University Physics IV · Angular Momentum and Spin · 9.9

Spin-Orbit Coupling & the Landé g Factor

Every alkali line you resolve is secretly two, and almost no Zeeman pattern has three lines. Both trace to the electron's spin moment sitting in the magnetic field of its own orbital motion. This lesson prices that internal coupling, then builds g_J — the one number fixing each level's response to a magnet.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Transform to the electron's rest frame and the nucleus becomes a circulating charge, generating a magnetic field of order a tesla in hydrogen and tens of teslas in heavier atoms — a field the atom supplies itself. The electron's spin moment sits in that field, so an interaction ξ(r)L⋅S appears with no apparatus involved; its coefficient carries a factor of one half that a naive rest-frame derivation misses, because the electron's frame is accelerating and a sequence of boosts amounts to a rotation — Thomas precession.

The coupling costs the atom its old labels: L_z and S_z no longer commute with H, so mₗ and mₛ die as quantum numbers and the coupled pair j, mⱼ replaces them, with L⋅S = ½(J² − L² − S²) turning every shift into arithmetic. What it buys is fine structure — each l > 0 level split into j = l ± ½ by an interval scaling as Z⁴/n³ — and, hanging on it, the Landé factor g_J: the projection onto J of a magnetic moment that is not parallel to J, because spin counts double. g_J prices every weak-field Zeeman pattern, and it is only the weak-field answer: once μ_B B beats the fine-structure interval, the field decouples L from S and mₗ, mₛ come back to life.

Simple definition: Spin-orbit coupling is the interaction between an electron's spin magnetic moment and the internal magnetic field of its own orbital motion; it splits each l > 0 level into fine-structure components with j = l + ½ and j = l − ½.
Example: Sodium's yellow line is really two: 589.0 nm and 589.6 nm, one 3p level split by 2.1 × 10⁻³ eV because the 3p electron's spin moment sits in an internal field of roughly 18 T.

Spin-orbit Hamiltonian (Thomas ½ included)H_SO = (1/(2mₑ²c²)) (1/r)(dV/dr) L⋅S

The ½ is Thomas precession: the electron's rest frame rotates, and the naive derivation gives exactly twice the observed coupling.

V(r) = potential energy in J; mₑ in kg; L⋅S carries ħ², so H_SO lands in joules

L⋅S in the coupled basis⟨L⋅S⟩ = (ħ²/2)[j(j+1) − l(l+1) − s(s+1)]

Turns the operator product into arithmetic — the shift of every fine-structure level without a single integral.

from squaring J = L + S; j, l, s are pure numbers and ħ² carries the units. p electron: +ħ²/2 (j = 3/2), −ħ² (j = 1/2)

Hydrogenic doublet splittingΔE_SO = 13.6 eV · Z⁴α² / [n³ l(l+1)]

The Z⁴/n³ law: doublets widen fast down the periodic table, while a level's full fine-structure shift scales as Z⁴/n⁴ and depends only on j.

α ≈ 1/137.04; gap between j = l + ½ and j = l − ½; hydrogen 2p: 4.5 × 10⁻⁵ eV

Landé g factorg_J = 1 + [j(j+1) + s(s+1) − l(l+1)] / [2j(j+1)]

The projection of the tilted moment μ = −μ_B(L + 2S)/ħ onto J — the only part that survives the fast precession of L and S about J.

dimensionless; assumes gₗ = 1, gₛ = 2. ²S₁/₂: 2 · ²P₁/₂: 2/3 · ²P₃/₂: 4/3

Weak-field (anomalous) Zeeman shiftΔE = g_J μ_B B mⱼ

Each level fans into 2j+1 sublevels spaced g_J μ_B B — a different spacing per level, which is what makes the patterns anomalous.

μ_B = 5.79 × 10⁻⁵ eV/T; B in tesla; valid while μ_B B ≪ the fine-structure interval

Strong-field (Paschen-Back) shiftΔE = (mₗ + 2mₛ) μ_B B + ξħ² mₗ mₛ

The field decouples L from S; the remnant ξħ²mₗ mₛ is the small print left of fine structure.

good numbers mₗ, mₛ once μ_B B exceeds the fine-structure interval; ξħ² is the spin-orbit scale

Ride with the electron and a magnet appears

In the lab frame the electron orbits a stationary nucleus. Transform to the electron's instantaneous rest frame and the charged nucleus circulates instead — a current loop threading the electron's position with a magnetic field Bᵢₙₜ ≈ (v/c²) × E, built from the same Coulomb field E that binds the atom. Its size is not small: hydrogen's 2p fine-structure splitting of 4.5 × 10⁻⁵ eV implies Bᵢₙₜ ≈ 0.4 T, and sodium's 3p doublet implies about 18 T — fields you would need a serious laboratory magnet to match, supplied free by the atom's own motion. The electron's spin moment μₛ = −gₛ μ_B S/ħ then has an orientation energy −μₛ⋅Bᵢₙₜ. Since Bᵢₙₜ points along the orbital angular momentum L, that energy is proportional to L⋅S: the atom couples its own spin to its own orbit, with no external apparatus anywhere in the story.

