Problems and Solutions on Quantum Mechanics
Major American Universities Ph. D. ~ualifying Questions and Solutions -
.
Problems and Solutions on Quantum Mechanics Compiled by: The Physics Coaching Class University of Science and Technology of China
Edited by: Yung-Kuo Lim
World Scientific Singapore New Jemey London Hang Kong
PREFACE
Solving problems in school work is exercise of the mind and enhances understanding of the principles. In genera1 amhation questions w u d y pardel such problems, Thus workrng out problems forms an essential and important p& of the study of physics. Majw Arnm'can University Ph.D. Qptdifging Questiow and Sol~tlons is a series of seven volumes. The subjects d each volume and the respective refem (in parenthem) are as foUows:
1. Mechanics (Qiang Yuan-qi, Gu En-pu, Cheng Ji&fu, Li Ze-hua, Yang
D*tian) 2. Electromagnetism (Zhao Shu-ping, You Jun-ban, Zhu Jun-jie) 3. Optia (Bai Gui-ru, Guo Guang-can) 4. Atomic, Nudear and Particle Physia (Jin Huai-cheng, Yang Baoxhwg, Fan Yang-mei) 5. Thermadynamim and Stat isticd Fhysim (Zheng Jiu-ren) 6. Quantum Mechanics (Zhang Yong-de, Zhu Dong-pei, Fan Hong-yi) 7. Solid State Physim, W t i v i t y and Miscellaneous Topics (Zhang Jklu, Zhou You-yuan, Z h m g SM-ling]
This series covers almost all aspects of University Physics and contains 2550 problems, mast of which are solved in detail. The problems have been carsfully &asen from a collection of.3100 problems, of which some m e h r n the China-U.S.A. Physics Examination and Application (CUSPEA) Program, some were selected from the PBD, Qudifyiag Examination on Experimental High Ener$y Physics sponsored by Chao Chong Ting. The rest came h m the graduate preliminary or qualiiing mamination questiom of mven world-renowned American universities: Columbia University, University of California at Berkeley, Mass~,&usetts Institute of Technology, University of Wisconsin, University of Chicago, Princeton University and State University af New York, Buffalo. Generally speaking, examination problems in physic8 in American universities do not involve too much mathematics. Rather, t h y me to a large extant characterized by the fallowing three aspects. Some problems hmlving variaus h n t i s r subjects and overIapping damains of science are selected by professon d i t l y from their own research work and thus have an %up t&ten flavor. Some probIms invoIve broad Eel& and require a quick vii
I. BASIC PRINCIPLES AND ONE-DWNSIONAL MOTIONS
Quaat- phen~m%n& am ~ W M@g&ls I in Cite d p s World. Show this numehdy for the ful~wiryy (EL] Tbw amnlihde d the z+mwlnft@ t & b for .apendulum d length 1=1rtiand-m=1kg.
(bjThe tmmhg4probability for .a m b l e of m = S g m&# at s ~ d l O ~ m / w ~ a ~ o ~ o P h ~ $ H = 5 c r n a n d w i d w = l s
( ~ ) ~ $ i ~ t i o n o f a ~ ~ o f ~ m = O . 1 & m ~ & q x d u=0,§ mi* byawhhtlfsk 1 x Ma2. ( WisWwiA)
T ~ I tHb -pint ~ O X of aT -pie ~ penddtim is mgLQ$bf, (b) If we regard the widthand hei& of the rfgld obhcle as the width d ~ a f @ ~ E a ~ ba&e$, i t ~ the p ~ op ~r o b a M U ~ i s
marble is mWly mm. That Is, the' tunding pwbatr'i br ~ c ) T h & B ~ ~ o f ~ ~ b a l l i s
(*I ~
teffa.
~
c
(b} Bkdcbadg.-*(c) h m & H m b expmimmt. (d) exptshmk (el ampton -&zing-
-
Wutfoflr
(a)
E f f d
T h h ~ ~ ' & a m i a g i a s o f ~ ~ ~ w ~ o n e ~ ~ a
m e t d u n d t # : ~ ~ ~ftwssfwndtbt l & ~ . thema@ku&t of theahchic cum& tSws producedis proportional ta the oftbe ~ ~ ~ ~ k h e h q ~ o f ~ ~ i s g w a
t
m
t
b
u
a
~ ~ e & a r a e ~ o I ~ ~ w h t l e t b ~ u f t h e ~ dinsnstdepandohthe@t intedy,butonitshquencf. Theeremlts * o ~ f b o ~ b e ~ b ~ ~ p ~ . ~ i n X g Q S ~ ~ ~ b y a a s r t m i a g ~ , h ~ i n t p r r -
~ n w i t h n t & t t e r , ~ d d f ~ D f ~ ~ , ~ p ~
~ a , ~ ~ ~ # n i ~ ~ r r m & * ~ I t i a d t h e e l ~ , * d ~ t ; h e ~ b , ~ a n ~ d w p & .W&feibbEEI~~~l~m&t~~~dkves~hah&~~
--
AWbndyksaewMJl~*&tbe~n~~nit,~b ~ ~ ~ a n o £ t h e d l a t b n . e m i t ~ ~ n ~ ~ ~ a a b e & v e d f i x l m t h e ~ h o f ~ b e t m m m a t k a a d r a d i r r t i oThe a
a
t
~
~
~
w a ~ ~ ~ ~ r m e d ~ h e t e e t r o n r r b y ~ ~ d &r~~~iP&&~teare&able ~ ( ~ ~ anddo&emit&ht. The problem them Mdeat hem wsrr obteJlred by tm?.bng-m thro*aa&md ~ b ~ a ~ w ~ o d d d ~ t ~ & o m f r o m putentid. K n o a t i n g t h e ~ d t h e ~ b ~ i t ~ ~ t e t o light. d a e 3 w m a i n l a i t w . d deducean a@m&dPalrn for tbg ehhm wttvelqthtrnd A b d c prhciph of Qmnhm M d m i w that, *but @ x t ~4W~ ? l , t h e ~ ~ a h a a a E o n r T i h S iWi ~~ t ~b t w e r e i n ~ ~ d w i ttb he d e B m & e ~ i c m A = h/pl '1~- hie P ~ t s ~ m d p i t h e ~ aShnkaperf t b ~ s n a & m i n a a d ~ ~ t g ( f l a & ~ s E a t e ) w u t d * yonandnot emit light s p o a - e . h d t y , w, spnWious & d t b n of gill b t s erare W performed by o t b with beams of helium atam and ~ ~ d ~ ~ t ~ t ~ ~ sxdtedatQmdaesocmrr8adlightia~ s t r u ~ ~ m ~ Aucmdhg to Q w m m m m d y m , the Enhrw4ion d the d b tode&mm. t i o n M d d ~ & c b m i n a n a t o m ,whi&bmtwuqumtum8ystam, (e) Oompkon Scattering ~~ a krln Of fbe Fihgbphtoa c r & h opera* of, w w doss not C0mpt-mo b m d the wt&n& d X-rays by frPa (or d y bound) ~ h ~ i f t h e r e i s n r p ~ ~ It~Ebistermfhat&ahma i r t l i ~ a m u s andhund tbe araveIeqg& 0 f t b e mdMOfl~x& ~ ~ tw ape- t r d * ~ . in a a a staw u t EM, of the h&bt radiation. The difke~mAA nuied as a fm&n of* mgie 8bW~mtkeinddefftand~&e&bm~ ~~an~entinwhkhabeamof&t~ol3akdimctda~a p h t e c o r r t ~ ~ s l i t s , I ~ A w a seyO~,dthapl&laasar;een dB.
Xothe~~re~tum~abigtheorebtmlpmbIemwask %pa an atam from emitting li$ht. &pldn+ Mar Quantum Merddm, a bii b m t W p m b h waa to make atoms in mdt9d stam em& lighE. -lain. w l & ~ ~ ~ & m 8 P r m i t ~ ?
(w-1 Sol*: h t h e d a J r s h h r e ~ M ~ , ~ t a ~ R u t h m h r d atomic modd ektmllpl rnwe mutldthe d e w r in ellI@cal orW, C l a d d & t P o ~ ~ ~ ~ n t o b e a m i t t e d w h a ~ p a a t i a e dw-. Thus the atom mu& W Eght. TbJs W the h t m m would lclEse wntSdmw?y and ui,tiaataly capturd by the mdms+
~io~ual~Eheelsctrons&dfslJ~*Pnclstrs~
~ ~ ~ h a n ~ o f d e b e c t o r s w M & e n a b ~ ~ n theektmmhit h s m b e~a c ~h o f~t h e f o ~ ~ & w a m u ~ pa& of tb dative n u d m a£incident ehcbns as a f&n dp d t h along %hem e a and g i w a b M aplanatth. (a]SlitAopm,&Bcbd (b) Nit B o m ,alit A c b d [c) B&h &%a opea. the s l i in ~ su4 a muwar Id) 9kn4kIaehv appamtw &adied
t M ~ a W - r # i & n s = A / 2 a p ~ t ~ A s a d ~ ~ ~ t h= -h/2 can pam through B. {a}Ody~Wbhs,=&/2cal3~thrOugEtAmctonly~ t h - h R* Sri& am = h/P Can
w w t a ~ ~ o f ~ t h e b e a m ~ % y # h t h a t ~ o r ~ i ~ ~ t h m q h t h e a p p a F a t l l s & a time? rry (Cobribin)
1 m
A partide af mas m i~ mbj@ ta a force F(r) = -W(r) auch that the w m WCtjoEt *, 6) eLti&es the m v t m w S c h a h g m
*-
[ . p B / ~ av;r I P ~t), = *[pt
t)W,
1 m
tIsl+ia&pmdmt &h&bger a q d i a &r 8oale arbitrmy p a W Vtx). Fkw h t if a wlntbn @ (s]bps the p p q t b n r t $(=I + Q 0 x 4 Am, the&Iutionm& bemndegmConstdm the W
n
r
j
&ate d themhe d, apart b m a possible overdl p k £actor, r n k mow that the ccmbry mminpkn 1 4 tQ a diction.
*(I
C
o
~
a
~
n
r
r
l
b
s
M
d
~
.
(a) Show h t
(Ib nged not be a shtkmwy state). @)Show that, i f t b ~ ~ h 3 i o ~ ~ t i D n a r y 2 5 ~ a ~ & * g i v e n t i i r n s k i t will a l w a ~ ~ astatbwystate, i n
[ ~ ) I f g t t = ~ t b g ~ ~ ~ b ~ k ~ ~ t b t bs e a@ ~ n a n d m e h m b , ~ t h o o m p l e E B w m ~ f t u r c t i oatn amhqumt
Ba.*i,air: Prinriplra and Onp-Dimm~ianalMotions
For the one dimc~isioncave we I~nvc,integrating over all spacp,
wbere 6 is a constant.
c=
I
N~rrnalizat~ion J:,
q/~'$~d.7: = 1 rrquires that,
1
{5;;)l.
Snpposc tl~reigcnfunetion of the hound state is (r 1 a) and H n ) =
En I n). Thrn IF TIP is a honnrl state, then
-+ fm) = 0 and henrr
d~(3:
and
(h) Supposit~gt h e partic:lp is
it1
a stationary statc! with Pnergy E at
t = to, we have f l $ ( x , t [ , )= E $ ( z , t * ) , where fi does not d~pendon t explicitly. At any later time t , t h SchrS~ d i n g ~ reqi~ation ih3IJl(z, t ) / D t = H $ ~ ( xt ,)
Hence
applies. As fi does not depend on I explicitly, the Sd~rijdin~rr ~qrlation has the formal solution
Multiplying both sides by 61 from the IPR and noting thc comrrlutability lretwe~nk and exp [-i ( f - f o ) f I / r i ] , IW finrl
$CZ, t )
is R s01lition of t h Schrijdingcr ~ equation for a frw part id^ of rn;Lqs m in one dimmsbn, and Hence (c)
( x , t ) represents a stationary state at .my Iatcr time t. The w a w function given for i = 0 can he written as
$Jl(x,0) =
6, 1x1 < a , 0,
otherwise,
*j, (a,0) = A
LZXP1-r2/n'}.
(a) At time I = 0 find t-ht? probability m p l i t t i r l ~in mo~ncnttlrns p a c ~ .
(b) Find dl (2,1.). (Berkeley)
14
Problrmi and Solutions on Eleclmrnnqnetkrn
Bmic Principles and One-Dtmpn~ionalMotinns
15
Solt~tion: (A)
At tirnc t = 0 the probability nmpliturEc in momentum space is
=-
A
m
*I-.
J
exp (-x"n2
-
tpx/ti) ~ f x
-m
i-J I ~ ,
Ant --
rn
PXP
(-n2p2/4h2)
<
(b} The Srhriidingcr rqt~xtionit1 momentum spacc for a frm particle, which a g m with the pr~viousr ~ u l t .
$(p, I )
=B
rxp
($1
A particle of rnws m is confind to n one-dimensional rcgion 0 as shown in Fig. 1.2. At t. = 0 its normnlized wave function is
At rime t = 13, WE have B = d ~ ( p , O ) Hence . rj,(r,
I):(
t = o ) = && [l+eos
<x 5 n
sin (../a).
R later tirnt. 1. = lo? (b) What is the average enern or the system at t = O ant! at t = to? (c) Irllxt is t h e prohahility that the particle is folrnd in t h left ~ haif of the box (i.e.,in the region O < 3: 5 n / 2 ) a t i! = to? ( MfT)
(a) What is the wave filnction at
SaIution: The timcindependent Schriidinger equation for 0
< r < a is
v,
1% can also expand the wave tunctiorl as FI linpnr superposition of plane waves and get
It has solution $ ( r )= A sin ks,where k is given hy k2 = satisfying $(O) = 0. The borindary condition $(a) = O then rcquises ba = ma. Hence t be mmnlized eip;cnfilnctions are
Bmtc Pnnciplas and One-Dimrnwand Motmns
17
(a) Tl~lls
$,, =
+ Pxp
Ji (P) ,
with
- sin - +
-\i~a
\iG
"in-,
a
II~VP A,(Q) =
2 fi'
m(12
)
(I)) The xvprage enprgy of thr: systcm is
Any wnvc function + ( x , t) can be expanded in $,:
.VP
-i2a2hto
sin
and the enerEy ~igenvalumnw
-
(
1 AaEO) = -
6'A,(n)
=O
for n
#
1, 2 .
cos
]
EX
sin n
RRsic Principles and One-Dimefllionnl Motions
19
1.013 A particle of mass 7n moves in a one-dim~nsionalbox of Iengtli 1 with the p o t ~ nia1 t
v=w, v=o,
Z
O<x
l. At a certain inst,ant, say t = 0, thc. WXVP runction nf this partick is known to have thr! form
4 = ~ x ( E - T ) O, < X < ~ ! , dl = 0,
otherwise
Write dowrl an rxprcssinn for $ (1,il > 0) for the coefficients in thc serirs.
n 7 .
A rigid body with rnnmrnt of inertia of IZ rotates ffrwiy in the r-7) plan^. Let Q lie the angIr hrtwwn the x-axis anr! ttip rotator axis. Find the energy eigertwlurs and r he corwponding eigenfunctions. At t,ime i: = 0 the rotator is d~scriherlhy a wave packet +(O) = A sin". Find t,h(t)for t > 0. ( Wisconsin) (a) (11)
Solution: (a) The Hnmiltonian of n pFat~eroktor is
H
a series, arid expr~ssio~is
and aa the Srhriiding~rquation is ( 'GVisconsin)
Solution:
- ( d / 2 I , ) ri2$/d& = Ed). Setsting n2 = 21,E/h2, w~ write t , h ~solution as
Tllc cigenfi~nr.tionsand the corresponding pnergy cigcnvalucs XF! +n(.r,=
= -(li2/21,) d2/dh2
8
sin ( g n ) ,
,
(T
=
Tt)2
n=l,2,3.
Thus
..
xlr = A eta*
+B
where A, R are ,zrbit*rnryconstants. For the wave function to lw singlev d t ~ e d ,i . ~ .711 (4) = d~( 4 + %), we reqtiirv a=rn=0,
& l q k,2. . . -
The ~igenvalltesof allergy art- then
{TI.
1 $ ( t = n)} =
[ 8 (7 sin
n)
.
Cc(l-
s)dr
Em=m%'/~l,, m = O ,
,... ,
awl thr corwspondinl: ~igenhmctioarcarc
after normalization :J
d$ ~!+,,dd= 1.
(h} At t = O
which corresponds to m = 0 and rn = f2. The angular spmd is given by Hence we h a w far time r Em = 1.ck2, or =
4 F.
Pmblemq and .Tuttlttons an E i r c l m m ~ . p c ~ i ~ m
20
A -n [rtY&-fif/I.j ,/,(t) = 2 4 -
+ p-12b+hf/fal
I
*
For the grnrrnd stnte, n = 1 ant1
An r l ~ cmn t is cnnfinetl in tlw grni~nrlstate in a onc-climensror~it~ t ~ o xnf width 10-lo m. Its enprgy is 3H rV. Cnlculatrt: r
(a) Thr energy nl the elrctrnrl irr its
first rlxcited state, (11) The avprage force on t h walls ~ of tlw Imx whm the ~ l r r t m nis in thr ground state. ( Wisconsin)
I
1015
b-,
.
-r d ,- I
Give the ancrgy lev~ls E!:) of thr om-tlimensional pot~ntial in Fig. 1.3(~) as wll as the energy lwels ~1;' nf thc potential in Fig. 1.3(h) ( Wi- consi in)
(a) An clrctron confined to a o~ic-dirnunsiurialbox can have cilPrKv I P V ~ I S (Problem 1011)
Thus for the first excit~dstate ( T L = 2), the energy is El = l E i = 152 eV. (b) The average force on the walls of thn box is Fig. 1.3
Diffprentiating the equation of a stationary stnte ( H - En)z(,, = fl, I
Solution: (;I)
tion is
-
at1, dn
us,
.
lnt~gratingthe Idt-hand side of the above, we have
wl~irhis zero sinc~fi is raxl. Integrating t hp ri~ht-hands i d of ~ the quat ti an t I l ~ r lR ~ V P S
(ari/oa)= a ~ jr3a. ,,
'IJsr: c n o r d i n a t ~system as,show11in Fig, 1.4. T ~ Schriidinger P ~
I I R -
Problems and Solutions on Hectmnngnrctism
Basic Principles nnd One-Dimmsiot~alMotions
2A sin K:o = (C- D }el';'",
2B For x > n (reginrl I ) , €or - n < 3: < a {regjoii TI} , Erjr 2 < -a (rrgio~lTTI) .
V =0 V =- 4 V
=O
cos
ku = (C+ D)e-""
,
2Ak cos ka = -(C - D)k' ePk'",
2BL sir1 kn = (C t D)kt c - ~ " .
For bour~ds t n k wr' tpr~uirp-Vg < E < 0. Lei
For solutions for which not all A, B, C,D vanish, wt?must have either 0, C - D giving ktankn = k t , or B = 0, C = -D giving X: rutka = -kt. Thus two classcs of soli~tionsarc possible, dvir~gr i s ~to bnutid st,at,ps. Z,rt, f = ka, q = P a . C~(I"?S I:
A
Thr SchrZjdinger rcluntion 1,~romrs d2$! I dx"
-
=0
"+I
-
5 tan [ = r l ,
for regioti 11,
c2
d1ltl
d21) dr2 which have solutions
--
47
=
k1'q'> = 0 for rt'!:ic,n I nnrl I11 ,
A sir1 k r t B cns kz
,j= cC-k'r+ ~ ~ k ' r .
WherPYP = k Z 0 2
$J
for - n < :r: < n , lor x < --n and s > n .
and
4~'b~ coirtinuous at x
-
2
= T2,
mv"~Z ha
+
Since { and r~ arp rwtriut~lto positive values, the m w g y Icvels are Iountl from t h ilitenections ~ in thr first rluaclrant of thr: cirrle of radius 7 with the rurve of 5 tan plutt,rd a ~ a i n s t5, as shown in Fig. 1.5. The ar~rnbcrof rliscret~lwels ~ P ~ P I on I ~ VU S and n , which dctermirrc y. f i r small 7 only nne soli~t~ion 'IS j)c~ssiJ~l~.
<
The rcq~iirrmmtthat rl, -> O w :r: -t dm drmands that
Thr hou~hdaryconditions that
+ kan2
I- q2
=
Ln then
givr
A sin ko
.C
- A sin ko 4-
B cos k n
= ce-"l",
n ccrs kn = ~
Ak cos X:n - Bk
sill
e ' " ~ ,
ka = -Ckt e - k ' n ,
Ak cos Lo -I Bk sin Lo = Dk' ehki" ; Fig. 1.5
21,
Prn b kmv and So[ut,ioru on EL~1~ctmrnrrgn~ hm
Cla.is 2:
Thc SclirGdinger eqtlatinn has solutions
[cot.< = - q ,
["- 'rl"
y.
d? = A sin kr $ = cCe-k*x dj = 0
A similar twnstri~c:tianis shown in Fig,
1.c. Hcre the s~nallr-stmlur of gives nn snZution whilc the larger two givu nnp solution ca<:h.
vfla2
+B
for 0 < x < n , for z > a , int a < 0 ,
cos kz
-,
0 as x satisfyitlg t . h ~ rcqirirelncnt x = 0 and :I: = n then give B = 0,
-t
m. The boundary condit.ions at
,
A sin kn = CP-"".
dk
cos
tn = -
4 cot (2 1-
c ~ ' P - ~ ' :~
<= 7j2
-T),
= ?-2
I
for t,hc class 2 snlutio!~snlmvc.
N0t.e that: = O,tl = O is a ~01~tior1 of( l.an = rl and so no l n a t t ~ rhow s~rially is, t;her~is RIW~LYS a class 1 snlution, w21ercas 7 1 1 ; ~to~ k)~: above a minimum fnr a clzss 2 sollition to xis st, given by mt. = n which has a .. rninlmlim solllt.ion 5 = ?, i.e. 7 = 2 nr V0oZ= Hvn ' 2 ( b ) Use coorditlntes 7.a show11 irr Fig. 1.7.
<
nz
pot etrt,i;~l(Fig. 1.8)
<
v
-f
00,
.7:
(a} Show thal. the bout~dst,a,t,e r n e r ~ i ~( E s
< Vo) arc ~ i v c nby the
q u a t ion
(b) Witholit solvi~lgany h r r t h ~ r skctrh , ?,hcground state w w r htnction. (Dltfil0)
Solution: (a) Thr Srlitiiding~requations for 1,hr: t,wo rcgiolis arr:
~ U S Frincrples ~ C
with resprctiw boundary conditions 9 = 0 for x -+r-oo. TIIFsolutions for E < Vo are t.lten
2
= O anrI 111
where A is a canstant. The requimment that VJ and
-t
0
and One-DtmensionnE MutWns
for
2 are c~ntinun~tsat
whereHr= H f , , p' = p + A . and $,(P'. The eigcnfuncbiona and eigenvalues of Hhart: respeetiwly ( p + A), the new eigenfi~nrtionsare As the wave number is k' = $- =:
d, = ,b(0)
p-iXfJh
,
x = a gives tm
~
~
(
L
J
=A- [ E ) /(& - E ) ] " ~ .
;tnd the corraponding eigenvillucls are
(b) The ground-st.ate w a v ~funr.t,ionis aq shown in Fig. 1.8.
Considrr the one-dim~nsiondwavr function
whew A, n and so <weconstants. (a) Using Schrijdingcr's equation, find t,he potential V { r )and energy 6 for which this wave function is an eigenfutrction. (Assume that as x + ~ 3 t' (x) 4 0). (b) What, sonnect.ion do yo11 see ketwwn t.his pok~nt~ial and the effctiw r;tdid potential for a l~pdrogenir:state of orbital angular rnomcnt~iml? ( Wisconsin)
Fig. 1.8
The dynamics of a p:trticlc moving one-dimensionadly in a potpntial V ( x ) is governcti t)y tthp Hnmiltonian Hfl = $/22m + V ( r ) , wh~rc!p = -efi d / d x is the mornerltum operator. Let. ,Tio), n = I , 2, 3, . .. , he the dgenvalu~s of Ho= Now mnsider a npw Hamiltonian H = Ho Ap/m, where X Is a given parmetm. Given A, rn and I;:!:): find thc eigenvalues of H . (Princeton)
+
The npw Hamiltonian is
Solution: (a) DiRerent,iating the given wave hlin~t~ion,
,
28
Pmbtems and .Solutions o n Electmmagnet~~m
and si~hst~it iitirig it in the time-indepentl~ntScliriiding~reqllatiori
(a) Can cach well slipport a bourld statr for mi arbitrarily snrall dnpth Explain qualitatively. (1)) For Vl = V2,what is the rclaiionship brtwcen thr energies af the t)ound states of t l r ~Iwo wells'! (s)For collti~iutl~rl states of a givcn PIlergy, hnw many irldependctlt snlubionu car1 r:ac.h wcll have? ((1) Explain qualitativrly how i t is possibl~to h;~vt!bound states for whidi z,hr partic.1~is Inore likcly k, hp oul.side the wcll tl~a~i i~lsirlt.. ( Wisconsin)
IS ( i = 1, 2)?
As V (z) -+ 0 w11r11r -t m, WF: h a w E
= -A"2mz;if
and hcnce
(h) Tllc cffectiv~radial potential For a llydmgen atmorn is e 2 / r - 1 ( 1 + 15 h 2 / 2 ~ r 2Comparing . this with V (r) wc! see that t,hc l/rz trrm is fnrrvially idrnt,iral with tllc I / x 2 term wit,h t h nrigulnr ~ moment,mrr1 taking t8heplacc OF ra. The $ brrm of V ( x ) depmrls on 7 1 = I , while the (Cnulornb) term in t.hc ~ffectivrpotcntiitl for thr! hydrogen atmomis inclrp~ndcntof Ihe orbital nn.!q;ljarmommtum I. Tlib is the diffrsetirr*1,ctwot.n Ihr! twcl pot~ntials.
Solution: (a) Far l~oilrldstnt~s.we must have -Y < E < 0. 1,ct
-
whcrr V = V , , V2 for i,hr i.wo r i c n s , and silt 5 = k o . r) = k'a, y ?-a. 'rhr disrussion i l l Prablcrn 1015 shows rlrat for the pni,~ritialin Fig. I.!),
the so111tiaiis nrr givm i ~ y 1019
< cot 5 =
Consider t,he following one-dirne~qionalpairntial wt*lls:
-71,
c2 447 = -fZ -
<
The cnr:rgy lrvnls arp given I)y the iritc:rsection of the curve cot
<-
--TI
la
with a cirrlt. of sadil~s7 wit,h cpntcr at, 1-he origin {Fig. 1.6) it1 tllc first llnadrant. As t,lir: fifilrrp shows, 7 mlist, tit' g r ~ a t ~r.ha11 r the WLIUP nf ( For Hitnct: fur a borlnd stntr to exist, wt! require which rot = [I, i.c. ( > 7r2fiZ "Fx 2 +, nr V, 2 m. Yor t.he pot,entid show11 in F ~ E1.10, . two clms~s of soli~lionsare possiblr. !hie class arr thc sarnc as thost. for t , h c~qlst: ~ of Fig, 1.6 slid are not possible libr arl~itrarilysmall V2.The oihrr clnss of solubions are givtn by
Fig. 1.10
As Ihr curvp of tan 5 = q~stari,~from 1,he origin, y may be arbitrarily :;~tlalland yrt nri intmsectian wit,h (11r c u r v ~exists. Howcvrr small Vz is, I,l~c*re is always R bound slate. (h) For Vl V2,(he \pound statm nf the potential of Fig. 1.9 arp also I*l,i~nd states of the poter~kialOF Pig. 1.10.
z.
<
Fig.
<
-
30
( c ) For continuum stat,^ of a giwn energy, there is ordy one independent sol~~tion for well I , which is a stationary-wave solutinn with $ = 0 at x = 0; thetr are two independent sol~itionscomspontling tfo trawling waves in - t x and -n: directions lor well 2. (d) Let ply pz denote resprctively t h e prot)ahilities that the particl~is irisidc and outside thc wcll. Conqider, for example, the odd-parity solution
II, = A sin k.7: far 0 < 3: < a , , I / > = C ~ - ~ ' [or " n
e- $04 Pz
A2 sina kx
kta
~2
ti,
Obtain the binding enerm of a particle of mass rn in one dimension due to the lullowing short-range peten tial:
Solution: The Schriidingcr equation
sin 2k:n
JamCZe-"'.P rix
The colitinllity of JJ at
on setting
-
:c = (1 fives
C
31
Bmic Princtple~and One-Dimefsstnnd Mottom
PwLlt-m and Soldiohs on Electromagn~trsm
Uo =2.m,v0/h2,
(.;Inbc written a s
&'"
-
,
sin ka
Sctting, as bcfore, 7 = k'a., = ka, wc! have
Tntcgr;~tingLot t~ sides of the above equation fiver z from is an arbitrarily snlnll positive numbcr, we get
sin 2<
-
c to E , whcrc
1 -- cos 25
Thc odd-parity solutions are given by ( rot
4 = -q,
E2 + rI2 = y2 whereT2=
(i= lj2)-
4)
An analytic solution is pos~ibleif y -+ (n -i-
for which thr solution is J
4
( n-1 $)n,q
4
At T,
or
0, and
The particle is then more likely oi~tsidethe well than inside.
x
+ O (d(x)
--
O) the SchrGdinger eq~lationhm soluti~ns
-
$(T) - c x p ( - k a j $(x) cxp (kzj
for T > 0, for z < 0.
It, lollows frnnl Erl. (1) that
A comparison oC the two results gives k = rbrla:rEyis - E = R2k2/2m= mV:
UnJ2. Hence the
binding-
Prnb1em and Sol7~Llionson B e c t r o m o p c t i s ~ n
Consider a partirle of r n w m in the one-ditne~lsional 6 Surictior~ pot~nti;d V ( T }= K, h ( r ) .
Show that if Vo is negative there cxist,s a bound atatr, and bhat the binding energy is mVi2/ 3 h 2 . ( Cotzimbin)
Sol~lt ion:
we sct
E < 0 for a. bound statc ns well
Raqzc Principles and (he-Dimensiorr,al Motinn*
wherp thp arttitrdry constant A has beru ohtair1t.d by thc normalization '' J-, 4 j 2 ds + T / J ~ d r = I.
Jr
A pnrticlr of mass m moves nor]-wlativist,imlly in onp dimcnsiorl in a ~r~t~entialgivrllby V ( s )= a 6 ( z ) ,whcrefi(3:) istheusualDiracdclt~afunction. The pnrticlr is Gonnd. Find t t c value of zn such t tiat the pro11:thility 1 < ~0 is exartly equal to 1 /2. t j f findink thc particle with I ( Columbira)
as
and obtain rf%,h/d~"
-
lnt,~gratirigbnth sides ovcr 2: from small positive number, obtain
dl'
(5)
- +'(-F) - k
II:LS for s # 0 t.hc .wIi~tio~~s fin it^ at
Uo h(z)$ -- 0 -5
to
+E,
1; n:,:d
wherc
F
is an arbitrarily
=
-
and hcnce
$'(o+)
-$'(0-)
= -2k+(fl) = LT0$(O).
Thus k = -Uo/2, whirh requires Vo to Ije negative. Thr energy of t h ~ hound state is then E E = -mV:/2h2 and the binding energy is El = O - E = rnVz/2h2. The wavc function of the hoiind stat^ is
-%
:I*
=
ccx; as Tallows,
< 0,
A rk"
for n:
A .-*r
far :r > 0 ,
Undl(0)= 0
With E 3 0+,this ht-c'oinrs $((I+) - d)'(flW)= r/oy'~(O), For z # O the Schrijdinger rqiiation haq t h r formal solution d)(z) rxp (-k I z 1) with X: positive, which gives
Y3
and A is an arbitrary coilstant. Applying lirn,,n+ rvllcrr k: = ' dz to thr Schriidin~eretlnst.ion fiivcs
Basic: finnplts and fir-llimensto~zalMotions
On account of symmetry, the pro1)abilit.i~are
35
Solution: L/Z+a I,A,2d:': to both sides of
apply in^ lim,,o
a
he Schrbtiinger equation,
wr get
+bt(L/2-k E ) - d* ( L / 2 - E )
=
(2n-d/fi2)d 7 { ~ / 2 ) ,
(1)
since
Subject to the boundary coilditions $(0) = @(C) = 0 , the S d ~ r i i clinger eq~~ation has soiutio~lsfor s # 4 :
we have
where k = J- -2);--Z and E is iln ilrtilrarily small positive number. Tlw r.rmtitmitv of ttic wavc frlnction at L / 2 requires A , s i n ( k L / 2 ) = -A2 sin ( k L / Z ) , or A, -/I2. Substituting thr waw fu~ictionin (I), wp get -;
A p~rticlcof mass t n nroving i l l onp tlirn~nsionis c i ~ n f i n ~tod t.hp s~aiart 0 < r < L by an infi~litesquaw we11 ~>ot.elitial.Ztl addition, thp pnrticlr experiences a. delta filnctiott pot~ntialof stren~tt)X Iocntrcl at t h center ~ nf t,he welE (Fig. 1.1 I). T ~ Scl~riiding~r P equation which dr:srribr?s this systetn is, within ttic well,
A2k cos ( k L / 2 ) - A , k cos ( k L j 2 ) = ( 2 r n ~ /Ah~a ) sin ( k L / 2 ) ,
-%, or
wllence tan $ ' = 18-ntal erjuat ion for I
- f i 8 , which is the transeon-
= 1 ~ 1 enprgy 1 ~ cijc,~nvaltscE . tat1
zh
An infinitely [Imp one-dimensional square well potential confines a par1 id-le to the region 0 < 3: L. Sketch thp wave fu~ictionfor its lowest rnl#.rgveifft.nst.atc. I F a wpi~lsivc! drl ta hinction potm~tLI,H' = N ( : r - L / 2 ) 4 \ > O), is a d d ~ dat the cmter of the well, sk~tchthe nrw wave funetinn ,111t I statr whethrr the encrgy incrcascs or d c ~ r t ~ s t s If. it waq originally Eg, iul ~;ttdoes it b ~ a m when e A -1 m? ( Wisconsin)
<
Fig. 1.11
Find n tra~iscendmtalcqlhation for the energy eigcnvalues E in tem~sof the mass vn, the pot~ntialstrength A, and the sizr L of the system.
( Caltrmbin)
For the squaw w~l1potential thr eigenfi~nctioncorrmponding to the Ir twrst c.aergy state and its tnergy ~ ~ I arc I Prcspertivcly &n (a)= sin (lrz/l),
Baqic Princrplrs and Onr- I l i m c n s i n ~ ~ oMottonn I
A sketrlt of this wave 1unr:tion is shown in Fig. 1.12 With the drlition of the delta potential f i t = Xd(:r dingcr equation I?etomcs
titt +1k2
-
-
Condition (2) then stlows that dental quntjorl
L / 2 ) , (.he Sr:llrii-
ud(x - L / 2 ) ] (1 t- 0 ,
As rr>t,I?)
where IC2 = 2 m E / d , a = 2rnAlTr.3. The bouridary tonrlitin~lsare
i~ r~rgativr.,7;/2
< Ail
37
is t.hc s~nallestroot of the t,rauscen-
L / 2 5 r, nr x / L 5
< 2 x J L . The
new grourld-st21tr wave function is sltown Fig. 1.13. The carsrsporldi~~g rr2h' , sincp ko > .; Thns the energy of energy is F = h" ki/2m. Fa, = the ncw grtlilnd stat.(? inr:rrases. - 1 2s/Larid t h new ~ groutid- stat^ energy Furthmm~ort..ilX -> +oo, I3 4 riE,,.
>
Pig. l.t2 Fig. 1-83
A nonrt~lai,ivisticp;vt.iclr of miss rn unrlesgoes on<.-dimensiotlaltnot.ion 111
Nute that (2) ariscs from taking lirn,,~~ JJ:-" I r dx owr 110th sides of the Schrijdingrr eqt~ationattd (3) arises from tho rr,nLimliw nf @(r) at r = $. Thr sol~rtionsfor rr: $ sat.isfying I) arc
0 < 1: 5 L / 2 ,
do(.) =
A sin ( b x ) ,
0 < 3 : _ < LJ2,
- A sin[kr~Ix- L)], L / 2 < x < L .
-
-916/3:
--a)+ 6 ( x + a)!
St1Eution:
<
sinTk(3: - L)], L / 2 X I L Let k = kko lor the grot~nclstate. Condition (3) rquirrs that A1 = A:! = A, say, and thc wavc funrtian for tlic grnuncl stale beronl~s A2
v(r)
wl~ert:9 > O is a coristnnt and a ( a ) is the Dirat. tlrlt;l lunctiori. Find the 1:round-state energy cig~nft~nr.tior~ and obtain an quatio11which rclntrs !,he 1 . r >rrr.sponrljngcnprkv ~oigr~ivnlu~ to thr ronst,ant g. ( Berkeley)
+
A r sin (kx),
thc potential
3
Since V (s) = V (-T), t l i ~energy cigenf~nctionshave drfinite parity. ' I ' l ~ eground state ha- even parity, 11; (-r) = 11) (T).It is a houlld stntr and 81s enrrgy is negabive, fi < 0. E'or 3: '> 0, the Sr\lriiding~requation is
3#
Problems and Sobriims on FIec1mgnciurn
-
whose solutions for x # a art- ?(I exp (AT), w h ~ r ek = m / h . With the canditiurr that the wave h~nctianremains tinitr as x -5 oo md Itas e v m parity, we obtain
Bostc Prmctples and Orw-Drmmeonnl 4fotion.s
39
where d (r) is the so-called 'd~1t.aIunction" . (a) Find tbc: modification ol tbr lotlnd-state enrrgy calzcred by the w d l when it is far away. Explain aIso haw far is "far away". (h) What is the exact condition on Voand d for the existence of at leh~t one boltnd st.at~? (Bt1ffnEo)
The canl.imrity of .JI at r = n .I/)
(x )
r ~ q i ~ i r Irhat s A
-
Beh rrsh (ka). Tlrllr;
Bckn cosll (kn)
x > o,,
13 cosh (k.r.1,
Q
=
n=
( 2k
-t-
I
Zkn) 2k
-112
-
4
At. z = n, thrtrc is a
Fig. 1.14
Solution: (a) The pot~nt.ialis as slbown in Pig. 1.14. I n the Sd~rodingerquation
k
I1+ tan h (knj]= 2ntg/Ti2,
which is to br satisfied hy k. B y symmetry the w a v ~fi~nt.t,ionfor the ~ n t i r e
Ivt
k
=
J-2rn.R//i, wlww E < 0. This has the formal snlulions
space is
(
?/I f)
=
n p k U COSII ( ~ I I . )
1x1> a ,
B cosh (ka),
1x1 s n .
P-~~ZI,
a ~ k z + b f ? - ~for ~
~/r(:r) :ISd ! ( x ) i s
-
V ( x ) = 00,
:r<
d,
d < z <0,
for x > 0,
Bnitp for s -r oo. The ront,inuity of thp wave function and thc
rlisc~nt~inuity of its derivativ~at r = D (Eq. :LT the r~juiremcnf~ J ( X= - d ) = 0, give
An apptnximatc rnodrl for thp prahl~rnof ;tn atom near n walI is to considcr a particl~moving llnder the infltrence of i,lw one-dimensional potential gilvm by V ( T ) -V0h(x), 3-> - d ,
--
(I)of Prahlem 1020),as wrll
to have a solution is that the slope of c ~ t m 2 at the origin is great~rthan that o l linp I:
"1
dk ,,=,I
The wall is "far away" From tllr particle if kd >> 1, for which t z mv0/h2. A I ) e t t ~ approximation r is k = (rr,.l/n/h2)jl - ~ x (-?mV0 p d/fi2)], which givcs the bound-stab energy
= 2rnv,d/h2
,1 .
Thr wavc Cuucticln a l the grorlt~dstat^ of a hnrrt~onicoscillat.or uf force ~:ottslantk
anrl rnzxs rrl is
Oht.aiil an ~xprcssionfor the p m h d ~ i I i tof~ finding t hc particlr olltsiric the t*l=~ical r ~ g h n . Wismnsin) Solution:
The second tcrtn ih t.hc last exps~ksioniq t.he trlotlifiratio~lof c3r1ergy cat~scdI>y the wall. Thus for thr: modifici~tinnn l PocrEy to I>(* small wp requirr! d >> i / k = h 2 / mti. This is !,he mcaning or being, "far awayF'.
T ~ I~jart.it:I~ P is said 10 l~t?outsidr the dxssiwl reion il E < V ( e ) . < Ihr the pnuntl state, E = hr*ro/2 and the uancla5sicd rrgion is $
The prnlxthility offinding thr! partick iri t h i s noncl:~ssic;ll r q i o n is there irwc
Fig. 1.15
(11)
Figure 1.35 shows l i n ~1rcpr~seriting11 = k and crtrvc 2 rcpr~lswting ( - 2 k d ) ] , w l ~ c r?je~ = n1V0/!1~. Thecnnditioir Cor th~squatian
!I = ?/e 11 -cxp
43
RmIc Principles and One- Dirnrrnsionnl M n l t ~ ~ s
Then t , h ~Hanliltonian of a harmonic oscillat.or, I-I = p2/2tn Consirier a l i n ~ a harmonic r oscillator and let GO ant1 b r its real, normnlizcd ground arld first rxcit,rd state energy ~igcnh~lictiom mpectiwly. L P ~Adto + +$I with A and B real numlrers Ije t h wave ~ function of the oscillator at some instant, of t,im~.Show t,hat t,hc average value or x is in grneral diffcr~tit,fro111 zero. What valucs ol A and U msurirnize (s)and what mluw minimize it'!
