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


Download 59.37 Kb.
bet4/4
Sana16.06.2023
Hajmi59.37 Kb.
#1496162
1   2   3   4
Bog'liq
angularaddition

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





Download 59.37 Kb.

Do'stlaringiz bilan baham:
1   2   3   4




Ma'lumotlar bazasi mualliflik huquqi bilan himoyalangan ©fayllar.org 2024
ma'muriyatiga murojaat qiling