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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 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) the along the current loop.

Here, I get stuck.

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

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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.

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but shouldn't Physics be the same in all frames of reference? –  Harold Apr 9 at 9:36
Absolutely, and the proper implementation of that is the Dirac equation. It can be solved exactly for the hydrogen atom and to order $\alpha^4$ its eigenvalues agree with the corrections from the Schrödinger equation. I think the problem here is that an electric field transforming into a magnetic field is a relativistic effect, while the Scrödinger equation is not relativistic. (It has two space derivatives but only one time derivative, relativistic equations should have the same number of space and time derivatives.) –  Robin Ekman Apr 9 at 11:09
Doing it from the Dirac equation you also find that $g =2$, whereas otherwise this has to be put in by hand and is unintelligible. –  Robin Ekman Apr 9 at 11:18
Robin thanks for your answer. You say that relativistic equations must have the same number of space and time derivatives, why? –  Harold Apr 9 at 14:48
In special relativity space and time are treated on equal footing, and in moving from prerelativistic physics to special relativity the spatial gradient $\vec{\nabla}$ is promoted to $\partial/\partial x^{\mu} = (\partial/\partial t, \partial/\partial x, \partial/\partial y, \partial/\partial z)$ –  Jordan Apr 9 at 15:43

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