University Quantum Mechanics II · The Fine Structure of Hydrogen and the Lamb Shift · 8.3
Spin-orbit coupling from the electron's rest frame
Hydrogen's nucleus makes a purely electric field, yet the 2p level splits by 11 GHz as though the electron sat in a magnet. Build that magnet from the proton as the electron sees it, find why the turning of the electron's frame takes half of it back, and keep track of which steps are derived and which are quoted.
Build the model
Connect the measurement to the mechanism.
In the lab the proton's field is purely electric, so a spin moment at rest in it would feel nothing. The electron is not at rest. Transform the Coulomb field to its instantaneous rest frame and, to first order in v/c, a magnetic field appears: B′ = −v × E/c², which works out to (1/(4πε₀))(e/(mc²r³))L, the vector Biot–Savart gives for a proton circling the electron. It lies along the orbital angular momentum, so the energy of Unit 3's moment −(ge/2m)S with g = 2 is proportional to S⋅L/r³, and spin parallel to L costs energy.
All of that is derived here, and it comes out twice what hydrogen shows. The electron's frame accelerates, and boosts along a curving path compose into a rotation, the Thomas precession, quoted here from special-relativistic kinematics: for the Coulomb orbit it subtracts (e/2m)B′ from the spin's precession, turning g into g − 1 and, because g = 2, halving the coupling to H′ₛₒ = (e²/(8πε₀))(1/(m²c²r³))S⋅L. The Dirac equation, also quoted, delivers the same operator with the half already inside. Its size is of order α⁴mc², α² below the Bohr energies: one α is v/c, the other the reduced Compton wavelength over the orbit size.
The picture stops where its approximations do. It is first order in v/c, which is Zα/n, so it fails for heavy hydrogen-like ions, and near r = 0, where 1/r³ diverges, s states need the Dirac equation's Darwin term instead.
- Simple definition
- Spin-orbit coupling is the energy of the electron's spin magnetic moment in the magnetic field the moving nucleus produces in the electron's rest frame, reduced by Thomas precession to H′ₛₒ = (e²/(8πε₀))(1/(m²c²r³)) S⋅L.
- Example
- In hydrogen's 2p level it puts j = 3/2 above j = 1/2 by α⁴mc²/32 = 45.3 μeV, or 10.95 GHz. The measured doublet is 10.97 GHz; the rest-frame energy without the Thomas half would give 21.9 GHz.
From B′ = γ(B − v × E/c²) with B = 0 in the lab and γ ≈ 1. The lab holds no magnetic field at all; the electron's frame does.
B′ in T; v, the electron's velocity, in m s⁻¹; E, the proton's field at the electron, in V m⁻¹. First order in v/c.
Putting E = er/(4πε₀r³) into the card above turns −v × r into r × v = L/m. On the n = 1 Bohr orbit B′ = 12.5 T.
L = r × p in J s, r in m, m in kg, e in C. It equals Biot–Savart's (μ₀/4π)eL/(mr³) because μ₀ε₀c² = 1.
Spin parallel to L points the moment against B′, the costly orientation, so j = l + ½ lies higher. Derived here, and twice the splitting hydrogen shows.
μ = −(ge/2m)S in J T⁻¹, S in J s, H′ in J. The coefficient is positive.
With a = −eE/m it equals −(e/2m)B′: a turning of the electron's frame that opposes the Larmor rate g(e/2m)B′ and does not depend on g.
ωT in rad s⁻¹, a in m s⁻², v in m s⁻¹. Lowest order in v/c; the exact prefactor is γ²/(γ + 1) in place of ½.
The rest-frame energy with g replaced by g − 1 = 1. The Dirac equation, quoted, yields this operator with the ½ built in. It splits 2p by α⁴mc²/32, or 10.95 GHz.
H′ in J; e²/(8πε₀) in J m; S⋅L in J² s²; m in kg; r in m. With S⋅L ≈ ħ² and r ≈ a it is α⁴mc²/2 = 0.72 meV.
Thomas subtracts one from g rather than halving it. The measured g = 2.00232 raises the 2P doublet by 0.23%, which hydrogen confirms once its 0.054% reduced-mass correction is included; g/2 would give only 0.12%.
V(r) is the electron's potential energy in J; g is dimensionless. V = −e²/(4πε₀r) with g = 2 returns the card above.