+ mw2a2/2,
giws the aifcrageenergy as
d q
lor a, b real positive wc have
(6 - fi)2 2 0,or ct + b 1 2&6,
{ Wisconsin} Solution:
The ortbonormltl condition
~
V
n2~ i. Sn2 = 1.
d{(~
Gen~ra1IyA and R are not xcro, sr, bhc: average mluc
or x , (x)=
An electron is confind ~ I tlir. I grol~ndst& of a one-dimnnsianal harmonic oscillatnr such hat - { s : ) ) ~ }= 10-lo m. Find t , I i ~ energy ! (in PV)rrqrlir~dto cxcitt! it t,o its first axcited st.;lte. \Hint: Thr virial thmrem can help.] ( Wi.~consin)
J + ~ J+ ~nqj)%
=ZAD(+~
dl) Solution: The virial t,hmrem lor a onr-dimr~~sional harmonic oscillator s t a t ~ sthat, ( T ) ( V ) . Thus Eu (I$) = (T) c ( V ) = 2 { V ) = mewZ {a:'), or, b r the gonnd stat^, tw 2 = rn,*wZ(2))
is not equ~alto zem. Rcwriti~igthr! almvr a=,
and consider in^ f = A n = A(1- A2 ) .?-,which 11m.s$xtr~mums at A wc s r c that i T A = R = I/&, {z} is maxinlizrtl; if A = -L3= I/&, minirnizecl.
-
-
-k,'
\/z
giving
(r)is As
-
Show t,hat irhe minimlinl rnrrgy of a sinlplr harmonic oscillator is h / 2 if A:rAp R / 2 , wharr! ( A P ) = ~ { ( p I;D))'J. ( Wi~consin) Solution: For a harmonic oscillator, ( r )= (p) = O, and so
-
-
(2)= 0
fnr n harmonic oscillator, wt! haw
The energy q n i r e d to mcite the clsct,rnn bo ibs first cxcited st\at,c is th ~ w f o r ~
A2
h2cZ
A E = T w = 27n, {x2)- 2m,P ( x 2 )
-
TElr wavc frlnctian at, time 1 O b)r a parzitlr in x It;~rn~onic: oscillator potrl~tialV = kr" is of thc. lorrn
~ h p r p;ij and A nrr r~a1constants, 0% mials are nnrrnalixai HI that
& Z / h , ,md tllc- Hrrmitr! plyn+ n..
-
.V,,F-''
2 2 2
f
H,,( n : r ) . ~
[3 Hn (or) (a) Write an expression tor
valum'! ( c ) What is
IT) at t = 07
A s t llr functions P X P
I
sin fl 2 J2
+ - N.r ( n r )
d~(r, t ).
(b) \i7hnt are tile p t j . ~ ~ i l ~rw~llts lf* OF it InrmtfrernF~~t of f ho F I I ( T ~ Yof ~ I I P part irk in this state, ~ n rwhat l are thc rrlativu proba>abilitimnf c~tting: thesr
e - " ~ ~ ~ ' ~
cfr
- 5 r 2 )11" ( x ) arv ort h o ~ o ~ ~alla lnn , = O excrpt = .&V0
Hnw rlcr~s{ T ) rhmgt* w i ~ htime?
J:-
ros R,
( 1CJlsmnsira) Solution: ( a ) The Schrijdingcr c q ~ ~ a t i ofor. n r Itc systrln is
whrrr 7) (r,t )take tllr ~ i v m value $ ( r , o) .IL t = 0. As nn t cxpliritly, & ( x , t ): d~,,{ x ] c - 1 I ? - f / r l wllerr
fi dnta not depend +-
1f21 /ti1 2 h f i sin Ot/~?(:r)r-' 1 .
(5)'[ I
r
-A
rb, (x) is the enrrw r i ~ ~ n f i ~ n c t isatisfying on as
1.05
I'"
d%tro(~) p t v 4 sin fi$,(r) p - ' m2
+.
1
.tr=
givrn Ey , r i t ' ~ ~ ( xdn. )j~ 7 1 tn hr No= I;;-) Nz= 7[ ~ * r t (b) 7'i1tk1)11c~vabl~ P I I ~ T ~ Y~ I I I P S Inr this state arc )30 = l1w/2 and = 5 l w / 2 , i d t,hr rrlatiw prolrability of ptting t h e e V ~ I I I ~ iTj ~
N,
irrp
(r,0) is a linrar ro~nt~ination of orlly have evrrt parity, ti,(-x, 0)T 7/,(r,0) . (e)
dj
( r )and $2 (r)wllicll
46
P r o l l ~ m sand Sdutioar~on E ! ~ r t r o r n a ~ t - l i s m
Rencr: for d
= 0,
Daqicl Prineaples and One-Dlmehsional Motions
and b, (z)are t.he eigenfunct.ions of
It folbws that tbe nvcrage vnlur of 2
docs lint
c h n n g ~with time.
(a) For a particle of maw m in a oti+dimetlsional harrnnnic osciIlalar p o t ~ n t i d = ~nw%'/2, writr down t,hc most gerirral so[utioll t.0 the t i m c dependent Schrijding~r~quatiotl,d~(x, t ), in terms nf harmonic oscillator ~igenstatesd, ( x ) . (b) Using (a) show that the rxpectation valm of 3:, ( x ) , as a filnctio~lof timc can be written as A rns wf. + B sin w i , where A and B arr mmistants. ( c ) Using ( a ) show rxpticjtly that, the timr: n,vrragp of thr! ~rotentkd enerRy satisfies { V ) = ( E ) for a genera1 $) (x,t ) # not^ thp qua1it.y
(b) Using the given equality we h a w
v
Solution: (a) From r he timfidey~ndentSchriidinger equation
= A cos wt.
+ 13 sin wt. ,
where a q
Tj docs :slot d v e n d on timp mplisitly, wr get
We can expand $ ( x , 0)in terms of
where
IT): i ~ n dwe have u s 4 En+I - E,, = b. ( r ) The time average of the pntpntia1 energy can he ronsiderd as the time averagp of t,he ~nselnhleaverage of the operator @ on +b, (x, t ) . It is sufficient tn take t,irnr: average ovpr o m period T = 2nlrj. Let ( A ) and A
48
Pmhlerns and Sotr~taonson Elcctmmngnt~!iFm
denote t h~t imc: avtyragp and ensem hle average ofan operator A respectively.
On bhe other hand,
As
and { E ) = E . Therefore
( V )= { E ) / 2 .
1033
Considrr a par~icleof mass m in the on+dirnensional potrntial
i.r. a harmonic* oscillator pntcatial with a high, 4 >> h2/rrtIi2 > thin, nrilrly i~tlpcnrtrtlbl~ I~arriurat T = 0 (see Fig 1.1 [i).
whcrv
wt?have =($1~14)
Fig. 1.16
(a) What is the low-Iying currgy spwhrurn under tlic approximation that the bartier i s romplctrly itripcnetral~le? (b) Dmrrille rlilalit at i v ~ l ythe effect on thr spertrum of the fin it^ pmrtrability of t f i r harrier.
+ 51 fw
I n,,,n., 1
J ( ILI1 ) ( I L -t
2)
cos ( 2 ~ t - 6., t 1.
n=O
wherr 6, is the phaw ol o;, n,, . Avrraging ii o v ~ ra pcriotl, a.5 the second tcrm h~mrnrszcro, we ~ c t
/MIT) Solution: (a) Far the low-lyiying cnrtgy spertrum, as the ljarrier i- comp!ctrly irnpenstmblc, the potential is etluivalmt to two qeparate halv~qof a harmonic osrillator potential and the low-lying r?igc.nfunctians must satisfy thp conrlition Il, (a)= O a t T = 0. The law-lyir~genrrgy spw.trum thus rorrkqponds
60
Prnhlernr and Solutions on Electromagnetism
Ilusic Princtples and h e - D i m ~ r a r ~ i m nMotrmq l
toothat of a normal h<arrnonicoscillator with adtl quantum numhers 2n -t1, for which ?I, ( x ) = 0, a t .7: = 0 and En = (2n-i-3/2)liw,n = 0, 1, 2, .. . with a degeneracy of 2. Thirs only the ncId-parity wave fimctions are allowed for the Low- lying levels. (b) Thcre will be a weak penetration of toheIjarrier. Obviously the pmbnbilit,y for the particle t o be in IT] < b, where the barrier ~xists,becomes less than that for thc! ease of no potential barrier, while the prohnhility outside the harrier becomes rclatiwly larger. A small portion of the even-parity soli~tionsis mixed into t.he particle.states, while near the origin the probability clistribution of ev~n-parityst-ates is greater t,hari that of odd-parity states. Correspondingly, a small pnrtinrl ol the energy EL = ( 2 ~1/2)lw ~ is rnixed into the energy for thc casc (a). Sinct. ( 4 I~barrier potential (@) > 0, the energy levels will shift upwards. The level shirts for the even-parity states are greater than for odd-parity stat,es. Rirthermore, t,hr:energy shift is smaller for greater energies for s t a t ~ sof the same parity.
* - +$li,.
nu
(r-
&) ps3
-2
-3)r
mi
we have n = 3. Hence the eigenfunctions closest, in enerbv to $a have n = 2, 4, the unnormalizerl wive fimctions being
+
The Harniltonian Tor a harmonic oscillator can be w r i t t ~ nin dimensionless units (m = h = w == 1) as
1 (z+ $)
=2
'I
I
where the unirnport;mt constants [lave been omitted.
where
&=(5-+ifi)/q5, ~ + = ( ? - i $ ) / & .
I
One r~nnormalizedenergy eigenf~nct~ion is Find two other (unnormalized) eigenfunctions which are closest, in ergy tn q:,.
pn-
(MITI Solution: In the Fock representation of harmonic oscillat,ion, 6 and C+ are the
annihilation <mtl creation operators such that, J
=f
i n1 ,-
1
=
,
6&+.&, = (n + 1) G,,
At time t = O a particle in the potentid V ( r ) = m w 2 x Z / 2 by thc wave funct,ion
described
I (a) are eigenstates of the energy with cigenvalues En where = 6,,*. 1/21 Tiw. You are given t h a t (@, ,
= (n
+
(a) Find the normalization constant A. (h} Write an expression for 11, ( a : , 1 ) for 1 > 0. (c) Show t h a t I v) (a,t ) l2 is a periodic function of time m d indicate the
longest period T. (d) Find the expectation mlue of the energy at 1
=
0.
(Berkeley)
Basic Principles crnrl One-Dimenszonal Motions
or, Ily rliffrrrntiatinn,
Selut ion:
CC
-fL
(a)The norrrialization condition
-1
~ i vA ~= s 1/&, t a k i n ~A ns prnit,ivr rr;d. (h) T h e tim-del>~nrlmtw m r hlnt:tion is
Not,c that the timr factor cxp 1-iw(n 271
-
m ) l ] is a fi~nct~inn wit,11 period
thr mcaximimi p ~ r i o dIlring ~ T / L J . Id) The expectation value of energy is
( n - m ) ~p
wllrrch.I) is :L positive ir~ttng~r and Vo > U. Discuss quiilit;ltivrlv t11r rlistrit)r~tic)l~ c ~ rrrr*rEy f rigi~rlv;zli~rs itnrl thr pi~riti~s, if ;my, of tlw c.rlrrraspc>~~dir~g P ~ W I Iti.1~ i~o tI~Is . U s r ~t 1 1 ~irit~~?rl ~ : ~ i ~ p~ rt iy~ l c i ! ~to! ~~r-t.a n orOt?r-of-~~bibgnit I I ( ~ cst,iliiatc*[or tTia IOHY'S~ rti(-rRy ~ i ~ e n ~ i i l \Spcciali.~ru ~(*. t II~K rstj t r ~ i ~t,o t , ~I he ~ cases rz -. 1 i l ~ r lr~ -3 m. S t n t r what V ( r ) b ~ c o m t will tbcst? c:ascs a~trl roniy,ilrr tilt- r s t iniatrv with p jur prcvir)w r x p r r i ~ ~ i c : ~ . ( D l ~ j f ilo) l Solution: Si~lc+r t,l1(1~ ~ ~ l t c - t iV(.r) t ~ i d + m xi :r -> M, tlicrv i~ mn infiriitr ntlrrtb~rof in t llr pott:r~tial n r d the eilcrgy eiger~villucsart- tlist-rrtr!. Also. thr a r tIi r x r i t n l sti)tcvs l ~ c r t ~ l It l i ~ v rn r tiodrs in t h~ rtyjclt~nf ff > V ( r )g i v ~ n by kA:r ( ~ tn I ) T . A:t: i r ~ r r ~ ; l s cslowly s xs rn, i~lcrrmrs.Frnrli thr virii~l thenrcm 2 T N 271,V, ~vhhitv~ boullrl stilt rs
-
Generally, ,z.; 71 incrc;~ws,t h d~i l k r e n c ~hetwcrr~ad-iaccnt Pnergg Icw!ls i!~trea.qes too. Since V(-.c) = V ( x ) ,the eigmstnt cs bave drfinite parities. The grourtrl stmatea11dthe secortd, fourth, . .. exciter1 states !lave even parity while t,Ile ottlcr statr3s have odd parity.
M
Basic Prinecples and Onr- Mmensional Mations
Prnblemn and Solrrttans on Glectrrmmynrt~um
The energy of the particle can he estimated using t . 1 unc~rtainity ~ pri11riplr
Solution: Mcthorl 1:
p,
-- h./2b,
(a)as shown in Fig 1.17. WP fake a squnrp-well poteut.i;kl V"(:c) in t hr! potr.tltia1 V (z)srirh that Assumr
where
R
pnt;ezlti:il V ( r ) =
Thus
Vt(3') = - &,, V ' ( : E= ) 0,
1x1 < 0 ,
V' ( a ) 2 j (25
for a11 r .
1x1
> a,
For thr! lnwcst ettrrgy let d E / d l ~= 0 and obktiri
For n = 1, V ( X )is thv 1)otcntinl of a har~rmoniroscillator,
Tn this case I;; equals ttw/2, c.onsist~ntwith the resrllt of a precise cnlrulation. For n, = m, V ( : c )is an inrtnit~srltlare-well p a t ~ nitd , nnrl
Wv k ~ ~ otlirtrr: w is far which
to IIP compared with thr! accurak rtsult h ? ~ ~ / 2 ~ i a "
at Ir;rst. n lwuntl stntr p p ~i11 ) tilt! wrll pcltrrltiid V' ( r )
1037
Cmwid~ra pnrtirlr in clnr? tlilnension with Harniltonian
where V ( x ] 5 0 for all r, V ( k m ) = 0, and V is not rverywherc zern, Show that t h t w is at least one hnurtd state. (One rn~thotlis to us^ t,hr Raylaigh-Ritz mriat,ional principle with a trial wave Ermctitien ?I,(x)= (h/r3'/'
exp (-1>::r:~/2) .
Let -
Howevrr, yori [nay use any m d h o d yorr wish.)
(Colztrnbira)
- T / ) ~ - ,
(x),
denote the rigcnfunctions of H , and expand
56
Problems and Solutions on El'lrctmmnpnc~ism
<
-4s V(:C) 0 for all s,I/(&CG) = 0, and Lr is tint c*vrrywherr zaro, WP have { V ) < O, (.r+"l) < O and hmt-t*I? < 0, h > 0. In fact., u1tdt.r thc ronditiw Ihat t hr tot.aI cancrgyhas ;I r.c-rtair~ n ~ g nive t mlrlr* (wl~iclin111st Lr pratrr than { V ) to ttlnktm (T) ~x~qitivr). w l ~ n t ~ thr v~r fon~ or~V ;b y;trtir.lv i l l it I * ~ \ I I I I ~IIIOVP F~ t.n irifiuily ;u~tl111uststay hl a Ilotnld statr.
I l r n t ~t h ~ r cxists r nQleast one hound stntr Metl~od2: t ~ tile l wilve f ~ ~ n s t i obvl ~
where
b is
i11
V (7).
an und~tcrrninc!rlparatr~c!ttmr.I%'(* hnvr:
wh~re
{ V ) = (~rjn)''"
V ( x ) i:xp(-hx2) d x ,
i:ili3rgy E
({:)
~ , ft hc- p;lrtic-lr'! t I~rm lowrsi ~ ~ l p s gr y * i j y a ~ i v i ~ l u i($1 h V ( r ) irr
I.'illll
limn f - r 4 r u
Suhstitut.ian in the exprE5sion for {If) yields
h2
-
E = (H)= -
4m
--
~I(.F,
t ) = liin r-r
t
Y
tt-rlt~so f
nrc-i'T~17tlf1 =0
thr* ~:ivt:r~
58
Pmhlems and .VoIaliorr.r an El'lccEmnp.lnrlism
Bear: h n c t p k s a d
whcnce the pot~ntiillfor x > 0:
59
Chic-Dirnmional Motions
is just. onr sinusoitlal loop ("IinIl a sin^ wave") with nodrs just at the edges of the left half of the potential as shown.
As thr statinnary wavt. function of t tlr: partirlr in V (XI sat.isfit.s
z-+o
4 0 , wr wr: that thr, almve cqllatio~his thc! same as t.hat sat,isfi~dby the tarlid w:ivr funrtirm of a hytlrogrn ntam with 1 = 0. Tllr corresponding Dohr rwlius is n = fi"lh.12 = 1 / ( j , wllilr t h envtgy ~ Ievcls ntwl QE (Y)
are
Ek
= - 6 { ~ ' ' / 2 h ~ a= t ~ -fi2h2/2h,ft12, ti =
1, 2,
... .
Henw ,g - - r -t (p2rL'L/2nl)(1 - lj,i2),
.I!
= 1, 2,
ant1 ronscque~itlythv lowest Pnrrgy eigcnvnlur ir; El funrtion ~J(:I',
-
.. . ,
- y with tlie
1 ) = nx exp (-Or) cxp ( i y t / h ) u: qhgl (r) exp ( - i E l t / h )
wavr:
.
(a)F i ~ i ~ the l avcrqy valur nf thc t!ilergy at, t = O (in terns of symbols drfinrd above). (b) Will tlhe avcriqe v a l u ~o l th(! ClwrKy lie constant for times subsequc~lt to the relc~wof t.hr piwticlr.'! Wily? (c) 1s this a state of definite enatgy? (That is, will R mPxuremcnt. el the Pnfvgy in this s t n t ~always givr tht! s;imr! villu~?)Why? ((1) Will the wave function rhatrge with ti~nr!from its V H ~ U Cat, t = O'! If "y~s",cxplnin how you w t ~ l t attempt l to ca1eulat.c thr ctiarrge in thc wave Function. If "rro", cxphin why not.. (r) Is it possible that thr particle could =ape Itom the potential wd't (from tb.wholc potmtial ~ 1 1 from , both halvrs)? Explafr~. ( Wisronsin)
Solution: (a) Thp n~rrnalix~rl wave f~nlrtionat t = O is $(r. 0) =: Ttius
1039
A patticle of niass m is r~leawdat t = 0 in thc one-dimensional doubl~ squarr? well shown in Fig. 1.14 in mr:h a way that I t s wave function a t t = 0
&
sin
F.
(0)is a constant for f > O sincr a{h)/at = 0. (c) It is not a stmateof dpfinite encrcy, Lwause the wave function 01the initial st ate is t,he eigenfi~nctionor an infinitely deep square well potential (h)
Basic Pwncipies and Ihc-:-D~mcnmonalMotions
w i t l ~width n , and 1101.of the ~ivc.np o t ~ n ~ ~ i1 a1.li.s a sup~rpnsition stat^ nf the riiffr?rt?ntencrSyy eigcnstaks of t.he givm potential. T'her~Sorrdifferent rnmsurrmpnts nf thc plirrRy in this statc will not give thc! sanw valuc, b t l t a gnlup of ~liergiwaccording to their prnbahili t ics. ((1) The shape of thr w:tv~hr~~rt~ion is time d ~ p ~ t l r l c sinw ! ~ ~ t the solutio~t satisfyil~gt.hp given rondi t ions is a sul?rr~)osit.jo~~ str~tr:
Solution:
(a) As
The shape of dl (:c, I ) will c-hangr.wit11 t,irr~r*hrr.;rrrsc~E,, c.h~ttgrswith n. ( P ) !'he pi~r~iclec;ln cTcilIjc'fro~r~ tAhr:wl~t~lc ~ ) r ~ t c ~ ~wvll ~ t , iii l~ tl h f~o l l o w i ~ ~ ~ I-ondillln is satisfirrl. h2a2/2~nrr> Ifo. That i: hn say, il t IIP ~ i c l t 1 1u(. tlw potential wt.1 t I s small r r ~ o a ~(i.e., h l l ~ rkinrt.ic. CtIrrKy of t l ~ rpnrtirle is I;\rcc* mmigll), the depth is nnb w r y l a r g ~(kc., t,l~r. v i r l t l ~nf I.'o is riot v ~ r ylargr), and tha rrkcrgy of t h 11nrt.ic.1~ ~ is l~osi tivt*. bl!il 1 ~ ~ r t ~ iinnri c . 1 P~ S C ' R ~ Pfro111 1111nhnlr gotrritjal wrll.
1040
A free partirl~of Inisq nr mows in olte di~nritsirr~~. At. t i m ~t uormalhed w n v ~ftrtic'tian of thr part~iclris
=O
tl~r
wtl~rrn: = !x2) .
J?G,0 ) -
' (2rh)1 !z
J
c-~~rr/fh
x r x p (-:r"dn:
d
m
(a) Computr the ~llornentumspread rr,, = it.~wr,iatcd witli this wavp ilinction. (h) %ow that, nt t imr 1 > O t,hr ~,mi~al~iJit,y dcnsit,y 01 the particle I~an t,lW form tt/,,(r,1)12 = [ d ~ ( 1 0, : , rr: + a~t.'/~ti')1~ +
(c) Iaterpl+el.the r ~ s ~ ~nf l t1mrt.s s (a) anrt (I)) a b o v ~in t r m s uncertainty principlr.
of t . 1 ~
(Colum bin)
$(x,
(I)
dx
1 tt
= 1 ( 2 i ~ a ? ) ' J ' / f i a tcxp t] ~ - a ~ j i ' / h ' ~ l .
62
h h l r m s and Solutiotw m E i ~ c f m m q n e t i s m
For a free partich. l3y inverse Fnrrrier transformati011
(d) Find the prohahility $ ( p ) d p that the p;trticle's monheatutn is h e lween p arid p dp. ( Wiscoaqzra)
+
Soltzt ion: (a)
Tllr Illran positinn of t , h particle ~ is
(h) Thr
incan rnomc!tltum is
(p) =
J
-'7J
(c) Disrussion: (i) The rt~~rlts imlic::ltr: t JIP wi(lt !I of thr. Gnusqiirn wave packvt at t i ~ n r 1 (wltic.!~W~W~ r i ~ i l l i k nT ! ! ~at. t = 0) is
h
m
{b*
(XI
t
d
1
- I/J(.~) d r
(dr
can he written as
wt~t~n r:r = !?/.in:.
(ii)
As n,a,
=
lrJ2, rhr z~n(.rrtainty~~rinriplr* is satisfid. we have
Find t h rnParl ~ posi~,ionol' thc jmrtirlr. (b} Find the Inran mornrntum oC tbt. particle. ( c ) Find V (zj.
(a)
hZ E - V ( X )= -2m. (-yZ
+ y4xZ),
(d) Tht! Schriidirlgnr qililtinn in mompntt~mrcprmc~ltationis
= 0, I IIP llot.~t~tial s l l d r l ~ n disiip~~cars, l~ SO that t h p:*rt.i~l~ ~ is h r for t i m ~ > 0. Givr a fornlt~lnfor t h r prclha!,iLt,y p r ~ ~ r tjrnr ~ i t that thr pi1rtic.1~ ; l r r i v s ;kt t i r ~ ~ 1 eill nrl ol,sc-qrvrr who is n d i s t a ~ ~ rLranlay. I
I
and stihstituting it ititn thr i ~ h a m~uation. v~ wr ~ r t
/ Mf~.pron.van)
Solution: Lrt r l ( ~x )~ I)(* t h wnvp ~ fnurt inn at I = 0. Thrtr
As thc paramrt~r(1 is indcp~rldrntn l p, t.lrr ;lljnvc? relation isfi~dby a =. 1 / 2 h ~ ? ~ t. l c n c ~
car1
hr
sat-
Thiq is t h rijir~lh~~rt:tion ~ of thr statr nrit11 rrlrrgy tt'y2/2rrt ill t.lv t y r o mrnturn r~pr~sml.;~l.ion. Nortndiznf.km ~ i v f iA' -. ( l /hZy2nj 14. Thzw t hr: prohahi1it.y that thp partidr niomPnt.llnl is I ~ r t w c np and p t rip is
'
Note thnt
I,'?
l p ) ciln
Irr. UIILIIIIIPCIdirrctly
1 ) ~t h f~o t ~ r i ~trarisfor111 r of
\h (a) :
111 nnr! dimmsion, a part.ic1~OF mass 7tr is in t h ~ grot~~td stat(%r ~ fa potrntial which r o ~ ~ f i n t11r r s partirlr to a small rrgion r)T spirr. At t i ~ n ~
Basic P n n c i p l e ~nnd Dm-DitnmtonaI Motions
Represent, t.he pahide as a Ga1iss;si~nwave packet of d i m m s ~ aa: ~~
tvtlere tp is t h Fourier ~ transform nf the initial wave function:
Tha I%.;$ integral I hcn givm
(b) Givr a plausible physical i n t e r p r ~ t ~ t i oof n
I /)(T,
thr lirniti~~g value of
t}I?
Hint: N0t.r that when a
+ m,
Sollttion: (a) Tllc Schriidingrs q u a t i o n is
BVFourier t.ransfr~r~n, wc rim
whence t.he currerit density
writ,^
and the quation km-a~nm
l3y putting r = I,, wt= g ~ .the t prolability per nnit t i m ~ that the ~~a~.tic-lr? arrivrs ;kt the o t j s ~ r v ~n rdistance L n w w
~ ( kt ), = $ ( l i , 0) exp
A I r e yartir.!e of mass ln moves in function of t h particle ~ i~d){xI O).
tine
,
dimemion. T ~ initial P wavr
( a ) Show that ~ R P Ta slificicntly long t.irnr t t llr wave Ii~uctitionof the pmtirle spreads to r~
(-ig)
exp I- ii7/4) mcp (im$'/2h.t) Ca(rnz/h.i) ,
d)(k, 0) = -
"%
I"
-m
(:c, O)
= cp (k).
68
P r o b l e m a i d Sohtions m Elcdroma~pe!~sm
giving
)n the other hand, we have
With $ = k. - ~mx/Tkt,this ber.omcs
which sllows thc t.onst!rv;~tior~ r rf total prnl)al,ility. For I -t m, wc IJILVC I$(r,t)]" 0 l l p { O ) 1 2 = 0,
tlic-
1Iriitirig c:nst?of
-
For a
-t
which inllic:;ltcs t.tt:rt. t EIP navr fu tlction of t h ~mrtirlr ~ will diffi~sr:infinitclv.
m, 7
artd S(F after a long time
1044
t ( t 4 m),
The onr-dimensional q~rantumrnccharlicd potc*rlbial cl~crgyof a particlr of mass ?la is given by
and
5)
X ~ ~ (€+ G P
enp ( - i a )
as shown in Fig. 1-19, At tirn~t = 0,the wavr hilrrt,ior~of tht? particle is cornpl~t~ly confind to the region -n < .r < 0. [Define e h quant,it,ies ~ k = m / h and rx 2rnvU/h2]
-
(a) Write down t h ~normalized ! l o w s t - e n ~ rw~a v ~filnction nf the patticre d t i m ~t = 0. (1,) Givtr thr boundary conditions which the energy eigenfi~nctions
must, satisfy, wberc the region 1 is -n < r < 0 and the region 11 a
> 0.
(h) The Schroding~rcclr~atinnfor
2
> -a
is
8 dl+ Vn6(z)d)= Ed?, --h" 27n t
d.~"
)r $" ((J
with
Fig. I.I!1
(c) Find the (real) soli~tionsfar the racrgy rigtmhliictions in the t w o regions (up t o XI cwera!! cor~statlt)which satisfy thr* lvonndary condi tiatw. (d) The r = O wave hmctiou cixti hp rxl~rrssrtl an intrgtal nvPr mrrgy eigenfrlnctior~s:
[l
~ Y (= x)
+ k24t( r )= aS(z)'lit ( r ) 2m.E
k2 = h' '
cy=
2vt Vo -. It'
The holmrlary c:onditinns nnrl the dim)nti~l~lity cotlditiot~to be satisfied R
re
I,!,'(-n) = 0,
3' (n) =
(01,
r/,"
( 1 0 0 ) = finite,
j ( k ) t , b k ( ~ )dk.
Show how f (k) can btl ck.tr.rlinrd fnrn thc snl~~tior~s d~~(:c). (e) Give an exprrssit)n for the timr devt*loprnt!t~t of t.hr w i w hrnrtiot~ill terms or f ( I ) . What vdli~esof k iLrP rxprc:t~!dt.0 KDVT~IIt.hp timr: 1whaviricv at large birn~s? W~SMTAS~TL)
Solution:
The Iz~\t.rqiratiori is ol~tninrtlby iritcgnting the Sclltiitlinger cqnirt i o ~ i fiver a snlall interval 1-E, E ] and ]ratting E 3 fl ( s r r Problem 1020). (c) IH hnth thr regions I arid LI. tllc warn rqllation is JI" (a)
+ k24,( : r ) = 0 ,
whose r p l solutioos ;irr sinusoitlal hmctiorzs. The snll~t.ionsthat satisfy t h ~ hountiarv condition.: ;tnW
(a)The rcquiml wave h~nctiond)(:r) m~sstsatisfy the botiadary cotltlitiom $(-a) = I / ) (I)) = 0. A r o ~ n p I c tort.hnnnnna1 ~ s ~ nl t WWP Iunctitl~~s ckfinrd in -a i.c < 0 irnd satisfying the Sd~riirringcrequation co~aists(kt.
.d,:, (5)= ck
sin k ( r
+ rt),
(T) = rl; sjll k ( x f a )
-a
< x < 0,
+ Ak sin kz, x > 0, r < -a.
The disuontimlity and ~iormalization cnndit.ions then g i v ~
Ia*
outsidel-a, 01 ,
wherr n = 1, 2, . . . , with
Ak
CkCY
=
k sin ka ,
The normdized Inw-t-energv wave htnction is given by n = 1 a? ( d ) Expntld the wave hrnction $ ( x ) in terms of
4jk
(x),
72
P m b k w ~ snnd Solutioxq on Rectromnpctism
and obtain
-
M
1,
f lA.1 d(k: - k') dk = f ( A : ' ) ,
At t i n ~ rI = 0. tllr parliclr is in tlir grourltl stilt(. of an infinitely fll-q, well potential of width n, it, is ;I wmrm~ ~ i ~ t : kWhe11 e t . t > 0. s i l l r r t hr 6 {z) potential Ijarrj~ris pe~lPtrabl~, the wavr ~ ) i ~ ( : will k ~ t spre;trl uvcr to tltt: region r > 0. Quitntit;~tivr!y, wr cnmpiztc! first SCIURTF:
where Ek= h2k"2m. Iu t h last ~ expression the upper and tower rows are for refions 1;uld IT r~spertivrly. Whm f + w, tilt? rjsrill;rt,ory I;I,~-t,nt ~ x l ) ( - ' i E ~ i / rhl)u u l g ~rvtw ~ mort riipirll~~ wl~ilrt hr otltrsr Iitnrtir>r~s of tllr intrgrand b~l~avr: qilite ~ ~ o r m a l l y (k = ?r/n is t ~ o itt pol(:). Thl~s4: I ) trtul to err0 h r ally giv(*13X . W h ~ n t is very lnrgr, cbt>inpntlr>llt wilvns of snli~llwavr ~ ~ ~ i l ~ iA:hpl;~y ( s r tIte principal role. At that, t,in~rtlw particle has practir~llycscap~tlfrom t h ~ r~gir~~l ! - a , 01.
IT,
(R) 111an :ttl~inpttrr 4 ' ~ l l ' t l l i l tthr ~ lift:tiinc:, Erst c.orlsitlt*rthr fiuitr POtmtial 1,arrirr shawl1 in Fig. 1-20. CaI(.rlliit(. t llr. trirrlsit,ii>u~~ml~aliility 7' for n pi~rt~irlr of 111iths T K L i l ~ ~ ' i d t ! fro111 ~ ~ t , t.hr Irft with rin!rhT E in tt~rlimit T << 1. (b) Using the altlhov~rrsult, obtitir~il, rorigh ~iui~lc:ric:al cqtimi~t~ for t l ~ c lifetim~o f the mlt'lmts Dig? CGllc>t,sr.sc?usihlr-Z>xrrierpar:imc?trrsto approximatt- tht trut?alpho, pnrticlr j)~t,c~itiaI.
(errs)
" " "L d * ( ~1 *)
sin ~ . n
/I
=T
ck
16
($
Ck
Fig. 1.20
ka k2sin - (;12
sin k ( x -tn)
{sin
A:(2
+ a ) + f sin k m sin k~
e-'Ek'/h
dk,
Sal~t~ion: (a) If T g 3 , the incirlctd wave is rrtlwt~dat a = 0 M if t , h pat.~~~tiill ~ I~arricrwere infinitply thick. We thlis haw $ ( X I = chk z ( I I - 1) e"k", s < 0, ,/! (T)= t1c-k'=, O
71
where
f.l
P r u b I m and .Coii~Eio~ on Eiectmmnp.tt~etism
Baste Prfncrples
and One-Dlrnenqianal Motim~
is the a m p I i t u d ~trmsrnis~ionco~lficient,and
TIir ~ont~inuity of plr' ( x ) :kt r = fl giv~s
Consid~rthe reflection a I b. WP h a w Fig. 1.21
The rarlius or the nucl~usor T1 is
-
n = 1 x 10-l3 x Z Q H ~
and so
-k'P
Ez) = iktz,
6 x 10-l3c m ,
or t 2 = 2 j k J / ( k + i k ' ) .
IIencc thr t railsition p~ol>nbilit,~ in givcu by
An rk-part;ic.lc, moving with speed ?I in the nur1~u.s o l Tl,rn;lkEs & collisions pm wtw~rrlwith the wails. H ~ n c rthe IiT~timer of a313i"' is giv~rr
(h) To r s t i m a t ~t h ratme ~ t ~ nf - d r r q ol H3Bi2"2,WP treat, in first itpproximation, the Coulomt) potential mperienred t ~ ytltr a-particlr in the nlTl n~~clrus as a rwtangular potential l~arsi~r. As shown irk Fig. 1.21, t h wirltI1 ~ nf thr b;irric!r m call br! take11 to ht.
Takiug lor t h r~ctangular ~ potential hnrricr a height V,fi:- (31)- 6)+G = 22.5MPV,h = ro- R = 3 3 1~~ - ' % c m (sw Pig. 1.21). 71 O.lc, WA find
4%
=fie
=
7G
Prohlmq and b~oluliona an EI~c!nomagnetssm
w t t e r ~I/o is ~,nsibivc.I f a particle wit11 mar;
An p ! ~ t . t r owit11 ~ ~ mezg E - 1 r V is irrcident upor1 a rrctangubr bartin of potrntial energy Vo = 2 CV (W Fig. 1.22). Allolrt !low wide nlrlst t hr lx~rrirrbt so that t lhr trat~s~rrissior~ pmhahility is 10- 3'?
m is incidret from the left wit11 Isrllrt-lativistic k n ~ t i cPnPrgy I;;", what is its pro\,x\iilit.y for t r a l l ~ t l l i ~ 5 i ~ r ~ r ltmlIghthc potcntid? f i r what vallucs c)f E will this probal~itit~ 1)r urritv?
{ C'n[ttmhin)
( Wi?tmnsjn)
Tllr
t rnnsmission
Tz:
proha1lilitv is (Peohlern 1 04 5 )
lSE(V, - E ) y1)2
[-
- ~ x p 11 J-]
Thr r.n~~stnats I?. ? !., A, II aro t o t b r tlrtrr~ninwlfrom tlir hnundary r.nfiditinrls tllat @ ( 7 ) md $'(z) arthIwth c-nnti~lzinusilt 3- = 0 and x = a, w11it.11
1047 Considcr a onr-dhnensiond squarr-wcll pot~ntiai(ser Fig 1.23)
-
a < 0. V ( z ) 0, V(P)=-Va, D < x < n , V(z)=0. T > n,
78
P m h l m s and .Colulians o n Elrrctrmnappnp~~sm
l3cwzc Prinrtplcg and Orw- Dmmsimml Motinw
and the probability for transmimion is
St~lution:
The St:hriidinger etltlationa for thr diRerent rrgians are
Rrmnarlce transmission accirrs when k'o = nr, i.c., when the kirlptir: cnergy ! ,of t h iririilr~~t ~ particle is (a) E < 0.
The prntlahility
Tclr
( i} Cn~witI~r first t l CZW* ~ ~nf Vn < - I?, for whirh the wave [unction is
trananissirru, P, thm bccomrx ~ n itv. i
1048
Consider a on~rlirn~nsian J squ;u~+w~11 p o t ~ n t i i ~( w l e Fig. 1.24):
As coth r > 0 fnr 3: > 0, t,h(:r~is 1 1 0 ~0111t.ion for this I:>LSP. {ii) For Vfl> - B , il. -> I , k = J-/h, and the eqilnrlon ~lct~rminiag thr: taergy bn:~tursb rot ( k m ) = - k t - The wave function is
Fig. 1.24
(a) For E < O, find the wave it~nctionnC a partide horrnd in this putential. Writc an quation which determines thr allowed values of E . (b) Suppose a part,islr*with rnrrgy E > O iq ittcident upon trhispotential. Find thr phase wlnlion ~~~~~~n the incidmt ar~dthe out~oingwave,
( Columbea)
IJrom thtr continuity and normalization
of
t h wave ~ fu~ictionwe g ~ t
Probletnx and Sol~atlomon Electmrnngnefx.rm
' I& (k.1 sin2
=
(I>) F= > 0. The
1
L
+ fi - &,
ek"O sin (kn).
sin ( 2 k n )
wave fi~nct.ionis
I~tareVD is a pc>sitivr constant. If n beam or pnti,icle-5 with mrrgy I3 i s i r l ~ l ~ From t. tlw lplt (i.r., from r = -m). what iract.ion rrl the bran1 is I I .~rismitt,erli l l l d what ir;l(:kior~rrfln:t.rrl?ConsirIrr all possibl~va111cnof fi. ( Cobm,hia) v
1111
where
A q
i3 1ndyJ O h z is ~ani~inuous at. :E = n, Pip,. 1-25
(kn) cot.(kn)= (k'a) rot (kh+ +), whence = arccat
(k
Sr~It~tion:
For x < 0,tht* Sr:l~riitlulgc.rc*cluntirw is
cot (h)) - k'n.
For r > n. whose snlutioil hits thc* 6>rrii where
H e n c ~t the pllase shift o l the outgoing wavr! in r~lationtto the i~lcidcnt wavc is d
= 2p
= 2 [arr.cot
(b
cot(b))
-
k'o]
. {i} If 6 < b, writr ~ I I V~imvr!i l s
1049 Consider a one-dimensinnat syst.ern with potential energy (sce Fig. 1.25)
As
I,!I
(a)must h r finite lor it: + clo, the solution lias ttie €om) ( x ) = beLk",
Ri
D m b l r * m s and Snluiions n EI*rtnnnmetrsm
/ { i k - Gy = ((1 - %kt/k)/(F a ikf/k). T h r r ~ f o r rt l t ~ Iract.ior~wtlrr:tc.d is Ji! = - j,r/j,,, = I r 1" 1 thr frnt.t.ir>a Ernns~nit t d i-
whrnrr r
-
(b'
+
iA-)
T=l-R=O. (ii) I; > Vn. khr rr. > O. w havr
tB- -1 2tra , ( B - V")
,Lrp
hL
uf (r) =
-
Noting t h a t thrrc a r OIIIV ~ ~ l t t g n i lwavw l~ for 1 - ) m, W P havr I c r = I , ik iks tk'l, ant! thus r -- ( k t - t ) / ( k ' - 1 k). H ~ n t - rt h rnrtinll ~ tdlrrtr.
-
;1kY J(k
- t')'.
3 1356
A pnrtir:k. oi nims rn nlid rnnrlw~llurrl p is int:idmt potrnli;hl strp shown in Fi& 1.26.
h 1 1 1the
IPR
011
thc.
Calr-ulatrm t hc pro!l;tlhifi tv t l ~ a lt t ~ rwrtirlr i s srat trrrtl backw,ud h~ tllr pnt~ntialif
< Vn. (b) p2 J2nt > l/l.
(a1 p1J2frl
(Cahrmbin)
&h(3') -- O
f~)r l l ' 2 X O .
f ~ k ( r - ~4 ~ lr )p - * k ( r - ~ n ) ,
:r
<
l p - k ' ( ~- T V I
f
> Xu,
1
b
t:
-
that t1ai.r~is ntdv oatj:oi~~g w * l v ~for x > J ~ I Tllr . ~ - r b r r tir~uit,?tc.t,rldil inns givr I t r t , iA. - ikr --. ~ k ' lartd , 11cb1tc.c r = (k - A.')/(k t A!,:'). Tllr rrrrrhallility r r f r~ftrctior~ is t1~r11 I? = ( r .1' - [(k- k r j / ( X .+ kl)ll. I I I I ill: ~
I 6%
1'lic.r~arc only ~.ransmitiedwaves in (,he rrgion x Inrirlingm quatifin i s
r l 11
< 0. wherr thr
kz = m/li and .* ha- tha soti 1t;iotl
(king ~ h m v ~ t i ~ ~t-o~~r\itirtns ~ ~ i t y o r the w;lw frrnrt.iori nl, x = 0,IVY ~ c t II = S. Fkortl tlll. r.otttin~~lt,v o f the tirst. rli*riv;rtivrclt thr wavr Ilinrtion, I . i:t8t k1(1 = k-, -5'. Ht.11t.rR = (k,- k7)j'(kl t k2). givirq: tllr rrJI~rt inn ,
m)
I052
Find the rrflrr-tiorr;u~rltrn~lsrnissic~t t c-o~ffic.ir*trt::for 1 Ilr n ~ a c - c l i t ~ ~ r ~ i h i o ~ ~ a ~ prrl ~ n t i i l ls t.pp s l i n ~ in l ~Fig. 1.27 if f . 1 1 ~p;ut kL.s ;re itit-iclr*nt.frrlm thcl r i ~ h.i ( !Visrnns.trr) VFwl
+" + p d r -- o
+" F l?l'(:t.)
w h t w kl = &??n(~.:
4
k;
d l { ~ )
= 0,
- v,)/r'l.has snlt~tions~ r ithr~form 3 = cxp ( - i k l z ) t R ex11( a kl r ) .
k - &E/h. wlut ions ;ueof t kr wtlrre
for I n , ( k / 2 ] 2 $ = 0 lor :r > 0 ,
AS for r < 0 there will also bc rcbdectd w a v ~ s the , forill
$ = ex13( i k ~ 4 ) r m p ( - i k z ) .
x
50,
u" = t rxp ( i b x / 2 ) ,
x
> 0.
