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Spin-orbit coupling perturbation

  1. Nov 3, 2007 #1
    1. The problem statement, all variables and given/known data

    An electron in a hydrogen atom is in the n = 2, l = 1 state. It experiences a spin-orbit interaction [itex]H' = \alpha \mathbf{L} \cdot \mathbf{S}[/itex]. Calculate the energy level shifts due to the spin-orbit interaction.


    2. Relevant equations

    Degenerate perturbation theory.

    3. The attempt at a solution

    This n,l state is triply degenerate due to the three possible values of m = -1,0,1.

    The unperturbed Hamiltonian is just what goes in the Schrodinger equation right? In which case the eigenfunctions of the unperturbed hamiltonian are just the spherical harmonics [itex]Y_{lm}[/itex] multiplied by strictly radial functions. So I put

    [itex]\psi^{(0)} = \alpha Y_{10} + \beta Y_{1-1} + \gamma Y_{11}[/itex]

    So I then write down the matrix [itex] \langle Y_{1,i} |H'| Y_{1,j} \rangle[/itex] and find the eigenvalues.

    Am I getting warm?
     
  2. jcsd
  3. Nov 3, 2007 #2

    Meir Achuz

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    You have to evaluate [tex]L.S[/tex] for each angular state (J,L,S) by using
    [tex]J^2=L^2+S^2+2L\cdot S[/tex].
    You don't need the explicit wave functions.
     
  4. Nov 3, 2007 #3
    Hi Meir Achuz,

    Thanks for your reply.

    So for n = 2, l = 1 we have two possibilities for the total angular momentum corresponding to j = 1/2 and j=3/2 right?

    But we also have some degeneracy coming from the possible values of m = -1,0,1.

    Does this mean there will be a total of 6 energy shifts?
     
  5. Nov 4, 2007 #4

    Meir Achuz

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    L^2 and S^2 are known.
    There are two energy levels.
    One for J^2=(3/2)(5/2), and one for J^2=(1/2)(3/2).
    There are 6 states, but because of rotational invariance, the 4 J=3/2 states are still, degenerate, as are the l2 J=1\/2 states.
     
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