Problems and Solutions on Quantum Mechanics
Major American Universities Ph. D. ~ualifying Questions and Solutions -
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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
$Jl(x,0) =
6, 1x1 < a , 0,
otherwise,
*j, (a,0) = A
LZXP1-r2/n'}.
(a) At time I = 0 find t-ht? probability m p l i t t i r l ~in mo~ncnttlrns p a c ~ .
(b) Find dl (2,1.). (Berkeley)
14
Problrmi and Solutions on Eleclmrnnqnetkrn
Bmic Principles and One-Dtmpn~ionalMotinns
15
Solt~tion: (A)
At tirnc t = 0 the probability nmpliturEc in momentum space is
=-
A
m
*I-.
J
exp (-x"n2
-
tpx/ti) ~ f x
-m
i-J I ~ ,
Ant --
rn
PXP
(-n2p2/4h2)
0. ( Wisconsin) (a) (11)
Solution: (a) The Hnmiltonian of n pFat~eroktor is
H
a series, arid expr~ssio~is
and aa the Srhriiding~rquation is ( 'GVisconsin)
Solution:
- ( d / 2 I , ) ri2$/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~.
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 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 - ~ ' :~
+oo, I3 4 riE,,.
>
Pig. l.t2 Fig. 1-83
A nonrt~lai,ivisticp;vt.iclr of miss rn unrlesgoes on o,,
13 cosh (k.r.1,
Q::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
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~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 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. 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
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
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 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 $'(xl, 1 ' ) i ~ ~ o r r l i nto g o b s ~ r v ~0' r wit11 coordinates (r',i t ) re1at.d to ( r ,E ) I>y the Gidilean translornlatian
Supposing B = c x p ( - I E l l h ) ~ x (itr). p wr iravt. k.' = ( 2 n ~ / f i ' (1 ) +
i V / E ) . As V
> 1 tllr I,orln 1 npgtcr+t,~l. Takr~itog~tllcr,wr g ~ tI.hnI . thr! nt~mhrroi llnuntl st aim I,; h r t , w r ~ (~nl w n 2 / 2 h ] tinrl {2rrlwnz/irhj.
~ . , L I be I
( i D l ( A 1) = A ( 3 ) (1 - A) lil ( r ) . Ollvio~~sly V ( 0 ,r) = Vf ( X I ,V(1, :r) = 112 ( r ) ,aV/iIA = V2(:c)- V,( x ) > fl, Tlir Ha~niltonionis t bell
and the ~ l ~ ~ ~ q t ~isa t i o n
m)In,A) = E,,(A) 1 wherc EtL(A) = ( 7 1 , A I ~ T ( xI),I$ 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.
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> 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, -
-
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
+
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 + $210 + f i ~ z J j 2 1 1+ h421-1 1 2E1~J)100 Then
As r 5 a 0.
;L
splic.ric~;illysyrnirict,ric pot,c:rlt,i;~lV
=
(a) Use the unccrta.int,y 1)rinciplc to cstirri;~t,c:t,lic: groiintl st,;~tcc:iic.rgy. (b) Use t h t Bolir Somrnctrfcltl clu;~ntiz;it,ionrill(: t,o c:al(:ulat,c:t h t gro~incl st,ate energy. (c) Do the sa.nlc: usirig t,hc variat,ioii;~l1)riiic:iplc ;r.rltl a trial w;wc fi1nc:tiori of your owri clioicc. (d) Solve for tlic energy c:igc?rivahie arid c:igcnf\irlction cxac:tly for t,hc: ground st,atc. (Hint: Use Fo11ric:r trarisforrrls.) ( e ) Write down t,lle effective pot,cnti;~lfor nor~zcroangular rnomcrlturn states.
3
kjfi2 113
'=1(4m)
.
