University Quantum Mechanics II · The Fine Structure of Hydrogen and the Lamb Shift · 8.4

Evaluating spin-orbit coupling in the coupled basis

Topic 08.03 handed you a Hamiltonian; this is where it becomes numbers you can check. Pick the basis Hₛₒ leaves diagonal and every hydrogen level with l ≥ 1 splits, each member moving by one eigenvalue times one radial average: 2p by 10.95 GHz against 10.969 GHz measured, and the s levels not at all.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Spin-orbit coupling is a perturbation Hₛₒ = ξ(r)S⋅L on hydrogen's level n, which holds 2n² degenerate states, with ξ(r) = e²/(8πε₀m²c²r³) quoted from topic 08.03, Thomas factor included. First-order theory must diagonalise Hₛₒ inside that level, and the operators it commutes with do the job: L², S², J² and J_z, but not L_z or S_z. The good states are therefore the coupled |n, l, ½, j, mⱼ⟩ of topic 05.07, on which S⋅L = (J² − L² − S²)/2 is the number (ħ²/2)[j(j + 1) − l(l + 1) − ¾].

What remains is radial, ⟨1/r³⟩, quoted from topic 06.08, and the shift follows in two lines: j = l + ½ rises by α⁴mc²/[4n³(l + ½)(l + 1)], j = l − ½ falls by α⁴mc²/[4n³l(l + ½)], and the pair splits by α⁴mc²/[2n³l(l + 1)] about a weighted centre that does not move. One power of α comes from the Coulomb strength and three from 1/a³, which puts the splitting a factor of order α² below the Bohr energy, as a magnetic effect at v ≈ αc should be. For l = 0 the shift is exactly zero, because S⋅L annihilates an s state before any radial integral is taken.

The model stops where its inputs do: at l = 0 the closed form reads 0/0, at large Z the expansion in (Zα)² fails, a many-electron atom needs its own ξ(r), and the last 0.2% of hydrogen's 2p splitting is quantum electrodynamics.

Simple definition: The first-order spin-orbit shift of a hydrogen level is the expectation value of Hₛₒ in a coupled state |n, l, ½, j, mⱼ⟩: the eigenvalue of S⋅L, fixed by j and l, times the radial average of e²/(8πε₀m²c²r³).
Example: For 2p, S⋅L is +ħ²/2 in j = 3/2 and −ħ² in j = ½, and ⟨1/r³⟩ = 1/(24a³), so 2p₃/₂ rises 15.1 μeV and 2p₁/₂ falls 30.2 μeV: a 45.3 μeV, or 10.95 GHz, splitting against the measured 10.969 GHz.

Spin-orbit Hamiltonian, quotedHₛₒ = [e²/(8πε₀m²c²)] (1/r³) S⋅L

Quoted from topic 08.03, its Thomas factor from relativity. It commutes with L², S², J² and J_z but not L_z or S_z, and that is what fixes the basis.

J; e in C, ε₀ in F m⁻¹, m in kg, c in m s⁻¹, r in m, S⋅L in J² s². The Thomas factor ½ is already inside.

S⋅L in a coupled state, quotedS⋅L |l, ½, j, mⱼ⟩ = (ħ²/2)[j(j + 1) − l(l + 1) − ¾] |l, ½, j, mⱼ⟩

Quoted from topic 05.07 through S⋅L = (J² − L² − S²)/2. On good states this eigenvalue is the whole angular and spin content of the shift.

J² s²: +lħ²/2 for j = l + ½, −(l + 1)ħ²/2 for j = l − ½, and exactly 0 for l = 0, where only j = ½ exists.

Mean inverse cube radius, quoted⟨1/r³⟩ = 1/[l(l + ½)(l + 1)n³a³]

Quoted from topic 06.08, Feynman-Hellmann plus Kramers. It carries (mcα/ħ)³, three of the four powers of α; for l = 0, R(0) ≠ 0 and the integral diverges.

m⁻³; a = 52.92 pm, so 2p gives 1/(24a³) = 2.81 × 10²⁹ m⁻³. Finite only for l ≥ 1.

