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What is the degeneracy of the ground state

  1. Oct 12, 2006 #1
    [tex]H=\frac{p^2}{2m'}+V_0(r)+V(r)
    [/tex]

    where

    [tex]V(r)=2[3\frac{(S \cdot r)^2}{r^2}-S^2]
    [/tex]

    and V_0(r) is a rotationally invariant potential, p=p1-p2, the relative momentum and m' the reduced mass. S=S1+S2 spin operator.

    Assume first that V(r) is zero; what is the degeneracy of the ground state assuming that each of the dipoles are spin 1/2. After turning on V(r) how is the degeneracy split; what are the "quantum numbers" i.e. eigenvalues J and S of those new states.

    ________________________

    Any hints on how to start on a problem like this?
     
  2. jcsd
  3. Oct 14, 2006 #2
    Whats confusing me is how to handle

    p^2 |a>

    V_0 |a>
     
  4. Oct 15, 2006 #3
    I solved it
     
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