(b) Thc Bolir Sorrirric:rfcltl quantization rule gives
Choose polar coordinates such t h a t the particle is moving i r ~the plane = 1, and the orbit is circular wit,li radius a . The second integral gives
O = 7 r / 2 . T h e ground state is given by n,,. = 0, 71.4
Solution: (a) T h e uncertainty principle st;~tc:sthat A P A ~2
fl
2,
where
nP = [(P
-
p)yll2
=
2 112 42-P ),
[(p2
-
zpfi4-F ) ] l 1 2
'['he central force is
144
Problems and Solu,tzor~,so n E~ect~orna,qnetis~r~
Combining we have a = (hI2/7nk)'/< and hcrlce
II
11 [
hence
1/3
(d) The, Srl~rridirlgcrc,cluatioll for the radial motion can be written as (c) T h e notion in the grolmci state does riot dcpclnd on 0 ant1 (j Take a trial wave functior~$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 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,
0ho~)~)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 ~~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(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,). 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
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 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(:~LIISrgy 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;/(: : rBrt,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 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.lr 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
where 4 is t,lic: rliagiictic flux through t h e sol(:iioid. Thus t h c introductioii of t,hc solclioid gives a ph:~se fi~c.t,orccblc t,o t h e proba1)ility alnplitud(: a t poii~t,sor1 t,llc sc:rc:r:rl c:orltril)utc:d l)y t,hc, lowcr slit. Usii~ga, nicthotl aiialogolis t,o the t,rcat,rncnt of Youlig's irltc~rf(~rt:lic,c in opt,ic:s, wc: sec tllat, tlic i~lt~rfor(>n(.(: ~ ) i ~ t , t , (is: rs1iiftc:d ~~ by AS. Assurriing 1 >> d ant1 1 >> A S , wc: liavc: wlii(.h is
t,llv
S(~l~rii(liiip,(:r (:(l~ii~t,iol~ for w r o iii:~gii(:ti(:fic,l(l. 11(:ii(~ t
,$,(x,L )
:z
,,I,() ( x , L)
: (:l";ill
,(/I()
(x) ,
(I)) This is a ~)rol)l(~ili oil tlicl Ali;iroi~ov Uoliili ( % l l i ~ . tWli(~i1 ,. I -.- 0, for :~iiypoilit oil 1.11(: s(.r(,(:lit.11~: ~)rol);~l)ilil,y ; i l ~ i ~ ) l i t , ~,fi (isl ( ,f ~ f + t ,f . , w1ic:rc: f I ;~11(1f r(spr(~s(:iit,t,lir (:oiit,ril)~it,ioi~s of t,lir I ~ I ) I ) (i1,1i(1 ~ ~ low(,i- slits r(~s1)(:(~t,iv(sly. W I I ( ~tlic ~ I (:urr(,ilt, is on, i.v., I # O ? w(, l i i ~ vtali(: ~ prol)i~l~ilit,y ; ~ i ~ i ~ ) l i t f, ~' i-(Llf'+ ( ~ t f / with
I k bcirig tlic: wnvc: ii1iiril)c:r of thc p;l.rticlcs, ;~rldso
Not,e t,hc: t,rc.;~t,~rient is only valid rlonrrl;~t,ivistica.lly
-
( i ~Wha.t ) arc: t,llf: cmergies a n d energy eigenfunct,iorls for a rlonrelativistic particle of mass 7n, rnoving on a ring of ra.dius R as shown in the Fig. 4.7.?
where c+ anti c _ dc1lot.c iilt,cyy-nl 1);~tlis;~l)ovc:i ~ i i ( 1l)(:Iow tlic soI(:iloi(l r('spectively. T h u s Fig. 4.7
(b) W h a t are t h e energies arid energy eigenfunctions if the ring is doubled (each loop still has radius R ) as shown in Fig. 4.8?
Problems and Solutions o n Q ~ ~ a n t u . m Mechanics .
0 Fig. 4.8
(c) If the particle has charge q , what a.re the cilc:rgics and energy eigcrifunctions if a very long solenoid containing a iriagrictic flux pa.sscs tllc rings in (a) as shown in the Fig. 4.9.? and ill (b)'! Assurrlc the systcrll docs not radiate electromagnetically.
H(:ric:c tlic c:igcxifli~lc:tio~is a.rc
ct-, 3
14'ig. 4.9
II
Solution: (a) As
(1)) 'I'll(: s;I.II~(: H;~.lililtoiiiil.lia.~)~)lics, i ~ 1 1 ~so 1 wc still have tlic s;~riic Scllriiciiiigcr c:clliiat,ioii
2d\,tI (P (8) = EQ (6)) 21 dt)" - -
we have the Scliriidingcr equation
I-Iowc~vc.r,tlic sirigl(, v;rlliedness of the solutions now requires where
I =~ , L R ~ , Hciic:c tlic iioriiializcd cigenfunctions and the energy eigenvalues are now
with and Thus the solutions are (c) T h c H;~rniltoniailin the prcserlct of
For single-valuedness we require
a,
r~iagnc,ticficltl is
M o t i o n i n Electrwmagnet.lc F ~ e l d
Problems and S o l u t ~ o n so n Q u a r ~ t u mA4echantcs
290
111 t,he region where t h e part,ic\e movcs, B = V x A = 0 and we can choose A = V p . From the syininetry, we have A = Aoeo, A. = (:onstant. Then
AHRdQ = 2 r R A o Thus
A
- 4% ----
27r R
- 4,
SiLY.
