Spin-Orbit Coupling from Relativity

Spin-orbit coupling is where an electron's spin first stops being a spectator. Sodium's yellow light is not one line but two — the famous D-doublet — and the same splitting patterns the fine structure of every atom heavier than hydrogen. Historically the coupling was guessed: Goudsmit and Uhlenbeck's spin, plus the intuition that an orbiting electron sees the nucleus circling it as a current loop, gave the right shape $\mathbf S\!\cdot\!\mathbf L$ but the wrong size — twice too large. The missing factor of two, Thomas precession, was patched in by hand. This article shows there is no patch to make: the coupling, with its correct coefficient, falls straight out of the Dirac equation. The mechanism is the elimination of the small spinor components. Writing the Dirac equation in a static potential as a $2\times2$ block system in the large spinor $\psi$ and the small spinor $\chi$ , and solving for $\chi\approx\tfrac{\boldsymbol\sigma\cdot\mathbf p}{2mc}(1-\tfrac{E'-V}{2mc^2})\psi$ , one substitutes back and expands to order $(v/c)^2$ . A single Pauli identity, $(\boldsymbol\sigma\!\cdot\!\mathbf A)(\boldsymbol\sigma\!\cdot\!\mathbf B)=\mathbf A\!\cdot\!\mathbf B+i\boldsymbol\sigma\!\cdot\!(\mathbf A\times\mathbf B)$ , splits the correction into a spin-independent piece and a spin-dependent one — the latter is $H_{\rm SO}=\tfrac{1}{2m^2c^2}\tfrac1r\tfrac{dV}{dr}\mathbf S\!\cdot\!\mathbf L$ , coefficient and all. No rest frame is ever invoked; the Thomas $\tfrac12$ is simply present. Once you have the operator, the physics is angular-momentum bookkeeping. Because $\mathbf S\!\cdot\!\mathbf L = \tfrac12(\mathbf J^2-\mathbf L^2-\mathbf S^2)$ , the coupling is diagonal in the total-angular-momentum basis $|n,\ell,j,m_j\rangle$ , and each $\ell\neq0$ level splits into $j=\ell+\tfrac12$ (higher) and $j=\ell-\tfrac12$ (lower). Feeding in the hydrogen radial average $\langle 1/r^3\rangle$ — owned by Spin-Orbit Coupling — gives the numerical splitting. This article owns the relativistic origin of the operator and the eigenvalue algebra; the assembly with the kinetic and Darwin terms into the $n,j$ -only fine-structure formula is developed in Fine Structure of Hydrogen and Relativistic Corrections to the Hydrogen Atom.

Key takeaways

Spin-orbit coupling $H_{\rm SO}=\tfrac{1}{2m^2c^2}\tfrac1r\tfrac{dV}{dr}\,\mathbf S\!\cdot\!\mathbf L$ ties spin to orbital motion; this article owns its Dirac origin (elimination of the small spinor components), with the Thomas $\tfrac12$ automatic.
Owned result: the $\mathbf S\!\cdot\!\mathbf L$ eigenvalue $\tfrac{\hbar^2}{2}[j(j{+}1)-\ell(\ell{+}1)-\tfrac34]$ splits each $\ell\neq0$ level into $j=\ell+\tfrac12$ (above) and $j=\ell-\tfrac12$ (below); for $\ell=1$ , $+\tfrac12\hbar^2$ and $-\hbar^2$ (checked by explicit diagonalization).
The classical rest-frame estimate overshoots by exactly $2\times$ ; the Dirac coefficient $\tfrac{1}{2m^2c^2}$ needs no Thomas insertion.
Hydrogen numeric (verified): with $\langle 1/r^3\rangle = 1/[a_0^3 n^3\ell(\ell+\tfrac12)(\ell+1)]$ , the $2p_{3/2}$ – $2p_{1/2}$ splitting is $4.53\times10^{-5}$ eV $= 0.365$ cm $^{-1}\approx10.9$ GHz.
Good quantum numbers become $n,\ell,j,m_j$ : $L_z,S_z$ are no longer separately conserved, only $\mathbf J=\mathbf L+\mathbf S$ and $J_z$ .
Because $H_{\rm SO}\propto Z^4$ , the coupling is a whisper in hydrogen but dominant in heavy atoms, driving $jj$ -coupling and relativistic chemistry.
Key idea: Eliminating the small components of the Dirac spinor produces $H_{\rm SO}=\tfrac{1}{2m^2c^2}\tfrac1r\tfrac{dV}{dr}\mathbf S\!\cdot\!\mathbf L$ with the Thomas half built in, and its eigenvalue $\tfrac{\hbar^2}{2}[j(j{+}1)-\ell(\ell{+}1)-\tfrac34]$ splits every $\ell\neq0$ level into a $j=\ell\pm\tfrac12$ doublet.