The lab sees only an electric field
Fix the proton at the origin. At the electron its field is E = (1/(4πε₀)) er/r³, purely electric, and a moment at rest there would feel no torque. The electron moves, though, at about αc = 2.19 × 10⁶ m s⁻¹ in the ground state, and the fields in its instantaneous rest frame follow from the Lorentz transformation, B′ = γ(B − v × E/c²). The lab B is zero and γ − 1 ≈ α²/2 = 2.7 × 10⁻⁵, so to first order in v/c the electron sees B′ = −v × E/c². Put in the Coulomb field: −v × r = r × v = L/m, so B′ = (1/(4πε₀))(e/(mc²r³))L. Two things follow at once. The field lies along the orbital angular momentum whatever the orbit's shape, because only the part of v perpendicular to r contributes; purely radial motion makes no field. And at fixed L it falls as 1/r³, so the inner part of the orbit dominates. Dimensions: e/(4πε₀) is V m, L/(mc²r³) is s m⁻³, and V s m⁻² is a tesla.
The same field from a circling proton
Stay in the electron's frame and let the proton do the moving. It sits at −r and moves with −v, so the Biot–Savart field of a moving point charge at the electron is (μ₀/4π)e(−v) × r/r³. Both signs flip, the result is (μ₀/4π)eL/(mr³), and because μ₀ = 1/(ε₀c²) this is the transformed field term for term. The circular orbit makes it concrete. On the n = 1 Bohr orbit the proton circles at a = 52.9 pm with speed αc, a current I = ev/(2πa) = 1.05 mA, and the field at the centre of a loop is μ₀I/(2a) = 12.5 T, about four times the field of a 3 T hospital scanner. The same number comes from vE/c² with E = 5.14 × 10¹¹ V m⁻¹. The two routes are one physical statement written twice, and both hold only to first order in v/c; for hydrogen the neglected terms are near α² ≈ 5 × 10⁻⁵ of the result.
The spin moment in the rest-frame field
Unit 3 gave the electron a moment μ = −(ge/2m)S with g = 2 to the precision needed here; the minus sign is the electron's negative charge. Its energy in B′ is −μ⋅B′ = (ge/2m) S⋅B′, and with the field of the first step H′ᵣₑₛₜ = (e²/(4πε₀))(1/(m²c²r³)) S⋅L. The coefficient is positive, so spin parallel to L, which points the moment against B′, is the high-energy orientation: the argument predicts j = l + ½ above j = l − ½, the ordering hydrogen shows. Dimensions: e²/(4πε₀) is J m, S⋅L is J² s², m²c²r³ is kg² m⁵ s⁻², and the quotient is a joule. Size: with S⋅L near ħ²/2 and r near a, H′ᵣₑₛₜ is about α⁴mc²/2 = 0.72 meV, a factor α² below the 13.6 eV Rydberg. One α is v/c, from the motional field; the other is ħ/(mcr), the reduced Compton wavelength over the orbit, because the spin's moment is quantum-sized. Everything so far is derived, and every energy it predicts is twice what hydrogen shows; the 12.5 T field itself is not in question.
The electron's frame turns: Thomas precession
The rest frame used so far is inertial for an instant only. The electron accelerates towards the proton, and boosts in directions that keep changing do not compose into a pure boost: each pair leaves a small rotation behind. Along a curved path the electron's frame therefore turns relative to the lab at the Thomas rate ωT = (1/(2c²)) a × v to lowest order, a result this lesson quotes from special-relativistic kinematics rather than derives. For the Coulomb orbit a = −eE/m, so ωT = (e/(2mc²)) v × E = −(e/2m)B′. In its own frame the spin precesses about B′ at the Larmor rate g(e/2m)B′; seen from the lab, the frame's turning is added, giving (g − 1)(e/2m)B′, and the energy carries the same factor. The correction is not small. On the n = 1 Bohr orbit ωT = 1.1 × 10¹² rad s⁻¹, only α²/2 of the orbital rate 4.1 × 10¹⁶ rad s⁻¹ but exactly half the Larmor rate it corrects. With g = 2: H′ₛₒ = (e²/(8πε₀))(1/(m²c²r³)) S⋅L.
Holding the operator to hydrogen's 2p doublet
A clean test needs a splitting that spin-orbit coupling alone produces, and hydrogen's 2P₃/₂–2P₁/₂ interval is one: the kinetic correction of the previous topic depends only on n and l, the same for both levels, and the Darwin term vanishes for l ≥ 1. With ⟨1/r³⟩ = 1/(24a³) for 2p from Kramers' relation, and S⋅L = +ħ²/2 for j = 3/2 and −ħ² for j = 1/2, the doublet is (3/2)ħ² times the coefficient times ⟨1/r³⟩. The rest-frame energy with g = 2 gives α⁴mc²/16 = 90.6 μeV, or 21.90 GHz. The Thomas-halved operator gives α⁴mc²/32 = 45.28 μeV, or 10.949 GHz. The measured interval, the 9911.2 MHz from 2S₁/₂ to 2P₃/₂ plus the 1057.8 MHz Lamb shift, is 10.969 GHz. Experiment settles the factor of two; the 0.18% left over is the electron's anomalous moment, measured and computed in QED, less a reduced-mass correction. Evaluating every (n, l, j) this way is the next topic's work.
What the Dirac equation adds, and where the picture fails
The Dirac equation, reduced to order v²/c² and quoted here from relativistic quantum mechanics, contains H′ₛₒ with the ½ already in place, together with the kinetic and Darwin terms, and never mentions a rest frame. That it agrees is the strongest evidence that the Thomas factor is right. The rest-frame argument also generalises: for any central potential energy V(r) it gives ((g − 1)/(2m²c²))(1/r)(dV/dr) S⋅L, which is how alkali doublets such as sodium's D lines, 589.0 and 589.6 nm, are treated with a screened V. It fails in two places. It is first order in v/c, and v/c ≈ Zα/n, so for hydrogen-like uranium, Zα = 0.67, the α⁴ doublet of 3.24 keV lies 29% below the 4.57 keV the Dirac equation gives. And at r → 0, 1/r³ diverges: for l = 0 the operator gives zero because L does, but the region where the potential changes within a Compton wavelength belongs to the Darwin term.
Change one variable at a time
Make the relationship visible.
Thomas off, g = 2: the rest-frame field alone splits 2p by 21.90 GHz, twice the dashed doublet. Switch Thomas on and it lands on 10.95 GHz. Then drag g to 1, where the coupling vanishes, and to 0, where it inverts: Thomas subtracts one from g; it does not halve it.
NET PRECESSION FACTOR2.0
PREDICTED 2P DOUBLET21.90 GHz
IN ENERGY90.6 μeV
PREDICTED ÷ MEASURED1.996
Live interpretationNET PRECESSION FACTOR: 2.0. PREDICTED 2P DOUBLET: 21.90 GHz. IN ENERGY: 90.6 μeV. PREDICTED ÷ MEASURED: 1.996
Catch the common trap
Explain before calculating.
In hydrogen's 2p state the electron, viewed from its own rest frame, has the proton circling it. Which statement about the rest-frame field B′ and the order of the 2p doublet is correct?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyTreat hydrogen's ground state as the n = 1 Bohr orbit: radius a = 52.9 pm, speed αc. Find the magnetic field the electron sees in its rest frame, check it with Biot–Savart for the circling proton, and compare the energy μB B′ of a Bohr magneton in that field with α⁴mc².
- The proton's field at the electron is E = e/(4πε₀a²) = 5.14 × 10¹¹ V m⁻¹, radially outward. The speed is v = αc = 2.19 × 10⁶ m s⁻¹, perpendicular to r.
- Rest-frame field: |B′| = vE/c² = (2.19 × 10⁶ × 5.14 × 10¹¹)/(8.99 × 10¹⁶) T = 12.5 T. Direction: with r along x and v along y, −v × E points along +z, and so does L = r × p. The field lies along L.
- Biot–Savart: in the electron's frame the proton circles at radius a with speed αc, a current I = ev/(2πa) = 1.054 mA. The field at a loop's centre is μ₀I/(2a) = (4π × 10⁻⁷ × 1.054 × 10⁻³)/(1.058 × 10⁻¹⁰) T = 12.5 T, the same field by a second route.
- Energy: μB B′ = 5.788 × 10⁻⁵ eV T⁻¹ × 12.52 T = 7.25 × 10⁻⁴ eV. And α⁴mc² = (7.297 × 10⁻³)⁴ × 5.110 × 10⁵ eV = 1.449 × 10⁻³ eV, so μB B′ = α⁴mc²/2, a factor α² = 5.3 × 10⁻⁵ below the 13.6 eV Rydberg.
AnswerB′ = 12.5 T along L by either route, and μB B′ = 0.72 meV = α⁴mc²/2. This fixes the scale only: the real 1s state has l = 0 and no spin-orbit energy at all.
MediumHydrogen's 2P₃/₂–2P₁/₂ interval is measured as 10.969 GHz. Using ⟨1/r³⟩ = 1/(24a³) for 2p, predict it first from the rest-frame energy with g = 2 and no Thomas term, then from the Thomas-halved H′ₛₒ, and decide which the measurement supports.
- Only spin-orbit coupling separates these two levels: the kinetic correction depends on n and l alone, and the Darwin term is zero for l = 1. So the whole interval tests the S⋅L operator.
- From Unit 5, S⋅L = (ħ²/2)[j(j + 1) − l(l + 1) − ¾], which is +ħ²/2 for j = 3/2 and −ħ² for j = 1/2. The difference is (3/2)ħ².
- Rest-frame energy: ΔE = (e²/(4πε₀m²c²)) × (3/2)ħ² × 1/(24a³). With e²/(4πε₀) = αħc and a = ħ/(αmc) this is α⁴mc² × (3/2)/24 = α⁴mc²/16 = 90.57 μeV, or 21.90 GHz.
- Thomas-halved: α⁴mc²/32 = 45.28 μeV, or 10.949 GHz. The measured 10.969 GHz is 45.36 μeV, so the rest-frame value is 99.6% high and the halved one 0.18% low.
- The last 0.18% is quoted physics: the measured g = 2.00232 makes the factor g − 1 = 1.00232 (+25.4 MHz), and the reduced mass lowers the result by 0.054% (−6.0 MHz), giving 10.9687 GHz, within 0.4 MHz of experiment.
AnswerRest frame alone: 21.90 GHz. With the Thomas half: 10.949 GHz. Measured: 10.969 GHz. Experiment demands the half; the residual 0.18% is the anomalous moment less the reduced-mass correction.
HardHydrogen-like oxygen, O⁷⁺ (spectroscopic O VIII), emits Lyman-α near 19.0 Å as a doublet. From hydrogen's measured 2p doublet, 10.969 GHz, and its Lyman-α photon energy, 10.199 eV, use the Z-dependence of the rest-frame field to predict the O⁷⁺ doublet in energy and in wavelength. Compare with the tabulated O VIII lines at 18.9671 and 18.9725 Å, and say where the argument fails.
- Scaling: the nuclear field at fixed r grows as Z; orbits shrink as a/Z, so ⟨1/r³⟩ grows as Z³; L stays a multiple of ħ. So B′ ∝ Z⁴, and the Thomas term, set by an acceleration also ∝ Z/r², scales the same way: H′ₛₒ ∝ Z⁴. Gross energies scale as Z².
- Doublet: ΔE = 8⁴ × 45.364 μeV = 4096 × 45.364 μeV = 0.1858 eV. Lyman-α: E ≈ 64 × 10.199 eV = 652.7 eV, so λ = hc/E = 12 398.4 eV Å / 652.7 eV = 18.995 Å.
- Separation: Δλ = λΔE/E = 18.995 Å × 0.1858/652.7 = 0.00541 Å. Because ΔE ∝ Z⁴ and E² ∝ Z⁴, Δλ = hcΔE/E² does not depend on Z at this order: hydrogen's own components, 1215.6682 and 1215.6736 Å, are also 0.0054 Å apart.
- Table: 18.9725 − 18.9671 = 0.0054 Å, agreeing to the figures the table supports. The pair itself sits 0.025 Å below the scaled 18.995 Å: 0.05% from oxygen's smaller reduced-mass correction and 0.08% from relativistic and QED shifts of the levels, which move both lines together.
- Where it fails: v/c ≈ Zα/n = 0.029 for O⁷⁺ 2p and (Zα)² = 0.0034, so first order in v/c is ample. The error grows as (Zα)²: for Fe²⁵⁺, Z = 26, the Dirac equation, quoted, puts the doublet 2.3% above its α⁴ value, and by uranium, Zα = 0.67, the rest-frame argument has failed outright.
AnswerΔE ≈ 0.186 eV and Δλ ≈ 0.0054 Å, the same separation as hydrogen's because ΔE grows as Z⁴ while E² does too; the tabulated O VIII lines agree. The argument degrades as (Zα)² and fails as Zα approaches 1.