Spin-Orbit Coupling from the Rest Frame

Derivation

Spin-Orbit Coupling from the Rest Frame

D-299 Home PU-306 Threads energy · fields · matter · symmetry Depends on Bound-State Spectrum of the Hydrogen Atom, spin-half-algebra, biot-savart-law

Statement

For a single electron of mass \(m\), charge \(-e\), moving with velocity \(\vec v = \vec p/m\) in the central electrostatic field of a nucleus of charge \(+Ze\), the coupling between the electron spin \(\vec S\) and its orbital angular momentum \(\vec L\) is \[ \hat H_{\text{SO}} = \frac{1}{2m^2c^2}\,\frac{1}{r}\frac{dU}{dr}\,\vec S\cdot\vec L \;=\; \frac{Ze^2}{8\pi\varepsilon_0\,m^2c^2\,r^3}\,\vec S\cdot\vec L, \] where \(U(r)=-\,Ze^2/(4\pi\varepsilon_0 r)\). The overall factor of \(\tfrac12\) is the Thomas correction; the naive rest-frame magnetic argument alone gives twice this.

Why it matters

Spin-orbit coupling is the leading term that lifts the accidental \(\ell\)-degeneracy of the hydrogen spectrum and produces the observed fine structure: the sodium D doublet, the \(2P_{3/2}\!-\!2P_{1/2}\) splitting of hydrogen, and the \(Z^4\) growth of fine structure across the periodic table all trace to this single operator.

It is also the cleanest place in undergraduate physics where a purely kinematic relativistic effect — Thomas precession — is not optional decoration but a measurable factor of two. Getting the coefficient right is a genuine test that special relativity has been applied consistently, and it foreshadows the Dirac equation, which reproduces exactly this operator with the correct \(\tfrac12\) built in.

Assumptions

The potential is central and electrostatic, \(U=U(r)\).If dropped, \(\vec E\) is not radial, \(\vec E\parallel\vec r\) fails, and the compact \(\vec S\cdot\vec L\) form is replaced by a general tensor coupling. The electron is slow, \(v/c\ll 1\), so fields transform to first order in \(v/c\) and energies to order \(v^2/c^2\).If dropped, higher Lorentz-transformation terms and the full \(\gamma\) factors survive; one must solve the Dirac equation instead of expanding it. The electron rest frame is momentarily comoving but non-inertial; successive boosts compose into a rotation (Thomas precession).If dropped — i.e. if one treats the rest frame as a single inertial frame — the Thomas \(\tfrac12\) is lost and the fine structure comes out twice too large, in flat contradiction with spectroscopy. The spin gyromagnetic ratio is \(g_s=2\).If dropped, the coefficient carries \(g_s\); QED gives \(g_s=2.0023\), a per-mille correction that matters only at high precision. The nucleus is infinitely heavy and structureless.If dropped, reduced-mass and nuclear-moment (hyperfine) corrections appear; these are separate, smaller effects.

Derivation

\[ \vec E = \frac{Ze}{4\pi\varepsilon_0\,r^3}\,\vec r,\qquad U(r)=-\frac{Ze^2}{4\pi\varepsilon_0\,r},\qquad \vec E=-\frac1e\frac{dU}{dr}\,\hat r. \]

Coulomb field of the nucleus in the lab frame; write it so the potential-energy gradient is explicit for later generalisation. A

\[ \vec B' = \vec B - \frac{1}{c^2}\,\vec v\times\vec E \;\xrightarrow{\;\vec B=0\;}\; \vec B' = -\frac{1}{c^2}\,\vec v\times\vec E . \]

Lorentz transformation of the electromagnetic field to the electron's momentary rest frame, kept to first order in \(v/c\); in the lab \(\vec B=0\) so only the motional field survives. B

\[ \vec B' = -\frac{1}{c^2}\,\frac{\vec p}{m}\times\frac{Ze}{4\pi\varepsilon_0 r^3}\vec r = -\frac{Ze}{4\pi\varepsilon_0 m c^2 r^3}\,(\vec p\times\vec r) = +\frac{Ze}{4\pi\varepsilon_0 m c^2 r^3}\,\vec L . \]

Insert \(\vec E\) and \(\vec v=\vec p/m\), then use \(\vec p\times\vec r=-(\vec r\times\vec p)=-\vec L\). The field in the electron frame is parallel to \(\vec L\). A

\[ \vec\mu_s = -\,g_s\,\frac{e}{2m}\,\vec S \;\xrightarrow{\;g_s=2\;}\; \vec\mu_s = -\frac{e}{m}\,\vec S . \]