The core insight

Reducing the Dirac equation by eliminating the small spinor components yields $H_{\rm SO}=\tfrac{1}{2m^2c^2}\tfrac1r\tfrac{dV}{dr}\mathbf S\!\cdot\!\mathbf L$ with the Thomas factor $\tfrac12$ automatic; its eigenvalue $\tfrac{\hbar^2}{2}[j(j{+}1)-\ell(\ell{+}1)-\tfrac34]$ splits each $\ell\neq0$ hydrogen level into $j=\ell+\tfrac12$ (above) and $j=\ell-\tfrac12$ (below).

Where this lives today

Spin-orbit coupling is no longer just an atomic-physics effect; it plays a central role throughout condensed-matter physics. In crystals, the same relativistic interaction gives rise to Rashba and Dresselhaus spin-orbit couplings, enables electrical control of electron spins in semiconductors, and drives band inversions responsible for topological insulators. Its approximate $Z^4$ scaling makes relativistic effects increasingly important in heavy elements, contributing to phenomena such as the colour of gold, the liquid state of mercury, and subtle changes in chemical bonding. In atomic clocks and precision spectroscopy the coupling defines the fine-structure intervals against which everything smaller (hyperfine structure, the Lamb shift) is measured. In quantum-information devices, spin-orbit coupling is both a resource and a limitation: it enables all-electrical control of spin qubits while simultaneously providing a channel through which spins couple to their environment and decohere.

Figure

The Z⁴ law: why hydrogen's fine structure is a whisper and heavy-atom doublets shout

The physical picture

Ride along with the electron. In its instantaneous frame the proton swings around it — a moving charge, a current loop — and makes a magnetic field $\mathbf B$ at the electron's location, pointing along the orbital angular momentum $\mathbf L$ . The electron carries a spin magnetic moment $\boldsymbol\mu\propto\mathbf S$ , and its orientation energy $-\boldsymbol\mu\!\cdot\!\mathbf B$ depends on whether spin points with or against $\mathbf L$ . That is the $\mathbf S\!\cdot\!\mathbf L$ structure, and it is why an $\ell\neq0$ level splits: spin "parallel" to the orbit costs a different energy than spin "antiparallel." The naive version of that ride overcounts. The electron's frame is not inertial — it is being whipped around a circle — and a chain of infinitesimal boosts along the curving trajectory composes into a slow net rotation, the Thomas precession, running at half the orbital rate and opposing the spin's precession. It cancels exactly half of the magnetic interaction. In the heuristic story you must know this and multiply by $\tfrac12$ ; in the Dirac story you never boost to a rest frame, so the $\tfrac12$ is already in the coefficient the reduction produces. The Dirac equation does the special relativity for you, silently and exactly. The energy bookkeeping is clean because $\mathbf S\!\cdot\!\mathbf L=\tfrac12(\mathbf J^2-\mathbf L^2-\mathbf S^2)$ . For a $p$ -electron ( $\ell=1$ ), the two outcomes are $j=\tfrac32$ with $\mathbf S\!\cdot\!\mathbf L=+\tfrac12\hbar^2$ (higher) and $j=\tfrac12$ with $\mathbf S\!\cdot\!\mathbf L=-\hbar^2$ (lower). The weighted average of the two shifts vanishes, so the center of gravity of the multiplet remains unchanged while the degeneracy is lifted — which is the figure below. And the steep $Z^4$ growth (from $\tfrac1r\tfrac{dV}{dr}\propto Z$ and $\langle 1/r^3\rangle\propto Z^3$ ) is why gold is yellow and mercury is liquid: in heavy atoms the coupling is no longer fine structure but a leading reorganization of the levels.

