Solutions to Problem 11.4
4. [15] An electron is in the + state when measured in the direction S  S z cos   S x sin  ,
so that S    12   . However, the angle  is uncertain. In each part, it is
probably a good idea to check at each step that the trace comes out correctly.
(a) [3] Suppose the angle is    13  , with equal probability for each angle. What is the
state operator in the conventional  z basis?
We have, from a variety of sources, the states in terms of this basis, which is
  cos  12     sin  12   
The state vector is then simply taken by averaging the results for the two angles, so

1
 
2   3
 
2
1  cos 6
 
2  cos 6 sin 6
1  cos 6 
cos 6


 
2  sin 6 
 cos 6 
sin 6   
cos 6

 
  sin 6 
cos 6 sin 6   cos 2 6

sin 2 6    cos 6 sin 6

 sin 6  

 cos 6 sin 6    cos 2 6
  
sin 2 6    0
0   34 0 

.
sin 2 6   0 14 
The result has trace one, so it’s probably right.
(b) [4] Suppose the angle  is randomly distributed in the range  12     12  , with all
angles equally likely. What is the state operator in the conventional  z basis?
Instead of adding two angles, we need to integrate over all angles, and divide by the
range of angles, which is  , so we have
 cos  12   

  cos  12   sin  12   d
  12 
  12   sin  12   
cos 2  12  
cos  12   sin  12   
sin  
1 12  
1 12   1  cos 
d

  1 



 d
2
1

1
1
1

sin  2  
1  cos  
2  2   sin 
  2   cos  2   sin  2  


1
1
2

 d 
1
1
2

1    sin   cos   2
1  2  1



 
2   cos    sin     2  0
2
0    2  1
0    12  1

  

0
 2  1   0
2  1 
0 
.
1
1 

2
 
Once again, the trace is one, so it’s probably correct.
(c) [4] Suppose the angle  is randomly distributed in the range      , with all
angles equally likely. What is the state operator in the conventional  z basis?
This is identical to the previous part, except the range is twice as big and of course the
limits of integration change, so we have
1

2

1
4



1
 d   
4

sin  
1  cos 
1

  sin  1  cos d  4


   sin   cos  


  cos    sin  
  1   1    12 0 


   0 1 

1


1





 
2

So this is a completely unpolarized state operator. It again has trace one.
(d) [4] In each of the cases listed above, what is the expectation value of Sz?
The expectation value can be found via
S z  Tr   S z   12 Tr   z 
In every case, this trace is easy to work out.
Sz
Sz
Sz
a
b
b
 3 0   1 0   1 3 1
1
 12 Tr  4 1  
   2   4  4   4 ,
0
0
1


4 

 1  1
 12 Tr  2 
 0
1
2
0  1 0   1 1 1 1 1
1
         

 1   0 1  2 2  2 
 1 0   1 0   1 1 1
 12 Tr  2 1  
   2   2  2   0.
  0 2   0 1  
Perhaps not surprisingly, the first two cases have a positive expectation value, while the third
vanishes. This is because in the first two cases, though the spin is random, it’s definitely at an
angle that is closer to +z than –z, but in the third it is equally likely at all angles.