f

t,

L(xCl

,XCI;

t)dt l

Assume now that the coefficients a(t), bit) and c(t) do not depend on time, i.e. they are constants. Show that the Feynman propagator then is given by the formula . t2 K(X2 ,t2 ,Xl ,t,) = A(t2-t,) exp[~ f L(xci ,X" ,t)dt l

tl

In both cases A (.) denotes an unknown function and Xc I denotes the classical path.

The above quadratic Lagrangian includes for instance the following cases: (a)

A free particle:

(b)

The harmonic oscillator: L

(c)

A forced oscillator:

Before we use fOnmlla (2.24)

L

L

1sIluF 1snci' -i II1W 2 X2 1smx2 -i mw 2 X2 - '-it) x

let us point out another feature of

the propagator. By definition K(X2 ;t2lx l ;t l ) is the probability amplitude for finding the particle at the pOint X2 at the time t2 when we know that it was emitted at XI

at time t,. Let us keep XI and tl

fixed for at moment and regard K(X2 ;t21

XI ;tl) as a function of Xl

46

I

and t2: K(X, ,t,) (x,t) = K(Xitl x, ;t,) Then K(x, ,t,) (x,t) is simply a Schrodinger wave function since it denotes the amplitude for finding the particle at x! But it is a Schrodinger wave function corresponding to a very special situation, because at the time t, we know exactly where the particle is: It is at x = x,. The amplitude is not smeared out at t = t,! This can also be seen from the integral equation (2.18)

for an arbitrary Schrodinger

wave function 1jJ (X2 ,t 2 )

= fK

(X2 ;t2 I x, ;t,) 1jJ (x, ,t l ) dXl

If we put t 2 = t, we simply get 1jJ (X2 ,tl)

= fK (X2 ;t, I x, it,) 1jJ (Xl ,t l ) dXl

But then we can immediately read off that

Hence K(x, ,t,) (x,t) reduces to a 6-function at the time t,: K (x, ,t,) (x,t) = 6 (x-x') Since K(x"t, ) (x,t) itself is a Schrodinger wave function it must furthermore satisfy the integral equation (2.18). We have thus deduced the following important group property of the propagator: (2.25)

2.5

ILLUSTRATIVE EXAMPLE: THE FREE PARTICLE PROPAGATOR

We now are in a position where we can calculate the propagator for a free particle. Here L(x;x) ~mX2 and we can use exercise 2.4.1 about quadratic Lagrangians:

=

. t2

K(xz ,tzl x, ,t, ) = A(tz-t,) exp

[li- f

~ x ~L dt]

t,

(2.26)

= A(tz-t,) e vn [i m '0",

Ii 2"

(The classical path is a straight line, i.e.

is a constant!)

(xz-x,)z tz-t,

47

I

To finish we must determine the function A. Substituting (2.26)-into the intergral equation (2.25) we get: 2

i. ~

A(t3-t , )exp

Using the

h 2

(X3- X' ) = t3-t,

alge~raic form~a:

2

2

(Xl-X2) + (x,-x,) _ t3-t, [X2 _ Xl (t2 -t, )+Xl (t3 -t,)] + (X3 -x, ) t3 -t, ~ - (t, -t2 ) (t, -t, ) t3 -t, t3 -t, we may rewrite the last integral:

X2

=+00.

f

2

I

2

(X2 -x,) 1 dx2 t, -t,

[1 m (X3 ':"x,) + exp ft2 t3 -t,

Xl=_CO

where we have introduced

to get rid of a lot of junk. The total integral formula is thus reduced to: im(x3-x,)2] A(t,-t, ) exp [ ft2~ u=+oo

) exp ( ) ( At3-t2At,-t,

[ i m (x3-x,)']

f

F2~ exp u=-oo

[i m (t3-t,) 2 ft2(t3-t 2)(t2-t,)U du

Notice that X2 has completely disappeared! The integral does not depend on X3 and x, so both sides have the same dependence on X3 and x" Everything fits beautifully. Using the formula:

x=_ (2.27)

f exp x=_oo

2 [-ax ]dx

=/f

which may be used for complex a too, we finally get: A( t3 -t, ) _ / 211ih (t3 -t2 ) (t2 -t,) _ A(t3-t,)A(t2-t,) m (t3-t,) - /

h1lih(t3 -t, ) m

,;",----"",mr.----:--,-

\r211ih ( t3 -t,)

211ih (t, -t, )

From this we see that,.up to a trivial phase factor exp(iat),which we normalize to 1, the function A(t) is given by A(t)

= ~1T~flt

So now we have computed in all details the

~ee-partiaZe p~agator:

48

E~raise

2.5.1 .. Problem: Show by an expl~c~t calculation that the propagator of a+free particle reduces to a 6-function in the limit where t ~O .

Let us try to get familiar with the free-particle propagator. We know that K(X,tlo,o) is the Schrodinger wave function of a free particle emitted from x = 0 at the time t = 0: (2.29)

1/J(x,t)

.rrll

K(x,t /0,0) = ~ 'exp

rimx2]

f!"2 t

Let us focus the attention on a specific point (Xo ,to). If the free particle is observed at x = Xo at time to, we would say that it classically had momentum po= mvo= m Xo to and energy

2

Eo= ?;mvo = ?;m

x~ t1

we investigate the variation of the phase (Xo; to) we find: If

rrn-

.

2

1/J(x,t) = ~ exp r~ ~ ~

211m tx

2

in the viscinity of

] +

a

x2

a-t(t Ht-to)

+ ... ]}

Ixo ,to =

m 21Ti t exp ~

.

[~

.f!

r."'to' X2 x

2

-

'-- Xo ] .,m tr t

Thus, very close to (Xo,to) the wave function varies like (2.30)

1/J(x,t) ""

~1T~ht

exp

[~

[Po x - Eot] for (x,t) "" (Xo ,to)

This is closely related to the Einstein-de Broglie ruZe, according to which a particle ~ith momentum p and energy E is associated a wave function with wave lenght A = and frequency V = E/h:

P

exp [i

(2~X_~)] = exp [~(pX

- Et)]

While 1jI(x,t) represents a particle which is located at x = 0 at the time t = 0, we would also like to have a Schrodinger wave function representing a free particle with specific energy and momentum. At the time t = 0 we suggest the following amplitude: 1jI(x,0) = exp

[k px]

where p is a constant. At a later time t we can compute 1jI(x,t) using the integral equation (2.18):

49

I

I

i m (X_X )2] [Ii "2 - - t - exp [~xll w dx l

I/I(x,t)

Using the algebraic identity: pxl+

~ (X_~l

~~ rxl-(x_~l]2

)2

+ px _

~

t

we get: 1/I(x,t)

Using the substitution: u

= Xl_ (x-~) m

.

1/I(x,t) =

Thus

exp

{fi

2

[px - L

2m

t]}

this may be reduced to:

~ u=_

V 2~ilit I

exp

u=-oo

[i m

2

11 2" u ldu

. z exp {[ [px- ~ tl}

the free particle is represented by a solution of the form: . z exp{ 2:. [px - E- t]} h 2m

which according to Einstein-de Broglie's rule must be interpreted as a particle with the momentum p and the energy: 2

E

=

L

2m Let us summarize: We have played a little with the free propagator and discovered two exact wave functions: ~

(2.31)

1/1 ( x, t)

= ..; 2lTilit

. z exp [[ • ~ • ~ 1

which represents a free particle, strictly located at x = .

(2,32)

l/J(x,t) = exp [

fi [ px -

° at the time t

0, and

2

~tll

which represents a free particle with a specific momentum p and specific energy: E =

2.6

~ 2m

- LORENTZ FORCE

BOHM - AHARONOV EFFECT

Having described the basic principles of the quantization procedure we will return to our original problem: in an external electromagnetic field. A:

The motion of charged particle Consider a path

(xl,tll and

r

connecting

B: (x ,t ). Before switching on the external field 2 2 this path contributes with the phase factor

~Or(BIA)

= exp

[~ Itz ~ fl

tl

mV 2dtl

I

50

to the propagator, but after we have switched on the electromagnetic field we must add to the action the interaction term: (2.13 )

S

I

=

Consequently the phase factor is changed into:

~(B

SO

=

exp[*[So(r} + SI(r)]]


exp[* So(r}]exp [* SI(r}]

IA)

you see that the external field produces a change obtained simply

by multiplying the original amplitude with the gauge phase faatop:


{2.33}

J

Ao.dxo.]

This is the quantum meahaniaal law, which replaaes the Lorentz forae!

Now we are ready to give a qualitative discussion of the slit experiment. We first look at a situation where there is no external field.

(see fig. 23).

Fig. 23

If the electron gun emits electrons with a characteristic momentum p, then the electron wavefunction has the de Broglie wavelength where

A

=

hlp.

Construct~ve

A,

interferences are expected when the dif-

ference between the two pathlengths is an integer multiple of A. If the difference between the pathlengths is a, then the difference of the phases in the electron wavefunctions is l!.0 =

But if we assume that

L»

I . 2'IT A, where

L

is the distance between slit

and screen and d is the distance between the two holes, then simple geometric considerations show that

ta x g = L (see fig. 24).

an d tg a-ill - d/2

= da.~.e.

1£ L

g,

d

51

I

(1)

d

2 d

2

(2 ) Fig. 24 Consequently we may replace difference at the point

(2.34)

nO(x)

x

If

=

=

a

by

x, thus obtaining for the phase

on the screen:

xd

LA 2~ nO

is zero, then

AL

x

nO

= =

0, and we have constructive interference.

When

x

etc!

Thus we have obtained a qualitative understanding of the wave

2d '

then

n

,

and we have destructive interference,

pattern on the screen! Now suppose there is a pure static magnetic field between the slit and the screen.

We know that this will produce a change of the phase fac-

tor associated with a "classical" electron arriving at the point

o

0

o

0

x

(Gee fig. 25).

x

PIx)

G) 0

0

0

0

Fig. 25

(Strictly speaking we should use the classical trajectories which are circles. If the field is not too strong and the momenta of the electrons are sufficiently high, then the radii of the circles will be great compared with L and we can safely replace the arcs of the circles with straigth lines). Now the main contribution to the phase comes from the paths and

f2

shown on fig. 25.

Let us look at

rl

•

on the external field, a "classical" electron !lOVing along

ated with a phase factor

exp[iarl, i.e. a

phase

a1 •

fl

Before switching fl

wili be associ-

52

I

But according to (2.33) switching on the external field will change this phase into

(Observe that

~

=

0 , as there is no electric field.)

a "classical" electron moving along

r 2 will

In a similar way

rcw be associated the m::xiified

phase:

Therefore the phase difference between the electron wavefunctions is given by: (2.35) Here

nO

is the phase difference, when there is no external field

present, and it was calculated in (2.34). The loop integral is found by integrating forwards along the path But since we are in a static situation magnetic flux enclosed by the

r2

-

r1

r2

and backwards along

p A. loop!

dr

r1

•

is nothing but the

This is the formula we

are looking for. We will "use" a very special kind of magnetic field.

Under certain

circumstances iron crystals can grow in the form of very long, microscopically thin filaments called whiskers. These whiskers can be magnetized and they then act like very tiny solenoids! Hence we have a purely static magnetic field concentrated in the whiskers (the field outside is extremely small and we may neglect it), but even though there is no magnetic field outside the whiskers, there is a nontrivial 'gauge potential

it.

This is exactly what we are after.

x

D Fig. 26

53

I

Now we can show that there is a measurable effect when you let electrons move outside the whiskers.

Phase factors corresponding to

paths lying on one side of the whiskers interfere with phase factors corresponding to paths lying on the other side of the whisker, and at the screen, the whisker (i.e. the gauge potential), has theref=e produced the following change of phase (2.36) Here is the magnetic flux through the whisker. Cbserve, that the phasedifference is independent of the position x on the screen. Hence the total wave pattern On the screen will be shifted a constant amount! This effect is called the

o

Bohm-Ahcmmov effect. (See fig.

27).