Figure

The p-level splits by S·L, center of gravity fixed by the 2j+1 sum rule

The math

The spin-orbit operator is not postulated here — it is extracted from the Dirac equation by removing the small spinor components. We take the standard $2\times2$ block form and reduce, keeping every factor so the coefficient (and its Thomas $\tfrac12$ ) is honest. The four-component structure of the Dirac spinor and the identity $\mathbf S=\tfrac{\hbar}{2}\boldsymbol\sigma$ are developed in Spin from the Dirac Equation and Dirac Spinors and Plane-Wave Solutions, recalled by name. The endpoint is the operator and its Coulomb specialization,

H_{\rm SO} = \frac{1}{2m^2c^2}\,\frac1r\frac{dV}{dr}\,\mathbf S\!\cdot\!\mathbf L \;\equiv\; \xi(r)\,\mathbf S\!\cdot\!\mathbf L, \qquad H_{\rm SO}^{\rm Coul} = \frac{1}{2m^2c^2}\,\frac{Ze^2}{4\pi\varepsilon_0}\,\frac{1}{r^3}\,\mathbf S\!\cdot\!\mathbf L \qquad (1)

where: $V(r)$ is the central potential energy, $\mathbf S=\tfrac{\hbar}{2}\boldsymbol\sigma$ the spin, $\mathbf L=\mathbf r\times\mathbf p$ the orbital angular momentum, and $\xi(r)$ the radial coupling. Read the coefficient physically: it is $1/(2m^2c^2)$ , not $1/(m^2c^2)$ — the factor of two smaller than the naive estimate is the whole point, and the Coulomb form makes the steep $1/r^3$ (hence $Z^4$ ) dependence visible. To turn the operator into energies you diagonalize $\mathbf S\!\cdot\!\mathbf L$ . Squaring $\mathbf J = \mathbf L + \mathbf S$ (with $\mathbf L$ and $\mathbf S$ commuting, since they act on different spaces) gives $\mathbf J^2 = \mathbf L^2 + \mathbf S^2 + 2\mathbf S\!\cdot\!\mathbf L$ , so

\mathbf S\!\cdot\!\mathbf L = \tfrac12\big(\mathbf J^2-\mathbf L^2-\mathbf S^2\big), \qquad \langle\mathbf S\!\cdot\!\mathbf L\rangle = \frac{\hbar^2}{2}\big[j(j{+}1)-\ell(\ell{+}1)-s(s{+}1)\big] \qquad (2)

where: $j=\ell\pm\tfrac12$ is the total angular momentum quantum number, $s=\tfrac12$ , and $\ell$ the orbital number. This is manifestly diagonal in $|n,\ell,j,m_j\rangle$ : the operator commutes with $\mathbf J^2,\mathbf L^2,\mathbf S^2,J_z$ , so those are the good quantum numbers, while $L_z$ and $S_z$ are not. For $\ell=1$ the two branches are $+\tfrac12\hbar^2$ ( $j=\tfrac32$ ) and $-\hbar^2$ ( $j=\tfrac12$ ).

Derivation ASpin-orbit coupling from eliminating the small Dirac components

We derive the spin-orbit interaction by reducing the Dirac equation in a static potential to an effective two-component equation valid through order $(v/c)^2$ . The procedure is straightforward: eliminate the small component of the Dirac spinor, substitute the result into the large-component equation, and isolate the spin-dependent term using the Pauli identity. The resulting Hamiltonian already contains the Thomas factor of $\tfrac12$ ; no additional relativistic correction needs to be inserted by hand.

Step 1. Write the stationary Dirac equation in $2\times2$ blocks, with $E'=E-mc^2$ the nonrelativistic energy:

\begin{pmatrix}V & c\,\boldsymbol\sigma\!\cdot\!\mathbf p\\ c\,\boldsymbol\sigma\!\cdot\!\mathbf p & V-2mc^2\end{pmatrix}\begin{pmatrix}\psi\\\chi\end{pmatrix} = E'\begin{pmatrix}\psi\\\chi\end{pmatrix}.

The lower-diagonal $-2mc^2$ is the rest-energy gap that makes $\chi$ small.

Step 2. Solve the lower row for $\chi$ and expand the energy denominator to first nontrivial order in $1/c^2$ :

\chi = \frac{c\,\boldsymbol\sigma\!\cdot\!\mathbf p}{E'-V+2mc^2}\,\psi \approx \frac{\boldsymbol\sigma\!\cdot\!\mathbf p}{2mc}\Big(1-\frac{E'-V}{2mc^2}\Big)\psi.

Why $\chi$ is small: it is suppressed by $\boldsymbol\sigma\!\cdot\!\mathbf p/2mc\sim v/c$ relative to $\psi$ — the "small component" name is literal.

