Quantum Mechanics A Paradigms Approach 1st Edition McIntyre Solutions Manual

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1. Ch. 2 Solutions
3/20/19 2-1
2.1 Let
S
x
ab
cd






and write the S
x eigenvalue equations in matrix notation
ab
cd






1
2
1
1







2
1
2
1
1






ab
cd






1
2
1
1







2
1
2
1
1






which yields
ab
2
            cd
2
ab
2
            cd
2
Solve by adding and subtracting the equations to get
a0      b
2
      c
2
      d0
Hence the matrix representing S
x in the S
z basis is
S
x
2
01
10






Let
S
y
ab
cd






and write the S
y eigenvalue equations in matrix notation
ab
cd






1
2
1
i







2
1
2
1
i






ab
cd






1
2
1
i







2
1
2
1
i






which yields
aib
2
            cidi
2
aib
2
            cidi
2
Solve by adding and subtracting the equations to get
a0      bi
2
      ci
2
      d0

Quantum Mechanics A Paradigms Approach 1st Edition McIntyre Solutions Manual
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2. Ch. 2 Solutions
3/20/19 2-2
Hence the matrix representing S
y in the S
z basis is
S
y
2
0i
i0







2.2 Solve the secular equation
detS
x
I0

2
2

0
Solve to find the eigenvalues

2

2






2
0

2

2






2

2
which was to be expected, because we know that the only possible results of a
measurement of any spin component are 2 . Find the eigenvectors. For the positive
eigenvalue:
2
01
10






a
b







2
a
b






yields
ba
The normalization condition yields
a
2
a
2
1
a
2

1
2
Choose a to be real and positive, resulting in
a
1
2
b
1
2

3. Ch. 2 Solutions
3/20/19 2-3
so the eigenvector corresponding to the positive eigenvalue is

x
1
2

1
1






Likewise, the eigenvector for the negative eigenvalue is
2
01
10






a
b







2
a
b






ba
The normalization condition yields
a
2
a
2
1
a
2

1
2
Choose a to be real and positive, resulting in
a
1
2
b
1
2
so the eigenvector corresponding to the negative eigenvalue is

x
1
2

1
1







2.3 From Eq. (1.37), we know the S
z eigenstates in the S
x basis:

1
2

x
    
x 

1
2

x
    
x 
Let the representation of S
z in the S
x basis be
S
z
ab
cd






and write the S
z eigenvalue equations in matrix notation
ab
cd






1
2
1
1







2
1
2
1
1






ab
cd






1
2
1
1







2
1
2
1
1







4. Ch. 2 Solutions
3/20/19 2-4
These yield
ab
2
            cd
2
ab
2
            cd
2
Solve by adding and subtracting the equations to get
a0      b
2
      c
2
      d0
Hence the matrix representing S
z in the S
x basis is
S
z
S
x2
01
10






Now diagonalize:

2
2

0        
2

2
2
0        
2
as expected. Find the eigenvectors:
2
01
10






a
b







2
a
b






        ba
yielding

S
x
1
2

1
1






Likewise
2
01
10






a
b







2
a
b






       ba

S
x
1
2

1
1






Hence the eigenvalue equations are
2
01
10






1
2

1
1







2
1
2

1
1






            S
z

2
    OK
2
01
10






1
2

1
1







2
1
2

1
1






            S
z

2
    OK

5. Ch. 2 Solutions
3/20/19 2-5
2.4 The general matrix is
A
ab
cd






The matrix elements are
A10
ab
cd






1
0






10
a
c






a
A10
ab
cd






0
1






10
b
d






b
A01
ab
cd






1
0






01
a
c






c
A01
ab
cd






0
1






01
b
d






d
Hence we get
A
AA
AA









2.5 The commutators are
[S
x
,S
y
]
2
01
10






2
0i
i0






    
2
0i
i0






2
01
10






2






2
i0
0i







i0
0i














2






2
2i0
02i






i
2






10
01






 iS
z
[S
y
,S
z
]
2
0i
i0






2
10
01






    
2
10
01






2
0i
i0






2






2
0i
i0







0i
i0














2






2
02i
2i0






i
2






01
10






 iS
x

6. Ch. 2 Solutions
3/20/19 2-6
[S
z,S
x]
2
10
01






2
01
10






    
2
01
10






2
10
01






2






2
01
10







01
10














2






2
02
20






i
2






0i
i0






 iS
y

2.6 The spin component operator Sn is
S
n
Sˆn
S
x
sincosS
y
sinsinS
z
cos
Using the matrix representations for Sx, Sy, and Sz gives
S
n
2
01
10






sincos
2
0i
i0






sinsin
2
10
01






cos
2
cos sincosisinsin
sincosisinsin cos






2
cossine
i
sine
i
cos







2.7 Diagonalize Sn:
S
n
2
cossine
i
sine
i
cos






Now diagonalize:
2
cos
2
sine
i
2
sine
i

2
cos
0

2

2
2
cos
2

2
2
sin
2
0        
2

2
2
0        
2
as expected. Find the eigenvectors:

7. Ch. 2 Solutions
3/20/19 2-7
2
cossine
i
sine
i
cos






a
b







2
a
b






acosbsine
i
a        bae
i1cos
sin
The normalization condition yields
a
2
a
21cos
sin






2
1
a
2

sin
2

22cos

4sin
2
2cos
2
2
4sin
2
2
cos
2
2
yielding

n
cos

2
e
i
sin

2

Likewise
2
cossine
i
sine
i
cos






a
b







2
a
b






acosbsine
i
a        bae
i1cos
sin
The normalization condition yields
a
2
a
21cos
sin






2
1
a
2

sin
2

22cos

4sin
2
2cos
2
2
4cos
2
2
sin
2
2
yielding

n
sin

2
e
i
cos

2


2.8 The 
n eigenstate is

n
cos

2
e
i
sin

2
cos

8
e
i53
sin

8

The probabilities are

8. Ch. 2 Solutions
3/20/19 2-8

P
y

y

n
2

1
2

i
2
 cos

8
e
i53
sin

8
 
2

1
2
cos

8
ie
i53
sin

8
2

1
2
cos

8
sin

8
sin
5
3
isin

8
cos
5
3
2

1
2
cos
2
8
sin
2
8
sin
25
3
sin
2
8
cos
25
3
2cos

8
sin

8
sin
5
3 

1
2
12cos

8
sin

8
sin
5
3 0.194

P
y

y

n
2

1
2

i
2
 cos

8
e
i53
sin

8
 
2

1
2
cos

8
ie
i53
sin

8
2

1
2
cos

8
sin

8
sin
5
3
isin

8
cos
5
3
2

1
2
cos
2
8
sin
2
8
sin
25
3
sin
2
8
cos
25
3
2cos

8
sin

8
sin
5
3 

1
2
12cos

8
sin

8
sin
5
3 0.806

2.9 The expectation value of S
z is easy to do in Dirac notation:
S
z
S
z

2

2

2
The expectation values of S
x and S
y are easier in matrix notation:
S
x
10
2
01
10






1
0







2
10
0
1






0
S
y
10
2
0i
i0






1
0







2
10
0
i






0
To find the uncertainties, we need the expectation values of the squares:
S
z
2
S
z
2

22

2






2

2






2
S
x
2
10
2
01
10






2
01
10






1
0







2






2
10
1
0







2






2
S
y
2
10
2
0i
i0






2
0i
i0






1
0







2






2
10
1
0







2






2
The uncertainties are

9. Ch. 2 Solutions
3/20/19 2-9
S
zS
z
2
S
z
2

2






2

2






2
0
S
x
S
x
2
S
x
2

2






2
0
2






S
yS
y
2
S
y
2

2






2
0
2







2.10 These expectation values are easier in matrix notation:
S
x
1
2
1i
2
01
10






1
2
1
i







4
1i
i
1






0
S
y

1
2
1i
2
0i
i0






1
2
1
i







4
1i
1
i







2
S
z
1
2
1i
2
10
01






1
2
1
i







4
1i
1
i






0
To find the uncertainties, we need the expectation values of the squares:
S
x
2

1
2
1i
2
01
10






2
01
10






1
2
1
i







1
22






2
1i
1
i







2






2
S
y
2

1
2
1i
2
0i
i0






2
0i
i0






1
2
1
i







1
22






2
1i
1
i







2






2
S
z
2

1
2
1i
2
10
01






2
10
01






1
2
1
i







1
22






2
1i
1
i







2






2

The uncertainties are
S
xS
x
2
S
x
2

2






2
0
2






S
y
S
y
2
S
y
2

2






2

2






2
0
S
zS
z
2
S
z
2

2






2
0
2







10. Ch. 2 Solutions
3/20/19 2-10
In the vector model, shown below, the spin is precessing around the y-axis at a constant
angle such the y-component of the spin is constant and x- and z-components oscillate
about zero.



2.11 The commutators in matrix notation are
[S
2
,S
x
]
3
2
4
10
01






2
01
10






    
2
01
10






3
2
4
10
01






3
3
8
01
10







01
10














0
[S
2
,S
y
]
3
2
4
10
01






2
0i
i0






    
2
0i
i0






3
2
4
10
01






3
3
8
0i
i0







0i
i0














0
[S
2
,S
z
]
3
2
4
10
01






2
10
01






    
2
10
01






3
2
4
10
01






3
3
8
10
01







10
01














0

11. Ch. 2 Solutions
3/20/19 2-11
In abstract notation, the commutators are
[S
2
,S
x
]S
x
2
S
y
2
S
z
2
,S
x




S
x
2
,S
x




S
y
2
,S
x




S
z
2
,S
x




S
x
2
S
x
S
x
S
x
2
S
y
2
S
x
S
x
S
y
2
S
z
2
S
x
S
x
S
z
2
S
x
3
S
x
3
S
y
S
y
S
x
S
x
S
y
S
y
S
z
S
z
S
x
S
x
S
z
S
z
0S
y
S
x
S
y
iS
z S
y
S
x
iS
z S
y
S
z
S
x
S
z
iS
y S
z
S
x
iS
y S
z
S
y
S
x
S
y
iS
y
S
z
S
y
S
x
S
y
iS
z
S
y
S
z
S
x
S
z
iS
z
S
y
S
z
S
x
S
z
iS
y
S
z
0
[S
2
,S
y
]S
x
2
S
y
2
S
z
2
,S
y




S
x
2
,S
y




S
y
2
,S
y




S
z
2
,S
y




S
x
2
S
y
S
y
S
x
2
S
y
2
S
y
S
y
S
y
2
S
z
2
S
y
S
y
S
z
2
S
x
S
x
S
y
S
y
S
x
S
x
S
y
3
S
y
3
S
z
S
z
S
y
S
y
S
z
S
z
S
x
S
y
S
x
iS
z S
x
S
y
iS
z S
x
0S
z
S
y
S
z
iS
x S
z
S
y
iS
x S
z
S
x
S
y
S
x
iS
x
S
z
S
x
S
y
S
x
iS
z
S
x
S
z
S
y
S
z
iS
z
S
x
S
z
S
y
S
z
iS
x
S
z
0
[S
2
,S
z
]S
x
2
S
y
2
S
z
2
,S
z




S
x
2
,S
z




S
y
2
,S
z




S
z
2
,S
z




S
x
2
S
z
S
z
S
x
2
S
y
2
S
z
S
z
S
y
2
S
z
2
S
z
S
z
S
z
2
S
x
S
x
S
z
S
z
S
x
S
x
S
y
S
y
S
z
S
z
S
y
S
y
S
z
3
S
z
3
S
x
S
z
S
x
iS
y S
x
S
z
iS
y S
x
S
y
S
z
S
y
iS
x S
y
S
z
iS
x S
y
0
S
x
S
z
S
x
iS
x
S
y
S
x
S
z
S
x
iS
y
S
x
S
y
S
z
S
y
iS
y
S
x
S
y
S
z
S
y
iS
x
S
y
0

2.12 For S
x the diagonalization yields the eigenvalues
S
x
2
010
101
010









2
0
2

2
0
2













0     
2

2
2
2

20 

2

2
0     1,0,1
and the eigenvectors

12. Ch. 2 Solutions
3/20/19 2-12
2
010
101
010








a
b
c








1
a
b
c








      
ba2
acb2
bc2
a
2
b
2
c
2
1      b
2
1
2
1
1
2 1      b
1
2
,a
1
2
,c
1
2
1
x

1
2
1
1
2
0
1
2
1
2
010
101
010








a
b
c








0
a
b
c








      
b0
ac0
b0
a
2
b
2
c
2
1      a
2
111      a
1
2
,b0,c
1
2
0
x

1
2
1
1
2
1
2
010
101
010








a
b
c








1
a
b
c








      
ba2
acb2
bc2
a
2
b
2
c
2
1      b
2
1
2
1
1
2 1      b
1
2
,a
1
2
,c
1
2
1
x

1
2
1
1
2
0
1
2
1
For S
y the diagonalization yields
S
y
2
0i0
i0i
0i0









i
2
0
i
2

i
2
0
i
2













0     
2

2
2
i
2

i
20 

2

2
0     ,0,
and the eigenvectors
2
0i0
i0i
0i0








a
b
c








1
a
b
c








      
iba2
iaicb2
ibc2
a
2
b
2
c
2
1      b
2
1
2
1
1
2 1      b
i
2
,a
1
2
,c
1
2
1
y

1
2
1
i
2
0
1
2
1

13. Ch. 2 Solutions
3/20/19 2-13
2
0i0
i0i
0i0








a
b
c








0
a
b
c








      
ib0
iaic0
ib0
a
2
b
2
c
2
1      a
2
111      a
1
2
,b0,c
1
2
0
y

1
2
1
1
2
1
2
0i0
i0i
0i0








a
b
c








1
a
b
c








      
iba2
iaicb2
ibc2
a
2
b
2
c
2
1      b
2
1
2
1
1
2 1      b
i
2
,a
1
2
,c
1
2
1
y

1
2
1
i
2
0
1
2
1

2.13 The commutators are
[S
x
,S
y
]
2
010
101
010







2
0i0
i0i
0i0








    
2
0i0
i0i
0i0







2
010
101
010








2
2
i0i
000
i0i









i0i
000
i0i


















2
2
2i00
000
002i








i
100
000
001








 iS
z
[S
y
,S
z
]
2
0i0
i0i
0i0








100
000
001








    
100
000
001







2
0i0
i0i
0i0








2
2
000
i0i
000









0i0
000
0i0


















2
2
0i0
i0i
0i0








i
2
010
101
010








 iS
x

14. Ch. 2 Solutions
3/20/19 2-14
[S
z
,S
x
]
100
000
001







2
010
101
010








    
2
010
101
010








100
000
001








2
2
010
000
010









000
101
000


















2
2
010
101
010








i
2
0i0
i0i
0i0








 iS
y

2.14 Using the component matrices we find
S
2
S
x
2
S
y
2
S
z
2
2
010
101
010







2
010
101
010









2
0i0
i0i
0i0







2
0i0
i0i
0i0









100
000
001








100
000
001








2
2
100
020
001









2
2
100
020
001









100
000
001








2
2
100
010
001








The eigenvalue equation is
S
2
smss1
2
sm
For spin-1 this is
S
2
1m2
2
1m
Hence the S
2 operator must be 2
2 times the identity matrix:
S
2
2
2
100
010
001









15. Ch. 2 Solutions
3/20/19 2-15
2.15 a) The possible results of a measurement of the spin component S
z are always 1, 0, 1
for a spin-1 particle. The probabilities are

P
1
1
in
2
1
2
29
1
3i
29
0
4
29
1



2

2
29
11
3i
29
10
4
29
11
2

2
29
2

4
29

P
0
0
in
2
0
2
29
1
3i
29
0
4
29
1



2

2
29
01
3i
29
00
4
29
01
2

3i
29
2

9
29

P
1
1
in
2
1
2
29
1
3i
29
0
4
29
1



2

2
29
11
3i
29
10
4
29
11
2

4
29
2

16
29
The three probabilities add to unity, as they must.
b) The possible results of a measurement of the spin component S
x are always 1, 0, 1
for a spin-1 particle. The probabilities are

P
1x
x1
in
2

1
2
1
1
2
0
1
2
1 
2
29
1
3i
29
0
4
29
1 
2

1
29

3i
229

2
29
2

1
58
23i
2

11
58

P
0x

x
0
in
2

1
2
1
1
2
1 
2
29
1
3i
29
0
4
29
1 
2

2
229

4
229
2

36
58

P
1x
x1
in
2

1
2
1
1
2
0
1
2
1 
2
29
1
3i
29
0
4
29
1 
2

1
29

3i
229

2
29
2

1
58
23i
2

11
58
The three probabilities add to unity, as they must.
c) For the first measurement, the expectation value is

S
z
mP
m
m
 1
4
29
0
9
29
1
16
29

12
29
For the second measurement, the expectation value is

S
x
mP
mx
m
 1
11
58
0
36
58
1
11
58
0
The histograms are shown below.

16. Ch. 2 Solutions
3/20/19 2-16


2.16 a) The possible results of a measurement of the spin component S
z are always 1, 0, 1
for a spin-1 particle. The probabilities are

P
1
1
in
2
1
2
29
1
y

3i
29
0
y

4
29
1
y




2

2
29
11
y

3i
29
10
y

4
29
11
y
2

2
29
1
2

3i
29
1
2

4
29
1
2
2

1
58
23i
2

11
58

P
0
0
in
2
0
2
29
1
y

3i
29
0
y

4
29
1
y




2

2
29
01
y

3i
29
00
y

4
29
01
y
2

2
29
i
2

3i
29
0
4
29
i
2
2

1
58
i24
2

36
58

P
1
1
in
2
1
2
29
1
y

3i
29
0
y

4
29
1
y




2

2
29
11
y

3i
29
10
y

4
29
11
y
2

2
29
1
2

3i
29
1
2

4
29
1
2
2

1
58
23i
2

11
58
The three probabilities add to unity, as they must.
b) The possible results of a measurement of the spin component S
y are always 1, 0, 1
for a spin-1 particle. The probabilities are

P
1y

y
1
in
2

y
1
2
29
1
y

3i
29
0
y

4
29
1
y




2

2
29y
11
y

3i
29y
10
y

4
29y
11
y
2

2
29
2

4
29

P
0y

y
0
in
2

y
0
2
29
1
y

3i
29
0
y

4
29
1
y




2

2
29y
01
y

3i
29y
00
y

4
29y
01
y
2

3i
29
2

9
29

17. Ch. 2 Solutions
3/20/19 2-17

P
1y

y
1
in
2

y
1
2
29
1
y

3i
29
0
y

4
29
1
y




2

2
29y
11
y

3i
29y
10
y

4
29y
11
y
2

4
29
2

16
29
The three probabilities add to unity, as they must.
c) For the first measurement, the expectation value is

S
z
mP
m
m
 1
11
58
0
36
58
1
11
58
0
For the second measurement, the expectation value is

S
y
mP
my
m
 1
4
29
0
9
29
1
16
29

12
29
The histograms are shown below.


2.17 a) The possible results of a measurement of the spin component S
z are always 1, 0, 1
for a spin-1 particle. The probabilities are

P1
2
100 
1
30
1
2
5i








2

1
30
1
2

1
30

P
0
0
2
010 
1
30
1
2
5i








2

1
30
2
2

4
30

P-1
2
001 
1
30
1
2
5i








2

1
30
5i
2

25
30
The expectation value of S
z is

S
z
PP
0
0P


1
30

4
30
0
25
30

24
30

4
5
b) The expectation value of S
x is

18. Ch. 2 Solutions
3/20/19 2-18
S
x
S
x

1
30
125i 
2
010
101
010








1
30
1
2
5i









1
302
125i 
2
15i
2









1
302
2215i5i2 
2
15


2.18 a) The possible results of a measurement of the spin component S
z are always 1, 0, 1
for a spin-1 particle. The probabilities are

P
1
1
in
2
1
1
14
1
3
14
0
2i
14
1



2

1
14
11
3
14
10
2i
14
11
2

1
14
2

1
14

P
0
0
in
2
0
1
14
1
3
14
0
2i
14
1



2

1
14
01
3
14
00
2i
14
01
2

3
14
2

9
14

P
1
1
in
2
1
1
14
1
3
14
0
2i
14
1



2

1
14
11
3
14
10
2i
14
11
2

2i
14
2

4
14
b) After the S
z measurement, the system is in the state 1 . The possible results of a
measurement of the spin component S
x are always 1, 0, 1 for a spin-1 particle.
The probabilities are

P
1x

x
1
in
2

1
2
1
1
2
0
1
2
1 1
2

1
2
2

1
4

P
0x

x
0
in
2

1
2
1
1
2
1 1
2

1
2
2

1
2

P
1x

x
1
in
2

1
2
1
1
2
0
1
2
1 1
2

1
2
2

1
4
c) Schematic of experiment.

19. Ch. 2 Solutions
3/20/19 2-19
2.19 The probability is

P

f

f

i
2

1i
7y
1
2
7y
0
i
7y
1 
1
6
1
2
6
0
i3
6
1 
2

1i
7
1
6y
11
1i
7
2
6y
10
1i
7
i3
6y
11
2
7
1
6y
01
2
7
2
6y
00
2
7
i3
6y
01

i
7
1
6y
11
i
7
2
6y
10
i
7
i3
6y
11
2

1i
7
1
6
1
2

1i
7
2
6
i
2

1i
7
i3
6
1
2

2
7
1
6
1
2

2
7
2
6
0
2
7
i3
6
1
2


i
7
1
6
1
2

i
7
2
6
i
2

i
7
i3
6
1
2
2

1
168
1i22i3i3220i26i23
2

1
168
5222ii3i26
2

1
168
522
2
2326 
2
 

1
168
64824386 0.524

or in matrix notation

P

f

f

i
2

1i
7
1
2
i
2
1
2

2
7
1
2
0
1
2

i
7y
1
2
i
2
1
2 
1
6
1
2
i3










2

1
16812222i2122 
1
2
i3










2

1
168
12242ii3i26
2

1
168
64824386 0.524


2.20 Spin 1 unknowns. Follow the solution method given in the lab handout. (i) For
unknown number 1, the measured probabilities are

P
1
1
4
P
1x
1
4
P
1y1
P
0
1
2
P
0x
1
2
P
0y0
P
1

1
4
P
1x

1
4
P
1y
0
Write the unknown state as

20. Ch. 2 Solutions
3/20/19 2-20

1
a1b0c1
Equating the predicted S
z probabilities and the experimental results gives

P
1
1
1
2
1a1b0c1 
2
a
2

1
4
      a
1
2
P
0
0
1
2
0a1b0c1 
2
b
2

1
2
      b
1
2
e
i
P
1
1
1
2
1a1b0c1 
2
c
2

1
4
      c
1
4
e
i
allowing for possible relative phases. So now the unknown state is

1

1
2
1
1
2
e
i
0
1
2
e
i
1
Equating the predicted S
x probabilities and the experimental results gives

P
0x

x
0
1
2

1
2
11 
1
2
1
1
2
e
i
0
1
2
e
i
1 
2

1
22
1e
i

2

1
8
1e
i
1e
i

1
8
11e
i
e
i
 
1
4
1cos 
1
2
cos1
Giving the state

1

1
2
1
1
2
e
i
0
1
2
1
Equating the predicted S
y probabilities and the experimental results gives

P
1y

y
1
1
2

1
2
1
i
2
0
1
2
1 
1
2
1
1
2
e
i
0
1
2
1 
2

1
4

i
2
e
i

1
4
2

1
4
1ie
i
1ie
i

1
4
11ie
i
ie
i
 
1
2
1sin 1sin1

2
Hence the unknown state is

1

1
2
1
1
2
e
i

2
0
1
2
1
1
2
1
i
2
0
1
2
11
y
(ii) For unknown number 2, the measured probabilities are

P
1
1
4
P
1x
9
16
P
1y0.870
P
0
1
2
P
0x
3
8
P
0y0.125
P
1

1
4
P
1x

1
16
P
1y
0.005
Write the unknown state as

2
a1b0c1
Equating the predicted S
z probabilities and the experimental results gives

21. Ch. 2 Solutions
3/20/19 2-21

P
1
1
2
2
1a1b0c1 
2
a
2

1
4
      a
1
2
P
0
0
2
2
0a1b0c1 
2
b
2

1
2
      b
1
2
e
i
P
1
1
2
2
1a1b0c1 
2
c
2

1
4
      c
1
2
e
i
allowing for possible relative phases. So now the unknown state is

2

1
2
1
1
2
e
i
0
1
2
e
i
1
Equating the predicted S
x probabilities and the experimental results gives

P
0x

x
0
2
2

1
2
11 
1
2
1
1
2
e
i
0
1
2
e
i
1 
2

1
22
1e
i

2

1
8
1e
i
1e
i

1
8
11e
i
e
i
 
1
4
1cos 
3
8
cos
1
2

2
3
,
4
3

P
1x

x
1
2
2

1
2
1
1
2
0
1
2
1 
1
2
1
1
2
e
i
0
1
2
e
i
1 
2

1
4

1
2
e
i

1
4
e
i
2

1
16
12e
i
e
i
 12e
i
e
i
 
1
16
64cos2cos4cos 

1
16
64cos2cos4coscos4sinsin 

1
16
52cos4sinsin 
9
16
which yields
cos2sinsin2
2sinsin2cos
4sin
2
sin
2
44coscos
2

41cos
2
 
3
4
44coscos
2

4cos
2
4cos10
2cos10
cos
1
2


3
,
5
3
Equating the predicted S
y probabilities and the experimental results gives

P
1y

y
1
2
2

1
2
1
i
2
0
1
2
1 
1
2
1
1
2
e
i
0
1
2
e
i
1 
2

1
4

i
2
e
i

1
4
e
i
2

1
16
12ie
i
e
i
 12ie
i
e
i
 
1
16
64sin2cos4sin 

1
16
64sin2cos4sincos4cossin 

1
16
76sin4cossin 0.87
which gives

22. Ch. 2 Solutions
3/20/19 2-22
3sin2cossin3.4623
2cossin233sin
4cos
2
sin
2
12123sin9sin
2

41sin
2
 
3
4
12123sin9sin
2

sin
2
3sin
3
4
0
sin
3
2
0
sin
3
2


3
,
2
3


3

2
3
Hence the unknown state is

2

1
2
1
1
2
e
i

3
0
1
2
e
i
2
3
11
n

2
,

3
(iii) For unknown number 3, the measured probabilities are

P
1
1
3
P
1x
1
6
P
1y0.0286
P
0
1
3
P
0x
2
3
P
0y0
P
1

1
3
P
1x

1
6
P
1y
0.9714
Write the unknown state as

3
a1b0c1
Equating the predicted S
z probabilities and the experimental results gives

P
1
1
3
2
1a1b0c1 
2
a
2

1
3
      a
1
3
P
0
0
3
2
0a1b0c1 
2
b
2

1
3
      b
1
3
e
i
P
1
1
3
2
1a1b0c1 
2
c
2

1
3
      c
1
3
e
i
allowing for possible relative phases. So now the unknown state is

3

1
3
1
1
3
e
i
0
1
3
e
i
1
Equating the predicted S
x probabilities and the experimental results gives

P
0x
x0
3
2

1
2
11 
1
3
1e
i
0e
i
1 
2

1
6
1e
i

2

1
6
1e
i
1e
i

1
6
11e
i
e
i
 
1
3
1cos 
2
3
cos1
Equating the predicted S
y probabilities and the experimental results gives

23. Ch. 2 Solutions
3/20/19 2-23

P
1y

y
1
3
2

1
2
1
i
2
0
1
2
1 
1
3
1e
i
01 
2

1
3
1
2

i
2
e
i

1
2 
2

1
3
1
i
2
e
i
1
i
2
e
i

1
3
3
2
2sin 0.0286
sin1
3
2
Hence the unknown state is

3

1
3
1
i
3
0
1
3
1m
n
(iv) For unknown number 4, the measured probabilities are

P
1
1
2
P
1x
1
4
P
1y
1
4
P
00P
0x
1
2
P
0y
1
2
P
1

1
2
P
1x

1
4
P
1y

1
4
Write the unknown state as

4
a1b0c1
Equating the predicted S
z probabilities and the experimental results gives

P
1
1
4
2
1a1b0c1 
2
a
2

1
2
      a
1
2
P
0
0
4
2
0a1b0c1 
2
b
2
0      b0
P
1
1
4
2
1a1b0c1 
2
c
2

1
2
      c
1
2
e
i
allowing for possible relative phases. So now the unknown state is

4

1
2
1
1
2
e
i
1
Equating the predicted S
x probabilities and the experimental results gives

P
0x

x
0
4
2

1
2
11 
1
2
1e
i
1 
2

1
2
1e
i

2

1
4
1e
i
1e
i

1
4
11e
i
e
i
 
1
2
1cos 
1
2
cos0

2
,
3
2
Giving

4

1
2
1i1 
with no more info from Sx, Sy, Sz measurements. In the SPINS program choose n at
angles  = 90˚,  = 45˚, 225˚ to see that the unknown state is

4

1
2
1i1 0
n

2
,

4
0
n

2
,
5
4  

24. Ch. 2 Solutions
3/20/19 2-24
2.21. The spin-1 interferometer had an S
z SG device, an S
x deveice, and an S
z SG
device. The S
z eigenstates are 1,0,1 . The S
x eigenstates are
1
x

1
2
1
1
2
0
1
2
1
0
x

1
2
1
1
2
1
1
x

1
2
1
1
2
0
1
2
1 .
Let 
i be the quantum state after the i
th
Stern-Gerlach device. The first SG device
transmit particles with S
z
 , so the state 
1 is

1
1 .
The second SG device transmits particles in 1, 2, or 3 of the S
x eigenstates 1
x
,0
x
,1
x
. To find 
2 , we use the projection postulate:

2

P
n

1

1P
n
1
where Pn is the projection operator onto the measured states. For example, if the second
SG device transmits particles with S
x
 , we get

2

P
1x

1

1P
1x
1

1
x  x
11
11
x  x
11
1
x
as expected. In matrix notation, the S
x eigenstates are
1
x
12
12
12










0
x
12
0
12










1
x
12
12
12










and the projection operators are

25. Ch. 2 Solutions
3/20/19 2-25
P
1x
1
x  x
1
1
2
1
2
1
2










1
2
1
2
1
2 

1
4
1
22
1
4
1
22
1
2
1
22
1
4
1
22
1
4










P
0x
0
x  x
0
1
2
0
1
2










1
2
0
1
2 

1
20
1
2
000
1
20
1
2










P
1x
1
x  x
1
1
2
1
2
1
2










1
2
1
2
1
2 

1
4
1
22
1
4
1
22
1
2
1
22
1
4
1
22
1
4










The probability of measuring a result after the third SG device is
P
2
3
2 . We
want to calculate the three probabilities

P
1
1
2
2
P
0
0
2
2
P
1
1
2
2
for all possible (7) cases of 1 beam, 2 beams, or 3 beams from SG2.
(i) When the second SG device transmits particles with S
x
1 only:

2

P
1x

1

1P
1x
1

P
1x
1
1P
1x1
P
1x1
1
4
1
22
1
4
1
22
1
2
1
22
1
4
1
22
1
4










1
0
0









1
4
1
22
1
4










1P
1x
1 100 
1
4
1
22
1
4










14

2
1
14
1
4
1
22
1
4











1
2
1
2
1
2










which is 1
x as expected. The three probabilities are

26. Ch. 2 Solutions
3/20/19 2-26

P
1
1
2
2
100 
1
2
1
2
1
2










2

1
4
P
0
0
2
2
010 
1
2
1
2
1
2










2

1
2
P
1
1
2
2
001 
1
2
1
2
1
2










2

1
4
(ii) When the second SG device transmits particles with S
x
0 , 
2
0
x and the
three probabilities are

P
1
1
2
2
100 
1
2
0
1
2










2

1
2
P
00
2
2
010 
1
2
0
1
2










2
0
P
1
1
2
2
001 
1
2
0
1
2










2

1
2
(iii) When the second SG device transmits particles with S
x
1 , 
2
1
x and the
three probabilities are

27. Ch. 2 Solutions
3/20/19 2-27

P
1
1
2
2
100 
1
2

1
2
1
2










2

1
4
P
0
0
2
2
010 
1
2

1
2
1
2










2

1
2
P
1
1
2
2
001 
1
2

1
2
1
2










2

1
4
(iv) When the second SG device transmits particles with S
x
 and S
x
0 in a
coherent beam

2

P
1x
P
0x 
1

1
P
1x
P
0x 
1

P
1x
P
0x 1
1P
1x
P
0x 1
P
0x
1
1
20
1
2
000
1
20
1
2










1
0
0









12
0
12










P
1x
P
0x 1
14
122
14











12
0
12











34
122
14










1P
1x
P
0x 1 100 
34
122
14










34

2
1
34
34
122
14











32
16
123












The three probabilities are

28. Ch. 2 Solutions
3/20/19 2-28

P
1
1
2
2
100 
32
16
123












2

3
4
P
0
0
2
2
010 
32
16
123












2

1
6
P
1
1
2
2
001 
32
16
123












2

1
12
(v) When the second SG device transmits particles with S
x
 and S
x
1 in a
coherent beam

2

P
1x
P
1x 
1

1
P
1x
P
1x 
1

P
1x
P
1x 1
1P
1x
P
1x 1
P
1x
1
1
4
1
22
1
4
1
22
1
2
1
22
1
4
1
22
1
4










1
0
0









14
122
14










P
1x
P
1x 1
14
122
14











14
122
14











12
0
12










1P
1xP
1x 1 100 
12
0
12










12

2
1
12
12
0
12











12
0
12










The three probabilities are

29. Ch. 2 Solutions
3/20/19 2-29

P
1
1
2
2
100 
12
0
12










2

1
2
P
00
2
2
010 
12
0
12










2
0
P
1
1
2
2
001 
12
0
12










2

1
2
(vi) When the second SG device transmits particles with S
x
0 and S
x
1 in a
coherent beam

2

P
0x
P
1x 
1

1
P
0x
P
1x 
1

P
0x
P
1x 1
1P
0x
P
1x 1
P
0x
P
1x 1
12
0
12











14
122
14











34
122
14










1P
0x
P
1x 1 100 
34
122
14










34

2
1
34
34
122
14











32
16
123












The three probabilities are

30. Ch. 2 Solutions
3/20/19 2-30

P
1
1
2
2
100 
32
16
123












2

3
4
P
0
0
2
2
010 
32
16
123












2

1
6
P
1
1
2
2
001 
32
16
123












2

1
12
(vii) When the second SG device transmits particles with S
x
 , S
x
0 , and S
x
1
in a coherent beam.

2

P
1xP
0xP
1x 
1

1P
1xP
0xP
1x 
1

P
1xP
0xP
1x 1
1P
1xP
0xP
1x 1
P
1x
1
1412214
12212122
1412214












1
0
0









14
122
14










P
1x
P
0x
P
1x 1
14
122
14











12
0
12











14
122
14











1
0
0








1P
1x
P
0x
P
1x 1 100 
1
0
0








1

2
1
0
0








which is 1 as expected. The three probabilities are

31. Ch. 2 Solutions
3/20/19 2-31

P
11
2
2
100 
1
0
0








2
1
P
0
0
2
2
100 
0
1
0








2
0
P
1
1
2
2
100 
0
0
1








2
0
The cumulated results are


2
P
1
P
0
P
1
1
x
1
4
1
2
1
4
0
x
1
2
0
1
2
1
x
1
4
1
2
1
4
1
x
&0
x
3
4
1
6
1
12
1
x
&1
x
1
2
0
1
2
0
x
&1
x
3
4
1
6
1
12
1
x
&0
x
&1
x
100

2.22 a) The probability of measuring spin up at the 2
nd
Stern-Gerlach analyzer and spin
down at the 3
rd
Stern-Gerlach analyzer is the product of the individual probabilities:

P
nzP
nP
nz
n
2

n
2
The 
n eigenstate is

n
cos

2
e
i
sin

2

so the probability is

P
nz
cos

2
e
i
sin

2
 
2
cos

2
e
i
sin

2
 
2
cos
2
2
sin
2
2

1
4
sin
2

b) To maximize the probability requires that 2 , and the probability is

P
nz

1
4
sin
2
2

1
4
c) If the 2
nd
Stern-Gerlach analyzer is removed, then the probability is

32. Ch. 2 Solutions
3/20/19 2-32

P
z

2
0
because the two states are orthogonal.