-t-52;;~~~P (x)

----

Classically you can only find out whether or not there is a whisker if you hit i t directly. But quantum mechanically you can detect the whisker because the gauge potential changes the phases of electrons passing by. This remarkable effect was predicted by Bohm and Aharonov (Bohm-Aharonov [1959,1961])

and later verified experimentally (Hollen-

stedt and Bayh [1962]). D.Bohm and Y.Aharonov, "Significance of electromagnetic potentials in the quantum theory", Phys. Ref., 115 (1959) 485 and - "Further considerations on electromagnetic potentials in the quantum theory", Phys. Rev. 123 (1961) 1511. G. Mollenstedt and W.Bayh, "The Continuous variation of the phase of electron waves in field free space by means of the magnetic vector potential of a solenoid", Phys. Blatt. 18 (1962) 299. You might still feel uncomfortabJ e about the whole analysis. We have studied the quantum mechanical interaction of charged particles with the Maxwell field and seen that it produces strange effects. We will now derive something wellknown. We will show that this theory reproduces the Lorentz force too. For simplicity we consider only a very simple situation. We consider the slit experiment once more, but this time we put a homogenous magnetic field behind the slit. This homogeneous ,static field is concentrated in a region of width W, where W « L (See fig. 28). If we had no magnetic field there would be a phase difference: (2.34)

54

I

x

P(x)

D Fig. 28

for electron s arrlvlng at the screen . But the presenC e of a magnetic field will produce a phase shift proport ional to the magnetic flux. In our case the magneti c flux is approxim ately BWd. Therefo re the actual phase differen ce is given by: A "

~~

21f +

~

B Wd

The point of maximum intensit y will be determin ed by:

o " xd 21f LA

+ fiq BWd

i.e.

x = _ EWALq 211'h I~the

classic al limit we would interpre t this in the followin g way: We would see electron s tra velling through the magnetic field shift their directio n of movemen t. Consequ ently they must have experien ced a force. As the electron s have the velocity v =

.E

m

they experien ce the magneti c field during the time:

Fig. 29

At =

!iv

If t denotes the time it takes to travel from the slit to the screen, then t is given by: t

=L v

As they hit the slit at the point x = _ B~~ the passage through the magneti c field has produce d a velocity in the x-direc tion given by: Av = ~ = - BWq x t

I

55 Consequently, we would say classically that the electrons experienced a force in the x - direction given by F

x

=

~

At

= - Bvq

Ap

h

Using Einstein - de Broglie's rule

A we

= ~p

finally get: Fx

=-

qvB

=q

(-;;xB)x

and this is exactly the Lorentz force!

2,7

GAUGE TRANSFORMATION OF THE SCHRbDINGER WAVEFUNCTION

Let us now return to the general theory! We have seen that we can represent the electromagnetic field by the gauge potentialAa and also discovered the gauge symmetry: Physics is unchanged if we replace the gauge potential Aa wi th the new potential

Ah = Aa + a(XX Now, how do these ideas fit together with our quantum mechanical description of a particle ineracting with the Maxwell field Aa? Consider a specific path r

1

in space-time connecting the space-

time points A and B. We have seen that the external field Aa produces a change in the original amplitude corresponding to multiplication with the gauge phase factor (2.33)

but the interaction term S1 is not gauge invariant! If we perform a gauge transformation, it is changed according to the rule (2.14)

S1 (rJ ~s' I (rJ

=

S1 (r) +q [X (B)

Consequently the gauge phase factor

-x (A) ]

~r(BIA)

is not gauge invariant, but

transforms according to the rule: (2.37)

But this has consequences for the propagator! We know that the propagator is given by the path integral formula K(BIA)=

J

~r(BIA)

D[r]

We can decompose the phase factor
~~(BIA), which describes the interaction with the gauge potential. Co~

I

56

quently we get

I ~~(BIA)~~(BIA)

K(BIA)=

D[r]

Performing the gauge transformation we then get the new propagator: K' (BIA)

J ~~(BIA)~rI(BIA)

=

I

But the phasefactors exp[ do

not

¥

D[r] J

o i.9. I ~r(BIA)exp[l1X(B)]~r(BIA)exp

X (B)]

depend on the path r

exp[I

¥

[!SI - flX(A)]

X (A) ]

so we can pull them outside the

path integral: K' (BjA) =

eXP[irX(~)]{f~~(BIA)~~(BIA)D[r]}

exp[- irX(A)]

Thus we have shown: K(BIA) ... K' (BIA)

(2.38)

eXP[1rX(B)]K(BIA)exp[- irX(A)]

so that the propagator of a charged particle transforms in exactly the same way as the gauge phase factor! This has consequences for the Schrodinger wavefunction! The Schr6dinger wave function ~(r,t) satisfies the integral equation

(2.18)

and since we have seen that the propagator is changed under a gauge transformation we conclude that the Schr6dinger wave function must change too:

~(~,t)

+

~'(~,t)

To find the gauge transformed wavefunction

~'

we observe that it

must satisfy the integral equation:

~'(;2 ;t2) =fK I (;2 ;t2 1;1 ;til ~ I (rlltl) drl

.... -+ + )] I (+ + ex P [ - !SI =f exp[ .!9:'" fl x(r2it.2)]K(r2it2Irlittl flx(r1itl ~ rli t 1) drl The first phase factor exp tegration variable obtaining: exp[-

¥

r

1,

[iiI X (;2 ,t2)] does

not depend on the in-

so we may pull it outside the integral, thereby

X(i'2it2)]~'

(r2it2)=

I

57

Comparing this with the integral equation for 1jJ we inunediately obtain: exp[- ¥

X(~,t)]1jJ·cr,t)=1jJ(r,·t)

i.e. (2.39 )

This is really beautiful! Remember that the only physically measurable quantity associated with the wave function 1jJ is the square of the norm:

;1jJ(;,t) I 2, but this is

aompletely unahangedby the gauge

transformation. Hence we see ti,at quantum mechanically too, all gauge potentials, differing by a gauge transformation, represent

the

same physical state. You also see that when a charged particle interacts with an electromagnetic field, then the phase of the wave function:1jJ=I1jJiexp[i~] can be chosen completely arbitrarily at any space-time point!

2.8

QUANTUM MECHANICS OF A CHARGED PARTICLE AS A GAUGE THEORY

We have found a lot of formulas which describe the effect of performing gauge transformations. These formulas may be cast into a more transparent fOrm if we introduce some fancy language. Observe that the wave function 1jJ(r,t) is changed simply by multiplying

it with a phase factor -+

1jJ(r,t)~

.!..9.

exp[ 11

-fiI

-+

x(r,t)]1jJ~,t)

If you perform two gauge transformations: 1jJ

l

eXP [¥X}1jJ :

exp[¥x}[exp[~Xl]

1jJ

=exp[~(Xl+X~]1jJ

then this is equivalent to a single gauge transformation, with the combined factor: exp[igx ].exp[igx ]= exp[~ (X +X )] n2

nl

Thus the phase factors constitute a

~21

group, the group of all com-

plex numbers of modulus 1. This group is denoted U(l) led the unitary group in one dimension.

and it is cal-

(Remember that a unitary

58

I

matrix is a complex matrix with the poperty:

l. At= At A = I The unitary 2x2 matrices constitute the group U(2), the ,unitary 3x3 matrices constitute the group

U(3), etc. Now a lxl

matri~

is nothing

but an ordinary complex nUmber z, and the hermitian conjugate is just the ordinary conjugate: z=x+iy~z=x-iy. Hence U(l)

consists of

all complex numbers with the property: zz=l,i.e. it consists exactly of the phase factors.} We have now attached a group G to our gauge transformations.

gauge g:l"Oup,and in this case G=U(l). The group U(l)

It is called the

is an Abelian group, i.e. g.g = g.g 1

2

2

for all 1

(contrary to U(2)!) and therefore electromagnetism is called an

a belian gauge theory. If we put x=(r,t) for abbreviation, then to

every spaae-time point X

lJe have attaahed a group element: (2.40)

g(X)

exp[iiI xIX) 1

This group element is responsible for the gauge transformation. First we consider

~~e

Schrodinger wave function

~(X).

It transforms

according to the rule: 1jJ (X) ~1jJ' (X) = g (X)

(2.41)

w(X)

A quantity transforming in this way is called a gauge vector,

E:x:ercise:2.8.1 Problem: Prove that

aa x = - ~(a g)g-1 1'1 a where g is given by (2.40).

T~en

there is the gauge potential A" (X). According to (1. 31) and the

above exercise the gauge potential transfoDtlS

according to the rule:

(2.42) We shall explore the geometrical significance of this strange formula later on. Finally we have the field strengths FaB(X), which transform according to the rule: (2.43)

F aB (X) ~ F ~S (X) = F as (X)

59

I

and a quanti ty wi th this property is called a

gauge saalaP.

As a consequence of the gauge symmetry of our theory we immediately see that any physically observable quantity must be a saalaP. Let us playa little with gauge scalars! Schrodinger wave function

~(X)

~

~(X).

g(X)1jJ(X)

gauge

First we consider the

As

and

we see that ~~ is a gauge scalar. Actually it is observable. It is the

probability density. Then consider the Maxwell field A a.(X). Here

~le

can form the gauge

scalar:

But this is not the only one. Consider a closed loop f

in spacetime.

Then the integral

tji A dxa. r a is a gauge scalar. To see this we perform a gauge transformation:

mere aa. X does not contribute to the integral because we consider a alosed loop) .Is the above l=p inte-

gral

t-axis

vo!'l

~le

? Well, consi-

der two space-time points A

B

and B connected by the paths

.rY.,Aa(X )

r l and r 2 • Then an external field will produce a phase

r1ri yf2 B

shift between electrons moving from A to B along fl or along

r2 •

This phase shift is

given by:

A

y-axis

Fig. 30 where f=f2-fl' This phase shift is in principle measurable, being nothing but the Bohm-Aharonov effect!

Thus we conclude that

exp[i;! tji AadxCl ] is in principle measurable quantum mechanically.

I

60 Worked exercise: 2.8.2 Problem:

Consider the parallellogram shown on the figure below. Show that the a field :trength FaS is related to the loop integral ~A dx through the followlng formula: a xa+oxCi. . 1 I A &c a 11m "E2 !!'

as (X)AxSo x a

F

E->0

r

CI.

E

xa+D:x.a

Finally we look at the Schrodinger wavefunction O/(X) and the Haxwell field Aa(X) at the same time. What can we make out of them? aa 0/ of 0/. This is not a gauge vector as

Consider the derivative

T~e

last term spoils the game, but containing aag(X), it reminds us

of Aa' Consider Aa'V. It transforms in the following way: A 0/ a

+

A' o/'=[A -i~ (a g)g-l]go/= a a q a

g(X)A (X)o/(X)-i~ [a g(X)lo/(X) (l q a Hence the same spoiling term occurs! Consequently: aa o/-i* Aa 0/ = (aa -it Aa) 0/ is a

gauge vector, i.e. it has the correct transformation properties: )ljJ' "'g(X) (a

(2.44)

_l
a

A )0/ a

The differential operator Da=aa¥ -"a is called the

gauge aovariant derivative, and it plays a very important

role when you want to construct gauge invariant theories! The discovery of the gauge covariant derivative allows us to write down non-trivial

gauge scalars based upon 0/ (X) and Aa (X). E.g. liD()o/,

(Dao/) (DSo/)'

F

as (Dao/)

(D

8

0/)

or written out explicitly: i(x) [aao/(X)-i* Aa(X)o/(X)] [aai +

¥ Aai]['aljJ -

¥

ASo/]

etc.

61

Exeraise: 2.8.3 Problem: Let o/(X)

I

be a gauge-vector. Show that [D 'D

)1' V

10/= - ~ 0/ 1'1 )1V

So we have walked right into our first gauge theory! We summarize the most important points in the following table: THE

BASIC INGREDIENTS OF A GAUGE THEORY:

A gauge theory aonsists of a gauge group

G

and a gauge potentia l

Electromagnetism is a gauge theory based upon the gauge group Maxwell field Aa(X) •

U(

1) and the

To perform a gauge transformation you must attach a group element g(X)

to each space time point

(2.45)

gauge covariant derivative

(2.46)

gauge vectors

(2.47)

gauge scalars

(2.48)

gauge potential

(2.49)

gauge phase factor

X.

a = aa -

D

~Aa

o/(X) Dao/(X)

0/' eX) = geX)o/eX) D' 0/' eX) = geX)Dao/(X)

Fai3=aaAS-a Aa S

F'as = Faa

~Aadxa

~A~dxa = ~Aadxa

Aa eX )

tPr(BIA) = exp[-\rJrAa dXa ]

A~eX)

= Aa eX ) -

i:[

aa g ] g-1

tPreBIA) = g(B)tP r e B1A )g-1(A)

I

62

2.9

THE SCHRO'DINGER EQUATION IN THE PATH INTEGRAL FORMALISM

At this point we will get in contact with the conventional a]JproQc11 to

quantum mechanics

based upon the Schrodinger equation.

