This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Physics 731 Assignment #9, Solutions 1. We are given the Hamiltonian H = A ~ S ( e ) Â· ~ S ( e + ) + eB mc ( S ( e ) z S ( e + ) z ) , and the spin part of the wave function of the system: Ï‡ ( e ) + Ï‡ ( e + ) =  +i = 1 âˆš 2 (  10 i +  00 i ) . (a) For A â†’ , the  m 1 m 2 i states are energy eigenstates. The state  +i is an eigenstate of H in this limit with eigenvalue e Â¯ hB/ ( mc ) . (b) For B â†’ , 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 ) Â· ~ 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 3Â¯ h 2 2 (  10 i +  00 i ) = A Â¯ 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 / Â¯ h = ( J 1 + J 2 ) / Â¯ h to each side, we have J Â¯ h  22 i = 2  21 i = J 1 + J 2 Â¯ h  + + i = âˆš 2(  0+ i +  + 0 i ) . Thus,  21 i = 1 âˆš 2 (  + 0 i +  0+ i ) . Similarly, J Â¯ h  21 i = âˆš 6  20 i = J 1 + J 2 Â¯ h 1 âˆš 2 (  0+ i +  + 0 i ) , such that  20 i = 1 âˆš 6 (  +i + 2  00 i +   + i ) . By symmetry, we have  2 1 i = 1 âˆš 2 (   i + i ) , and  2 2 i =   i . Using h 11  21 i = 0 , we have  11 i = 1 âˆš 2 (  + 0 i   0+ i ) . 1 Here we have used the CondonShortley phase convention. Again, we apply J / Â¯ h = ( J 1 + J 2 ) / Â¯ h to each side: J Â¯ h  11 i = âˆš 2  10 i = J 1 + J 2 Â¯ h 1 âˆš 2 (  + 0 i   0+ i ) , such that  10 i = 1 âˆš 2 (  +i    + i ) ....
View
Full Document
 Fall '09
 Everett
 23 Enigma, 11:11, dy, JZ

Click to edit the document details