An electron moving through the electric field of a nucleus sees, in its own rest frame, a magnetic field, and its magnetic moment feels it: an energy proportional to $\hat{\mathbf{L}}\cdot\hat{\mathbf{S}}$. This spin–orbit interaction does not commute with $\hat{L}_{z}$ or $\hat{S}_{z}$, since it turns orbit and spin into each other, but it commutes with $\hat{L}^{2}$, $\hat{S}^{2}$ and the total $\hat{\mathbf{J}}=\hat{\mathbf{L}}+\hat{\mathbf{S}}$. So its eigenstates are the coupled states, and a level with $\ell>0$ splits in two, $j=\ell+\tfrac12$ and $j=\ell-\tfrac12$. The yellow light of sodium is split this way into two lines 0.6 nm apart.

A spinning top carried round a circle on a turntable trades angular momentum with the turntable through friction: neither keeps its own, only the total is fixed. Spin and orbit trade the same way under spin–orbit coupling, and only the total $\hat{J}$ keeps its value.

A $p$ level of one electron, six states, split by $\xi\,\hat{\mathbf{L}}\cdot\hat{\mathbf{S}}$ with $\xi>0$, in units of : four states with rise by and two with fall by , so the centre, weighted by the number of states, stays put.

For an electron in a central potential $V(r)$ the spin–orbit interaction is

• The form of $\xi(r)$ comes from relativity; it is derived from the Dirac equation in the chapter on relativistic quantum mechanics and taken here as given.
• $\hat{\mathbf{L}}\cdot\hat{\mathbf{S}}$ commutes with ${\stackrel{}{}}^{}$, , and but not with or : the good quantum numbers are , , and .

Square the total angular momentum. Because $\hat{\mathbf{L}}$ and $\hat{\mathbf{S}}$ act on different parts they commute, and the cross term is twice their dot product. Solving for the dot product writes it through three squares, all diagonal in the coupled basis; putting in $j=\ell\pm\tfrac12$ gives the two shifts.

$\hat{\mathbf{L}}$ and $\hat{\mathbf{S}}$ commute, so the two cross terms are equal.

Solve for the dot product.

All three squares are diagonal in the coupled basis, with $s(s+1)=\tfrac34$.

$(\ell+\tfrac12)(\ell+\tfrac32)-\ell(\ell+1)-\tfrac34=\ell$.

$(\ell-\tfrac12)(\ell+\tfrac12)-\ell(\ell+1)-\tfrac34=-(\ell+1)$.

In the coupled basis the dot product of orbit and spin is half the difference of three squares, each a number.

Spin–orbit coupling turns orbital angular momentum into spin and back. The separate $z$-components are no longer fixed, but the magnitudes and the total are.

With $\ell=0$ the only total is $j=\tfrac12$, and the dot product of orbit and spin is zero: s levels are not split by spin–orbit coupling.

For $\xi>0$ the level with $j=\ell+\tfrac12$ rises by and the level with falls by .

Great work! You now know how to:

• write the spin–orbit interaction through the total angular momentum
• compute its energies for $j=\ell\pm\tfrac12$
• say which quantum numbers stay good and why s levels are not split