1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook