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# hw4a44 - PHYS 507 Answers to Homework IV Assigned December...

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PHYS 507 Answers to Homework IV Assigned: December 3, 2003, Wednesday. Due: December 12, 2003, Friday, at 5:00 pm. 1. (a) h J x i = h j, m | J x | j, m i = 1 2 h j, m | ( J + + J - ) | j, m i = ¯ h 2 p j ( j + 1) - m ( m + 1) h j, m | j, m + 1 i + p j ( j + 1) - m ( m - 1) h j, m | j, m - 1 i · = 0 h J y i = 1 2 i h j, m | ( J + - J - ) | j, m i = ¯ h 2 i p j ( j + 1) - m ( m + 1) h j, m | j, m + 1 i - p j ( j + 1) - m ( m - 1) h j, m | j, m - 1 i · = 0 Coupled with h J z i = m ¯ h these relations can be written as D ~ J E = m ¯ h ˆ z . This is a good rule to keep in mind. In general if a state is an eigenstate of J n , then D ~ J E is a vector along ˆ n . For calculating h J 2 x i and J 2 y fi let us use the following trick for a faster calculation J 2 x fi = h j, m | J 2 x | j, m i = h φ | φ i where | φ i = J x | j, m i , and similarly for J 2 y fi . We have J x | j, m i = 1 2 ( J + + J - ) | j, m i = ¯ h 2 p j ( j + 1) - m ( m + 1) | j, m + 1 i + p j ( j + 1) - m ( m - 1) | j, m - 1 i · J y | j, m i = 1 2 i ( J + - J - ) | j, m i = ¯ h 2 i p j ( j + 1) - m ( m + 1) | j, m + 1 i - p j ( j + 1) - m ( m - 1) | j, m - 1 i · which gives J 2 x fi = J 2 y fi = ¯ h 2 2 ( j ( j + 1) - m 2 ) This result is easy to understand. We have h J 2 x i = J 2 y fi by symmetry. Since J 2 = J 2 x + J 2 y + J 2 z we should have J 2 x fi = 1 2 J 2 - J 2 z fi = ¯ h 2 2 ( j ( j + 1) - m 2 ) . 1

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Note that, again by symmetry, we should have h J 2 n i = h J 2 x i for any unit vector ˆ n on the xy -plane (which is proved in part (c)). (b) Start with h J x J y i = 1 4 i h ( J + + J - )( J + - J - ) i = 1 4 i J 2 + - J 2 - + J - J + - J + J - fi It can be seen easily that J 2 + fi = J 2 - fi = 0. For the remaining terms, we either calculate the commutator or use the application rules for J ± . Following the second h J - J + i = ¯ h 2 ( j ( j + 1) - m ( m + 1)) , h J + J - i = ¯ h 2 ( j ( j + 1) - m ( m - 1)) . The final result is h J x J y i = i 2 m ¯ h 2 . For h J y J x i we either follow the same procedure or use the relation ( J x J y ) = J y J x . h J y J x i = - i 2 m ¯ h 2 . Since both of J x J y and J y J x is not hermitian, their expectation values do not have to be real. Note that J x J y + J y J x is hermitian and we have h J x J y + J y J x i = 0 which is real.
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