Problem Solutions for

Problem Solutions for Modern Quantum Mechanics, 3rd Edition J.J. Sakurai and Jim Napolitano Jim Napolitano Contents

• Fundamental Concepts 2
• Quantum Dynamics16
• Theory of Angular Momentum 39
• Symmetry in Quantum Mechanics 68
• Approximation Methods 72
• Scattering Theory104
• Identical Particles 117
• Relativistic Quantum Mechanics 127

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Copyright, Cambridge University Press.2 Chapter One 1.See the accompanyingMathematicanotebook.

2.ACfD; Bg=ACDB+ACBD,AfC; BgD=ACBD+ABCD,CfD; AgB =CDAB+ CADB, andfC; AgDB =CADB+ACDB. ThereforeACfD; Bg+AfC; B gDCfD; AgB + fC; AgDB =ACDB +ABCDCDAB+ACDB=ABCDCDAB= [AB; CD] 3.Recall that

(A) 2

=hA 2 i hAi 2 andjSx; +i= 1 p 2 j+i+ 1 p 2 ji(1.4.17a). Since both parts of the problem deal withSy, recall also thatSy= h 2 [ij+ihj +ijih+j]. Therefore SyjSx; +i= h 2 h i p 2 ji i p 2 j+i i andS 2 yjSx; +i=

h 2

2 h 1 p 2 j+i+ 1 p 2 ji i . ThushS 2 yi=

h 2

2 andhSyi= 0, so

(Sy) 2

=

h 2

2 . This would seem to be a problem because

(Sx) 2

= 0 and the left side of the inequality statement is zero. However [Sx; Sy] =ihSzandSzjSx; +i= h 2 h 1 p 2 j+i 1 p 2 ji i sohSzi= 0 and both sides of the uncertainty inequality are zero.For second part,S 2 zjSx; +i=

h 2

2 h 1 p 2 j+i+ 1 p 2 ji i sohS 2 zi=

h 2

2 and

(Sz) 2

=

h 2

2 and the left side of the uncertainty inequality is

h 2

4 . The right side needs [Sz; Sy] =ihSx so is 1 4 h 2

h 2

2 and the uncertainty relation is satised by the equality.

4.(a)Tr(X) =a0Tr(1) + P ` Tr(`)a`= 2a0sinceTr(`) = 0. Also Tr(kX) =a0Tr(k) + P ` Tr(k`)a`= 1 2 P ` Tr(k`+`k)a`= P ` k`Tr(1)a`= 2ak. So, a0= 1 2 Tr(X) andak= 1 2 Tr(kX). (b) Just do the algebra to nda0= (X 11+X22)=2, a1= (X12+X21)=2,a2=i(X 21+X12)=2, anda3= (X11X22)=2.

5.Since det( a) =a 2 z(a 2 x+a 2

• =jaj

2 , the cognoscenti realize that this problem really has to do with rotation operators. From this result, and (3.2.44), we write det

exp

i^n 2

= cos

2

isin

2

and multiplying out determinants makes it clear that det(a

• = det( a). Similarly, use

(3.2.44) to explicitly write out the matrixa

and equate the elements to those ofa.With^nin thez-direction, it is clear that we have just performed a rotation (of the spin vector) through the angle.

6.(a)Tr(XY) P a hajXYjai= P a P b hajXjbihbjY jaiby inserting the identity operator.Then commute and reverse, soTr(XY) = P b P a hbjYjaihajX jbi= P b hbjY Xjbi=Tr(Y X).(b)XYji=X[Yji] is dual tohj(XY) y , butYji j iis dual tohjY y hjandXji is dual tohjX y so thatX[Yji] is dual tohjY y X y . Therefore (XY ) y =Y y X y .(c) exp[if (A)] = P a exp[if(A)]jaihaj = P a exp[if(a)]jaihaj (d) P a

a(x

• a(x

00

• =

P a hx

jai

hx 00 jai= P a hx 00 jaihajx

i=hx 00 jx

i=(x 00 x

) 2 / 4

Copyright, Cambridge University Press.3 7.For basis ketsjaii, matrix elements ofX jihjareXij=haijihjaji=haijihajji

.For spin-1/2 in thej zibasis,h+jSz= h=2i= 1,hjSz= h=2i= 0, and, using (1.4.17a), hjSx= h=2i= 1= p

• Therefore

jSz= h=2ihSx= h=2j

:

= 1 p 2

• 1
• 0

8.A[jii+jji] =aijii+ajjji 6= [jii+jji] so in general it is not an eigenvector, unlessai=aj.That is,jii+jjiis not an eigenvector ofAunless the eigenvalues are degenerate.