First-order spin-orbit shift, l ≥ 1Eₛₒ⁽¹⁾ = (Eₙ²/mc²) n[j(j + 1) − l(l + 1) − ¾] / [l(l + ½)(l + 1)]

Derived here as eigenvalue times ⟨1/r³⟩. It is the same for all 2j + 1 values of mⱼ, positive for j = l + ½ and negative for j = l − ½.

eV; Eₙ = −13.606 eV/n², mc² = 0.511 MeV, so Eₙ²/mc² = α⁴mc²/(4n⁴). 2p₃/₂: +15.1 μeV; 2p₁/₂: −30.2 μeV.

The j = l ± ½ splittingΔEₛₒ = α⁴mc²/[2n³l(l + 1)]

Derived. It is α²/[n l(l + 1)] of |Eₙ|, falls as n⁻³ and l⁻², and grows as Z⁴ in a hydrogen-like ion; the relativistic term shifts both members alike.

J; α⁴mc² = 1.449 meV, or h × 350.4 GHz. 2p: α⁴mc²/32 = 45.3 μeV = 10.95 GHz; measured 10.969 GHz.

The weighted centre stays put(2l + 2) Eₛₒ(j = l + ½) + 2l Eₛₒ(j = l − ½) = 0

Derived: S⋅L is traceless at fixed l. The upper member moves l/(2l + 1) of the gap and the lower (l + 1)/(2l + 1): a third and two thirds for p.

Energies in J, weighted by 2j + 1 states: 4 and 2 for p, 6 and 4 for d.

Let the perturbation choose the basis

Before any correction, hydrogen's level n holds 2n² degenerate states, so first-order theory must diagonalise Hₛₒ inside that subspace, not merely read off diagonal elements. Unit 6's theorem does the work: find operators that commute with H₀ and with Hₛₒ and whose eigenvalues tell the degenerate states apart. Because S⋅L = (J² − L² − S²)/2 and ξ(r) is rotationally invariant, Hₛₒ commutes with L², S², J² and J_z; it fails to commute with L_z and S_z, because S⋅L moves a unit of ħ between orbit and spin. Every state of the level carries a different triple (l, j, mⱼ), so the coupled states |n, l, ½, j, mⱼ⟩ are good states and each first-order shift is one diagonal element. The uncoupled basis fails the test: its diagonal elements mₗmₛħ² for 2p are ½ħ², 0 and −½ħ², not the eigenvalues ½ħ² and −ħ². The relativistic term of topic 08.02 commutes with L and with S separately, so it is diagonal here too, and the two shifts add state by state.

The angular factor and the radial factor separate

A coupled state is Rₙₗ(r) times an angle-and-spin part. S⋅L acts only on the second and 1/r³ only on the first, so ⟨Hₛₒ⟩ = [e²/(8πε₀m²c²)] ⟨1/r³⟩ (ħ²/2)[j(j + 1) − l(l + 1) − ¾]. Insert ⟨1/r³⟩ = 1/[l(l + ½)(l + 1)n³a³], quoted from topic 06.08, then write e²/(4πε₀) = αħc and a = ħ/(mcα). The constants collapse, (αħc/2)(ħ²/m²c²)(mcα/ħ)³ = α⁴mc²/2, and with the ½ from S⋅L the shift is α⁴mc²[j(j + 1) − l(l + 1) − ¾]/[4n³l(l + ½)(l + 1)]. Since Eₙ²/mc² = α⁴mc²/(4n⁴), that is the formula card's (Eₙ²/mc²)n[…]/[l(l + ½)(l + 1)]. Check dimensions before any number: J m × J² s² × m⁻³ ÷ (kg² m² s⁻²) = J³ s⁴ kg⁻² m⁻⁴, a joule because J² = kg² m⁴ s⁻⁴. Nothing depends on mⱼ: all 2j + 1 states of a member move together, as a rotationally invariant perturbation requires.

