
Advanced Quantum Mechanics course sequence · Interactive lesson
Spin-orbit coupling with the Thomas factor built in
Quantum Mechanics II got the spin-orbit energy right by taking a rest-frame answer and cutting it in half on Thomas's authority. Here the Dirac equation hands you the operator with its coefficient already correct, and tells you which part of it is magnetism and which part is the geometry of an accelerating frame.
UAQM-06.07 · Lesson 7 of 9 in this unit. Position is not a completion score.Build the model
The half Quantum Mechanics II borrowed from Thomas comes out of the Dirac equation.
The key idea
Spin-orbit coupling is the order-(v/c)² energy (1/(2m²c²))(1/r)(dV/dr) S⋅L that a Dirac particle acquires in a static central electrostatic potential, with the Thomas half already part of the coefficient rather than applied to it.
Example. In hydrogen's 2p state the operator lifts 2P₃/₂ by 3.65 GHz and lowers 2P₁/₂ by 7.30 GHz, a doublet of 10.95 GHz, or 4.53 × 10⁻⁵ eV; the measured interval is 10.969 GHz.
The complete physical model and assumptions
The argument of this topic is that the spin-orbit operator is a consequence, not an assumption. Topic 6.06's Foldy-Wouthuysen reduction left one spin term at order (v/c)², written with the electric field; for a static central potential it becomes (1/(2m²c²))(1/r)(dV/dr) S⋅L with nothing chosen, and for hydrogen it is exactly the operator Quantum Mechanics II reached only by computing the rest-frame magnetic energy and halving it on Thomas's authority. Give the particle an anomalous moment and the same reduction returns the coefficient g − 1: a magnetic part proportional to g, which is the rest-frame calculation, and a g-independent −1, which therefore cannot be magnetic and is the precession of an accelerated frame. Experiment agrees: hydrogen's 2P doublet, measured at 10 969.0 MHz, is 10 943 MHz times g − 1 to 0.003%, while halving would leave it 13 MHz short. The model stops in three places. It assumes the binding is the time component of a four-vector, and a Lorentz-scalar potential, entering as a position-dependent mass, reverses the sign. It is leading order in (Zα)², so heavy ions need Unit 7's exact levels. And it cannot predict a: the anomaly is quoted from radiative corrections.
1From the field form to the central form
SI units throughout; where four-vectors appear the metric is (+, −, −, −). Topic 6.06 left one spin term at order (v/c)², −(qħ/(8m²c²)) σ⋅(E × π − π × E), quoted from there. Specialise in three steps. Static fields: ∇ × E = −∂B/∂t = 0, and since (π × E)ψ = −iħ(∇ × E)ψ − (E × π)ψ, the antisymmetric pair collapses to 2E × π. No magnetic field: π = p. Central potential energy V(r) = qΦ(r): qE = −∇V = −(1/r)(dV/dr) r. Then −(ħ/(4m²c²)) σ⋅(qE × p) = (ħ/(4m²c²))(1/r)(dV/dr) σ⋅(r × p) = (1/(2m²c²))(1/r)(dV/dr) S⋅L, derived, with S = ħσ/2 and L = r × p. (1/r)(dV/dr) depends on r alone and commutes with every component of L, so the operator is Hermitian. Dimensions: J m⁻² × J² s² ÷ (kg² m² s⁻²) = J. Limit: as c → ∞ it vanishes, as a relativistic correction must.
2The Coulomb operator and the doublet it predicts
For hydrogen V = −e²/(4πε₀r), so (1/r)(dV/dr) = e²/(4πε₀r³) and HSO = (e²/(8πε₀))(1/(m²c²r³)) S⋅L, derived. For energies take Quantum Mechanics II's expectation values, quoted: ⟨1/r³⟩ = 1/(a₀³n³l(l + ½)(l + 1)) and ⟨S⋅L⟩ = (ħ²/2)[j(j + 1) − l(l + 1) − ¾]. For 2p, ⟨1/r³⟩ = 1/(24a₀³), and ⟨S⋅L⟩ is +ħ²/2 at j = 3/2 and −ħ² at j = 1/2, so 2P₃/₂ rises by mc²α⁴/96, which is h × 3.65 GHz, and 2P₁/₂ falls by mc²α⁴/48, h × 7.30 GHz; weighted by 2j + 1 the shifts cancel, 4 × 3.65 = 2 × 7.30. The doublet mc²α⁴/32 is h × 10 949 MHz, or h × 10 943 MHz with the reduced mass (quoted), against the measured 10 969.0 MHz. The kinetic correction depends only on n and l, and the Darwin term vanishes for l = 1, so this interval is the spin-orbit term alone: a clean test of its coefficient.