Step 3. Substitute into the upper row. The correction of interest lives in the $\boldsymbol\sigma\!\cdot\!\mathbf p\,(E'-V)\,\boldsymbol\sigma\!\cdot\!\mathbf p$ piece; using $[\mathbf p,V]=-i\hbar\nabla V$ , it contains the term $(\boldsymbol\sigma\!\cdot\!\nabla V)(\boldsymbol\sigma\!\cdot\!\mathbf p)$ . Apply the Pauli identity $(\boldsymbol\sigma\!\cdot\!\mathbf A)(\boldsymbol\sigma\!\cdot\!\mathbf B)=\mathbf A\!\cdot\!\mathbf B+i\boldsymbol\sigma\!\cdot\!(\mathbf A\times\mathbf B)$ with $\mathbf A=\nabla V$ , $\mathbf B=\mathbf p$ . The cross-product term $i\boldsymbol\sigma\!\cdot\!(\nabla V\times\mathbf p)$ is the seed of spin-orbit.

Step 4. Specialize to a central potential, $\nabla V=\tfrac{\mathbf r}{r}\tfrac{dV}{dr}$ , so $\nabla V\times\mathbf p=\tfrac1r\tfrac{dV}{dr}(\mathbf r\times\mathbf p)=\tfrac1r\tfrac{dV}{dr}\mathbf L$ . Collecting the coefficient (the $1/2mc^2$ denominator expansion together with the factor from proper normalization of the two-component wavefunction) gives

H_{\rm SO} = \frac{\hbar}{4m^2c^2}\,\frac1r\frac{dV}{dr}\,\boldsymbol\sigma\!\cdot\!\mathbf L = \frac{1}{2m^2c^2}\,\frac1r\frac{dV}{dr}\,\mathbf S\!\cdot\!\mathbf L,

using $\mathbf S=\tfrac{\hbar}{2}\boldsymbol\sigma$ . Why the coefficient is $\tfrac{1}{2m^2c^2}$ and not $\tfrac{1}{m^2c^2}$ : the naive rest-frame argument gives the larger value, and the missing half is the Thomas precession — here it is already contained in the reduction, so the coefficient comes out correct with no external factor. *The reduction's coefficient is exactly half the naive rest-frame coefficient (ratio $\tfrac12$ ), and the coupling scales as $\alpha^4 mc^2$ on Coulomb states.*

Go deeper · Derivation B · 1 minThe doublet split and the hydrogen 2p splitting

Derivation BThe doublet split and the hydrogen 2p splitting

We evaluate the spin-orbit energy shift for hydrogen and show explicitly that the $j=\ell+\tfrac12$ level lies above the $j=\ell-\tfrac12$ level. Because the spin-orbit operator is already diagonal in the coupled basis $|n,\ell,j,m_j\rangle$ , the calculation factorizes into a radial expectation value and an angular eigenvalue. The required hydrogenic expectation value $\langle r^{-3}\rangle$ is taken as known and is not rederived here.

Step 1. Separate radial and angular parts:

\Delta E_{\rm SO} = \langle\xi(r)\rangle\,\langle\mathbf S\!\cdot\!\mathbf L\rangle = \frac{1}{2m^2c^2}\frac{Ze^2}{4\pi\varepsilon_0}\langle r^{-3}\rangle\cdot\frac{\hbar^2}{2}\big[j(j{+}1)-\ell(\ell{+}1)-\tfrac34\big].

Why the separation is exact: $\xi(r)$ acts only on the radial wavefunction and $\mathbf S\!\cdot\!\mathbf L$ only on the angular/spin part, and the coupled state factorizes into a radial function times a spinor spherical harmonic.

Step 2. Insert the hydrogen radial average $\langle r^{-3}\rangle_{n\ell}=1/[a_0^3 n^3\ell(\ell+\tfrac12)(\ell+1)]$ (requires $\ell\neq0$ , consistent with $\mathbf S\!\cdot\!\mathbf L=0$ at $\ell=0$ ) to reach the compact form

\Delta E_{\rm SO} = \frac{(E_n)^2}{mc^2}\,\frac{n\big[j(j{+}1)-\ell(\ell{+}1)-\tfrac34\big]}{\ell(\ell+\tfrac12)(\ell+1)}.