86
Pmblmn5 and Soltitions on Elcctrompnetrsm
Rorn the ront.iar~itycondit,ions of thr? wave function at x = 0, wc* oht.air~ 1 + r = f , k ( l - r) = k t J 2 . and henm I. = 1/3. Thrw oncrrintIl of the particleu art- r~flert,~tI at, .r = 0.
I 1,
a
I-.,rl
I
~ ~ i + i dae rone-rlimcnuional protllc~nnf A particle of mass m incident , I ~ i o t , ~ n t of i d n shape shnwr~ill Fig. 1.28. Assume that t,he energy r + --m is gr~aterthan IIi, where I6 is tht. icsymptotir value oi the
, . . ' I l , l l l 1:11 as T ' !II r
Conuirler a part-itrle beam :1pprnoxi~1~atrt2 I>y a ~~lnuiz wavc direr:tocl along t.hr x-axis from t.he left and incitlrtli, I I ~ B O I Iit pnttr~tii~l V ( : c )= 7 6(:v), y > n, 6 ( x ) i s the Diradx rlr1t.n I~~nc:tio~l. (a) Givt. thw forin of t,ht. w a v ~hlnrt-jot) [or x
H7
Basic Snnrrplcs and Onr-Dirncr~sionalMotions
I
11.
+ oo.
rw t,hf~tt , h st1111 ~ ~ , frrflwt,rd and transmitted i~it,ensitirsdividpd
by
lt~cident.intensity is otle.
(Princeton)
< 0.
Give t h form ~ ( ~ Ihr f wave lulrctioli for z r 0. ((:) Give t h~ ~ o l l i l i t ~ i oon t ~ti19 ~ wavt: fut~ct~ion at t hr iloulli1;vy Ii(~twrt*ll tlir rrgions. (rl) Clalcu1;ll.r:t.hc prr>l>nl>ility o f tr~nsr~tissic~~l. ( Br rkrby) (11)
Solution:
(a) For :r < 0, t . 1 1 ~ L~T~CinskIr~ntwitvrs t)f thr fmnt r*xp(tkr) at~tlrrflect.4 waves o l t,ht*Form R c:xp ( - i k s ) . Tlius q)(z) = c!xp (ikz) t. R
(-ik:s).
:r < 0.
(Ir) For .c > 0, t.1a.r~rr~ilymist tralksrnitbc(1 wiiw:5 of tla. f o r ~ S ~ t rlxp ( i k s ) . Thns t/j
(c) T h r
(:r) = S ex 1)
(i kt:),
3:
> 0.
Sshriic!j~tgcs~ q ~i011a tis
t I s rIi*til~rrlas the k, nrp ctrnst,ant.s, Tllc: u w i d ~ l ~int,rrisii,y 111111il)~r of partjrl~sincirletit prr m ~ i ttime: I = h k h . Similarly, the ~~~llt.r.t.pd and trnrisnlittetl intcnsitirs zrc rwpcrtivrly
IYII~-rfa I', t,
ant1 its solrlt.ior~ssatisfy (Problem 1020)
.ELs the wavr f~~nct~ion is ~wnt.inuolmat x = 0, 'L'I(0') = $(0-). (d) From (a), (b) and (c) w have 1+ R. = S ,i k S - i k ( l - R ) = 27n,-yncijha, giving S = I/(]. + irny/h2k). Hence t,hr t.ransrnissiotl ra~ficientis
h? --2rm $*v2t/) +
dl*
;I tit1
where 6 = hzk"/2m.
1/11) = K,:.d)*d),
the conjrtgrte Schriirlinger equatiot~by 711,
AR
Problmns and Solutions nn E1wtrmna~trtt.m
IJUIC Pnnciplcs and One-T)imanriondMoliotla
and taking t h diff~rcnce ~ of t . h ~rwo ~ q u a inns, f wr! Ilave
+*v2$- ~,/tv'.tb' = V
+
89
'I'bis ~ q ~ ~ a t isi osntisfirrl u il wr wt K = -ik and F = k2. H ~ n c e
{qb*Vd, - t/?V.J,*) = 0.
$(z) = cikx (taith r
- 'ik)
This mrarls that.
is a ronst.ant. Then equating
.I
I
f (r) = $*tith/d.T - T/)dd>-/IJ:r
solitt iu11nf t h r - rqrg;ikion ;anc1 I lic t:arrc.sl,ondin~cSnrrgy is k2. TIIPH ~FS 1 for .I' -3 b~ il111I t i l l k h ( - T ) = - I . ; L H ~ :C HT C ~ ~ V P
1 . 1 1 1 1 1 .r 3
f (4 m) a11cIf (-m), wr find
kt1 - l r I 2 )= flj~tv. Mi~ltiplyi~ig hnt,ll sidrs l,y
2 ~iws
<
I
I-R+T.
Silx-e Vi(.r} t ) , ~ > 0, thr t,ri~nsniissiunc.nrffirir*rit is ' 7 = I ;unl tlir ~,llcc.tionc-orfic-irtttis R = 0 ils thr: p;lrtir.le trilvcls thmrrgh V ( a ) . So 1 t ~ r *
': lrlatrix is
-(I
t
-
zk)/(l
* ik)
(1
A SchrGdinger erl~tatioriin one ~ I ~ M P ~ I K ~ Orc:~(Is II t . 1 1 ~Sc.IiriitIi~l~~r rttil;it ion wr t ~ n w :- I / J = F $ . i.hi* W ~ I O ~ P J ~ P C : ; ~ ~MTII ~ S P .r is ;L ~ic~tl-~it>clt* l)nilt~rlst:ttti
(b) L c t t i ~ ~ g = s r r h r
i11
1 lt.tlcp E = -1, ~.oordinaipspilrr, it, ~tir~st llr t l ~ rpaulrrl xt.;~tr.
(a) Show t.h;lt ex11 (ikz)(tanh :r: t r;r,ust) is ;r st,l!~tianfor a pari.irul;rr val~~c! oi tlw rnnstiu~t,.Calrulate t h S-rniit ~ rix (trtutsmiusion : ~rvHrrtiorl d c o c f i r i ~ n t s )for this prol>l(vn. (1)) The w w e function s c t ~ h :I~apptens ~ to satisfy thr Schriitli~~gcr rqllation. CaEc~llat.r!the energy of the correspnti(linR Imunrl s1nt.e artd givr ;I s i n l p l ~agulrirnt t it nutst be tllc g~011ikdY tat^ 01 the pot~ntial. ( c ) Ontlinc I~owyo11 ~niqhthave proc~erlnlto at inint,~ ~rour~rl-strrtc! meray if you did tint ktlow t llr wave Ir~nvtion.
(n z l n d O ) Solution:
1 . h ~consttan?.in the givrn aolirtion 71 I.>(? IC and su!~stit,itt,it~g t,lw Schr;jdi~tgcr~qlrntion, we ohtain
(a) Letting I!) i t ]
alight ~ > m t wby d nsslu~iill~ ;I iitui-nndr 1w11urlevrrr fil~lr.tirrt~ wil,l~ (c) pararntltr-r ;iritI ol,t,nin ao ;~lq)r(~xitn;~tt% wtltw* crf t l ~ r~ror11lt1 stidr* rnprgv I,y the v;triatioual ttir*t,hurl. ;I
1050
,4 rno~zo~nr-rgetir 1m-iillr1l hcarn o r nnrlrr*lnt ivist ic. t~c.utrn~rs of ellr*rcp E is incidrnt ontn tlte I ~ H ~ stlrfart* I P of ;L [?iat,rof mitt lur of r hir.kl1r.s~I . 111 thc. rnattrr. flip 1~111tr0ns IIIOVP ill a lltlifonn :ittr;vbtivepr>t~ntiir! V. Tht. i15riclr:nl l>cn~nlllilkes a11 angb H wit.11 rcsprrt. i n the ~lortnidt,cl t.11~111n1le sl~rfaccxis slmwn in Fig. 1.2!). (a) What Iriw,c.tionnI t31r it~cidrnlhmm is rcflect,cd il I is iulinitc*'! (11) Wlwt frnr.tion of thr incident I ~ r a ~isnreflected if Y is repuhive and V = E? Colrsirler 1. finite. (ClJS)
91
Rasic Pn'ncaples and One-Ditnensiond Motions
As bhe potcntid does not vary with 8 , k, = ko sin B and the abovc t na.c.r lnlP
1+R=T,
-
t ~ i e hgive R (kC,tos 8- k,)/(kO cos O+ A,). The prob;tliiIity nT refl~et~ion P = [ R l2 = (ko MS t9 - k , ) " { k ~ cos 8 t k,)" with k: = 2m(E C ' )/h2- k$ sin2 0, k;;!= 2 7 1 ~ E / f i , ~ . (b} For LC < 0 the wave function has the same form as that in {a}. For 4 I .: 7 < t, E - V = O, and tho Schriirlingcr quat ti on is tt
I
+
: 1 hen
Solution: (a) Let kfl 1)e thp wave nurnl,rr o l arl inrirlr~~t nnit,ror~,given lry 27nE
F. For
.c 4
0, thr
k:
=
W ~ V I fiirlcti~n *
nsslllnr $ = exy) ( i k ' y ) rxp (k:c), wt~erek' = Lo sin 8. Sr~bstit~ution g i v ~ s- k'% kk'= O, or k = fkt. F ~ ~ I I C P I III* wave lunvtian for 0 < R: < t is As the potc~itialis unifnrm in
Writjng ?b(r,?I}-- +(r)
e'k'", we
WP
For t 1 1 three ~ reginns
~ R V P
With t infinit4eand thr potrritinl ncgaiivt!, for x > 0 thc St:hrijdir~gr.rptlrl;ltion is
Assuming a solutinrl
ant1 rtihstiti~tin~ it tho cquatinn. wc nbtuin The boundnry conditinns at 3: -- O
k:
+ kt = 2rn ( E + V ) / l i z .
'l'hp! boundary canditinns
give
1+ r = n + b , ixl,(l-r)
then givp
= k'(n
- b),
c mp(ik,t) = n rxp(klt)
ik,c
+b
exp{-ktt),
exp (ik,t) = k'n exp (k't) - k t t exp (-
k't)
,
whose solution is
with
sthe ahovr mrp (-6) is thi: at trnuitt,innrncfirir-lit wltrre
I II
-
A*' ikx {{= R.' t tkr ' ararI thc traction of n ~ urotls i rcfltrted is
A1 ternativc Solution:
The sniutic~t~ C,W also 1)r ol)ti~i~w(l l*,y s ~ q ~ ~ r ~ ~ oofs i ntf ~i t ~i tio. ~~; i ~ ! ~ [ ) l i f u r l s , similar L o t.hr aiac elf a Fal~ryPrnrl i s t rrfm~lntatc*~ i l l c,l,t Fig. 1.30).
with
A = I;'/.
ic-s
A.'
=
tlf
n w;tw
J 2 7 n ~( I - rr,i2 B)/h. =
ill
rn~rliurn3,
sin B / h .
As
k,,
(scbck
=
SE cns BJh,
we find
=
(kzr
i s i t l O - ~ . o ~ f l -zio
-
X.I~)/(~++ I ~ k?x) = i sin 0 + cos fl
Fig. 1.30
We nwtI only ~ ~ n s i d tlw c r x-componmnt nl thr waves. Let TI?,Rill ~lrn n t ~the cneffisiwts nT ;unplitllrlr transmission and r~flrrtionas a i + r a ~ ~ * RnPs fro~nnicdiu~ll 1 to rnetliurn 2, r~;.pwtiv~Iy.Let T21, RZI de11r)tc c:o~fficicnt.sof arnyhilutle transn~issionanrl reflrrtion from rnerliuln 2 t c l
-- 4i cos10-sin p-4aUt!-'2kft Be-'" -
eTkft
=ci
,
94
Pmhl~rnanrtd Solnitions m E1rctromngltptr.m
The transnlissivity is tl~rrefnrp
and T/J- r e f ~ rto the ~xpon~lntially i~t.tenrt:itedight- and leftl 1 4 - t ~@ r 8 .tr,r.ling wwrs resp~rt~ive'ly. The pr~l)ahilityc ~ l m n is t br
T~P.sF: arc t hr! exponent iaIIv attcni~at~l rwsrrnts in the r~spcctivr:dirccI I, 111s.T ~ aljsorption F cnt?fticirnt is thm
J 057 Find thr w a w Er~ilr.(.iolr I t ~ ra ~>nt.tic.lrrnnvil~l: iri nlrr* rtirumSiollit, ;I ronst;lnl i ~ n n g i ~ i ; ~1)nlc:nti;ll ry -- iV whrrr* V << E. Cdrnlat f.l~t*~woI);tMlityt n i ~ m * t ~m~cl t . show I . l ~ i;a, ~ t irtisgiEmv,lnf w ~ i ; d r ~ t r r w l l t s;il)mq>tioaor p;ati(.lw. Find an (rprnsioll h,r thi. irl,a,rl)( inn cboeffic:ienti i i (.rmnsof V . Wismv;.~itt)
Sol trtion:
T h r Srhriiclinger vql~ntiouis
The i~riagirlary~)otrllt,i;~l iV is r t ' s ~ ) ~ ~ ~for ~ itl,~hl~abtior~~t~ic~n e of thc* I~:bdide,s i n t ~thi* P X ~ ~ ~ I Iin P Ij I woillrl ~ l ~ eirr~ia~irlary.Ilenrc? 1.11rrpwt~rilcl Is. m absorption iF V wrrr rcn!.
I Lpt the wiolnt,ir)nto kht: ol1r!-di1r~c.nsiotla1 Free-prticle tinrr-drpt:nd~nt Sr.hriirFinger equation of ~t~finitt* w n v ~ l r n ~ thl l he 4 (x,i ) as d ~ ~ ~ r by i l ~ ~ d Mitmar ohservcr 0 in a lramc with coorclinat~s (2,1 ) . NOWcorlsidcr tlic same ,~;vhir!eas desrrib~rlbv wave funrtiorl $'(xl, 1 ' ) i ~ ~ o r r l i nto g o b s ~ r v ~0' r wit11 coordinates (r',i t ) re1at.d to ( r ,E ) I>y the Gidilean translornlatian
Supposing B = c x p ( - I E l l h ) ~ x (itr). p wr iravt. k.' = ( 2 n ~ / f i ' (1 ) +
i V / E ) . As V
<< E ,
DO $(z, I ) , dlr (z', 1') iIesrri11~waves nf t h same ~ wawlength? (b) What is t,lw relationship between dj(a, t ) arlrl l / l ' ( x l , t f ) if both satisfv thc Schriidingpr equation in their rrspectiv~coordinates? (it)
(Berkeley)
96
b6lem.r
and Soluttntu
MI
BIcrtmai~r~rt~.pm
B a ~ t ch n r i l , i c s and O ~ ~ ~ - U i r n m s t o nMotton* nE
(a) Tile on~dimensiollalt.irrtr-rlep~~al~nt Schriidirlgrr equal ion for a f r r ~
181.1king USP of Eq. ( I ) i11wl thr drfiniticrr~sof k and
1 ~ WP , SPC
that
particle has a snltlt.ic~ncnrrcspolwling to a clrfinitc witvrlcrrgth A T h i s is just thr. SrhrLcliltgrr rqu:rtion tfra,t 4) ( x , 1 ) satisfirs. l i ~ t ~ r t . , . ~ l ~ r - u r rtn t t ~:t phwr fartor, wr E I ~ V P tht* rrl;~tiori
A%tt~lrpnrtirlr innnimtutn p is diflirrtlt ie Lhr twcl rrhrcncr f r a ~ ~ aria* r, wnvflcn@Il X is alsu different. (1 r) Applying the Gn1ilr;ui t r;rusforn~at it111 ; i 1 ~ 1~ n x k i tt~l~s :~ nf t lv- Sr.llriid i ~ l ~ equat r ti on in t I i ~( s t , t ' ) frmmr! WF fi11t1 h, part,ic.I(*(II ltwss 7 1 ~lmrll~t!in ;t ot1r-rlin1rnsir.ln;~1 harti~at~ir r)sr'illnt.c~r 2~,~tential of Frvqilt~~iry u mirl in t.110 grnllnd s t , i ~ ti s, ~slll~jr~t~cvl t.0 ;tn ia~pnlsivr. Iclrr~p h ( I ) . Find tlrr (~rolmlrilityit. r t * ~ ~ ~ nini r its ~ s grrrnwl SEittC. ( Wiscons?n)
h:1
= - - ' - " , t ) - t ' - , , t ) . 27n
- jh,,i(krand
wl)
(dtIb1-
(I)
Tht: ptirtirlr. trc~ives; L ~ I i ~ ~ s t : t ~ ~ t ; r urnolnr~ltrl~n rt~l~s p nt t = I1 ;~ntlits vt*locit,yc - l ~ i ~ l ~t,n jim y / ? i r i ~ ~ s t ; m t i ~ t ~ t r Thr ~ ~ ~ sd~iriition ly. of tllc* impulsr is, II~IWPV~T, tot1 ~111)rt r[)r tht* W;IVP ft~tlc.t,icmto rhangc. Hrrlc.r>,~ I Itlic virw crf :I rralrir K' ~ i o v i u gwit 11 LIIV pnrtir.)~, lltc l i ~ l t ~isrstill in ihr grr)und statr t d tile hilnnanir osrilli~trrr1 1 (~5 ~' ) . B I I ~in , t . 1 1 ~virw of a stationary Frarnt~ I<. it is in t.hr st;tt,r t/jo ( x t ) exp ( - i p r / h } . Wr may rrastlna1,ly t r ~ a tt,he ~>ositiooof t1lc 1,zrtit:lr ns rnnsbitnt {luring the procws, so that a l Ike end ( I It l l ~irnl,lllw tIlc* ~umrrlln;~tr: of I f t r pitrt.icl~is Slip same for hot.11IC w1r1 li'. TIenr-tl Iitp initial w n v ~fi~nct~inr~ in ' A is
l'lnrs, t.hp prol)al)ilii.ythat the particle rrmains in its ground state aft.er the impubr is
Problen~sand Saltibm~an L T ~ ~ t m n p n c t i , m
Bnsic Prsnrriplcs nnd Onc-Dimensiond Motions
99
the tilhle surfare as tllc r~f~rrncr point of grnvitat,ional potentialI I % :, ground $4 a4.e w m funrtion ~ of thr h r m @ = x exp (-h:'/2), whrre 1 r . 11, he detcrminml. C O T V S ~ ~ P ?
I
whrre n
=:
11,
$ 1 :1
J G h .
An idraliz~tl pin^ ~ l r > r ~1)nll r: of mass n r L is t~cwrrcitigiil i t s grollnrt stat(* (EII a reroilless t ahlr in a orir-tlimcelsiot~i~l wc~rltlwit11 ty i k v ~ r t i v acl l i r i r t i o ~ ~ . (a) Provr tlwl thr rs-rgy r l ~ ~ > r r son l r t t l , , q , b ;a,t.r)nlia~10: z =
liary
(rn"/f~lr,Z)" mvrlt dr~trnrtinrtr. E
i~~l tht' (h) By ~ a r i i ~ t i ~ l l~i;i t! l f ' l l \ ~ l t rstil~li~t(' for nh -- 1 grilnl ill rrgs.
( . ~ l l s t i l l l th. i ~ ~ l r~illltitt(! t l
(Princctmn)
Sol~rtion: (a) By t l r ~nirlhotl of tlirnnl~sirlnnlanalysis, iF wr have*
{h) Take t , l ~ pz coordinate in the vertical up rlirrction with origir~at the tahlr. The HamiItonihn is
Thr following t.hmrnn mncrrns thc energy sig~nwkluesE,, (GI< E2 < E3 < . . . ) of b h ScI~riidinger ~ ~ql~ation in one dim~nsioa: Vl ( r )gives the rigenmluw Etnand the p* Thmr~m:1E the tentid V,(:r) gives the c i g c n v n l ~ ~F2, ~ a and VI(2)5 K(r)for all r, then Gin
I E2n.
(a) Prove
this theorem.
-
Hint: Consider a p o t ~ nial f V(X, r ) ,w l l e r ~V{O,: r ) Vl (r)and V(1, r) = Vz (z)and BVJBA 1 0 (For a11 :r), and ~Jc.11btci?E,,/8A. (h) Now considcr t h pntcwtinl ~ ( F ~ E 1.31) .
,
1 ) = , ,I I t 1 r 1 is r Noh) t ! ~ , t t W P havr 11wr1(,tcd! 1 T ~ A -= ) 1. { L } - 2 . Th-a V(:r) t l ( : r ) . 11 IT,, is all iturrgy level I . I I i,hr potential L J ( T ) , i 1 1 ~ 1 1E,, < ( 1 1 I 1 / 2 ) b ~ ~ l, ~ c . rw( := h r il l ~ ~ ~ lstatr. t t ~ d El, 5 k n 2 / 2 . SoIvitt~( N + 1/21) hw k n 2 / 2 , wrmfitld
,
,
=
A
>
<
31j t\
2
1 1 c . r ~ jAl
in~1it;l.t.~~~ t Ilc o~:rrtimil~nint rg(*rt.11;it ir 11:s~t ~I:LII A. Wp now (+llcr(,sr* for V(:J:)a StllliWP wrll of f i ~ ~ itlrpt.lr, tr
WP want to drttwriitir: kbc r~i~rnhrr nl l~oi~nrl st.;lt~xt.lr;tt tliis ~rrbt~rltial can t~otd. Ass~lmrt.llis nr~mbcrN is > 1. It I I I J ~1 ~ 1 hrlplul ~ to tlmw 11. qitditalivr: picturr. of the. wnvc fitnvtiot~for t.hr. h i ~ h r s tbotrnd nt.atr.
Cllnosc a sulvatlc: c+trlnpitrisorl pot~trtialm(1usr! t l l ~t llmrp~n1~11ovc to d ~ t e r m j ~ ri ei t h ~ ra rigoro~~s itpper t~ourrrlt,o N or x rix{:c,rous 1wt.s I,orrral tcr N . (Dnt k rml ltc tlolrr luit p r l ate i ~ s k for ~ l (111lyo ~ l p . )
fi.
11tr
TIw ~ n m b r rof Imnl~dstatrbs n i l l ( : r ) is I r s s than that (XIV ( : I . )wliicmll , thc tnt.t.rr is (2mr~n"~hl4- 1. WE ran t n k 11zc ~rpprrI~rn~tlrl to tliv
r1111d,cr of 1~c~1intl sti1t.e~nf l I { x j 11s [ 2 f n u i t 2 / ~ ;IS t ~ jklr N >> 1 tllr I,orln 1 npgtcr+t,~l. Takr~itog~tllcr,wr g ~ tI.hnI . thr! nt~mhrroi llnuntl st aim I,; h r t , w r ~ (~nl w n 2 / 2 h ] tinrl {2rrlwnz/irhj.
~ . , L I be I
( i D l ( A 1) = A ( 3 ) (1 - A) lil ( r ) . Ollvio~~sly V ( 0 ,r) = Vf ( X I ,V(1, :r) = 112 ( r ) ,aV/iIA = V2(:c)- V,( x ) > fl, Tlir Ha~niltonionis t bell
and the ~ l ~ ~ ~ q t ~isa t i o n
m)In,A) = E,,(A) 1 wherc EtL(A) = ( 7 1 , A I ~ T ( xI),I$ A). AS
71, A),
{n In'} = &,,; .F=o anrl W are pawhpre J r l ) arr an twthowrmal rwnetrrs, Asstnne p~riotIic:hoi~ndaryconditiou~qso that 1 N -1 j ) = I j ) . Cdc~ulatetlie rwrgy levels and wave runc.tions. ( Wkcoasin)
102
Problems and Solutions o n E l e c t r o m a g n r t ~ s m
Solution:
Basic Frinczples and O n e - D i n ~ e n s i o n a lMotions
and so
- A 1
From t h e fact t h a t 111) form a cornplcte set of o r t h o ~ ~ o r m functions, al where n = 1, 2 , 3 , . . . . N, arld
0
.
.
.
.
U
A-Xi=
i.e.,
or
NxN
(let (A - XI) = ( A ) +~( - I ) ~ + ' = ( - I ) ~(XN
giving with
A..I =(:"O-',
6,
27l
.
- - J ,
N
.
J X O , l,'L,
.,,
-
1) = 0 ,
,N-1.
and then
=
we know t h a t H , A ant1 A'- IL:LVC thr s m r eigc:llvt:c:tfors. Heucc wrx ollly n e ~ dt o find t h r c8igc:ilvec:tors a n d eigcnvahlcs of t h c operator A+ to solve: the pro1)lnn. As
Akfk = ( k ' ) A 1 k ) =
(23" + 'LW c:os 6.j ) 1 E j ) .
Hrllc:c thc cigc:nv;ilucs of H are
E, = Eo:o+2Wros B,,
with
27l 6, = N J,
(j
=
0, 1, 2, . . . , N - 1)
6 / ; 1 , k . ~ ,
T h c corrcsl,ondi~igcigcnfunctiolls can he obtained from the matrix rqnations (A-X,)\Ej)=O
105
Baszc Principles and One-Dzrnensional Molrons
(b) Let thc lowest energy of a wave that can p r o p a e t e through this ~,otentialbe Ea = hL2k~/2i11 (this definw ko). Writ,(:down a transcendental ,,(pation(not a ilifferentisl i:(luatio~l)that can h:solved to give ka and thus It;(,. (c) Writr (low11t,he wavc fliri(:tio~lat energy EOvttlid ill tllc region 0 5 a (For ullif(ariiit,y, let 11s cliooso nors1i~lisi1t,io11 ;111(l pliaso sllch that , / I ( ~ = 0) = 1). What h a l q ~ n ~tos t,lir: wave hil~(.t,io~l I)ct,w(:(:~lI. = a and .i: = a + I)'! (d) Sllow that there arc rimg(:s of v:~l~i(:s of E l gr~at(:rtllitil Eo, for which Illwe is no cipo~ifilli~:t,io~~. Find ((:xit(:tly) tllc t:Ii(:rgy at w1iic:ll t.11~first sllch 1;ap begins.
<
Give a brief discussion of why tlicre arc energy ballds in a crystallinc solid. Use the ideas of quantum ruechanics but do rlot att,em~,tt,o carry out any complicated calculatiolls. Yo11shollltl assllrnc that anyone reading rnc:cha~lic:sI)~ltdocs rlot ulldcrstar~~l your discussion undrrstands (~uallt~llm anything abolit the tl~eoryof solids.
( Wiscorrsrn) Solution:
Solution:
A crystal rnay l)e regarded as all infirlitc, ~)criotlic:array of j)otc:llti;tl wells, such as thc latticc structure givt:n in Problem 1065. B1o(:lllstllc:o-
(a) Tllcx S(:llriitlillgm aclu;~t,iollis
rem states that the solution to the S(:hrijtliilgcr c:q~l;~tiol~ then 11;~sthe: forlll v ( z ) e x p ( i K z ) , where K is a cori~t,anta 1 ~ ,~I.(z) 1 is j)eriodic wit11 t11~pcriodicity of the lattice. The contillrlit,y c:olitlitiolls of ,(, (3:) all(1 &I,( 3 : )Id3: at the well boundaries lirnit the energy of t,hc propagating particlc t.o ce:rtain ranges of values, i.e., energy 1)alltls. All cx;il~~t)le is givcrl in dctitil ill Problem 1065.
.+'
((1,'
) - ,I/)' (a- )
= 'Lb2,Jt (a,),
whercb 12 = ,IILV;~I,/IL~. This ; ~ n dt,lic, o t l ~ c r1)oundi~rycondition
A particle of mass nlovc:s in one dilncllsion in a periodic potcrltial of infinite extent. The potential is zero at most j)laces, Ijut in narrow re&. mns of width b separatcd by spaces of length a (b << a ) the potential is Vo, whcrc Vo is a large positivc potential. [One may think of the potcntial as a sun1 of Dirac delta functions:
apply to the wavc f~llictiorlat
C
(b) For equation:
3:
# ,rm, tllcre arc two fundalrielltal sol~lt,iossto tllc Sc:lirijdiager %kz
,111 (z)= C.
,
lb2
(z)= c - " ~ ~ ,
Vobd(x-r~a,).
n.=-DO
the corresponding energy being
Alternatively onc call arrive at the same answer in a somewhat more messy way by treating the intervals as finite and then going to thee li~nit.] (a) What are the appropriate boundary condit,ions t,o apply to the wave function, and why?
, -1, 0, 1, . . . ,
fm.
00
V(2) =
= rln, wllcrc n = c o ,. . .
:I:
Let
$(z) = AeikZ+ ~ e - ' ~ " ,0 < z 5 a,.
106
Prob1m.s and Solutions on Elcctromagnetzsm
According to Bloch's Theorem, in the region a
Basic Principles and One-Damensaonal Motions
5 x 5 2a
,,>()
whcre K is the Bloch wavc riurnl)cr. Tlle boundary contlit,ions give
iKo
$( ) = (e
koz e-ikox , e--ikoa) sin 7 sin k o n +
0<x.
For x E [ a , a I b], t,he wave function has the form exp ( % k l , 2 ) , where
(d) For k a = 71.x + 6 , wh(:rc b is a small positivr number, we have For norlzcro solut.io~lsof A and I3 wc rerluirc. ( ; ~ . l i a. .cika.
ikr'""
--
e%Ka- (:-ik(~
( I k + 212)
+ (ik -
261) i:-iku
=
0.
or cos k n
+ 11k sill. -
-- (:OSK ~ L ,
which determines t l ~ cBloc:11 wavc 11111111)(:1. K. Cor~scquclltly,tli(: :~llowc(I values of k are limited t,o thc: range givc.11 I)y
ko is the iriir~irnurnof k t,hat satisfy t,his inclcluality. (c) For E = E o ,
where ko = J zrn~. Norn~alization$ ( x = 0 ) = 1 givcs $ ( s ) = 2 i A s i n kox+e-"o",
0i.rFu.
The boundary conditions at x = a give ezKa -
2i A sin Icon
+ e-lkoa,
When cF is qliitc s~riall,the lcft side z 1 + 626/k > 1. Therefore i r ~ c:crtain region of k > ~ L K / ( J ,t,h(:rc , is 110 cig(:1!fllii(:t,io11. 011 thc other hand, k(l. = 7rK corresI,olids to c:igciivalues. So th(: c:nergy at which the first cncrgy g:~p begins satisfies t,hc rc:lat,ion k a = K , or E = n2h 2 / 2 n t a 2 .
We wish t o study partid(:-wavc: pro~)agationi11 a 011e-dir1iensiona1pcriodic potc:ntial co~istruct,ctlby itcrating a "single-potentia1" V ( x ) a t intcrvals of length 1. V(3:)V ~ L I I ~ S ~for C S / 3: 1 112 and is symmetric in z (i.e., V ( x ) = V ( - 2 ) ) . The scat,terirlg propcrties of V ( : L ) can be summarized as follows: If a wave is incidcnt from the Icft, $+(z) = exp ( i k z ) for x < -112, it produces a transmitted wavc ,$+(z)= e x p ( l , k x ) for x > 112 and a reflected wave + - ( x ) = e x p ( - i k x ) for x < -112. Transmitted and reflected coefficients are given by
>
108
Basic Principles and Orte-Dan~ensionalMotions
Problerns and Solutions on Electromagnetism
and 6, and 60 are the phase shifts due to t,he potential V(z). Take these results as given. Do not derive them. Now co~isideran infinite periodic potential V,(z) constructed by iterating the potential V(z) with centers separatcd by a distance 1 (Fig. 1.32). Call the points at which V , ( z ) = 0 "interpot,ential points". Wc shall at,t,ernpt to construct. waves propagating i r ~the potential V,(z) as superpositions of left- and right-moving waves ++ a r ~ d4_.
I
For the wave incident frotll the right, let the t,ransmission and reflectiorl (,c4licients be t' and r' respect,ively. It can be shown that t' = t , r' = r.*t/t*. In the periodic potcrltial, the transmission and reflection co-
~,llicient,sat adjacsr~tinterpote~ltialpoints have relations t,, = tn-l and I . , , = rn-1 exp ( i 2 k l ) . So the transnlission coefficient can be denoted by a :;ingle notation t . (a) Tllc waves at adjacent inter~mtential points are as shown in I:ig. 1.33. Ot)viously, only thc reflectiotl tern1 of @ and the translnission t.cwn of 4:-' contribute to
a;:
Similarly,
8:
= r,,4!1
r~+l + t 4I
Fig. 1 .:12
(a) Write recursion relations w1iic:ll rclatc t,he an~plitutlesof the rightand left-moving waves at the n t h illterpotential point,, @$, to the alnplitntlcs at t,he (n - 1)th and (n, 11th intcrpotclltial poillt,s, dl;-' a.nd (b) Obtain a recursion relation for 4- or 4+ alone by elirninatirig tlic other from part (a). (c) Obtain an expression for the ratio of amplitudes of 4+ to c j - at successive interpotential poirits. (d) Find the condition on X:, 6 , and 60 such that traveling waves arc allowed. (e) Use this result to explain why it is "normal" for conductioll by electrons in metals to be allowed only for bands of values of energy.
+
@it'.
(MITI
109
Fig. 1.33
(b) With
7%
replaced by n
+ 1, Eq.
Solution: For the wave incident from the left, the pot,ential being V(x), let Equations (3), (4), (5) then give
(1) givcs
110
Problems and Solutions on Electromagn~etism Basic Principles and One-Dimensional Motions
Let r o = r . Then r, = r exp(22nkl). Assume r:, rk = r1 exp ( - i 2 ~ ~ k l )Hence .
=
-r,
tit*
Then
Similarly,
(d) The necessary condition for a stfable wave to exist iri the infinite ~~oriodic field is +n++l/47~ - f:16~
(c) As the period of'the potential is 1, if 7/1(3:) is tllr w11ve fu~lctio~i in the region [:c,,,-1, x,.], then $(r - 1) exp (26) is the wave fuii(:tion i11 tlir region [x, , 3:,,+1]. Tlilis ($:+I ez(6-kl) I
,$:+I Let c,, = 4';/@.
4n+,
= f:~(6+kl)
62.
+-
is rcal and independent of n. If this were not so, when r~ -t co where one of @ ; ant1 457" would bc infinite. Fkoin (7), we sec that 61 = 6 k1. I'rom (6), wc obtain -
(7)
From (4) and (5) we obtain rcs~)c<:t,ivr?ly
4n+ + 4:
1 -=T;-
(6"l
42
+ tc,, .
Using (7), (9) can be written a.s
Substitlitiiig 7.' we obtaiii
+
---r*t/t* iri t l ~ ca.t)ovccqliation arid using rr* tt* = 1, tr:ikl + t * c - i k l
=
2tt* cos 6,
which mcans
or, using the definitioil of t,
+ kl) + (XIS (2h0 + k1) + (:OS [2(6, do)]
cos (26, 1
or, using (8),
-
+ + kl)
i.e., r.c;
+ (t2eZZki rTIr:,ei2 k l - 1) c, + rLriZki= 0.
Solving for c, we have
-
cos (6, cos (6,
-
do)
In general, only some of the values of k satisfy the above inequality, i.e., only energy values in certain regions are allowed while the others are forbidden. Thus we obtain the band structure of energy levels. ( e ) In metals, the distribution of positive ions is regular and so the conduction electrons move in a periodic potential. (d) Shows that the
112
Problems and Soli~tior~s on E l e c t r ~ m a ~ n c t i ~ ~ r ~
113
Basic Principles and One-Dimensional Motions
electron waves can only have certain Ic values, corresporlding to bands of electron energies.
You are given a real operator A satisfying the quadratic equation
This is the lowest-order equation that A ol~eys
I
II(: "chargc: corijuga.tionl' operator C a.pplied to I $,) leads t,o an eigenstate of Q c:orrcsponding to cigenvalue -4:
1 rl,-,)
+
(a) Find thc cigc~lvalucsof t,hr opcra.tor C Q Q c . (b) Call a state si~~lllltali(:or~sly 1 ) an ~ cigcnstate of C and of CJ'? ( Chicasgo)
Solution:
(a) Wha.t are the eigenvalucs of A'! (b) What are tjhe eigenstates of A'! (c) Prove that. A is an observat,le.
(a) L(:t
(B I~fl(il.lo) Solution: (a) As A sat isfirs :i quadratic equation it can t)(, rcprc~sc~itc~tl by a 2 x 2 matrix. Its eigenval~lcsare the roots of the rlu;~(lratic.c,clr~ntioliX":jX+2 = 0, X I = 1, X2 = 2. (b) A is represented by the inatrix
(CCJ -1QC)
I ,,!I,)
-=
qC
~I/I,)
.+ C,!
1 4',L,1)= (1 I ,!I-~(!)
-
(1 I JjLrl) = O .
Thus t,tic: (:ig~11~:~111(~ of t,li(bop(:ra.tor CC) + C!,C is zero. (b) As C is the: c.llargc: (:oi!j1igi~t,io11 t,ransfOrn1at,io11,C&C-- = - (2, or C Q + ($3 - - 0, i.c,., C ; L I I ( ~(2 (lo not cornmut,e (they ;~nt~ic.oinililitc) t,l~cy c.;~nnothi~v(:(:011111101i (ig(:l~sti~t,(:s. (Urlless q = 0 , in which case it is cpitc: ~iieaniilgl(~ss t o i~lt,rotluc.c, c.hnrgc: c:oiljugatioll.)
'
The eigenvalue equatiori 1068
then gives (L = 1, b = 0 for A = 1 and n = 0, b and eigenstates of A arc
(h)
(y).
(c) Since A = A+, A is Hermitian and hence
=
1 for X
=
2. Hence t h r
a11observable
If 1 II,,) is any eigenstate of the electric charge opcrator Q corresponding to eigenvalue q, that is to say,
A
cl~l;~l~ttiill-l~l(:(:lli~lli(:syst,clll al
is known t.o possess only two energy eigensti~tc:~ tl(:ilotctl I I ) i~li(112). Tlic: systc:ln ;ilso inchides three other observablcs (t)csiclos t,hc c,ilc:rgy), known as P, I) aild R. The st,ntes 11) and 12) arc ilornlalisccl t)iit, t,llry are not necessarily (igc:nstji~t(?s of P, (2 or R. Det,c:rrriinc: ~ L S111illiyoE t,hc cigcnvalues of PI I2 i ~ n dR as possit~leon the basis of t,hc followi~lgsct,s of "cxperi~nental(latan. [Waruiilg: one data sct, is unphysical.]
(a) (1 1 P 1) = 2 (1 ( P" 1) = 114. (1)) (1 1 Q 1 1) = 112, (1 1 Q2 1 1) = 116. (c) (1 1 R 1 1 = 1 (1 1 R~ 1 1) = 514, (1 /
n" I 1) = 714.
114
Basic Principles and 01l.e-Dim.ension,al Motions
Y r o b l c n ~ sand Solutions o n Electromagnetism.
Solution:
~v lwre
A is to be determined. Then
We are given three observables P, Q, R, which satisfy the Herrniticity (n1 P ( m )= ( m1 P 1 n)*,and that the mechanic:al system has a complete set of energy eigenstates 11) and 12). (a) The cornpletcness of thc two statcs and thc "experimental dtxta'!, give
.Ilowing that
R 12) = A* 11) wliere 77 is to be determined. Consider
where a is a constant to be determined. The orthogonality of the eigenstt~tes and the Hermiticity of P give
Then as So we have
P ( 2 ) = tr* ( I )
+ /3(2),
we have
where /3 is to be determineci. The11 ;md so
As P y l ) Therefore,
-
accordi~lgto expcrimtiit, cr"n
-
and
0 and hencc a = 0.
i.e., at least one of the cigenvalues of P is 112. (b) Let
-
where y is to be determined. By a similar procedure, we get y'y 116 114 < 0. So this data set is unphysical and the cigenvalue of Q collld not be determined. (c) As ( 1 ( R( 1 ) = 1 , we can write
It follows also
+ 77 12).
116
Basic Principles and One-Dzmenszonal Motions
Problems and Soli~tionson Electrornagnetzsrn
2+
Experimentally (1 1 R" 1) = $. Thus (1 Hence on the bases 11) and 12) the matrix of R is
+ ,r,) = $, giving q = 1.
1070 Using the coordinate-momentum commutation relation prove that
To find the rigrnvalucs of R, sotve where E,, is thc c:ncrgy c:orrc:sponding to tllc: cigrnstatc \TI.). Ol)t,ain the l l = p 2 / 2 ~ V (z). value of thc (:onstarit,. Tlic: I-Iarriilt,orlia~lhas the: f i ~ r ~H
+
(Be~ke1e:y)
i . . 1 A)" ofR=+ 3 -
$ = (1 - A - ?) (1 - X + A) 2 2
= 0,
and ot)t,ain tlic cigc~iv;l.l~ic:s
Solution:
As
2 ' 2'
we havr
1069 For a charged 1)artic:lc in a. inagnctic field, fintl t,llo corrimutt~tiolirlil(:s for the operators corrc~s~)ondirig t,o t,hc c:o~rlporlent,sof tlic: vc:loc:ity.
(B(:~kel(:?y)
and so
Solution: Suppose t,hc inagnctic: field :irises froln a vc:c.t,or potc:rltia.l A. Tli('11 t,ll(: velocity cornporlcnts of the: ~)art~iclc a.rc
O n thc: o t l ~ t rhand,
Hence
-
9
A
1
, A *-
[A,A,]} =
where ~ i j kis the Levi-Civita density, use havirrg been made of the correspondence rule pi -i) $
&.
