Chapter 3 - ANGULAR MOMENTUM - TWO SOURCES
Problems with solutions
3.1 An electron in an atom is in an uncoupled spin and orbital eigenstate given by
m ; ms  1 1; where we have omitted s  1/ 2 from the ket because all electrons
have s  1/ 2 . What is the probability that a measurement of the total angular momentum
will yield the value j  3 / 2 ? j  1/ 2 ? j  5 / 2 ?
Clebsch-Gordan coefficients for j2  1/ 2
1
j1 , ; m1 , m2 j m j
2
m2  1 / 2  ms   2
j
j
j1  1 / 2
j1  1 / 2
1

m2  1 / 2  ms   2
 m j  1 / 2 / 2 j1  1
j
1
Solution:
We need to find the wave function
 m j  1 / 2 / 2 j1  1
j
 m j  1 / 2 / 2 j1  1
1
 m j  1 / 2 / 2 j1  1
 1, m  1,
j
1
in terms of the coupled basis set
1
1
  (recall that m is given in
2
2
the problem and ms  1/ 2 ). Therefore, for the coefficient of the j  3 / 2 coupled state, we
need the top coefficient in the first column with j1  1 and m j  1/ 2 . That is, the wave
j , m j . In our case j1   1 and m j  m  ms  1 
function that we have, 1  1; , can be written as a linear combination of coupled
eigenfunctions j , m j :
m ;  1  1;
3
1
1
1
 A j  :m j    B j  ; m j  
2
2
2
2
We do not need any  's because our eigenstate is one of  . It is also one of m j , namely
m j  1/ 2 so we don't need any other values of m j in our expansion. From the table
  1  1
1    2   2 


  1
A 
 2 1  1
3
Probability of finding the electron with 3 j  3 / 2  A  1/ 3
2
Probability of finding the electron with j  1/ 2  B  1  A  2 / 3
Probability of finding the electron with j  5 / 2  0
Note that the other coefficient, B could have been evaluated from the table.
2
2
  1  1
1    2   2 


  2
B 
3
 2 1  1
and we have A  B  1 as it must.
2
2
Solution to Chapter 3 problems
page 1
Chapter 3 - ANGULAR MOMENTUM - TWO SOURCES
Problems with solutions
1 1 1
 ;1
. What is
2 2 2
the probability that a measurement of the z-component of the orbital angular momentum
will yield the value m  0 ? m  1 ? m  1 ?
Solution:
We must expand the coupled eigenket as a linear combination of uncoupled kets
3.2 An electron in an atom is in an coupled eigenstate j m j ; s 
1
m ; s ms . In our case these are the 1 m ; 2 ms . We use the table of Clebsch-Gordan
coefficients in the previous problem. Since j1   1 and the state that we must expand
has j  1/ 2 , we need the bottom row, i.e. the row that gives j  1/ 2 . Further, m j  1/ 2 ,
as given in the coupled ket in the problem. The coefficients are
 1 1
1    
 2 2   2

2 1  1
3
 1 1
1    
 2 2  1
and 
2 1  1
3
so that the coupled ket is given by
1 1 1
2
11
1
1 1
 ;1

1 m  1;

1 m  0; 
2 2 2
3
22
3
2 2
Note that the values m  1 and m  0 in each of the uncoupled kets are consequences of
the first one being spin up, +1/2, so that m  1 is required to make m j  1/ 2 . The
second one is spin down, –1/2, so that m  0 is required to make m j  1/ 2 .
We read off the probabilities:
2
 1  1
Probability of m  0 : 
 
 3 3
Probability of m  1 : zero
2
2
 2 
Probability of m  1 :  
 3
3

1
3.3 Show that Jˆ1  Jˆ 2  Jˆ1z Jˆ2 z    Jˆ1 Jˆ2  Jˆ1 Jˆ2
 2
Solution:
Jˆ1  Jˆ 2  Jˆ1x Jˆ 2 x  Jˆ1 y Jˆ 2 y  Jˆ1z Jˆ 2 z


  1   Jˆ
 2
  1   Jˆ
 4
  1   Jˆ
 2
But Jˆx   1  Jˆ  Jˆ
 2
Jˆ1  Jˆ 2




and Jˆ y   1  Jˆ  Jˆ . Therefore
  12   Jˆ
 2i 


 Jˆ2   1  Jˆ1  Jˆ1
  21i   Jˆ

1
 Jˆ1
1
Jˆ2  Jˆ1 Jˆ2  Jˆ1 Jˆ2  Jˆ1 Jˆ2  Jˆ1 Jˆ2  Jˆ1 Jˆ2  Jˆ1 Jˆ2  Jˆ1 Jˆ2  Jˆ1 z Jˆ2 z
1
Jˆ2  Jˆ1 Jˆ2  Jˆ1z Jˆ2 z
2
 2i 
2
 Jˆ2  Jˆ1z Jˆ2 z


3.4 An eigenfunction of Ĵ 2 and Jˆ z (total angular momentum and its z-component) is given
Solution to Chapter 3 problems
page 2
Chapter 3 - ANGULAR MOMENTUM - TWO SOURCES
Problems with solutions
in terms of the uncoupled eigenfunctions j1 m j1 ; j2 m j 2 as
3 3 1
2 3 3
 ;1  1 
 ;10
5 2 2
5 2 2
Using the operator identity of the previous problem, find the value of the total angular
momentum and its z-component.
Solution:
We must operate on  with Ĵ 2 and Jˆ z and find the eigenvalues.
 