Stu-

dying dynamical problems we nave been using the propagator which governs the dynamical evolution through the integral equation:

(2.18) We now want to show tnat this integral equation actually is equivalent to t:1e Schrodinger equation. For simplicity we will only consider the one-dimensional motion of a particle in a potential V(x). In this case the Lagrangian is given by, (2.5)

L(x,x,t)= ~mx2- V(x)

and the propagator by

X(t2)=X2

(2.21)

J exp{ X(td=Xl

K(X;t2Iy;t0=

,

i

t2

K f [; i

2- V(x)>>t} D[x(t)l

tl To convert the above integral equation (2.18) into a differential equation we put tl=t and

t2=t+~t:

~(x;t+~t)= fK(x;t+~tly;t)~(y;t)dy

Since

~t

is going to be

very small, we expect the

t-axis

phase factor to be approximately equal to: i[m ~2 1 exp{fi 2 (~t) - V(x) J~t} Of course this is not true for a very "wild"

Fig. 31

x-axis

path, but the wild paths kill each other due to the rapid oscillations.

Consequently the "infinitisemal" propagator is given by i K(xit+AtIYit)~ A exp{fi

[m2 ~2 At -

1

V(x)~tJ}

in the limit of very small At. A is a normalization constant, which

arises from summing up all the amplitudes in the path integral and

I

63

we will determine its value in a moment. Inserting this approximation into the integral equation we get: +'". [ m ( -x ) 2 1. (2.50) ~(x,t+~t)~ A _~ exp{fi 2 ~ i

A exp[- fi

V(X)~t]} ~(y,t)dy

V(x)~t]

where we have put y=x+n. Observe, that the integral is wildly oscillating for great values of n. Hence the main contribution comes from values of n where n 2 is comparable to ~t. We may therefore expand

~(x+n,t)

to second order in n (second orGer corrections in n

corresponds to first order corrections in

~t

because of the phase

factor

Doing this we get oljJ

3X + %n 2

~(x+n,t)~ ~(x,t)+n

o2ljJ axz

Inserting this into (2.50) we now obtain ~

A exp[-

~

(x,t+t.t)

V(X)t.tf!

exp[~

i ~~]{~(x,t)+n* +

%~}

dn

~ does not contribute to the integral because it is an The term nax

odd function of n

Thus we have shown: -too

i ~(x,t+tlt) '" Aexp{- fjV(x) tit} [ ~(x,t)

J exp[2fi ntit] dn + 2

im

2

a

• \Ji ~

+""

f

Here the first integral is a Gaussian integral of the standard type, cf.

(2.27), while the second is obtained from the formula

f +~2exp[-ax2] -00

which follows from respect to

(2.27)

dx

=....!....f!I. 2a a

when you perform a differentiation with

a. In this way we finally obtain:

(2.51) ljI(x,t+tlt) '"

A/21T~t exp[- ~V(x)At]ljl(x,t)

+

Aj21T~t ~t

exp[-

ff(X)At]~:t

First we observe that this fixes the value of A. Because when fit

+

0

64

I

1jJ(x,t). But this is consistent

the left hand side approches only if:

A=! 2TTi~t.t

( 2.52)

This should be conpared with the free particle propagator (2.28). In fact the "infinitisemal" propagator can now be decomposed as follows (2 • 53)

~

i

K(x;tHtly;t) '" V~t exp[fi

m (x-y) 2 i lit lexp[-fiV(x) lItl

2"

KO(x;t+lItly;t)~I(x;t+lItly;t) where KO(BIA)

is the free particle propagator, and

factor corresponding to the interaction

~~(BIA) = Inserting (2.52)

~~(BIA) is the phase

cf. (2.33)

exp[-ifv(x)dtl

r

into (2.51) we now get

i iht.t 1jJ(x,t+t.t)<>< exp[- h V(x) t.tl1/l(x,t) +

2

a $ 2ffi"axz-

i.e.

g

1jJ (x,t+t.t) -1jJ (x) ~ ih . t.t - 2m ax

which in the limit t.t a,I> it

ih 2m

+ exp[ -(l/h)V(x)lItl-1 ,I>(x,t)

t.t

'j'

0 reduces to:

~

a2$

ax7 -

KV(x)1jJ(x,t) .

Finally this may be rearranged as:

(2.54)

ih a1/l

at

h2

- 2m •

a2'4> axz-

+ V (x) 1/1 (x, t)

So we have captured the Schrodinger equation! By using exactly the same procedure we can find the Schrodinger equation for a charged particle in an external field. However, it becomes technically more complicated, partly because i t is now a three-dimensional motion, and partly because the generalized potential (2.11)

has a more complicated structure. We will leave this as an exercise: (Details can be found in Schulman (1981». E=l'aise: 2.9.1 Problem~

Consider a three-dimensional motion of a particle in a ordinary potential V(x). Use the path integral formalism to derive the Schrodinger equation: .

ih

~

at

h2

= - 2m

ai ai 1/1

.... + V(x)ljJ

65

I

Exercise: 2.9.2 Problem: Consider the three-dimensional motion of a part·'.cl p in the genel'alised potential:

Use the path integral formalism to derive the Schrodinger-equation:

ih

2.10

a,h if =

h2 - 2m

(

ai~

iq

(i

q

11 Ai) a - inA

i)

ljJ +

q
THE HAMI LTON I AN FORMALISr·1

The other method you can use when you want to quantize a system is the

canonical quantization. Here the starting pOint is the Hamiltonian

formalism. Consider a Lagrangian

L(qi,qi,t),i = l, ..... n. With each of the

generalized coordinates qi we associate a

conjugate momentum Pi defined

by: (2.55 )

i=l, ... ,n

Observe, that Pi is a function of qi,qi,t: i

·i

Pi= Pi (q .,q ,t); i= l, ..• ,n Sometimes we may invert tnese equations and express q· i as a function i

of q , pi,t. If this is the case, we say that the system under consideration, is a

canonical system.

For a canonical system we then define the (2.56)

\"Ie

Hamiltonian:

i

H(q 'P.i,t) =

shall often abbreviate this as H = P . qi - L but you should al1

ways remember that q i should be regarded as a function of the canonical variables p. ,qi. Observe, that we are using the Einstein sum 1 . convention! An expression like Pi q1 implies a summation over i =l, •. ,n. The Hamiltonian is very

us~ful.

We may use it to cast the equations

of motion into a very beautiful form. Consider the partial derivatives: and

66

I

Let's work them out:

a

aE 3Pi

[po qL L ]=6~ q j+ o.

3Pi

J

aqj

3L

3q j

• J 3p i

J

aqj dpi of the momentum conjuc;ate to q j.

by the definition (2.55)

.i q

l'lext we have: 3H =_3_. [p.qj -L]= p. ~_ aq). aq). J J aqi From the Euler-Lagrange equation (2.6) we

~~ aqj aqi

3L

d

dt

'i

aq

leading to

(2.57 )

and

These equations are referred to as

Hamilton's equations and they

are completely equivalent to the Euler-Lagrange equations. To check these ideas in a trivial example we consider a one-dimensional motion in a potential V(x). From the Lagrangian (2.5) L ( X,x• )

21 mx• 2

- V(x)

We see that P

= 3L = mx

3x

so that the momentum conjugate to x in this simple case cone ides with the usual kinematic momentum. The system is canonical and we have: X

1

mP Hence we can find the Hamiltonian: 2

(2.58)

H(p,x) = px - L = ~ + V(x)

which reduces to the usual formula for the mechanical energy. Finally we get Hamiltons e~uations: x ::; 3H 1 3p =

p

which are obviously

3H 3x

mp

=_

V'(x)

e~uivalent

to Newton's

e~uation

of motion.

For a time-independent system we may identify H with the total energy of the system. TO establish this we consider a system characterizec a Lagrangian, which does not depend explicitly on time:

~y

67

I

Then the Hamiltonian too does not depend explicitly on time: H = P/li(qi ,P ) _L(qi;qi(qi 'Pj)) j Consider a trajectory, qi= qi(t), which satisfies the equations of motion. Then H is

co~stant

along this trajectory:

dH dt

o

according to Hami lton' s equations! Hence

H is a constant of motion

long the classical

pat~.

total energy follows

from the illustrative example above.

a···

That this constant of motion is actually the

Let A(qi;p.) and B(q\p.) be functions of the generalized coordinates ql, ... ,qn and their ~onjugate mom§nta p , .. . ,p . We define the Poisson bracket of A and B as follows: 1 n dA aB {A; B} (2.59) -. -a-- dB aA aqi OPi aql Pi where a summation over i is implied as usual

Exercise.' 2. 10. 1 Problem: Show that the poissoncrackets satisfy the following simple rules: 1. {A;B}= - {B;A} (Skew symmetry)

2.

3.

i {A ;c}= 0 whenever c is a number, i. e. a constant function of q and Pi'

{Al+A2 ;B} = {AI ;B} + {A 2 {AA;B} = )..{A;B}

4.

{A1A2 ;B}

5.

{A;{B;C}} +

;B}}. . Llneanty

{AI ;B} A2+Al {A2 ;B} {B;{C;A}} + {C;VI;B}}

0 (Jacoby identity)

Exercise: 2.10.2 Problem: Consider the i'th coordinate function qi(ql, .... ,qn; pi, ..... ,p) = ql Similarly we can consider the i'th componenf of the conjugate momentum as a special function: Pi(ql, .... ,qn; Pl, .... 'Pn) =Pi i Show that q and Pi :ati.sfy the rule: (2.60 )

{ql;q J} ={Pi ;Pj}

=0

and

{qi ;Pj}

= oi j

Exercise: 2.10.3 Problem: Let F(ql, ... ,qn;Pl," .,p ,t) be a function.of the generalized coordinates, their conjugate momenta and ~he time t. Then '11 and p. themselves depend on time and their time dependence is governed by Hamilton's eqliations (2.57). Show that the total time derivative of F along a trajectory in phasespace is given by: (2.61)

dF dt

= at dF

+ {H.F}

'

(Note that if F does not depend explicitly on time, then F is a constant of motion if and only if the Poisscn bracket {H,F} vanishes).

I

68

Exercise: 2.10.4 Problem: Let F(ql, ... ,qn) be an analytic function. Show that: { Pi,F (qI , ... ,q) n } = -.aqi aF (qI , ... ,q n) 1. 2.

(q\F(ql, ... ,qn)} = q'F(ql, ... ,qn)

(Hint = expand F(ql~ ... ,qn) in a powerseries and show that: {Pi,q~}

n-i - nqi

where we exceptionally do not imply a summation over i!)

CANONICAL QUANTIZATION AND THE SCHRODINGER EQUATION

2.11

Now the canonical quantization runs as fOllows: The generalized coordinates (ql, ..• ,qn) specify a position in configuration space. Quantum mechanically we characterize the system by its Schrodinger wave function W(ql, ... ,qn;t) which gives us the probability amplitude for finding the system at the pOint (ql, .•. ,qn) as a function of the time t. The dynamical evolution of the wave function is governed

~y

the Schrodinger equation which you construct in the fol-

lowing way. Consider a Hamiltonian:

E

hand side you substitute, substitute: Pi +

iii.

-

a~i

+

E

= H(qi,p.). 1

On the left

ih~t' and on the right hand side you

Then both sides are converted to diffe-

rential operators and the Schrodinger equation reads:

(2.62)

. Ii.

1

a

at

-,10 't'

To check these ideas in a trivial example we consider a one-dimensional motion in a potential V(x). We have previously found the Hamiltonian: H(x,p) = p2/2m + V(x) . Thus we get the ordinary Schrodinger equation: (2.54)

l'li.