= V(4Q/27r),
ant1 we call takc p = 4Q/27r, ricgl(:c:t,irig possibly ;L ~oiist,i~iit ph;ts(: fiu:tor ill t,he wave fuiictioris. Thc Schrijdir1gc:r cy~~at,ioii is
cb
2 (1
V---V() 27rfl . -
?
and
Q
h."
-2711
Siiriilarly for ttic ring of (I)), wc 1i;~vc:
On writing and it bcco~ncs
where with solut,ioiis
For the ring of (a), the single-valuediiess condit,ion
rt,
0, *Il *2> . . . .
Pert~rrbation Theory
5. PERTURBATION THEORY (1)) For the statioriary statc,
(a,) Show that in the usual stationary statc pcrt~~rl)at,ioii thcory, if tjhe Harniltoniari can t)c writtcn H = Ho + H ' with f1040 = Eo4(),t h ~ nthe correctiorl A E o is AEo % (4olH'l4o) . (b) For a spherical nuc,l(:~is,the rlucleons may bc assr~nlotlto 1 ~ in : a 0, 7 . < Z Z , syhcric:al potcritial wcll of ra(liris R giv(:t~by K p =
{
oo, 7. > n. For a, slightly dcforrrietl nuc:l(:r~s,it iriay bc c:orrcs~)olldingly:LSSIILLI(:(~t h i ~ t the nucleons are in arl c:lliptic:al wc:ll, again with irifirlit,~w;~11liciglitj, tallat,
where
..
+Y2
V
0
=
inside t11c (:llipsoi(l
2
R ~ l ) l i ~ ~ ithe i l g vi~ri;~l)lcs Z,j j , z by ctioris in the relixt,ion between T and p and comprltc the g r o l ~ ~ stat(: l d lcvcl shift A E to order $ ( c =speed of light). Solution: 111rcsla.tivistic:lriotiori, the kiiictic cncrgy T is
Solution:
3 1 ~ 1 ,
The, c,rlc,rgy c.orrcctiori of first order perturbation is given by
I
E(') = ($;O)(x), whcrc V
L
VO
=
v$iO)(x)),
Vo for -a 5 .c 5 a. Thus
for - a < z < a ,
as shown in Fig. 5.4. Consider the Vo part a s a perturbation on a flat box ( V = 0 for 3 a < x < 3a1 V = cx for 1x1 > 3jal) of length 6a. Use the first order perturbation method t o calculate the energy of the ground state. ( Wzscor~s7rl)
Hence the cncrgy of ground state given by first order pert,urbation is
Problems and Solutzons on Quantum Mecha~izcs
302
Pertrrrbation Theory
5008
5009
A one-dimerisior~alsimple harmonic oscill;lt,or is sr~bjcctedto a small pertlrrbirig potential 6 V ( z ) , protlucing a "diinl)len a t thc: ccntc!r of the rnotion. T h u s
A perfect,ly plastic ball is bouncing between two parallel walls. (a) Using classical mechanics, calculate the cliaiige in energy per unit time of the I d 1 as tllc walls arc slowly and unifornily moved closer together. (b) Show t h a t this c:harige in energy is the sarrlc ;IS the quantum rnecharlical result if the ball's quariturn number does riot c1i;~ilge. (c) If tlic ball is iri tlie quanturn s t a t e with 7). = I, uiltler what conditions of wall nlotion will it rcrnairi in t h a t state'? ( CIr icugo)
!I
?
Calculate the corrcc:tiori t,o thc ground st,i~tccnorgy of the oscilli~tort,o first orcler in X ill the cvcrit that (a) n >
JW.
;L
siin1)lc lli~rrrlorii(:
(x)exp(-711~1'/2h,),
d ~ ~ ( :=r )
1
II
7r
( CoI1~7n.biu) Solution: T h e energy corrcc:tiori for the ground state given 1)y first ortlcr 1)c:rtrrr 1)atiorl is
(:tri(:fi(:l(i tential wt.11 V(3:) = A k:c2 illid s111)jcctcd ~ L ~ St oO a l)(:rt,llrl)i~ig
F
=
FX.