3What Quantum Mechanics II did, and why it needed a half
Quantum Mechanics II took another road. In the electron's instantaneous rest frame the nucleus moves, and its field transforms into B′ = −v × E/c²; with v = p/m and qE = −(1/r)(dV/dr) r, the moment μ = g(q/2m)S has energy −μ⋅B′ = (g/(2m²c²))(1/r)(dV/dr) S⋅L, derived. At g = 2 that is exactly twice the Dirac operator, and QM-II halved it by quoting Thomas. The half is not cosmetic: unhalved, the operator predicts a 2P doublet of 21.9 GHz, double the measured 10.97 GHz. The Dirac reduction needed no halving because it never left the nucleus's frame; the 1/(2m²c²) came out of topic 6.06's generator and its double commutator. The rest-frame argument is not wrong, only incomplete: the rest frame of an accelerating electron is itself rotating relative to the lab, and the energy of that rotation must be added before the two roads can agree.
4Reading the coefficient as g − 1
Topic 6.04 carried Pauli's term through the same reduction and found the spin-orbit coefficient multiplied by 1 + 2a, derived there: the term's odd electric half, crossed with cα⋅p, adds 2a times topic 6.06's spin term. Since g = 2(1 + a), 1 + 2a is g − 1, and that rewriting is the point. The rest-frame count above gives a magnetic energy g/(2m²c²) for every g, so subtracting it from the Dirac result leaves −1/(2m²c²), a term with no g in it. It cannot be magnetic, because it survives when the moment is switched off. It is kinematic, the precession of the accelerating rest frame, which topic 6.09 derives from non-commuting boosts. Neither piece was inserted: the reduction produced their sum, and the dependence on the moment sorts them. Hydrogen agrees: (g − 1) × 10 943.32 MHz = 1.002 319 × 10 943.32 MHz = 10 968.70 MHz, against the measured 10 969.0 MHz. Limits: at g = 1 magnetism and Thomas cancel and the coupling vanishes; at g = 0 a spinning body with no moment still precesses, with coefficient −1.
5A scalar potential reverses the sign
All of this assumed V = qΦ, the time component of a four-vector. A Lorentz-scalar potential W(r), written W so that S stays the spin, enters as a position-dependent mass: H = cα⋅p + β(mc² + W) + V. Eliminate the small component as in topic 6.01: χ = [2mc² + E′ − V + W]⁻¹ cσ⋅p φ, with E′ = E − mc². Expanded once, the spin-dependent part is −(1/(4m²c²)) σ⋅p U σ⋅p with U = W − V, and σ⋅p U σ⋅p = Up² − iħ∇U⋅p + ħσ⋅(∇U × p) gives HSO = (1/(2m²c²))(1/r) d(V − W)/dr S⋅L, derived. The renormalisation topic 6.05 needed adds only spin-independent terms, so the coefficient stands. The bookkeeping agrees: a scalar field has no rest-frame magnetic field, but it accelerates the particle, so Thomas's −1 acts on V + W while g acts on V alone. Nuclear mean fields, a scalar near −400 MeV and a vector near +350 MeV, quoted, add in V − W, which is why nuclear p₃/₂ lies about 6 MeV below p₁/₂ in ¹⁶O while hydrogen's 2P₃/₂ lies above 2P₁/₂.
6Power counting and where the operator stops
The operator is order (v/c)² relative to the Bohr energies: hydrogen's 2P doublet over its n = 2 binding is (mc²α⁴/32)/(mc²α²/8) = α²/4 = 1.3 × 10⁻⁵. The reduction kept that order and nothing else, so it holds while Zα is small. For hydrogen-like uranium Zα = 0.67, and the leading-order doublet Z⁴mc²α⁴/32 = 3.24 keV falls 29% short of the 4.57 keV that Unit 7's exact Dirac levels give, quoted. For s states L annihilates the wavefunction, so the operator gives zero even though ⟨1/r³⟩ diverges; the s-state energy the n- and j-only formula needs comes from the Darwin term, topic 6.08. And the anomaly is quoted, never derived here: Schwinger's α/(2π) = 0.001 161 and the measured 0.001 159 652 come from the quantised field, which this operator does not contain. At hydrogen's precision the 0.34 MHz left after (g − 1) × 10 943.32 MHz is the size of the terms this order drops, α² × 10 943 MHz = 0.58 MHz.