Intrinsic magnetic moment of the electron with \(g_s=2\); the sign reflects the negative charge. A

\[ \hat H_{\text{mag}} = -\,\vec\mu_s\cdot\vec B' = \frac{e}{m}\,\vec S\cdot\vec B' = \frac{Ze^2}{4\pi\varepsilon_0\,m^2c^2\,r^3}\,\vec S\cdot\vec L . \]

Zeeman energy of the moment in the rest-frame field. This is the "naive" spin-orbit energy — correct in structure, wrong by a factor of two. A

\[ \vec\omega_T = -\frac{\gamma^2}{\gamma+1}\,\frac{\vec v\times\vec a}{c^2}\;\xrightarrow{\;\gamma\to1\;}\; -\frac{1}{2c^2}\,\vec v\times\vec a,\qquad \vec a=\frac{\vec F}{m}=-\frac{e\vec E}{m}. \]

The comoving frame rotates because two non-collinear boosts compose to a boost times a rotation (Wigner rotation); \(\vec\omega_T\) is the Thomas precession rate, purely kinematic — no field enters its definition. C

\[ \vec\omega_T = -\frac{1}{2c^2}\frac{\vec p}{m}\times\!\left(-\frac{e}{m}\vec E\right) = -\frac{Ze^2}{8\pi\varepsilon_0 m^2 c^2 r^3}\,\vec L,\qquad \hat H_T=\vec\omega_T\cdot\vec S = -\tfrac12\,\hat H_{\text{mag}} . \]

Evaluate \(\vec\omega_T\) with \(\vec a=-e\vec E/m\), reusing \(\vec p\times\vec r=-\vec L\). A spin precessing at \(\vec\omega_T\) carries energy \(\vec\omega_T\cdot\vec S\); comparing with Step 5 gives exactly \(-\tfrac12\hat H_{\text{mag}}\). C

\[ \hat H_{\text{SO}} = \hat H_{\text{mag}} + \hat H_T = \left(1-\tfrac12\right)\hat H_{\text{mag}} = \frac{Ze^2}{8\pi\varepsilon_0\,m^2c^2\,r^3}\,\vec S\cdot\vec L . \]

Add the field energy and the kinematic Thomas energy. The half survives; this is the physical spin-orbit Hamiltonian. B

\[ \hat H_{\text{SO}} = \frac{1}{2m^2c^2}\,\frac{1}{r}\frac{dU}{dr}\,\vec S\cdot\vec L,\qquad \vec S\cdot\vec L = \tfrac12\!\left(\hat J^2-\hat L^2-\hat S^2\right). \]

Rewrite via \(\frac1r\frac{dU}{dr}=\frac{Ze^2}{4\pi\varepsilon_0 r^3}\) to expose the general central-potential form, and use \(\vec J=\vec L+\vec S\) so \(\vec S\cdot\vec L\) is diagonal in the coupled basis. B

Result

\[ \hat H_{\text{SO}} = \frac{1}{2m^2c^2}\,\frac{1}{r}\frac{dU}{dr}\,\vec S\cdot\vec L,\qquad \big\langle\vec S\cdot\vec L\big\rangle = \frac{\hbar^2}{2}\big[j(j{+}1)-\ell(\ell{+}1)-s(s{+}1)\big]. \]

Reading. The electron feels an internal magnetic field \(\vec B'\propto\vec L\) because in its own frame the nucleus circulates around it (Biot–Savart). Its spin moment couples to that field; the Thomas precession of the accelerating frame halves the result. States of different total angular momentum \(j\) are split, with \(j=\ell+\tfrac12\) raised and \(j=\ell-\tfrac12\) lowered.

Units check. \([Ze^2/(4\pi\varepsilon_0)]=\)J·m, so \(Ze^2/(4\pi\varepsilon_0 r^3)\) is J·m\(^{-2}\). Dividing by \(m^2c^2\) (kg\(^2\)m\(^2\)s\(^{-2}\)) and multiplying by \(\vec S\cdot\vec L\) (J\(^2\)s\(^2\)) gives \(\frac{\text{J·m}^{-2}}{\text{kg}^2\text{m}^2\text{s}^{-2}}\cdot\text{J}^2\text{s}^2\). With J\(=\)kg·m\(^2\)s\(^{-2}\) the kg and s cancel to leave J. Energy — correct.