(a) Deterrrli~ict,hc shift in t,lre e1ic:rgy l(:vc:ls of this systc,~riciuc, t o thc: el(,c:tric: field. (1)) T h e clipole rnoincnt of this systctm in stntc 71, is tlcfiiictl ;I.S P,, = -e(:c),,, whcrc (:I:),,, is t h e oxpcc:tatioli V ~ L I U Cof :I: ill tlic stat(, 71,. Fiiid the: dipolt 1lionic:rlt of t h e systcnl ill thc: 1)rosf:iic:c: of the cl(:c:tric: ficltl. ( Wisconsin)
Solution:
f
:
sph(:r(: of charge is irl a n clcc:t,rost;it,ic If i~ v(:ry SIII;LII ul~ifor~~i-(l(:~isity + ... , potc:iit,i;l.l V(,r), it,s ~)ot,(:~it,i;~l CIICL-gy is U ( r ) = V ( r ) -t riV"(r) whc1.c. r is t,lic, ~)osit,ioiiof t h e c.c.11tc.r of thc c.lii~rgci~ntlro is its vc,ry siiiitll radiiis. 'rlio " I i ; i ~ ~ shift" il) can 1)c t,lioiiglit of ;LS tkic: s~riallc:orrec:tioii t,o t,lic' energy l(~vc,lsof' t,t~c:hytlrogc:n ; L ~ , ~ Il)(:(.ails(> IL t,l1(: ~)hysi(:i~l c:I(~;trolldocs IELVC this l,rol)cr.t,y If t,hc r; t,c,r~nof U is trc$at,c:d as ;L vcxry srnall pcrt~irl),zt,ior~ com1)arc:tl t,o the, Coulonll) illt,c:rac:t,ionV ( r ) = -c"r., what arc t,hc: L i ~ ~ r i t ) shifts for tlic, I s ; L I I ~21) I(IV(~IS of th(: hy(lrog(:ii ;~t,o~ri? ExI)~(,ssyour rcslilt it1 tcr1lls Of r.0 iL1ld fllll(~i~lll(:llt~i~.~ (:011~t,;Lllt,~. Ttic: uii1)(:rt,lirl)(:(iW ~ L V I 'fiin(.t,io~isarc
(a) T h e Ha~niltoninnof the system is -312
.$ls
(r) = Z a ,
.e
-
Solut ion: T h c st;~tc:I s is rio~idegelier~~t,c, so the: crLcrgy correct,ion is where x' = x
-
$.
f'mble7ris and Solutrons o n Q u a 7 ~ t u nMecha~rlcs ~
Perturbation Theory
339
For electrorls a n d positrons, lgj = 2. Using first order perturbation theory colllpute the energy differcncc hetween t h e singlet and triplet ground states. L)etcrminc which s t a t c lies lowest. Express t h e crlergy splitting in GHz (i.e., energy divitlcd hy Plarlck's constant). Get a il~rmber! (Berkeley)
Solution: (a) By analogy with t h hydrogerr ~ torn t h c norrnalizcd wave function for t h c 1s ground s t a t e of positrolliurll is we have
AE =
/ 2:
-
r ~ ~ ~ b ( r ) l * $ ~fix ,~(x)1~
3,
with (LO = 7r1. t)cing tlrc clcctri)i1 rrst rriass. Not,(: tlmt the f;ic:tor 2 in front of no is t o ac:courrt for t h c fact t,lr;lt t h e rcd~rc:c:tlrriass is p = (h) T h e ~lleari-squarcradi~rsfor tlic: 1s statc: is
inr.
5025
Positroiii~riliis a 1iydrogc:n ; ~ t o ~but l l with i i p o ~ i t ~ r oas n ' L ~ i u ( : l e ~in~~ll, strntl of ;L prot,oii. 111 tlic ~ionr~lativisti(: lirnit, t,lrc: clrorgy 1cvc:ls alrd wave functions arc t h c s1~111t:;LS for hydrogr:rl, c:xc:c:~)tfix sc:i~lc. ( a ) Fro111your krlowlcdgc of t t ~ clrydrogcn at,orn, write tlowrr thc: norrrlalized wavc furiction for tlic: 1s gro~rrltlstatc of positrorriurri. Us(: s~)lrerical coordirlates a n d thc hytlrogeliic Bohr radius no as n sc*al(:1)ar;~rnetc:r. ( b ) Evaluate thc root.-nlcari-stluare ra.tlius for t h e 1s stsat,(:in urlits of no. Is this a n estimate of t h e physical diameter or the radins of positrorli~inl'? (c) I n t h e s states of positrorriunr there is a conta.ct 11y1)crfiric:iilt,c:raction Hint =
87l pe . p p b ( r ) , 3
--
where pe and p, a r e t h e electron a n d positron rn;~gnetic:nloilrrrits s). (p=gzil~c
a n d tlic: root-rrl(:;~ii-sq~ri~r(: racli~rsis 1
9 This (:an be considcrccl a physical c,stirrr;ltc: of t h e ratli~rsof positroniurn. (c) Ta.ki~igt h e spill into :~cco~lnt a st;~t,cof t h c systerrr is t o he tlesci-ibed hy In, 1, 711, S, S,), whcrc S aricl S, are respectively t h e t,otal spin and t h e z-component of t h e spiri. Thus
340
P1.ob1ern.s and Solutions on ( 2 ? ~ a n t 1 Mechanzcs ~m
Solution:
where we have used
S = s,
+ s,
If u7t,consider the protoil to be a spherical shell of radius R and charge e , the potelltial clicrgy of thc t:lcc:tro~i,of charge -e, is
For the singlet stat,(,, S = 0, S, = 0,
Takc: tl~c:tliff(:rc!nc:c bc:t,woc:ri tlic ;~l)ovc:V(7.)i111cl t,li(>~ ) o t , e ~ i t energy ii~l due t,o a poi~~t,-c:h;irgc: protori ;LS p c r t ~ ~ r l ) i i t i o ~ ~ :
-
$
S == 1, S, = 0, k 1 , For thc trip1c.t strat(%,
Considcr tho protor1 t o 1)e a sp11eric:al sl~cllof c:hargc of ratiius R. Using first order perturbation theory calculate the c:hallgc: ill tl~c:l ) i ~ ~ c lc:riorgy i~~g of hydrogen due to the 11011-poirit-like nature of thc: pl.otori. Doc:s the, sign of your answer make sense physically? Expla.in. Note: You may use t h e approximation R 0 , t,hc: grou~ltlstatc crlcrgy lcvcl of thc hytlrogcr~a t , o ~ nwol~ltl irlcrc:isc: ( 1 1 1 ~to t,li(: r ~ o ~ i - ~ ) o i ~ i t -riaturc l i k c of the prot,orl, i.e., thc 1)inding ellergy of the 11ytirogc11atoll1 would tlccrcase. Physically, cmrr~paringt , h ~ point,-likt: a11t1sllcll-sliapc ~nodclsof tlic proton r~uclel~s, we see t h a t ill the lattcr 1llot1t.l thcre is a n additional repulsive action. As the hydrogen a t o m is held together by attractivc forcc, ally non-point,-like niiture of the proton would wc;ikcn the attractive intcrac:tion in the system, thus rcduc.iilg the binding cncrgy.
5027
-
Ass~l~rit: t h a t the prot,on has a nonzero radius r, lop1" cm and t h a t its charge is distributed uniformly over this size. Find the shift in the energy
342
Pcrtwbntron T h e o ~ y
Problems and S O ~ ~ U ~on Z OQuar1.1urn T~S Mechnnics
of the I s and 2p states of hydrogen due t o the tliffcrc,nce between a point charge distribution and this extended chargt. (Colu~nb,ia)
Solution: T h e Coulomb for(:(: a n c1cc:tron i~lsidct h e spht:re of tllc ~ ) r o t , ocxpcril~ ences is
The e1ectric;tl l)otc'~itinlc,llcrgy of the electron is
v,=-/,. (12
,,
kC
for
- .
r5r1,,
2r7j
v,=
(:2 --
-
I'
for
r
> rp ,
with ( L = $ . wo (:;III t;r.k(: (:-'/'" As .r.,, hc:g1.01ind st,at,eenergy when the nucleus has a physically rcalist,ic size : L I I ( ~t h grourid ~ state energy for a point-nucleus. Express tllc rcsult,: a ) in elr>c.t.ronvolts, 1)) a s a fraction of the ionization energy of this at,om. (Berkeley)
(Z= 13, A
and treat V' a s a perturbation. The cnersy correction for the, 1s state t o first order is
Solution: If wc: t n w t t,hc: nlic~lrusas a uniformly c:harged sphere, tllc cl(:c:t,rical potelitrial energy of thc electron is Vl = -zc2/7 for
7-
>p,
346
I'roblems and Solutions
071.
Quur~turnMec11,ur~zc.s
Perturbatton Theonj
p b e i ~ ~the g radius o f the nuclcus. Inside the 1itic:llics t,he cdectron suffers a
Coulomb forcc F
=
-Ze2
(:) .