In the above we have used
Basic Principles and One-Dimensional Motions
Problems and Solutrons on Electromagnetism
H 1 m) = Em 17n) , (711 j
z2 7n)
=
x x x x x
(~rn1 z 1 71,)(n1 2: I rn)
11
=
1
I(1rt. :C 7 ~ ) '
1,
(srrt,
1 :LHX1 711,) =
(rr~,1 :I:H1 71,) (,rt, 1
:I;
1 ~rt,)
ET1, (srrt, 1 :L:1 T I , ) (TI,
I:]:
+
As U++ = U, Cf = C. Therefore C,,,,, = U , ,.,, U:,, is the matrix represerltatioll of a. Hermit,iarl opcrator. (c) Tlic cigcrlkcts of n H(:rlnitian opcrator for111a c:orrlpl(:tc: nrlcl orthonorma1 set. Thus any I 11,,,,) c;m 1)c ~ x p i ~ ~ i di11 ( : dthc (:o~npl(:tcs(:t l,~,,):
11.
=
1
srr~,)
71
=
ET1 1 (7rL 1 :L: 1 Tb) 1' .
which d(:firlc:s VkTrr1
71.
Equat,ing the t,wo rc:sult,s and setting
7r1. =
0, we 01)ti~ili
Similarly,
(a) Givcil a, H(:rli1itii~11Op<:ri~t,ol. A wit,h (:ig~ll~i~111(:~ an RIKI (:ig(:llfilll(:tions 11, (:c) [TI, = 1, 2, . . . , N ;O 5 :I: 5 L ] , show tlii~tt > l i01)~'rat~or ~ (\xp(,iA) is unit,ary. (1)) Conversely, give11 the matrix U,,,,, of a iinitary operator, c:ollst,r~lc:t the matrix of ;L Hermitian operator in t,crlrls of U.,,,,, . (c) Given ;I scconcl Hcrmitian opcrat,or B with cigcnvalues b,, and c:igcrlfunctions 11,, (3:), (:onstrli(:t a rcprc:sc:rltatiori o f thc 1irliti3,ryoperator V tallat transforms the eigcr1vcc:tors of B into t,hose of A. ( C/~,ica.go) Hence
Solution: (a) As Af = A, A being Hermit,ian, {exp (iA)If = exp (-iAf ) = CXI) (-iA) = {exp ( i A ) I 1 Hence exp(iA) is unitary. (b) Let
Cmn = Umn
+ U,TTn= Umn + ( U + ) m n ,
120
Problems and Solutions o n Electromagnetism
Baszc Prznczples and Or1,e-Dirnensio~l,nlMotions
showing that V is unitary. Thus V is a unitary opcrator transforlrlirlg the cigenvectors of B into those of A
'"PO)
-
dt
1 [(p) cos wt,HI - ( ~ ) sin a wt iJL 1 +[~nw(.:l.) sin wt,H] + rrw"(:r.) cos wt l,h
1072 =
Corlsider a or~e-dirnensior~al osc:illat,or with thc Halniltonian H
1ifi
- [ p , II] c:os wt - w(p)sill wt
+
+ ,rrw2::l:"/2.
= p7/27n
(a) Find thc tinw d~prnd(t11ceof the cxpec:tatioii values of thc "init,i;~l position" and "init,i;~linomenturn" o1)t:r;Ltors
(b) Do these operators c:o~iiln~~t,c with tl~c:F I i ~ ~ r i i l t , o ~ ~ i ; ~ ~ ~ ' ! (c) Do you fiild your rc:s~llt,sfor (;I) arid (1,) t,o I)(, c:ou~patil)lc'!Discy~ss. (d) What are tllc 111otioli~ ( l ~ i : ~ t iof o i th(: ~ s op(:r;~torsill th(: II~is(\~~l)(,rp, picture? (e) Coinp~itcthe (:ommutator [ p o l :c:o]. W l ~ a is t its sigiiific:;~~~c.o for III(:;Lsureinelit theory? ( I'rincr: t on.) Solution:
7TLW .7 JL
[ x , HI sin
wt + rrwyx) c:os wt = 0
Thus t,lic cxpcct;~t,ionv;~luc:s of t,hc:sc: operators ;xc: ii1d~:p(:11(1(:11t of time. (1)) Consider
Thlls t,l~c:op(:r;~t,ors:KO,po (lo not c:oll~rlilltrwith H . ( c ) Tl~c,rctsult,~of (;I.) a i ~ d(1)) ;i.rc: still c:o~npatible. For whilc thc cxp r e s s i o ~ ~for s a:() ; I I I ( ~pg ( : o r ~ t i ~ti(xl)li(:itly, i~ their non-commutation with H does i ~ o tc:xrlutl(: t>hc:ir l)c?irlg c:oi~sc:rvc:cl. I n fact
(a) Making ~ s ofc thc rclatiorl
df
1
dt
a"f
ih, If. H I + % ,
- -- -
we have
"xO)
--
dt
-
1 ih.
- [(:L.) cos wt,1 11 - W(X) sill wt -
[-
L (") it?
mw
sin wt,H
] ?!
- -
cos wt
showing that t1ic:y ;m: ac:t,llnlly c:o~~servetl. ((1) 111 t,he Hciscn1)c:rg pic:ttirc:, thc rnotion cqll;it,ion of a n opcrat,or is
1 i)A dA/dt= - [A, H] + lfr
1-
=-
.iti
Bt
[x,H] cos wt - w (z) sin wt Thus the rnotion ccl~iationso f
1 1 1 [p, H ] sin wt - - (p)cos wt = 0 ,
mw ih
731.
TO
and pg are respectively
122
Ploblems and S o l r ~ t i o n so n Electromagnetism
2. CENTRAL POTENTIALS
(e) Using t,he exprcssions for zo ant1 po, we have [po , :Q]
+ m w z sin w t , zo] = [ p , :I:o]cos w t + [:c, . L ~ ] sill w t =
[ p c.os w t
2001
7 1 1 ~
+
[
:r, :I: c:os w t
= [ p , I:] =
(:osaw t
- [ 3 : , p]
=
-
P
- -
711w
sin wt
I
An e1cc:tron is confined in a three-di~ncllsio~lal infinite potential well. 'l'he sidcs p;irallc:l to t,hc z-, y-, ;~rldz-axc:s arc of lcilgth L each. (a) Write the ;ippropriatc Schriitlingor cqliation. (b) Writc the tirne-iildcpent1t:rlt wave frlnctioil corresponding to t,he stat,e o f the lowest possil)lc cnc:rgy. (c) Givc ail exprcssiori for the rllirnl)cr of states, N , having crlcrgy less I.han somc: givc:ri E. Asslnrlc N >> 1. ( Wzscon,s,in)
711w sin w t .
[ x , p] sina w t
--,if).,
as [x, = [ I ) , = 0, [:L., p] = ill,. In gcncr;~l,if two obsc:rv;il)lc:s A arid
Solution:
B satisfy tlic: c:clriiition
(a) The S(.hriiclingclr cqriatiori is
[A, B]= if,., then their root,-meail-sclliare deviiltiolis AA,AB,whc:~~ t1ic:y simultaneously, nlust satisfy t,hc: ~iric:c:rt,;~irlt,y priiic.ipl(:
zhi)d~(r,t)/i)t=-(h,"l'L~rr)V".J/(r,t),0 5 2 , y , z < L , ,d) = 0, otherwise.
riic:;~slirc.tl
(b) By sc:parat,iori of vxri;iblos, wc: (:a11take t,hat tlic wavc fiir1c:tion to hc: prodoc:t of t,hrcv wave: frinc.t,iorisoac:h of a orle-dimensional infinite wcll ootential. Thc: w;ivc: fiin(:t,io~i of t h lowest ~ energy level is I lie
111 the prcsc~ltcitsc:, thc: sirrlull,arlc:orls rllc;Lslirc:irlc:rlt,s of posit,ion ;~rltl inomc~rit~im in t,hr s;unc tlircc:t,iorl inrist rc:slilt in
where
J/
-
=
2
. sin
(:
z)
, etx.
Thus The relation shows $,,,(z,
It is a relation between possi1)lc. upper limits to the precision of the two quantities when we measure thcrn simlultanc-ously.
y, z ) =
(E)
312
sin
(9)(7)(7) sin
sin
The corresponding energy is Elll = 3 f i ' 1 ~ ~ / 2 ? n , L ~ . (c) For a set of quantum numbcrs n,, n,, n, for t,he three dimensions,
Hence the number N of states whose energy is less than or equal to E is (qua1 to the number of sets of three positive integers n,, n,, n, satisfying t,he inequality
Problems and S o l ~ ~ t i o non s E1ectromc~gneti.s.m
Central Potentials I (mperaturc.
125
Give a numerical estirnate for the longest wavelength of elecstrongly by t,liese c1ec:trons. (MITI
I I-omagncticratfiation absorhetl
Consider a Cartesian coordinate system of axes I),, n,, n,,. Thc number N requiretf is numerically equal t,o the volurnc in the first quadrant of a sphere of radius (2m,L2E/fi27r2)'l",yrovitled N > 1. Thus
Sol~t~ion: The txic?rgy lcvc~lsof all elcct,ron in a. cubic:al 1)ox of sidcs a are given by
A,
A 'qn2~rk' (mass -- n l p / 3 ) is confinetl in a (:11l)ic:t~1 box wit.11 si(1cs of lengt,h 2 ferinis = 2 x ni. Fintl the c:xcitatiori c:ric:rgy frorn thc: grourltl
wherc 71, , r r ~and k ;\re positive int,egcrs. Taking a -- 1 t,he grorlrltl state tmergy is El = 31r27r"22ma% 112 cV. Ei)r a c:rystal at room tc?rripcr:rt8~ire, (.he elcctl.oiis arc ;~lnlostall in tlic g-ror~~itl statc. The lorigcst wavc:lollgt;h c:orrespontls to L: tri~~lsitiorl fro111 thc: grourid stsate to the nt:arc:st c:xc:it,(:ci state:
state tlo the first cxc:itcd st,at,cin MeV.
( Wist:on.sin)
for wliic~li
Solution:
r. lt,(, X=-=--ll0ii.
v
AE
The energy lcvcls irl the, cnl)ic.al box arc, 8ivc.n 1)y
~ h u the s energy of thc grouii(i statc is E~~~ = 3/!.2~2/2,rrb(~,2, t,lli~tof the first cxcitetf st,a.te is E2" = ~h.%~/2rnc~" 31tlb'L.rr2/nr.n2. Hc:rl(:(: t,h(: excitation energy from the ground stat,(: to t,iio first c:xc:it,otl statc: is
An c:l(:c:Lrori is c,oiifi~~c:tl to t,hc interior of a hollow spherical cavity of walls. Find an expression for the pressure radius R with iilij)c~llc~t,r~~l)l( exert,etl or1 tho walls o f tlic: c:;ivit,y by tlie electron in its ground state. ( M I T1 Solution:
-
1 . 5 ~ 7 6 . 5 8x
10,")~ x
For tho gro~iritlstat,(:, 1 = 0, arlcl if we set tlie radial wave function as R ( r ) = - y ( r , ) / r , t,hcn ~ ( ris) given by ((3
(F)x ( 2 x 10-'5)2
x 10')"
= 461
MeV.
A NaCl crystal has some negative ion vacancies, each containing orle electron. Treat these electrons as moving freely inside a volume whose dimensions are on the order of the lat,tice constant. The crystal is a t room
wherc p is the electron rest mass. R ( r ) is finite at r = 0, so tha.t ~ ( 0=) 0. The solutioris satisfying this condit,ion are
126
Problems and S o l ~ ~ t z o non s Electromagnetzsm
Central Potenlzals
for which
\ ( a ) = 0 requires the solution to havc thc form x(r) = A sin [K(7. - u,)] .
The average force F acting radially on the walls by the electron is given by
Then from ~ ( b= ) 0, we get the possiblc v;~luc:s of Ir':
As the electron is in thr grouiicl statc,
=
1 and
For thc particle in thc ground state, i.e.,
=
1, we obt,ain the erlcrgy
The prcssure rxcrtrd on the walls is
d m . Hr~iccfor t h r gro~lildstate, tllc nornlalizcd radial
A particle of mass m is c:oi~st~ri~incd to move I)c:t,wcc~lltwo c~orlc,c:nt,ric: impermeable sphcrcs of radii r = a and r b. Thrrc; is iio other 1)otc:lltial. Find the ground statc ciicrgy ailtl riorinalizcd wave fuiict,iorl.
-
(MIT) Solution: Let the radial wave fililction of t,hc pi~rti(:lcbc R(r) ~ ( r satisfirs ) thr cquatiori
1
b-(L
-
,
Slll
T
~(7.)/7..Tllc11
7r(,r - a ) , b-a
and the norinalizcd wave fi~nc.tiorlis 1
=
For the ground state, 1 = 0, so that only the radial wave function is h ~reduce , the cquation non-trivial. Since V(r) = 0, letting K 2 = 2 n ~ ~ / wc to + K" = 0 , with
we get A = wave fu1ic:tiori is
-
. 7 r ( ~- a) sin -.
b-a
(a) For a simple harrnonic oscillator with H = ( p 2 / m-t k x 7 / 2 , show t,hat the cilcrgy of tlie ground state has the lowest value compatible with tjhe uncertainty principle. (b) The wave function of tlie state where the uncertainty principle miniinum is realized is a Gaussian function exp(-ax2). Making use of this fact, 1)ut without solving any differential equation, find the value of a. (c) Making use of raising or lowering operators, hut without solving any differential equation, write down the (non-normalized) wave function of the first excited state of the harmonic oscillator.
128
Problen~sand Solutions on Electron~agnetzsm
Central Potentials
(d) For a three-dinlensional oscillator, write down, in polar coortlinates, the wave functions of the degenerate first excited statre which is an c:igenstate of I,.
129
= 0 we see that when a = &/ah = mw/2t, the energy is From I liinimum. Therefore tr = 1nw/2h. (c) In t h e Fock rcprr:sc!~it,ationof harrnonic ost:illation wc define
(Berkeley)
fi
Solution:
=
iit =
(a) T h e ground st,at,e of the harnloriic oscillator has even parit,y, so that Then [ci, G,+]
=
i(
-i
imw?;)/J2mfrw,
-
(e+ Z T ~ I , WJ'GZZ. ~)/
1,
H
= (ti+&+
1/2)Iw.
Denot,ir~gthc: gro~irlclst.i~.t,<~ wave fi111c:tion 1)y 1 0). As H 1 0) I,he last c:cl~latiougivcs ci,+ci 10) = 0. I t also givt:s
a n d so
-
i h w 1 O),
T h e uncertainty pririciplc rctluircs -
I!? > -
-
Ap2 . Ax2
4
It follows that
'1,
--
0
6
as = w Thus tlie cncrgy of the ground st,i~tchas the lowcst v:~lac compatible with the uncertainty principle. (11) Using the given wave function we calcula,tc
v
t
)
a
(
i)
-iI~,+imwz
~
&I:
(-% 2)
For the grountl state, (n,I, 112, 113) = (0, 0, 0). For t h e first excited stat,es, (711, n2, ,113) = (1, 0, 0); (0, 1, 0); (0, 0, 1).
,bolo( r ) = N: N12uV exp and hence 1 -. 4a
exp
in the c~oorclinatcrcprescntation. (d) For i~ 3-tli1ne1lsiollal osrillator, the wave function is
(r) = N i N12crx exp
li2 k E=-a+-. 2m 2
)
~ J O (Or )~
= N; N12az exp
(-i #) , (-i , (-i a%z] a2r2)
130
Central Potentials
P~oblemsand Solt~tionson. Electrornc~!gnetism
Expanding z, ?I,z in spherical harmonics and recornbini~lgthe wave fiinctions, we get the eigeristates of 1,
where N7z1= Note t,lii~thcrc tu = from t,hxt given in (1)).
Jq, which is tlic usual
Fig. 2.1
definition, diff(?rc:r~t
2007 Thc iliagra~n(Fig. 2.1) sliows tall(:six lowest. onorgy 1i:vcls and tlic :I.SSOciat,ed ang~llar~lioineritnfor :L spirilcss pi~rticlcmoving ill a c:crt:~irithree?dimensional c:cntral pot,i,~iti;ll. T1ic:rc arc: iio "ni:c:idi:litfnl" d(.gc?ilc:rai:ics ill this energy spcctr~im.Give tlie il~iinl)c~r o f iloilcs (i:linrigc:s in sign) i r ~i.hc radial wave furictioii assoi:iat,i:tl witli i::~c:hl(:vcl.
(MITI Solution: The ra.dial wavc fimc:tioil of a. pnrtii:l(: ill a thrcc-tlimrnsioi~i~l c:cntral poterltial c:a,n hc written as R(r) = ,Y(T)/T.Witli a givc:n a.nglrlar clu;~ntrlrri number I, t,he equation satisfic:d 1)y X(T) h i ~ stali(:for111of a. orie-dinl(:risiollal Schrodingcr eqriation. Bcncc, if all cncrgy spci:tr~iinhas no "accidt?nt;~l" degericr:~cics,the rolc of the nodes ill thc radial wavo filnctioll of thc partii:lc is t,he same as that in t,he one-dimensional wavc function. For 1)olindstates, Sturm's theorc:m rcinains applic:at)lc, i.e., ,Y(T) obeys Sturin's t,hcorerri: the radial wave func:tion of the ground state has no node, while that of the 7rt,h excited state has n nodes. Thus, for a. 1)ouncl statc: of energy E n , wliicli has quantum number n = n, r! 1, the radial wave has 7r,. nodes. For angular quant,~imriumbcr I = 0, the numbers of nodes for thc three energy levels (ordered from low to high energy) arc 0, 1 and 2. Similarly, for 1 = 1, the numt)ers of nodes arc 0 and 1; for r! = 2, the number of nodes is 0. Thus, the numbers of nodes in the energy levels shown in Fig. 2.1 are 0, 1, 0, 0, 1, 2, from low to high energy.
+ +
A part,ii:le of Inass Tn, arid c:liargct q is bo~indto t,hc origin by a. s~)lii:ric:ally symmetric: linear rc:storing for(:(:. Tlic c:ilt;rgy levcls arc i~clna~lly spaced at iiltervals lrwo al)ove thc ground statc: energy Eo = 31iwo/2. The states can be dcscri1,c:d altcriiat,ivc:ly ill a. Cnrtcsian basis (three ollc-dirncnsio~ial Iiarmonic: osc:ill;~tors)or in i~ sp1icric:al basis (cent,r;tl fii:l(l, separatetl int,o i~ngularand radial motions). (a) I11 tlic Cartesian basis, ta1)le the oi:c:upation nurn1)crs of tllc various statc:s of thn oscillat,ors for the gror111d and first three e x i : i t ~ llevels. I>eterrninc t,llc tot,al di:gcncracy of each of these levels.
(b) In tlic spherii:al basis, write down (do not solve) the radial equation of motion. , where L' is (Note that in sp1icric;ll rooriliriates V2 $ $ ( r Z $) the oper;rt,or of tot,al orl)it,al arlg~ilarmoment~imscluared in units of ti'.) Identify the cRective pot,ential and skctcll it. For a given angular momentum, sketch the "ground state" radial wavc function (for a given 1 value) and also the radial wave functions for the next two states of the same 1. (c) For the four levels of part (a), write down the angular momentum content and the parity of the states in each level. Compare the total degeneracies with the answers in (a).
-
-
5
(d) Does tlie secorld excited stat,e ( E 2 = 7 h o / 2 ) have a linear Stark effect? Why or why not'! Compare similarities and differences between this oscillator level and the second excited level (n = 3 ) of the nonrelativistic hydrogen atom. (Berkeley)
132
Problems and Solutions on Electronragnt!tism
Crntral Potentials
133
Solution: (a)
Energy level
0ccup;~tionNumbers
Degeneracy
rl';ll)i(!
2.2
-
(b) Let +(r) = R(r)KT,,(0, cp). T h e radial wavc function E ( r ) satisfies thc cclu, 1' t,'IC)II
I
nr.
Eo E, E2
0
0 0,*1 0,*1,*2 0 0 , 1 1 , *2, *3 O]*l
E:{
so that the effective potential is
1 2 0 3 1
-N o t ~ :I' = parity, I1
which is sketched ill Fig. 2.2, where ro = [h"(l 4 l)/7n,"w"]114. The shapes of the radial wavc functions of the three lowest states for a given 1 are sllowri in Fig. 2.3.
P
E
=
+ -
-t -
D 1 3 G 10
degeneracy.
(d) Thc: sc:corld cxc:itcd state does not have a linear Stark effect because is a n opcrator of odd parity whilc all the degenerate states for Ez have w e n parit,y, wit,li the rcsult that thc matrix elements of H' in the subspace of the energy level E2 a.rc all zero. On t,he other hancl, for the second excited level of the hydrogen atom, = 3, its degenerate states have both even and odd parities, so that linear Stark effect exists. :I:
2009
(a) A nonrelativistic particle of mass Fig. 2.2
7?2
lnoves in the potential
134
Problems and Solut,i.ons o n Electro~magnetism
Central Potentials
where A > 0, B > 0,I X I < 1, kt, is arbitrary. Find the cnc:rgy eigcnvalues. (b) Now consider the followirlg modified problcm with a new potential qlcw:for z > - p and ariy n: and ?I, V,,,, = V , where V is the same as in = + m . Firicl the grourld part (a) a.bove; for z < p i ~ n dany :r: and :y, stfate energy.
135
Then Schrodirlger equation becomes
v,,,
( CUS
Let 4(p, t, z) = U ( p ) T ( t ) Z ( z ) . The above equat,ion can be separated Illto
Solution: (a) Wc c.hoosc two rlcw v;~riablvs/L, t dcfined by
with
a,nd writc tlir: pot,erltial as
El + E 2 + E3 = E.
By scttir~gz' = z + /L, E; = E3+ Bp2, all tlic: a1)ovr: t,llrr:e ecluatiorls car1 I)(: reduccxl to t,hat for a hi~rrnorlic:osc:illator. Tll~istllc enorgy cigenvallies
and the diffcre~lt~inls as
a
az
-
a
1
i)
1
ap
fi
at
&
'
a" (7bl, 7L2,
7123 =
0, 1, 2, 3, . . . )
(b) With a new potential V,,, such that for z < -p, q,,, = m and for is the same as tha.t in (a), the yave function must va.nish for : + -p. The 2-equation has solution
.:
> -p, V,,,,
Z
<
~ ~ ~ ~ ( < ) e - ~ ~ ~ ~ ,
where = (2mB/h2)'14(z + p ) , H,,, (<) is thc n3th Herrnite polynomial and has the parity of 713. Hcnce 713 rnust be an odd integer. The ground state is the state for rL1 = r~~= 0 and n3 = 1, with the corresponding energy
136
Central Potentials
Problems and Soli~tionso n Electro~naynetism
A particle of mass
7n
137
Assume that the eigenstates of a hydrogen at,onl iso1ntc:d in space are ;111known arid dnsignated ns usiia.1 1)y
moves in the logarithmic pot,er~tial
Show that: (a) All cigcnst,ates have the same meari squarc:d velocity. Find t,liis 1nc:;l.ri squared vclocit,y. (b) The spacing between any two levels is indcpcndcnt of thc: mass 7 r r .
(CUS) Solution: (;I) We have
Si~pposctllc: nucleils of a l~ydrogcriatoni is located a t a distancc $ from an iufinite potcnt,ial waH which, of c:ourse, tc:nds to distort the hydrogcn atom. (a) Find the exp1ic:it form of t,hc groulld state wave fur1c:tioll of this l~ydrogen;~tolrlas d ;~l)l)roi~(:h(:s Xcro. (b) Find all ot,llcr c:igc:nstiit,c:s of this l?ydrogcn atorn ill half-spa(:(:, i.c. t l t 0, in tc:rrns of the It,,,( and K ,,. (B,uflalo)
Solution: and, for a stationary statc, the virial law givc.s
( T )=
1 -(I-.
2
VV)
Hence
(a) Choosa a, c:oortlina,tc: syst,c:m with origin a t t,hc cclltcr of t,he rlllc:lci~s :I rld z-axis pc:rpc:ndi(:ul;~r. i,o t,lic will1 surfs(:(: i ~ ~110~11 s ill Fig. 2.4. AS (J! t 0, I Iic solutiolis of tlic: S(:liriicli~igcr(:quation arc: strill R,,(K,, in t,hc half-space > 0 i.e., 0 < 8 < 7r/2, I ) i ~ t111llst sat,isf:y tall(:c:olitlitioll $ = 0 21.t 0 = ~ / 2 where V = cm. T11a.t is, only solut,iolis satisfying I + ,m, =odd i1itc:gc:r arc :~.c:ceptablc.As I rrt. I 5 I , tlic: first silit,at)lc spherical harmonic is
YL0=
Jjlq, cos 8.
which is true for any cigenstate. (b) Since
C 27n ' a E , , / a m is independent of n. It follows that - --
C + - =C0 , am 2m 2m i.e., En - En-1 is independent of the mass 7n,. (
E
n
-
- --
1 V=m
Fig. 2.4
Problems and solution^
138
071
Electrornagnetisrr~
> +
Since n 1 1, the ground state wavc function is R21Y10. (b) All the other eiger~stateshave wave functiorls RnrK,, where 1 + 7r1, = odd integcr. For a. given 1, we have m,= 1 - 1, 1 3, . . . , -1 + 1 and hence a degeneracy 1.
Central Potentzals
;is El
= m e 4 / 2 h 2 , E2 = ;El for the hydrogen atom. (b) Since
-
we have 2012 At the time t = 0 the wave func:tion for hytlrogeli atorri is =
where the subscri1)ts a.rc va1uc:s of the? q~iant,lllilnuin1)ers n,,I, 7n,. Igilorc spin and radiative tr;rilsitioi~s. (a) What is the cxpccta.tioii V ~ L ~ I Ifor C t,he energy of this syst,c:n1'! (b) What is the probability of fintlillg the systcln with 1 = 1, 7n = +1 as a function of time'! (c) What is thc probability of findirig the e1c:ctrou withiii 1 O p I 0 (:in of the proton (a.t tilnc t = 0):) (A gootl :~p~)roxirilate rc:sult is ac:cc:ptal)lc: her(:.) (d) How does this wave f'ii11c:tioll c:volvc: ill tiinc; i.c., w1i:~t is $(r, t ) ? (e) Suppose a measurement is nla.tln w1iic:h shows t,hat L = 1 a i d L, = + l . Dcscribe t,hc wave fi1nc:tion ii1ilrlc:tli:~tc:ly aftm sllc:h a ~nc:asurc:lllc:~itill terms of the ~isodal)ove. (Bcrkclcy)
,,,
6
exp
(-am)
,
11singthe given wave function for t = 0. Hcnce thc 1)rol)a.bility required is
Thus if 71 = 2, P = 115; ot,hc:rwise P = 0. (c) Let cw = 10- l o tin. WP have.
Itlaking usc of the given wave function as in (a). Here for the hydrogen [.om
;I
Solution: (a) Making 11sc of the orthonorinality of thc wavc filnctiolis, thc: c:x~)c:c'tation valuc for the ellerg-y is
E
.111da = 5.29 x 10-"m.
1
=
(4 I H I I,) = % (2y"ioo + $210 + f i ~ z J j 2 1 1+ h421-1 1 2E1~J)100 Then
As r 5 a << a , we can make the approximation
Problems and Solutzons on Electro~~~agnetism
140
(d) T h e wave function a t time t is
~ I I J= , ~
9.11 x
1.05 x
-
where wl = E l l h , w2 = E2/h. (e) Sirice n 2 L 1, witli L required state vector has the for111
+
1 we have r~ = 2. Conscq1lc:iitly t,he
g, ,mp = 1.67 x erg sec.]
g, c
=
4.80 x 10-lo e.s.u., FL =
(a) Compute the grourlcl state energy and Bohr radius of the positroni11n1. (b) What is the t1cgerier;~c:yof the posit,rorliurn grollnd state clue t o ~~lcctrorl spirl'? Writr down the possible spill wavr functiorls which have 1 Icfinite values of t,lle tot,al spin togctllor witli t,lie corresponding cigenvalllcs. (c) Tlic grolllld state of posit,rolliurn can decay 1)y arlnihilation into ~)liotons. Calculate the enrrgy and arlglilar rrio~nentlimre1e;~srdin this 1)rocess a.rid prove: tallat there must bc a t laast two photons in th(: firial .,;I,;ite.
(Bl~#(i,lo) Using L, L+K,, =
(L+ + L-)/2, wit11 J ( I rn, ] ) ( I - *rn) =
+ +
L - K ,,,
+
+
J(1 I ) (L TI,) ~ , , ~ , - - l , we (:an write L,I) = 1 ) as =
-
Solution: I I:.,
(a) The rotllicctl mass ,rn of the positrorlilirrl is given 1)y ,rn. = 7rb,:/2. 1t.s 71s~:i11 th(: f o r r n ~ l l igivm ~.~
& =&+
and obtain C+ Hence
=
C- =
co -
Jz
1
1 ) = -2 c o (JZ 1211) + 2 1210) + JZ 121 - 1)) Normalizatiorl
(1) gives Co =
' JZ'
=
C,2 -(2+4+2) 4
-
(b) Thc d(:gcvicrac:y of thc ~)ositroniumground state is 4. Derlote the ~ ~ o s i t r oby n 1 ailtl thc? rlcc:tror~by 2, and let the spin eigenstates in the z lir.c:ction of a sirlglc p;~rticle1)e ru i~ritlP, corresponding to eigenvalues h/2 11lt1 -h/2 rc~spcc:tivcly.The11 the ground cigenstates with definite total spin ,, , : s l + s2 ant1 z-cornpollent, of tlic total spin S, = sl, + s2, are
1
1
Therefore
2013 T h e ground state energy and Bohr radius for the hydrogel1 atorn are Eo = -e2/2ao, a0 = h2/me2, where m is the reduced mass of the syst,em.
(c) The energy released in the anrlihilation process nlostly comes from rest inasses of the electron and positron, A E = 277r,c2 = 1.02 MeV. 'I'II(:released a.ngular momentum depends on the state of the positrollium I11,.
143
Central Potentials
Problems an.d Solutions o n Electro7rzagnetism
142
before the annihilat,ion. For the ground state S = 0, no angular momcntllm is released. For the stat(%S = h, the angular nloment,um releasccl is AJ= fi, = a h , . There must be at least t,wo photons in thc firial state of ;Ln annihilation, for if t,here wc:rc only one photon producccl in the arlriihilation of a positroniurn, the erir!rgy and rnoincntuln of tlie systcin could riot both be c:orlscrvctl. This car1 t)c shown by a sirnple argnincnt. If thc single phot,on has cncrgy E , it rrlust havc mornentunl E / c a t the sarne tirric. Thus the rnorncrit,urn of t,hc pliot,oii (:annot be zero in ally rcfercncc fra.rne. But in tlie posit,roni\rrri's rest frarnc thc monlerlturrl is zc:ro throughout t,llc ari~iihilation. 1Henc:c wc havc tJo t:oiic:liide that th:ro iil~ist be a t least two phot,ons in the firi;~ls t ; ~ tof, ~a n ariniliilat,ion wllosc inorilcrit,a cancel out in the rctst. fril~neof t,lic positronium.
The potent,ial is spherically symmetric, so we can take
J?. For a n estimate o f the energy
fi = 0, i.e.
Ap x
we shall also ttake
Then
For thc grouilcl st,at,c encrgy E, wc 1i;~vc: 2014
Consider a n r:lcc:tron inoviiig in k r , where k > 0.
;L
splic.ric~;illysyrnirict,ric pot,c:rlt,i;~lV
=
(a) Use the unccrta.int,y 1)rinciplc to cstirri;~t,c:t,lic: groiintl st,;~tcc:iic.rgy. (b) Use t h t Bolir Somrnctrfcltl clu;~ntiz;it,ionrill(: t,o c:al(:ulat,c:t h t gro~incl st,ate energy. (c) Do the sa.nlc: usirig t,hc variat,ioii;~l1)riiic:iplc ;r.rltl a trial w;wc fi1nc:tiori of your owri clioicc. (d) Solve for tlic energy c:igc?rivahie arid c:igcnf\irlction cxac:tly for t,hc: ground st,atc. (Hint: Use Fo11ric:r trarisforrrls.) ( e ) Write down t,lle effective pot,cnti;~lfor nor~zcroangular rnomcrlturn states.
3
kjfi2 113
'=1(4m)
.
(b) Thc Bolir Sorrirric:rfcltl quantization rule gives
Choose polar coordinates such t h a t the particle is moving i r ~the plane = 1, and the orbit is circular wit,li radius a . The second integral gives
O = 7 r / 2 . T h e ground state is given by n,,. = 0, 71.4
Solution: (a) T h e uncertainty principle st;~tc:sthat A P A ~2
fl
2,
where
nP = [(P
-
p)yll2
=
2 112 42-P ),
[(p2
-
zpfi4-F ) ] l 1 2
'['he central force is
144
Problems and Solu,tzor~,so n E~ect~orna,qnetis~r~
Combining we have a = (hI2/7nk)'/< and hcrlce
II
11 [
hence
1/3
(d) The, Srl~rridirlgcrc,cluatioll for the radial motion can be written as (c) T h e notion in the grolmci state does riot dcpclnd on 0 ant1 (j Take a trial wave functior~$J = cxp (-Xr) it~ldc,villuatc \vllc:re x = ,I.R, I2 1x:iug the r~itlialwavc: function. For the ground state, the 11gu1arwa.vc: fiinc:t,ion is coilstant. By the transformation
.I
where I llc.
Schriidiiigcr c:clllnt,ioii \)c:c~)illc~s tall(:Airy c:clu;l.tion
whose soliit,ions iL1.C: A%(-:I:) :111(1 Ai(:/:),wh(%r(: :C :=: -):1/1, for ;y < 0 a11(1?/ > 0 ~ri~spect,iv(~Iy. Tlw l)o~ili(li~ry (:o~idit,iollsthat R(,I.) i~n(1R1(,r) l)c c:o1iti111io11~ I' . ;\(, r = , i.(%. 7, :-O , s;~tisfi(:(l; ~ l i t o ~ ~ ~ i ~ t ,ais( :Ai(3;) ~~lly= A~,(-x), .2i1(z)= A,;'( - - : I : ) for :c - b 0 . Thc: c:ollcIitiorl t h a t R ( r ) is fir~it~c: a t r --t 0 I-c:quircs 1,lii~t.A i ( - s ) == ~I{(,I.)-+ O as 7. --t O. T h c first zero of Ai(-z) I )(:curs i~t. :I. : I : ~= '2.35. H(:lic:c t,llc: groirncl stat,e energy is
-
;1nd thc grourltl stiLt(?c:igc:rifiiric:tioll is
For stable motion,
H
is n minimum. Then taking
(e) Tlic: c:fl(:c:livc: potc:~lt,ialfor riorlzcro ;~rlgularmomentum is
we find T h e interactions of llcavy quarks are often a.pproxinl;~tctdby a spinindepcndcnt nourclativistic pot,ential which is a linear filnction of t,he radial
146
Problems nn,d Solutions o n Electronrugnetism
variable r , where r is the separation of the quarks: V(r) = A+ BT. Thus the famous "charmoniunl" particles, the ant1 $J',with rest energies 3.1 GcV and 3.7 GcV (1 GeV- lo9 cV), arc t)elicved t,o be t8hcn = O and 71 = 1 bound states of zero orbital angular mo~nerlturnof a "charm" quark of mass nr, = 1.5 GcV/c" (i.e. E = 1.5 GcV) ant1 ill1 anti-quark of the same nlass irl t,he above linear potentiill. Sirnili~rly,the rc:ccrltly cliscoveretl upsilo11 particles, the Y ; ~ n dY', are bclievcd to 1x1 the rt. = 0 and n, = 1 zcro orbital angular momc:ntuln bound statrs of a. "t)ottomn quark i ~ l ~anti-quark d pair in t,he same pot,c.~ltial.The rest rrlass of I)ot,torn quark is rnb = 4.5 GeV/c2. The rest eliergy of Y is 9.5 GcV. (a) Usirlg dilnc:nsiollal analysis, &:rive a rc:lat,ioll t)ctwocll the c:ncrgy splitting of the 4) ;r,nd ,lbl ant1 that of t,hc: Y a,nd Y', i~n(1t1l~ri:l)yOV:I.~II:L~(: the rest energy of tbc: Y'. (Exprc:ss all c:llc:rgic:s ill units of GcV) (h) Call t,he r), = 2, zero orl~itala11g111i~r 111orrler1t,1l1ll c l i i ~ r ~ n o l l i~);xrticlc ~~ln the . Usc t,hc. WKB a1)proxilrl;ttioll 1.0 t!st,illlat,etllc crlcrgy splittillg of t,lie $' ant1 thc $"I ill tc:rrns of thc cnorgy split,t,ir~g of t,hc: *(/, :LII(I thr (111,;~rld thereby give a n11lncric:i~lestilnilt,(: of thc: rc:st arlcrgy of thr: I/J". (PI.inr:cton)
(a) S u p p o s ~thc energy of a bound state depends or1 the principal quann, which is a dirncnsionlcss quantity, the collstarlt B in V ( r ) , I \ I ( - quark redllced mass / L , and Ti, rlamcly
I 11111number
$J"
. I I I ( I hence
E
=
f (vr) (EJh)"l" ( p - ' / " ,
Solution: In the centcr-of-I~I~SS systc:111of of relat,ive mot,iorl is
;L
(11li~rki~lldits arltiqtlark, tlu: c q ~ ~ ; ~ . t i o r ~
(Bfl)2/"
A E . , -- E,,y - Ed, = f (1) 1/3
-
(~h)2/3
f (0) 7 PC
ILc
where El3 is the rclativc. ~llotionc:liorgy, nr,, is thc rnass of t,he quark. Wk1(:11 t,hc angular mornentllm is zc:ro, tho ;~l-)ovc: c:cl~lil.tiollill spherical coortli~latc:~ can he simplified t,o
1 similarly
I I I ~
Let R ( r ) = xo (r)/r. Then yo (r) satisfies
E T ~ Er -
%
0.42 G e V ,
Erj = Ey
+ 0.42 = 9.5 + 0.42
ZZ
9.9 GcV.
Central Potrntials
P~vhlernsand SolutLons o n Electrumagnetism
148
(b) Applying the W K B approxilna.tion to the equa.tion for ;yo wo obtain the Bohr-Sommerfeld qua.ntization rule 21"
J2 p ( E ~ A -~
r ) d =r (71 t 314) h with
(a) Show t,li:tt for 1 = 0 the radial Schriidinger e q ~ l a t i o for ~ l this system can be reduced t o Bessel's differcntia.1 ecluation
d2 .Ip( x ) -+ 1 (it/,, (x)
n = EIZ - A
B
dl:"
'
:I;
t[ I :
+ (1
by ln(;alls of t,ll(: challg(l of v;~rial)lc3:
which gives, writing E,, for E R ,
of
Thus -
E,y
=
0.81 x ( E d , ,-
= 0.81
x (3.7
--
J.l)
z 0.49 GeV , and
2016 Two particles, each of Inass A f , are attractccl t o ~ a c other h by x potc~itial
V(7.)= - ( g 2 / d ) where d = fi/.mc with 940 MeV.
rric"
=
C X (+Id), ~
140 million electroll volts ( M e V ) , Ale"
-
$) J,
= t x (:XI) ( - / + I . )
i ~ ~ /lj.d
Application t,o the clicrgy splitting givcs
E,pi
149
Fig. 2.5
(z) = 0
for a suitak)le choice
150
Problems and Solutions on Blectro~rr~agnetism
Cen.tral Potentials
(b) Suppose that this system is found to have only one bound state wit,h 21 bincling eiiergy of 2.2 MeV; evaluate g"F,F,c nulncricaHy and stat,e its units. [Not,c: a gra,ph of values J,, (z) ill x (1 plaric: has been providetl witli the infor~rit~tion at t,he bcgiiining of tliv cxniiiir~ntion(Fig. 2.5)]. ((1) What wollltl tllc iniriirii~ilrlV ~ L ~ UofC <,"/h(. 11;~vr:t,o be in orclrr t,o have two I = 0 bound st,atc:s (d arid hl rcrn:~inii~g thc: salnc)?. -
(MT)
151
(b) For bound states we require that for T i m, R(T) + 0, or Jp remains fin~te.This demands that p 2 0. R(T) must also 1,c finite at T = 0, which mc;Lns that ~ ( 0 = ) Jp( a )= 0. This equation has an infi~iitenun1l)t.r of real roots. For E = 2.2 MeV,
Figurc: 2.5 shows the c:ont,olirsof .I,, (3.) for tliklorc:i~tvnl~ies(ir1tlic;~tedby ~ ) l i ~ ~Thtr i f : . 1owc:st zero right ant1 t,op n1iir11)crs)of the: fii11c:tioll ill t,ho :I: of J,, (:r) for p -- 0.65 is 3.3, thc: iic>xt,6.6. Thus for (1 z 3.3, t,lic: syst)(:il~ has ollct i! = 0 l)o1111dst,;~t(:,for wl1ic.h -
Solution:
(a) When 1 equa.t,ion
=
0, thc: r;uliiil wavr func:tio~~ R(T.)= X(T)/Ts ; ~ t i ~ f itllc (.~
the reduced mass 1)c:irig /L = M/2. By the, (:h;~llg(: of v;~ri;~l)l(: T-,z
and writing X(T)-- .J(.I;),WP
== tuc:-P",
.I:E [O,
-
w1iic:h is ;L tli~nc~ilsiorll(:ss c:o~~stal~t. ((:) Eilr tu (i.6, t,lic:rc: is iui at1dit~iori;~l 1 = 0 l)o~iiidst,i~t,c.Thus th: rllillillllllll ~ ; ~ l ~ofi c(u: for two 1 = 0 l)o~il~tl st;~t,(:s is 6.6, for whic:h (~~2/11,~:)111i,1 = ~ttt,(;2(r~lil,/4~4(:2 = 1.62 .
(XI,
~I;LV(:
we can reduce thc Schrijdiiig-ercqliatiorl to 13cssc:17sdifferer~t,ialcq11;~tiorlof order p
wlicrr ,./I(o) is t,la wavr fun(:t,iorii ~ the t origin, 711,ill(: p;~rt,iclcI I ~ R S S ,i~11i1L2 the squan: of t,hc: or1)it;tl anglilar ~ i i o ~ r ~ c n topc:r;~tor ~iiil (let IL = 1). Givc: a c:lassic:al ir1tc:rprc~tationof this ecluation for the c:a.sc: of a stat,(:wit,li ailgular
Solution: Thus the (~iiiriorinaliscd)radial wave fi1ric:tiorl is
(a) I11 the fitdcl of central force, the Schrodiuger equa.tion is
152
Problems and Sol~rtionso n Electro~m.aynetism
Let $(r, 8, p ) where u ( r )
=
=
R ( r ) x m (8, p )
-- 1 u ( r ) Zrn (8, P) -
Central Potentials
Hn-c F,.
