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Modern Quantum Mechanics答案.pdf
Solution Manual
Chapter 1
Chapter 2
Chapter 3
Chapter 4
Chapter 5
Chapter 6
Chapter 7
Modern
Quantum Mechanics
Solutions Manual
/. /. Sakurai
Late, University of California, Los Angeles
San Fu Tuan, Editor
University of Hawaii, Manoa
THE BENJAMIN/CUMMINGS I J3LISHING COMPANY, INC.
Menlo Park, California • Reading, Massachusetts
Don Mills, Ontario * Wokingham, U.K. • Amsterdam • Sydney
Singapore • Tokyo - Mexico City • Bogota • Santiago • San Juan
Copyright © 1985 by Addison-Wcslcy Publishing Co., The Advanced Book Program,
350 Bridge Parkway, Redwood C'rty, CA 94065
All rights reserved. Mo i*arl of this publication may be reproduced,
stored in a retrieval system, or transmitted, in any form or by any means,
electronic, mechanical, photocopying, recording, or otherwise, without the
prior written permission of the publisher. Printed in the United Slates
of America. Published simultaneously in Canada.
I SEN 0-8053-7502-3
4 5 6 7 MQ 9 5 9 4 9 3 92
Contents
1 Fundamental Concepts
2 Quantum Dynamics
3 Theory of Angular Momentum
4 Symmetry in Quantum Mechanics
5 Approximation Methods
6 Identical Particles
7 Scattering Theory
Chapcer 1
1. [A3,CO] = ABCD - CDAB - ABCD + ACBD - ACSD - ACDB + ACDB + CAD3 - CADB -
2.
3.
CDAB = A{C,B}D - AC{D,B) + {C,A}DB - C{D,A>B.
(a) X =
cr(c^X) • tr(£ a^o^o^) • I 3^2(5^ = 2a^ (where we have used
because tr(o
, cr(X) - 2a
+ Ea
o
Q
t
l
l
) - 0. Next evaluate
tr(c.o..) = CrOiCOjOj + o
o
j
>) = 2 6 ^ ). Hence a
i
* 4 tr(X), a^ •
Q
X).
k
H cr(c
(b) a
with X =
Q
- '^(X^ + X
), while
2 2
and i,j - 1,2. The result is *
can be explicitly evaluated from
) , a
- lj(X
+ X
l
12
2 1
• 4 trCo^X)
+
-. |(-X
21
2
X
), and a
1 2
• 4( X
- X
n
2 2
).
3
a.a = a a +c a + oa
zz
y y
xx
x y
z
a +ia
x y -J *
det (o.t) = -|1|2.
Without loss of generality, choose n along positive z-direction, then
exp(±io.n<)>/2) = 1 cos $/2 t ia
cos $/2 + isin 4>/2, then
sin /2, and if B is defined to be B =
z
exp(io 4>/2)o.a exp(-io $/2)
z
z
B*B
a
z
(a
-ia
x
)B2
y
(a +ia )B*2
—a B*B
Since B*B » cos2 <}>/2 + sin2 $/2 - 1, det [exp(ia^/Da.a x
ex?(-io */2)]» - (a,2 + a 2 + a 2) = -|a|2, that is determinant Is
z
x
y
z
invariant under specified operation. Next we note
/ a'
a'-ia'
V a'+ia*
\ x y
-a*
z
a
•
(a +ia )(cos* -isin$)
.
y
(a -ia )(cos<: + isin$)
* y
-a
z
hence a^ • a^, a^ = a^cos-J + a^sin*, a^ = a
cos<> - a^sir.^. This is a
counter-clockwise rotation, about z-axis through angle * in x-y plane.
(a) Note tr(XY) = E, - ,E „ (by
closure property) - . Z „ (by rearrangement) =
y
a , a
|„ . Since a" is a dummy summation variable, relabel a" •
har.c. tr(rf) - tr(YX).
(b) <(XY)+a'|a"> - = -
m . Therefore (XY)+ = Y+X+.
(c) Take exp[if(A)]|a> = (1 + if (A) -
= (1 + if (a) - t £2 °^ +
assume that A|a> • aja>. Therefore exp(if(A)] •
Eexp[if(a)][a>} has been used.
(d) Z, •
,(x") - |, * - J, x ^
" exp[if(o)]|o>. where we
*(x')*
)|a>
a
a
2
*
- E, <^'|
a
'xa'|x*> - <$*!$•>.
(a) |a><0| - |, |„ |a,xa"|o.xB|a, ,x
a
"|- J, |„ |a'xa"| x
(<«'|o*>. Hence |o><8| - [ < a( i )|
> < a( J )| B > * ], .here
a
expression inside square bracket is the (i,J) matrix element,
(b) ja> = js = K/2> - |+>, |S> = |s = tf/2> = k[\+> + j->].
Hence
<+|a><-!s>
i
<- a><- 8>
i *
Given A|i> - a^\i> and A|j> »" a^|j>. The normalized state vector |i> +
|j> is of form |t|/> »
aj|j>] where a^, a^ are real numbers if A is Hermitian; but for a^ t
|£> + |j>]. Hence A(*> = (l/ZlMaJ^ +
=
clearly r.h.s. is a state vector distinct from jij;>. However under the
condition that | i> and |j> arc degenerate (i.e.
- a), then A\-j>>
a[(l//2)(|i> + |j>)] -a|i(i> and |*> or |i>+ |j> is also an eigenket of
A.
(a) Let k> c { | a * >> and A|a'> - a'|a" * Then since II, (A - a')|£> is
product over all eigenvalues, and |c> * E, |a'> must
therefore satisfy n, (A-a')|£> - 0. Hence n, (A-a') is the null
operator.
n
m
n
| ,
(A-a") I ,
(*'-«"> ,
a'Va* (a'-a") |a
•mi
( b) a"fa' (a'-a") |a
Hence 8 J €> -
, ^ ^ ) U> " |a'>. The
operator therefore projects out of ket |s>, its
ja'> component.
„J
a
a
a l> m
'a *'
z
(c) Let A = S
a'2+K/2 ^Sz "
we have 8
-
+
, than ft, (S^-a') - (S
- tf/2)(S
z
z
+ K/2). Hence evidently
a' > lt
:> " °« T1*1-3 verifies (a) above. For case (b)
(%, 0_ =• -(S
z
-tf/2)/X and S - «/2( |+><+| -j -> x
<-|) while ket j£> -
e_|5> • < - j s > | -> and 8
states.
+
J£> + {_><_j
j-j-><
are the projertion operators of |£> to |±>
>. Hence 6 J5> - <+|?>|+> and
?
+
I
The orthonormality property is <+]+> - <-|-> - 1, <+]-> - <-[+> « 0.
Hence using the explicit representations of S^^ ia terms of linear
combinations of bra-ket products, we obtain by elementary calculation
[S
Let n » ni + n j + n k, then n » singcosa, n • sinSsina, n -
z
and { S ^} =
* S
y
] - i e
x
a2/2)S
,S
z
x
y
y
l j k
i
j
k
±
cos8 and j>.n • sinficosa S + singsina S + cosB S . Also due to
x
y
z
completeness property of the ket space |s\n;+> • a|+> + b|-> where- ]a|2 +
|b|2 = 1 (normalization). Therefore the relation s'.njli.-r';+> *
(ii/2) |£.n;+> [taking advantage of explicit representations S„ • j ( I+> *
l- x+ D,
< -! +
leads to :-
- M(-[+><-| + ! - > < + [ ), sz
sy
- 5<+| - !-><-!)]
(sin6cosc - isin8sina)b + cosg a • a
(singcosa + isingsina)a - cos3 b • b
2
2
Together with the normalization condition |a| + jbj • 1, we find
a • cos(B/2)e a and b - sin(g/2)e b. From equation (la) we have
a = —
rr— , hence e b a - e . Choose 9 • 0, then 6, » a, ana
i(9. -8 )
ia
r
.
_
.
sinB e~i0tb
(x—cosS)
a
o
js".S;+> - cos(B/2)|-«-> + sin(B/2)ela|->.
Modem Quantum Mechanics - Solutions
10. H = a(|l><2| + |l><2| + |2>. Let 11> = ( J) , 12> = (°) ,
<1| - (1,0) and <2| =» (0,1), H can be explicitly written using
outer product of matrices as
\ 1 -1/
The eigenvalues and corresponding eigenkets are obtained from
(H - XI)X = 0 where X * f ] are eigenvectors and X are corresponding
/Xl\
\x2/
eigenvalues determined from secular equation det (H - XI) * 0. This leads
to X » */2a and x
of X we have
= (i/2 -
= /2(2+/2) ' ^ °uS eigenvectors and eigenvalues are
U> + (/2 - 1)J2>
hence X » *if*/J*. l) a nd by
n o r m a l i2ation
2
1
'V = /2(2 - %
' X~ ^
|l> - (/2 + 1)|2>
1*2* = /2(2 + 7T)
' x= -/ 2a
11. Rewrite H as H = 4( H
+ H
U
22
)(|l><2|) + J
(H
S
11
- H
2 2
) (11> x
<2| + |2><+|) + 5 n (-i|+> x
<-j+i|-><+|) + \ n
(|+x+|-|-><-|). The analogy is:
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