;?I
=
-
-- 1 and
AE
,
tial ellcrgy bc:ing V = ?! $ r 2 t C , wliorc C is ;L constailt,. Tlre c:ont~iiillity 2 p. o f the potrritial a t tha surfac:e of the ril~c:lcus,Vl (0) V2(p), rcqllirc:~tllilt C = - '1 &, TIluS 2
we can take e - 2 z r / a
- z ~ ' T -tlir / ~mrrespondillg '~, poten-
= (1001H11100)
1
i
P
I :' Zc" 'I'
V(1.)=
":
[(a)
1. 2 /) , 2
-31
,
;/.
wlicrc, Vo(r) =
%(," -- --
I'
,
(oo > r.
> O),
is t,o l)c tre;itctl ;LS i1, 1)(;rt,llr1)at,ioii. Tlic tirst ortl(:r ciicrgy c:orr-(:(:tioilis thc:11
A propos;~lhas bccn inadc t,o study the propcrt,ics of ;LII atom c:ornposed of a z f (In,, = 237.2711,~)aritl ;L / L - ( I I ~ ,-, 206.77rnC)in order t o mcastirc? t h e chargc r ; t d i ~ ~ofs the pion. Assurrie t,liat all of t,lie pi011 c:hargc is sprrad uniformly or1 a sphc.ric:al slicll a t Ro - 10-'" crri and t h a t tlic. LL is a point charge. Express t h e potcritial ;LS a. Coulorrll) potcrltial for a. poirlt charge plus a pertlirbatio~iarid list: pcrtllrbatiori t,heory t o calculate a nurrierical valuc for tlie percentage shift iri t,hc 1s -2p energy diffcrc~iccA. Neglect spin orbit effects arid La.ml) shift,. Give11
( Wisconsin) Solution: where a = h2/m,e2 = 5.3 x 10-%.nl (Bohr ri~dills). As
T h e Coulomb potential energy of the muon is
348
Problems and Solutions on Quantum Mechanics
It can be written in t h e form
Therefore
where
for ,r 2 R ,
0
=
{ (:a)
Perturbation Theory
2
hr.5
R,
is to be trcatc>tl as ii p e r t ~ i r l ) i ~ t i ~ i i , T h e cncrgy levcnls arltl wave fririt:tions of the uiiportur1)c~tlsystc:rri arc
B,, = As spill orbit :tritl L;iiul, c:ff(~:t,sarc: to k)(! ilegl(:c:t,c:d, wc: iic~c:tl orily c:orisitl(:r
RTPL iii p(,rt,~rrt):~tio~l cal(~~ilt~tioiis. Tli~is
with
Hcrrce
T h u s a,o >> R arid t h e factor cxp(-2r/no) in the intcgrantl ; i l ) o v ~rnay 1)e neglected. Hence
Mrioiiic: i~toiiis(:oi~sisto f 11111 in(:soiis (niass mi, = 206711,:) \)c)lind to atoiiiic: 1iuc:loi ill liytlrogciiic ort~its. Tlic circrgics of t h c rrlu riicsic lcvcls are shift,(ylrc.lii.t,ivc: t,o tlicir valucs for ;Lpoilit r~ric:l(:tis1)c:c;~usc:the riuclcar charge. is tlist,ril)~lt,c~tl ovc:r a rcgion with rntlius 11. T h c c:fFcct,ivc: Co\ilonib potcwt,i;ll c:ii.ii 1)(1 ;ipproxiiiiatc:tl as
IS, 2 ~2p, , 33,3 ~3d, (a) St.iLt,f' (I~li~lit,iltiv(:ly how t k l ~~ i i ~ r g i c0fs i n ~ i o ~ i il(:vc%ls c will 1)c sl.iiftc:tl al~solutcly; ~ r l t l rc:l;~tivc: to c:;~cliotllcr, iilltl expl:iiii p1iysic:~lly :illy tliff(:rc:iic:cs irl tlic shiftas. Sk(:tch t h e uiipertul+l)ctl a n d ~ ) c r t ~ l r l ) ctworgy ~tl lcvcl diagrarns for thc?sc: states. (1)) Givc i ~ i (i ~ ~ ~ ( ~ for s s t~ h eofirst I I orclcr cliangc in crlcrgy of tllc 1s s t a t e assoc:iatc:(l with tho bict t h a t thc rlliclcus is not poirit-like. (c) Estiin;itc thc 2s-2p ericrgy shift under t h e ass~lrllptiorlthat Rln,, t,l~(:~i liitv~t,o go t,o s(:(:oii(l ort1c.r c:ric:rgy c:orrc:c:t,iori. T l i ~ i sthc: exitc~tlstab(,s of ntoniic: sociiiiiii show orily clliatlratic S t i ~ r kt?ff(,c:t,.