Two members, unequal steps, a fixed centre

For j = l + ½ the bracket j(j + 1) − l(l + 1) − ¾ equals l, and for j = l − ½ it equals −(l + 1). So the upper member rises by α⁴mc²/[4n³(l + ½)(l + 1)], the lower falls by α⁴mc²/[4n³l(l + ½)], and the gap between them is α⁴mc²/[2n³l(l + 1)]. The order is set by the sign of ξ(r): the Coulomb force is attractive, ξ is positive, so the positive eigenvalue, j = l + ½, lies higher. The steps are unequal, the lower member moving (l + 1)/l times as far, but it holds 2l states against 2l + 2, and the degeneracy-weighted sum vanishes, as it must for the traceless operator S⋅L. For 2p the four upper states rise 15.09 μeV (3.650 GHz) and the two lower fall 30.19 μeV (7.300 GHz): 4 × 3.650 = 2 × 7.300. Limiting case: as l grows both steps approach α⁴mc²/(4n³l²), and the pattern turns symmetric once the ½ in l ± ½ is negligible.

Count the powers of α before computing

Four powers of α, and each has a source. One comes from the coupling, e²/(4πε₀) = αħc, and three from the orbit, because ⟨1/r³⟩ ∝ 1/a³ = (mcα/ħ)³. So every spin-orbit shift is a pure number times α⁴mc² = 1.449 meV, and against the level it perturbs, ΔEₛₒ/|Eₙ| = α²/[n l(l + 1)], which is 1.33 × 10⁻⁵ for 2p. That α² is (v/c)² in disguise, for an electron moving at v ≈ αc/n: a magnetic interaction seen from a moving frame is that much weaker than the electric binding. It is also the perturbation's small parameter, so first order is right to relative order α². In a hydrogen-like ion of charge Z, e² becomes Ze² and a becomes a/Z, so the shifts scale as Z⁴α⁴mc²: He⁺ 2p splits by 16 × 10.95 GHz = 175 GHz. At Z = 80, (Zα)² = 0.34, the expansion fails, and the exact Dirac energy, quoted in topic 08.07, must replace it.

Why an s state feels no spin-orbit shift

For l = 0 every state has the angular function Y₀₀, which L annihilates, so S⋅L|n, 0, ½, ½, mⱼ⟩ = 0; the eigenvalue formula agrees, ½ × 3/2 − 0 − ¾ = 0. Hₛₒ acting on an s state gives zero identically, so the first-order shift is zero, and because Hₛₒ|s⟩ has no component along any other state, no higher order shifts it either. The radial factor is where the trouble hides. Rₙ₀(0) is not zero, so the integrand of ⟨1/r³⟩, |R|²r² × r⁻³, goes as 1/r near the nucleus and the integral diverges logarithmically. The order of operations settles zero times infinity: act with S⋅L first and the product vanishes at every cutoff. The closed form reads 0/0 at l = 0, and cancelling the l in the bracket against the l in the denominator produces a finite α⁴mc²/(2n³) that is not a spin-orbit shift at all. That it equals the Darwin term, quoted in topic 08.05, is why the slip so often goes unnoticed.

Against hydrogen's 2p, and where the model stops

Spin-orbit coupling alone predicts the gap between the two members of one (n, l), because the relativistic term of topic 08.02 depends on n and l but not j, and moves both members together. For 2p that gap is α⁴mc²/32 = 10.949 GHz, and the measured 2P₃/₂–2P₁/₂ interval is 10.969 GHz, 0.18% higher. The proton's recoil, quoted, lowers the prediction to 10.943 GHz; the remaining 25 MHz is quantum electrodynamics, chiefly the electron's anomalous moment: g = 2.00232, quoted, multiplies the Thomas-corrected coupling by g − 1, adding 25.4 MHz. In many-electron atoms the model fails outright. Sodium's measured 3p doublet is 17.2 cm⁻¹, while hydrogen's 3p gap, α⁴mc²/108, is 0.108 cm⁻¹, 159 times smaller: the valence electron penetrates the closed shells and feels nearly the full charge 11e just where 1/r³ is largest. And the coupled basis stops being the good one once μ_B B approaches the gap, 0.78 T for hydrogen 2p, the Paschen-Back regime of Unit 9.