=
-
dV(r)
.
IS
153
the c:cnt,ripc:tal forc:e, anti
1
7'
, ~ R ( T arid ) , we have for thc radial inotiol~ - Trl,(Lr
,
where l i t is t,llc: t , ; ~ i l g ~ ~ lv(:Io(:ity t i i ~ l iilolig t,l1(:sphcric:al surf:j.c:e of r, is mass ," " rnultiplic~tl1)y thc c:c?i~t.ripc:tali~(.(:(?l(:~iit,io~l-T11(: (:(l~ii~tioll thus expressc:s Nowt,or~'ssc:c:ol~tl1i1.wof iriot,iol~.
+.
Multiplying tJhc two sidw of the i~bovcwith u l ( r ) ancl iritegra.tiiig from r = 0 t o r = MI, we get
= .II,(O)--: 0. With For the eigcnst,atcs wc may assurne 7r1 (m) = 0, IL(CXI) zt~'(0)-- [R(T) + r R 1 ( T ) ] , . == ~ R(O), partial i ~ i t ~ g r a t i ogivw ri
Hence
L2
[R(,I.)F,,,(8, p ) ] 7[R(T)X,,, (8, p)]r"kr(1fl , 7"
(b) For I
#
0, 1 qi/(O)l2 = 0, and so
(a) W11i1,tis 1.11(,i i ~ i i ~ i i ~ i(l(:~)t,l~ l i i ~ i of t,ll(r ~)ot,o~it,ii~l ii(~~1(:(1 t,o i~(:lii(:~(: t,wo
1)ound st,i~t;osof mrro i i l ~ g ~ ~I IlI Oi I~I Ir( ~ I I ~ , I ~ I I I ' ! (b) Wit,h i i ])o~,(:llt,ii~.l Of t,liis (l(!~)I,ll, wlliit iLr(: tll(! (:ip,(~ll~iLlll(:~ Of tll(: ~~anlilt,oilii~.ll t,h;~f.\)(~1011g t,o m1.0~ , O I , ; Lil.llj!,llliLr ~ ~llolll(:llt.lllll'! (If ll(:(:(?SSiLI-y you may ox1)1.c~ss1ji~1.tof y o ~ i r; L I I S W ( : ~t,h~.ol~gh t.li(: sollit,ioi~of ;L ~ ~ ; L ~ I s ( : o I I clental ( : ( l ~ ~ i ~ t i o ~ ~ . ) (c) IT t,hc: pitrt,ic:l(:is ill t,11op,rol~i~(l st,i~t,(:, sk(:t,c:h tllc: wiivc fii~lc.t~ion ill t,Ilc: c:oordiili~t,(: 1)i~sisil.11(1 t,llo (:orr(:s~)o~i(li~~g c:oortliriatc proba.bility tlist,ril)ution. 15xplaiii (:i~r(,fiillyt,ll(, physi(.i~lsigilifi(:i~i~(:~ of the latter. of a (sirlglc) incasureinent of the particle kinetic: (d) Prc:tlic:t tlltr rc!s~~lt, cxergy iii t,c>rlllsof t.l~isW;LV(: fi111(:t,ioli.YOU may express your prediction I.11rougli oiic:-di~~~t:~isioi~id (l(>fi~~it,(: iiit.c:grals. ((3) 0 1 1 t,llc?1 ) i ~ ~ ofi st,11(: ~ i i l ( . ( > r t i ~ ~)rinciple, . i ~ ~ t ~ give a qualitative conriec(,ionbetwc:cli p i ~ r t((.) , ~ iiii(1 ((I) ; ~ l ) o v ( ~ . (Bcrkcley)
'iolut ion:
-w/;),
(a) Tlic: at;tr;~c:t,ivcl)ot.(:litii~liIliiy I)(: r(:~)r(:s~~it,c:(l by V = whrrc Vo i s a positivc ( : o ~ ~ s t i ~For i ~ t ,l)o1111(1 . st,;it(:s 0 > E > V o . Thns for 1 = 0 t,hc
I-;~.(lial wavc fuiic:tioli IZ(,r) = X(T)/Ts;~t,isficst,hc (:qui~t,ioils Its corresponding classical expression is
154
Central Potentials
Problems and Solutions on Elect~onlag~~etis~rl
with ~ ( 0 =) 0 and ~ ( o o finite. ) To suit these conditions the wave function may be chosen as follows: ~ ( r =) sin crr,
0
x(r) = B exp (-,LIT),
ro
a
+
where N = J 2 m ( E Vo), p = d -. Frorn the boundary conditiori that a t ti~iuoliswe get a cot LYTO= @. Defining ("
fI2
7.
<
--
< r < 00,
(F)iqggy [F
=
0.
cot
Solving for
p, we gct the sccolld eigcnvalue of the Ha~ililtonian,
x ancl
y' shollld be con= (rr0, q = pro, we have 7.0,
= 27r~v0r;/h~,
Each set of the positive nuriibc:rs t, 71 satisfying t,lic:sc: cclllat,ioils givcs a bound state. Irl Fig. 2.6 clirvc 1 rt:presents 71 -( cot ( arid curvc 2, t2 {' = 22, for example. As sllown in the figurc:, for a give11 vi~lllcof Vo, to have two intersections i11 tlic quatlrant wc require
-
+
TO
155
(c) Scttirlg t h e normalized gronrid state wave function as ~ ( r=) A sill
txr-,
~(9.) = A sin
are
0
-
r)], r
> ro ,
which is the mirliinli~ilpotential tlcpth ilec?dcd to ac1iic:ve two bound states of zero angular inoinc~rlt,um.
1 -47rA-2a
1 - -
(arO- sin
ar-0
1 cos firo)+ - sin2 al-0
a3
The wave function a ~ i dprobat~ilitydistribution are shown in Figs 2.7(a) and 2.7(b) respectively.
Fig. 2.6
(b) With a potential of depth given above, one intersection occurs a t q = 0, for which @ = = 0, i.e., E = 0. The other intersection and -t cot [ = q, i.e., occurs a t t2 q2 = ( % ) 2 , i.e., ( = 31
+
J*,
I t can be seen that the probability of finding the particle is very large for T < ro arid it attenllatcs exponentially for r > ro, and we car1 regard t,he particle as being bound in the square-well potential. (d) The kinetic energy of the article, ET = p2/2m, is a function dependent solely on the momentum p; thus the probability of finding a certain value of the kinetic energy by a single rneasurement is the same as that of finding the correspontli~lgvalue of the momentum p, 1 ,$(p)I? Here $(p) is the Fourier transform of the ground state coordinate wave function,
Problems und Solutions o n Electronr.aynrtisnr.
Central Potentials
X irj =
El
+ 2aV0A2
sin 2 a r o
(t:) From the above we see that the wave fiiiict,ioli in space coordinates in (c) gives thc space probability dist,rihutiorl, whcrcas thc wave function in pspac:c in (d) gives t,he momentum probability tlistribution. T h e product of unccrt;~initicsof one sim~lltarleousmcasurcrnent of t,hc position and the momc~lturrimust satisfy the unccrtaint,~principle
Tliat is to say, t,hc two corrlplcmerlt cw:h other.
2019 (a) Givcii a ouc~-tliinc~risio11;~l pot('ntiii1 (Fig. 2.8)
show that thcrc is always a.t loast olie bound state for attract,ive potentials Vo > 0. (You lrmy solve t11c cigcrlvalue condition by graphical means.) (b) Coniparc t,he S(:hriitlillgcr equation for the above one-dimensional case wit11 that for thc ra.tlia1 part U ( r ) of the three-dimensional wave function when L = 0 , $(r) = .-'U(r)
The integratior~call be affcc:tc:ci when the expression for ~ ( r ill ) ((:) is l . avcraee kinet,ic ~ I I P I - P Vl?v i s substituted in t,hc i n t e ~ r a ~ i tThci
Yr,~(o),
where $(r) is the sollit,ion of the Schrijdinger equation for the potelitis1
Why is there 11ot always a bound state for Vo sional case?
>0
in the tliree-dimen-
(MIT)
P~oblcmsand Solutions o n L?Zectromagne~ism
159
Central rotentials
V(x) A
The continuity of the wave function and its derivative a t x = *a requires k t,an kn = kt. Set ka = (, ktn = 7. Then the following equations determine the energy levels of t,he bound states:
-0
Q
>x
0
(tan<
=
q,
i2= 2 r n ~ ~ a ' l h ~ .
[" - "0
These equations must in general be solved by graphical means.
<
Fig. 2.8
Solutiori:
( a ) For the bvu~ldstjat,c:of a, partic:lc, E < 0. For / >: the S ~ h r i i d i l i ~rqllatiorl rr
1>
v
I
0 an(]
2VbE --- .J, = 0 dz 9,"
dx'l+h
-
+
has solutions
$(z)
=
whcrc
i
3;
B Ck'z,
LC'<-a,
kt=
For
> .,
AC+'Z,
Fig. 2.9(a), curve 1 is a plot of q = tan F , and curvp 2 plots t2+ q2 = 1. The dashed line 3 is t,he asymtotic curve for tlhe former with = r / 2 . Since curve 1 goes through the origin, thcre is a t least one solution no matter how small is the value of Voo2. Thus there is always at least one t~oundstate for a one-tiimensional symmetrical square-well potential.
<
rl 3
2 1
JT. Fig. 2.9(a)
1 1 < a , the S(:hrodinger equation 1("
227n
~ + ~ ( v o + E ) I = o
1 3
has solutions
$(x) .-coskx,
(even parity)
sin kx,
(odd parity)
$(x)
where k,
JY
provided E > -VO.Here we need only consider states of even parity which include the ground state.
In
2 1
0
1
2
3
Fig. 2.9(b)
L
k
160
Problems and Solutions on Electron~aynetrsm
d2U -+dr2
Central Yotentrals
161
~kxistsonly if the depth of the well has a t least a ccrtain minilnum value. ( lalculate that rninirnur~lvalue.
2mE U = 0, h2
(b) The analogous problrrrl in one dinlension leatls t,o a different answer. What is that ariswer'?
has s o l ~ ~ t i oUn( r ) = A exp(-K'T), where
(c) Call you skiow that the general nature of the answers to (a) and (b) .rl)ove relnairis the sarnc for a wcll of arbitrary sh;~pe'! For example, in the 1 ,lie-dimclisioll:~lc:a,sc (b) For r. < a , the equation is
-d2U + dr2
2m -(Vo+E)U=O h2
and the solution that sa.tisfies the boundary condition U(T)'-O B sin ~ ; r where ,
is U(r)
=
(
onsidcr v;-rriousvaluc~sof X wllilr krcpirlg f ( r )~iiicharlgcd.
( CUSPEA)
Solution: ti
The cont,inuity of the wave function and its derivative at r cot tia = -d. SetstingKa = (, n'n = 71 we get
= a,, rcq11irc.s
-
(a) Stil)posc: t,hi~t, t,hc:rr is a bouritl stat,c ?j,(r)nrrd ttlilt it is the groullti ,.t,ate ( 1 = O ) , so that, ,I/,(r) +(T). The eigcriequ;~t,ionis
(cot ( = -7,
t2+ 772 = 2rnv0a2/fi2. Thesc: arc: again to 1)e solved by graphical means. Fig. 2.9(b) shows curve 1 which is a plot of ( cot ( = -71, and the dashed line 2 which is its asymptotic if ( = 7r. It can be seen that only when
t2t 7 j 2 = 2
where E
V(T) = 0, V(T) = -Vo,
>
> R, 0 < r < R,
r
m ~ ~ a ~ / h ~ with Vo
or voa2
< 0, and
> 0, a.s shown in Fig. 2.10.
The solution is
> 7r2h2/8m,
can the cquations have a solution. Hence, unlike the one-dimensional case, only when Voa2 7r2h2/8m can there be a bound state.
>
2020 (a) Consider a particle of mass m moving in a three-dimensional squa.rewell potential V(I r I). Show that for a well of fixed radius R , a bound state
Fig. 2.10
Problems and Solutions orr E l e c t r o m a g ~ ~ r t z s m
wherc. A a n d B are nornlalizatio~lco~lst~ants. The continuity of .Ij>ancl 4' a t .r = R, o r equivalently
Fig. 2.11
gives For bound states, we requirc 0 equation has solution
k cot (k1Z) = - I ; ' ,
while tlic tlcfinitiorls of k, I;' rcquire
These equatiorls c:a.rl be solvccl grapl1ic:ally s.s in Problem 2018. 111a similar way, we (:all show t11a.t for thcrc tjo 1)c a t least i i bound st;l.tc? wc require
= V(-J:),
where k , kr hnvc tlic same dnfinit,iorls as in (a). T h c c:ontinllity of
(b) If t,he poterltial is a onc-dimensional rectang111a.r well pot,ontial, 110 matter how dcep the well is, there is always a b o u ~ l dstate. Tlle ground state is always syrnm~t~ric: about the origin which is t,he center of t h e well. T h e eigsnsqnat.ion is
tl~:
4 a n d $'
Since Vo > - E > 0 , there is always a bound-state solution for a n y Vo. (c) For a onc-dimensional potential well of arbitrary shape, we can alsuch t h a t ways define a rectangular potential well l/,(~)
I
where, as shown in the Fig. 2.11,
> E > -Vo As V(J;)
V,(X) = -Val V5(z) - 0 ,
z1
I
< R/2, > R/2,
a n d -Vo > V ( z ) always (see Fig. 2.12). F r o ~ n(b) we see t h a t there always exists a 1 ,t/,o(z)) which is a bound eigenstate of V,(x)for which
Problems and Solutions on Elcctromagr~ctis~l~
165
Central Potcntinls
V(x1 .r.
The Grcm's function is zcro for .I: 5 0 and for a (1). Let
> 0 is equal to the
> 0 part of the solution of
G(k, r',
G ( r , r', t ) = We have: il,,aLG= f,wG and H = cives
W)
dw .
(2)
e,
and thc sub~tit~utioll of (2) in (1)
Fig. 2.12
This means that thcrc is always a bound st,at,e for a oric-tliinciisioilal well of any shape.
2021 Calculate Green's function for a rionrelativistic electron in tllc potcnti;r.l
We first i~it,c:grat,c:with rc:spcct to w . Thc: pat,h r is choscll t o satisfy thc c.ausality c:ontlitio~~. Ca1is;~lityrc:quiros that wllc:il t < 0, G ( r , r', t ) = 0. First let thc polar 1)oint of w shift, a little, say by -ie, whcrc E is a small posit,ivc: iiurnber. Li7inally letting E -t 0 , we: get
G(r, r', t ) =-
)d3k
V(z, y, t) = m, x 1 0 , (any 11, z ) V(z, y, z ) = 0, x > 0, (any y, z) and evaluate ( G(r, r', t ) (5 Dcscribc thc cvolution in timc of tlic pattcrn of probability and interpret physically the reason for this behavior.
Hmce when both x aiid t arc, grcatcr than zero, the Green's functiorl is given by (5); otherwise, it is zero. Whr.11 n: > 0 and t > 0,
Solution: The potential in this problenl can be replaced by the boundary conditiori G(r, r', t) = 0 and a = 0. The boundary problem can then be solved by the method of images. Suppose a t r" is the image uT the elect.ro11aL r' about x = 0. Then
(ih& - H ) G ( r , r', t ) = d(t) [d(r - r') - b(r
-
r")] .
(1)
If the potential V(z, y, z) were absent, the Green's function for the free space, IG(r, r', t ) 12, would be proportional to t - 3 . But because of the I'resence of the reflection wall the interference terrrl occurs.
166
Gentrd Potentials
Problems and Solutions on Electro~nugnet~~sn~
2022 An rlectron moves atlove an inlpenctrahle colic1llc:ting surface. It is attracted toward this surface by its ow11 image chargo so that classically it bounces along the surface as shown in Fig. 2.13. alld crlcrgy (a) Write the Schriidi~lgcrcquation for the erlergy cigc~lstat,cs eigtnvalucs of the electron. (Call y the dista1lc:e above the surfac:~.)Ignore inertial cffccts of thc imagc.
I
Fig. 2.14
I'ig. 2. 13
with
(b) What. is thc :l: and z dcpcndcrlc:e of tlic eigel~statcs? (c) What arc the remaining boundary corlditions? ((1) Find the ground state and its energy. [Hint: they are closely rclated to those for the usual hydrogc:rl at,oln). ( c )What is the colnplete set of discrete and/or c:ontinuous crlcrp;y eigcmvalries? (Coh~7n,bia.)
Not(: that since V(y) = -2 4~ dcpends on y only, p, and p, are constants
I
I
Solution: (a) Figure 2.14 shows the electron and its i m a ~ e Prml-rl;nrrl.r . -c t,ric energy for the svstern is V ('r l = 4 5'. o ; V' : : -?P . 2 u", x,. 1 -. 2?/ t (-F) -e2/4y. The Scliriidinger equation is then
. = u
I
h2
Y, z) = $n (Y
442, z )
we can write the above equation as
-
,7bn, (y) ,j,,:,
of
+h-1-,b U L C1CL-
.
C ] =ZY
-
4,
).(
4,(.)
-
el(":z+"=")/"
and boundary condition is '$(x,Y, 2) = 0 for Y ( c ) The (d) N~~ consider a hydrogen-like atom of nuclear charge 2. The Schrbdinger eqllation in the radial directioll is h2 1 d 2m r2 d r
the type
4, ( z ) ,
(5)
h2 d2 e2 Sm dy2 d'n (Y) - - $TI(?/) = E v i n (y) , 4~
--
"I6LJ
,j,,(.,
4(2:, y, z ) -- E,+(ie,y, z ) .
(b) Separating the variables by assuming sollrtions $)(XI
I , \ . V IU
of the notion. He1ic.c
On setting R i
;
t
=
r
R+
'('2mr2 + 'Ih2
R
I
= E'.
xlr, the above becomes
h2 dZX 2m dr2
Ze2 r
+
+
~ ( 1 l)h2 27nr2
x = EX.
!
168
In particular, when I = 0 we have
(a) Writ,c down t,he appropriate Hamiltonian for the three-dimensional
i.
whicli is identical witli (1) with the rcp1;~cernents7. --t y, % i Hence the solutions of (1) are simply ?j multiplied by tllc, radial wavc func:tions of the ground state of the atom. Thus
where a =
169
Central l'ote7~llr~ls
P?.oblerns and Solvtzons on Electrornagnetzsm
lnotion of this electron. What boundary conditiorls must the electron's wave functiorl satisfy? (b) Firltl thc ctrlergy levels of the clectroli. ((:) h l r t l l ~st,ate of k)wcst energy, find the avcragc (listnncr of the dectron ;ll)ovc the c:ontluctor's surface.
(Col~m~bia)
Soll~tion: (a) Take Cartesian (:oordinatc:s with tllr origill on and the z-axis ))*Ipel1clic:ul;u t,o the cr)ndu<:torsurfa(:c s11c:h that the cond1a:tor orr:el)ies t'he poterlhalf-sl);lc:c < 0. As in Problem 2022, the elcctron is subjc(:t to tial V ( s )= H(:iicr thc Halr~iltoniarlis
&.With Z = 1,we 1laW
&.
4
Note that thc 1)onndary coildition iri (c) is satisfied by tliis wavc filnctio~i.The grorliltl-state energy due t,o y rnotion is similarly o1)tairictl:
E!I- z~~~~2 h2
me4
-
32Ft2
(e) The complete energy eigenvalue for quantum state n is
with wave Rlnction
Tllr w;lv(; fllll(:tion of the cl(:ctron satisfies tile \)olllltlary ~:OIltlition +(z, y, z ) = 0 for
Z
(1)) As sl~ownin Problem 2022 the energy eigenvalues are
(c) Thr grolirltl stilt? has cwrgy
where A is the normalization constant and wave function
A nonrelativistic electron rnoves in the region above a large fiat grourided conductor. The electron is attracted by its irnage charge but carinot penetrate the conductor's surface.
5 0.
where
170
f'roble7rl.
a n d Solutzons
OTL
E~ectromflgr~etism
and A is the llornlalizatio~lconstant. Hence
3. SPIN AND ANGULAR MOMENTUM
Consider four Hermit,ian 2 x 2 matrices I, o1, u 2 , and a3, where I is the unit matrix, and the others satisfy ai aj aj oi = 26ij. You must prove the followillg wit,liout using a specific representation or form for the matrices.
+
(a) Prove that T r ( o i )= 0. ( b ) Show that the eigenvalues of ai are 3z1 alicl that d e t ( a i ) = -1. (c) Show t,h& t,he four mat,rices are linearly independent and therefore that any 2 x 2 matrix can be cxpanded in terms of them. (d) F'roiri (c) we know that
whcre M is any 2 x 2 matrix. Dcrive an expression for rn, (z = 0 , 1, 2 , 3).
(Buffalo) Solution: (a) As
aa a 3. = - o j a i
(i
# j),
ajaj
=I,
we have
a, = rsiajaj = -ajoioj , and t,hus
Hence T r ( o i )= 0. (b)Suppose ai has eigenvector
4 and eigenvalue X i ,
Then
oiai4 = ajxi4 = x i f f i 4= ~ On the other hand, aiai4 =
14 = 4 .
54.
i.e.
172
Spin and Angular Momentum.
Problerns and Solutzons on Quantum Mechariics
(d) Given M a.ny 2 x 2 matrix, we can writ,e
Hence = 1,
x 3
M
or
=
inoIt
7niOi.
2= 1
Xi = & I .
T~ determirl(: th(: cocff-icicr~ts mo, rn,, take t,he trace of the above:
As Tr(a2)=X1+X2=0, the two eigenvalucs of ai are XI = t l , X2 = -1, and so Dct(ni) = X I X Z = -1. (c) If I , ai, ?: = 1, 2, 3, were linearly dependent, thcn four c:orlst,aiits mo, mi could 1)c fount1 such that,
TT(M) = 2 m 0 , 1
)I.
7n,o =
I T'T( M ) . 2
-
Considrr 3
m O 1 +m,ioi ~ =0.
0,
M = Trio n j
Multiplying by a, froin the right a.nd from the left wc 1ia.v~
(rn,;n j 0, ,
.3
3
+C
~n,.ni u:, .
i==1
+C
ffi f f j
=0,
Addillg the last t,wo ccluat,ioiis fiivcs
i= 1
ajM
and 3
C
:<
il.tld = ,,no T i
71kJ O j $
x
i= 1
1=1
m,o a,
+
nl.~ f f j CJi
=0
.
Thus
i= 1
Tr(g,, M
Adding t h r a.bove two equations gives (lr 2mon,iC i n i ( ~ . q i + ~ ~ ~ ) + Z m ~ I = O , i#j
inoaj + m j I
=
+ M n j = =mum j + 2mj I .
+Mnj) =2
T ' r ( m o a j + m j I ) =in0 T r ( g j ) + 2 m j = O . As Tr (uj ) = 0, we would have m j = 0, and so mo = 0. Therefore the four matrices 1 and a , are linearly independent and any 2 x 2 matrix can be expanded in terms of them.
,
I rn. - - T r ( v J b f ) . 3-2
0.
Thus
T ~ ( ~ ~=h4mj f )
3002 The tllree matrix opera.t,orsfor spin one sa,tisfy S, c:yclic permutations. Show that s3= ---
sZ,
S, -
s Y s z = is, and
(s,&i~,)~=O.
( Wisconsin)
174
For the same in, nl, can have the values i ,rrL, n~ has eigenvalue m(m t 1). Thus
Solution: The matrix forrrls of the spin angular momentrlm operator (s the (s2,s,) representation, in which s 2 , s, are diagonal, are
Sz
175
Spzn and Angular M o n ~ e ~ ~ t u n ~
Problems and Solutions on Qzrantu7n Mccl~anics
=
-
1, . . . , -nl, while
I? l 2
1) in
=
0
0
-1
3 x 27
Dircct calculatiori gives
=
81 times
2 x 48 = 96 t,imes
I
Three matrices M z , kI,, each with 256 rou7s and colllmns, arc known to olxy t~hccornmutation rules [b1,,M?,] = iA4, (wit,h c:yclic permutations of :r, y and z). The eigenvalues of the rnatrix M x are f-2, each once; f 3 / 2 , each 8 times; & l ,each 28 t~imes;t1/2,each 56 timcs; and 0, 70 times. Stsatethe 256 eigenvalues of the matrix M 2= hl: + M i + A.1,".
For this stat,e calculate (L,) and ( L ; ) .
Solutio~l: As L, is a Hermitian operat,or, we have
Then
( Wisconsin) Solution: comlnutes with
kfx.
So we can select the common rigenstate
IAf, Mx). Then
M 2 1 ~ , =~ m , ()m i - l ) ~M ,~) , , hl, I M , h.L) = ,m, ( M ,M z ) .
rr~h ih
= -(($
\ Ly\$) - ($1
f?,14)) = 0 .
Considering the symmetry with respect to X , Y, we have 1 1 (i:)= = - (1,; L", = - (L - L,), 2 2
(LC)
+
-,
A'L
176
and so
=
(
=
51 ( 1 L
-
1 = - [l(L
L 1)
+ 1)
-
177
.Sp,~nun,d Angular Momentum
Problems a n d Solvt~onson Quantum Mech,anics
m,"] h' .
d
m
,axid K and n are real const,ants.
(a) What is tllr t,ot,al culgular i n o m d t u n r of tjhr partaide? (I)) What is tla: exyn(:ti~t,imvr~,l~l(: of t,he r-componmt of angular momelltllrn'! (,:) If tho z-rolepoaallt of iu~guLrmolnel~tum,L,, wrrc mr.rrnred, what, is the prol)i~,bilitythat t . 1 result ~ would be L, = +K? ((1) Wli:~,tis t l a prohnl)ilit,y of finding tllc part,i(:1~ at 8, 4 and in solid anglo dil'! Herr H , y\ arc tlic U S U ~ I:~egles ~ of sph(aic::~lcoordinat,i-s.
2 It can also be calculated using the "raising" and "lowerillg" operators.
The spin functions for a free e1ect)rorl in a basis where d, is tliagonal can be writt,en as and ( ) with eigenvalues of 6 , being +1/2 ant1 1 / 2 respectively. Using this basis find a normalized eigerlfunc:tiorl of .?, with eigenvalue - 1/2.
(A)
Yoll 111:iy fintl t hc, followillg cxprcssiolis for the first f ~ w spherical harmonics uscful:
(MIT )
Solution: 111 the diagonal represc~ltat~ion of d2, .?,, we car1 rel)rcsmit. .??, k)y
Let thc rccluired cigenfun~t~ion of 6, be a,
=
Solution:
( i ) .Tlicn ;IS
The: wa,v(,f,lll(.t,i~)ll llli~y ,(/I =
K,r (c:os (1) sill H -1 sill 4sill H
it,s aligular p;~,rtbeing we havca. = ib, a n d s o u , = b ( i ) . Normalizat iori
r(:writ,t.(:ll in sp1ic:rica.l c:oortlin?~tc:sa.s
+ 2 cos 8) e-"'
,
-
,1/1(8,4) I(' (COS (I,sill 8 -tsin 4 sin 8 + 2 cos 8) , whc~rc.K' is thc: iiorlrlalizat,ioll c:ollstalit such tlhatj
gives b
=
K"
' v5'
I
Hence
lT IT d8
we have
Consider a spinless particle represented by the wave function
$ = K (z + y
+ 22) e-"l‘ ,
+ sin 4 sin 8 + 2 cos 812 d 4 = 1 .
Sinc:c (:os(1) =
3006
sill 0 ( c . 0 ~4sill B
L2 ((:i$ + e-i+),
1 'ib sin 4 = 7( e 22
- e-idl) ,
178
Prohlcms ond Solutzons
on
Quontum Mcchalazcs
T l ~ normalization r conditio~rand tho orthonormality of );'" ,
1
87r
,
1
X7r
.
Spin and Angular Momentu~n
t1lc11give
47rl
II
179
A particle in a central potential ha an orbital a~lgularmoment~im1 = 2fi .lud a spin s = lh. Find the erlergy levels and degen~r2rciesassociated with . I spin-orbit interaction term of the form Hso = A L . S , where A is a
and thus
Solution: choose { H , J" . J Z : L\ SS"} as a complet,e set of mechanical variables. 'The wave furlctior~associated with angle and spin is 4,.,,,l,s, for which J 2 ( " j I n j l s= h 2 j ( j 1 ) 4 j m J l s I L a d j r n j l s = h21(1 -1 1)4 3 n t i l s ,
+
S 2 $ j r n , , l , 7 = h 2 ~ ( ~ + 1 ) 4 3 7 1 - JzbjT)lj~,5 l,ls~ =h7r1zi4jmi~s,
with
(a) l h e total arigular ~nomentunlof the r)art,ir:lc i s
1 s2), 2 its J = L I S. Thus the energy lcvcls aiicl clageilcracies are respect,ively ti" E," = - A [ j ( j 1 ) - 1 ( 1 3 1) - s ( s + I)] 2
Hso = AL . S
as t,he wave fiinction corresponds t o 1 = 1. (b) 'I'hv z-component of the angular mornenturn is
=
-A(J~ - L"
+
=
d=2j+l=
{
ZAP,
j =3 ,
-Ah",
j =2 ,
-3Ah2,
j = 1,
7, j = 3 , 5, j = 2 , 3, j = l .
(c) The probability of finding L, = +fi is
(d) The probability of finding the particle in tlle solid angle d o at 8 , ip is
S
1 [sin @(sin4 y',*(B, 9)4 (0, g)09= 8 7r
+ ax 4 ) + 2 cos @I2 db) .
I
One can show that the "raising" and "lowering" operators for angular momentum, J* = J , 31 i J v , commute with J 2 , and that, if j , 7n are the eigenvalues of .I, .Jz, then
. ~ * / j , m ) = f i J j ( j + l ) - r n ( m 3 1 1l )j , m * l ) , for appropriately chosen phase conventions of the state vectors. Use these properties t o express those states 1 j , m) for which m = 1 - 112 in terrns of the states / 6 , mi; s , m,) with s = 112. (MIT)
180
5'pl.n a n d Ang.lllar Mvrnentwrn
Proble7n.s a n d Soluti,ons o n C ) ~ ~ u n t thlccl~~cmlcs m
Using t,he result of (a), Eq. (1) car1 be written as
Solution:
According t o the theory of angular niomentl~nlcoupling, the 1)ossible values of t.hr total angular nlonientum arc j = 1 112 tu111 j = 1 - 112 for
+
,s = 1
2.
Using the, propertips of J , L
and S-
Similarly, Eq. (2) can
bt.
written as
+ "b (1
-d
1,
The last t,wo c:clli;~t;iolisshow that. ( A , 0 wc: I)ot,h red I) H(:IICY
=
- d m ,
m
;l.lld (:a11
t,:~k(:~l i\S
.
Suppose nrl c:l(:ctroli is in a. st,at,cdcscrihed by the wave function
Q where
1 = - (e4 sin H
J471
+ cos 0) g ( r ),
00
lg(~)I2r2dr= 1, The orthonorn~al !ity of t.he eigenvectors gives
21
;rnd 4, 8 are the azi~rluthand polar angles respectively (a) What arc: the possil,le rc:sults of a measurement of thcr a-cornpoilent I,, of the angular mornent~lrnof t,he electron in this state? (b) Whal is the prok)ability of obtaining each of the possihle results in part (a)? (c) What is the expectation value of LZC! ( W~scon,sin,)
P.roblems and Solutions on, $?rant?~rnMechanzcs
183
Spin and Ang1~1a.rM o m e n t l ~ m
Solution:
..,.- polynomials of degree 2 j wit,lr respect t,o the va,riables sin (812) and (P/2). H ~ r t( :j , P ~ L )refers to a n eigcrlstate of t,he square and z-component o !I t.he angular momenturrl: (
t j:;
3" jj; 3, 1 j ,
the wave ftlnct,ion call 1)o writ,ten a s
~ 1 1 , )= 111,)
j(j t 1 ) hZ / j , na) ,
I
= 71th j ,
71b)
.
( Wisconsin.) Henc:e the possiblo values of L, arc +ir, 0. (1)) Sincc
:iolution:
We use the method of inathenla.tica1 intluctioll. If j ,)I,'
= 0 and
t,lle given wave ft1nc:tion is norrn;tliaocl. Thc proh;~I)ilitytlollsity is t;hc>llgiven l)y P = 1 j,
1'.
of L , = 0 is ('
Tlnls the probability of L, "5)
=
+h is
(fi)'
= 0, t,hrn 7n t.hc st;~tenic?ntis o1)viously c:orrcc:t. If j = 112, let,
Consicjcr P;ilili's ln;it,rict:s or;, where k = : I : ,
?j
Or
Z.
Sin(T
or 2/3 iultl tliat
or 113.
I lle
x
Lz
unit nlatrix, we hi~vofbr
(Y
=(:011~t,i~llt
[d'(--r~: + 1 Y))
I If U ( P , 9)refers to a rotation through an anglc /3 about t,hc y-axis, show that the matrix elcrncnts
= cos ct f ior; sin (Y .
U (p,
5) EE exp (
)
- i ~ ~ ~=/CXP h ) = cos
p
- I,
- -
2
-
2 z tl
P z.ovsin. -P
oY = cos 2
.
u , 2
jy sin -
-
2
-
186
Problems and S o l ~ i t i o n so n Q u o n t u m Meclran,tcs
187
S p i n and Angular M o m e n t u m
(a) Show that f , J2 and J z can be sin~ulta.neouslymeasured. (h) Derive the rna.trix representation for f irl the I .J, M , jl, jz)basis. (Label rows a.nd cohirriris of your matrix). (c) Derive thc nmtrix representation for f irl thc I jl, j2,,rrllbl,m , ~t)a.SiS. )
( Wisconsin) Solution:
where J,M are row labels, J', M ' are column 1at)cls. (c) Denote the state of J = 0 and t,he state of J = 1 and Jz = M as and x l ~ respectively. , Since jl = jz = 1 1 2 , we (:an denote the state ( ;jl, j2,m l , rn2) simply as I n ~ 1 m2). , Then as
(a) f , J~ and .Jz (can be measured sirilulta~leo~lsly if c!ac:h pair of them cornmute. We know tha.t J%nd J, comrnute, also that eit,her c:on~rrl,ltes with a , a corlstarlt,. From dcfirlitior~,
Now for ~ l - ~ cpa.rticle li
1
we have
Using the above expressiorls and the result of (b) we can write the matrix c,lernents ( m l , m2 I f (mi,711;) in t,he basis ( jl, j 2 , m l , ma) as follows:
Thercforc, f , J%alld J, can be measured simultaneously. (h) 111 the I J, M , jl, j 2 ) basis,
I ;: .I 0
a-b
2b
Spin u r ~ dAr~g.1~1ur Momentum
Problems and S o l ~ ~ t z o non s Quunt~irnM c ( - ~ ~ L ~ L ~ c s
1x8
3012
and sirnilarly
Corisider t11~following two-particle wave fiinctioil in position spare:
21
~ ; $ ( r , . r 2 ) = -h"rz =
where a and b a.re a.rbitrary coilstant vectors, f aiid {I a.rc ark)it,r;l.ryfi~ilctior~s,aiid a , and y are corista.nts. (a) What arc tlic eigcnvalues of the squarctl angu1a.r iriomciituiii for c;~ch part,icle (Lf and L:)? (11) With an ;~.pl~ropriate choice of a , p and y , ,,h2(rl,ra) (:ail illso be ail eigerifuiic:t,ion of thr: t,otal angular momcnt~ilrisclli;irc:tl J" (Ll + L?)'. What a.rc thr: possik~lt:v;xhics of the total anglllnr irioiiic:iit~llilscl~~;~.rc,tl 2i1id what arc: tlic appropriate V ; L ~ I ~ Cof S (Y, P a.nc1 y for (:;i(:li stat,(,?
(MITI
l(1
Tlius we rr:qliirc
L~ . ~ ~ y i (r 2r) ~=, 11,"rl x x (b =
-
. r2)
VI)
. (3x V2) {f (,ry)!/(rZ) [ ~ (. ar l )
+P(b -
+ @ ( b . r l ) (a -
= WP
~ : ~ + b ( r ~=, r ~fi2(rl ) x V l ) . (rl x V l ) {f(ry),q(~:) [((a . r l ) ( b . r2)
=-
+ B(b . rl) (a . r ~+)y ( a . b) ( r ~. r ~ ) ] xV1) . [a(rl x a ) ( b
=l ( l +
.r~)]
1) h2+(r1, r2) ,
-
=-
b ) ( r ~. rz)l>
+ y ( a . b) ( r ~. rz)] x Vl)
.
[(,(a . r1)(r2 X b )
x a) + y ( a . b) ( r 2
X
rill
( r i ) [a(r1x a) . ( r z x b )
x b) .
(r2
x a)
+ y ( a . b) (2rl
h2f (r?)!/ (r;) [-@(a. r l ) (b . r 2 )
-
. rz)]
a ( b . r l ) ( a . r2)
+ P + 27) (a . b) (rl . r 2 ) l
fL2X$(r1,
r2)
for $(rl, r2) to be an eigenfil~ict,ionof L1 . La. This denlands that r ~ )
+ p(r1 x b ) (a . r 2 ) + y(a . b ) (rl x r 2 ) 1 = 2h2f (r:)9 (r,") [a(a . r l ) ( b . r 2 ) + p ( b . r l ) ( a . r,) + y ( a . b ) (rl
h2f (r:)
t (a
r2)I)
h2f ( ~ f ) ~ ( (rl r i )x V1) . (rl x V l ) [tw(a r l ) ( b . r2)
=-h2f(r;)!/(ri)(rl
=
r2)
tl."(,rf)9(,r;)(r1
+ p(rl
have
+ B(b . r1) (a . r 2 ) + y ( a . b) (r1 -
-
.
f,,"(rf)!l(~i) (rl x V1) . (r2 x V2) [ ( ~ (- ral ) ( b . r2)
-I- a(b . rl) (r2
i l i r x V,
r l ) ( a . r2) +$a
1
r x V(a - r) = r x a ,
=
so each
(axb). (cxd)=(a.c)(b.d)-(a.d)(b.c).
=
r xp
x V z ) $ ( r l . r2)
(rxV). (a.r)e=(rxa) . c ,
and that
=
. (r2
Hcllct: tllc cigerivalucs of L'f a.nd L; arc cac:ll equal to 2fi" partic:lc 11a.s t,he quantum number 1 = 1. (1)) Wc further note that
(a) Wc first note t,lia.t
As L
x V2)
+ 1)h2$(rl, r 2 ) .
-
Solution: r V f ( r ) = .ff (7.1 1
189
-p
=
Xa,
-N =
Xp,
a
+ 0 + 27 = Xy
which give three possible values of A: -1
a=p=
X=+l, X=2,
a=-p,
3 --y; 2 y=o;
a=p=o.
1
190
Problems and Solutions o n Q,uanturn Mechanics
191
Spin and Angular M o m e n t u m
Therrfore the possl1)lc values of the total angular momenturri squared, J" L! Lz 2L1 . La, ant1 thc correspo~ldingvalues of a, [j aalid y are
+ +
(b) Under the assumption of L-S coupling what are the "good" quantum iiumbers? Give sets of values of t.hese for t,he configuration of the t,wopwave rlectroiis. (c) Add up the dtgencracies of the terms found in (b), and show that it is the samc as the number of terms found in (a). (Buffalo)
Solution:
+
3013
A qu;~~lt~iln-~lie<;lii~liic:al ~ t i ~oft a, ~particle, wit11 Cartesian c:oc~rdinittes :c, 71 i-\,ridZ , is dcsc:ril)c(lby the iiorrnalized wiwc filiic:t,ion
Show tliiit t,lic: system is in a stat,e of definite ailgrilitr ~noliiclit~i~n itlid g-ivcx tho vt~1uc.sof L2 and L, associated wit.h the st,iitcr.
( wis(:o,l,<~,tlI,)
(a) Each electron (:an occupy onr: of the (21 + 1) (2s 1) = 3 x 2 = 6 st,at,es,but it is not perniitt,ed t,hat two electrons occupy the same state. So t,he nulnber of psrmitt,cd states is Cg = 15. (b) Tlle "good" cluant,um riulrll~ersare L< S2,J2 a i d J,. Under the itssumption of L-S couplirig, t,he total spin quantum numbers for two elecI;rons, S = s l + 5 2 , are S = 0, 1 a11d the t,otal orbital quantum nlimk)ers, L = l1 4-12, are L = 0, 1, 2. C!onsitlering the symmetry of exchange, for t,he singlet S = 0, L shoulti 1)s rvcil: L = 0, 2, <:orresponding to 'So, 'Dz respectively; for the triplet S = 1, L odd: L = 1, corresponding "0, l , 2 . (c) The dcgencracy equals to 2.7 1. For J = 0, 2 and 0, 1, 2 iri (b), the total ii~lrnberof dcge11crac:ics is 1 + 5 1 3 5 = 15.
+
+ + +
Solution: ?i-arlsforirliilg to sp1icric:al coordinat,es by :c = 7, sill N c.os p,
y = 7- sin B sin p,
z = r c.os B ,
Hcilcc the partic:le is in a st,ate of definite angular rrioiric:ntllm. For this " l ( 1 + 1) h2 = 2h2, L, = 0. state, 1 = 1, L
(a) Det,erininct the energy lcvrls of a particle bound by the isotropic potential V (r) = kr2/2, where k is a positive constant,. (b) Dcrivc a forrnula for thc degeneracy of the N t h excited state. (c) Identify t,he angular momenta and parities of the N t h excited state. ( Col~~mbia) Solution: (a) The Hamiltonian is
3014
A frec atom of carbon has four paired electrons iri s-states :l.nd two more electrons with p-wave orbital wave funct,ions. (a) How inally states are ~)crmittedby the Pauli cxclusioll principle for the l a t h pair of electrons in this configuration'?
