I am reviewing properties of atoms and I find myself becoming uncomfortable with spin orbit coupling. For ease, let's consider the case of hydrogen. In particular, at very small $r$,
$$H_{\mathrm{SO}} = \frac{c}{r^3} \vec{S}\cdot \vec{L} \gg \frac{\hbar^2 l(l+1)}{2m r^2}=H_{\mathrm{Cent}}$$
This seems to me to be a problem when $\vec{S}\cdot{\vec{L}}$ is negative [i.e. when $j(j+1)<l(l+1)+3/4$, so when the spin and angular momentum are anti-aligned], as it means that attraction from spin-orbit coupling can overcome the centripetal barrier. For example, I would naively expect that one could variationally show that the energy is unbounded below. This leads me to the following question:
Is the lowest energy eigenvalue of the naive Hamiltonian for the hydrogen atom, including spin-orbit coupling as the only perturbation to the coulomb potential, in fact unbounded below?
If this is the case, I will ask a separate question about the resolution to this problem. I suspect the inclusion of perturbative terms of relativistic origin that will further penalize wavefunctions that are sharply peaked will be important. However, I would prefer to keep these considerations out of this problem and instead to focus on whether taking spin-orbit coupling in the form $H_{\mathrm{SO}} = \frac{c}{r^3} \vec{S}\cdot \vec{L}$ as the only additional term to the potential will cause the lowest energy to be unbounded below.
Best Answer
As you say, the spin-orbit interaction is a perturbation. This means that the perturbation Hamiltonian $H_{SO} = \vec{S} \cdot \vec{L} / r^3$ is such that the corresponding energy corrections are much smaller than the unpeturbed energy levels (i.e. the Bohr energy levels). This seems to be case for the Hydrogen atom, as effectively the electron is on average at a Bohr radius from the proton. It is also likely that the expression for $H_{SO}$ should not hold all the way to $r \to 0$, as the proton is not a point-like particle.