Valid to use <1/r3> to get spin-orbit correction to H? (perturbation)

[Happiness]

TL;DR Summary

<1/r^3> uses the standard wavefunctions ψ_nlm of hydrogen, which are not good states to use in perturbation theory because the Hamiltonian (under spin-orbit interaction) no longer commutes with L. So shouldn't we solve for the simultaneous eigenstates of L^2, S^2, J^2 and J_z first? And then use those to find <1/r^3>?

Below is the derivation of E¹_so, the first-order correction to the Hamiltonian due to spin-orbit coupling of the election in hydrogen atom. My question is whether it's valid to use [6.64] (see below). $<\frac{1}{r^3}>$ I believe is $<\psi_{nlm}|\frac{1}{r^3}|\psi_{nlm}>$, but $\psi_{nlm}$ is NOT a good state to use in perturbation theory, because $\psi_{nlm}$ is an eigenstate of $L_{z}$ but H'_so does not commute with $L$ (as mentioned in the paragraph above [6.62]-[6.63]).

For elaboration, the phrase "good state" relates to the following theorem:

Ordinary first-order perturbation theory means using [6.9] below, with $\psi^{0}_{n}$ replaced with a good state.

 

[PeterDonis]

Below is the derivation

Where is this from? Please give a reference.

 

[Happiness]

Where is this from? Please give a reference.

Introduction to Quantum Mechanics, 2nd edition, by David J. Griffiths

 

[DrClaude]

Indeed, the $\psi_{nlm}$ are not "good" functions, because of the $m$ index, and the "good" functions will be linear combinations of $\psi_{nlm}$ with different $m$. However, the value of $\braket{1/r^3}$ is independent of $m$, therefore you you know that the linear combination will have the given value, whichever $m$ states are combined.

 

[Happiness]

Indeed, the $\psi_{nlm}$ are not "good" functions, because of the $m$ index, and the "good" functions will be linear combinations of $\psi_{nlm}$ with different $m$. However, the value of $\braket{1/r^3}$ is independent of $m$, therefore you you know that the linear combination will have the given value, whichever $m$ states are combined.

But how do you know that the "cross terms" are zero?

Suppose a good state $\psi^{0}=\alpha\psi_{a}+\beta\psi_{b}$ , where $\psi_{a}$ and $\psi_{b}$ are some $\psi_{nlm}$ .

$\braket{\frac{1}{r^3}}=\braket{\alpha\psi_{a}+\beta\psi_{b}|\frac{1}{r^3}|\alpha\psi_{a}+\beta\psi_{b}}$

$=\alpha^2\braket{\psi_{a}|\frac{1}{r^3}|\psi_{a}}+\beta^2\braket{\psi_{b}|\frac{1}{r^3}|\psi_{b}}+\alpha^*\beta\braket{\psi_{a}|\frac{1}{r^3}|\psi_{b}}+\alpha\beta^*\braket{\psi_{b}|\frac{1}{r^3}|\psi_{a}}$

How do you know the cross terms $\braket{\psi_{a}|\frac{1}{r^3}|\psi_{b}}$ and $\braket{\psi_{b}|\frac{1}{r^3}|\psi_{a}}$ are zero?

 

[Happiness]

How do you know the cross terms $\braket{\psi_{a}|\frac{1}{r^3}|\psi_{b}}$ and $\braket{\psi_{b}|\frac{1}{r^3}|\psi_{a}}$ are zero?

Ok, I've tried calculating them. They are indeed zero.