~ _ Ii. a ()] ( ) at- -2m l a?"+Vx wx,t 2

2

Let us attack a more interesting problem. Suppose a particle is moving in an external electr6magnetic field, and let us try to quantize this problem in the canonical way: From (2.12)

we see that the canonical momentum is given by ~

p =

aL

~

=

~

mv +

~

qA

and hence they do not just reproduce the kinematical momenta! This system is canonical too and ~

l'~

S!. A

v=iiiP-m

I

69

Thus we can construct a Ham1ltonian:

H(x,p,t)

= p.v

- L

= 2k(P

- qA)2+

q~

i'ie may rearrange this slightly:

E - q~

(2.63)

=~ 2m

(p -

qA)2

Here you see something interesting: TO pass from a 2

where E

=~ ,

free particle,

to a particle in an external electromagnetic field

you have to make the following substitutions:

(2.64)

E

-+

p

E -

q~

p-

qA

and this simple rule is known as the

ruZe of minimal coupling. It is

of very great importance and we shall investigate it more closely later on. For the moment we want to pass to the Schrodinger equation: (2.65)

[ 1'Ii ~ at

~l ",(~ t)= [-ili~-qAl2 't' r, 2m

q",

Observe that the Schrodinger wave function interacts directly with the gauge potential (~,A) and not with the field strengths! You should also observe that the equation is gauge co-variant. we rewrite it slightly: 'Ii r 1

a + i g "'l n

'at

,10 't'

't'

_ -

-Ii 2

[V -

¥ Al2

2m

TO see this

,10 't'

Hence to pass from a free particle to a particle in an external field we make the following substitutions: (2.66)

a

at

-+

a

*

,

at + 1

V -+ V -


¥A

These substitutions can be written more compactly as:

(2.67)

aa

-+

aa

i

If

A

a

Da

Consequently we have simply converted the ordinary dErivatives to gauge covariant

dErivatives! This is also referred to as the rule of minimal coupling. Using this we can rewrite the Schrodinger equation as (2.68)

70

I

Here you explicitly see that it is a gauge covariant equation, since both sides transforms as gauge vectors. Thus if they are equal to each other in one particular gauge, then they are identical in all gauges! Consider a Hamiltonian operator A

H

i

A

=

H(q ,-

a

in ---.)

aql. If we introduce the following inner product for wavefuntions:

<1/J[

=

J ¢(x)
thenAit can be shown under very general assumptions that the operator H is Hermitian i.e.

" 1/Ji

where 1/J,

= 0).

Furthermore we may generally assume that the set of eigenfunctions {ljJn(x)} is complete i.e. that we may expand an arbitrary wavefunction as a superposition of eigenfunctions: 1/J(x)

=

~

anljJn (x)

Here the coefficient an is determined by the relation am



provided the set of eigenfunctions is normalized, i.e.

Worked exercise: 2.11.2 Let {1/Jnix)} be a complet: orthonormal set of eigenfunctions of the Hamil. tonlaJl operator H. Show that. Pro~lem:

(2.69)

~ 1/J (xd 1/J

n

n (x,) =0 (Xl- xz)

Consider an eigenfunction of H,i.e.a wavefunction 1/Jn(x) with the property:

This eigenfunction can immediately be extended to a solution of the Schrodinger equation:

I

71

In accordance with the Einstein-de Broglie rule this is interpreted as tl1e state of a particle with energy En. Observe that the probability distribution is time independent: l¢n(x,t)1

2

= jlj!n(x)i

2

and we therefore say that the wave function represents a

stationmy

state. Let lj!(x) be an arbitrary wave function. Then we can decompose it as a superposition of eigenfunctions: lj!(x) But the Schrodinger equation is linear so we can immediately extend this to the solution ¢(x,t) =

~

an1/Jn(x) • exp [-

~

Ent]

which reduces to 1/J(x) for t=O. Once we know a complete set of eigenfunctions we therefore

controll the dynamical evolution of Schr6din-

ger wave functions. This suggest that the Feynman-propagator it self can be expressed through a complete set of eigenfunctions:

Worked exercise: 2.11.:3 Problem: Let {lj! (x)} be a complete orthonormal set of eigenfunctions to the Hermitian operator ft . Show that the Feynman propagator can be expanded as: (2.70)

Let us make a final comment about the canonical quantization. In the preceding descussion the Schrodinger wavefunction has played a central role. However it is possible to avoid it. The physical quantities are then represented by Hermitian operators that do not necessarily operate on the space of Schrodinger wave functions. These operators cannot be chosen arbitrarily: .In the classical context the physical quantities are represented as functions A(q';p.) of the generalised coordinates and their conjugate momenta. In the quantum cditext these functions are rep raced by Hermitian operators in such a way that their Poisson brc:cket is rep raced by the commutator: (2.71)

{A;B}+-1-[~,~1 if!

Exercise: 2.11.4 Problem: Show that the commutator [ ~;~ ] ding to exercise 2.10.1.

= ~~ -

~ satisfies the rules correspon-

Especially the generalised coordinates qi and their canonical momenta p. must be replaced by Hermitian operators satisfying the socalled Heisenberg commlltation rures: (2.72)

Compare (2.60).

I\i ; I\j] 1\ 1 [q q = [1\ p.;p. = 0 1 J

ih6~

J

I

72

Exercise: 2.11 . .s . P:oblem: Show x~at if we replace the generalised coordinate q" by the multiplicatIon operator q = q • and the canonical momentum Pi by the differential operator

~.

= _ ih _a_.

I

then

and

aql.

satisfy the Heisenberg's commutation rules.

Exercise: 2.11.6 . Problem: Show that if ~l. and satisfy Heisenberg's commutation rules then their commutator with other operator§ satisfies rules corresponding to exercise 2.10.4.

£.

ILLUSTRATIVE EXAMPLE:

2.12

SUPERCONDUCTORS AND FLUX QUANTIZATION We finish this chapter by reviewing another experiment which shows how gauge potentials may produce unexpected quantum mechanical effects~

Consider a suitable piece of metal say aluminum. If we cool

it down it becomes superconducting and interesting things happen. (Feynman

[1964], de Gennes [1966]). Suppose we originally had an ex-

ternal magnetic field. This would penetrate into t!1e metal when the temperature is high, but it turns out that when the temperature falls down below a certain critical temperature Tc' then the external field is expelled. This is the famous Meissner effect.

Fig. 32

T>T

c

T
c

It turns out that there is produced currents in the outher layers of the metal and these currents prevent the magnetic field from penetrating into the metal. This is an interesting situation. We have a magnetic field

B which

stays entirely outside the -lump of metal, but there is a gauge potential

A too,

and it may very well penetrate into the metal!

73

I

We now change the experiment a little. We take a ring whose width is great compared to the penetration depth of the magneT>T

c

tic field.

i~e

place it in an ex-

ternal magnetic field

B

at room

temperature. Nhat happens? The magnetic field is spread throughout the whole space and especially it penetrates into the metal. Then we cool dm'Tn the ring, and when we pass below the critical temperature Tc' the r-ieissner effect occurs. The magnetic T
c

field is expelled due to surface currents in the superconducting ring. Hence some of the field lines pass outside the ring and some of them pass through the hole in the ring. But now we remove the external field. However, this does not mean that the magnetic field disappears completely. Part of the field passing through the hole is trapped by the surface

currents. So the superconducting ring now acts much like a solenoid!

B , only B which pass

Observe, that inside the metal there is no magnetic field a gauge potential A.Since we have trapped a magnetic field through the ring, there is a magnetic flux

~

through the ring. It is

this flux we are going to examine! Let us try to get a qualitative understanding of the situation. According to the BCS-theory, the electrons in the metal will form pairs in the superconducting state. They are called Cooper pairs. Now the electrons are fermions, but the Cooper pairs act like bosons! This has the important consequence that the Pauli principle no longer applies. Two electrons cannot

occ~py

the same state, but

74

I

there is nothing to prevent two Cooper pairs from occupying the same state! Actually hosons have a strong tendency to occupy the same state. Suppose ~(r,t) denotes the Schr5dinger wave fUnction for a Cooper pair. We may actually assume that there is a macroscopic number of Cooper pairs all described by the same wave function

~(r,t).

This has important consequences. The absolute square I~(;,t) 12

is

the probability for finding a specific Cooper pair at the point r . If N

is the total number of Cooper pairs, then I~(r,t) 12

N

is simply the dEnsity q

=

-+

of Cooper pairs at the pOint r. If furthermore

2e is the charge of a Cooper pair, we conclude that:

is the cha:t'ge dEnsity of the Cooper pairs. Since a solution to the Schrodinger

~tion

is only determined up to a constant we may redefine it:

~(r,t)

-+

~ ~(r,t)

=

~'(r,t)

So rcw we have a macroscopic number of Cooper pairs described by a

single wave function:

~' (r,t) and

-I ~

l

(r, t)

12

simply denotes the charge density at r . Observe

that although I~I

2

in general only has a statistical meaning,

I~'

12

will in this case denote a macroscopic physical quantity! In what follows we will drop the prime and simply denote the wave function by ~(r,t). We may decompose it in the following way: (2.73)

Here p (r,t) is the charge density, which has a direct physical meaning, and

-+ ~(r,t)

•

is the phase which has no physical meaning as we

can change it by performing a gauge transformation! In the superconducting state the number of Cooper pairs are conserved and so is the charge. If we denote the Cooper current by we therefore conclude that p and (1.17)

J

3

obey an equation of continuity:

75

I

We can use this to find an expression for the Cooper current. p

= -

ljilji

an __ a,,, at - at lji

implies that

-

aij;

at

lji

Using the Schrodinger equation we can rearrange this as:

which may be rewritten as

ap

at

in

~

~

(v +

= 2m v {lji

~ ~

n

-

A) lji -

-lji (v~

-

!g

n

~

A) lji }

From this expression we immediately read off the Cooper current: (2.74)

This show us that the Cooper current is a gauge scalar as it ought to be, and that it is a real quantity. In the following we will always assume that the density p of Cooper pairs is constant thoughout our piece of metal! This is not strictly correct at the boundary where it falls rapidly to zero. Except for that,it is a very reasonable assumption that the Cooper pairs do not "crowd" together but are evenly spaced.

Exercise: 2.12.1 Problem: Show that the expression for the Cooper current can be rearranged as: ~pn ~ qA] (2.75) [ v \1>- 11

J=m

provided pis a constant.

Using exercise 2.12.1 we can now derive the Meissner effect. Since there is no external electric field involved, the Maxwell equations (1.3) and (1.6) reduce to (2.76)

v. B

~

-t

J =~

= 0

But from (2.75) we then get Vx (VxB)

=~ EoC

Vx.... J

= - Eomc ~

VxA

where we have used (1.12) to get rid of the term

= - ~2 B Eomc Vx(V~).

Furthermore using (1.14) and (1.3) the left hand side can be rearranged as

I

76

Hence we end up with the simple equation (the London equation) (2.77)

ai3

=

1

I"f

it

where (2.78) is the socalled London-length. That equation (2.77) indeed implies the Meissner effect is left as an exercise, see below. Exercise: 2.12.2 Problem: Consider a semi-infinite superconductor occupying the half-space: z>O. Let us apply a constant field = Eo (1,0,0) parallel to the surface. Show that inside the superconductor the solution to (2.78) is given by: ... [ z B (z) = Bo (exp - ~1,0,0) z > 0

B

so that the magnetic field vanishes exponentially for z > AL.(A is typically a L few hundred Angstrom.)

Observe that (2.76) implies that not only

B vanishes

inside the

superconductor, the same thing holds for the Cooper current

3-

Consider now a superconducting ring as shown on fig. 34. Inside the ring at the curve r the Cooper current vanishes, as does the magnetic field!

Us~

(2.75) we

therefore get:

o

=

~ m

(Vc.p-

9. Ii.

lA)

But now we can compute the magnetic flux inside the ring beccluse: (2.79)

pit. r

dr

Ii..! q!ll

r

V,n' '"

dr

You might be tempted to say that it is zero! But let us look a little closer at the phase lP. The wave function ljJ(r,t)

=

jl p(r,t) I exp[

ilP(r,t) 1

is only non-trivial in the ring, and all we can demand is

that it

is single valued. But then nothing can prevent lP from making a jump of 2nn. If that is the case, VlP will contain a 6 -like singularity

77

and the integral need not vanish!

I

(Compare the discussion in sec. 1.4) •

TO compute the line integral we assume that V~ makes the jump at the pOint B. If B+ and B- are pOints extremely close to B on each side of B we get: h

q

P A· d r

B+

h 0<

[~(B+) - tp(B - ) 1

-+

V~

L

q

dr

B h

q

21Tn

+ as B ,B

+

B

consequently we get

(2.80)



where

n<1>o

So the trapped flux is

quantized~

<1>0

21Th

Iqf

This remarkable effect has been

established experimentally (Deaver and Fairbanks [l96l1)~)The quantity o is the fundamental fluxquantum. We have already seen that a vanishing magnetic field inside a superconductor implies a vanishing Cooper current. In the next exercise we will show that the opposite holds too: If t:1e Cooper current vanishes then so does the magnetic field. Exercise: 2.12.:3 Problem: (a) Show that the expression (2.74) for the Cooper current can be rearranged as: j. = ih iii D.1jJ (2.81 ) 1

m

1

provided the Cooper density p is constant. (b) Use this expression and exercise 2.8.3to prove directly that a vanishing Cooper current implies a vanishing magnetic field.