Step 3. Evaluate the two branches. For $j=\ell+\tfrac12$ , $\langle\mathbf S\!\cdot\!\mathbf L\rangle=+\tfrac{\hbar^2}{2}\ell$ ; for $j=\ell-\tfrac12$ , $\langle\mathbf S\!\cdot\!\mathbf L\rangle=-\tfrac{\hbar^2}{2}(\ell+1)$ . Since $\xi>0$ , the $j=\ell+\tfrac12$ level lies above $j=\ell-\tfrac12$ — the normal doublet ordering. Why the weighted average vanishes: $(2j_+{+}1)\langle\mathbf S\!\cdot\!\mathbf L\rangle_+ + (2j_-{+}1)\langle\mathbf S\!\cdot\!\mathbf L\rangle_- = (2\ell{+}2)(\tfrac{\hbar^2}{2}\ell)+(2\ell)(-\tfrac{\hbar^2}{2}(\ell{+}1))=0$ — the center-of-gravity sum rule.

Step 4. Put in numbers for $2p$ ( $n=2,\ell=1$ ). The $\mathbf S\!\cdot\!\mathbf L$ eigenvalues are $+\tfrac12\hbar^2$ ( $j=\tfrac32$ ) and $-\hbar^2$ ( $j=\tfrac12$ ), a gap of $\tfrac32\hbar^2$ . With $\langle 1/r^3\rangle_{2p}=1/(24a_0^3)=2.81\times10^{29}\ \text{m}^{-3}$ , the shifts are $+1.51\times10^{-5}$ eV and $-3.02\times10^{-5}$ eV, so the splitting is

\Delta E(2p_{3/2}-2p_{1/2}) = 4.53\times10^{-5}\ \text{eV} = 0.365\ \text{cm}^{-1}\approx10.9\ \text{GHz}.

Explicit diagonalization of $\mathbf L\!\cdot\!\mathbf S$ for $\ell=1$ gives eigenvalues $\{+\tfrac12$ (multiplicity 4), $-1$ (multiplicity 2) $\}\hbar^2$ , consistent with $\langle 1/r^3\rangle_{2p}=1/(24a_0^3)$ and the $2p$ splitting $4.53\times10^{-5}$ eV.

Limiting cases

case	result	meaning
$\ell=0$ ( $s$ -states)	$\mathbf S\!\cdot\!\mathbf L\,\vert n,0\rangle = 0$	no orbital current, no spin-orbit shift; the Darwin term carries the $\ell=0$ correction instead
hydrogen scaling	$\langle H_{\rm SO}\rangle\sim\alpha^4 mc^2$ ( $Z=1$ )	order $(v/c)^2$ below the gross structure — $\sim10^{-4}$ eV, hence "fine"
$j=\ell+\tfrac12$ vs $j=\ell-\tfrac12$	$+\tfrac{\hbar^2}{2}\ell$ vs $-\tfrac{\hbar^2}{2}(\ell+1)$	the upper ( $\ell+\tfrac12$ ) level lies higher; center of gravity fixed by the $2j+1$ sum rule
heavy atoms	$H_{\rm SO}\propto Z^4$	coupling competes with electrostatics, driving $jj$ - over $LS$ -coupling and relativistic chemistry
$c\to\infty$	$H_{\rm SO}\to0$	the $1/c^2$ prefactor confirms a genuine relativistic effect with no classical analogue

Common trapThe electron "sees a magnetic field in its rest frame" — so multiply by one-half

The heuristic that an orbiting electron sees $\mathbf B=-\tfrac{1}{c^2}\mathbf v\times\mathbf E$ in its rest frame gets the $\mathbf S\!\cdot\!\mathbf L$ structure right but the magnitude wrong by exactly a factor of two. The trap is to either not know that, or to "fix" it by fudging a constant without naming the physics. The correct fix is Thomas precession: the electron frame is accelerating, and the composition of boosts along its curved path produces a net rotation that halves the coupling. In the Dirac treatment there is nothing to fix — the equation is covariant, it never invokes a rest frame, and the coefficient $\tfrac{1}{2m^2c^2}$ emerges already halved. A second, quieter trap: once $H_{\rm SO}$ is on, $L_z$ and $S_z$ are not separately conserved — only $\mathbf J=\mathbf L+\mathbf S$ and $J_z$ are. Labeling states by $m_\ell,m_s$ after switching on spin-orbit coupling is a common error.

Worked examples

Example 1 — the S·L eigenvalues for a p-electron (confidence builder). Compute $\langle\mathbf S\!\cdot\!\mathbf L\rangle$ for the two fine-structure levels of a $p$ -state ( $\ell=1$ , $s=\tfrac12$ ), and verify the center-of-gravity sum rule.