Thomas precession halves the naive answer

The rest-frame derivation hides an error: the electron's frame is not inertial. It accelerates continuously toward the nucleus, and in special relativity the composition of two non-parallel boosts is not a boost but a boost plus a rotation. The electron's frame therefore rotates — Thomas precession — at a rate exactly half the naive spin-precession rate and with the opposite sense, and the net effect is to halve the coupling: H_SO = (1/(2mₑ²c²))(1/r)(dV/dr) L⋅S, the ½ being Thomas's. History pivoted on this factor. In 1926 the doublet intervals came out twice too large from the naive calculation; gₛ ≈ 2 and Thomas's ½ conspire so the observed splitting looks as if the spin g were 1, and the spin hypothesis of Uhlenbeck and Goudsmit was rescued within months. The Dirac equation later delivers the ½ automatically; at this level you insert it by hand and know why it is there.

Evaluate L⋅S where it is diagonal

H_SO does not commute with L_z or S_z separately — the coupling makes L and S precess about their sum — so mₗ and mₛ stop being good quantum numbers. It does commute with J² = (L+S)², L², S² and J_z, so the coupled basis |l, s; j, mⱼ⟩ diagonalises it. Square J = L + S to get L⋅S = ½(J² − L² − S²), and every matrix element becomes arithmetic: ⟨L⋅S⟩ = (ħ²/2)[j(j+1) − l(l+1) − s(s+1)]. For a p electron (l = 1, s = ½): j = 3/2 gives +ħ²/2 and j = 1/2 gives −ħ². Two checks come free. Degeneracy-weighted, 4 × (+½) + 2 × (−1) = 0: the coupling splits the level about an unmoved centre of gravity, so an average over the multiplet feels nothing. And the gap between adjacent j levels is proportional to the larger j — the Landé interval rule, your first diagnostic for whether a measured multiplet really is spin-orbit split.

Scale it: Z⁴/n³, and the term-symbol label

The coefficient ξ(r) ∝ (1/r)dV/dr averages to something proportional to Z⟨1/r³⟩, and for hydrogenic states that scales as Z⁴/n³ — giving doublet gaps ΔE_SO = 13.6 eV⋅Z⁴α²/[n³l(l+1)]. Hydrogen's 2p: 13.6 × (1/137.04)²/(8 × 2) = 4.5 × 10⁻⁵ eV, so Lyman-α is a 10.9 GHz doublet. Sodium's 3p doublet is 46 times wider at 2.1 × 10⁻³ eV because the valence electron penetrates the core and sees an effective charge well above 1. Keep the two scalings distinct: the j-splitting within one n grows as Z⁴/n³, while a level's full fine-structure shift — spin-orbit plus the relativistic kinetic correction — goes as Z⁴/n⁴ and, in hydrogen, depends on n and j alone. The surviving labels are collected in the term symbol n²ˢ⁺¹L_J: sodium's doublet is 3²P₃/₂ and 3²P₁/₂ decaying to 3²S₁/₂, and every fine-structure diagram you will meet is drawn in this notation.

Project the moment: where g_J comes from

A level's magnetic moment is μ = −μ_B(L + 2S)/ħ — spin counts double because gₛ ≈ 2 — and that 2 means μ is not antiparallel to J = L + S. In a weak field, L and S precess rapidly about J at the fine-structure frequency while J precesses slowly about B. The component of μ perpendicular to J therefore averages to zero, and only the projection along J couples to the field. Working out that projection gives g_J = 1 + [j(j+1) + s(s+1) − l(l+1)]/[2j(j+1)]. Test it on limits you already trust: l = 0 leaves pure spin and g_J = 2; s = 0 leaves pure orbit and g_J = 1. For sodium: g(²S₁/₂) = 2, g(²P₁/₂) = 2/3, g(²P₃/₂) = 4/3 — three levels, three different magnetic prices per unit mⱼ. That is precisely why alkali Zeeman patterns are anomalous: the normal three-line pattern needs every level to shift at the same rate, which happens only when s = 0 forces g_J = 1 everywhere.

One ratio decides the regime

Before computing any Zeeman pattern, form the ratio μ_B B / ΔE_FS. Weak field (ratio ≪ 1): j and mⱼ are good, each level fans into 2j+1 sublevels spaced g_J μ_B B, and Δmⱼ = 0, ±1 gives the anomalous patterns — sodium D1 splits into four components, D2 into six. Strong field (ratio ≫ 1): L and S decouple and precess independently about B, mₗ and mₛ are good again, and shifts become (mₗ + 2mₛ)μ_B B with only the small ξħ²mₗ mₛ remnant — the Paschen-Back regime, whose Δmₗ = 0, ±1 with Δmₛ = 0 lines collapse back to a normal-looking triplet. The boundary is an experimental fact, not a convention: hydrogen 2p, split by 4.5 × 10⁻⁵ eV, crosses over near B = ΔE/μ_B ≈ 0.8 T, while sodium's 2.1 × 10⁻³ eV holds to about 36 T. The same magnet is strong for one atom and weak for another; in between, neither basis diagonalises H and the levels bend — the regime the interactive figure below lets you walk through.