9.Since the product is over a complete set, the operator Q a 0(Aa

• will always encounter

a statejaiisuch thata

=aiin which case the result is zero. Hence for any stateji Y a

(Aa

)ji= Y a

(Aa

) X i jaiihaiji= X i Y a

(aia

)jaiihaiji= X i

• = 0

If the product instead is over alla

6=ajthen the only surviving term in the sum is Y a

(aja

)jaiihaiji and dividing by the factors (aja

• just gives the projection ofjion the directionja

• For

the operatorASzandfja

ig fj+i;jig, we have Y a

(Aa

• =

Sz h 2

Sz+ h 2

and Y a

6=a 00 Aa

a 00 a

= Sz+ h=2 h fora 00 = + h 2 or = Szh=2 h fora 00 = h 2 It is trivial to see that the rst operator is the null operator. For the second and third, you can work these out explicitly using (1.3.35) and (1.3.36), for example Sz+ h=2 h = 1 h

Sz+ h 2 1

= 1 2 [(j+ih+j)(jihj) + (j+ih+j) + (jihj)] =j+ih+j which is just the projection operator for the statej+i.

10.I don't see any way to do this problem other than by brute force, and neither did the previous solutions manual. So, make use ofh+j+i= 1 =hjiandh+ji= 0 =hj+iand carry through six independent calculations of [Si; Sj] (along with [Si; Sj] =[Sj; Si]) and the six forfSi; Sjg(along withfSi; Sjg= +fSj; Sig). 3 / 4

Copyright, Cambridge University Press.4 11.From the gure^n= ^ icossin+ ^ jsinsin+ ^ kcosso we need to nd the matrix representation of the operatorS^n=Sxcossin+Sysinsin+Szcos. This means we need the matrix representations ofSx,Sy, andSz. Get these from the prescription (1.3.19) and the operators represented as outer products in (1.4.18) and (1.3.36), along with the association (1.3.39a) to dene which element is which. Thus Sx

:

= h 2

• 1
• 0

Sy

:

= h 2

0i i0

Sz

:

= h 2

• 0

01

We therefore need to nd the (normalized) eigenvector for the matrix

cos cossinisinsin cossin+isinsin cos

=

cos e i sin e i sincos

with eigenvalue +1. If the upper and lower elements of the eigenvector areaandb, respec- tively, then we have the equationsjaj 2 +jbj 2 = 1 and acos+be i sin=a ae i sinbcos=b Choose the phase so thatais real and positive. Work with the rst equation. (The two equations should be equivalent, since we picked a valid eigenvalue. You should check.) Then a 2 (1cos) 2 =jbj 2 sin 2 = (1a 2

• sin

2

4a 2 sin 4 (=2) = (1a 2 )4 sin 2 (=2) cos 2 (=2) a 2 [sin 2 (=2) + cos 2 (=2)] = cos 2 (=2) a= cos(=2) and sob=ae i 1cos sin = cos(=2)e i

• sin

2 (=2)

• sin(=2) cos(=2)

=e i sin(=2) which agrees with the answer given in the problem.

12.Use simple matrix techniques for this problem. The matrix representation forHis H

:

=

a a aa

EigenvaluesEsatisfy (aE)(aE)a 2 =2a 2 +E 2 = 0 orE=a p

• Letx1andx2

be the two elements of the eigenvector. ForE= +a p 2E (1) , (1 p 2)x (1) 1+x (1) 2= 0, and forE=a p 2E (2)

, (1 +

p 2)x (2) 1+x (2) 2= 0. So the eigenstates are represented by jE (1) i

:

=N (1)

1 p 21

and jE (2) i

:

=N (2)

1 p

• + 1
• / 4