HW9sol2 - Physics 731 Assignment #9, Solutions 1. We are...

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Unformatted text preview: Physics 731 Assignment #9, Solutions 1. We are given the Hamiltonian H = A ~ S ( e- ) ~ S ( e + ) + eB mc ( S ( e- ) z- S ( e + ) z ) , and the spin part of the wave function of the system: ( e- ) + ( e + )- = | +-i = 1 2 ( | 10 i + | 00 i ) . (a) For A , the | m 1 m 2 i states are energy eigenstates. The state | +-i is an eigenstate of H in this limit with eigenvalue e hB/ ( mc ) . (b) For B , the | sm i states are energy eigenstates. The expectation value of H is given by h H i = 1 2 ( h 10 | + h 00 | ) A ~ S ( e- ) ~ S ( e + ) ( | 10 i + | 00 i ) = A 4 ( h 10 | + h 00 | )( S 2- S ( e- ) 2- S ( e + ) 2 )( | 10 i + | 00 i ) = A 4 ( h 10 | + h 00 | ) S 2- 3 h 2 2 ( | 10 i + | 00 i ) =- A h 2 4 . 2. (a) Sakurai tells us to use the notation + , ,- for m 1 , 2 = +1 , ,- 1 . Let us begin with the state | jm i = | 22 i = | + + i = | m 1 m 2 i . Applying J- / h = ( J 1- + J 2- ) / h to each side, we have J- h | 22 i = 2 | 21 i = J 1- + J 2- h | + + i = 2( | 0+ i + | + 0 i ) . Thus, | 21 i = 1 2 ( | + 0 i + | 0+ i ) . Similarly, J- h | 21 i = 6 | 20 i = J 1- + J 2- h 1 2 ( | 0+ i + | + 0 i ) , such that | 20 i = 1 6 ( | +-i + 2 | 00 i + | - + i ) . By symmetry, we have | 2- 1 i = 1 2 ( | - i + |-i ) , and | 2- 2 i = | - -i . Using h 11 | 21 i = 0 , we have | 11 i = 1 2 ( | + 0 i - | 0+ i ) . 1 Here we have used the Condon-Shortley phase convention. Again, we apply J- / h = ( J 1- + J 2- ) / h to each side: J- h | 11 i = 2 | 10 i = J 1- + J 2- h 1 2 ( | + 0 i - | 0+ i ) , such that | 10 i = 1 2 ( | +-i - | - + i ) ....
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HW9sol2 - Physics 731 Assignment #9, Solutions 1. We are...

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