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Spin orbit coupling

  1. Mar 15, 2006 #1
    I'm a little confused as to whether the following commutation relations still hold when spin orbit coupling occurs:
    [Sx,Lx] = 0
    [Sx,Ly] = 0
    [S^2, Lx] = 0
    [L^2, Sx] = 0
    etc.

    thanks very much for your help
     
  2. jcsd
  3. Mar 16, 2006 #2
    Err,

    What have you done to solve this problem yourself ?

    Helping you out does not imply that we will just "spoon feed" you the solution.

    How do you think you can tackle this problem ?


    marlon
     
  4. Mar 19, 2006 #3

    Meir Achuz

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    The CR's between the ops are not changed by the coupling.
     
  5. Mar 23, 2006 #4

    dextercioby

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    It doesn't matter how the hamiltonian looks, for a particle in quantum mechanics the Hilbert space has the structure

    [tex] \mathcal{H} =L^{2}\left(\mathbb{R}^{3}\right) \otimes \mathbb{C}^{2n+1} [/tex]

    ,where "n" is the spin of the particle.

    Daniel.
     
  6. Mar 23, 2006 #5

    nrqed

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    They are still valid. But that's not the point. The point is whether the *perturbation* hamiltonian commutes with these operators. One finds that the spin orbit hamiltonian commutes with L^2, S^2, J^2 and J_z, but not with L_z and S_z. Therefore, m_s and m_l are not good quantum numbers but must be replaced by m_j and j. So the states of definite energy when the spin-orbit interaction are taken into account are the states labelled by the quantum numbers l,s,j, m_j (instead of the usual l,m_l,s,m_s that one uses to label the unperturbed hydrogenic wavefunctions).

    Hope this helps. If it's not clear, write again.

    Pat
     
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