Sakurai Chapter-2
Solution
Masatsugu Sei Suzuki
Department of Physics, SUNY at Binghamton
(Date: October 12, 2011)
_
(2-1)
_
(Solution)
Sx, Sy, and Sz are the operators in the Schrdinger picture and Sx(t), Sy(t), and Sz(t) are the
operators in the Heisenb
1.5
(Solution)
y>
1 1
y
2 i
y
SG-y
1
2 1 ( i ),
i
2
2
1
1 i 1 ( i )
2
2
cos
2
n cos e i sin
2
2
e i sin
2
n cos
(a)
2
e i sin
cos , e i sin
2
2
2
+> y
-> y
cos 1
1
2
1 i
(cos ie i sin )
y n
2
2
2
2
e i sin
2
and
y z
1
1
1
HW-18
Chapter 7 (8, 10, 11, 12)
_
TWD 7-8
Sei 7-8
(Solution)
m = 1000 kg, f0 = 1000 Hz.
xM
2E
2
m0
1
2 0 ( n )
2 2n 1
2
m0
m0
(a)
xM n 0
= 4.09683 x 10-21 m
m0
Note that the size of proton is on the order of 10-15 m.
(b)
xM n 1 xM n
( 2n 3 2n 1)
m0
1
HW-14
Chapter 6 (2, 4, 5, 6)
_
TWD 6-2
Sei 6-2
(Solution)
(a)
p x p x x x dx
x p x x dx
1
i
x exp( px) x dx
2
where
x p
1
i
exp( px) ,
2
px
1
i
exp( px)
2
On the other hand,
i
p i p x x dx
p
p
1
i
i
exp( px) x dx
2 p
i
i
1
i
( ) x exp( px) x dx
2
HW-13
Chapter 5 (16, 18, 21, 24)
_
TWD 5-16
Sei 5-16
(Mathematica)
Clear"Global`"; x
0 1
; z a, b;
1 0
zc a1, b1;
z OuterTimes, z , zc
a a1, a b1, a1 b, b b1
1
U
2
1 1
; UH TransposeU;
1 1
x a UH .1, 0 b UH.0, 1
a
2
b
,
2
a
2
b
2
xc x . a a1, b b1
a1
2
b
Second Hour Exam
Quantum Mechanics II
11-1-13 (Friday)
9:40 10:40 AM
Open-book examination
_
1
(3-20) HW
_
TWD 3-20
Sei 3-20
(Solution)
We note that
1n
0
n
1 cos
sin
1 cos
1
0
1 ,
2
2
2
1 n
sin
sin
1 cos 0
1 ,
2
2
1 cos
sin
1 cos
1 ,
1
0
2
2
2
(
Fist Hour Exam
October 04, 2013 (Friday)
Phys.421 and Phys.508
_
1.
(33 points)
SP
Townsend 1-5
Sei 1.5
(Solution)
y>
1
1
y
2 i
y
+> y
-> y
SG-y
1
2 1 ( z i z ),
i
2
2
1
1 i 1 ( z i z )
2
2
cos
2
n cos z e sin z
i
2
2
e sin
2
n cos z e
Sakurai Chapter 4
Solutions
Masatsugu Sei Suzuki
Department of Physics, SUNY at Binghamton
(Date: October 12, 2011)
_
(4-1)
_
The energy is given by
2
2
2
E (nx , n y , nz ) E0 (nx n y nz )
where E0
2 2
( ) , and nx , n y , nz are positive integers.
2m L
Sakurai Chapter 3
Masatsugu Sei Suzuki
Department of Physics, SUNY at Binghamton
(Date: October 12, 2011)
_
(3-1)
_
We know that
y
1
1 1
,
[ i ]
2
2 i
and
y
1
2
[ i ]
1
2
2
2
, where 1
The probability of finding the state y is
2
P
y
2
1
1 i
Sakurai
Chapter-1 Solution
Masatsugu Sei Suzuki, Department of Physics
(Dare: October 12, 2011)
_
(1-1)
_
[ AB, CD] ACcfw_D, B Acfw_C , BD Ccfw_D, AB cfw_C , ADB
Proof:
Left had side = ABCD CDAB
Right hand side = AC ( DB BD) A(CB BC ) D C