5038 T h e Stark c'ffc(:t. Tlic crlc:rgy 1cvc:ls of the gerl arc ill~istr;tt,(:(lin Fig. 5.14.
TL =
2 statf:s of iit,oinic hytlro-
s. T h c l ~( j , r r ~ , ( j , j , r r r j ) = 7 1 ~ i~~1l ~~th(: d energy of tllc: systc:m is Ell
l,
+ m., f u r , ,
4 ,,,,, =
4
1 + ~ ~ I I L ~ ~ Jrrrj L , = *L ,
4
E,,),,,, =E2,;+:711,hWf,, 3
78,=*$.*$,
2 1 E , , + ,,,, = E , , i + - 3. m j h ~ r , , ~ n =j* , .
(b) Whc:n spi~i=O,t,llcrc is no ~pi~l-r(:li~t,ril i:ffci:t so t,hat
For
71,
-- 2 , --
E20,
EZIO = E21 ,
where
cB - . 271b,~ When t h e weak magnf:t,ic firld is applied, thcx contril)l~tior~ of the tc:rm 6, is (Problem 5057) WI,
-
(c) Svc. the, solut,ion of Problem 5042.
Stark showc.cl experinlentally that,, by applying a n external weak uniform elect,ric field, the 4-fold ctegeneracy in t h e n = 2 level of at,otnic hydrogen i:ould 1)e removeit. Investigate this effect by applying perturbation theory, neglecting spill a n d relativistic effect,^.
368
Perturbation Theory
Prob1e.m~and Solutions on Quanturn Mechnr~ics
Specifically: (a) W h a t are the expressioiis for the first order corrections to the energy level? (Do not atternpt to evaluate the radial integrals). (b) Are there any rernairling dege~ieracies'! (c) Draw a n energy level diagrarli for 11. = 2 which shows the 1evc:ls l~cfore and after applicat.ior1 of thc electric fieltl. Describe t,hc spet:tral liilcs t,l~a.t originate from these levels which can 1)r o1)servctl. ( Cl~it:n!lo)
369
Solvirig t,he secular equatiorl
I
-7u1
(O,OIH'/1,0)
0
0
I
we get four roots
Solution: Write the Ha111iltolli;~rlof t8hcsystc~ri~ as H
=
Ho
+ fl', whc1.e
taking tjhc dircc.tjiori of t,hc c!lcctric: fic:lcl E i1.s tlic: z t1irc:c:tioll. I7or ;I wc:a.k fieltl, H ' hcrest of the ( : i ~ l ( : l ~ l ; ~Thc: t i ~ ~same i. r i ~ i ~1 y )~: done for m y integral over angular wavc: filnction, orlccr you have: asc:crtai~lc:d t h a t it tlocs rlot vanish.)
371
(b) T h e energy lcvel for n = 2 without c'orlsidering spin is four-fold s respectively degenerate. T h c corrcsyondirig energy ant1 s t a t ~ arc
Supposc a uniforril electric field is apl)liecl alorig the z-axis. Take a s perturbation H' = eE7; = E o V r , w1ic:rc Eo = CEO,", V' = z/a" 2 . a,(, = . S l l l ~ ~
+
=
rcosB/ao,
(B c ~ r k ~ L : ~ )
-
Solution: (a) Take thc: dir.c:c:t;iori of t,llc: llli~gll(:t,i(: fi('l(l iLS t,h(: z (lir(:(:t,io~l. TII(:IIt,l~(: Hairiiltoniari of tllc: syst,orrr is
$.
whcrc V ( r ) = Corisi(1cring H' = filnc:tioris for thc: 1l1ij)c:rturl)c~d st,nt,c:s arc. -~
f2
H~l,rrL,,r,,lrrL # 0 for o ~ l l yA1 &1, Ant elemcr~tsof thc: pc:rturl)ation rlli~t,rixarc
=
o. Hcnc:c thc: noll-var~ishir~g
as ~ ) ( r r t , ~ ~ r t ) i ~tali(! t , i o(:ig(:rlli~ Lrt (H')200,zlo= (H')210,200= El, i.e., Ho1= Hlo = E', a n d solve the secular c'qu a.t lor1 det I H,,, - E(')b,, ( = 0 . '
As ( H , 1" ,1) are still corlsc:rvc:d cll~antitics,( ~ i . t , t n l)i, ~I I , ~ ~ I I = ,) energy splitt,ings t,o first ortlcr for 7,. = 2 arc givc.11 t)y
,rri,Ji,
i ~ n dthc
The roots a r c E(') = +El,0,O. Hcncc the energy state n = 2 splits into three lcvcls:
E2 & E ' , E2
(two-fold dcgcricracy for E 2 ) .