Ĵ 2 first: Jˆ 2  Jˆ1  Jˆ2

2
 Jˆ12  Jˆ22  2Jˆ1  Jˆ 2
From the previous problem we write the last term in terms of the raising and lowering
operators so that
2
 1 

Jˆ 2  Jˆ1  Jˆ2  Jˆ12  Jˆ22  2   Jˆ1 Jˆ2  Jˆ1 Jˆ2  Jˆ1z Jˆ2 z 
 2 

Now evaluate each of these terms separately and add them up at the end.
15 2
 3  3 
Jˆ12       1 2  
 and Jˆ22   11  1 2   2 2 
4
 2  2 




Note that  is an eigenfunction of both Ĵ12 and Ĵ 22 since each of the constituent basis
kets has the same value of j1 and j2.
3 3 1
2 3 3
Jˆ1 Jˆ 2   Jˆ1 Jˆ 2
 ;1  1  Jˆ1 Jˆ 2
 ;1 0
5 2 2
5 2 2
The first term vanishes because Ĵ 2 lowers it out of existence. We have
2ˆ 3 3
2
3 3
Jˆ1 Jˆ2   Jˆ1
J 2  ;10  Jˆ1
11  1   0 ..
 ;1  1
5
2 2
5
2 2
1
5
2
2
3  3   3  3  3 1
3
  1        1  ;1  1  2
2  2   2  2  2 2
5
2
3 1
 ;1  1
2 2
3 3 1
2 3 3
Jˆ1 Jˆ 2   Jˆ1 Jˆ 2
 ;1  1  Jˆ1 Jˆ 2
 ;1 0
5 2 2
5 2 2
The second term vanishes because Ĵ1 lowers it out of existence. We have
3ˆ 3 1
Jˆ1 Jˆ2   Jˆ1
J 2  ;1  1
5
2 2
3
3 1
11  1   1 0 
 ;10
5
2 2
 Jˆ1

3
6
5
2
5
3  3   1  1  3 3
  1       1  ;10
2  2   2  2  2 2
2
2
3 3
 ;10
2 2
3 3 1
2 3 3
2 Jˆ1z Jˆ 2 z   2 Jˆ1z Jˆ 2 z
 ;1  1  2 Jˆ1z Jˆ 2 z
 ;1 0
5 2 2
5 2 2
Solution to Chapter 3 problems
page 3
Chapter 3 - ANGULAR MOMENTUM - TWO SOURCES
Problems with solutions
The second term vanishes because the eigenvalue of Jˆ2 z is 0.
3 3 1
3 1 
2 Jˆ1z Jˆ2 z   2 Jˆ1z Jˆ2 z
 ;1  1  2  

5 2 2
5 2 
Now, to evaluate Ĵ 2  add up all these terms.

3 1
3
 ;1  1 
2 2
5
2
3 1
 ;1 1
2 2
15
3 3 1
2 3 3
3 2 3 1
Jˆ 2    2   2 2   2  2  ;1  1  3  2  ;1 0 

 ;1  1
4
5 2 2
5 2 2
5 2 2
or
 3 2 3 1

15 2
2 2 3 3
Jˆ 2  
  2 2   3
 ;1  1 
 ;10 
4
5
2 2
 5 2 2

35 2
 15

 5  5 
   2  3 2  
      1 2 
4
 4

 2  2 
Therefore, this is the eigenket corresponding to j  5 / 2 .
To obtain the value of m j we may operate with Jˆz  Jˆz1  Jˆz 2 or it is much simpler to
notice that the sums of the individual z-components add up to –3/2 for each of the
1
3
3
3
uncoupled kets that compose  , i.e.   1  
and   0   . Therefore, the
2
2
2
2
ˆ
eigenvalue of J z for this  is 3/2.
3.5 An eigenfunction of Ĵ 2 and Jˆ z (total angular momentum and its z-component) is given
in terms of the uncoupled eigenfunctions j1 m j1 ; j2 m j 2 as
2 33
1 31
8 3 1
;1  1 
;10 
 ;11
5 22
15 2 2
15 2 2
Using the table of Clebsch-Gordan coefficients below, find the value of the total angular
momentum and its z-component.
 
j1 
3
2
j1  1
j
5
2
j
3
2
j
1
2
m
m1
m2
5
2
3
2
3
2
3
2
1
2
1
2
1
1
0
3
2
1
2
2/ 5
1
1
0
1

2
3

2
5

2
3
2
1
2

1
2

3
2
1
2

1
2
3/ 5
1/10
2/ 5
1/ 2
1/15
 1/ 3
 2/ 5
3/ 5
3/ 5
Solution to Chapter 3 problems
page 4
Chapter 3 - ANGULAR MOMENTUM - TWO SOURCES
Problems with solutions
1
1
2

1
2
1

1
2
0

1
2
1

3
2
1

3
2
0

3
2
1
3/10
8/15
1/ 6
 8/15
3/10
1/ 6
 1/15
3/ 5
3/ 5
 1/ 3
2/ 5
 2/ 5
1/10
1/ 2
 3/ 5
2/ 5
1
Solution:
Since it is given that  is an eigenfunction of Ĵ 2 and Jˆ z it must be a linear combination
of the uncoupled kets as given in a column in the table. This particular  is that in the
column for j  3 / 2 ; m  1/ 2 .
Solution to Chapter 3 problems
page 5