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?
 A: 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.
A: I think this is a really good question, and one that unfortunately I haven't found addressed when learning about this stuff, and I think that avoiding it by saying that you need Dirac's equation neglects that the question can be interpreted classically. So yes, we need relativity, but we shouldn't need quantum mechanics to understand this effect.
To understand where this comes from, the first thing is to substitute the spin of the electron by a classical magnetic dipole (because spin is quantum mechanical). A classical magnetic dipole is equivalent to a very small current loop.
Now, you'll need to understand a bit of special relativity for the next step. This is all very well explained in Wikipedia and in a video by Veritasium. If you understand that, you can understand this. In the rest frame of the dipole (electron) it is just a normal current loop. However, in the rest frame of the proton, Lorentz contraction causes the densities of positive and negative charges in the wire loop to change in such a way as to cause a net torque in the proton rest frame, equivalent to the one in the dipole's (electron's) frame.
Now, indeed, Dirac's equation is needed to explain what really happens in an atom. But, this effect need not Dirac equation to occur, as indeed it just follows from the principle of relativity, satisfied by classical mechanics too.