Change one variable at a time

Make the relationship visible.

Interactive model

Orbital quantum number l1

Radial nodes nᵣ = n − l − 10

Start at 2p (l = 1, nᵣ = 0): the lower level sits twice as far below the dashed centre as the upper sits above it. Raise l and the steps even out while the gap sinks down the log axis; set l = 0 and the lone j = ½ level sits on the dashed line with no partner, since S⋅L annihilates an s state.

SHIFT OF j = l + ½3.650 GHz

SHIFT OF j = l − ½-7.300 GHz

SPLITTING ΔEₛₒ45.282 μeV

ΔEₛₒ / |Eₙ|13.31 ppm

Live interpretationSHIFT OF j = l + ½: 3.650 GHz. SHIFT OF j = l − ½: −7.300 GHz. SPLITTING ΔEₛₒ: 45.282 μeV. ΔEₛₒ / |Eₙ|: 13.31 ppm

Catch the common trap

Explain before calculating.

In hydrogen's n = 2 level, which set gives the first-order spin-orbit shifts of 2s₁/₂, 2p₁/₂ and 2p₃/₂, in that order and in units of α⁴mc²?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyFor hydrogen's 3d level (n = 3, l = 2), find the first-order spin-orbit shifts of 3d₅/₂ and 3d₃/₂ in μeV and their splitting in GHz, then check that the weighted centre does not move. Use α⁴mc² = 1.4490 meV and h = 4.1357 × 10⁻¹⁵ eV s.

The bracket j(j + 1) − l(l + 1) − ¾ is 35/4 − 6 − ¾ = 2 for j = 5/2 and 15/4 − 6 − ¾ = −3 for j = 3/2, so S⋅L is +ħ² and −1.5ħ².
Denominator: 4n³ × l(l + ½)(l + 1) = 108 × (2 × 2.5 × 3) = 108 × 15 = 1620, so Eₛₒ = α⁴mc² × bracket/1620.
3d₅/₂: 2α⁴mc²/1620 = α⁴mc²/810 = 1.789 μeV up. 3d₃/₂: −3α⁴mc²/1620 = −α⁴mc²/540 = −2.683 μeV.
Splitting: 1.789 + 2.683 = 4.472 μeV, matching α⁴mc²/[2n³l(l + 1)] = α⁴mc²/324. As a frequency, 4.472 × 10⁻⁶ eV ÷ 4.1357 × 10⁻¹⁵ eV s = 1.081 × 10⁹ Hz.
Centre: six 3d₅/₂ states and four 3d₃/₂ states give 6 × 1.789 − 4 × 2.683 = 10.73 − 10.73 = 0 μeV, so the weighted centre is unmoved.

Answer3d₅/₂ rises 1.79 μeV and 3d₃/₂ falls 2.68 μeV; the splitting is 4.47 μeV, or 1.08 GHz, and the weighted centre stays at the unperturbed 3d energy.

MediumWithout the closed form, evaluate the 2p₃/₂ spin-orbit shift in SI units from Hₛₒ, the S⋅L eigenvalue and ⟨1/r³⟩, check the units, then find the 2p₁/₂ shift and the splitting in GHz. Use e²/(4πε₀) = 2.30708 × 10⁻²⁸ J m, m = 9.1094 × 10⁻³¹ kg, c = 2.9979 × 10⁸ m s⁻¹, ħ = 1.05457 × 10⁻³⁴ J s, a = 5.2918 × 10⁻¹¹ m, e = 1.6022 × 10⁻¹⁹ C and h = 4.1357 × 10⁻¹⁵ eV s.