In the previous discussion we have assumed that the superconducting state of a metal only depends on the temperature. Actually it also depends on the strength of an external magnetic field. Let us discuss this in some detail for a special kind of superconductor known as a type II superconductor. Consider a superconduction cylinder. Outside the cylinder we have a solenoid. Suppose a weak current flows in the coil. Then it will produce a magnetic field of strength B which is expelled from the cylinder due to the Meissner effect. When we increase the current, B will reach a critical value BCl where the superconducting state starts breaking

U

~t

~l:--~ ---------------- Fig. 35

*) "Experimental evidence for quantized flux in superconducting cylinders", Phys. Rev.

Lett. 7 (1961) 43

I

78

down.

Thin vortices are formed where the normal state of the

metal is reestablished. The magnetic field starts penetrating into the metal through these vortices. As the magnetic field strength increases more and more vortices are formed and when we approach another critical field strength BC2 only small superconducting regions are still destributed throughout the cylinder. When we finally pass BC2 the superconducting state breaks down completely and the cylinder is now back in its normal state. If we decrease the current again then the same things happen in the reversed order. Now let us concentrate on a single vortex. Let us inclose it by a great circle

r as shown on the fi-

gure. Far away from the vortex the magnetic field and

thus

the Coo-

per current vanishes. The magnetic flux through the vortex is given by:

1111

Fig. 36

due to the same arguments as for the superconducting ring. But then we see that the magnetic flux q~tize~

~

through the vortex is necessarily

The existence of quantized vortices was predicted by Abri-

kosov (Abrikosov [1956 tland they are therefore ~ye

call~d

Abrikosov vortices.

have previously stated that charged particles may interact

with the gauge potential A in a spacetime region Q, even if the field )J strength F)JV vanishes identically throughout this region. This interaction was a pure quantum mechanical effect, the Bohm-Aharonov effect (sec. 2.6). We may throw light on this using our results concerning the flux quantization: Consider two identical rings: A and B. In the following experiment the two rings are placed at room temperature. Inside ring B we also place a tiny solenoid. In

L~is

solenoid we have a current

which produces exactly one half of a flux quantum. Hence in the beginning of the experiment the flux through A is 0, while the flux

through B is

%~o.

*) Soviet Phys. Jetp 5 (1957) 1174

79

I

A

Fig.37

Now we cool down the two rings and they become superconducting. !ihat happens? In the first ring nothing happens. But in the second ring the preceeding analysis concerning flux quantization is clearly valid. The superconducting ring will only allow a quantized flux through the ring.

Thus

a Cooper current is produced in the outer

layers and this contributes to the flux, so that the total flux becomes 0 or

~o

•

But what is the origin of this Cooper current? If the Cooper pair only interacted with

L~e

field strength , this would be a mystery

because there is only a nonvanishing field strength inside the solenoid. But as we have seen, there is a non-trivial gauge potential outside the solenoid and this penetrates into the ring. This suggests that it is the interaction between the gauge potential and the Cooper pairs

that is responsible for the Cooper current.

This is also in accordance with the schrodinger equation: 2 (2.65) ih( d~ + ~)~(r,t)= A)2o/(r,t)

:m (V-ir

ir

which shows us that the Schrodinger wave function interacts directly with the gauge potential.

SOLUTIONS OF WORKED EXERCISES: No. 2.4.1 Now let x = Xcl(t) be the cla~sical path. Then we can write an arbitrary path in the follow1ng form: X=XC1(t) + yet)

where

y(tll = y(tz) = 0

Let us expand L using Taylor's theorem: •• • dL dL· L(xcl +y;x cl +y;t) =L(xcixcht ) + dX ·y+ax y+

1. '2 [

d1:. •. d1:. • d1. ·21 dxdX y ... 2 dXdX YY + m~ J

This expansion is exact because L is qUadratic! Then we can rewrite the action:

I

80

s

t 2 • L(x l'x l,t)dt tl c C

=f

~2aL aL· (-- y+ - r y)dt tl ax ax

=J

where the term in the miMle vanishes t 2 dL

f (3"

tl

x

dL •

y +

ar y)dt x

t 2 =

f (

tl

aL ax -

d aL dt ax )y dt

o

because x satisfies Euler's equation! cl The last term can easily be computed explicitly:

substituting this into the path integral we get:

D[y(t)J

But Xl and X2 are not at all involved in this last path integral, so it can depend only on tl and t2 ! Consequently we have shown:

If furthermore the coefficients we can easily show that

Observe that:

a(t), b(t) and c(t) do not depend on time, then

t2

t2+At t2+At [ay2 (t )+by 2 (t )+cy:y.Jdt = f [ay2 (t -lIt)+ ... Jdt = f lay ,2 (t)+ ... Jdt t I t I +At t I + At

f

where we have introduced

y'(t) = y(t - At) . Thus

from which the desired result follows immediately when we put: At

-tl·

n

I

81 No. 2.8.2

A (X B+ f OX B)
A (x) +

B

2 £ oA


2

~ox + •••

o

axP

c

A

B

In the end of the calculation we are going to let £ + 0 hence we need only to compute the lowest order terms. In this approximation we get:

Interchanging the dummy indices a and

~

a

AadX

B in

the last term we finally get:

B a

= £2 [oaAB- "BAal Ax 6x

Taking the limit we obtain the exact result:

No. 2.11.1

(a)

Let E be an eigenvalue and let tion:

~

be the corresponding normalised eigenfun-

-E = -E~I~> = <~I~ = < H~I~> " Using the hermiticity of

Q this

is rearranged as

= ~~~> = ~IE~> = ~1~>E = E Thus

E = E and therefore E is real

82

(b)

Let E"Ez be different eigenvalues and let eigenfunctions. Then we get:

I

~"

~z

be the corresponding

No. 2.11.2

Let ~(x) be an arbitrary wavefunction. We can decompose it on the complete set of eigenfunctions:

Using this decomposition we get:

But this clearly shows that:

No. 2.11.3

For fixed (x, ,t I) we know that the Feynman propagator is a solution to the Schrodinger equation. Hence we may decompose it as i

K'X2; t 40
n

~.n (x,)exp[- hi

E t,J n

This forces us to put: an(xl;t,J = an(x,)exp[

~

Ent,l

since the left hand side is independent of t,. The above formula then reduces to:

but a comparison with exercise 2.11.2 then gives us:

and we are through.

n

83

I

ehapter 3 DYNAMICS OF CLASSICAL FIELDS 3,1

ILLUSTRATIVE EXAMPLE: LAGRANGIAN FORMALISM FOR A STRING

We have already discussed the Lagrangian formulation of the dynamics of a system with a finite number of degrees of freedom, say a finite number of particles moving in an external field. Now we want to include the dynamics of ~ields. We will start by considering a one-dimensional string. It has an important property: If you disturb the string at one place, then this disturbance may propagate along the string. We can understand this in an intuitive way. The string consists of "atoms". Each atom interacts with its nearest neighbours. Hence, if you disturb one atom, this disturbance has influence on the neighbour. But this disturbance of the neighbour has influence on the neighbour of the neighbour, etc! In this way a travelling wave is created which propagates along the string!

x

a

- axis

....-"-

• • • • • • • • • • • • • • • • •~ • -4a -3a -2a -a

Fig. 38

0

a 2a 3a 4a

xn

)-

na

In our model the string is composed of "atoms", which in the equilibrium state are evenly spaced throughout the x-axis. The important assumption is that the "atoms" are coupled to each other through forces proportional to their relative displacements (Hooke-forces). We will enumerate the "atoms" with an integer n so that the equilibrium position of the n'th "atom" is Xu = an. If we set the "atoms" in motion, then the n'th "atom" will be displaced an amount q from its equilibrium position. When the string is put into vibration~, its dynamical evolution is described by the functions: n = ••• , -2,-1,0,1,2 ... qn = qn(t) The kinetic energy of the

n'th "atom" is:

1

.2

2 mqn and the potential energy associated with the separation of atom 1

2

nand

2

[qn+l - qn 1 Hence the total Lagrangian for the system is:

2 mv

(3.1)

....

L =

L

n=-

Let us determine the equations of motion. Using (2.6) we get 2 (3.2) mqi = mv (qi+l - 2qi + qi-l)

n + 1

is:

84

I

It is easy to check that the equations of motion actually allow wave solutions. we put:

If

(3.3) then this will represent a travelling wave. motion, we get:

Inserting this into the equation of 2

mv Cos (kx -wt) (2Cos(k· a) -1) n or

(3.4) Thus (3.3) is a solution to the equation of motion provided (3.4) is satisfied. Relation (3.4) is called a dispersion relation. Observe that for small k, i.e. in the long wave length limit we may expand the cosine, getting: (3.5)

which is a linear dispersion relation. Now we want to investigate a continuous string, where there is "atoms" everywhere. We can do this by letting a ~ 0 in our discrete model. We say that we pass to the continuum limit. Now, instead of describing the displacements by the infinite set of numbers: qj(t), we will represent it by a smooth function q(x,t) giving the displacement of the "atom" with equilibrium position x .

Fig. 39 In the discrete model the mass density is: limit, we suppose that it approaches a constant

m When we pass to the continuum ~ , the mass density of the conti-

nuous string, i.e. m " .... p

for

a

~

0

Finally it will be necessary to make an assumption about v. In the disctete model the velocity of a wave in the long wave length limit is (compare (3.3) and (3.5» w k = va We assume that it approaches a constant c, the velocity of a travelling wave in the continuous string:

va ~ c for a ~ 0 With these preliminaries we can investigate what happens to the Lagrangian in the continuum limit. Let us take a look at the kinetic energy:

x= +1:0

! Xn=-OO

'! [q("n,t)]2axn .... J

x= -ao We can treat the potential energy in a similar way.

From the observation:

85

I

we get: Xn=+co

2

!mv [q n=-oo

l:

1 (t) - q (t)]2

n+

n

x=-c:o n x=+ao

Xn=+CO

l:

xn=-otI

l!! (va)2 [oq (xn,t)] 2 lIx a

ax

n

-+

Jl x=-c:o

p

c

2 [~ ax (X,t) ]2 dx

Thus we get for the total Lagrangian:

(3.6) showing that in the continuum limit the Lagrangian is expressed as an integraZ over The integrand is called the Lagrangian density.

space.

-

(3.7)

~ but also

Observe that it contains not only vative

~x come from? o

It came from the term

~ pc 2(~)2 ax

~~

Where did the space deri-

( qn+1-q)2 n

in the discrete model.

Hence, it reflects the property of ZocaZ interactions. Each point in space interacts with its nearest neighbours. In a similar way we may analyse the equations of motion. In the discrete model we have: We may rearrange the term on the right side:

= [q(x n+l't:}

qn+1 - 2qn + qn-1

- q(x n , t)] - [q(x n , t) - q(x n, t)]

a[~xq(xn+ ~,t) - 19. (x - g t)] o

ax

""

2'

n

Inserting this we get: (3.8)

-q(x ,t) n

~

2

a q (x ,t) v 2 a 2 --ax2

i. e.

-+

n

.1....Gt=is 2 2 c

at

ax2

As before, we may look for a solution representing a travelling wave:

(3.9)

= A Cos(kx-wt)

q(x,t)

If we insert this into the equation of motion (3.8), we get: -w 2A Cos(kx -wt) = -c 2k 2A Cos(kx-wt) i.e.

(3.10)

w

= ±

ck

Hence, (3.9) is a solution to the equation of motion provided ear dispersion relation (3.10).

w satisfies the lin-

I

86

3.2

LAGRANGIAN FORMALISM FOR RELATIVISTIC FIELDS You should now be motivated for the abstract field theory.

start with a field

~(t,~)

defined throughout spacetime.

of the field at a particular point stretching

q(t,~o)

We

The value

~o ' ~(t,~o)' corresponds to the The dynamics of the L, which we in analogy with the

in the preceding example.

field is governed by a Lagrangian preceding example write as: (3.11)

The Lagrangian density

depends not only on the time derivative

L

but on the space derivatives, as well:

L

= L(~,a\l~)

The presence of space derivatives

ai~

reflects the principle of local

interactions. If we choose two times

tl

and

t2 ,we may specify the field at these times.

Any smooth function

~(t,~)

which satisfies the boundary conditions:

~(tl'~) = ~l(~) and ~(t2,i) = ~2(i) represents a possible history of the field.

To each such history we asso-

ciate the action: t2 S

i.e. (3.12)

S

J Ldt tl

JL

t2 3 J JLd 1Cdt tl

(

d"x

n n

where

is the four-dimensional region between the hyperplanes: t=tl

As usual, we want to determine a history ~(t,~) which 2 extremizes the action. This, of course, leads to the equation of moand

t=t

tion for the field. action.

