Radial term in the spin-orbit coupling

Question

The spin-orbit interaction for the hydrogen atom is of the form $\hat{H_1} = A\frac{1}{r^3}\pmb{\hat{L}}\cdot \pmb{\hat{S}}$

Now in my course, we treated this interaction by working in the basis of total angular momentum $\pmb{J} $and from there calculated the energy eigenvalues of $\hat{H_1}$ and assumed that theses were the correction to the energy levels.

My question is, what exactly is $\frac{1}{r^3}$?. Because if we treat this term as being an operator, then it is not obvious at all that $\hat{H_1} $ should commute with $\hat{H_0} = \frac{\pmb{\hat{p}}}{2m}-\frac{e^2}{\hat{r}}$. This non commutativity implies then that you can't correct the energy eigenvalues of $\hat{H_0}$ with those of $\hat{H_1}$.

So my question is, are we actually doing some kind of perturbation theory where we assume that $\frac{1}{\hat{r^3}}$ is actually $<\frac{1}{\hat{r^3}}>$, i.e. the expectation value from $\hat{H_0}$?

In that case the two operators would commute and the corrections to the energies would make sense.

Thanks you.

Answer 1

You have uncovered one of the pitfalls of trying to treat spin non relativistically. If you had used the Dirac Hamiltonian rather than the Schrodinger Hamiltonian you would not encounter this problem. Frequently these $1/r^3$ singularities can be avoided in non relativistic treatments by using the fact that the nucleus is not really a point charge. It is also important to use the Foldy/Wouthuysen reduction methodology to make sure that you are being consistent in your non relativistic treatment. I addressed this issue in a paper long ago: L. D, Miller, "A Foldy-Wouthuysen reduction of the atomic relativistic Hartree-Fock equations", J. Phys B., At. Mol. Phys. 20 (1987), p. 4309-4316