Spin-Orbit Coupling and Thomas Precession

The relativistic kinetic correction accounts for the electron moving fast; it says nothing about the electron's spin. Spin carries a magnetic moment, and a moving electron in the nuclear electric field experiences, in its own frame, a magnetic field. The coupling of the spin moment to that field is the spin-orbit interaction. It is the largest of the three fine-structure terms for $\ell \ge 1$ states and the one that splits a single non-relativistic level into a doublet. Two ingredients need care: the magnitude of the internal field, obtained by transforming the Coulomb field into the electron's frame, and a factor of one-half that a naive frame transformation misses because the electron's rest frame is not inertial. The second is Thomas precession, and getting it right is what makes the spin-orbit energy agree with experiment and with the exact Dirac result.

The magnetic field in the electron's frame

In the laboratory frame the proton sits at rest and the electron orbits it. Boost to the instantaneous rest frame of the electron: now the electron is momentarily at rest and the proton circulates around it. A circulating charge is a current loop, and a current loop makes a magnetic field at its center, where the electron sits.

Quantitatively, the leading-order transformation of the electromagnetic field to a frame moving with velocity $\vec{v}$ relative to the lab gives, for a purely electric lab field $\vec{E}$,¹

The nuclear Coulomb field is radial, $\vec{E}=\frac{1}{e}\frac{\mathbf{d}V}{\mathbf{d}r}\hat{r}$, where $V(r)=-Z{e}^2/4\pi {ϵ}_{0}r$ is the potential energy of the electron and the factor $1/e$ converts it to the field seen by the charge. Using $\vec{v}\times \hat{r}=\vec{v}\times \vec{r}/r$ and the orbital angular momentum $\vec{L}=m\vec{r}\times \vec{v}$, so that $\vec{r}\times \vec{v}=-\vec{L}/m$,

The internal field is parallel to $\vec{L}$ and its strength is governed by the radial derivative of the potential. For hydrogen this field is enormous: at the Bohr radius it reaches several tesla, which is why the spin-orbit splitting, though tiny on the scale of the binding energy, is a real magnetic interaction.

The interaction Hamiltonian

The electron's spin magnetic moment is

with $g_s$ the spin g-factor, very close to $2$. The energy of a magnetic moment in a field is $-{\vec{\mu }}_s\cdot {\vec{B}}^{\prime }$, so the naive spin-orbit energy is

With $g_s=2$ the prefactor is $1/m^2{c}^2$. This expression is wrong by a factor of two. The error is not in the field transformation but in the assumption that the energy of the moment can be read off in the electron's rest frame as though that frame were inertial. It is not: the electron accelerates continuously toward the nucleus, and a sequence of infinitesimal boosts along a curved path composes into a net rotation, the Thomas precession, that the moment feels as an additional effective field.

Thomas precession

Two Lorentz boosts in different directions do not compose into a pure boost; their product is a boost followed by a rotation, the Wigner rotation. An accelerating particle is boosted successively along a turning velocity vector, so its rest frame rotates relative to the lab even when no torque acts. The angular velocity of that rotation, to lowest order in $v/c$, is¹²

where $\vec{a}$ is the electron's acceleration. For a bound electron the acceleration is Coulombic, $\vec{a}=\vec{F}/m=-(1/m)(\mathbf{d}V/\mathbf{d}r)\hat{r}$, directed toward the nucleus. The precession ${\vec{\omega }}_T$ is opposite to the orbital angular velocity and is exactly half the magnitude that a naive co-rotating frame would assign.

The moment precesses in the rotating rest frame at the rate set by the internal field, but the frame itself precesses at ${\vec{\omega }}_T$; adding the two, the net spin precession seen in the lab corresponds to an interaction energy reduced by the factor $\frac{1}{2}$. The corrected spin-orbit Hamiltonian is¹

The explicit factor $\frac{1}{2}$ is the Thomas factor. It cancels the $g_s=2$ that would otherwise double the result, leaving an effective coefficient equivalent to $g_s=1$ for orbital motion. The same $\frac{1}{2}$ emerges automatically, with no frame gymnastics, from the non-relativistic reduction of the Dirac equation; its appearance there is the strongest evidence that the Thomas argument is correct.

Reducing L·S with the good quantum numbers

The Hamiltonian contains $\vec{S}\cdot \vec{L}$, which mixes the separate orientations of spin and orbit. The individual projections $L_z$ and $S_z$ are no longer conserved, because $\vec{S}\cdot \vec{L}$ does not commute with them. What is conserved is the total angular momentum $\vec{J}=\vec{L}+\vec{S}$: since $H_{\text{so}}\propto \vec{L}\cdot \vec{S}$ is a scalar built from $\vec{L}$ and $\vec{S}$, it commutes with $J^2$, $J_z$, $L^2$, and $S^2$. Squaring $\vec{J}=\vec{L}+\vec{S}$,

On a simultaneous eigenstate $|n\ell j{m}_j⟩$ with $s=\frac{1}{2}$,

For a given $\ell \ge 1$ the total angular momentum takes two values, $j=\ell +\frac{1}{2}$ and $j=\ell -\frac{1}{2}$, splitting the level into a doublet. The two carry $\vec{L}\cdot \vec{S}$ of opposite sign:

The radial factor and the shift

Collecting the pieces, the first-order energy is the product of the angular factor above and the radial expectation of the coefficient. For the Coulomb potential $V=-Z{e}^2/4\pi {ϵ}_{0}r$, the derivative is $\mathbf{d}V/\mathbf{d}r=Z{e}^2/4\pi {ϵ}_0{r}^2$, so

The energy requires the radial expectation $⟨1/r^3⟩$, the hydrogenic result

The restriction $\ell \ge 1$ is essential: $⟨1/r^3⟩$ diverges for $\ell =0$, and indeed the $\vec{L}\cdot \vec{S}$ factor vanishes there ($\vec{L}=0$ for an s-state), an indeterminate $0\times \infty$ whose correct handling is the Darwin term. Assembling the angular factor $\frac{1}{2}[j(j+1)-\ell (\ell +1)-\frac{3}{4}]$ with the radial expectation gives the closed form:¹

Simplifying against $E_n=-Z^2{e}^2/8\pi {ϵ}_0{a}_0{n}^2$ collapses the constants to

The doublet and the Landé interval

The two rungs of a spin-orbit doublet are separated by an interval fixed by the angular factors. The energy of the upper member minus the lower is

using $⟨\vec{L}\cdot \vec{S}{⟩}_{\ell +1/2}-⟨\vec{L}\cdot \vec{S}{⟩}_{\ell -1/2}=(2\ell +1){\hslash }^2/2$. The classic case is the sodium doublet: the $3p$ level splits into $3p_{1/2}$ and $3p_{3/2}$, and the transitions to $3s_{1/2}$ produce the two yellow D-lines at $589.0$ and $589.6$ nm.

The interval rule generalizes: within a fine-structure multiplet the spacing between adjacent $j$ levels is proportional to the larger $j$, the Landé interval rule, because $⟨\vec{L}\cdot \vec{S}⟩$ increases by $j{\hslash }^2$ from one level to the next. For hydrogen the doublet is the whole multiplet, but in many-electron atoms the same $\vec{L}\cdot \vec{S}$ structure produces multiplets of several components, all governed by this rule.

The spin-orbit shift carries the spin content of fine structure, complementing the velocity content of the relativistic term. Neither alone gives the observed spectrum. Their sum, once the $\ell =0$ gap is filled by the Darwin term, reorganizes into a formula depending only on $n$ and $j$, which the next lesson derives and the Dirac theory confirms exactly.