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Unformatted text preview: PHYS 507 Answers to Homework VII (Fall ’05) 1. (a) We have W j = ∑ n‘ ² jn‘ V n U ‘ and therefore [ J i ,W j ] = X n‘ ² jn‘ [ J i ,V n U ‘ ] = X n‘ ² jn‘ [ J i ,V n ] U ‘ + X n‘ ² jn‘ V n [ J i ,U ‘ ] = i ¯ h X nk‘ ² jn‘ ² ink V k U ‘ + i ¯ h X nm‘ ² jn‘ ² i‘m V n U m = i ¯ h X nk‘ ² ‘jn ² kin V k U ‘ + i ¯ h X nm‘ ² jn‘ ² mi‘ V n U m = i ¯ h X k‘ ( δ ‘k δ ji δ ‘i δ jk ) V k U ‘ + i ¯ h X nm ( δ jm δ ni δ ji δ nm ) V n U m = i ¯ hδ ij X k V k U k i ¯ hV j U i + i ¯ hV i U j i ¯ hδ ji X n V n U n = i ¯ h ( V i U j V j U i ) = i ¯ h X k‘ ( δ ik δ j‘ δ i‘ δ jk ) V k U ‘ = i ¯ h X k‘n ² ijn ² k‘n V k U ‘ = i ¯ h X k‘n ² ijn W n Therefore, ~ W = ~ V × ~ U is a vector. (b) ~ J × ~ J = i ¯ h ~ J . 2. (a) Note J n = J x cos φ + J y sin φ = 1 2 (( J + + J )cos φ i ( J + J )sin φ ) = e iφ 2 J + + e iφ 2 J It is now a trivial exercise to show that h J n i = 0 , › J 2 n fi = ¯ h 2 2 ( j ( j + 1)...
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 Trigraph, Jx, Jy Jx, cos Jx Jy

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