# Why do we need to account for relativity to describe spin-orbit coupling

I am struggling to understand why one has to rely on relativity to explain spin-orbit coupling.

In fact, I always thought that the very nature of spin arises when one includes relativity into the Schrödinger equation. While it is true that spin naturally arises from Dirac's equation, I also read that alternative non-relativistic quantum formalism such as the Lévy-Leblond equation are able to describe the "emergence" of spin. Consequently, the spin is a purely quantum property and one does not need to rely on relativity to explain its nature.

However, the spin-orbit coupling is always refereed to as a relativistic effect. What I dont understand is that, since spin and (orbital) angular motion are non-relativistic and spin-orbit operator can be expressed based on classical argument, we still call spin-orbit a relativistic effects.

Although I understand that accounting for relativity is needed to FULLY describe the SO coupling, why is it so difficult to find a clear explanation that separates the classical part from the relativistic one.

I have to emphasis that I am chemist, working in the field of quantum chemistry. As such I have, for example, only little knowledge about representation theory. Hence, I am more seeking for an "intuitive" (if that is possible when mixing quantum theory and relativity ...) explanation. However I have some good understanding of quantum mechanics in general and some basic knowledge of (special) relativity.

Yes, you can just say an electron has a (quantised) magnetic moment as an observed fact, without involving Dirac or relativity.

As the electron orbits the nucleus of an atom it sees an electric field which keeps it in orbit. Hence (if you like) the Bohr model. But the spin is not affected by the electric field, as it's a magnetic effect. In the frame of the atom there is just a Coulomb (or screened Coulomb) electric field. So no spin orbit coupling...

... except that if the electron is travelling fast, then in the frame of the electron the nuclear electric field is transformed into a mixture of electric and magnetic fields. The magnetic field (there because of relativity) couples to the magnetic moment/spin direction, hence spin-orbit coupling.

• Thank you for your answer. But, since the orbiting electron is a moving charge, doesn't it produces a magnetic field (like in a solenoid) that can then interact with the spin ? What am I missing here ?
– Mils
Jul 4, 2018 at 15:45
• The electron doesn't interact with itself (unless you get into quantum field theory). The moving electron does indeed produce a $\mu e v/2\pi r$ field but that only effects other electrons and the nucleus and is a smaller effect. It's the transformed $Ze/4 \pi \epsilon r^2$ that gives the spin-orbit energy shift. Jul 4, 2018 at 16:05

I'm not sure if this answers your question properly, but the spin-orbit coupling can be described in non-relativistic quantum mechanics. You can add an artificial term to the Hamiltonian as described here and do the calculation. However, in relativistic Quantum mechanics this contribution appears naturally. This is probably the reason why people call it a relativistic effect, as it is intrinsically present in the Dirac equation. To arrive at the exact result you also have to take into account the Lamb shift which arises from radiative corrections in QED and could thus also be called a relativistic effect.

The spin-orbit interaction is a relativistic effect because it arises from the coupling of the magnetic moment of the electron,

$$\mu_s=\frac{g_s\mu_B {\vec S}}{\hbar}$$

to the velocity dependent, relativistic correction of the magnetic field which is given by

$$-(\frac{{\vec v}}{c}) x {\vec E}$$. (1)

You can re-express the electric field as

$$\vec E =-\frac{(\vec r)}{r} \frac{dV(r)}{dr}$$. (2)

Substitute this in (1) to get the relativistic correction of the magnetic field as

$$\frac{1}{m_e c}\frac{1}{r}\frac{dV(r)}{dr} {\vec L}$$ (3)

with

$$\vec L={\vec r} x m_e {\vec v}.$$

Take the scalar product of this with the electron spin and get the $$S\dot L$$ coupling as an effect of the relativistic correction to the magnetic field.

You may wish to check out Chpt.11in

https://www.researchgate.net/publication/337902107_Special_Relativity_at_the_Core_of_Quantum_Mechanics

• While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review Aug 6, 2023 at 20:56