Limiting cases

\(\ell=0\) (s-states): \(\vec S\cdot\vec L=0\), so the spin-orbit shift vanishes despite the divergent \(\langle 1/r^3\rangle\); s-state fine structure comes entirely from the separate Darwin term.
\(c\to\infty\) (non-relativistic): \(\hat H_{\text{SO}}\propto 1/c^2\to0\); the effect is intrinsically order \(\alpha^2\) relative to the Bohr energies.
High \(Z\): \(\langle 1/r^3\rangle\propto Z^3\) and the prefactor carries one more \(Z\), so \(\Delta E_{\text{SO}}\propto Z^4\) — fine structure explodes for heavy atoms and \(LS\) coupling gives way to \(jj\) coupling.
\(g_s\to2\) exactly: the coefficient is clean; the physical \(g_s=2.0023\) shifts it by \(\sim0.1\%\).

Breaks when

\(v/c\) is not small (high \(Z\), inner shells). The first-order field transformation and the \(\gamma\to1\) Thomas limit both fail; only the full Dirac equation (or a Foldy–Wouthuysen expansion carried further) is trustworthy.
Strong external magnetic field (Paschen–Back regime). When the Zeeman energy exceeds the spin-orbit energy, \(\vec L\) and \(\vec S\) decouple and precess independently about \(\vec B_{\text{ext}}\); \(j\) is no longer a good quantum number.
Many-electron correlation. The single-particle central potential is only an approximation; in open-shell atoms the coupling competes with electrostatic term splitting, and neither pure \(LS\) nor pure \(jj\) coupling is exact.
\(\ell=0\). The formula gives the indeterminate \(0\times\infty\); the relativistic Darwin correction, not this operator, describes s-states.

Failure modes

Dropping the Thomas \(\tfrac12\): quoting \(\hat H_{\text{mag}}\) as the answer, giving fine-structure splittings twice the measured value — historically the very discrepancy Thomas resolved in 1926.
Using \(g_s=1\): confusing the spin \(g\)-factor with the orbital one, halving the coupling.
Sign slip in \(\vec p\times\vec r\): forgetting \(\vec p\times\vec r=-\vec L\), which flips the sign of the field and inverts the multiplet ordering.
Treating the rest frame as inertial: transforming the field but ignoring that the frame accelerates — the conceptual root of the missing factor of two.
Writing \(\langle 1/r\rangle^3\) instead of \(\langle 1/r^3\rangle\): these differ by large numerical factors for hydrogenic states.
Applying the operator to \(\ell=0\): reporting a finite s-state shift from \(\langle 1/r^3\rangle\) rather than recognising \(\vec S\cdot\vec L=0\).

Discussion

The physical picture is deliberately frame-dependent and instructive because of it. In the lab the electron simply orbits a static charge; there is no magnetic field at all. Transform to the electron and the nucleus becomes a moving charge — a current loop — whose Biot–Savart field \(\vec B'\propto\vec L\) threads the electron. The spin-orbit energy is nothing more than the orientation energy of the spin moment in that field. Two observers thus attribute the same measurable splitting to different intermediate constructs, which is exactly what relativity permits.

The Thomas factor is the subtle part, and it is purely kinematic. It has nothing to do with electromagnetism: it arises because the sequence of infinitesimal boosts that track the accelerating electron does not compose to another boost but to a boost accompanied by a rotation. That rotation makes the comoving frame precess at \(\vec\omega_T\) even in the absence of any torque, and its energy \(\vec\omega_T\cdot\vec S\) is opposite to and half the magnitude of the magnetic term. The cancellation to exactly one half is a strong internal consistency check.

The deepest statement is that none of this need be assembled by hand. The Dirac equation, reduced to two-component form by a Foldy–Wouthuysen or non-relativistic expansion, produces \(\hat H_{\text{SO}}=\frac{1}{2m^2c^2}\frac1r\frac{dU}{dr}\,\vec S\cdot\vec L\) automatically, with the Thomas half already contained in the relativistic wave equation and \(g_s=2\) delivered as a prediction rather than an input. Together with the relativistic kinetic correction and the Darwin term, it yields the closed-form fine-structure energy \(E_{nj}\) that depends only on \(n\) and \(j\). The rest-frame derivation is best seen as physical intuition for a result the Dirac theory guarantees.

Common misconceptions. Spin-orbit coupling is not a magnetic interaction "in the lab" — in the lab there is no field; it is a relativistic effect that only looks magnetic in the electron frame. And the Thomas factor is not an ad hoc fudge to fit spectra: it is a rigorous consequence of the non-commutativity of Lorentz boosts, derivable with no reference to the atom at all.

Worked examples

Hydrogen \(2P_{3/2}\!-\!2P_{1/2}\) fine-structure splitting.