Prefactor: e²/(8πε₀m²c²) = 1.1535 × 10⁻²⁸ J m ÷ (8.2981 × 10⁻⁶¹ kg² × 8.9874 × 10¹⁶ m² s⁻²) = 1.1535 × 10⁻²⁸ ÷ 7.4578 × 10⁻⁴⁴ = 1.5467 × 10¹⁵ J s² kg⁻² m⁻¹.
Angular and radial factors for 2p₃/₂: S⋅L = +ħ²/2 = 5.5606 × 10⁻⁶⁹ J² s², and ⟨1/r³⟩ = 1/(24a³) = 1/(24 × 1.4819 × 10⁻³¹ m³) = 2.8118 × 10²⁹ m⁻³.
Product: 1.5467 × 10¹⁵ × 5.5606 × 10⁻⁶⁹ × 2.8118 × 10²⁹ = 2.418 × 10⁻²⁴ J, which is 15.09 μeV. Units: J s² kg⁻² m⁻¹ × J² s² × m⁻³ = J³ s⁴ kg⁻² m⁻⁴ = J, since J² = kg² m⁴ s⁻⁴.
2p₁/₂ has S⋅L = −ħ², twice as large and opposite, with the same ⟨1/r³⟩, so its shift is −30.19 μeV. The splitting is 45.28 μeV, and 45.28 × 10⁻⁶ eV ÷ 4.1357 × 10⁻¹⁵ eV s = 10.95 GHz.
Order check: 45.28 μeV ÷ 1.449 meV = 1/32 of α⁴mc², as α⁴mc²/[2n³l(l + 1)] requires. The measured 10.969 GHz is 0.18% higher; recoil and the anomalous moment g − 1 = 1.00232, both quoted, close the gap.

Answer2p₃/₂ rises 2.418 × 10⁻²⁴ J = 15.09 μeV and 2p₁/₂ falls 30.19 μeV; the splitting is 45.28 μeV, or 10.95 GHz, 0.18% below the measured 10.969 GHz.

HardA student argues that because ⟨1/r³⟩ diverges for s states, 2s₁/₂ must carry a huge spin-orbit shift. Cut the radial integral off at the classical electron radius r₀ = 2.818 fm, where the Pauli expansion fails anyway, and evaluate ⟨1/r³⟩ for 2s using R₂₀ = (1 − r/2a)e^(−r/2a)/√(2a³). Compare it with 2p's 1/(24a³), then find the true first-order shift and what the cancelled closed form would give.

Near r = 0, |R₂₀|² → 1/(2a³), so the integrand |R₂₀|²r² × r⁻³ = |R₂₀|²/r behaves as 1/(2a³r): a logarithmic divergence that only the cutoff tames.
With u = r/a and (1 − u/2)² = 1 − u + u²/4: ⟨1/r³⟩ = [1/(2a³)]∫ from u₀ to ∞ of (1 − u + u²/4)e^(−u) du/u = [1/(2a³)][E₁(u₀) − e^(−u₀) + ¼(1 + u₀)e^(−u₀)], where E₁ is the exponential integral.
Here u₀ = r₀/a = 2.818 fm ÷ 52.92 pm = 5.325 × 10⁻⁵, which is α². For small u₀, E₁(u₀) ≈ ln(1/u₀) − 0.5772 = 9.840 − 0.577 = 9.263. So ⟨1/r³⟩ ≈ (9.263 − 1 + 0.25)/(2a³) = 4.257/a³, 102 times 2p's 0.0417/a³, and still growing as ln(a/r₀).
The angular factor: S⋅L|2, 0, ½, ½, mⱼ⟩ = (ħ²/2)(¾ − 0 − ¾) = 0 exactly, because L annihilates Y₀₀. Zero times 4.257/a³, or times the value at any smaller cutoff, is zero: the first-order spin-orbit shift of 2s₁/₂ vanishes.
Cancelling l first, α⁴mc²/[4n³(l + ½)(l + 1)] at l = 0 and n = 2 gives α⁴mc²/16 = 90.57 μeV, or 21.90 GHz, twice the 2p splitting. It is manufactured by treating zero times infinity as a finite number, and it happens to equal the Darwin shift of topic 08.05.

AnswerWith a 2.818 fm cutoff, ⟨1/r³⟩ for 2s is about 4.26/a³, 102 times the 2p value and divergent as r₀ → 0, yet the spin-orbit shift is exactly zero because S⋅L|2s⟩ = 0; the 90.6 μeV from cancelling l is spurious.