Spin-orbit Interaction & Degenerate Perturbation Theory

[PhysicsKin]

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
$[\hat{S} \cdot \hat{L}, \hat{L_z}] \ne 0$
$[\hat{S} \cdot \hat{L}, \hat{S_z}] \ne 0$

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

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

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

Thank you in advance!

 

[DrDu]

Surely if $| nlm_lsm_s>$ is not an eigenstate of $\hat{S} \cdot \hat{L}$, then the matrix element cannot be evaluated?

I can't follow your conclusion.

 

[PhysicsKin]

Hiya! Thanks for the reply!

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

Thank you so much!

 

[DrDu]

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 $L_z|q\rangle=a m |m\rangle +b m' |m'\rangle$.

 

[PhysicsKin]

Ok I see you what you mean, but how could I evaluate the matrix elements $<nlm_l'sm_s'|\hat{S}\cdot\hat{L}|nlm_lsm_s>$ if I cannot use the eigenvalue equation?

 

[DrClaude]

Ok I see you what you mean, but how could I evaluate the matrix elements $<nlm_l'sm_s'|\hat{S}\cdot\hat{L}|nlm_lsm_s>$ 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.

 

[PhysicsKin]

Is there a quick way to show that it is not diagonal?

 

[DrClaude]

Is there a quick way to show that it is not diagonal?

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