(c,) Assutning t h a t the magnctic field is along the z-axis a.nd t h e electric field is along tile z-axis, the perturbation Harlliltorlian of t,he system is
where
372
Perturbation Theory
Problems and Solutions o n Q u a n t ~ ~Mechanzcs m
T h e non-vanishing matrix elements of z are 1-1,m
Heric:~:the: cnergy state 71 = 2 splits into thrcc levels, of energies
l,7r~-l
(")1,7n-l = ( X ) ~ - ~ , T T L
5043
3 74
-
-
l Z )(I + TTl, - l ) ( l + 'TI?,) ( 2 I + 1 ) ( 2 ! 1) (1.
(71" q
A iiorirc:l;~t,ivistic:liy(1rogc:ri a t o r r ~with , n s1)irilcss cl(:c:ti-ori,is 1)l;~c:cdin a n c.lectric fictltl E iri tlic: z t1ircc:tioir ailtl ;L iii:~gii~ti(: ficltl 'H irrl t,hc 3: dircc:tiori. T h e t.ff(:ct, of t,hc: t,wo ficlds o n the c:irlcrgy l(:vcls arc c:ornp;~r;~l)lc. (a) If tlic ;~t,oriiis ill iL state wit11 TI,, t l i ~~)riii(:i~);~l q~laiit,urniiiliril)(:r, (:qua1 t,o t,wo, stnt,c w1iic:h rliatrix clcirierits ill thc first-order 1)crturl)atiori ca1ciil;~tioriof tlic: c:iic:rgy shifts arc, zc:ro. (1)) Now ot)t;~iii;ill (:(~~i;~tioii for tlic crlc:rgy sliift,~;oricc yo11 liavc: tali(: deteriiiiii;~iit;~l (:(l~ii~t,ioii yo11 11(:(:(1110t go tliro~rglithc: ;~lg(:l)ri~ of ~ ~ i ~ l ~ i i ~ t i ~ i g the d(~t(:rr:iiiii~lit.Do iiot i r i s ~ r t l i ~pr(:(:is(: forills of tali(:r;~(lii~l W ~ V Cf i i 1 1 ~ tions; c:xI)rc:ss your rc:s~iltsiri tc:riiis of rnatrix (:l(:iii(:iits of rrl (whcrc 71. is ail appro~)ri;~t,c: 1)owc:r) 1)c:twc:c:ii r;~(lii~l W;LV(: fii~i(:tioiis.
Thus, for n = 2,
and t h r 1)t'rtilrl)i~tioii111;~trixis
(c, f ,LC,)IC, 711,) -- J{(c
f
rrt,) (e
*
711
+ 1 ) ) IC, ,ITL * I ) . (Berkeley)
Solution: ( i ~ Tlic ) pcrturbnt,ioii IIarniltouian is
L r t tlir stat(, vectors for T h e secular equatiori
0 det
-
0
-7 has roots
0
~ ( 1 )
-pY
Y Y
-
E(I)
-E(l)
-0
71 =
2 be 1200), /210), 1211), 121, -1). As
Pert,t~rbatior~ Theory
As z
=
r c:os H. we have
375
Solution: Both particles can stay in thc ground state because they are not identical. The energy and wave function arc rcsl)cct,ively
ST
with ( r ) = r 3 R m R21t/r, ot,ticr rrint,rix dc:~ilciit,sof z 1)ciiig wro. H m c e the perturbatiol~11li~tl.i~ is
If out p;~rt,icleis in the grountl s t i ~ t cthe , other ill t,he first c:xited state, t,he e~iergicsand c:orresponding wave fu~ic:t,ioiisarc
E21
=
2 . 2.ir:cl .
5fi2T2
T3:2
7,$21 , = - S l l l -Slll . trr~L L L L -
Who11 l)ot,h 1);~rtic:lcsare ill t,lic: singl(:-1)i~rtic:lt:ground st,i~tc:,i.c:., thc syst,cln is ill thc: g r o l ~ ~st,at,o, ~ t l wc: 11;~vcthc: c:llc:rgy c:orrc:c:tioi~
and tho wi~vc,fiiiic.t,ioli t.o xc:~-ot,liordcr in X
d m ,
wli(,r(: