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