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Spin-orbit Interaction & Degenerate Perturbation Theory

  1. Nov 7, 2014 #1
    Hello! This is my first time posting, so please correct me if I have done anything incorrectly.

    There's something that I don't understand about the spin-orbit interaction.

    First of all I know that
    [itex][\hat{S} \cdot \hat{L}, \hat{L_z}] \ne 0[/itex]
    [itex][\hat{S} \cdot \hat{L}, \hat{S_z}] \ne 0[/itex]

    so this means that [itex] \hat{S} \cdot \hat{L} [/itex] doesn't share a common set of eigenstates with [itex] \hat{S_z} [/itex] and [itex] \hat{L_z} [/itex].

    I know that [itex]| nlm_lsm_s>[/itex] is a common eigenstate for [itex] \hat{S_z} [/itex] and [itex] \hat{L_z} [/itex],
    so that would mean it is not an eigenstate for [itex] \hat{S} \cdot \hat{L} [/itex].

    However, I've read that [itex]<nlm_l'sm_s'|\hat{S}\cdot\hat{J}|nlm_lsm_s>\ne0[/itex] for all [itex] m_l \ne m_l', m_s\ne m_s' [/itex] i.e. the diagonal elements are non-zero. Surely if [itex]| nlm_lsm_s>[/itex] is not an eigenstate of [itex] \hat{S} \cdot \hat{L} [/itex], then the matrix element cannot be evaluated?

    Thank you in advance!
     
  2. jcsd
  3. Nov 8, 2014 #2

    DrDu

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    Science Advisor

    I can't follow your conclusion.
     
  4. Nov 8, 2014 #3
    Hiya! Thanks for the reply!

    I think my thought process went like this:
    Suppose I have an operator [itex] \hat{Q} [/itex], then the eigenvalue equation [itex] \hat{Q} |q>=q|q> [/itex] can only be evaluated if [itex] |q> [/itex] is an eigenstate of [itex] \hat{Q} [/itex].
    Then since [itex] |nlm_lsm_s> [/itex] is not an eigenstate of [itex] \hat{S}\cdot\hat{L} [/itex], then surely the matrix element [itex] <nlm_l'sm_s'| \hat{S}\cdot\hat{L}|nlm_lsm_s>[/itex] could not be evaluated? I feel like something went wrong, because all the books I've read managed to evaluate it to be [itex] \ne 0[/itex] for [itex] m_l'\ne m_l \& m_s'\ne m_s [/itex]. Or could you shed some light into how to go about evaluating it?

    Thank you so much!
     
  5. Nov 8, 2014 #4

    DrDu

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    Science Advisor

    Of course, |q> is only an eigenstate if it fulfills the eigenvalue equation. However, Q can act on any state within it's domain of definition, only that the resulting state will not be a multiple of the original state.
    E.g. take ##Q=L_z## and ## |q\rangle =a |m>+b|m'>## where |m> and |m'> are eigenvectors of ##>L_z## with corresponding eigenvalues m and m', respectively. Then ##L_z|q\rangle=a m |m\rangle +b m' |m'\rangle##.
     
  6. Nov 8, 2014 #5
    Ok I see you what you mean, but how could I evaluate the matrix elements [itex] <nlm_l'sm_s'|\hat{S}\cdot\hat{L}|nlm_lsm_s> [/itex] if I cannot use the eigenvalue equation?
     
  7. Nov 8, 2014 #6

    DrClaude

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    Staff: Mentor

    Either you express ##| n l m_l m_s \rangle## in terms of a linear combination of eigenstates of ##\hat{S}\cdot\hat{L}## (in this particular case, using Clebsch-Gordan coefficients), or you actually calculate the integral corresponding to the bracket, as you would do to calculate ##\langle \hat{x} \rangle## or ##\langle \hat{p} \rangle## for say an eigenstate of the harmonic oscillator.
     
  8. Dec 10, 2014 #7
    Is there a quick way to show that it is not diagonal?
     
  9. Dec 10, 2014 #8

    DrClaude

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    Staff: Mentor

    $$
    [\hat{L}_z, \hat{S}\cdot\hat{L} ] \neq 0
    $$
     
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