# How do you derive the spin orbit coupling classically?

In our lecture on atomic and molecular physics we are currently dealing with spin orbit coupling. Our prof showed us a derivation where he approximated the dirac equation to the non-relativistic case and said that the extra terms besides the schroedinger equation are correction terms and that one of them is the so called spin-orbit coupling.

$E_{SL}=-\frac{e}{2m_e^2c^2r}\frac{1}{r}\frac{dV}{dr}\vec{S}\cdot \vec{L}$.

The approach was kind of unintuitive and ad hoc to me. But one of his slides at the end of the lecture hinted at a derivation from a classical point of view. It was something along those lines:

We have an electron flying through an electric central potential V(r) and its spin $\vec{S}$ is coupling with its orbital momentum $\vec{L}$. We are then supposed to transform to the reference point of the electron and determine the magnetic field it conceives. And the actual spin-orbit coupling is then the coupling of that magnetic field with the magnetic spin moment of the electron.

But I have no idea how to approach this. Does anyone have any ideas? Or maybe some literature that dealt with the introduction of spin-orbit coupling in that way that I could read?

## 1 Answer

Yes, the spin orbit coupling can be understood classically as follows

The interaction energy of a fixed spin with a magnetic field is given by: $$E_{int} = \vec{\mu} \cdot \vec{B}$$

Where $\vec{\mu}$ is the magnetic moment, given by: $$\vec{\mu} = -\frac{g q}{2m c} \vec{S}$$ $g$ is the gyromagnetic factor, $q$ is the charge $m$ s the mass, $\vec{S}$ is the spin.

Considering an electron moving in a central potential $V(r)$. In order to apply the formula for the interaction energy, we need to work in the rest frame, because the spin is the angular momentum in the rest frame. In its rest frame, the electron feels a magnetic field obtained from the Lorentz transformation of the magnetic field: $$\vec{B} = − \gamma \frac{\vec{v}}{c} \times \vec{E} \approx -\frac{\vec{v}}{c} \times \vec{E}$$ (In the nonrelativistic approximation).

Writing:

$$\vec{E} = -\frac{dV(r)}{dr} \frac{\vec{r}}{r}$$ we obtain:

$$E_{int} =-\frac{g q}{2m^2 c^2} \frac{dV(r)}{r dr} \vec{S} \cdot (\vec{v} \times \vec{r}) = -\frac{g q}{2m^2 c^2} \frac{dV(r)}{r dr} \vec{S} \cdot (\vec{p} \times \vec{r}) = -\frac{g q}{2m^2 c^2} \frac{dV(r)}{r dr} \vec{S} \cdot \vec{L}$$ It is worthwhile to mention that there are many classical models which correctly describe the classical dynamics of spinning particles, such as the Bargmann-Michel-Telegdi equation. . These models describe all position and spin dynamics of an electron in the classical limit.

• Awesome, thanks. A nice and intuitive way on how to get this classically. – Rab Nov 30 '17 at 8:46