Now suppose that

~(t,~) where

~o(t,~)

really extremizes the

Consider another history:

n(t,i}

= ~ o (t,~)

+

£n(t,x}

satisfies the boundary conditions:

n(~,i)

I

87

Then the action:

I

SeE)

n has an extremal value, when

L

4

(o+£n, dll O+£dlln)d x

£=0 . Consequently we get

where we have neglected the surface terms due to the boundary conditions on n. But n (t,~) was arbitrarily chosen. 'llierefore the above result is consistent only if <1>0 satisfies the differential equation:

o

(3.13)

This generalizes the Euler-Lagrange equation for a system with a finite number of degrees of freedom. Observe that all the derivatives occur! This has an important consequence: The equation of motion is Lorentzinvariant provided

L

is a Lorentz-scaLar.

We would also like to discuss fields with several components: a' a=l, .•. , n (like the Maxwell field .Ace l. We leave the deduction of the equations of motions as an exercise. Exercise 3.2.1 Problem: Suppose the field has several components: <1>, a=l, ••• ,n. Show that each of the components must satisf~ the appropriate EulerLagrange equationl

o

(3.14 )

a=l, ... ,n

We may also discuss the energy-momentum corresponding to our field a A direct generalization of the Hamiltonian method suggests that the energy density is given by the HamiLtonian density:

(3.15)

H

(Compare with (2.56».

We expect energy to propagate throughout space,

so we must also have an energy current

5

associated with the field.

I

88

Since the total energy: E =

J Hd 3~

should be conserved, we must demand that

H

and

-+

s

(

satisfy an equa-

tion of continuity:

We can use this to determine an expression for the energy current

dL

doa (doa)

dL • dotb a + d(doa) dodoa -

-+

s:

dL dadoa

Using the Euler-Lagrange equations (3.14) we get:

Inserting this you find:

From this we can immediately read off the energy current:

(3.16)

We know that the energy density corresponds to the

TOO-component of

the energy momentum tensor and that the energy current si corresponds to the T Oi component (compare (l.3~), but then the above expressions for

Hand

s1

suggest that we put

(3.17)

If

L

is a Lorentz scalar, this is a tensor which reproduces

Hand

-+

s .

Exercise 3.2.2 Problem:

Use the equations of motion to show that the above energy-momentum tensor is conserved:

From exercise 3.2.2 we learn that momentum!

~aB

produces a conserved energy-

So everything should be all right:

~aB

is called the

89 canonica~

~aS

you should be careful:

I

However, there is one point where

energy-momentum tensor.

need not

be symmetric, contrary to our

previous demand (sec. 1.6) based upon conservation of angular momentum. Consider the expression (3.17) for the canonical energy-momentum tensor:

Obviously, the last term is symmetric, but the first term need not be so~

What can we do if the canonical energy-momentum tensor is not

symmetric?

We must repair it:

Suppose we can find a tensor

eaB

with the following properties: is symmetric (3.18)

Then

TaB

+

2)

dSe

aB

3)

Je

od

eaB

a

= 0 3

i =

0

is a symmetric tensor which is conserved and it.re-

~aS

produces the same energy-momentum as T

aS = ~aB + eaB

Hence, we might call

the true energy-momentum tensor.

eaS

.systematic method to construct linfante, see for instance:

There exists a

(the method of Rosenfeld and Be-

Barut [1964], ch. 3, sec. 4).

But the

method is very complicated and we will not have to use it. Later on (ch. 11) we shall see how it can be const"xucted fran a dEferent point of view. Okay, suppose we want to construct a free field theory. What should we demand about

?

L

That depends on what we expect of it!

Although we will not quantize the fields,

it might help to look at the

following very naive discussion: A particle of mass

m, energy

by the wave function:

~(PX-Et) e

(where we have put

E

1).

~

and momentum

p

is represented

iP~xP = e

When we quantize a field theory of a free

field, we expect that the field represents particles (quanta)

.~efore

we might look for solutions to the equations of motion on the form:

(3.19)

~(x)

=e

ip

P

x~

where

and we might interpret these solutions as the wave functions of the quanta of the field.

For instance, if we quantize the Maxwell field,

we expect it to represent massless particles

=

photons.

(You might be

worried about the fact that we allow comp lex solut'ions in a classical theory of fields. of the form

If you are very worried, you may look for solutions

.

Cost p ~XP)I)

90

I

Let us summarize our expectations:

We want to construct a free

field theory based upon a Lagrangian density with the following properties:

(3. 20)

11

L

is a Lorentz scalar

21

H

is positive definite

3}

The equations of motion allow solutions on satisfies the the form: e ip xlJ where P]J lJ 2 dispersion relation p plJ =_m • )l

Property 1) quarantees that we are constructing a relativistic theory, 2) guarantees that the energy density is positive, and 3) guarantees that if we quantize the field, it will represent free particles with a mass

3.3

m, where

HAMILTONIAN FORMALISM FOR RELATIVISTIC FIELDS

In analogy with the Hamiltonian formalism for a particle we can now associate to each component of the field a conjugate momentum defined by: a _ aL " - a(ao~a) Here "a should be considered as a function of ~a' ao~a' ai~a' If we can invert tne relation and obtain ao~a as a function of ~a' "a, ai~a we say that the system is canonicaL (3.21)

Exercise J. 3 • 1 Problem: Consider a canonical system. to the following HamiLton equations:

~ a"a

(3.22)

=

a~ 0

a

Show that the Euler equations are equivalent

aH

a~a

_ a "a =

0

+

a

aH

i~ ~ a

Consider a suitable function space M. A functionaL F is a map, F : M -+ R • that maps a function into a real number. If M is the set of all smooth functions f : [a,b]-+R then we can for instance consider the following functiona1s: b

F1[f] =

Jf(x)dx a

b

Fz[f] =

J!(f'(x»Zdx a

If the functional F is sufficiently nice, we can introduce a jUnctionaZ derivative. Consider first the case of an ordinary smooth function , f : Rn -+ R .

I

91

Let

Xo

be given.

Then we can expand f(x

O

f

in a neighhourhood of

Xo

+ y)

To formalise this we consider the map: y

This is a linear map.

Thus

J

af

aE

(xo + £Y)[£=O we can write it in the form: +

+ £y)

a. (x 1

depend on

Here the coefficients rivatives of f:

xQ

0

)yi

and actually they are the partial de-

at a.(x o) ~ -.(xo) 1 axl This motivates the following definition: consider a functional F. Let fa be given. Then we can expand bourhood of fa F [fa + g] = F£f o ] + ~: (x) g(x}dx + ••• If=fo To formalise it, we consider the map:

F

in a neigh-

J

of

This is a linear map.

g + aE [fa + £g] Thus, we canl~~te it in the form:

aa

J k(x)g(x)dx

F[fo+ £g]

£1£=0 The function k(x) depends on of F at the function fO :

and we define it to be the funationaL dePivative

fa

= k(x) 5F Oflf=f

o

N.B. When we perform the variation: g vanishes on the boundary.

fo

+

fo + £g

then we will always assume that

Exez>aise 3.3.2 Problem:

Consider the following functionals: t2 df 2 F} [f] = [jm(dt) - V(f]dt , F2 [f] = f(x O)

J t}

Show that the functional derivatives are given by:

Exeraise

of 2

_

Of

- 5(x-xO)

5F

,If =

-

5'(x-xO)

3.3.3

Problem:a)Show that the Euler-Lagrange equation can be written in the form: 5S _ 0 oq,a

(3.23) where

=

J

92

I

b) Show that Hamilton's equations can be written in the form: oH

(3.24)

o~a = -

6H aa 611a = at"

alia

at

where

Let

F

and

G be two functionals of the field components and their conjugate

momenta, i.e.

f F(<pa;ai<Pa;lI)a

F[a;lI a 1 =

d 3-+ x

R3 and a similar expression for logy with (2.59):

We may then define their Poisson Bracket in ana-

G.

{F;G}

Exercise J. J. 4 Problem: Show that the Poisson Bracket of two functionals satisfies the properties listed in exercise 2.10.1.

Exercise J. J.5 Problem: Consider a field theory. through the formula:

For fixed (t,lt}) we can define a functional a -+



a (t,~})

'S

This functional which depends on t and will simply be denoted as In a similar way we can define a functional:

~a(t';;:l)

lib -+ lIb(t;~ ) b

1

+

simply denote as 11 (t,x 1 ). Show that the functionals satisfy the rules:

{a(t'~l); b(t,X 2 )}

{1I

6~ Hint:

Rearrange the functionals

b .... 11 (t,x ) 2 o<pa(t,xl ) 6~b

b -+ 611 (t,x)

-+

(t,x 6

3

l

);

b-+

11 (t,x )} = 0 2

(it} - ~2)

and

a(t,~}) and use this to show:

a

and

in the form

-+3-+

-+

(x - Xl) f b -+3-+-+ f 11 (t,x)6 (x - x 2 ) ~a(t,x)o

d3~

d 3;t

+

ob 63(;t_~) a 1

0

oa(t,x 1 )

0

61!b 6l1 b (t,; ) 2

6a

611 -------

a

b 3 6 6 a

(x - it2 )

93

Exeraise

I

3.3.6

Problem: Consider a canonical system. Let F be a functional of the fields and their conjugate momenta. We will also allow F to depend explicitly on time:

The field components $ and their conjugate momenta themselves evolve in time according to Hamilton's e~uations (3.20). Show that the total time derivative along a field history is given by: dF of dt = at + {FIH} Let us make a short comment on how to quantize a field theory. quantum mechanics we can use two different strategies: (a)

As in the elementary

Path-integraL formaLism: This is closely connected to the Lagrangian formalism. configuration

~ (2)(~)

$a (])(;)

at time

Consider a field

t1 and another field configuration

We are interested in the transition ampLitude

at time t2

<: (2)1$ (1», i.e. the probability amplitude for finding the field in the a

state

a

$a (2)

at time

t2

when we know that it was in the state

To come from the configuration

at time

must have developed according to some history between $ (1) and $ (2):

$ (1) a

$a(t,~)

to

$ (2) a

$a (I) the field

which interpolates

a

a

~ ( +) "'a t 2 ,x

$ (1) 6~) a

= "'a ~ (2)(x+)

To each such history we have associated the action: t2 S[$a] = L ($a,oj.l$a}d3~ dt

f J tl R3

A straightforward generalization of Feynman's principle (2.21) then gives the result: ~a(t2,i)=~~21(~) (3.26)

J

<$a(2)1 a(1)>

exp{~S[~a]} D[~a(t,i)l

~a (tloi)~~ 1) (~) where we sum over all histories interpolating between (b)

$ (1) a

and

$ (2). a

CanoniaaL quantization: This is closely connected to the Hamiltonian formalism. Consider a canonical field theory. The physical quantities are represented as funationaLs F[$a,~b] of the field components and their conjugate momenta. In the quantum context these functiona1s are replaced by Hermitian operators in such a way that their Poisson Brackets are replaced by commutators: (3.27)

{F ,G}

-+

.." if!1 [~,G]

As we have seen the field components ~a(t'~l) and their conjugate momenta n b (t,i2 ) can themselves be interpreted as functiona1s. According to exercise 3.2.7 they must, therefore, be replaced by Hermitian operators satisfying the Heisenberg Commutation RuLes: ['a(t'~l) ~b(t'!2l1 = [wa(t,xll ; h (t,x2 )] - 0

n

[~a(t'~l}

;b(t'!2)] -

iflo~o3(;2-il)

I

94 As an illustration of the preceding ideas we take a look at the following Lagrangian density

(3.28) This is obviously a Lorentz-scalar and if we write out the first term explicitly, L(

, a)l <j»

= ~[(l!)2_ at

(l!)2j - U(<j» ax

we see that it is a direct generalization of the string-Lagrangian (3.7). The last term can be interpreted as a potential energy density. It can easily be included in the string model too if we simply assume that each »atom« in the string has a potential energy U(x) arising, e.g., from the gravitational potential. Notice too that the above Lagrangian has the same form as the non-relativistic Lagrangian for a pointparticle (2.5). With this choise of the Lagrangian the conjugate momentum is given by

1T=~ at

i.e. it coincides with the kinematical momentum. Furthermore the Hamiltonian density reduces to

(3.29) This is positive definite, as it ought to be, provided the potential energy-density is positive definite. In the following we will assume that the minimum of U is zero, and that this minimum is attained only when ~ vanishes identically. ~e Lagrangian therefore satisfies the first two conditions in (3.30) which must be valid for any field theory if we are going to make sense out of it. The third condition is specifically related to free-field theories. In our case the equation of motion is given by )l a2~ a2~ (3.30) a~a <j> = U'(<j» i.e. at2 - ax 2 = -U'(<j» which should be compared with Newton's equation of motion. If we make a Taylor eXpansion of the potential energy density, using that U(O) and U,(O) vanish by assumption, the equation of motion reduces to a a)l<j> = U"(O). + )l

~ U'"(O).2 + •••

This will only allow solutions of the form <j>(x) = Eexp[ip x)lj provide~the potential is purely quadratic. That follows immediately from the fo~ulas: a a)l<j> _p p)lEexp[ip x)lj )l )l )l U'(<j»

{U' '(0) + W"'(O)Eexp[ip x)lj +.;.}Eexp[ip x)lj )l )l

In the case of a quadratic potential we get the dispersion relation

_p p)l = U' '(O)l )l which shows that U',(O) is the square of the mass of the particle in the theory. In the general case Eexp[ip x)lj will never be an exact solution, but it will be an approximative solution pro~ided € is so small, that we can neglect the higher order terms.In the weak field limit we can therefore treat the field theory with a general potential as a free field theory, where the mass-square of the particle in the theory is still given by U·'(O). As the field quanta in the general case are no longer free they must exert forces on each other. We therefore say that they are selfinteracting. Notice that the free field case,where U is quadratic, corresponds to a linear equation of motion, while the self-interacting case corresponds to a non-linear equation of motion.

