# Spin-orbit interaction and retarded potential

Can anyone explain to me why, when the spin-orbit interaction is a relativistic effect, the Coulomb potential is then used in calculating the hydrogen energy levels, rather than the retarded potential? Surely if the electron is assumed to be moving at a relativistic velocity, the correction to the Coulomb potential will be significant?

• The Coulomb potential doesn't change in the laboratory frame, even if the electron is ultra-relativistic.
– Cham
Commented Feb 29 at 16:20

Spin-orbit interaction is not directly related to the velocity of the electron (in whichever way such a velocity is defined). However, the velocity can serve as a general indicator of the importance of relativistic effects (thus, the importance of spin-orbit coupling as well).

For low nuclear charges, the electron does not move at a relativistic velocity at all. For low $$Z$$ (like $$Z=1$$ in hydrogen), the importance of all (non-QED) relativistic effects can be ordered with the increasing powers of $$Z\alpha$$, where $$\alpha$$ is the fine structure constant. For example, the non-relativistic energy (assuming infinite nuclear mass) is $$E_n=-\frac{(Z\alpha)^2mc^2}{2n^2} \ ,$$ and all energy corrections carry some higher power of $$Z\alpha$$.

Let us take the squared velocity of the electron to be $$v^2=\frac{\langle\psi_{nlm}|\hat{p}^2|\psi_{nlm}\rangle}{m^2} \ .$$ Then, using the virial theorem, we find \begin{aligned} v^2 &=\frac{2}{m}\langle\psi_{nlm}|\hat{T}|\psi_{nlm}\rangle \\ &=-\frac{2}{m}E_n \\ &=\frac{(Z\alpha)^2}{n^2}c^2 \ , \end{aligned} so the velocity is just $$v=\frac{Z\alpha}{n}c \ .$$ Of course, this "velocity" is not unique (similarly to the multiple definitions for the "size" of the atom), but the only important thing here is the scaling $$v\sim Z\alpha c$$, showing that for low $$Z$$, $$v\ll c$$. This suggests that using the Coulomb interaction only (without any relativistic/QED corrections) is a good approximation. The leading relativistic corrections to the energy (including the spin-orbit interaction) scale as $$(Z\alpha)^4mc^2$$, and (again, for low $$Z$$) are small enough to be included perturbatively.

Here, the Coulomb-gauge is assumed, in which the photon exchange can be conveniently partitioned into instantaneous and retarded terms. Actually, only the instantaneous part contributes to the $${\cal{O}}((Z\alpha)^4mc^2)$$ correction. Retardation is a (non-radiative) QED effect, and it only shows up in the energy at higher orders; to see this effect the nucleus must be treated explicitly, with a finite mass.

For large $$Z$$, the "velocity" of the electron becomes comparable to $$c$$ (the electron must travel very fast to avoid falling into the nucleus $$-$$ a nice, but somewhat misleading picture), and the perturbative approach built on the $$Z\alpha$$ expansion breaks down. In this case, the exact solution of the Dirac equation must be used to include relativistic effects. The QED corrections can be treated in a perturbation theory built on this exact relativistic zeroth-order solution (although it is a very hard task).

• Why the downvote? Did I write something incorrect? Commented Feb 29 at 18:08
• This spin-orbit effect does not depend on the velocity of the electron. Commented Feb 29 at 18:33
• So the retardation can only be handled with QED? Commented Feb 29 at 18:34
• @JerroldFranklin I am aware of this, I made an edit to make my point (hopefully) more clear. Commented Feb 29 at 18:44
• @Mauricio Yes, you need to consider the quantized EM field. In Coulomb gauge, retardation effects come from the Feynman diagram of transverse photon exchange. To find the corresponding energy contribution, the finite-mass proton must be treated explicitly; the usual external field approximation (taking the nuclear mass to be infinite) is not sufficient. Commented Feb 29 at 18:52

Thank you very much for everyone's detailed help with my question. Perhaps I could just summarise where I am so far with my thinking on this, and see if anyone can point out any misunderstandings? From what I understand, in the reference frame of the electron, it 'sees' the nucleus is orbiting around it, and this leads to it experiencing a magnetic field in addition to the electric field of the nucleus. I had previously thought that this electric field would be retarded due to the motion of the nucleus, but I think that due to the fact that it is moving in a circular orbit, the 'acceleration' field is zero and the retardation effects in the 'velocity' field are cancelled out, so that the electron experiences simply an unmodified Coulomb electric field. Transforming back into the reference frame of the nucleus (i.e. the laboratory frame) simply preserves this exact same Coulomb electric field.

• Another way you can think of this is the following (but it is harder to calculate): in the frame of the nucleus, the electron is moving, a moving magnetic dipole behaves as an electric dipole, this electric dipoles couples to the Coulomb field of the nucleus. Whatever retardation effect here is negligible because the electron is close to the nucleus and the retardation effect is a higher order correction. Commented Mar 5 at 17:07
• For a charge moving in a circle of constant radius, what will be the electric field at the centre of the circle?
– dgwp
Commented Mar 6 at 12:02
• why do you ask? A charge moving at constant angular speed, would produce either zero field in the center (average out) or some more complicated time dependent electric field. But whatever field the electron is making is not important for spin-orbit coupling. Commented Mar 6 at 12:56

The spin and magnetic moment of the electron come out of the relativistic Dirac equation for the electron. Their appearance is unrelated to the velocity of the electron, and does not require high velocity for the electron.

• This just moves the equation higher up. Why we do not put the retarded Coulomb potential in Dirac's equation? Commented Feb 29 at 18:29
• I just answered the question, "Can anyone explain to me why, when the spin-orbit interaction is a relativistic effect?" Whether the electron's motion is relativistic or not does not matter for the spin-orbit effect. Commented Feb 29 at 18:29
• "Why we do not put the retarded Coulomb potential in Dirac's equation?" Ask Dirac. Commented Feb 29 at 18:31