Spin-orbit Interaction & Degenerate Perturbation Theory

  • #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!
 

Answers and Replies

  • #2
DrDu
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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?
I can't follow your conclusion.
 
  • #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!
 
  • #4
DrDu
Science Advisor
6,076
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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##.
 
  • #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?
 
  • #6
DrClaude
Mentor
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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?
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.
 
  • #7
Is there a quick way to show that it is not diagonal?
 
  • #8
DrClaude
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Is there a quick way to show that it is not diagonal?
$$
[\hat{L}_z, \hat{S}\cdot\hat{L} ] \neq 0
$$
 

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