Define \(\zeta_{n\ell}=\dfrac{Ze^2}{8\pi\varepsilon_0 m^2c^2}\,\hbar^2\,\langle 1/r^3\rangle\); then \(\Delta E=\tfrac{\zeta}{2}[\,j(j{+}1)-\ell(\ell{+}1)-\tfrac34\,]\). Symbols first, numbers after.

\[ \Big\langle \tfrac{1}{r^3}\Big\rangle_{n\ell}=\frac{Z^3}{a_0^3\,n^3\,\ell(\ell+\tfrac12)(\ell+1)} \xrightarrow{Z=1,n=2,\ell=1} \frac{1}{a_0^3\cdot 8\cdot 1\cdot\frac32\cdot 2}=\frac{1}{24\,a_0^3}. \]

Standard hydrogenic radial expectation value (assumed prior result on hydrogen bound states).

\[ \zeta_{2p}=\frac{e^2}{8\pi\varepsilon_0}\cdot\frac{\hbar^2}{m^2c^2}\cdot\frac{1}{24 a_0^3},\quad \frac{e^2}{8\pi\varepsilon_0}=0.720\ \text{eV·nm},\ \ \frac{\hbar^2}{m^2c^2}=\frac{(\hbar c)^2}{(mc^2)^2}=1.49\times10^{-7}\ \text{nm}^2. \]

Group constants into eV·nm and nm\(^2\) using \(\hbar c=197.3\) eV·nm, \(mc^2=5.11\times10^5\) eV.

\[ \langle 1/r^3\rangle=\frac{1}{24(0.0529\,\text{nm})^3}=281\ \text{nm}^{-3},\quad \zeta_{2p}=0.720\times1.49\times10^{-7}\times281\ \text{eV}=3.02\times10^{-5}\ \text{eV}. \]

Insert \(a_0=0.0529\) nm and multiply; units nm·nm\(^2\)·nm\(^{-3}\) cancel to leave eV.

\[ \Delta E = \tfrac{3}{2}\zeta_{2p}=4.5\times10^{-5}\ \text{eV}\;\approx\;0.365\ \text{cm}^{-1}\;\approx\;11\ \text{GHz}. \]

Reading. The \(j=\tfrac32\) level lies \(\zeta/2\) above and the \(j=\tfrac12\) level lies \(\zeta\) below the unperturbed \(2p\) energy, a total gap \(\tfrac32\zeta\). This matches the measured hydrogen fine-structure interval, and it is the value that would be doubled had the Thomas factor been omitted.

Internal magnetic field seen by the electron in hydrogen \(2p\).

From Step 3 of the derivation, \(\vec B'=\dfrac{Ze}{4\pi\varepsilon_0 m c^2 r^3}\vec L\); take \(|\vec L|=\sqrt{\ell(\ell+1)}\,\hbar=\sqrt2\,\hbar\) and \(\langle 1/r^3\rangle\) as above. Symbols first.

\[ |\vec B'| = \frac{Ze}{4\pi\varepsilon_0}\cdot\frac{\hbar\sqrt2}{m c^2}\,\langle 1/r^3\rangle,\qquad \frac{e}{4\pi\varepsilon_0}=1.44\times10^{-9}\ \text{V·m}. \]

Collect the Coulomb constant times charge; the rest is expressed in SI to land in tesla.

\[ |\vec B'| = \frac{(1.44\times10^{-9})(\,1.49\times10^{-34}\,)(2.81\times10^{29})}{8.20\times10^{-14}}\ \text{T}, \]

with \(\hbar\sqrt2=1.49\times10^{-34}\) J·s, \(mc^2=8.20\times10^{-14}\) J, and \(\langle 1/r^3\rangle=2.81\times10^{29}\) m\(^{-3}\).

\[ |\vec B'| \approx 0.74\ \text{T}. \]

Reading. The electron in a hydrogen \(2p\) state experiences an effective field of order one tesla purely from its own motion through the nuclear Coulomb field. The Zeeman energy \(\mu_B|\vec B'|\approx(5.8\times10^{-5}\,\text{eV/T})(0.74\,\text{T})\approx4\times10^{-5}\) eV sets the fine-structure scale; halving by Thomas and resolving into \(\vec S\cdot\vec L\) reproduces the \(\sim10^{-5}\) eV level shifts of Example 1. Units: (V·m)(J·s)(m\(^{-3}\))/J \(=\) V·s·m\(^{-2}=\)T. Correct.