Choose (H, i2,f,) t,o have comrnon eigenstates. The energy levels of I)ound states are given by
192
Problems a n d Solutzons o n Q ~ ~ a n t uMechanzcs m
S p i n arrd A~ryularMorncn,tu:m
(b) The degeneracy of the stat,es is determined by n,, and I. As N = 2n, 1 , the odd-evcn feature of N is the same as that of I. For N even (i.e., 6 even), the dcgerleracy is
+
N
N
1-0 ( 1 even)
l=O (1even)
1 (N 2
=-
L'=O
+ 1) ( N + 2 ) .
For N odd (i.e., 6 odd),
x
N
f=
(2I+l)= 1 = 1 (1 odd)
[2(I-1)+3]=1
l = l ( l odd)
x
Solution: The two rlcw p;t.rtic:lcs arc Bosons; thus the wave function must be ;ymmetrical. 111 the grouritl state, t t ~ ct,wo particles nll~ststay in 1s orbit. I'hen tllc spacc wavc hlnct,ion is synlrrlc:trica.l, and conscq~lcntlythe spin w;Lve functiorl is syrr~mctric;tltoo. As s l = 1 and s 2 = 1, the total S has i.l~reepossible v;~lncs: S = 2, tll(: sl,in w;wc funct,ioll is syr~lrnctric:;l,rld its d[:gcnera.cy is :!,sf 1 = 5. S = 1, tllc: sl)i11 wave fullc:tion is arltisymrnc:tric a,ritl its tlrgencracy is :!S 1 = 3. S := 0, the spin wavc functiol~is syrnrrictric: and its tlcgcrlora.cy is :!is 1 = 1. If tllc s~)i~~-rlcl)nndr.nt forcrs arc nrglcrt,cd, t,hc dcgcnera.cy of t,he ground statre is 5 3 + 1 = 9.
+
( N - 1)/2
IV
193
(if+:)
I'=O
+
+
+
Hcrlce the dcgc~lcracyof the N t h excited state is f = ( N 1) (N + 2)/2. (c) 111the c:oinmon eige~~st~ates of (H, 12, l , ) , thc: wa.vc func:tiorl of the syst,c:~nis T h r z-c:o~rq)cx~(~r~t of t,l~cspill of nrl c:l(.c:troli irl frcle spa,(:(:(no c?l(:ctro~r~agnctic: fic:ltls) is ~r~c;~surcitl i~rltlfolllld t,o t)c +fi,/2.
TL,L 0, P)= R(7.1Km (0, P), and thc eigcnenergy is
As N
= 271,.
+ 6, the angular momentum I of thc Nth cxcit,cd sttttc? 112s
$ + 1 values 0, 3, 4, 6 , . . . , N
for N even, or N for N odd. Furthermore, the parity is
f ( N + 1) valuir, 1, 3 , 5, . . . ,
(a) If a sllk)sc:clucrit rr1c:asurcrnorlt is illad(? of thc, z:-c:nrril~~r(:nt of t , l ~ ~ spin, what arc: tlic possiblc rrs~~lt~s'! (b) What is the I)robal)ility of fir~tlii~g thesc various results? (c) If the axis tlefinirig tllc rrlcasured spirl directior~ma.kcs an a.~igl(:H with rcspcc:t, t,o t,hc original z-axis, what is the probability o f t11c v;~rious possiblc rcsl~lts'! (d) What is thc: cx~>cc:tat.ion valuc of the spill measurernellt i r ~(c)? (Berkeley)
Solution: The ground state of the realistic helium atom is of coursc ~iotidcgcncrate. However, consider a hypothetical helium atom in which the two electrons are replaced by t,wo identical, spin-one particles of negative charge. Neglect spin-dependent forces. For this hypothetical atom, what is the degeneracy of the grour~dslate? Give your reasoning.
( a J s)
(k),
(a) 111the (T; rclprcselrt,ation, thc spin wa,w function is the cigcnI 1 ), corres1)onding to eigenvalues + l , - 1 functiorls of cr, are JZ ( 1 ), (
' 5 Expanding (h) in these two stat,es, wc see that the possible
respectively. results of n rneasuremerlt of s, are fh./2 since 5, = ;a,, the mean value being zero:
194
Spin and Angular M o m e n t u m
Prohlen~sand Solutions o n Quan,tum Mecl~nnics
(b) T h r probabilitirs of finding tllc rcsult t o be P- respectively:
+$ ant1 -4 are P+ and
(c) Supposc? tlie spill axis is n = n(H, cp) =. (sin H c:oscp, sin Hsin cp, cosH). Then tlie cigcrlstates for s,, = s . n are
Hrncc the two eigcnva1ut.s of T arc
In t,hc rc:prcscrltation of G" and .irz,
c:orrcspondiiig to c:igc~llvahics+ h / 2 ancl -h/2 rcs~)ect,ivc:ly.T h e ~)rol);~.l)ility of filidirlg t h r cigcnv;lhlos +h/2 t1.1ltf -h./2 arc? cos"0/2) and sirl"(~/2) rcsptct,ively. (d) The cxpc<:tiltioli vi1.1(1(! of tlic: spin is
Let t,hc cigrnvcc:tor
t)t:
(t).Thcll the cigcncqlla.tion is
where
(a) Consider il system of spin 112. What are the t?ig~nvaluesa.rltl rlorn~alizc:cicigenvtrc:tor of thc opc:ra.tor A g u t B g, , whcrc &,, ?., are tlir: a11gtlla.r morncntum opcra.tors, ant1 A and B are real constants. (b) Assume t1)a.t the system is in a state corresponding to the: tipper eigen\aluc. Wliat is the 1jroba.bility that a measurclricnt o f iY will yic:l(1 thc va.lue h/2? T h e Pauli matrices are
0
2 or
Hence a. :
11
= iA : B
td
m ,
and so the normalized eigenvect.or is
(CUS) Solution: (a) Using t,he definition of angular momentum operators let
and i,, the cigenvector of iy is (b) I11 the representation of i2
196
Spin and Ang?~larMomentum
Problems and Solut~onson Q ~ l a i ~ t M ~ ~r r~nh a n ~ c s
Now a, t h c expectation value of s is s(s + I ) , etc., we have
Hence the proi>ahllity of finding s,, = h / 2 is
Note t h a t P- is thc probability correspo~ltlillgto t,l~csystem in t,lic state &lid P+ is that corrcspe>lltliii~ tjo thc: state
of eigelivaluc: T = h,d-/2,
I ;
\ild hence
A system of three (norl-itI(:ntical) spin one-half l,art,ic:las, whoso spin operators ;Lrc: s l , s 2 ailtl s:$, is govcrlletl 11y thc: Haulilt,olliali H = As1 . sz/h% B ( s l
+ s2) . s : ~ / / ? .
Firld the c:iic:rgy I<:v(:ls illld t1llc:irtlegclicracios. (Pri7rcc t on,)
Solution: IJsing a c:clrripletc? set of tlyriarriica.1 vari;~bl(?sof t,llc? s y s t c u ~(H, s:~, s 2 , ~ : iwhcrc ), siz = sl S.L,s = ~ 1 2 s:( = S ] s:! t s:], th(: c:ig(:lifilrlc:tio~l is ( slz sa S , I T L , ) , nlld the stationary state cquat,iol~is
+
+
I t follows ,,hat, for s12 = 0, s = 112 : E = -3A/4, t,hc dcgc:nc:rac:y of I,he cricrgy l(:vc:l, 2 s -+ 1 , is 2; for s l 2 = 1 , .s = 11'2 : E = A / 4 - B , t'he t1egeneru:y of t,hc c:ne:rgy Irvc?l, is 2; for s l . = ~ 1, s =; 312 : E = A / 4 -tB / 2 , [,hedego~lcrac:yof t,lic: c1ic:rgy 1nvc:l is 4.
+
A particle of spill one is s1111jec:i t o t h e Hamiltonian H = As,
+ Bs:,
where A and B arc c:onst,;~nt,s.Calculat,e t,he energy levels of t,his syst,en~. If a t t,ili~(! ~('1-o t,li(: spin is ill ail cigenstate of s with s, = +h,, calculate the cxpect,atio11 vahic: of t.he spiii at t,inlr t .
(Princeton.)
Solution: we have
a r e first fiiid the stationary energy levels of the s y s t r n ~ .T h e stationary S c h r i i d i ~ l ~ eequation r is
where
198
1
is a vector in the spin space. As
s,=
199
Spin and Angulal. Moinentu~n
Problems arid S o l u l i o r ~ so n Q t i a r ~ t u n Mecl~unics ~
(:: : ::) 0
0
-i.
1
li, wave furlctiol~of t,he system is therefore
The
cps( t ) = C l q s o exp
+ c:j9
[ ;' [
- - t] +c2y,s+ erp [
s rxp
i
% t]
Ei -I_ t ] .
Initially, s,
cp, (0) = trcps (0)
Let whvrc A' - All, B' = Bfia. The erlelgy levels ilrc. givctil 1)y thct cigcxl~v,il~~e!: of thc a l ~ o v rr~lntrix,which are roots of thc cc~untiori
I.V.,
(E - B') ( E ~ B ' E - A') = O -
Thus t h e cmcrgy 1t:vcls are Eo = B', wherc: w = \/BIZ Functio~isare
+ 4AI2/k, B"
E* =
=
(BI1f w ) h / 2 ,
Thus we can take the initial wave fiinc.t,ion (~iormalized)as
B1/fi= Bti.. Tlle corrcspondi~~g eigc:n-
Equating 9, (0) with Clpsa
+ Czcps+ + C 3 9 s gives
\
200
f'~o6lrms and Solz~tionso n Quantum M C C ~ U ~ L L C S
Sprn and illryt~lnrMornent.um.
We call now find the expc:cta.tion value of t,he spin: (sz) = p; (t) s z p s (t) = 0 ,
where we havc usc:d the orthogorlalit,y of pso, ps+ arid p s - . Similarly, (~1,)
=0 ,
,s,Xoo
-
A systc.111of two particles each with spill 1/2 is d(xsc.rii)ctlby all c,ff(,c.t,ive H
= A(slZ
0.
Hence for thtr t,rij)I(:t stsate,thc cllc,rgy lcvcls h" E &IShA B , with M s 4
(sZ)= p g (t)s, ps (t) I-
EIarliilt,onia~l
-
+ ~ 2 , )+ B s ~. ~
+
-
= 0,
f1 ,
For t,lic:siriglct st,i~t,c:,ttlc clicrrgy lcvel c:orlsists of only ollc: linc:
2 ,
whcrc st alitl S L ~ 1 -teh e LWC) spins, slZand 52, a.re tliclir z-c:ompoi~c:rits,inid A arid B arc c:oiista.l~ts.Find all the energy levcls of t,his II~~rnilt,oni;~ii.
( Wisconsin,) Solution:
Wc choosc z.sn4, as the common eigenstate of S2 = (s, + ~ 2 rid ) ~ S, = s i , + S ~ PFor S = 1, ill, = 0, it is a t,riplct niid is syii~ii~etric whcr~the t,wo c:l(:c:trorls arc exchanged. For S = 0, h.Is = 0, it, is a si~lgl(xt arid is ;ultisymmct,ric:. For stationary st,at,es we use t h t~~ i ~ r ~ e - i n d ~ : ~ ~ c l l t l c r l t Sclirij(lingcr equatiori
Hxsnrs As
we have
I=
E~siLr,.
Suppose ;in ;~t,olnis iriitii~llyin i ~ ntrxc:itcd 'So s t i ~ t c(Fig. 3.1) ~ L I Is~1k)sr(~ (luently tl(:c:xysiilt,o it iowcr, short-1ivc:d 'PI state wit,h c~uissiollo f iL photo11 y, (Fig. 3.2). Soor) aftcr: it tlcci~ysillto the 'So gronnci s t a t c by ~ ~ l l i t t ~ i l l g ;L ~ ~ ~pllOt,Oll ( ~ y2~ (Fig. ~ 3.3). ( 1 Let, B I)(: t,he aiiglc l)etwcc:ii thc: tjwo c:rriittt.d photons
Problems and Solutions o n & u a r ~ t u mMechanics
S p i n and Ang7ilnr M o m e r ~ t u m
203
the eigenstate of the z-component of anglilar mornenturn on the z1 direction is equivalerlt to multiplyirlg the initial stat,e wit,h the d,,~, matrix. Only atoms that are in states 7r1.', = f1 (:all elllit photo11 (as L', = fh must be satisfied) in z' tlircctio~land rnixk(: the at0111 decay into the 'So state ( m i = 0). Then the t,ra.nsit,ionsara from 7~1,= f1 to 7n1 = *I, and we have
(a) What is tlic rc:lat,ivc: prok)a.l)ilit,yof 8 in this proc:ctss? (I)) Wliat is t l ~ rat,io : of firidi11g l)ottll pliot,olis wit,li t,kl(: ~i~111(: (:ir(:lll;lr ~)olarizi~t,iori tjo tJhat of fiiitli~lgtill(: ph0t$011~ wit>ho~)~)osit,(: (:ir(:llli>,r I)ol;lrizll,t,io~ls'! It 111;l.yI)(: of solnc: 11(:11) tlo k110w the: rot,;lt,ioii ili;r.t,ric:c:s (i,,,,,,,,,wllic,ll rc:lat,c on(. i~riglila.rnlorn(?lltlilli~.(:~)~.(:~(:~lt~i~t,ioli ill o11(: (:oordi~li~.t,(: S ~ S ~ , ( : I t,o II arlot,hc:r a~lglilarniolric~ltrirnrc:~)rc:sc:ril,i~t~io~i ill a rot,;~t,(:(l (.oor(iilliLt,(: syst,(:~,l, given below:
whcrc: tr is t,he angle bc:t,w(:(:ri the: z-axis of o~ic:syst,c~rlant1 t,llc: the othctr.
of
( Cok~rt~biu.) Solution: The atom is initially in thc cxc:itctl statc ' S o . Thus the proj(:c:t,ion of the atomic angular momentum on arl~itraryz directio~lis L, = 0. We can take the direction of the first photo~lc~nissionas thc z tlirct:t,io~l. After the emission of the photon and the iit,onlgoc:s into the ' P I statc, if the angular momentum of that photon is L, = fh, correspondingly the a11gula.r momentum of the atomic state 'PI is L, = ~ hi.e., , m, = $1. If we let t,he direction of emission of the second photon be the zl-axis the projection of
(a) The rclativc: 1)rol)ability of 8 is
(b) The ratio of the probability of fi~ldirlgboth photons with the same circular polarizatiorl to t,llat of firlciirlg the photons with opposite circular polarizations is
Problerris and S o l ~ ~ t z o no sn Quankum. Mechar~zcs
205
Spzn c ~ n dAngular M v m e n t u m
Co~lsidera n electron in a llriiforlrl magnetic: ficl(l in t,he positjive z clirc:ction. T h e result of a measurcmrnt has show11 that, t,he elcc:t,ro~ls~)irlis alollg tlie l)osit,ive :IT dircctiorl a t t = 0. Use Ellrc11ii:st's thc:oreirl to c.o~rrl,~ltc: the l,n)b:~l)ilit8yfor i > 0 tjll:\t t,lle electroll is i l l t,llr stilt? s , = 112, (1)) sz = 1 / 2 , (c) s, = 112, (d) s , -- -1/2, ((:) s Z = 112, (f) = -1/2. Elircnfcst's theorem states that thc c:xl)c:ct,;itiorl v;~lllc~ of a ( ~ l ~ i ~ . l l t l l ~ i i mccharlical optr;r.tor obcy the classical crllli~tiollof rllotioll. [Hint: Rccall the r:onntx:tion betwccll (:~1)c!c,tationv:~lr~c:s :~11(1prol)al)ilitg co~~sidcratiol~s].
in agn:m:llt with the above. Note that use has been ma.dr of t h e commuti~tiolirelations [s,, s u ] = i h . ~ , ,etc. ~ ~ l i t , i ~(,sl,) l y = 112, (s,) = (s,) = 0, and so we can write for t > 0, (.sJ:) = : ((:oswt)/2, (.sll) = (sinwt)/2, (s,) = 0. 1r.t tla: i)n)t)at)ility for t > 0 of t,lre rlec:tn)l~being in s h t c 3% = 1/2
r ;Lnd \)(:inC ill the sta,t,r s ,
=
-112 be 1 - P since t l l e ~ carc the only
t,wo statcs of s,:. Then
( W~,.q(:o~t,,si~rt,) Solution: 111 the (:Iassic::~l ~ ) i ( . t ~ lill1 r ( ~cl(:(:tro~l , S I ) ~ I ~ I I wit11 ~ I I ~ i ~ ~ l ~ t Il Il Ii O ~ Ir~ ~ ( ; I ~ ~ ; ~ ~ ~ I I s ill n ma.gnc:tic: fic:ld B will, if tllc tlirc:c:tiolls of s i~11(1B (lo 11ot (:oill(:i(l(:, I)n?c:ess; L ~ ) O I It,h(: ~ dir(:(:tioll of B wit11 i1.11 i~llgllIi~r ~ ( ' l o ( . i tw, ~giv('11lly
wIlc,rc: w = statcs tl1a.t i r l
B,
711. l)(:illg t , h ~ (:l(:(:trol~IllilSS. ( I U ; L I I ~ ~ I I IIIIC(:~GLII~(:S I w(! ~ I ; L V ( I
4. (1 - )
(4)
1 .
1
2 1 1 1w = 2
0
(
-
wt) ,
El~r(:~~S<xst's t,l~~:or(:lll t,I1(~1~
-
d (it (s) 'rrI,(' (s) x B This (:an l)c d(:rivc:tl t1irc:c:tly i ~ follows. s AII cloc:tJroll wit,h spill a.llglll;~r11ioi11(:llt~11lll s 11ils i1 ~llil~~l(:t,i(: 1110111(:11t, p = $ s a nd c:ol~scclric~~~tly n IIarniltorli;~11 -
()
1 - p = 1,'' (
-
)
La,stly for ((:) i l , l ~(lf ) , we llavc:
I
A Ix~,rti.:lawith in;rg~irtinrnoir~cntp = pas and spill s , wit,ll magllitude
taking the z axis along t,he direction of B. T1lc.11
d (s) --
dt
1
A
- z h, [s' HI =
PB
-
-ts , y
7 [s,~ 1h r r ~ f:
+s,i,
s,)
112, is 11l;nrxl ill a (:o~lstimnt,m,?gnct,ic ficlcl pointing alollg the x-axis. At t = 0, tia: pnrti(:lc is found t o l ~ a v eii= + 1 / 2 Find the probahilit,ics a t ;my 1:rtor time of firidirlg tlic: 1)articlc with s , = ~ +I/?..
(Col~t~mbia)
Solution: Tllc: IIa.~liilto~lian (spin part) of t,hc systeln is
11
(I I
8;
:I
206
Problems and Solutions on Q.t~c~r~tum Mechanics
Span and Ar~.qularMomentum
as s = hc,, being in the x direction. In the a, representat,ion, tha Schrodinger equation that the spin wave function ) satisfies is
,,
(a1
r
the sollltion A1 = A2 . I ,~.cssionsfor a,l,2 results in
- t h--c- r. i v.~- n.s- -t a-t eof .s,.
As
--
.
..-
Y
=
B1
=
207
- B 2 = 112. Substitution in the
= + 1 / 2 is '
i
tlltl that of s , = -112 is
(1"
dt" a1,2 +
p; B~ (11,2 =
0,
which havc solutions
where
I
=
3
~ constants. As and Al,2, B I , arc
A (1 + sin 2wt) . 2
Similarly thck ~)rol)al)ilityof finclirig s ,
P
) = 1 ( ( ) 1 ( t ) )1
=
-112 is
1
=
2 ( 1 - sin2wt).
-
and 3025
the initial spin wave function is Schrodirlger cquat,ion then gives
da l ( 0 ) = 0 , dt
(A),
i.e.
nl(0)
=
1 , a 2 ( 0 ) = 0. The
H=
g = iw
----dnz(O) - i dt 2ii
=
1,
E s B. . 2m.c
Calculate the operator d s l d t if B = By. W h a t is s,(t) in matrix form? ( Wisconsin)
Solution:
These four init,ial conditiorls give
AI+BI
The Hamiltonian for a spin-; particle with charge +e in a n external I I 1;tgnetic field is
A2+B2=0,
W ( A I- B I ) = 0 , w ( A 2 - B Z )= w ,
In the Heisenberg picture,
ds dt
-
-
1
-[s,
ih
HI
ge 7 Is, s . B] =-
r2mhc
Problems and
As [s, s
.
B] = [s,, s .
S
. B] = [s,, s t ]B, = i A (s,
B,
-
+ [ s ~s?/] , By+ [ s , , ,sz] B, S,
= ifi,(B x s),,
B,) etc
we have
(a) How lrialiy enorgy lcvcls doc-\s thc, syst,cni llavc:'! Wlint arc their <.~l!:rgics?W h a t is t,llc d(:go~lcrnc:yof thc tIii~f(:rc:lltlcvcls? (b) Now ;~.d(la ni;tgiict,ic fic:l(l ill t h e z t1irc:c:tioii. What arc? tlic ncw ~11c:rgylcvc:ls'! Draw ail c:licrgy l(:vc,l tli;\.gr;rri~as a f11nc:tion of L3,. ( Clj.icngo)
[s, s . B] = i h s x B , a n d hence
If B
Two electrons are tightly bound t o different neighboring sites in a cerThcy arc?, th(:roforc, distilig~iishablep;rrticl(:s which can be de.,.~-ibed in tc:rnls of their respective Pttuli spin matrices n ( ' ) and d2). The I 1.1iniltonia1iof t,licsc: el(:c:t,rolis takes the forin
I .1111 solid.
and
=
B y , thc: almva givc:s
:iolut ion: (a) The, H;~liiilt,oiiitriiof t,11(,syst,(~~li is
_
,
[ ( l)
- u(l)'
2
a n d so
with the solution sz(t) = c~ cos (gwt)
+ ca sill (gwt) ,
where w = e R / 2 ~ n , cAt . t = 0 we have
and hence s z (t)
:;( ) +
= sZ(0) cos
-- t
s,
(01 sin
(@ 2mc
t)
--
u(1)2
210
Prob1e.m~nrrd Solutions
OIL
Q1~antu.mMechanics
21 1
S p i n and Angular M o m e n t u m
is t,hc total spin of the systern and
is its total z-c:ornponcnt. s2,,s, and H arc c:orrimutablc. Usirig the above arid the coupling thcory of angular mome~lturrl,we havc! (noting the eigenvalue of s 2 is S(S + l)h2)
(a) As seen from tJhc table, thc systcm has thrcc cricrgy l(:vf:ls -25, 0, 25, each wit,li a dcgcricracy of 2. Notc that if thc: clcct,rons arc: i~~tlistinguishable, thc seco~ldarid fo~irthrows of t l ~ et,able wolll(l 1 ~ diffi:rc:~it : from the above.
3027
(b) 111 thc presenc:c of a ~ n ; ~ g ~ i efield t i c I B 1 = B,, H =
up) up) + uJ1)(T2)] - p . B ,
where C
-e
and m bring the electrori charge arld Inass respcctiv~ly.Thus
A frcc ator11 of cart)o11 has four paired elect,rons in s-st,ates and t,wo ,.lectrons in pstatcs. Assume there is the finc structure coupling L . S , i.e., lJ2,S2 and .J2 are "gootl" q u a n t u ~ nnumbers. (a) Give t,hc values of S, L aud J of possible states by a table, indicating I,he correspollding ml11til)licitics. (b) Which stntc has the lowest energy? Give your reasoning. (Columbia)
Solution: (a) A carbon atom has two I s elect.rons and two 2s electrons, which form 1 wo closed shells. Thus the atolliic st.ates are det,errllined by the combination c )f the two 2p electroris. In L S coupling, as 11 = l 2 = 1, sl = s2 = 112, we have L = Ill + l2 1 = 0, 1, 2; S = 1 s l + s2 ( = 0, 1. Taking into a.ccount I'auli7s exclusiori principle and the antisymmetry of the total wave funct,ion, we obt,ain the following table. --
s2,s, and H are still commutable, so the new energy levels are: 25, eB,fi/mc, -eB,fi/mc, -25. The energy level diagram is shown in Fig. 3.4 as a function of B, (lines 1, 2, 3 and 4 for the above levcls respectively).
(b) According t,o Hund's rule, the
3 ~ state o
has t,he lowest energy.
Spin and A q u l a r Molnentwrri
P,roble~nsand .~01u1zons on Quantum. Mech,(~ni(:s
I,
S
j
'"+'L,,
0
0
0
IS"
1
1
2
3
I
rnr~ltiplicity
I
1
~
2
1
1
1
1
0
" P ~ "I'"
3
1
2
0
2
IDa
1
'orilomb potenti;~lbefort, the reaction, L1 p(n,) p(n,) = 1, we have ( - q L 2 = -1
=
213
0. Sillrr p ( n - ) = -1, p(d) =
,
T h e dcutron has ,J = 1 ant1 n - 1i;ls xcro spin, so t h a t .I = 1 I~rforethe (,actiont,akcs p1ac:c:. Tht: coriscrvat,ior~of a.ngular rnornt?nttlrrl requircs t,hat, . , I ~ . ethc r reaction, Lz + S = J . T h e itleutity of n ant1 n dc11la.nds that t,he I, ,l,al wave fuilct,ion be antisyinrnctric. The11 sirlce tllc spac:i:~lwavc function I . . ;~ntisyii~i~lc~t,ric:, t,he spin wavc furlct,ion rnust 1)e syrnrnc:t,ric, i.e. = 1 . 1 1 1 ( i so L2 = 2, 1 or (I. As L2 is odtl, wc must have Lz 1, S = 1. I
3028 A negatively c:hargc:tl n meason (a psclldosca1a.r part,ic:l(:: zoro spill, odd parity) is iilitia.lly l)ou~lrlin tllr, lowost-cnt:rgy Coulo1111) wi~vofiirlction around a deutc:ron. It is c:a.pturc!d by thc tlclitc:ro~~ ( i protoil ~ ii~l(li L iic~itron in a st,ate), which is c:onvc:rtctl into a pair of 11c:utrons: n
+ d + + IL. 71.
(a) What is thc! orl)it,al angular n l o i r l c ~ ~ t ~ofl ~t,lic: r i ~ic:ut,ronpair'! (1)) What is t,lic:ir t,ota.l spill 2~ngula.r~ n o i l i c ~ ~ t u l n ? (c) What is tali(:1)rol)al)ility for finding I)ot,h 11ci1tro11spins (lir(:(:t(:(lopposite t o the spin of the dcutc:ro~i? (d) If thc tleuteron's spi11 is iilit,ially 100% po1arizc:d ill thc R tlircc:t,ion, what is the angular tlist,ril)ut,ioil of the nc:utron e~rlissiorlproI)il.I)ilit,y (per unit solid angle) for ;L lielltron whose spin is opposit,c. t o t,lli~tof t,lic, initial deuteron? (not normalizc~tl)splic~ric~,~l liarliionics You lnay find some o f the first f ~ w useful: Ydl=l, ~:~=~sirlHe*", YF
=
cos 6, Y , ' = sin 26 pi'&.
(CUS) Solution: (a), (b) Because of the coilservatior~of parity in strong interactions, we have p(.-)
-
~ ( d(-l)L' ) = p ( n ) p ( n ) (-1IL"
+
where L1, Lz are the orbital angular momenta of n + d and n n respectively. As the n - , being captured, is in the lowest energy state of the
Fig. 3.6
T h e t,ot,al orbital anguli~rrnonlenturil i ~ n dtllc total spill i~rlgular1110h. 111entumarc both /L = (c) Assuine that the dcut,cron spin is in the direction .J, = 1/ibeforc the I (:action. If both neutrons had spills in t,he revcrsetl tlircctioli, we would I ~ i ~ vS, e = -1fi,, L, = 2h, which is in~possiblesince L 2 = 1. Hence the [,robability is zero. (d) Take the z-axis along the R direction. Then t,he initial state is 1 .J, J,) = I I , 1). In the nollcoupling representation, the st,at,e is lL, L,, ,C;, S,), with L = 1, S = 1. Tlius
J m q
214
Spin and Angular Mome.nt.c~m
P,rolilems and Solutions o n Q U U I L ~Mechanics TL~
The stsate 1
1l , O ) = ~ ( B , ( ~ O ) = - ~ s i n ~ e ~ ~ " ( 1 , 0 )
has S , = 0 a.nd so thcrc lnust 1)r one neutron wit,h pro1)al)ility distribution required is dP(B, 4)/d62
=
S,
=
-h)./2. H c I ~ (the :~
1 3 , 3 . - sin B = -- siu2 0 . 2 87r 1Gx
-
3029 An 62- hypc?ron (spiri 312, mass 1G72 MeV/(:< intrinsic parity +) c;~ridecay via the wcak iilterac:t,iorlinto a A llyperon (spin 112, rrr;~ss1116 Mc:V/c2, intrinsic parity +) and ;t K - rnesorl (spin 0, mass 494 M(:V/(:" iiutrinsic parity -), i.c., 62- -t A + K - - . (a) What is the nlost gcrlcr;~lfor111 of tjhc nilgu1:~r tlistril)~rtior~ of t,he
K - ~ncsorlsrc!l;~tiveto t,hc spin t1ircc:tiorl o f tllc: 6 2 for t,llr: (:;IS(: w l l c ~the R- has a rn;~xini~nn possi1)lc c:olnpo~lcritof angular r~lolrlc>~~t~lrll i ~ l o ~ the lg +:3/2 z-axis, i.v., t,llc i11it~i;l.lst,atjc 152"')3 = 1 6 2:3,2 ). (Assllll~c:t,l~a.t, t,ho 5 2 ~ - is at rest). (1)) What rc,st,ric:t,io~ls,if any, woul(1 1 ~ illlposc,d : oil t,llc: for111 o I the angular distrit)utiorl if parit,y wcrc: c:ollscrvc:tl in tllc: d(:c:;~yproc:c:ss'! (Bcr.kcley)
Thrrcforc
i.c., thc intcrisit,y of the: c ~ n i t t c dpartic:los is
where tu =
(a) Tllc illitrial stat,(: of the systc:in is ( 312, 3/2), whorc: thcl v;~luc:s are
the ort)ital and spin rno~nc?rlt,aof t;he 62-. Thc s1)in 1):~rtof the fiil;~lstate is 1 112, s,) 10, 0) = ( 112, s,), and t,be orbital part is Thus thc tot,al firm1 stat,? of this systcrn is
(8, p)
.=
I I, ,m).
By conser-vation of angular Inomerlturn I = 1, 2; r n = 312 s,. Thus the final state is a p wavc if 1 = I , t,he state t)c:ing I I , I ) / 112, 112); a d wave if 1 = 2, the statc 1)eing a coml)inat,ion of 12, 2) 1112, 1 / 2 ) and
) a,l 12).
- 2 R c ujncl/()(L,
'l'his is tlic: no st general forrii of t,he anglllitr dist,ribtlt,iorlof the K mesons. wcre c:onscrved in the decay process, tllc final sta.t,ewould (b) If I)t~rit,y lmve positive parit,y, i.e.
-
121 1) 11/21 112). Hence t,he wave fiinctions are
Since
PKPA = (-1) ( + I ) we get I
=
1. It would follow that
=
-1
1
216
P l v b l e m s and Solutions o n Q l ~ a n t u mMechanics
S p i n and A n g ~ ~ M l ao m ~.entl~m
and
'I,')jli/f
3 87r
2
= - sin 0 =
3 (1 - cos") 87r
-
(b) Defining tile opcrat,or J- = J1- r J 2 - , wc have
.
t
11,
using tile 1jropc:rties of J_ (Problem 3 0 0 8 ) ,
I 0: (1 - cos2 6 ) .
h a I 312, 1 / 2 ) . II
1312, 112) =
Give11 two Wfplar nlorrlcrlta J1 a.nd J2 (for c x a m ~ ~ lLc ,arid S) the corresponding wave ful~ctions,where jl = 1 and j2 = 112. ~ o l r l I , , l t the c Clebsch-Gordan coefficients for the states wit11 J = J~+ J ~rrl,, = where: (a) j = 312, n~, = 312, (b) j = 3/2, m = 1/2. Considcr tho rci~ctiolls
= h& 11, 0) 1 1 / 2 , 1 / 2 )
,
mil, 0) 1112, 112) + mil, 1) 1112, -112).
1 K-p)
= 1112, -112) 11/2, 112) = Jlp 1, 0)
\ E - T + } = 11, -1) ( 1 , 1 ) = = 1
1) 1 , I ) = Jllti
C-T',
~C07r")= ( 1 , 0 ) / 1 , 0) = m 1 2 , 0)
i
Cfv-,
IK-n} = : ( 1 / 2 , -112) 1112, -1/2}
I E - T ~= ) 1
-? x07r0,
1 o) = m
-
(4,
11, 0 )
-
(I), 11, - - I } ,
l 2 , -1) 2 , -1)
-
~
I
I- I ) ,,
+ 011, -1).
c-TO,
)
x0r- .
To ICpp rc:ac:t,ioris going thro~iglitlie resonxnt:~st,a.te I = 1, t,he fi-
Assume they proceed through a resonance arid hcn<:ca, pure I-spin sta,te. Find t,he relative rates based on I-spill collservatiorl: (c) for an I = 1 resonance state, (a) for an I = 0 resonancc stat,e. Use the clebsch--Gordan cofficicnt,~supp1ic.d. Tile ~ - ~for~K ,, 7r,,i C,~ and .rr are 1 / 2 , 112, 1 , a,nd 1 rcspectiveiy.
=
1 , 0) 1 -1)
v'Dl(),
-.
+ dl13 ( 0 , 0) , i 2, 0) + mi1 , 0) + ,hp 1 0 ,0) ,
( 2 , 0)
i
4
-1/2)
To ca.l(:r~latc:tllc? rc4;~tivc rcaction c:ross sectior~s,wc use the couplil~g I ,.l)rese~~tati to o rd(:scriI)c ~ tllc i ~ l i t i i ~ ~ lL I final I ~ I-spill states:
)
I c - 7 ~-?
+ h (1,1)I l / 2 ,
1(1 herice
3030
K-p
+ .J2-) 1 1, 1)1112, 112) .
.J-l3/2, 312) = ( J -
Hence parity con~ervatiollwould impose an angular distribution of the form
-&I
1,:,1 st,a,t,es1 Cp7rf) c:o~~tl-iljutrs 1, O), ivllile 1 E o r 0 } docs riot (:ontribute. Hence
1 x + ~ -contributes )
fi)
1, 0))
a ( C p 7 r f ) : u(Cf7r-) : a(x07r0) = 1 : 1 : 0 . ~
Similarly for
K-71, rea.ctions
we have
GJ(C-TO) : a(C07r-) = 1 : 1 .
Solution: (a) AS jl = 1, j 2
=
i,we have
1312,312) = 11, 1 ) 1 1 / 2 , 1 / 2 ) .
Only the I<-?) reactions go through the I = 0 resonance state. A similar ,.ollsideratiorl givcs t,he following react,io~lcross sect,ion rat,ios: U(C-7rt) : a ( C + 7 r p ) : a(C07r0) = 1 : 1 : 1 .
218
Prob1e.m.s and Solutions on Quantum Mechanics
Spin and Angular Momentum
3031 J1
!v(
have
(a) Compute tjhe Clebsch--Gordan coefficients for the states wit11 J = = rr~1 rrl.~, where jl 1 and j2 = 112, anti j = 312, M = 112
+J2, M
-
+
I T'P)
for the various possible rrll arid In2 values. (b) Consider thc rc ac.*t' ions:
.ir-p .ir-p
11, 1) 1 112, 112) = 1312, 3 / 2 ) ,
=
11, -1) 1112, 112) = a 1312, -112)
(.iron)= 11, 0) 1 112, -112) = c 1312, -112)
+ .ir+p + 7--p + 7-Ovn
71+p
=
+ b 1112, -1/2), + d 1112, -112).
m,
From a t,able of Clebsch-Gordan coefficients, we find a = b = c= d= For the A resollance state, I = 312 and the I .~l,ios of the cross sections are
m, m, m.
These reactions, which conserve isospin, can occ:ur in tllc isospin I = 312 statje (A resonance ) or the I = 112 state ( N * rc,sonarlc:c:). C;~lt:ulatethe ratios of thtse cross-~ect~ions, ai : aii : aiii , fc)r iLll ellc:rgy c:orrc:sporltlillg t,o a A resonance ant1 an N * rc:sorianc:c rcspcc:tivcly. At :L r(:son;Lll(:c cn(:rgy you can neglect the effect due to t,llc?otllcr isospill stntc:~. Notc: that the pion is an isospin I = 1 stat,(: and thc: nuc:lcorl a11 isospil~I = 112 state.
ai :
1 2 o i i : u i i i = l : la14: IacI2=1: 9 : -9 .
For the N* rcsonancr st,at,e,I = 112, aild the ratios are ai
: a i i : niii=O :
4
(b14 : l b d I 2 = O : 9
:
2 -9.
Solution: (a) As M = srbl Consider
+
7 1 1 , ~=
112, (7111,
1312, 312)
=
nr2)
can o111y I)(: (1, - 112) or (0, 112). Consid(:r an c:lect,rorl in a uniform magnetic field alorlg the z direction. I,c:t the rc:sult of a incasurernellt be that the electrorl spin is along the ,,ositive y dircctiorl a t t = 0. Firld the Schrodiilger state vector for the :,l)in, and the average polarization (expectation value of s,) along the z clirection for t > 0. ( Wzsconstn)
11, 1) 1112, 112).
As
M- 1312, 312)
=
& 1312, 1 / 2 ) ,
and
M- 1312, 312) = (MI=
+ M2-)
11, 1) 11/2, 112)
Solution:
1112, 112) + 11, 1) 1112, -1/2),
d'5 I l , O )
As we are only interested in the spin state and the magnetic field is 111liformin space, we can leave out the space part of the wave function. 'I'hen the Hamiltonian can be taken to be
we have (1, 1, 112, -112 1312, 112) = I / & , (1, 0, 11'2, 112 1312, 112) =
J2/3.
1;
where p = -pea, w = peB/R = eB/2mc, pe being the Bohr magneton 2mc As the electron is initially along the y direction, the initial spin wave '
.ir+
= 11, l ) ,
.irO
P = 1112, 1/2),
=
/ 1, O),
.ir-
=
11, - I ) ,
n = 1112, -1/2),
220
I
Problems and Solutions o n Q u a n t u m Mechanics
Let the spin wave function at, a later tiine t be equation ih d+/dt = H$
(F).
I
Thc Schrddinger
Spzn and Angular Mom.entzlrn
wliere p, = 1 e 1 h/Zm,c is the inagnitudr of the magnetic moment ~,l(xctron.As B is along the z direction, the abovr b e c o m ~ s
then gives
( )
ih ih,
or
(i) &, o (i =
-1O )
(l
=
-iwcy,
(i)
)
=
0
>(abj:l
-1
,
iiiat ( ~ ( t= ) IL
i l l a t b(t) = -k~,Bb(t). I(j =
iwp,
Thc solutiorls are
with the solution
(
(
=
n(t) = *,(to)c-t v.."(t-"o) -
)
iwt
= ( , i w
)
=
(
c-'iwt (;-"wt
)
b(t)
b(to) (:B
. . ;
""D(t-"n)
,
'
At tirnc t o , t,hc clcctron s1)iri is ill thc positivc y t1ircc:tion. Thus
Hence
1
i n(t0)
=
b(t,) .
Thc normnlizat,iori c.oilciitiorl
I (~(t,,) l 2
+ ) l)(tO)
2
= 1,
l hen gives
Ia(to) l2 = I b(to)
2
1
lZ = 1
'---I
As "0 = i, wc, can take l~(t,l)
(
3033
Hence for time t
Consider a n clectron iri a uiliform magnetic field pointing along the z direction. The electroil spin is measured (at time to) to be point,ing a.long the positive y-axis. What is the polarization along thc z and z direct,ions (i.e. the expectatiori valucs of 2s, a.nd 2s,) for t > to:)
( Wi.scon,sin,) Solution: The Schrodinger equation for the spin state vector is
;,
$ 1 f
15 '1
t
o
)
,
b(to)=i/.\/Z.
> t o , thc polarizations along
z and
2
directi(111s are
I
;
respectively ( 2 . ~= ~ )h(a*, b*)
( i) (:) [
=fj,(a*b+b*a)= - f i s h
2~ 1 , ?B(t-to)
222
Problerns and So1,utzons on Quantum Mechanics
3034 Two spin-; particles form a corilposite systern. Spill A is in the cigenst,ate S, = +1/2 and spin B in the eigcnstate S:, = +1/2. What is the probability that a rneasuremellt of t,he total spin will give t,hc value zcro?
I
Spin and Angular Momentum ;I
223
ud hence
(CUS) Solution: 111 the uncouplirlg rcpresclntation, t,hr statc in w1lic.h tlic total svin is zero can be writ
I Therefore
P
=
1 ([)I&)
1
1"
- = 225% 4
where SA, and SH, tl(:rlotc: tllle z-co~npoiic:r~t,s of t,llc?s~)inso f A i111d B respectively. As t,hc:so t)wo sl)iri-$ ~)art,ic:l(xs arc riow ill t,llc: st,atc,
j
IQ)
=
ISA,= t 1 / 2 ) ISnZ = +1/2),
the probability of fill(1iiig t,he tot,al spill to I)(: m:ro is
P
=
l(0 ) I J )1 2 .
S, is clefined its
5,
of we find thiit its cigrrlfunction 1 S, = Solving thc rigcnecl~i:~tiorl +1/2) call bc expressed in the reprcsc11t:ttion of S%aricl S, as
Thus
I
!