Step 1. Use $\langle\mathbf S\!\cdot\!\mathbf L\rangle=\tfrac{\hbar^2}{2}[j(j{+}1)-\ell(\ell{+}1)-s(s{+}1)]$ with $\ell(\ell{+}1)=2$ and $s(s{+}1)=\tfrac34$ . For $j=\tfrac32$ : $j(j{+}1)=\tfrac{15}{4}$ , so $\langle\mathbf S\!\cdot\!\mathbf L\rangle=\tfrac{\hbar^2}{2}(\tfrac{15}{4}-2-\tfrac34)=\tfrac{\hbar^2}{2}\cdot1=\boxed{+\tfrac12\hbar^2}$ .

Step 2. For $j=\tfrac12$ : $j(j{+}1)=\tfrac34$ , so $\langle\mathbf S\!\cdot\!\mathbf L\rangle=\tfrac{\hbar^2}{2}(\tfrac34-2-\tfrac34)=\tfrac{\hbar^2}{2}(-2)=\boxed{-\hbar^2}$ . The gap between the branches is $\tfrac12\hbar^2-(-\hbar^2)=\tfrac32\hbar^2=(\ell+\tfrac12)\hbar^2$ , as expected.

Step 3. Check the sum rule. The degeneracies are $2j+1=4$ ( $j=\tfrac32$ ) and $2$ ( $j=\tfrac12$ ). The weighted sum is $4(+\tfrac12\hbar^2)+2(-\hbar^2)=2\hbar^2-2\hbar^2=\boxed{0}$ . The multiplet's center of gravity does not move — spin-orbit coupling redistributes energy among sublevels without shifting the mean. (An explicit $6\times6$ diagonalization of $\mathbf L\!\cdot\!\mathbf S$ gives eigenvalues $+\tfrac12\hbar^2$ four times and $-\hbar^2$ twice, confirming both values and multiplicities — verified.)

Example 2 — the hydrogen 2p fine-structure splitting (exam level). Compute the $2p_{3/2}$ – $2p_{1/2}$ energy splitting in hydrogen in eV and cm $^{-1}$ , from $\langle 1/r^3\rangle$ .

Step 1. Radial average. For $n=2,\ell=1$ : $\langle 1/r^3\rangle_{2p}=1/[a_0^3\cdot 8\cdot 1\cdot\tfrac32\cdot 2]=1/(24a_0^3)$ . With $a_0=0.5292\times10^{-10}$ m, $a_0^3=1.482\times10^{-31}\ \text{m}^3$ , so $\langle 1/r^3\rangle_{2p}=1/(24\cdot1.482\times10^{-31})=\boxed{2.81\times10^{29}\ \text{m}^{-3}}$ .

Step 2. Coupling strength. Write the shift as $\Delta E_{\rm SO}=\tfrac{(E_n)^2}{mc^2}\tfrac{n[\,j(j{+}1)-\ell(\ell{+}1)-\tfrac34\,]}{\ell(\ell+\tfrac12)(\ell+1)}$ . With $E_2=-3.4014$ eV, $(E_2)^2/mc^2=(3.4014)^2/(5.110\times10^5)=2.264\times10^{-5}$ eV, and $n/[\ell(\ell+\tfrac12)(\ell+1)]=2/(1\cdot\tfrac32\cdot2)=\tfrac23$ .

Step 3. The two shifts. For $j=\tfrac32$ , the bracket $j(j{+}1)-\ell(\ell{+}1)-\tfrac34=\tfrac{15}{4}-2-\tfrac34=1$ , so $\Delta E_{3/2}=2.264\times10^{-5}\cdot\tfrac23\cdot1=+1.51\times10^{-5}$ eV. For $j=\tfrac12$ , the bracket is $\tfrac34-2-\tfrac34=-2$ , so $\Delta E_{1/2}=2.264\times10^{-5}\cdot\tfrac23\cdot(-2)=-3.02\times10^{-5}$ eV.

Step 4. The splitting. $\Delta E(2p_{3/2}-2p_{1/2})=1.51\times10^{-5}-(-3.02\times10^{-5})=\boxed{4.53\times10^{-5}\ \text{eV}}$ . Converting, $4.53\times10^{-5}\ \text{eV}\times 8065.54\ \text{cm}^{-1}/\text{eV}=\boxed{0.365\ \text{cm}^{-1}}\approx10.9$ GHz. This is the internal-magnetic-field splitting of the $2p$ level — as if the electron sat in a field of order $0.8$ T — and it matches the exact Dirac result (verified). The full fine-structure assembly with the kinetic and Darwin terms is developed in Relativistic Corrections to the Hydrogen Atom.

Historical development

Origins · 1 minspin-orbit coupling and how this was discovered

spin-orbit coupling

introduced 1925–1926 · Leiden, Copenhagen, and the factor of two

Spin-orbit coupling was born in the crisis it caused. In 1925 Samuel Goudsmit and George Uhlenbeck, graduate students in Leiden, proposed that the electron carries intrinsic spin to explain the doublet structure of alkali spectra and the anomalous Zeeman effect. The idea worked beautifully — except the fine-structure splitting came out twice too large, a factor-of-two error serious enough that Pauli and others doubted spin entirely. The rescue came in 1926 from Llewellyn Thomas, who showed that the electron's rest frame, being accelerated, precesses relativistically, and that this Thomas precession halves the naive coupling. Two years later Dirac's equation absorbed the whole story: spin, the $g=2$ moment, and the correctly halved spin-orbit coupling all emerged from a single relativistic wave equation, with no factor to insert. Sommerfeld had in fact obtained the right fine-structure formula back in 1916 from relativistic old-quantum orbits — but for the wrong, spinless reason; it was the Dirac equation that made the coupling mean what it means.

History

1916

— Sommerfeld's relativistic orbits give the correct fine-structure formula, but with no spin

1925

— Goudsmit and Uhlenbeck propose electron spin; the doublet splitting is a factor 2 too large

1926

— Thomas explains the factor 2 by the relativistic precession of the accelerating electron frame

1928

— Dirac's equation yields spin-orbit coupling with the Thomas one-half automatic

1929

— Foldy–Wouthuysen-type reductions later make the order-by-order extraction systematic

Going further

References. Griffiths, Introduction to Quantum Mechanics, 3rd ed., §7.3 · Sakurai & Napolitano, Modern Quantum Mechanics, 2nd ed., §8.2 · Bjorken & Drell, Relativistic Quantum Mechanics, §4.3 · Shankar, Principles of Quantum Mechanics, 2nd ed., Ch. 17, 20 · Cohen-Tannoudji, Diu & Laloë, Quantum Mechanics, Vol. II, Complement $A_{XII}$ · Thomas, Nature 117, 514 (1926). This article derives: (i) the extraction of $H_{\rm SO}$ from the Dirac equation, with the coefficient $\tfrac{1}{2m^2c^2}$ equal to half the naive rest-frame value; (ii) the $\mathbf S\!\cdot\!\mathbf L$ eigenvalue split $+\tfrac12\hbar^2$ ( $j=\tfrac32$ ) and $-\hbar^2$ ( $j=\tfrac12$ ) for $\ell=1$ , confirmed by explicit $6\times6$ diagonalization giving eigenvalues $\{+\tfrac12$ ( $\times4$ ), $-1$ ( $\times2$ ) $\}\hbar^2$ ; (iii) the level ordering $j=\ell+\tfrac12$ above $j=\ell-\tfrac12$ and the $\sum(2j+1)$ center-of-gravity sum rule; (iv) $\langle 1/r^3\rangle_{2p}=1/(24a_0^3)=2.81\times10^{29}\ \text{m}^{-3}$ and the $2p_{3/2}$ – $2p_{1/2}$ splitting $4.53\times10^{-5}$ eV $=0.365\ \text{cm}^{-1}$ . The perturbative $\langle 1/r^3\rangle$ (Kramers) is owned by Spin-Orbit Coupling by name; the assembly into the $n,j$ -only fine-structure formula is developed in Fine Structure of Hydrogen and Relativistic Corrections to the Hydrogen Atom. Sodium D-line wavelengths (588.995/589.592 nm) are NIST values.

Builds on: The Dirac Equation · Spin from the Dirac Equation · Nonrelativistic Reduction and the Pauli Equation · Dirac Spinors and Plane-Wave Solutions

Leads to: Relativistic Corrections to the Hydrogen Atom · Fine Structure of Hydrogen · The Anomalous Zeeman Effect · Hyperfine Structure

Isn't spin-orbit coupling just an electron feeling a magnetic field in its rest frame?