95

3.4

I

THE KLEIN-GORDON FIELD. Now we are in a position where we can construct our first explicit

field theory.

We start by constructing the equations of motion.

sider the energy momentum relation: p~p~ =_m 2 . stitution

p~ =-ia~

Con-

If we perform the sub-

, this leads to the equation of motion

(3.31) This is known as the invariant.

K~ein-Gordon

equation and it is obviously Lorentz

To check that it has the solutions we want, we observe that

(a~a~ - m2 ) eXp[ip~x~l = (- p~p~- m2)exp[iP~X~1 and this shows us that Gordon equation provided

(x) = exp[ip~x~l is a solution to the Klein2 p satisfies: p p~ =_m • ~

~

Now we must find the Lagrangian density

L.

We know that the

equation of motion

must reproduce From this

we immediately read off: a

(~~

.... L

=-~ (a~
+ terms involving

~=_m2 .... L = _1 m2 2 + terms involving o 2 So we have reconstructed the Lagrangian density:

a~

(3.32) Since



is a Lorentz scalar, the Lagrangian density is a Lorentz

scalar, too.

Finally, we should check that the energy density is po-

sitive definite:

Le.

(3.33) This is obviously positive definite, so everything is okay! Let us look a little closer at the Lagrangian density. dratic in



and

a~

It is qua-

, which is typical for a free fietd theory.

I

96

Furthermore, it consists of two terms: a) A term,- ~ (a l = 1 (1.1.) 2 _ 1 (V2 ,which is quadratic in <j>. It is called the mass term, because m gives the mass of the field quanta. Observe the sign! It will be crucial later on.

3

b)

E:x:eraise

3.4.1

Problem: Determine the conjugate momentum of the Klein-Gordon field and show that the theory of the Klein-Gordon field is a canonical field theory. Determine the Hamiltonian functional

H[,n] and verify by an explicit calculation that Hamilton's equations reproduce the KleinGordon equation.

At this pOint we have only looked for solutions on the form: (x)

= exp[ip 11 x).l]

that had an important significance on the quantum mechanical level. We might look for other solutions. The simplest possible classical solution is a static spherically symmetric solution (t,x). '= (r) . Using this ansatz, the Klein-Gordon equation reduces to: 2

m (r)

1 d

2

= r dr 2 (r
i.e.

From this we find the solution: (3.34 )

= (r)

(t,i)

±mr

£: e

r

This solution is singular at the origin. Only the decreasing solution is physically acceptable because the contribution to energy from infinity otherwise explodes. How can we interpret the solution (3.34)? In electromagnetism we have a similar solution, the Coulomb solution. There we use the ansatz: (t,~) = (r),A(t,x) = 0 where and A denote the scalar potential and the vector potential. Using this ansatz the Maxwell equations reduce to

o

2

=

1: ..i!..2

r dr

[rep]

(r) = a

+

e

r

This corresponds to a pure spherically symmetric electric field, the

I

97

Coulomb field: The field

aA E=-VCP-at


of singularity, r

!1 =

r

b r2

-+

r

r

is interpreted in the following way:

At the point

0, we have an electrically charged particle, say

a proton, acting as a source for the electromagnetic field.

The field

itself is interpreted as a potential for the electromagnetic force

E = - Vcp

which other charged particles will experience.

On the quan-

tum mechanical level two charged particles will interact by exchanging photons.

The photons,

i.e. the quanta of the electromagnetic field, will transfer momentum and when the momentum of particle changes, it experience a force. Let us make the same interpretation of the static spherically symmetric solution of the Klein-Gordon equation.

The electromagnetic field

is responsible for the electromagnetic interaction between protons.

Let us assume that the

Klein-Gordon field is responsible for the strong interaction between protons.

When you quantize the electromagnetic field, you get mass-

less photons.

In the same way we assume that you get

you quantize the Klein-Gordon field.

neutron) then interact strongly by exchanging

This exchange produces the strong forces. derived from the potential: cp(r) In other words:

n-mesons when

Two protons (or a proton and a

c r

e

w-mesons:

The strong force is then

-mr

At the position of the singularity, we have a par-

ticle, say a proton, acting as the source of the Klein-Gordon field. It produces the potential cp(r) = ~ e-mr and hence the force which r

other strongly interacting particles will experience. mr cp(r) = ~ eis called the Yuka~a potentiaL.

The potential

Contrary to the Coulomb potential we see that the Yukawa potential is exponentially damped! Hence, it has only a finite range. Let us estimate this range.

If

y

is a length measured in meters and

x

is

98

I

the mass measured in MeV, then in our units where

=~

c

1

it can

be shown that

x The observed mass of the

~-meson

is

140 MeV, hence, the typical

m

range of the force is:

2-10- 13 140 meter

= 1,4-10 -15

meter

Now that is exactly the typical distance between protons and neutrons in the atanic nucleus

am

the Yukawa potential can therefore very well account for the

strong force binding the nucleus together. force was never observed classically.

It also explains why this

In fact, the extranely short range

means that the force operates on spacetime regions so small that the quantum mechanical effects dominate completely.

3.5

THE MAXWELL FIELD. The next field we attack is the Maxwell field

the equation

Au

Here we know

of motion:

(1.33)

Can we find a simple Lagrangian density which leads to this equation? Since the equation of motion only involves derivatives of,' Au expect that

only involves

L

' we

a~Av

L

and we should determine it so that: all

3L _ U v v II II v v II a(aA}-olldA - a (oA)=a (aA - o A ) II v II II

This is a mess, so we will use a trick. gauge invariant,_

The equations of motion are

Now the Simplest way to construct a gauge-invariant

theory is to use a gauge invariant action. ever

All

All + o~X

This guaranties that when-

extremizes the action, then the same thing will be true for Note, however, that it is not the only possibility when

you want to construct a gauge invariant theory (compare the discussion in section 2.2), but let us try it! The starting point is a gauge invariant Lagrangian density Since

L

only involves derivatives of

Aa

a gauge invariant quantity out of the derivatives. the field strength:

L.

we must try to construct But this must be

99

Thus

we expect that

L

I

only depends on L = L(P

CtB

F CtB

)

But the simplest quantity you can construct out of

P

aB

which is gauge

and Lorentz invariant is the square: P

CtB

p CtB

So we expect (3.35) where

k

is an arbitrary constant.

Worked Exercise Problem:

3.5.1

Show that

4F\l\l

Using exercise 3.4.1, we see that the ansatz (3.35) leads to the correct equations of motion:

To determine ¥CtB

k

we investigate the energy-momentum tensor:

a(~~A\I)

aCtA\I

+

But here something is wrong:

Ct nCtBL =-4k[p B\la Av -

~nCtBpyOpYO}

It is not symmetric, and in fact it is

not gauge invariant, due to the term:

aCtA \I

This suggests that we

should make the replacement: aCtA\I

aCtA\I - avACt = pet

+

v

because the new term is gauge invariant. Therefore we try to repair the canonical energy momentum tensor a CtB = Is this legal?

Remember that

properties (3.18).

¥CtB

with the correction term:

4kFB\l a ACt

v

aCtB

should possess three characteristic

Pirst we observe that

¥CtB+aCtB

is symmetric:

~CtB + aCtB =_4k[pBVpCt Secondly we observe that

a CtB

_ ~ netBp pYa} v 4 YO is conserved:

4ka a{ pBVavACt]

..

4k{(a pav)a ACt \I

a

+

pBva

a

B v

ACt}

o

I

100

because the field's equation of motion kills the first term, and the antisymmetry of the field strength kills the symmetric tensor

e aB

Pinally we observe that

0Sov Aa !

does not contribute to the energy-

momentum of the Maxwell field:

4kJ povovACld3x

4kJ F

Oi

ai Aa d 3 x

4kJ (opFo~}Aad3X = 0

4kJ (oiFoi)Aad3X

where we have neglected the surface terms because the field strength vanishes at infinity.

The last integral vanishes due to the field's

equation of motion. So we have succeeded in repairing the canonical energy-momentum tensor.

Compairing the energy-momentum tensor:

¥a B +

pYa} e aB =-4k[p BV p Cl V _ :1411 aBp Yo·

with the true energy-momentum tensor for the Maxwell field: (1.38)

we finally obtain the result

k

~!

Thus we have constructed the Lagrangian density for the Maxwell field:

L

(3.36)

Exercise

EM

3.5.2

Problem: Determine the conjugate momenta of the Maxwell field A~ the theory of the Maxwell field is nota canonical field theory.

As we have seen

LEM

and show that

is Lorentz and gauge invariant, and it pro-

duces a positive definite energy density:

(Compare with (1.41)).

It remains to investigate the equations of

motion:

o

(1.33)

We look for solutions of the form: (3.37)

Here

£~

is a Lorentz vector, called the poZarization vector.

insert the wave function into the equations of motion, we get:

If we

I

101

o=

(-P)Jp)J£v + pv(p)J£)J»

exp[ip)Jx)J]

£)Jexp[ip)Jx)J] is a solution of the equations of motion pro-

Hence A)J vided:

(P)Jp)J)£v = (p)J£)J)Pv . This algebraic condition has two kinds of solutions: and Let us examine the condition

al.

b)

If

p )J p)J ~ 0 , we conclude:

(p)J£)Jl (p)JP)J)

Pv

Consequently £v is proportional to Pv and we see that the equations of motion allow solutions of the form: (3.38)

They are not observable because we

But these solutions are trivial. can gauge them away! AV' We therefore

=

Choosing

i.cL exp[ip)Jx~],we get:

X

Av + avx

disregard them!

Then we are left with the case

b).

If

P)JP)J

o ,

we conclude

but this is only consistent if p)J£j1 = 0 Thus , we conclude that the equations of motion allow solutions of the form:

A

(3.39)

Since

P)JP)J

with

v

= 0,

the photons are massZess!

o

mation. X

=

However, it

£)J,)J = 0,1,2,3 , which carry physical infor-

To see this, we make a gauge transformation choosing

iaexp[ipjJx)J]

o

These solutions are

non-trivial, i.e. we cannot completely gauge them away. is not all the components

and

• This leads to the transformed potential:

102

I

+ a x = E exp[ip xll] - ap expUp xll] = (£ - ap lexp[ip xll] v v v II v II v v II So if we perform a gauge transformation, we can change the polarization vector according to the rule: A' v

= A

(3.40l

To investigate the consequences of this gauge freedom, we consider a wave travelling along the z-axis. pll where

w

= -k

=0

pllEll

Then the four-momentum is given by (w,O,O,kl

because the photon is massless.

Due to the condition

, we conclude: £

o

= £ 3

Thus , the wave function of the photon has the form:

=

All

[E 3 ,E 1 '£2'£3] exp[iw(z-tl]

But we are still allowed to make gauge transformations:

Hence, i f we choose pletely! A =£ eiw(z-t) II

A =~ q,

pf

II

eiw(z-tl

=£Oeiw(z-t)

longitudinal

r...

part

=

£3/W

,

we

£11

tion vector into its different components, then:

+

t

£1 £2

£.1.

~=~

y-axJ.S

is called the scalar part

£0

0£ ,+

£ll= (£o,~ l

a

EO and £3 comIf we split the polariza-

have gaugEd away

z-axis

}

is called the transverse part is called the longitudinal part

£3

The above result then shows that we can gauge away the scalar photons and

Fig. 41

the longitudinal photons, but we cannot change the transverse part. Consequently all the physical infoonation is con-

tained in the transverse parts

E:r:eraise

£1

and

£2 !