(a) An electron has becri obscrvetl t,o have its spin iri t,hc direction of z-axis of it rec:t:l~lgularc.oordirlate syst,em. What is the probt~bilityt,llat ;I second ot)servat,ion will show the spin to be directed ill z - z plane itt, an .~.ngle8 wit,h reslxct to the z-axis? (b) The total spin of the nclltroil and proton i r ~a deuteron is a t,riplet :;t,ate. Tllc: resliltailt. spin ha.s been ol)servcd to be parallel to the z-axis OF a rectariglilar coordinate system. What is the probability that a second ~l)servationwill show the proton spin to be parallel to the z-axis'? h (Berkeley) Solution: (a) The initial spin state of the electron is
The st,ate whose spin is directed in x - z plane at an angle 8 with respect l,o the z-axis is /
8\
Thus the probability that a second observatioil will show the spin to be directed in z - z plane at an angle 0 with respect to the z-axis is P(B)=/($1$0)J2=
I( :
:) (;)
COS-,sin-
= co,
(q)
.
l2
~ ~ ~ t e r a c t i ao nquailtuln , state that is initially an S state cannot have a P ::Latecomponerit at any later moment. (b) The possible spin vi~lncsfor ;I system corrlposed of a proton and a ~icutronarc 1 and 0. We are given J = L + S and J = 1. If S = 0, L = 1, (.hesyst,rm would bo iri a P state, which rrlust be excluded as we have seen I 11 (a). The allowed valucs are the11 S = 1, L = 2, 1, 0. Therefore a G state ( I, = 4) ~i~11110t coiit,ril)~it~. (c) The total spill is S = s,, s,,. For a pure D state with J = 1, the , rrbital arlgular inorncilt,~irn(reli~tivc:to thc center of inass of the 7 , and p ) is L = 2 arid tllc t,otal s~)iliiilnst k)c S = 1. The total m:~grletic rnornerit ;r.rises frorn the coupling of tllc nlagnctic nloment of the total spill, p, with ILllat of t,he orl~italang~ilariiiornenttlrrl, p ~wherc , p = p , p , ~p,, , pn lrcing the spill rliagrictic: inonlc:rits of p it11d 71, r(~s~)ectively. The a,vcragf:valllc: of tlic coi~1l)orlcritof p in the direction of thc: totaiL1 :;pin S is
+
(b) The initial spin st,a.teof the neutron-proton systein is
J ~ o=) 1 1, 1)
225
Spin an,d Angular Mornen,tum
Problems and Solutions on Q u a n t ~ ~ rMechanics n
224
1 112, 1/2)n I 1/21 1/2)p.
Suppose a second observation shows the proto11 spill to be paritllcl to the z-axis. Since the neutron spill is ~)arallclto the protoll spirl ill t,hc dclrtcron, the final state remains, as l)cfore,
1 '$f) = 11/21 1 1 2 ) 1~1/21 ~ 1/2)p.
+
Her~ccthc proba1)ility that a sccoiid obsc:rvation will show tllc proton spin to be parallel t,o tlic z-axis is 1. where ILN
3036
+
(a) Explain why a P statc cannot contribute. (b) Explain why a G statc carinot contribute. (c) Calculate t,he magnetic moment of the pure D state 71 - p system with J = 1. Assume that tlie n and p spins are to be coupled to make the total spin S which is then coupled to the orbit,al angular momentum L to give the total angular momentum J. Express your result ill nuclear magnetons. The proton and neutron magnetic momellts are 2.79 and 1 . 9 1 nuclear magnetons respectively.
(,/I,
-,
gJ,=:5.58,
27rb7,c
!I,=-3.82,
s p = S T , = S. The r~iot,ionof the: r rot oil relative to t,he center of mass gives rise t,o ;I. magnetic: irioiiici~t,wllilc tllc motioii of the neut,ron does not as it is llnchargcd. Thus PI, = P N L I~
;IS
The deutcrori is a bound state of a protoil ant1 a ileutron of toti1.1nngular rnomentum J = 1. It is known to be principally all S ( L = 0) statc with a slna.11 admixture of a D ( L = 2) state.
-
where L, is the angular lnolnenturrl of the proton relative to the center of Illass. As L,+L,, = L and we may sssurnc L, = L,,, we have L, = L/2 (the (.enter of mass is at tllc mid-point of the connecting line, t,aking m, rn m,). <:onsequently, p L = pNL/2. The t,otal coupled magnetic: moment along t,he direction of J is then
( CUS Solution: (a) The parities of the S and D states are positive, while thc parity of the P state is negat,ive. Because of the conservation of parity in strong
4
Since J = L + S, S . L = (J" L" S ? With J = 1, L = 2, S = 1 ;Ind so J L2, L L 6 , S2 = 2, wc have S - L = 3 and thus L . J = 3,
S .J
-
-1. Hence
I
Problems and S o l ~ ~ t a o nosn Q u a n t u m Mechanics
227
S p i n anti Angular M o m e n t u m
where A and C are arl,it,rary constants. From the initial condition a(0) = 0 and b(0) = 1 as the initial spin is in the -z direction, we get A = -112, C = 112. Hence 1 (e--"t - & t ) = -2 slrl wl , a='
Taking the direction of J as the z-axis aiid lcttiilg J , ta.ke thc inaximuni valuc .I, = 1 , wc have 1- -- 0.31bN
)
=
'
coswt,
and the wave funct,ioil at time t is
component of thc spin of ail rlcctron is - F L / ~ .A l~iiiforrnlnilgrlrtic firld in the 2-directiol~of rnagnitudc B (us(, cgs units) is thcii switc llc~l011 at time t = 0. (a) Predict the rrsult of a sirlglc iilc:nsrircrncllt o f t,llc z-c:oilipoilcnt of the spin after clapsc of tiii~cT . (b) If, irrst,c:i~dof iricas~lriligt,lic z-c:orll~,orlciltof tllc s~)ill,t,lic :c-c:omponent is measured, ~,rctlic:t tho rc:slilt of slic11 i L single i ~ i ~ i ~ S i I r ~ i iafter ~(:ilt elapse of t,imc T .
11 1
I
(a) At t
=
T,
As ((: ) and ( ) a.rc the c:igcnvcctors for a, wit,h cigc~lvillues+1 ancl 1 respectively t,hc prol,at)ilit,y that the mcasurcd z-c:orn~)o~~cnt of t,hc sl)in is positive: is sin2wT; t,hc probabilit,y that it is ~lcgativeis (:os2wT. (1)) 111 the diagonal representation of a,, the cigellvcctors of a, im: -
(Be~.keley)
Solution: Method 1 Thc spill wave functio~i(:)
satisfies
As we can write p i
sin wT
w
&
() a +
w
.
fi (-i)
-
where w = cB/2mc, or it~, = wb, i b = wn.
{. Thus
ii
=
iwj, =
-w2a
the probabilities that thc measured 2-component of the spin is positive and is negative are equal, being
the solution being
Method 2 The Hamiltonian for spin energy is
228
1
Problems and Solutions o n Quantum. Mechanzcs
H
= -p .
B
= eBha,/2nrc.
229
S p i n and Angular Mom.entum
Solution: The Hamiltolliall
The cigenst,ates of u, are
I;T=--V. We can writr t,he init,ial wave function as
!,*,
An alkali atom in its ground state passes through a Stern Gcrlncil a p paratus adjusted so as to transmit atoms that have thcir spins ill the +z dire~t~ion.The atom then spends t,ime T in a magnetic ficld H ill the x direction. At the end of this tirnc what is the probability that thc atom would pass through a Stern-Gerlacll selector for spins in the z direction? Can this probability he made equal to unity? if so, how?
(Berkeley)
I
"-
~
IelfiH
lelH
w = - 2, r r ~
H = - ( T27nc ,=/~W(T,,
~
0 st
()
-
i sin wt
()
0
In t,he a,bovc: ( , ) is the eigenvector a, for eigenvalue -1. Hence the probability that t,he spins are ill the z direction at time T after t,hc at,om passes through the St,crn-Gerlach selector is
I(0 1)
( ~1~~ ) 1 2 = s l n ' t sin W T
. 2
wT=
1 - COS 2 ~ 7 2
Span and Angular M o m e n t u m
Problems and .Solut~ons o n Q u a n t u m Mecll.anics
230
u . n =a, sinHcoscp
The probability equals 1; if
cos 19
~ - c o s ~ w T = ~ ,
231
+ cr,sillHsi~lcp+ azcosH
sin 19 e-''
where
Hence t,he probability will becolllc, llliity at t,inlc,s T
=
(271 + 1) 7rrcnl
are Pauli's spin matrices. Lct its eigenfunction and eigcnvahie be X rcspec:tively. Then
( i ) and
IelH.
A bean1 of I);~rt,ic.lcs of spin 112 is scrlt t,liroligll ;L St,c:rn Ger1;~c:happaratus, which tlivid
a (cos H - A) n ew sir1 H
+ 0 e-" sir1H = 0 , b ( X + cos 0) = 0 .
-
For a, O 11ot tto vanish idcritically,
X sin H
cos 6' cL'
-
epZ' sin 19 -(A + cos 6')
=
X2 - cos2 H - sin2 H = 0 ,
or 1, i . ~ .X, = I f l , (.orresponding to spin angular momenta If; h. Also, norm;llization requires
Solution: For an arbitrary dirc,ctiou i r ~space n spin operator is
=
(sin 19cos cp, sin H sin cp, cos 6') tlle For X = + I . we have
O -
a
-
1 - cosH sine
0
-
and, with normalization, the eigenvector B
A
Fig. 3.6
sin 5 e'l' cos 0
'
S p i n and Angular M o m e n t u m
Problems and S o l ~ ~ t i o nosn Q v a n t u m Mechanics
232
and t l ~ cratio of the numbers of the two bea.ms of the particles is
b a
- - -
+
cos e 1 e - ~ vsin 6
0
-
cos 5 e-"v sin
and, with normalizatioll, the eige~lvcctor
For the first Stern~-Gerlaclla p ~ ) a r a t ~t,ake i s the dirc:c:t,ioir of the inagnc:t,ic: field as thr: z direction. Ta.kc n ;dong tlre rrlagnct,ic ficrlcl irl tkic scc.orid St,c:rri Gerlach apparnt,us. Tlrcn cp = 0, 0 = n ill thc above. If th(: part,iclcs which are sent iiito t,llc sccond S - G ; ~ p p a r ; ~ t 11;~vc: ~is spill up, we havti
c=
(T 71,)
=
2)
(i)
(COS((r/2), sill ( ( ~ 1 2 ) )
= cos (ct/2)
,
Thc~i~ii~gri(:ti~ nio111ent of a silvcr at,onl is csscrit,ially equal to the mag11c:tic: ~i~orric:iito f its ~inpaircdvalcncc c:lcctron which is p = -ys, where y = (!/?ILc i~11(1s is t l i ~ (:lc(:t8ro~1's spin. Si~pposet,lia.t a bear11 of silvcr at,oms having. vrloc:it,y V is passed t,llroligh a Stc:rri Gor1;lc:h ; ~ ~ ) I ) a r a tllaving us its field gradicnt iri the z dircctiori, ant1 t,lint only tjlic: 1)carrl with 7rr, = I L / is ~ t,o 1x1 c:orisidered iri what follows. This 1)c:ain t,hcn c>ritc:rs a region of l(:rigt,li L having a c:oristailt 111ag11c:tic fic:ltl Bo tlirc,ctly along thc axis of the: 1)cain (y-axis). It rrcxt, erltcrs anot,lic!r Stc:rri G(:rlnc:l~apparatlis idcrit,ic:nl to tlic first as sliown in Fig. 9.7. D(:sc:ril)t:rlvarly what is sccn wlicrl tkic bcarrl c:xits from the second Stcr~l-.Gor1ac:li n ~ ) p a r a t ~ iExpress s. t,lic: iiltc:~isitic:sof t,l~c:rc:sult,ing 1)eams in t,crrrls of V, L, Do and tkic: c:oiistarlt,s of the ~)rol)lcin. Us(: cl~i;~.i~t,urn ri~ecllariic:;ilccluatioris of motion to derive your result. (Berkeley)
Therefore, aft,er they lcavc the second S - G ap~)aratus,t,lic ratio of thc nunlbers of the two boains of particles is
I
c I\(:OS"Q/~) --
( d l2
sin" ( a / 2 )
=
c:ot"
(I
2
--.
If the particles which are sent int,o thc second S spin down, we have
-
G apparatus have
Fig. 3.7
Solution:
c = (T n 1 J z)
=
()
(COS( a / 2 ) , sin ( ~ 1 ' 2 ) )
=
sin ( a / 2 ) ,
If we took a picturc a,t the exit of the second S - G apparatus, we would see two black lines arising from the deposition of the two kinds of silver atoms with nt, = h.12 and r r ~ , = -h/2. Denote the st,ate of the syst,eill in the rcgion L by It). If we consider only atoins of m,,= h/2 in the beam t,hat enters the region L a t t = 0) then
Problems and So1~utionso n Q u a n t u m Mechanics
Spin and Angular M o m e n t u m
235
where pi = giposi is the magnetic momerlt of the i t h particle. T h e spin wave functions for the 4 states of the system, in terms of the eigenstates of the z-component of the operators ui = 2 s i , are
The Hamilt,onian of thc systeli~in the region of length L is
arid so
Hence a t the cxit. of the. region, t h r irlt.c~r~sit,ic!s of t)carris with rn, and m , = arc rc:spcc:tivcly
-2
=
2
whcre I. is thc irltcnsity of t,hc t)carn that entrrs the rc:giorl. Whcn the bearn leaves the rcgion L, t = L/V. So tho ratio of int~nsit~ics is c : o t 2 ( y ~ o ~ / 2.v )
(a) Find the cnergy cigenvalues associat,ed with cach stratey'ii. Discuss thc liiriiting cases p o H / A E >> 1 and p o H / A E << 1. (11) Assuine that an initial st,atc $(0) is prcparcd in which particle 1 is po1arizt:d along t,ha fit:ld directior~z , but particlc 2 is unpolarized. Firld the tin(. dcpcntfcllc:c of the polarization of partide 1:
Again discliss tllr limiting cases p o H / A E << 1 and p o H / A E >> 1. (Columbia)
Solution: (a) u ,
(1,
p may be represented by matrices
The splitting o f the 1)earn is scen when it exists fro111 t,he sccolltl StcmGerlach apparatus.
Two oppositely charged spin-; particles (spins s l and sz) are coupled in a syst,em with a spin-spin int,craction energy A E . The system is placed in a lllliform magnetic field H = H z . The Hamiltonian for the spin interaction is
for which the following relations hold: uix cr.i = pi,
uix p.L= a.i ,
uiy ai = iPL, uiVPi = - i a i , uiz ai = CYi,
uiz Pi = P
L.
Problems a n d Solutions o n Quan,tvm M e c h w ~ i c s
236
Spin and Angular Momentum
By the same procedure we obtain
Then as
H
= ( A E / 4 ) (a1 = (AE/4)
where we have used p
. az) - (PI
+p2)
.
H
( a l , ~ ~+, al,az, -1 a l , ~ , )
= gpos =
4 g p o a , we havc
(6) As particle 2 is ulipolarizetl and can be considered as in a mixed stat(,, its s t i ~ t cE2 (.a11 1 ) cxpancled ~ iri terirls o f (12 and PL:
wlicrc. 1 a and hellce El =
AE/4
1 2 (gl
- -
l2
=
11) l2
=
112. Tlicri thc initial total wavc functiori is
+ g2)
Similarly,
Then as
Thcn using the relations we have ~
$
= 2
( A E / 4 ) ( 2 d S - 1)$ 2 ,
S p i n an,d Angular M o m e n t u m
Proble,ms and ISolut~101~s o n Q u a n t u ~ nMechanics
239
we obtain A hydrogen aton1 is in a 'PII2 state with total angular rnomentum up along the z-axis. In all parts of this prol,lern show your computations and reasoning carefully.
(c) I11 thc: lirnit p,"H/AE
>> 1, i.e., :I: >> 1, we havc: 1
(a) With what probability will the e1cc:tron be found witahspin down? (b) Coinpiite the probat~ilit~y per unit solid anglc P ( 0 , p) that the electron will be found a t spherical angles 6, p (indepcndcnt of radial distance arid spin). ((:) An expt:rimcnter applies a weak ~nagneticfield along t,he positivc z-a.xis. What is the cffc:ct,ivc magilctic rriolnent of the atom in this field? (d) Starting from t,he oriRibal state, the cxperin~cnterslowly raiscs the magiictic: ficltl until it overpowers the fine struct,ure. What are the orbital and spir~quailtl~mriurn1)ers of t,llc filial state'? [Assn~ntrthe Ha.inittoriia.rlis lincar in tllc rriagnetic field.] ((:) What is tlic c:ffcc:t,ivt: nlag~lctic:i~iornc~iit of this final stat,e? (Berlcelcy)
AE g1+(/2 E1 : poH = -5 (1' 1 + '12) 1 ~ o H, -Soll~tion: 4 2 (a) I'or the statc 2 P , / 2 , 1 LIE 1 E2 =
-
(2fi17
I) = ( A E / 4 ) x 2:r
= - ((12 - 91) , " O H ,
2 1 = - -2 ( 9 ' / l ) p O H , E:3 1 A E- + 91L+ f i E 4 = poH = (91 -192) p o H . 4 2 -
4 AE (1 + 2 4
-
-
= 1, s = 112, J = 112,
Jz = 112. Trans-
forming thc colipliiig rcprcsc~ritatioiiinto the uncoupling representation, we ~ ~ L V C
-
5
PI, (t) z5 1 . When p o H / A E
<< 1, i.c., z << 1, we have
Thcrcforr~PL = 213. (b) As
we have 1 P ( 0 , p) dfZ = ; 3 (2YA Yi1
+ YA Yio) df2
Hence the prob;~hilityper unit solid angle is
pi, (t) = 1 - sin2 (AEt/2h) .
240
Prob1em.s and Solutions o n Quantum Mechanics
Spzn and An,gular M o m e n t u m
(c) In the weak magnetic field, J and J , are "good" quantum numbers and the state remains unchanged. T h e effective magnetic mosnent is
24 1
At tiirie t =; 0 the state of the system is J$(t = 0)) = I+). The particle moves along the y-axis through a uniform magnetic field B = By oriented i~l011gthe y-axis. (a) As expressed iri thc I+), I - ) hasis, what is 1 .J,(t))? (t)) What would be t,he expc~ctationvalues for rneasurcnlent,~of the o1)servnblcs s, s,, .s, a.s functions of time?
where m is t,he electron mass and
(CUS) Solution: Hcnce p = e\i/6rnc. (d) In a strorig niagnetic field, thc ir~tcractiorlof t,llc 111:rgnctic inorncilt with the field is rnuc,h stronger than thc cou1)ling intoraction of spill a i ~ d orbit, so that the latter can be negl(:ctcd. Here 1 ; ~ r i t l s arc good qliaiitliir~ nurnbcrs. T h e Ha~niltoriianre1atc:d to the magnctic ficltl is
When thc: magnct,ic fic:ltl is increnscd slowly from zero, tlic: stat,(:remains at the lowest, energy. From the exprcssio~lof W, wn sc:c: that when thc magnetic field becoincs strong, only if 1, = h , , .s, = -1t,/2 can t,he stlate remain a t the lowest energy. Thns tlic: clusi~tulnnulnbcrs of t l ~ cfilial stat,(: are 1 = 1, 1, = -1, s = 1/2, s, = -1/2. (e) the effective magnetic momcr~tof the filial state is
(a) The H:~iniltoniaiiof thc, particle is
In t,llc: rcyrc~seritationof i2,d,,
t ~ i i t 1so
th(: two (:ig(:iist,at(:sof .qll 1~rc
1
~ I . ; , , = ~ h ) =2 - ~ ~ ~ h l ~ , = i ~ ) , fils, =
3043
-$
n) =TB
a
hisy =
-+
h),
ally ~ t i of~ the t ~part,iclr car1 be expressed as
d m ,
Consider a neutral part.icle wit,h intrinsic angular rnoment~i~n where s = h/2, i.e., a spin-1/2 particle. Assurne the particle has a magnetic moment M = ys, where y is a constant. The quantum-mechanical state of t.he particle can bc described in a spill space spanned by the eigenvectors I+) and I-) representing a1ignment.s parallel and ant(iparalle1 to the z-axis:
I
I d ( t ) ) = c l .s,
=
i -
+ c ~S1/
fi) exp
(+
- - h) exp
i
(
1
-
The11 the initial condition
I $J(t= 0)) = / s,
=
h/2)
y ~ t ) i
a
y ~ t .)
Spin w r ~ dAt~qularM o m e n t u m
Problerns and Solutions 0.n Q,uantum Mechanzcs
242
(s,)
= 0,
because (s,)
=
0 at t = 0 and s, is conserved.
and so
+ 1
=
(i)
i 1)
1
=
.
+
=
1 -
fi
(i) fi
(-i 1)
=
1
--
. z.
Therefore? '
,I, 1 ,$(t)) = 4 fi
(
vxp (,i
;
s
(
t
)
1 =
,
(1)) (,5,)
=
(,,, (1
,~t)
fi(>l(l
B = B o z + Blcoswtx- Blsinwty,
fi 0
A partic 1e of spin 112 aiid inagrictic r~lonlcntp is placed in a ~rlagrictic
y ~ t )
-
sin
)
(a
s i n
.~t))
(1
Is,
? ~ t )
TL ( lo
')
-1
which is oftc.11 crnployctl in ~nagncticresonance experiments. Assume that tlic particle has spin up along the +z-axis a t t 0 ( n ~=, +1/2). Derive thc probability to find the particle with spin down (m,, -112) at tirnc t > 0. (Berkeley)
-
=
2
-
Solution: Tlie Haniiltoliian of the system is
Letting wo = pRolh,
W1
= PBlIf2.,
we have
-sin
(;
=
y ~ t )
-
2
h cos (yBt) .
H
=
-
p(Boa,
+ Blu,
cos wt
-
Bla, sinwt)
2 44
Spin and Angular M o m e n t u m
Problems and Solutions o n Quantum Mcchanics
giving =
-
(
-
wo
+
); *
Tlicreforc the gtncral foriri of
Let the wave functiori of t,he systcrn be
J(-W,
+~
0 is
p = A2+ exp (i62+t) + A2- exp (if)-t) , t111tl that of a is rr
Tlir Schriidiriger equation ihat I t ) = H I t ) , or
=
fi cxI, j i ( 2 w o + w) t ] iwl
+ f 2 . _ A 2 (:XI) (if2Lt)l . Iiiitially tlic spi11 is
111)
alorig t,lic z-axis, so
Try a sollltioii of tlic, type,
r
S ~ b s t i t u t ~ i oirin t,he abovc oquatioiis gives
I lic sollitiori is 7
Asslilrlr t,hat cu ant1 ,O liave the forms h(t) = exp (-iwot) P ( t ) = exp (-iwot)A2+
x [exp (i62+t)
-
exp ( 6 - t ) ]
where A l l A2, i~11df 2 are constants. Suhstitutiori gives
+ w + 0)Al -wlAl + RA2 = 0 . (-2wo
-
w1A2 -- 0 ,
For this set of equations to have ilontrivial solutioiis t,he det,errniriant of the coefficierits of A l , A2 must be zero, i.e.,
-
iwl exp [-i(w/2)t] J(w0 ~ 1 2 )w:~
+
-
x sin ( J(wo
-
,;.
1 2+) ~
~ 1 2+)W: ~ t )
Problems and Solutions o n Quantum Mechanzcs
246
Spirt. a n d Angular Momentum
9 (i)
The probabi1it.y that thc particle has spin down along the z-axis a t time t is
sin 2
(1)) The Hainiltoilian ill tlic interval 0
A spin-; syst,c~nwith ~nagrict~ic ~ n o ~ n c ipi t = pea is loc:at,c:d ill ;i ui~iforrn til~~c-il~clc~)crldcl~t magnct,ic: ficltl Bo in t,lic. positive: z tliroctio~~.R ) r tlic tilnc interval 0 < t < T arl atltlitioli;il illi if or in t , i l ~ l c : - i i i t l ~ ~ ) c ~ ~ fi(!l(l ~ c l ( ~B i ~ It . is ;ipplictl ill thc: positive, :I: t1irc:c:tioil. During t,liis iiit,(:rv;~l,t,110syst,(:ln is i ~ g ; i iin~ :i ~iliifornl(;ollst,:ilit, iliag~l(:t,i(~ fi(:l(l, but of (liff(:rc,iit il~i~giiit,il(lc~ :1,11cl dirc,c.tioli z' fro111 t,llo init,ii~loil(,. At ;i11(1 I)(:for(: t -- 0, t,li(: syst,(~iliis i l l t,hc I ~ I= , 112 statc with rc:spc:c:t to t,llc: z-iixis . (a) At t = O+, what arc: thc: ailiplit~ltlcsfor Iilitlilig t . h syst,(>ili ~ wit11 spin projc:c:t,ions ,rr~.'= f112 with rcspc:c:t to thc z' clirc:c.tioi~? (1)) What is the t,imc t1cvclopmc:nt of t,ho clicrgy ( \ i p , ~ l l ~ t , iwit,li ~ t , ( ' r(:sp(:ct ~ to t,lic z' dirc:c:t.ioii, tlurilig the t,iln(: i~it,(~rv:il 0 < t < T'! (c) What is t,lw prol)al)ilit,y:~111plit,li(l(: :it t = T o f o1)sc~rviilgt,ll(l systcili in t,hc spill st.;itc 7r1. = -112 along t,lic: origillal z-axis'! [Express your answers in trrills of t,lic. i ~ i i g 8l ~l)~t,wc:c~i t,li(: z i ~ i 1 c 1z' ilx(:s and thc frcqllellcy wo = poBo/lr.] (Bf!~.kf:/f!!/)
=
247
8 2,
< t < T is
The initial cigcrifiiilctio~isarc>
Slil)st,it~itioliin t;hc: Scliriidinger cquat,ion H x * (0)
At a later tiiiic t ill 0
=
*Ex+ (0) gives
< t < T, the eigenstates are
Solution: (a) 111the rcprcscntation of sZ, the eigcnvrctors of s , ~ arc:
x*(t)
=
cxp ( ~ i E t l h ,x*(O) ) = exp (*ipoBt/h,) x*(O).
(c) The probability amplit.ude a t t = T is
C- ( T )
corresponding to the eigenvalues s,, = 112 ant1 -112 respect,ivcly. Then the probability amplitudes for *m,' = *1/2 arc respectively
= (0
=(
-
1)exp ( - 2 H T l h )
i ~ iJ/
m
) sin ( p o J~ m / j i )
i sin 8 sin (poB T l h ) .
Spin and Angular M v m e n t u m
Problems and Solutions on Qvantum Mechanics
248
In the Schrodinger equation
An alternative way is to make use of
+(0) =
(A)
= cos
e
X + ( 0 ) sin -
e
X(O),
-
2
setting
and so
$(t)
= X+
( 0 )cos
e -
2
exp ( i p g B t l h ) we gct
e
- X - ( 0 )sin - exp (-ilroBtlh)
2
,
to get
C-(T)
= P+.J,(T)=
cos
8 . 0 sin -
-
2
2
Try a so111tionof tlic typc
x {exp (ipoB T l h ) - exp (-ipo B T l t l ) ) , = i sin 0 sin
poBT , f
-
wllere
=
wllcrc A, B i~iid12 arc c:orlst;r~its.Sul)stit,ut,iorigives A spin-+ system of magrietic rrlorrleilt p. is place
wliere
-
Why is the energ; of tllr spin s y s t ~ mnot consc~rved'~
(Columbia)
Solution: T h e Hamiltonian is
For riontrivia.1 sollitions we require
= - p(Hgz =-
(
cosw0t + H g y sinwot
+ Hog,)
Ho H exp (-iwot) H exp (iwot) H o
t
Problems and Solutzons on Quantum M~chanzcs
Spin and Angular Momentum
251
giving
Thcrcfore thc wave function of thc system is w'
$(t) = - isin (Qt) exp
where
Q
+ the subscripts 1, 2 c.orrcspondirlg to the valucs of 62 with tively, arid
At t = O the systeni is in the, B z = 1, Al + A2 = 0. Then as
I - +) state and
+,
(1 =
-
signs rc,sI)cc-
I
(w
cosQt
2
--
+
;
Q
wo) sin ~t
and the energy of thc system is
(0
Thus Bl
+
Note that as c=(+ilpz+ipy/-i)
we have
=p
C
( :+I -
p=-.
2
I:)
nz+iny - -
l
exp (i
t)
p,
Problenis and Solutzons o n Q u a n t u m hlech,anics
Spin and Angular M o m e n t u m
cos w2t
As the energy of the system changes with time t , it is not conserved. This is because wit,h regard to spin it is not a n isolated system.
i
sin wz t
(b) The 111ei~r1 value of the spin in the state / x ) , i.e., the polarized vector for ricutron. is
A beam of neutrons of velocity v passes from a region (I) (whcrc the magnet,ic field is B = Ble,) to a region (11) (whcrc the field is B = Bze,). In region (I) the beam is completely polarized in the +z direction. (a) Assuming that a givcn particle passes from (I) t o (11) a t timc t = 0, what is the spin wave function of tha.t particle for t > 07 (b) As a function of timc, what fractioii of the part,iclcs wotild t)c observcd to havc spins in thc +J: dircc:tion; the +?j tlircct,ion; t8tlt + z direction'? (c) As a practical ma.t,tcr, how abrupt must the transition l)ct,worrl (I) and (11) be to havc t,hc above dcscript,ion valid? ( Wisconsin,)
Thus in the region (11), the ncutrori spin is in thc ?jz plane arid precesses about the 2: direction with angli1a.r vt?lot:it,y2w2. ( r ) For the dcscriptioris in (a,) arid (1)) t,o t)e valid, the time of the t,rarisitiori b(.twccrl (I) arid (11) rnust satisfy
-
For example if B2 10" Gs, the11 t << 0.7 11,s. If tlie kinetic ericrgy of tlic i~icid(:iitn(:litrons is givcn, we riLn c;tlciilate thc 1ipp(:r lirnit of t,he widt,li of the transitioil rc:gion.
Solution: (a) Coi~sidcringoilly thc spin wavc fiinct,ioii, tho St:liriiclillger oclu t,'1011 CL
is
ihi3 I x ) / a t
/
= H S) ,
Tlic Harniltoniari for a (p+ F ) atorn in the n = 1, 1 = 0 state in an extc~riialrnagr~cticficltl B is
where
H=-p
.
B
= - / L , ~ Bzu,
,
with p,, = -1.9103 p~ l~eingt,he a.iiolnalo~isma.gnctic nioinent of t,lic!rlelitron, LLN = eh/2,tn,, c t,hc nuclear magneton, m, thc ma.ss of a prot,ori. Thus
where wz
=
7 1 I R2 . Let I S ) = hLpL
(i). The above givrs
The initial condition is a(0) = 1, b(0) = 0 as the beam is initially polarized in the +z direction. Solving for a and b and using the initial condition we obtain for t > 0
(a) What is the physical significance of each term? Which term dominates in the interaction with the external field? (t)) Choosing the z-axis along B and using the notation (F, M), where F s, +s,, show that (1, + l ) is an eigenstate of H and give its eigenvalue. ( c ) An rf field can be applied t o cause transitions to the state (0, 0). Describe qualitatively how an observation of the decay p+ + e+v,v, could be used to detect thc occurrence of this transition. ( Wisconsin)
-
Solution: (a) The first term in H is due to the magnetic interact.ion between LL and e, the second and t,hird terms respectively account for t,he magnetic
Problems and Solutions o n Q u a n t w n Mechanics
254
-
int,eractions of / L and e with the external field B. Of t,he latt,er, the term 1 e I s, . B/m,,c is dominant as m., m,/200. (b) As
4. M O T I O N I N E L E C T R O M A G N E T I C F I E L D
W(: Itlay gc:ric:ri~lizcthe st:ini-c:l;~ssic:alBolir-Sonlnlcrfcl(1 relation
Corisidcr t h e state
AS tlic t:igciivalucs of F" s,; s:, s,, s,, l y , hi^^(: 1) i i ' , h, 11. r c s ~ ~ c c t i v ~w(:
1 1
5( 5
+
4
arc: 1(1 + 1) TI:"
,
( + + l)lr.",
(where t,lic i11tt:gral is aloiig a c:losetl orbit) t,o apply t o thc cast: where elect,rorliagiictit: fic:ltl is prcscirit 1)y rc1)lac:irig P with p - f:A/c! w1ic:rc: r: is t h e c:lii~rg(:of tlic: 1)a~rtic:le.Usc this aritl thc cquatiori o f iiiot,ioii for thc: 1ino;rr ~lloiric:nt,riirip to tlcrivc: i~ qriimtixation c:oiitlitiorl 011 tho ~ i i i ~ g i ~ tflrix ~ t i (of : a seirii-c:l;~ssic:alcrl(:c:t,roii w1iic:li is ill ;L tn;~gric:tic:fieltl B ill i ~ r ii ~ r b i t r ; ~ rort)it,. y For t , l r c t r o ~ ~ills solids this c:oiitlitioii (:nu t)ci rc:stntctl ill tc:riris of t h t ~size .S of t , l ~ cort)it i l l k-s1)i~c:c:. 01)tiliii t,li(: ( l l i i ~ ~ l t i ~ i ~(:oii(litioii t,iori or1 S in tt:i-111s of B (Igilorc: s ~ ) i ic,tF(:c:ts). i ( C1ric:ngo) Solution:
w1ic:rc p is t,lic: c:i~lioiiic:a.lrnoinentum, e is the charge of particle. T h e genera1ixc:tl Dolir Soillrrlc:rf(:ltl relation becomes (c) T h e decay p + + c+v,:v,, (:an h(: d(:tcc:tctl t l ~ r o ~ i gthc l i o1)scrvation of the alinihilation of the positroili~ime+e- + 27. For t,he stat,c (1, + I ) , the total angular rnornc:nturn of the e+e- system is 1 , a.rld so c+e- cannot decay into 2 7 whose total angular ~nomentumis 0. For the state (0, O ) , the total angular morncnturn of the e+e- system is 0 a n d so it can decay into 27. Hence, detection of e+e- + 27 implies the dct:;ty p + + e+v,v, of t h e (p+e-) system in t h e state (0, O), as well as t h e transition (1, +1) + (0, 0 ) .
whcrc
& = /;B d s = L ( V x A ) - d s =
f
A d r ,
using S t o k ~ stlieore~il T h e classical equation of the rnot~oliof a n clcctron in a constaiit magnetic field B,
gives p = - e r x B / c and
PI-oblerns and Solutions
or1
hfotion i n Electroinng71,~t.l~ Field
Quantum hfech(~n,Z(:s
257
Solution: = ~ L 2 ~ d s =
(i~T ) h e Schrij(1ingcr equa,tion for the particle is
8.4 at
i l v 2 V q) -- q E . r$ 2,) 11, (1)) Tllc, I-Iaiililtoriiari o f t,lic: p;~rtic:l(> is
Heiicc
ill, - =
4 = (71 -t 112) 4 0 , whcrc $n = hclci. Defining k by p to B,
=
N
/t,k = -c:r x B / c , wc havc, ;tssli~rlirlgr is ~)c:r~)c:iitlic:lilitr hAk =
-
p2 27rt,
I : :
-
([E . r
E
,!..I . ] ------
-
.
Bc AT/(:,
or
(ips
A T = -hi:Ak/Bc.
1 -
Thcr~:forc,if the, orl)it oc:c:upic:s i r i i arct;~S,,,irr k-sp;tc:c: ;~iid:1.n in r-spa(:(:, wc: hi^^(: t,li(: rclilt,ioil
-
A,,
(/K:/
Art
tlt
i~.~(!it
aiitl
-
il,
I[ zit,
/
Bd*,,
- (k)' /
(s(r)
n ~s ). ~
4 )
S ,r,,
=
=: (1Ex
,
~
~. .
(P)
(E)"""
('"
:~-:
'
't?rBc(rt,+ 1/2)/It,c
.
([E
(11
112) hi:/(: ,
( )
:
tlt2
we havc:
[TI:,; , : I ; ]
,?,,
2
B'".TTL =
%h
1ic~iic:c~
AS =
(1E,,
1
[
.
4002
A pa.rticlc of c:h;~rgc:q arid 111;~s~ 7n is sul)j(:c:t t,o ;L uiliforrn c:l(:c:t.rost;~tic field E . (a) Write dolvii t,tlc tirrle-dcpcridcnt S(:liriidii~gcrc:quiltioii for this system. (b) Show t h a t t,hc cxpcctatioil value of tlic posit,ioil operator o1)t:ys Newton's second law of inotioii whcn the particle is in ; L I arl)it,r;~ry ~ sta.tc $ ( r , t ) . (c) It can be showri that this result is valid if tlicrc is also a. ~iiiiforrrrmagnetostatic field present. Do thesc rcsult,s have airy practical applicatioli t o the design of instruments such as mass spectrometers, particle i~(:(:~lerators, etc.? Explain.
-
In
dt '
ti" (r) = y E , dtL w1lic:li is just N ( : w t o ~ ~sc:c:oricl 's law of 11-lotion. ((:) T1ic:sc: rc:slilts show t,liat we could use classical mechanics directly wlic~riwc: i ~ r c(:oi~i~)lrtirig t,h(: t,ra,,jc:ctoryof a charged part,icle in instruments sucll a.s 111;~~s s1)(:(:tjro1ii~:t~:rs, particle accelerators, etc. TIL
4003
The: H;~iniltolii;~r~ fix a, spinless charged particlc in a m;lg-netic: field B=VxAis
where e. is the charge of particle, p magnetic ficlci B = Bee,.
=
(p, p,, p,) is tjhc: morncrltlliii c:o~ijugate
,
Motion i n Electro.mag7~eticField
Proble.n,s and Sol,~rtionson Q u ( ~ ~ t t 7 ~M7~n c ~ , ( L ~ L ~ c s
259
Hcllce t h c ellergy levels for the system are
( a ) Prove tfhat p, ant1 p, a r e constarlts of motZion. (b) Find t h e (cl~la~ltum) energy levels of this systcni.
E ,=&2rn+
(,n,+ 1 / 2 ) T w 1 n = 0 , 1 , 2 ,
(M I T ? Solution: T h e I-I;r,n~iltoniarl for t,lie particlc ca.rl l)c writ,t,c:ii ;IS
-
Ari clcctroii of Inass rrl arlcl c:hargc -c rnoves in a region where m;rgiic:tic ficltl B V x A exists in z direction. (;I) As H docs liot tl(~pcndon z a.iltl z oxl)lic:itly, the? l~asic:c:oirlirlut,;~tio~i rel;ltions iri qlli~1it11111 iri(:(:liani(:s
[x:.,p j ] =
%!ik,i, [ p i ,p:,]
=: 0 ,
rc:qliirc: [ p E HI ,
=
0, [p,, H] :~: 0 ,
w1iic:h show t,liat p,, 11, 'LK: (:onst:tiit,~oftlic 11iot;ioli. (1)) In vicw of (a) wf: (.ilrl c:lioosci {H, 71.. , pz } i1.s :I. (:01111)1(:t(: s(:t of I~I(:(:~I;LII~(.;L~ v;~rial)les.Tlic: c:orrc:sporitlirlg cigc:~lhiiic:t,ioiiis (:I;, ,,(,, z) := ( : % ( : I . T ) " + ~ P)"I,~ *(?J) , wli(:rc: p,, p, arc IIO lorigcr opc:r:~t,ors I)lit ; ~ r ciiow c:oiist;iiiits. S(:liriidirigcr cqli:rtio~i H$(x:, y , z ) =: E$(:I;, yj, z)
:L
uniform
(a) S(:t 111) thc: Sc1irijtliiigc:r c:cluat,iori iri rcc:tanglilar (.Oordi~li~t(:s. (11) Solvo the, (:q~i;~tioii for i l l 1 c:ri(?rgy lcv(:ls. ((:) Ilisc:~isst,lic: iiiotioii of t,lic: c:lcc:tron. ( Br~ffalo)
Solution: (ir) T h c I~I:~iiiiltorlii~~l is
'?/J
DA, 2.z aAy
wc ( ~ 1 1ti~k('A, = A, Sc1iriiciirlgt.r equ at .ion as
=
dA, =0, dx aA, -B1 az ay 0, A?/ = Bs,i.e. A
=
B z y , a n d write the
(b) As [R,,H I = [ k H , ] = 0, Py and P, a r e conscrvrd. Choose H , P,,,P, :rs a c o i n p l ~ t eset of mechanical variat~lesa n d write the Schrod-
we can write t,kic cqll at 'ion as
inger equatiorl as which is the energy eigclicql~ationfor a oiic-di~ncrlsio~~:~l liarirloriic osc:illator. T h e energy eigenvalues are therefore El
=
E -p2/2m
=
(n, + 112) iiw, n = 0 , I , 2 ,
. .. .
Let ( = :t. + c P , / c B ,
& = pT. T h e n [[, &] = ih and
B = V x A Tl~cxScliriitlingc~requation is t h r n
A(r) BC(:~LIIS<: t,hc (:xpr(:ssion of L;il ~ O C S11~)t ( ~ O t i t i ~I?!) i ~ ic:xplic:it,ly, t;lic: tl(tg(:rieracies of t h c cnc>rgy lcvcls are iiifiriit,c,. ((:) Iri tlic coortlinatt: fr;~iii(i(:liosc11,t11c t:ilc:rgy c:igc~~st,iit.c~s c:orrc:spor~(lt,o free inotiorl ili t 1 1 z~ cliscctio~ia11(1cis(:uli~riliot,ioll ill t , l ~ :I;r .- ;I/ 1)1;~1i(:, i.c. :L he1ic:i~Imotion. 111 tlit: z tlisc,c~t,io~l, tho ~ i ~ ( : ( : h i ~ i ~111011i(:iit;11111 i(:;~l , ~ I L ~ I , P , is corlsc~rvc:tl,tl(~sc:sil)iiig ;L ~ r i ~ i f i ) slinci~r ~ i i niotiori. 111 t,lic: 3: tlirc:c:t,ioll t,llc>rc:is i~ siml)l(~liil~~llO~ii(: os(:illiitioli so11ii(l t,Ii(:(:cl~lilil)ri~lin poitit, :I: .-:: -~~~(:I'!,/(:B. 111 tht; (lir(:(,t,io~~, tll(1 ~ ~ i ( : ( ~ l ~ ; ~Ir~iIi~(I:I;I~( :lI ~ ~ , I is I I I TI ~ L . O ! , := I:,) .I- (;B:I;/(: : rB
+ A r ( r ) = A ( r ) -1-
V,f (r) ,
--
Write (low11 tllc 1ft~1riiltoliit~1i for i~ spirll(:ss (:llarg(:(l pt~.rt,i(:l(: i11 i i I I I ~ L R netic ficltl. Sliow t,hat thc ~ ~ L I ~t ,Kr Ca ~ i ~ f ~ r ~A~(ir i) ~--1t iA~ (~r )l 1. V / ( I - is ) equivialcrlt t o rri~ilt,ii)lyi~ig tlit: wave filnct~ioli1)y thc: fiictor (:XI) [ i c : / (r)/l~,c:]. W h a t is t 1 1 signifi(:;ili(:(: ~ o f t.11is rcs~llt'! Corlsi(l(>rt,lic: (.;Ls(:o f :L 1iilifor111ficl(1 B dircc:tcd i~1011gt,li(: z-i~xis.Show t,lii~tthe (:~ic>~gy levc:ls c:tLrl I)(: writ.tc:il ;LS
T
L
I~,~ki B+-. 'rr~c 2111 Discuss t , l ~ e~llliilitiiti~e fcatllrcs of t.he U ~ R V Cfini(:tiolis. Hint: use t,lir gauge where A, = -B?j, A, = A, = 0.