That picture gives the right structure and the wrong size. It correctly predicts an $\mathbf S\!\cdot\!\mathbf L$ interaction — the electron's moment coupling to the field of the orbiting nucleus — but overestimates its magnitude by exactly a factor of two, because it treats the electron's rest frame as inertial. The frame is accelerating, and the relativistic composition of boosts produces the Thomas precession, which halves the result. So the heuristic is a useful mnemonic for the form, but you must either add Thomas by hand or, better, derive the coupling from the Dirac equation, which never invokes a rest frame and gives the correctly halved coefficient automatically.

Why does spin-orbit coupling vanish for s-states?

Because $H_{\rm SO}\propto\mathbf S\!\cdot\!\mathbf L$ and $\mathbf L|n,0\rangle=0$ — an $s$ -state has no orbital angular momentum, so there is no orbital current to make a magnetic field and nothing for the spin to couple to. The radial factor $\langle 1/r^3\rangle$ formally diverges for $\ell=0$ , but the angular factor $\mathbf S\!\cdot\!\mathbf L$ vanishes faster, and the operator gives zero on $s$ -states. The relativistic correction that $s$ -states do feel is the Darwin term, which is a contact interaction proportional to $|\psi(0)|^2$ and acts only where the wavefunction is nonzero at the origin — precisely the $\ell=0$ states.

What does "center of gravity" of a multiplet mean, and why is it fixed?

When a level splits into its $j$ -components, the degeneracy-weighted average energy stays put: $\sum_j (2j+1)\Delta E_j = 0$ . For a $p$ -level, $4$ states shift up by $+\tfrac12\xi\hbar^2$ and $2$ shift down by $-\xi\hbar^2$ , and $4(+\tfrac12)+2(-1)=0$ . Physically, spin-orbit coupling only redistributes energy among the sublevels — it does not move the mean, because it is traceless in the degenerate subspace (its matrix has zero trace, and the trace is the summed shift). This sum rule is a strong check on any fine-structure calculation and a handy way to catch sign errors.

If the operator has 1∕r³, and ⟨1∕r³⟩ diverges for ℓ = 0, how is the theory consistent?

The apparent divergence is harmless because it multiplies a factor that vanishes. Kramers' relation gives $\langle 1/r^3\rangle\propto 1/[\ell(\ell+\tfrac12)(\ell+1)]$ , which blows up as $\ell\to0$ — but the $\mathbf S\!\cdot\!\mathbf L$ eigenvalue $\tfrac{\hbar^2}{2}[j(j{+}1)-\ell(\ell{+}1)-\tfrac34]$ goes to zero there. Their product is well-defined and gives no spin-orbit shift for $\ell=0$ . The lesson is that you must keep the operator whole: $\mathbf S\!\cdot\!\mathbf L$ enforces the selection to $\ell\neq0$ , and the $s$ -state relativistic correction shows up in the Darwin term, not here. A naive "just plug in the divergent $\langle 1/r^3\rangle$ " is meaningless without the angular factor.

Why is the fine structure of hydrogen so much smaller than sodium's D-line splitting?

Both are spin-orbit doublets, but sodium's is much larger because of screening and effective charge. In hydrogen the $2p$ splitting is $4.5\times10^{-5}$ eV. In sodium the valence $3p$ electron penetrates the inner-shell core, where it sees a much larger unscreened nuclear charge; the effective $Z$ in $\langle 1/r^3\rangle$ is boosted, and since $\Delta E_{\rm SO}\propto Z^4$ the splitting jumps to about $2.1$ meV — nearly fifty times larger, the visible yellow doublet at 589.0/589.6 nm. The lesson generalizes: alkali fine structure is a sensitive probe of how deeply the valence electron penetrates the core, and the $Z^4$ scaling amplifies small differences in effective charge.

Does the Dirac derivation change the answer, or only the story?

For the coefficient it changes only the story — but it is the story that makes the coefficient correct. The perturbative and Dirac routes agree numerically on $\tfrac{1}{2m^2c^2}$ , but they get there differently: perturbation theory must import Thomas precession as an external relativistic correction, while the Dirac reduction produces the halved coefficient with no extra input. The deeper payoff is conceptual — the Dirac derivation shows spin-orbit coupling is not an optional add-on but a forced consequence of relativistic quantum mechanics, locked together with the $g=2$ moment and the kinetic and Darwin corrections. When you go to the exact Dirac spectrum, that unity becomes a single closed formula depending only on $n$ and $j$ .