2.23 (a) The commutator is
A,BABBA
a
1
00
0a
2
0
00a
3










b
1
00
00b
2
0b
2
0











b
1
00
00b
2
0b
2
0










a
1
00
0a
2
0
00a
3










a
1
b
1
0 0
0 0a
2
b
2
0a
3
b
2
0











a
1
b
1
0 0
0 0a
3
b
2
0a
2
b
2
0










0 0 0
0 0 b
2
a
2
a
3
0b
2
a
3
a
2 0










0
so they do not commute.
(b) A is already diagonal, so the eigenvalues and eigenvectors are obtained by inspection.
The eigenvalues are
a
1
,a
2
,a
3
and the eigenvectors are
a
1
1
1
0
0








,   a
2
2
0
1
0








,   a
3
3
0
0
1








For B, diagonalization yields the eigenvalues
b
100
0b
2
0b
2











0     b
1

2
b
2
2
0 
   b
1
,b
2
,b
2
and the eigenvectors

33. Ch. 2 Solutions
3/20/19 2-33
b
100
00b
2
0b
20










u
v
w








b
1
u
v
w








      
b
1ub
1u
b
2wb
1v
b
2vb
1w
      wv0
u
2
v
2
w
2
1      u
2
1      u1,v0,w0      b
1
1
1
0
0








b
1
00
00b
2
0b
2
0










u
v
w








b
2
u
v
w








      
b
1
ub
2
u
b
2
wb
2
v
b
2
vb
2
w
      u0,  wv
b
2
b
2
1     v
2
w
2
1     u0,v
1
2
,w
1
2
     b
2

1
2
23
0
1
2
1
2










b
1
00
00b
2
0b
2
0










u
v
w








b
2
u
v
w








      
b
1
ub
2
u
b
2
wb
2
v
b
2
vb
2
w
      u0,  wv
b
2
b
2
1    v
2
w
2
1    u0,v
1
2
,w
1
2
    b
2

1
2
23
0
1
2

1
2











c) If B is measured, the possible results are the allowed eigenvalues b
1
,b
2
,b
2 . If the
initial state is 
i
2 , then the probabilities are

P
b
1
b
1

i
2
12
2
0
P
b
2
b
2

i
2

1
2
232
2

1
2
P
b
2
b
2

i
2

1
2
232
2

1
2
If A is then measured, the possible results are the allowed eigenvalues a
1
,a
2
,a
3 . If b2 was
the first result, then the new state is b
2 and when A is measured the subsequent
probabilities are

34. Ch. 2 Solutions
3/20/19 2-34

P
a
1
a
1
b
2
2
1
1
2
23
2
0
P
a
2
a
2
b
2
2
2
1
2
23
2

1
2
P
a
3
a
3
b
2
2
3
1
2
23
2

1
2
If -b2 was the first result, then the new state is b
2 and when A is measured the
subsequent probabilities are

P
a
1
a
1
b
2
2
1
1
2
23
2
0
P
a
2
a
2
b
2
2
2
1
2
23
2

1
2
P
a
3
a
3
b
2
2
3
1
2
23
2

1
2
d) If two operators do not commute, then the corresponding observables cannot be
measured simultaneously. Part (a) tells us that the operators A and B not commute. Part
(c) tells us that measurement B "disturbs" the measurement of A so the two measurements
are not compatible (cannot be made simultaneously).

2.24 (a) The eigenvalue equations for the S
z operator and the four eigenstates are
S
z

3
2

3
2

3
2
S
z
1
2

1
2

1
2
S
z
1
2

1
2

1
2
S
z
3
2

3
2

3
2
(b) The matrix representations of the S
z eigenstates are the unit vectors

3
2
1
0
0
0












,  
1
2
0
1
0
0












,  
1
2
0
0
1
0












,  
3
2
0
0
0
1












(c) The matrix representation of the S
z operator has the eigenvalues along the diagonal:
S
z

3
2
0 0 0
0
1
2
0 0
0 0
1
2
0
0 0 0
3
2














(d) The eigenvalue equations for the S
2 operator follow from the general equation S
2
sm
s
ss1
2
sm
s

35. Ch. 2 Solutions
3/20/19 2-35
S
2

3
2

15
4
2

3
2
S
2

1
2

15
4
2

1
2
S
2

1
2

15
4
2

1
2
S
2

3
2

15
4
2

3
2
where we have suppressed the s label.
(e) The matrix representation of the S
2 operator has the eigenvalues along the diagonal:
S
2
15
4
2
0 0 0
0
15
4
2
0 0
0 0
15
4
2
0
0 0 0
15
4
2















2.25 The projection operators P
 and P
 are represented by the matrices
P

10
00






,     P

00
01






The Hermitian adjoints of these matrices are obtained by transposing and complex
conjugating them, yielding
P

† 10
00






,     P

† 00
01






Since the Hermitian adjoints are equal to the original matrices, these operators are
Hermitian.

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