E
=
(n.+ 112)
-
( Wisconsin.) Solution: T h e Hamiltoniaii for thc partickc is
where A is related t o the magnetic field by
S~il)st,it,~rtiori iri S(:1iriitliilgcr1scquat,ion gives
This sllows tllat ~rrltlrrthe gauge transformat,ion A' = A + V f , t h e Schriitlirigrr cclu;~tiorlroiriains the same and t h a t t,fiere is only a. phase differcricc 1)ctwc:cri tllc: original a ~ the d new wave f~lnctioris.Thus t,lic syst,rrn has g;llige irtvariarlc:c:. Now c:onsiclcr tlrc c;ise of a uniform field B = V x A = Be,, for which we have A,=A,=O. A,=-By, T h t H a r r ~ i l t o i ~ i(:an i ~ ~ be i written as
M o t i o n zn Electromc~gr~etic Field
H dot's 11ot (1(:1)(:1id011 3 : , z ~xpli(.it,ly, Sitlcc, [ f i x , H] = [$,, H I = 0 we may t:hoosc tt~c:co~riplctcset of rricct~ailit~al v;~ri;il)lcs(fi,, $,, H). Tllc c:orrc*s~)o~i(li~~g (:ig(:~~st;~t.(> is l/,(f,;;
;lj,
z) = (
263
Solution: (a) Cl~oosc,IL galigc A = Bo : ~ yp, -V p = Eo. T h r n
=
E o x so t h a t V x A
=
Boi,
x (!/I
; J ( I ~ - ~ ~ ~ I J z ~ ) / / ~
S ~ ~ l ) s t , i l ; ~it~ ti;~i ~ ~tt ,g1o1 ~S(:lit-ij(li~~~;(:r (~(111;~tio11, WP t1;ivo
As II tlocs not tlcpc:~lclon y aritl z r:xl)licit,ly, p, ant1 p, r:;ic:h commutes so t,hi~tpy ; L I I ( ~p, iirC COIIS(:~-VC(I. T ~ I ~thr:y I S (:all I)(: rcplaccd by with t,hc:ir ~ i g c ~ ~ v i ~(lirc(:t,ly. l ~ i c s HCI~CC
wlic~i-c.X:,
-
p r / l ] , , ;111(1 tali(,W;LVO f i ~ ~ ~ ( : t i~.rr ,io~~s
,q,
(:c,
;lj,
z)
:f ; 1 ( 7 1 c
.r t l J z z ) ' / ' ,
,Y,II,(!I
- ?jo)
I
arc. a 11c:w pair of c:o~!j~~gatc: v;~ri;~l)lcs. L(:t w :- / q I Bo/rrrc. B y corril)arirlg thc cxl)l.t-ssio~~ of H with t11;it for a o r l c ~ - t l i ~ r i c ~ ~ i shi~r~rlorlic: i o ~ ~ i ~ l oscillator, I C SH: we gct tlw O ~ ~ ( ~ I I V ; L ~ I of
E,, for c.lic,rgy tloc:s 11ot HI, t)c,i~igEI(:r111it,cp o l y ~ ~ o ~ ~AS ~ i t,h(, i ~ l (:x~)rossio~~s ~. del)entl ~ I p,,: I a ~ i tp, l c,x~)lic:it,ly, them ;~.rc,i l ~ l i ~ ~(l(~gv11(~1.~(.ios it(: wit,h r t , s ~ ) ( ~ : t , t o p, autl p,.
=
(r,
+ 1 1 2 ) i l i ~-1- p"/2,rn
-
, r r t , C " ~ i / 2 ~ ; c p W E ~ /, B n ~ = 0,1,2, . . . -
Thc: fii(:t t , l ~ aonly t p, ;r.ricl p,, hut not y and z , appea.r in t,he expression for c,ticrgy i~~(li(:;itc:s illfillit(: tlf:g(~rl(:racyexists with respect t,o p, a.nd to Pz.
(1)) A st,atc: of zcro I I I O I I I ~ ? I ~ ~ \sLiIgI r~ ~ i f i ~~ sI I Cill which tkic: eigenvalues of p , ;x11(111, as well as tlic, e x l > c ( . t i ~ t ivalue ~ ~ l of p , arc all xc,ro. As vc.loc.ity is
4006
A point l);~rti(.l(~ Of l11;Lss 7n iilld (:h;irg(: f] I I I O V ( ' S ill ~1)iltially(:Oll~til.ll~ crossed nla.gnctic ant1 electric ficlds B -:Uoz, E = E O X . (a) Solve for thc co~nplctcenergy spct.t,rru~l. (b) Evaluate the expectation value of t h e velocity v in a st,ate of zcro rnomerlt,um. (P.r.irt,ct:ton,)
.
value is its expc:c:tatio~~
Then as
cp, mc"Eo nlc2Eo ( L ) = ( ( ) + - + ~ = - , qBo qB, qBo"
I'~.oblcrris U
264
since
(6) -- O
I L Solu,tior~s ~ 071
Motron. in, E 1 c c t ~ o m n g n e . tField ~
Q u , a n , t ~ ~Mechnrrrcs in
for a llnrinonic oscillator ant1 p, = 0 , we 11;tve (v) =
CEO y Bo A
--
whc:rc: HI, i ~ r (H . ~ ~ r ~ i polyt~oillii~ls. ~it,c~ As 110 o!, t.c,riiis oc.cxr it1 t,tlo oxprc~ssiorifor c:rlc,rgy lcvds arid py can be ally :~rl)itraryrc,al 11111111)(:1., t l ~ ccl(\fi(>~~(:r;~(:i(:s of (:11crgy l(:v(:ls itrc infinite. 111c: c:igc:rihii~c:liolls for 1,11(:o r i g i ~ l i s~yl s t ( : ~iLr(: ~ th(:r(:for(: r 7
4008
H.l/!=
1
.
fi4
m/,
[ 4.-fix -1 @!I - F ~ ~ : r xc): /2] ,l/, = , 2,/11, whcrc: c is thc: c:lc:c,t,roi~c:l~a.rgc:((: < 0 ) . f i e ] = 0 , w(: (:all <:~Ioos(: il, ;i,?,, fi, iLS ;t AS [fI,fi,] [fI,f i z ] = 0, corr~plc>t,(: sc:t of' ir~c;(:l~t~i~i(;;~I v:tri;~l)l(:s,tho (:or~(:s1)o11(li11g (:ig(:~if~rr~(;I;ior~ l)(,ir~g ?/) ( , ~ ( / J ~ , ! / - ~ / J $0 * ~ )(:I.) / ~ ~,
-
y z
whcrc p,, p, arc: arl)it,rary red rlll1111)c~rs.Slil)st,it.lltioliof dingcr ccl~ii~tiori g-ivcs 1
[ fi: t ( cH o / c ) "2: 2711. where Eo = E p ' i / 2 ~ ~ 1or ,, -
/ e HO)"
,l/io
=
,1/)
in t,hc: S(:hrii-
( ' o ~ ~ s i t l:L ( lool) t ~ of t,hiri wirc~ill tlic: slit~pc:of ;t c:irc:lr of riitlil~sR (Fig. 4.1). A ( ~ ) i ~ s t l~li~grl(:l,i(; , i ~ ~ ~ t , fi(:l(l~ ) ( ~ r ~ ) ( ! ~ i (t,o l i t,l~(: ( ~ ~1)lii11(: ~ l ; ~ of r t,l~(: loop pro(l~~(:(>s 2i rn:igr~(~Li(~ kI11x 1)i~ssillg tliro~rglit11(:loop. 111ii~giri(: tll;it, tall(,wiw (:oiltairis orlly on(, c:l(tc,t,roi~ whic.11 is fr(:(x to 111ovc:. This ol(.c:trorl has ;L wavc furictiori $ ( O ) whir11 tl(:pc:ritls orlly 111)orlI,hc: ;tl~glllitr(.oor(liil;~t,c 0. Ncglect all interactions bct,wc:c:li tali(:c:l(:c:tror~ spill a r ~ t lt,hc i ~ ~ ; i g l ~ ( fic:ld - t i ( : a s well as all magnetic ficltls l)rotluc.c:tl 1)y t,hv c,l(,ct,rorlitsc,IE.
Etlc/,(~ ,
-
Fi2
- --
-
2
tl",(i/o/d:x
2111,
+ 711,2 (H0e/crrt,)"(:c -
-
x o )2 ,dio = EIJ'l/jo ,
where z o cp,,/cHa. T h e last equatiou is thc cr1c:rgy c:igc:nccluittion of a, orie-tlilnc:llsio~~iil osc:illat,or of na.tural frc:quc:rlcy wo = H o e / r n c arid cquilibri~nrl~ ) o s i t i o3:i ~= :~:o, t,he energy eige~lvalllcs1)c:irig
E O = ( n + 1/2)Fiwo, or
E
= p;/2m
-
71
= 0,
(n+ 112) Hoeh/rnc,
Fig. 4.1
1, 2, .. . , n
=
0 , 1, 2, .. . .
(a) How tloes thc grorlrid sta.te energy of the electron t1epc:ntl 111)nn t,hc v a h v of t,hc, :~pl)lic:d~llngl~ctic: ficld in t h e approximi~t,ior~ wc have drscribrd? Derivc it forr~iulai ~ t 1 ( 1givf, a rough picture of the result.
Problems and Sol~~t,Lon,s o n C)~~ar~trirn Mt~cl~ixrrzcs
266
Motion in Electromagnetic F,ield
(11) Iiliagine that we sirart with the wirc in its ground state ill the prr:sence of a magwt,ic flux 4. Next the nlagiletic field is tllrrictl OR. Calculate thc: currc:lit in the loop. (c) C;tlculat,e the currelit ill airips assurliiiig R = 2 (,ill anti tp 0.6 gallss clrl '. ( Clt.i~tl~l0)
Note t h a t
267
+' = +'(O) and $ = -$ $. Thus we have
-
Solut,io~~:
with solution
+'($) -- elC1' , where cl is a constant given by E
-
(11) 111 (:ylin(lri(:111coo1.(iilia.tcs 7-, 0, Z , HS V x A - - B gZ wlierc: 11 is a (:onstalit, wc: ~1111t,ak(: A,,. A, = 0 , A , = $, i.t:., A =60, and c:olisicicr the Scliriidiligc~(:(llii~t,ioii for t , l i ~(:l(:(:t,l.oli,
9
=
A'c' &.
Thus
+ eAR/ch) $1 . For single-valuedness, +(O) = $(B + a n ) , i.e., q(O) -- exp [i(cl
where n is zero or a n integer (0, *l, *2, . . . ). Solving for cl we have
and hence
:= (:XI)
(2
A
/'r
-
tlx
)
where q5 = 7 r R 9 , $0 = -ch/e. It is seen that t,he dependence of E,, on the external rnagnet,ic field B or the flux i+ is+ parabolic, a s shown in Fig. 4.2. @/)'
;
the Schriidirigcr cquatioli k)ccoli~cs
Since the electron is cor1finc:d to
;L
loop of r ~ ~ d i lR, l s w(: h i ~ ~ c : Fig. 4.2
As n is a n integer, the ground state (lowest energy level) energy Eg is given by h2 E [n* e ~ ' B / 2 c h ],~ "2mR -
-
, ,,,",,,",*
Motzon I n E k c t ~ O m ( L 9 n e t zField ~
P r o b l e n ~ sa n d Solut7,on,s o n Qq~antzrmMechnnzcs
where TL* is the integer nearest to e ~ % / 2 c f i (or e$/cll), w1iic:li is i i ~ g ; ~ t i v ( : as e is negative for a n electron. (1)) Suppose we start with a s t a t e E,, w1iic.h is tlic: grourltl stat,(:, 71, will rc,niniil the sanlc whcn B is trlrned off. Thlis thc: wav.vc: hiiic:t,iori will ,JI = Cc:xl) (i71H) iiil(1 th(, (+lectricc ~ i r r ~ rd(>~isity it is
where C is tlir: norn~i~lizi~tioli c o r ~ s t i ~ i iLet t . S (1(:11otc th(: t h e thin wirc:. W(: 1i;~vc:froiii t l ~ ~iorin;~lizatioii : c:ontlitioii
(:TOSS
s(:(:t,ioli of
( i ~ )Ass~nniilgt h i ~ t11o11r(:lativisti(: qliantlinl i n c ~ h a n i c sis invariant undtrr t i ~ n crcvcrsal, ~ tl(:rivc: thc: timc: rc:vcrsctl forrri of t,he Schriidinger wave fr~llctioli. (1)) W h a t is thc ( l l ~ i ~ i l t l l1riec1i;aiiic:al ~li H;~iriiltoniarlfor a frec electron wit11 iti;~.giict,ic~nolnciitp in t,hc c:xtc:rllal c:oiist,;~ntrnagrwtic: ficld H z in the z-tliroc:t,ion, ill t l ~ rc:Scrc:lic:c: : fri~mc:of tlic: c:l(:c:t,ron'? ((:) Sripposc: t11a.t (:xtr;~(:onstt~litnlagii(:ti(: ficltl FIY is iinposctl ill t h e y-tliroc.t,ioll. L)(,t,(:rliiiil(:th(: for111 of tall(:(11i:~lltlnriril(:(:ha.rlic:a.l o p ~ r i ~ t ofor r thc: tiill(' rat,()o f (:lii~ilg(: of p ill this (:;Ls(:. (B , I L ~ ~ ( I , I ~ ) Sollltion:
(;L) Colisi(l(~rt , l ~S(:lirii(liiig(~r : (:cl~iat,ioii that
1 1cl2=--------
~TRS
Notc t h a t j has 1)r.en c:onsidcrc:d t,o 1)~:~iiiili)l.iiiill ilily (.ross s('(.(.io~i ilS t h c wire is thin. Br,callse the electron is iriitia.11~in t,hc groliiltl st;~t,c~, for wliic.li E,, is t,l~c, miniiillinl c:nc,rgy, wr, lit~vc: If H * ( - t ) ~ ( t ) then ! t h e Schriidinger equation is covariant under timc r(:vc:r~i~l i~n(1tall(:t>inlereversed form of the wave function is $* (-t). (11) 1,ct c 1)c tlic: c:liarge of the electron. T h e n p = a ailtl in t ht: ref(+rc~ncc fr;i111(>of tile r,lectron, -
where [A] d(,notes t,he greatwt iiitcgcr w1iic:h is not grc:atc,r t,hi~iiA. For the case of macroscopic i n a g ~ l e t u d r :~~L Sill p i ~ r t((:), tli~:(llli~nt,ll~li number is numctrically largc: :\.rid we can sirriy)ly ilsc: 71, c(b/(:l~,in w1iic:li case I (."/4n2R27r~,r
-
(c) For R = 2 cnl,
--
4 = 0.6 gauss ern2, wc have iri SI rillits
&
Problems a n d Sol~ttonswn
1
.
(
271
1 f (!Ti \
>-.
2
=
Motion in. Electrornngnetzc Fzeld
( J ~ ~ n ~ i t uMzc/I.(L~,~cs rn
[(n, H ,
-
n I )
--
n,,H,:,;
II
4010
A ~)art,ic:l(> 11;~s111;~s~ lit,, c.liirrgc: (1, ii~t,l.i~~si(. ~ I . ~ I K I I ~ ~L L~ I I. O ~ I I ( ~ ~ Is~ (S , I ~ is I~I not ~ic(:(:ssi~rily ( Y ~ I I ~t,o L ~11,/2) : ~ ~ i ;L( lI I I ; I , ~ I I ((lil)ol(! : ~ ~ ( : ~ ~ i o ~ t /L ~ o -i ~ lt (, ~ s / 2 ~ ~ t , ( ~ . T h e part,i(:l(: I I I O V ( ~ S in t i t i ~ ~ i f o r nrrii~gn(:t,i(. i ii(>l(lB wit,l~i~ v(:lociLy s11ta11 C:C)IIII);L~(:(~ wit,li ('.
a
(:L) Writ(, (low11 th(: H ; L I I ~ ~ ~ ~for , O tallis I I ~ ~s,yst,(,~r~. III ('Ill(: vv(:t,orpot,(:~~t,ii~l P ! for thc uriifor~n~il;ign(:t,i(.ii(\l(l111;lyI)(' w r i t , t , r 2 ~1s~ A = B x 1-12) (b) Dcrivc: thc: ( I I I ~ L I ~ ~ I II ~ I I (~ \ ( . ~ I ~ I .(II(~is(:nl)(:rg) II~(:;L~ ( : ( l l ~ i ~ . t of , i o~~i~l o ~t , i o ~ ~ Ero~rithis Ba~nilt,o~lia.il, for t110 l i ~ ~ ( lIiI~I Or I I ~ ( : I ~ ~ ~PI ~ I; L~ ILI ( ~ for i,11(' ~ I . I I ~ ~ \ I ~ ~ L T rr~ornc:iitt~rns. Ttic A2 t(,r1111 1 1 2 ~t~w ~~(:gl(:(:t,(:(l ill t,l~isi~o~~r(~lir.t,ivist,i(: tqproxi~na.t,i(>~i. (Notc t,hat t,llc rc:sults look cxac:t,ly lik(1t,l~c,c.li~.ssic:;rlt!cl~iatior~s of 111otior1.) (c) Witllout solvir~gtli(:sc: c'cltlntio~~s, itlvl~tify1 . 1 1 ~va.luc: of t,l~c:( : o ~ i ~ t j i ~ ~ l t , g for which t11(1 11c:lic:ity will rc:ni;r.in corlst,i~.r~t. (tIolic:it,y is tl(:fii~c:ci1lc:rc. ;is t h e cosine of thcl ;~rigl(> t)etwc,c:~ithe vcc.t,ors P a11t1s) (d) W h a t is thc a.c:t~in.l value of' thc: c:ol~st;r.l~t (1 for ~ L I OI ~~I Cof tllc following particles: e , p, 71, T? (Berkeley)
Solution: (a) T h e Hamiltoriian for the s y s t e ~ nis
- -
E ' ( B x s),,
2mc as [s,, s J ] = z ~ L ~Notc s ~ .t h a t we have used the convention t h a t repetit,ion of a sr11)sc:ript irrip1ic:s summatiorl over that subscript. (c) As P ant1 s rornmute we can consider the problem in the c:ornmon eigensta.tcs of P, s 5 1 n d s, . T h e helicity 11, is tlcfincd as
a n d as
Problems and
ficlrl of str(:rigt,li B for i t (list,i~rl(:e 1. Assli11i(: t h a t t , h two ~ pat,hs from A t o D ;Lrc i(l(:lil.i(:i~I ~:x(:cl)tfor thc rcgioil of thc fi(:l(l.
r1si 771,ClS
I
PI I IP I [PL A:, l
'I'liis is :I, l)l-ol)l(~i~i oil spi~ioriilt,(;rf(:r(:11(.(:. Co1isi(lsr :L ii(~itroi1111 t,lic bciar~i. Tli(:r(! is i a i i i ; ~ ~ i i ( ~fi(~l(1 t , i ( . B iii t,li(: rcgioil wki(:r(: t,kif: S(:lirijdingcr eqll;it,iori Lor t,li(\ (~~i~(:kii~rg(:(l) 11(:11t,roriis
which rcquircs !I = 1. (d) T h c vtalucs of for t,lic: va.rious 1);~rticlc'sarc5
S111)1)osirigB t,o kx: c:oilstailt and uniform, we have
wlicbrc t o , t a r c rc:spc:ctivcly the instants when t h e lieutroll critcrs i~11(1I(>~LVCS t h e irl:ag~l(:ti(~ fi(:l(l. Writ,(. $(t) = ,c/)(r, t ) $ ( s , t ) , where $ ( r , t ) a.nd $ ( s , t ) arc: rcsl)c:c:tivc4y the spac:c t ~ i l ( 1spill 1);~rtsof ,$. T h e n
In a recent classic tahlc-top experiment, a rrlorroc:hro~rltLticnc~ltroill)(~a111 (A = 1.445 was split by B r i ~ g greHc:ction a.t point A of ail iiitctrfc-rorllc~tw into two beams which were rc:corilbiiied (after anothcr rc:Acctioii) ;at poirit, D (see Fig. 4.3). One beail1 passes through a region of t,rtailsvc:r-sc:irli~giict,ic
a)
which is tlrc s t ~ i n ca s thc: wave f~irict,iorio f a. frc:c! 1):artic:lc, aild
ti.r-oblc~nsund So1r~tlon.so n & ~ ~ a n , t , M u ~c rc ~l ~ u n z c s
T h e i1iterferenc:e ;~risc!sfrom the a.ction of B on tlic, spill w;avt: fi111(:ti011. As ,$(I-, t ) is t11c wave filrictio~~ of a, f r c ~p:art,ic:lc, wo h;avc t I - to = 1/~1-: ~n,l/hk; ~ n d
A neutron int,c:rferonleter bean1 splitter plus r ~ ~ i r r o ras s shown in Fig. 4.4 has lleer~1)uilt out of a sil~glec:rystal. out thin plate
"
'I'll(: i~it,c:r~sity OF where k = X = A is tht: wa.ve nuln1)cr of thc: ~-lc:~~t,ror~. the interfcrcnce of tlic two 1)t:i~msa t D is t,lic:n proporl;ioll;d I,o
(1)
cx I ,dl,]
(s,
t) -F
,I/,{;)
( s , t ) 12
=
1 ,dl(2) (s, t o ) + ,,,,( 2 )
t , ) 12 Fig. 4.4
1 +(')(s,
to) + ,d,(')(s, t i ) 1'
(;I) By v i ~ r y i ~thv ~ gt,hic:kl~(:ssof il tallill pli~~t;i(: sh(:(:t pl:ac:~(Ii11 t 1 1 ~ ~O~LIII in OIIC ;arIii of tlic i~~t,(-rf(!roii~(!t(:r on(! (.;LII vary t l ~ r(!liativc: : pll;~,st:;III(~ ~ I C I I ~ P shift t,hc fri11gc:s. Givc: ;L 1)ric:f ( l i l ; ~ l i t , i ~CtXi ~I )~~ ~ L I I ; of L ~ t~ ~hI ~ o~r i g i ~of~ t,l~cx ph;isc shift. (1)) By illsc~tingiri o ~ ~ nrrn c : ;L rr1:agrictic: ficltl which is normal to the be;~rl~ ti~r~ct , i ~ ~ t l e p c r ~ d (i ~: i~1 i( 1tvory 11(:iarly ul~iformSO t h a t the force on the n e l i t r o ~ ~(:;LII s 1)t: ~iegl(:c:t,cd, rid l)y choosiilg t,he field so each neutron spin vector prcc:csscs tlirougll just one rotat,iori, one finds t h e relative phase of t,he t,wo t~carnsis shift,cd 1)y .ir radians, or one-half cycle. Explain, with appropriate c q u a t , i o ~ ~why s , this is so. (P~inctrton)
=
= 1
+ cos
2 ~ i i ~ XB n,l /I"
+-"llis
f i a sill
I
2.irk~71~1XB fr2
2.irp:rr~lX B .ir/~,,rr~,I XB = 4 cos" h, /L2
"
Therefore, the int,erference iliterisity :at D cx c:os2 (.rr/~~rrr.lXB/h"), w11c:re p. is the intrinsic magnetic moment of the neutron ( / I < 0).
Solution: (a,) W11c11;I neutrori passes through the t,hin plastic sheet, it is under t,lie action of ;a11 additio~ialpotential, and so it,s moment~imcharlgcs togcther with it,s clc Broglie wav~lengt,h. The phase change of the neutron when it pa.sst,s through t11c plastic sheet is different from t,liat when it passes through a vacuum of the same thickness. If the tliirkncss of the plastic: sheet is va.ried, t,l~erclative phase of the two bcarlis (originating from t h e sallle beam) also changes, causing a shift in the fringes. (1)) The neutron possesses a n anomalous m;xgnc,tic: moment p,, = -p.,, u and its S(:l~riidingerequation is
276
hlotinn irr Elcctro7nc~gnrtzcFzeld
Problerrrs und S01ution.s o n Q u a n t u m M C ( : / I ( L T L ~ C S
We l i ~ i ~neglect y the reflection t h a t occurs whcr~a ~ic:lltror~ w;~vc: is ilic:itlcnt orr tllc "surfiicc:" of t,h(: shr:c:t-likr: rrlagl~c%t,ic: fic:lcl ;I.S tlrc: ;~c:t,iolrof thct fioltl or1 thc: l l c l ~ t r o ris~ rather w ~ i t k .Ulit1c:r sl~c:l~ ;LII ;i~)L)roxirl~;~t,iori w(: llliLy show (by solving thc ;~.l)ovct:wo-spirl-co~rll,o~~clit, S(:lirij(li~~g(:r ( : ( ~ t ~ ; ~ t for ,ior~ o~~c:-dirrrc:rrsio~~i~l sql1;Lrc: wcll): Tlic wt~vc:I~lr~c:t,ior~ ,1/4,,, for ;L r~c:t~l,ror~ irlc:i(l(:nt l ~ o r ~ r ~ tor1 ~ l tall(: l y sli(:c:t,-like r r ~ ; ~ ~ l ~ (fi(:l(l : t , i (is: r(:lat(:(l t,o t.,11(, l , ~ ; ~ ~ ~ s ~ r ~ i t , t , ( , ( l wave fur~c:tiori,,/,J,,,~ out. of thc: lic:l(l by i i 1111it,;~ry I;r;~~isfor~~iiit~,ioii
where p = w r , r e n , wit#hwr, = 2 p r ,B / h 1)c:ing the, L ; L ~ I I I fr(:(lll(:li(:y, OI. r L ~ n ~ / / it ,1k1 t,i111(: ~ tak(:11 for t,11(,I I ( > ~ I ~ ~ tOoI I~ L S t,11ro11gli S t,l~(: ~il;~g~i(:t,i(: li(:l(l of thic:krlc:ss I,, e n thc lll~itv(:(:t,or i l l t l ~ ((lir(x:t,ioll : of B, k: th(: w;I,v(,IIIIIIII)(:I. of the ill(:i(Iorit,r i ( : ~ ~ t r o ~ i . If i i 11(:11t,ro11 is poIt~riz(~(1 in 1;11(: (Q,p) (lir(,(.t,io~~ l)(:for(: (:ilt,(:ril~glLli(\ li(~l(1, i . ~ . it,s , ~)ol;~riz,,c:tl vc:c:t,or is
277
By acljllstilig B (or L) so t h a t /) = 27r,we rr~akethe polarized vector of a nclltror~prcccss t,hrough orlc rot;itiorl as it traverses the region of t h e rn;~gr~etic: fic~l(1.T11c:ri
(21) A l ~ y ( l r o g , (il.t,0111 : ~ ~ is ill it.s 21' ~ t , i ~ t ill , ( >;L , s t i ~ t ,of ~ : I,, - ~ L I - ~ I , . At t,i111(: 0 :L s l , r o ~ ~ ll~iig,r~(:t~i(: g f i ~ l ( 1of st~,r(~ligt,li 1 B 1 p o i l ~ t i ~i11~ gl,h(: z (lir(:(:t,ioi~is switc:hc~l011. h s s ~ l r ~ ~t,h;at i i ~ gthc: c:fk:c:t,s o f c:l(,c,t,roilsl)i~l(:;a11 1 ~ :rlc:gl(:c:t,cvl, cal(:lll~l~tl(~ t,ll(! t,illl(! (l(:p(:ll(l(:rl(:(:of t>IlC~ : x ~ ~ ~ : ~ : t v~Llll(~ ~ L t ~ iofo lLx:. l (1)) 1 1 0 ~tl.Ollg ~ 11nlst t.h(: llli~gll(~t,i(: fi(:l(l ill 1)ill-t (ii) I)(' SO t l l i ~ tt,h(: (:lf(:(:t;s of c:l(:cl,rolis1)ili (:;a11;~(:t,~i;~lly 1~ ~~(:gl(:(:t,(:(l'! 'I'll(, :~r~sw(:r slio111(11)r: (:xpr(:sse:(l in s t , ; i ~ ~ ( l i111;~(:ros(:o1)i(: ~r(l 1111it,s. ((.) Srlpposc: t , l ~ i ~ilist,(:a(l, t, t,hr: 111i.~grl(:ti(.field is very wrak. Suppose, furtlicr, t,h;it i ~ t ,t - . 0 thc: i~t01lih i ~ sL:c = +fi and s, = fi, a n d the maglict,ic: lic:ltl is st.ill oric:rltc:tl ill t h r z tiirectiori. Sketch how you would calclil;at,c: thc: I.,illl(:(~(:I)(:II(~(:II(:c of t h expectation ~ value of L, in this case. You rlc:c:tl riot (lo t,llc: flill c:al(:~ll;~t,ion, b u t explain clearly what t h e main steps wolll(1 bc. Not,(:: All c:fk:cts of nuc1c:ar spin are t o be ignored in this prol)lcn~. (P?-in.c:cto,^^)
t
=
4
wherc: 0 is thc: i~rlglct,hc: pot;irizctl vc:ctor illiik(,s wit,l~t,h(: (lir(:(:t;ioll of t,h(: ~n:~gric,t,ic: field. ' h k i l ~ gt,li(: lilt,t(:r 11s th(: z (lir(:(:t,iori,w(' ~ ; L v (p: . u . : om,. T h c ~ as i
Solution:
( a ) Thc: il~ititilWAVC fii~ict~iori of t h e atom is we have
wherc
Motion i n Rlectro7r~agnet~c Fzelil
is the cigc~ist,;~tc of L , = / I . At t = 0 a st,rong inagslctic: ficlcl Be, is switc.lic~tl011. Tlicii for t Harniltoiiian of t,lic s y s t c ~ nis
For
2.
not. too st,rosig ~ i ~ i ~ g i i ( ficl(1 ' t ; i ( ~B
- 105 Gs, w(:
(.ill1
2 O tlic
279
T h c cxpcctation value of L , is given by ($(r, t ) L , i $ ( r , t ) ) . As L , (L+ + L-112,
=
ii('gl(~,tt,h(: B2
T h c Scliriiclisigc~rc3clr~. ,Lt '1011 and
k~(:ti(~,
-
L,(t)
(,c/~(r, %)I I,,, ,e/)(r, t))
-
11.c.os
rlBt L//l.,:c.
; -
(1)) Ttic (:fI(~.t,sof (,l(;(~trosi spill (:ill1 I)(, I I ( ~ ~ I ( (!(I T ~ if' tlic i~(l(1itiosl:~l (,Ii(,rgy due to the, st,rorig iii;igii(~t,i(. ficl(1 is inri(.h gr(';it,(,rt , l i i ~ tli(' l (:o~spli~ (,li(:rgy ~g for spiri-or1)it iiit,c,r;ic.t,ioii,i.c,.,
or
For t
=
0,wc: tlic~rihavo
U
> 10" G s .
Tlilis wlic~i~ t,lic, iriagrlc:t,ic: fic~l(1B is greater than l o 6 Gs, the effects of elec:t,rori spin can I)c ilcigl(~ctctl. ((:) If t,lic, ii1;~g1i(:t,i(:ficltl is vcry weak, the effect,^ of electron spin must 1)e ta.l
w1iic:ti is tlic H;iriiilt,oriian for anomalous Zec?rriari effcct ant1 we can usc the coupling representation. When calculating tlie ad(lit,ional energy due to thc .s', tcrin, we can rcgartl 5, as approximately tliagonal in this representatioii.
PI-oblrms and Solr~tzonso n Q n a n t u m hlrchnnzc?
hlolior~zrl Electl-omagnetzc Field
(ii) Writ,e tlowl~the tirne-depeiident wave f u ~ ~ c t which i o ~ ~ satisfies t,llc: initial condition L, = +h, a n d s, = h,. At tiirle t = 0 the wave hiiiction is $o(r, s,)
=
R 2 1 ( ~(3 ) (0, 9)0 , s ,
-
wllerr (3 arid qbs arc t,hc: eigcl~func:tio~ls of L, 11.;~ricls, rtyrc:sclitations ( 1 2 , I , ) , (.s< .s,) i-ctspec:tivc~ly.Explicitly,
=
h,/2 ill tlic:
281
Coiisidcr t,hc onc>-dirn~nsion:L1mot,ioil of ,211 uncharged particle of spin 112 aild inagilct,ir n l o l n c ~ ~/Lt = 2 b o s / h , , , . Thc: part,icle is confined in a n infiriitc S 0) t,hc:re is i~ 1111ifori11 fi(:1(1 of tllc SiLllie iriagi~itli(l(:t ) l ~ pointing t in th(: :r: (lir(x.tio11B = 130e,. I I m : e , ulcl e , ;~rc:uiiit vc~.t,orsill tllcx :r a n d z dircc.tioiis. ( i ~ )Usr
~ ) ( % r t , ~ ~ r f )t,li(:ory i ~ t , i ot,o ~ l fiii(1 t 1 1 grolln(1 ~ st,tltc w(:rgy ai1(1grollncl wixv(\ S I I I I ( , ~ , ~ O(I)ot,l~ II SI);I,(:Pi ~ , l i ( ls1)ili ~ ) i ~ r i ,iii s ) t,li(: w(:i~kfi(:l(l liinit, B0 << (l,,/I,)'/2~1r.~r~~. (I)) Now c:ol~sitl(>r 1ic:ltls wit,li Do o f ;~rt)it,riiry st,r(>iigt,t~. Firi(l t,tw g(:n(\r;~l for111of t,li(x (!ii(!rgy ( ~ i g ( ~ ~ i l ~ l i i ((t)ot,l~ ~ ~ , i oS I~) ; ~ L ( Yi111(1 ~ spill pi~rt,s)i i ~r(>gioi~ I wlli(,l~s21,t,isfi(:stlic l(~Si:-lii~~~(l l)o~ii~(lilry ( , o ~ ~ ( l i t , iFoi~i ~~(.~ l l s oi,l~(> for111 that tllc c ~ i g c ~ ~ ~ f ' l i ~I I~~ LcS.illt , ir(:p,iol~ o ~ ~ I1 wlli(91s;lt,isfi(\st l i ~I-igtit,-lii~i~(l t)olii1(1a r y c.o~~tlit,iori (Fig. 4.5). ((:) Ol)t,2~iri2\11 ox~)li(:it, (l(xl,(~r~~ii~ ('(111i~t,iOl1 ~ i l ~ ~ t , wl~os(, ;~l sol~it,ioi~s wol11(1 give t,llc>ct1iorgy (:ig(~iiv;~l~i(~s k;. tat,^
(MIT)
+
b ' 3 ( j2L L2
+
(/)A2
- 22 ) )
whcrc: (j,,,,,, is tho c:igellhirictioll of (j2, j,) for tllc cLrlorgyl(,vc:l IT,,^,,,,, Tlic:rc:forc, t,hc: tirrlc.-t1t:l)c:iitlr:llt wii.vc f ~ l ~ ~ ( . tfor , i o t,ll(: i ~ syst,c:i11 is
+ hcj;;
cxp
(
-
1
h
L 2
Fig. 4.5
+ 4 +2 - + 2
exp
(
-
Ex i- t ) ] z
li a
(iii) Calculate the expectation value of L, in t,hc ~ l s u a inal1lic.r: l ($(rl
S1
t ) I Lz I ICl(r, S , t ) ) .
Solut.ion: ( i ~ 1)11 t,l~(: ~LI)S(~~I('(:of ~rl:~glict,i(: Ii(:l(I, H = : ~ r i ( lt h r (!ilcrgy eigcnf~iiictioils (sl)iL(.(:1)iLrt) iL1111 ~ig~n~:L111(:~ iLT(' rcspe('t,iv(:ly
JI,, =
J1/L sin
'TJT(:I;
+ L)
2L
'
Proble7n.s and S o l ~ ~ t z o ron ~ s Quant,urn Mecha.nzcs
Motion i n Electromagnettc Field
-
-ir2fi2,r12
11 1 , 2, 3 , . . . . 8nlL2 ' As for t h e spin p a r t , we know t h a t each eriergy lcvr?l has a tlegcncracy of 2. When a rnagnetic field is present, H Ho + H ' , whr:rc
E,,
=
-
we get t h e ground s t a t e wave functiori
-
- L 5 n: 5 0 ,
'L"B"u,,
B=
2pOs . B
If the ficld is wcnk, lct vectors. Then H i l = (,isl / H'
I ILL)
=
{(A),
p o u . B =
Ti
,IL~
,$I(z)
=
(b) T h e space part of t h e wave fiinction in regior~I is
~ L ~ B ~ ( 0T5, :, 1 : < L , r:lscwhcre
0 ' 1 ~ 2=
(CI)
'(/1(3:)
.
be t l t)as(> ~
T h e coritir~uityc:o~lditionof t h e wave ful~ctioirgivcs B = 0. 111 rcgioil I , t h e spin is aligrred t o t h e z direction, t h e cigenvcctors t,c:i~ig for z 4, ;111(1 for z T. I-Irnce
pOB O ( l 0)
(y)
(i)
2
H.Li = H i 2 =L
Hi2
= (711
B (1 0)
= (112
I HI
I
'
17.42)
(o
1
( ) 1''
(:I:)
,I/,, (2:)
(Lr: =
ILO
Do
-
2
)
1 ~ 2 )
In a siriiilar way we obtain t h e eigenfunctions for region I1 (0
0 arid froin det (HI
-
-1
2
,
E ( ' ) I ) = 0 wc: get
2
2
or
E(')=
=
sin k 2 ( x
$11 k z Z f =
sin k 2 ( x
I ~ I I
*--1 'L O B ~
-
-
L)
2m
L)
2m
P O B O;
+ poBo
.
(c) Considering t h e whole space t h e energy eigenfunction is
T h e ground s t a t e eriergy levcl is thcrcfore
E(, =
-ir"2
A $ I ~ , ~ J + B $ I ~ L; ~S ~ X, ~ O ,
1
+ D$IIq Z f ,
-- -
8mL2
fi PO Bo .
k2z.~
O ~ X < L ,
elsewhere. Thus
< x < L)
284
Problems a7~dSolutions o n Quan,tum. Mechanics
Motron i n Electron~ugneticField
285
4015 Consider a n infinitely long solenoid w11ic:h carries a current I so t,hat, thcrc is a, constailt 1na.gnctic:field inside t h e solenoid. Suppose in t h e region outsiclc thc solenoicl t,he ~ n o t i o iof~ a. partic:l(: witah c,ha.rgc c a n d mass m is dcsc:rit)cd by t,he Schriidingcr ccll~ntiorl.Assulnc t h a t for I = 0, the solution of t h e cqua.tion is give11 l)y T/,"(x, t )
=
cL""",J,o(x) .
(h = I)
Writ(: down a.nd solvc thc: Sc:liriidirlgt~r( ~ q u i ~ t i oilln the rc,gio~lo~lt,si(le t,hc solc~ioiclfor tlie ca.sc I # 0 . (1)) Co~lsi(lc:ra t,wo-slit, clifiri~ct,io~~ (:xp(:ri111(!1itfor t,li(: [)i~rticl(:s (l(~s(~ril)c:(l t h a t t,l~c:(list,i~11(.(: (I! ~)(:~wc(:II t h two ~ splits abovc: (sc:c: Fig. 3.6). Assu~~lc: is large, c,o~~il);~rcycl t,o t,l~rxtlit~lllc:terof thc: sol(:~~oid. C o ~ ~ i p u t,hc: t ~ e sliift A S (;L)
ancl so L l -: k.L = k:, k.', = k; .= k'. Thew t.hv c.oiitiiiriity of t,llc: wavc: hiric:t,ioi~a t
:I:=:.
0 givcls
1 Bk:' c:os k ' L
-
Fig. 4.G
C:k c.os k L -1- Dk:' c:os k ' L ,
Ak c:os k:L .-= ---Ck:c.os h L t DX:' c.os k l I J . To solvc: for A , B, C , D , for nollzcro sollit,iol~swcx rc,cli~irc, O
sin k ' L
sill k L
sill k ' L
sin k L
O
s i n kL
sill k ' L
O
k:' c:os k ' L
k: c:os k L
0
-
k c.os k L
k ros k L
-
k:' c.os k:'L
solenoid
of tho (liffri~(.t.ioli ~ ) i ~ t t ( ,011 r ~ it h e screen due to t h e presence of t h e solenoid with I f 0 . Assrl~nc:1 >> AS. Hint: Let + ( x , t ) = +o(x, t ) + (~ 4 ,
=
0,
whm: e
7 / ~ , ~ (=O. x)
(Ti= 1 ) .
k ' cos k ' L
(Chicago)
l.e.,
k sin k L cos k'L
-
k.' sill k ' L cos k IJ
=
0
Solution: (a) In the prcsciicc: of a vect,or pot,ent,ia.l A , p 4 p eA/c. In the absc11c:c of clcctro~nagiieticfield t h e Schriidinger c:quatioll is -
det,ermine t,he eigenvalues E
I
+ V(X)
+o(x, t ) ,
8
98,
'~-
Problems a r ~ dr i l ~ l u t i 071 ~ n ~ t ~ a n t u rf vnl ~ ~ t ~ n r i z c s
286
Motion in Elrctromagnetzc Fzeld
287
sc:
where, a.s l)c4ow, we slla.11 use units such tha.t li, = 1. T h c S(~l~riiding-er equation in the prescncc of a n electroluagnctic field t h ~ i sl)r ot)taiiied (using thc. rrliriiiii~iili(:l~(:tr~llii~glleti(: coupling thc:ory) ;LS
on dividing t h e two coiit,ributions by a common pllase factor exp(i :A d x ) , which does not affect t h e iiiterference pattern. T h e closed line int,egral, t o b r taken conilterclockwise aloiig a n arbitrary closc~lp a t h around t,he solenoid, gives