3.4.3

Problem: Compute the complex field strengths corresponding to the above travelling wave and show explicitly that they only depend on £1 and £2

I

103

3.6 SPIN OF THE PHOTON - POLARIZATION OF ELECTROMAGNETIC WAVES Let us investigate the physical meaning of closer.

Each polarization vector

£1 and £2 a little may be decomposed in the follow-

£11

ing way: [£0'£1'£2'£3]

£0[1,0,0,0] + £+[O,l,i,O] + £_[O,l,-i,O] + £3[0,0,0,1]

where £ ± = b(£ 2 1

+ i£2 l

We may decompose the wave function All

£ll exp[ iw (z-t) ]

in exactly the same way.

We want to show that this decomposition is

closely related to spin!

To see this we must investigate what happens

if we perform a rotation about an axis, say the z-axis. Now a rotation is nothing but a special kind of Lorentz-transformation: X,ll

= aY

v

XV

represented by the matrix (allvl where

all

Here

e

o o o

v

o

o

1

o

Cos 8

Sin 8

o

-Sin 8

Cos 8

a

o

is the angle of rotation.

a

1

In the new coordinate system we

have a new coordinate representation of the wave:

with

£'=£aY v II v

and (~llvl the reciprocal matrix of (all v}. Hence, the rotation only attacks the polarization vector. We now get,

[1,0,0,0] ... [1,0,0,0]

: 0 [0

o

o

Cos 8

-Sin 8

Sin 8

Cos 8

°

°

~1

[1,0,0,0]

104

I

and similarly for [O,O,O,lJ. Finally [O,l,±i,OJ transforms into

1 [O,l,±i,OJ

o

o

o

o

Cos 8

-Sin 8

o

o o

Sin 8

Cos 8

a

[0,Cos8±iSine,-Sin8±iCose,OJ

o

o

1

= exp(±ie).[O,l,±i,O] Thus, if we let

RS

denote the rotation operator, then:

[1, 0 , 0 , a ] exp[ ihl (z-t)] and

[ 0,0,0,1] exp[ihl (z-t) ]

are eigenfunctions with eigenvalue 1, and [0,1, i, 0 J exp[ihl (z-t)] and

[0,1, -i, 0 Jexp[ihl (z-t)]

are eigenfunctions with eigenvalues:

e

±is is

But the rotation operator is given by: the operator of angular momentum around the z-axis.

You then see that

the wave function for a scalar or a longitudinal photon carries spin projection

0 , while the wave functions for the transverse photons

carry spin projection ±l.

You should, however, remember that only the transverse photons are observable!

Thus, a

photon is a spin l-particle, and i t has

spinpro-

spinp')ojection +1 either spin projection 1 in the direction of the momentum or it has spin projection -1 ~ p photon in the direction of momentum, but i t never

------.. jecti~ I

...

has spin projection O!

We therefore say

that the photon has heZicity ±l.

The preceding discussion may seem to bear little resemblance to what you have previously learnt about electromagnetism. Maybe the following remarks will clarify this;. We have looked for solutions on the form: (3.37)

~(x) = £~ exp[ip~x~]

The above solution is complex valued and hence, it can have relevance only on the quantum mechanical level. Rut clearly the real and imaginary part will solve the Maxwell equations too, and they are real solutions so that they may have relevance On the classical level. We want to find out how we can interpret the above solution classically and quantum mechanically. Let us discuss the classical interpretation first. We know that Maxwell's equations allow plane waves as solutions If we introduce a coordinate system, where the 3-dimensional wave vector along the z-axis, this formula reduces to: A~(X)

= £~Cos(wz-wt)

k

points

105

I

We may decompose it into a scalar part and a vector part: AP (~,it) ~ gOCos(wz-wt)

it

=

;:Cos (wz-wt)

But as we have already seen, we can gauge gO and g3 away! If we work in the special gauge, where gU = g3 = 0, the scalar potential drops out, and the vector potential is represented by:

A=

(3.41)

;Cos(wz-wt)

k.

where ;: = [,1,,2,0] is orthogonal to the wave vector We can also calculate the field ,strengths. Observe that since it is time dependent, this wave will also represent an electric field:

R= g x it - -(;Xk) Sin(wz-wt) E = -~~ =;w Sin(wz-wt)

(3.42)

Here the electric field is pointing along the polarization vector and the magnetic field is orthogonal to bot~ the wale vector and the polarization vector. As B and E are pointing in constant directions in space, we speak about linearly E polarized light. ' The density of momentum g is pointing along E Poynting's vector (C=EO=l): 2 ~~~~~~~~~~~~ g = Sin (wz-wt) Hence, there is a simple connection between the momentum density g and the wave vector k which represents the momentum of a single photon (P--hk). Similarly, the energy density is given by (c='o=l) ~ ±2 ~2 dx,t) = i(l> +B )=w2 Sin 2 (wz-wt) Let us consider a big box at a certain time, say t=O. Assume that the box contains N photOns. Each has an energy given by ~w and a momentum given by ~k. i.e. in relativistic units, where 'o=c=~ = I, we see that the total energy and momentum is given by: E = Nw , P =Nk tot tot Now let us compare this with the similar classical calculation. Here the total energy is represented by: w2V E = Sin 2wzdxdydz w2 ,(to)d 3x -2tot

...

EXR kW

~

f

15tot

Jg(~,t)d3x

J wk J S'in2wzdxdydz ~

~

wkV -2-

where V is the volume of the box. Thus, you see that the classical and quantum mechanical calculations are consistent provided you put: N = wV 2 We have also found waVes representing spin: [O,l,i,O]exp[i(wz-wt)] We extract the real part and get the classical wave: (3.43)

o

A- [

;:i~:::l ]

106

I

which solves the Maxwell equations. In this case the vector potential is circulating around the wave vector with a frequency w • Let us compute the field strengths:

....

(3.44)

E -

....

aA

- -

at

= w

l;!~i~:::~

x

A=

-cos(wz-wt)] w

Sin(~z-wt)

[

Clearly they are rotating too, and we speak about circularly polarized light. Then we turn to the quantum mechanical interpretation of the complex~wave solution: A]:I(x) = £Pexp[i(~-wt)] which we interpret as the quantum mechanical wave function for a photon with the momentum =~k and energy E·= ~w • Again we work in the special gauge where £0 and E3 disappear:

p

A' (xl -

[l~

"",(i (ki-ll

We may decompose this wave function along the x-axis and the y-axis:

A' (x l

• "

[i 1

,,,,wki_ll'"

[! ],""

(i

(ki_) l

As the solutions are only determined up to a factor, we may normalize the solution so that: (£1)2 + (£2)2 = 1 i.e. £1 = Cos a ;£2= Sin e, where 9 is the angle between £ and the x-axis. Since all states can be obtained as a linear combination of A~ll) and AP(2) , these states represent a frame for the linearly polarized state~. AP (1 ) repr~sents a photon polarized along the x-axis, and AP(2) represents a photon polarized along the y-axis. In the general state, A).! (x) = cose A).! (1) (x) + Sin a A).! (2) (x), Cos e will give the probability amplitude for finding the photon polarized along the x-axis and Sin e will give the probability amplitude for finding the photon polarized along the y-axis. An experiment may he performed in the following way: If we have a beam of photons, a light beam, which passes through a polarizer then some of the photons will be going through the polarizerwhile some of them will be absorbed. After the passage, the light is polarized. All photons are now in the same state, they are polarized along the characteristic axis of thepolarize~ Now, suppose we insert a second polarizer which is rotated an angle a relative to the first polarizer. What is the intensity I of the beam after the passage of the second polarizer in terms of the intensity I of the polarized beam? Let us introduce a coordinate system whePe the x-axis points in the direction of the axis of the second polarizer: Then the polarized beam is characterized by a polarization vector:

; =

I :::: I

and the wave is decomposed according to

A~ - Cos

e

AP(I) + Sin

e

AP (2)

107

I

Detector

I~I

Unp:>1arized ligth beam

Fig. 43 Polarizer 2

Polarizer 1

(Observe that we have a macroscopic number of photons all occupying the same state. This is possible because photons are bosons. Compare with the discussion of superconductivity). Consider a single photon: The probability amplitude that it is polarized along the x-axis is Cos a. Thus, there is a probability cos 2 a that a given photon goes through the second polarizerl But as there is a macroscopic number of photons in the same state, we may actually interpret cos 2 a as a macroscopic physical quantity. If Ni is the total number of photons before the passage of the second polarize~ and Nf is the number of photons after the passage of the secondpolarize~ then: Nf ~ Ni cos 2a . As the number of photons are proportional to the classical intensity of the light beam, we may rearrange this as, (3.45)

and that is a formula which you Can test very simply in a classical experiment.

3.7

THE MASSIVE VECTOR FIELD We have seen that the Lagrangian density of the Maxwell field only

contains a kinetic energy term and no mass term. photons are massless!

Correspondingly, the

Now, suppose we add a mass term to this Lagran-

gian density and consider the following field theory:

(3.46)

L

The first thing we observe is that we have spoiled the gauge invariance.Therefore ACt

this field

ACt

is not a gauge field.

and

So in what follows:

ACt + 0Ct X represent different physical situations. the equations of motion. Using (3.14) we get:

m2 A f! These equations imply that:

m2aSAa = aadetFetS= 0 •

Let us find

108

Hence, the Lorenz condition, for the massive field.

l

=

aaAa

0, is automatically fulfilled

But then the equations of motion simplify con-

siderably: (3.47)

(a)Ja)J -

m2 )Aa = 0

and

aaAa = 0

Each of the components Ail thus satisfies the Klein-Gordon equation, and the field Aa will clearly represent free particles with mass

m.

We call Aa

the components the Loren<: pendent. Ao

All a massive vector fieZd.

Observe, however, that

are not independent, as they are connected through

condition. Therefore only three of the components are indeAs we shall see in a moment, this means that the scalar part

can be eliminated completely.

dom correspond to spin.

The reamining three degrees of free-

Using exactly the same method as we did in

connection with the Maxwell field, we can show that the massive vector particle is a spin

1 particle.

The projection of the spin onto the

z-axis assumes the values -1, 0, +1 part also contributes!

Worked exercise

and this time the longitudinal

The massive vector field is not a gauge field.

3.7.1

Problem: (a) Calculate the canonical energy-momentum tensor for the massive vector field and show that it is not symmetric. (b) Show that we can repair the canonical energy-momentum tensor by adding the following correction term: aaS =-0 (FSPAa ) P

leading to the true energy momentum-tensor (3.48)

(c) Show that the energy-density momentum tensor is positive definite. 'lb conclude

we have presented

TOO

corresponding to the true energy-

very naive semi-classical arguments to mo-

tivate three kinds of fields:

,

Name

Lagrangian density - ~ (0)Jq,) (allq,) _ ~ m2q,2

(3.49)

Klein-Gordon field

(3.50)

Maxwell field

- '4

(3.51)

Massive vector field

- 4

1

1

F

F

Ila

F aS

as

Equation of motion (O)J all - m2 )q,

=

0

(alldll)Av- dv(OpAll)

FaS _ 1 m2AaAa

'2

(0

II

=0

all - m2 )A Il = 0 Il

aaA

=0

and we have seen that they represent spin 0 particles, massless spin 1 particles (photons) and massive spin 1 particles (vector particles).

109

3;8

I

THE CAUCHY PROBLEM In what follows we will treat our fields as classical fields, i.e.

we will only admit reaL solutions to the field equations. The first thing we will study is the Cauchy problem. If we have a single particle moving in a one-dimensional space, it has the equation of motion:

d 2x m dt 2

Now, fix a time change of

x

tl

= -v

1

(xl

and prescribe the value of

x , and the rate of

at this particular moment: x(tll

=

dx dtlt

xl

=

VI

l What can we say about the dynamical evolution of the system once we have prescribed these initial data?

This is the Cauchy problem.

In this case there is a simple answer.

The above differential e-

quation is an ordinary second order equation, hence, it has a unique solution once we have specified the value of of

x

at a given moment.

x

and the rate of change

It is because of this that we say that

x

is a dynamical quantity and that the system in question is uniquely characterized by this single dynamical t-axis

quantity. Now let us look at the Klein-Gor-

SPACE-TIME DL~l

don field. Again we fix a time tl and we prescribe the value of the

'

field and the rate of change at that moment:

y-axis

)r Fig. 44

(3.49)

What can we say about the dynamical evolution? Well, let us write out the equa tions

o

or

0

f motion :

110

This

I

is a second order equation anl thus it has a unique solution once

we have prescribed

$

and

O