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Unformatted text preview: 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 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 xyplane (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)) ....
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This note was uploaded on 02/11/2012 for the course MATH 435 taught by Professor Starg during the Spring '11 term at Al Ahliyya Amman University.
 Spring '11
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