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hw203a - Physics 512 Homework Set#3 Solutions Winter 2003 1...

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Physics 512 Winter 2003 Homework Set #3 – Solutions 1. Let S ( k 1 ) q 1 and T ( k 2 ) q 2 be two irreducible spherical tensor operators of ranks k 1 and k 2 , respectively. We may form the tensor product W ( k ) q defined by W ( k ) q = q 1 + q 2 = q k 1 k 2 ; q 1 q 2 | k 1 k 2 ; k q S ( k 1 ) q 1 T ( k 2 ) q 2 Using the definition of spherical tensors in terms of their commutation properties [ J z , T ( k ) q ] = q ¯ h T ( k ) q [ J ± , T ( k ) q ] = k ( k + 1) q ( q ± 1) ¯ h T ( k ) q ± 1 prove that W ( k ) q is a spherical tensor operator of rank k . To prove that W ( k ) q is a spherical tensor, we show that it satisfies the commutation properties above. We start with the J z commutator (which is the simpler one). [ J z , W ( k ) q ] = q 1 + q 2 = q k 1 k 2 ; q 1 q 2 | k 1 k 2 ; k q [ J z , S ( k 1 ) q 1 T ( k 2 ) q 2 ] = q 1 + q 2 = q k 1 k 2 ; q 1 q 2 | k 1 k 2 ; k q [ J z , S ( k 1 ) q 1 ] T ( k 2 ) q 2 + S ( k 1 ) q 1 [ J z , T ( k 2 ) q 2 ] = q 1 + q 2 = q k 1 k 2 ; q 1 q 2 | k 1 k 2 ; k q q 1 ¯ hS ( k 1 ) q 1 T ( k 2 ) q 2 + q 2 ¯ hS ( k 1 ) q 1 T ( k 2 ) q 2 = q ¯ h q 1 + q 2 = q k 1 k 2 ; q 1 q 2 | k 1 k 2 ; k q S ( k 1 ) q 1 T ( k 2 ) q 2 = q ¯ hW ( k ) q In the penultimate line, we have used the fact that both S ( k 1 ) q 1 and T ( k 2 ) q 2 are spherical tensors in order to compute the commutators with J z . And in the final line, we note that since q 1 + q 2 = q we can pull out an overall factor of q ¯ h . For the other components, we proceed in a similar manner [ J ± , W ( k ) q ] = q 1 + q 2 = q k 1 k 2 ; q 1 q 2 | k 1 k 2 ; k q [ J ± , S ( k 1 ) q 1 ] T ( k 2 ) q 2 + S ( k 1 ) q 1 [ J ± , T ( k 2 ) q 2 ] = q 1 + q 2 = q k 1 k 2 ; q 1 q 2 | k 1 k 2 ; k q k 1 ( k 1 + 1) q 1 ( q 1 ± 1) ¯ hS ( k 1 ) q 1 ± 1 T ( k 2 ) q 2 + k 2 ( k 2 + 1) q 2 ( q 2 ± 1) ¯ hS ( k 1 ) q 1 T ( k 2 ) q 2 ± 1 Since we wish to relate the right hand side as much as possible to W ( k ) q , we shift q 1 q 1 1 in the first term, and q 2 q 2 1 in the second term. The two terms then 1
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combine, giving the result [ J ± , W ( k ) q ] = q 1 + q 2 = q ± 1 k 1 ( k 1 + 1) q 1 ( q 1 1) ¯ h k 1 k 2 ; q 1 1 q 2 | k 1 k 2 ; k q + k 2 ( k 2 + 1) q 2 ( q 2 1) ¯ h k 1 k 2 ; q 1 q 2 1 | k 1 k 2 ; k q S ( k 1 ) q 1 T ( k
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