Key equations
Read the units, assumptions and limits before substituting.
Spin-orbit operator, static central potential
HSO = (1/(2m²c²)) (1/r)(dV/dr) S⋅L
One operator for any central field, bare nucleus or screened core: only the slope of V enters, and (1/r)(dV/dr) commutes with L, so the product is Hermitian.
V(r): potential energy in J; r in m; m in kg; S, L in J s. Derived from topic 6.06's spin term with ∇ × E = 0.Coulomb form, hydrogen-like ion
HSO = (Ze²/(8πε₀)) (1/(m²c²r³)) S⋅L
Exactly the operator Quantum Mechanics II reached by halving its rest-frame result; here the 2 in 1/(2m²c²) is the reduction's own.
V = −Ze²/(4πε₀r); e in C, ε₀ in F m⁻¹, Z = 1 for hydrogen. Derived, with no factor of ½ inserted anywhere.Hydrogen's 2P doublet
E(2P₃/₂) − E(2P₁/₂) = μc²α⁴/32 = 4.526 × 10⁻⁵ eV
10 943 MHz against the measured 10 969.0 MHz: the leading-order operator is right to 0.23%, and the anomalous moment is that gap.
μ = 0.999 456 m, the reduced mass (quoted); uses QM-II's ⟨1/r³⟩ = 1/(24a₀³) and ⟨S⋅L⟩ = +ħ²/2 or −ħ².Moment in the rest-frame field alone
−μ⋅B′ = (g/(2m²c²)) (1/r)(dV/dr) S⋅L, with B′ = −v × E/c²
At g = 2 this is twice the Dirac operator. Used unhalved it predicts a 21.9 GHz 2P doublet, double the measured one.
μ = g(q/2m)S in J T⁻¹; B′ in T, the field in the particle's instantaneous rest frame; v = p/m. Derived, as QM-II did.Coefficient with an anomalous moment
HSO = ((g − 1)/(2m²c²)) (1/r)(dV/dr) S⋅L, g = 2(1 + a)
The part proportional to g is magnetic; the g-independent −1 is kinematic, Thomas precession, which topic 6.09 derives from boosts.
a: anomalous moment, measured 0.001 159 652 for the electron (quoted). The factor 1 + 2a = g − 1 is derived in topic 6.04.Four-vector against Lorentz-scalar potential
HSO = (1/(2m²c²)) (1/r) d(V − W)/dr S⋅L
A scalar field makes no rest-frame magnetic field, only Thomas precession, so equal slopes give opposite spin-orbit: nuclear mean fields rely on it.
V: four-vector time component, as qΦ; W: Lorentz scalar, entering as β(mc² + W), a position-dependent mass. In J. Derived.Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Start at g = 2, f = 0, where the solid levels sit on the measured ones. Drop g to 1 and the doublet closes: magnetism and Thomas cancel. Then raise f and watch 2P₃/₂ cross below 2P₁/₂, the reversed sign a scalar potential gives.
Live results
- MAGNETIC PART g(1 − f)
- 2.00
- SPIN-ORBIT κ
- 1.000
- 2P₃/₂ − 2P₁/₂
- 10.94 GHz
- WITHOUT THOMAS
- 21.89 GHz
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Spin g-factor g: 2.00; Scalar share of binding f: 0.00
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The tempting summary of Quantum Mechanics II is that Thomas precession halves the spin-orbit energy, and students carry it forward as a rule: take the rest-frame magnetic energy and divide by 2, which for a general moment gives a coefficient g/2. That is wrong for every g except 2. Thomas precession is kinematic: the rotation of an accelerated rest frame depends on the acceleration, not on the moment, so it subtracts a fixed (1/(2m²c²))(1/r)(dV/dr) S⋅L whatever g is, and the reduction with Pauli's term returns g − 1. Halving and subtracting agree only at g = 2. Hydrogen tells them apart: with the measured g, g − 1 predicts a 2P doublet of 10 968.7 MHz and g/2 predicts 10 956.0 MHz, against 10 969.0 MHz measured. A body with no moment, g = 0, shows it starkly: halving says no coupling, while its frame still precesses.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
In Thomson's model of hydrogen the electron's potential energy inside a uniform sphere of positive charge e and radius R = a₀ = 52.9 pm is V(r) = −(e²/(8πε₀R))(3 − r²/R²), and Coulomb's outside it. Use the derived operator to find the largest spin-orbit splitting any l = 1 level can have, compare it with hydrogen's measured 2P doublet of 10.969 GHz, and decide whether any l = 1 level reaches it.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- The sphere's potential is electrostatic, the time component of a four-vector, so the four-vector form applies. dV/dr = e²r/(4πε₀R³), so (1/r)(dV/dr) = e²/(4πε₀R³), the same at every radius inside, and there HSO = (e²/(8πε₀m²c²R³)) S⋅L.
- Outside, (1/r)(dV/dr) = e²/(4πε₀r³): equal to the inside value at r = R and smaller beyond it. So e²/(4πε₀R³) is the largest the slope term gets anywhere, and only a level lying wholly inside could reach it. Dimensions: e²/(4πε₀) is J m, so the slope term is J m⁻², as for any (1/r)(dV/dr).
- For l = 1, ⟨S⋅L⟩ is +ħ²/2 at j = 3/2 and −ħ² at j = 1/2, a difference of 3ħ²/2. So the ceiling is ΔEₘₐₓ = (3/4)(e²/(4πε₀R))(ħ/(mcR))², and with R = a₀, ħ/(mca₀) = α, ΔEₘₐₓ = (3/4)α²Eₕ, where Eₕ = e²/(4πε₀a₀) = 27.211 eV is the Hartree energy (quoted).
- ΔEₘₐₓ = 0.75 × (7.2974 × 10⁻³)² × 27.211 eV = 1.087 × 10⁻³ eV, and ΔEₘₐₓ/h = 1.087 × 10⁻³ eV ÷ 4.1357 × 10⁻¹⁵ eV s = 263 GHz, which is 263 ÷ 10.969 = 24.0 times hydrogen's measured doublet: exactly 24 at leading order, because the Coulomb 2p electron samples only ⟨1/r³⟩ = 1/(24a₀³).
- No level reaches it. An l = 1 wave held inside r = a₀ costs at least 4.493² × Eₕ/2 = 10.1 Eₕ of kinetic energy, 4.493 being the first zero of the l = 1 spherical Bessel function, while the well is only 1.5 Eₕ deep. So every bound l = 1 level lies mostly outside the sphere, where the potential is Coulomb's.
- A numerical solution of the radial equation, quoted, puts the lowest l = 1 level at −0.1245 Eₕ with ⟨(1/r)(dV/dr)⟩ = 0.798 of hydrogen's 2p value, so its doublet is 0.798 × 10.95 GHz = 8.7 GHz: below hydrogen's, because inside a₀ the slope term stops growing as 1/r³.
Answer. At most (3/4)α²Eₕ = 1.09 × 10⁻³ eV, or 263 GHz, 24 times hydrogen's measured 2P doublet, and only for a level wholly inside the sphere. None fits there: the lowest l = 1 level lies mostly outside, with a doublet near 8.7 GHz.
MediumExample 2
Sodium's D lines come from 3p → 3s, and its 3p doublet, measured, is 17.196 cm⁻¹. Treat the valence electron as moving in a central V(r) produced by the nucleus and the ten core electrons, with g = 2. Find ⟨(1/r)(dV/dr)⟩ for the 3p state and compare it with a hydrogen 3p electron's value.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- With g = 2 the derived operator is (1/(2m²c²))(1/r)(dV/dr) S⋅L, and the two 3p levels differ in ⟨S⋅L⟩ by 3ħ²/2, so ΔE = (3ħ²/(4m²c²))⟨(1/r)(dV/dr)⟩, giving ⟨(1/r)(dV/dr)⟩ = (4/3)(mc²/ħc)²ΔE.
- ΔE = hc × 17.196 cm⁻¹ = 1.2398 × 10⁻⁴ eV cm × 17.196 cm⁻¹ = 2.132 × 10⁻³ eV.
- (mc²/ħc)² = (5.1100 × 10⁵ eV ÷ 1.9733 × 10⁻⁷ eV m)² = 6.706 × 10²⁴ m⁻², so ⟨(1/r)(dV/dr)⟩ = (4/3) × 6.706 × 10²⁴ m⁻² × 2.132 × 10⁻³ eV = 1.906 × 10²² eV m⁻².
- Hydrogen 3p: (e²/(4πε₀))⟨1/r³⟩ with ⟨1/r³⟩ = 1/(81a₀³), quoted, gives 1.4400 × 10⁻⁹ eV m ÷ (81 × 1.4818 × 10⁻³¹ m³) = 1.200 × 10²⁰ eV m⁻², a doublet of only 0.108 cm⁻¹.
- Ratio: 1.906 × 10²² ÷ 1.200 × 10²⁰ = 159. A hydrogenic ion scales as Z⁴, one Z from the slope and three from the shrunken orbit, so sodium behaves as if Zeff = 159 to the power ¼ = 3.6: neither the charge 1 seen outside the core nor the bare 11, which would give 92 times the measured doublet. The 1/r³ weight lets brief excursions into the core dominate.
Answer. ⟨(1/r)(dV/dr)⟩ = 1.91 × 10²² eV m⁻², 159 times hydrogen's 3p value: an effective hydrogenic charge of about 3.6, set by the 3p electron's penetration of the core.
HardExample 3
Suppose a fraction f of the proton's attraction were a Lorentz scalar: V = −(1 − f)k/r and W = −fk/r with k = e²/(4πε₀), so V + W = −k/r and the Bohr levels are unchanged. With g = 2.002 319 and the measured 2P₃/₂ − 2P₁/₂ interval of 10 969.045 MHz (quoted), find the spin-orbit coefficient as a function of f, then f.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- The Dirac part goes with d(V − W)/dr = (1 − 2f)k/r², and Pauli's 2a acts only on the electromagnetic part, 2a dV/dr = 2a(1 − f)k/r². So κ(f) = (1 − 2f) + 2a(1 − f) = g(1 − f) − 1: magnetism g on the vector share, Thomas −1 on the whole force.
- Since V + W = −k/r, the wavefunctions and ⟨1/r³⟩ = 1/(24a₀³) are hydrogen's, and every other term at this order is the same for both j. The interval is κ(f)Δ₀ with Δ₀ = μc²α⁴/(32h) = 10 943.32 MHz, μ = 0.999 456 m (quoted).
- Measured: κ = 10 969.045 ÷ 10 943.32 = 1.002 351. Then 1 − f = (1 + κ)/g = 2.002 351 ÷ 2.002 319 = 1.000 016, so f = −1.6 × 10⁻⁵.
- Error budget: Δ₀ omits terms of order α²Δ₀ = 0.58 MHz, worth δf ≈ α²Δ₀ ÷ (gΔ₀) = α²/g = 2.7 × 10⁻⁵, so at this order f is zero within about 3 × 10⁻⁵. Unit 7's exact Dirac interval, quoted, is larger by (5/8)α²Δ₀ = 0.364 MHz; with Δ = 10 943.684 MHz, κ = 1.002 317 4 and f = +8 × 10⁻⁷.
- Contrast: f = 1, all scalar, gives κ = −1, putting 2P₃/₂ 10.9 GHz below 2P₁/₂. The measured order, 2P₃/₂ above, rules that out before any arithmetic.
Answer. κ(f) = g(1 − f) − 1. The measured doublet gives f = −1.6 × 10⁻⁵ at leading order, zero within about 3 × 10⁻⁵, and +8 × 10⁻⁷ with exact Dirac levels: the Coulomb binding is a four-vector time component to a few parts in 10⁵.
Quick check
Select an answer and explain it before checking.
Review & continue
Spin-orbit coupling is the order-(v/c)² energy (1/(2m²c²))(1/r)(dV/dr) S⋅L that a Dirac particle acquires in a static central electrostatic potential, with the Thomas half already part of the coefficient rather than applied to it.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.