Quantum Mechanics II angular Momentum II : Addition of Angular Momentum Clebsch-Gordan Coefficients

Example 5: Adding j₁ = 1 with j₂ = 1

Here in the j₁, j₂, m₁, m₂ basis, there are 3 × 3 = 9 members. In the coupled represen- tation the possible angular momenta are j = 2, 1, 0 with respectively 5, 3 and 1 states, making it 9, as expected. In this case we start with

| 1, 1; 2, 2⟩ =| 1, 1; 1, 1⟩

where the coupled representation is on the left. we apply J₋ on the left and J₁₋ + J₂₋ on the right. we get
^√(2 + 2)(2 − 2 + 1) | 1, 1; 2, 1⟩ = ^√(1 + 1)(1 − 1 + 1) | 1, 1; 0, 1⟩+^√(1 + 1)(1 − 1 + 1) | 1, 1; 1, 0⟩
Thus

| 1, 1; 2, 1⟩ = r¹ | 1, 1; 0, 1⟩ + r¹ | 1, 1; 1, 0⟩ (19)

2 2
Further application of J₋ = J₁₋ + J₂₋ gives

2

2
^√6 | 1, 1, 2, 0⟩ = r¹ (^√2 | 1, 1; −1, 1⟩ + ^√2 | 1, 1, 0, 0⟩) + r¹ (^√2 | 1, 1; 0, 0⟩ + ^√2 | 1, 1, 1, −1⟩)

| 1, 1, 2, 0⟩ = r¹ [| 1, 1; 1, −1⟩ + 2 | 1, 1, 0, 0⟩) + ^√2 | 1, 1; −1, 1⟩] (20)

6

r r
The state with j = 1, m = 1 is orthogonal to the state (19). Using the sign convention discussed above
| 1, 1; 1, 1⟩ = ¹ | 1, 1; 1, 0⟩ − ¹ | 1, 1; 0, 1⟩ (21)
2 2
An application of J₋ = J₁₋ + J₂₋ on (21) gives

2

2
| 1, 1; 1, 0⟩ = r¹ | 1, 1; 1, −1⟩ − r¹ | 1, 1; −1, 1⟩ (22)

It may be noted that in the above state there is no contribution from m₁ = m₂ = 0 state to m = 1 state.

Now the state | 1, 1; 0, 0⟩ is orthogonal to both (20) and (22). Let the state be written as
| 1, 1; 0, 0⟩ = a | 1, 1; 1, −1⟩ + b | 1, 1; 0, 0⟩ + c | 1, 1; −1, 1⟩
where a, b and c are constants to be determined. Since the state is orthogonal tio (22), we have a = c. Orthogonality to (20) gives a + 2b + c = 0. Solving, we get b = −c. Thus the state is
| 1, 1; 0, 0⟩ = a[| 1, 1; 1, −1⟩− | 1, 1; 0, 0⟩+ | 1, 1; −1, 1⟩]
1
The constant a is determined to be ^√₃ by normalization of the state. Thus we get
1
| 1, 1; 0, 0⟩ = ^√₃ [| 1, 1; 1, −1⟩− | 1, 1; 0, 0⟩+ | 1, 1; −1, 1⟩]
Collecting the above results together, we get the Clebsch-Gordan coefficients to be given by
m = 2, 1

⟨1, 1; 2, 2 | 1, 1; 1, 1⟩ = 1

1
⟨1, 1; 2, 1 | 1, 1; 1, 0⟩ = ^√₂
1
⟨1, 1; 2, 1 | 1, 1; 0, 1⟩ = ^√₂
1
⟨1, 1; 1, 1 | 1, 1; 1, 0⟩ = ^√₂
1
⟨1, 1; 1, 1 | 1, 1; 0, 1⟩ = −^√₂

m = 0:

1
⟨1, 1; 2, 0 | 1, 1; 1, −1⟩ = ^√₆

3
⟨1, 1; 2, 0 | 1, 1; 0, 0⟩ = r²

1
⟨1, 1; 2, 0 | 1, 1; −1, 1⟩ = ^√₆
1
⟨1, 1; 1, 0 | 1, 1; 1, −1⟩ = ^√₂
1
⟨1, 1; 1, 0 | 1, 1; −1, 1⟩ = −^√₂
1
⟨1, 1; 0, 0 | 1, 1; 1, −1⟩ = ^√₃
1
⟨1, 1; 0, 0 | 1, 1; 0, 0⟩ = −^√₃
1
⟨1, 1; 0, 0 | 1, 1; −1, 1⟩ = ^√₃