Problems

Evaluate \(\langle\vec S\cdot\vec L\rangle\) (in units of \(\hbar^2\)) for the \(2P_{3/2}\) and \(2P_{1/2}\) states of hydrogen.
Solution
Use \(\langle\vec S\cdot\vec L\rangle=\tfrac{\hbar^2}{2}[j(j{+}1)-\ell(\ell{+}1)-s(s{+}1)]\) with \(\ell=1,\ s=\tfrac12\). For \(j=\tfrac32\): \(\tfrac{\hbar^2}{2}[\tfrac{15}{4}-2-\tfrac34]=\tfrac{\hbar^2}{2}(1)=+\tfrac12\hbar^2\). For \(j=\tfrac12\): \(\tfrac{\hbar^2}{2}[\tfrac34-2-\tfrac34]=\tfrac{\hbar^2}{2}(-2)=-\hbar^2\). The difference \(\tfrac12\hbar^2-(-\hbar^2)=\tfrac32\hbar^2\) is why the splitting is \(\tfrac32\zeta\).
Show that for a fixed \(\ell\) the energy interval between adjacent fine-structure levels obeys the Landé interval rule \(E(j)-E(j{-}1)=\zeta\,j\), and evaluate it for the \(2p\) doublet.
Solution
With \(E(j)=\tfrac{\zeta}{2}[j(j{+}1)-\ell(\ell{+}1)-s(s{+}1)]\), \(E(j)-E(j{-}1)=\tfrac{\zeta}{2}[j(j{+}1)-(j{-}1)j]=\tfrac{\zeta}{2}\,j[\,(j{+}1)-(j{-}1)\,]=\tfrac{\zeta}{2}(2j)=\zeta j\). For \(2p\) the only interval is \(j=\tfrac32\): \(E(\tfrac32)-E(\tfrac12)=\zeta\cdot\tfrac32=\tfrac32\zeta\), consistent with Problem 1.
Using \(\Delta E_{\text{SO}}\propto Z^4\), predict the \(2p\) fine-structure splitting of the hydrogenic ion He\(^+\) from the hydrogen value \(4.5\times10^{-5}\) eV.
Solution
The coupling carries one explicit \(Z\) and \(\langle 1/r^3\rangle\propto Z^3\), so \(\zeta\propto Z^4\). For He\(^+\), \(Z=2\): factor \(2^4=16\). Hence \(\Delta E=16\times4.5\times10^{-5}\ \text{eV}=7.2\times10^{-4}\) eV \(\approx5.8\) cm\(^{-1}\). Heavier hydrogenic ions show correspondingly larger fine structure — the origin of the strong \(Z\)-dependence of X-ray line splittings.
Compute the spin-orbit constant \(\zeta_{3p}\) for hydrogen and the resulting \(3P_{3/2}\!-\!3P_{1/2}\) splitting.
Solution
\(\langle 1/r^3\rangle_{3p}=\dfrac{1}{a_0^3\,n^3\ell(\ell+\frac12)(\ell+1)}=\dfrac{1}{a_0^3\cdot27\cdot1\cdot\frac32\cdot2}=\dfrac{1}{81\,a_0^3}\). Numerically \(\langle 1/r^3\rangle=\dfrac{281\ \text{nm}^{-3}\cdot24}{81}=83.3\ \text{nm}^{-3}\) (scaling the \(2p\) value by \(24/81\)). Then \(\zeta_{3p}=0.720\times1.49\times10^{-7}\times83.3\ \text{eV}=8.9\times10^{-6}\) eV, and the splitting \(=\tfrac32\zeta_{3p}=1.3\times10^{-5}\) eV. Note \(\zeta_{3p}/\zeta_{2p}=24/81=(2/3)^3\cdot\ldots\) reflecting the \(n^{-3}\) scaling of \(\langle 1/r^3\rangle\).
Suppose an experimenter, forgetting Thomas precession, predicts the hydrogen \(2p\) splitting from \(\hat H_{\text{mag}}\) alone. What value do they obtain, and by what factor do they disagree with experiment? Explain physically why the correct answer is smaller.
Solution
\(\hat H_{\text{mag}}=2\hat H_{\text{SO}}\), so every splitting doubles: they predict \(2\times4.5\times10^{-5}=9.0\times10^{-5}\) eV, a factor of two too large — the historical discrepancy of 1925–26. Physically, the electron's rest frame is accelerating, so it precesses kinematically at \(\vec\omega_T=-\tfrac{1}{2c^2}\vec v\times\vec a\) even with no magnetic torque. The associated energy \(\vec\omega_T\cdot\vec S=-\tfrac12\hat H_{\text{mag}}\) partially cancels the magnetic term, leaving exactly half. The Dirac equation contains this automatically, confirming the factor is not adjustable.