quantum mechanics - Spin-orbit coupling from the rest frame of the proton?

When we calculate the spin-orbit interaction in a Hydrogen atom we just work in the electron's frame of reference: the proton is moving and produces a magnetic field which the electron's spin interacts with.

We can show here that the answer is $$ \Delta H = \frac{2\mu_B}{\hbar m_e e c^2}\frac{1}{r}\frac{\partial U(r)}{\partial r} \mathbf{L} \cdot\mathbf{S}$$ where $U(r)$ is the potential energy = $eV(r)$ with $V(r) = \frac{1}{4\pi\epsilon_0}\frac{e}{r}$ for a proton.

NOW: I want to get the same answer from the reference frame of the proton, where the proton is stationary and the electron is moving. Since Physics must be the same in all frames of reference, we should get the same answer.

I guess that the only way this can happen is if the electron's magnetic field (due to its motion, i.e. charged particle moving around) interacts with the electron's own spin.

We can calculate the current density $\mathbf{j}$ of the electron in Hydrogen, and it is given by: $$ j_\phi=-e\frac{\hbar m}{\mu r\sin\theta}\left|\psi_{nlm}\left(r,\theta,\phi\right)\right|^2 $$ (derivation found here on page 6)

I could use the Biot-Savart law to calculate the magnetic field due to this current density: $$\textbf{B} = \frac{\mu_0}{4\pi} \frac{1}{r^2} \int \textbf{J}d^3\textbf{r}$$ where the integration should be (at least classicaly) along the current loop.

Here, I get stuck.

Does anyone know how to get the $\textbf{L}\cdot\textbf{S}$ factor from this approach?

Answer

The problem here is that you're looking at the magnetic field $\textit{at the proton}$. Using this approach you can't derive the spin-orbit coupling that you want. Because of the symmetry of the system you would expect the magnetic fields at each particle to have the same magnitude, but the energy of spin-orbit coupling comes from the $\textit{electron's}$ spin interacting with the magnetic field. Usually self-interaction isn't included in these calculations. The important thing to note is from the rest frame of the nucleus there is no magnetic field due to the nucleus, ignoring contributions from the nuclear magnetic moment. You're actually very very close to another part of the hydrogen spectrum, which is the hyperfine structure. There will be another contribution to the energy of the hydrogen atom due to the nuclear spin interacting with the magnetic field from the electron. In the case of the hyperfine structure you need also to consider the magnetic field caused by the electron spin $\vec{\mu} \propto \vec{S}$, which, once combined with the calculation you're in the midst of doing, will yield another perturbation to the Hamiltonian $\propto \vec{I} \cdot \vec{J}$. To solve your problem I believe you'd need to use Dirac's equation.