Change one variable at a time

Make the relationship visible.

Interactive model

Spin-orbit constant a1.40 meV

Applied field B8.0 T

Set a = 1.40 meV (sodium 3p) and sweep B to 30 T — the ratio barely passes 1 and the levels stay near their Landé tangents. Then drop a to 0.03 meV (hydrogen 2p): 1 T is already past crossover, and all six levels straighten onto Paschen-Back slopes of 0, ±1, ±2 times μ_B B.

FS INTERVAL (3/2)a2.10 meV

ZEEMAN SCALE μ_B B0.463 meV

RATIO μ_B B / a0.33

LANDÉ ERROR, mⱼ = +1/20.021 meV

Live interpretationFS INTERVAL (3/2)a: 2.10 meV. ZEEMAN SCALE μ_B B: 0.463 meV. RATIO μ_B B / a: 0.33. LANDÉ ERROR, mⱼ = +1/2: 0.021 meV

Catch the common trap

Explain before calculating.

An alkali excited level carries the term symbol ²P₃/₂, so l = 1, s = ½ and j = 3/2. What is its Landé factor g_J, the number that sets its weak-field Zeeman spacing g_J μ_B B?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA single p electron has l = 1 and s = ½. Find the allowed j values, evaluate ⟨L⋅S⟩ in each, and show that spin-orbit coupling leaves the degeneracy-weighted centre of gravity of the level unmoved.

Allowed j run from |l − s| to l + s in integer steps: j = 1/2 and j = 3/2, holding 2j+1 = 2 and 4 states — six in all, matching the (2l+1)(2s+1) = 6 uncoupled products.
Square J = L + S: J² = L² + S² + 2L⋅S, so ⟨L⋅S⟩ = (ħ²/2)[j(j+1) − l(l+1) − s(s+1)], with l(l+1) = 2 and s(s+1) = 3/4.
j = 3/2: ⟨L⋅S⟩ = (ħ²/2)(15/4 − 2 − 3/4) = +ħ²/2. j = 1/2: (ħ²/2)(3/4 − 2 − 3/4) = −ħ². With H_SO = ξL⋅S and ξ > 0, j = 3/2 rises and j = 1/2 falls, split by (3/2)ξħ².
Weight each shift by its degeneracy: 4(+ħ²/2) + 2(−ħ²) = 2ħ² − 2ħ² = 0. The coupling splits the level symmetrically about its old position — a check worth running on any multiplet.

Answerj = 3/2 with ⟨L⋅S⟩ = +ħ²/2 (4 states) and j = 1/2 with ⟨L⋅S⟩ = −ħ² (2 states); 4(½) + 2(−1) = 0, so the centre of gravity is unshifted.

MediumSodium's D1 line is 3²P₁/₂ → 3²S₁/₂. In a weak field B = 0.60 T, find g_J for both levels, the energy spacing of each level's Zeeman sublevels, and the number of components D1 splits into. (μ_B = 5.79 × 10⁻⁵ eV/T; the 3p fine-structure interval is 2.1 × 10⁻³ eV.)

Check the regime first: μ_B B = 5.79 × 10⁻⁵ × 0.60 = 3.47 × 10⁻⁵ eV, about 1/60 of the 2.1 × 10⁻³ eV fine-structure interval — weak field, so j, mⱼ are good and shifts are g_J μ_B B mⱼ.
²S₁/₂ (l = 0, s = ½, j = ½): g_J = 1 + (3/4 + 3/4 − 0)/(2 × 3/4) = 2. Its mⱼ = ±½ sublevels sit 2μ_B B = 6.9 × 10⁻⁵ eV apart.
²P₁/₂ (l = 1, s = ½, j = ½): g_J = 1 + (3/4 + 3/4 − 2)/(2 × 3/4) = 1 − 1/3 = 2/3. Its sublevels sit (2/3)μ_B B = 2.3 × 10⁻⁵ eV apart — same j, one third the response.
Δmⱼ = 0, ±1 allows all four transitions between the two doublets. Line offsets are (gᵤ mᵤ − gₗ mₗ)μ_B B = ±(2/3)μ_B B and ±(4/3)μ_B B: four distinct components, at ±2.3 × 10⁻⁵ and ±4.6 × 10⁻⁵ eV from the zero-field line.

Answerg(²S₁/₂) = 2, g(²P₁/₂) = 2/3; sublevel spacings 6.9 × 10⁻⁵ eV and 2.3 × 10⁻⁵ eV; D1 splits into four components — anomalous, because the two levels shift at different rates.

HardHydrogen's 2p level is split by ΔE = 4.5 × 10⁻⁵ eV between 2²P₃/₂ and 2²P₁/₂. (a) Estimate the internal magnetic field the 2p electron experiences. (b) Find the applied field at which μ_B B equals the fine-structure interval. (c) Describe the level pattern at B = 3.0 T.

(a) Model the splitting as a spin flip in the internal field: ΔE ≈ gₛ μ_B Bᵢₙₜ ≈ 2μ_B Bᵢₙₜ, so Bᵢₙₜ = 4.5 × 10⁻⁵/(2 × 5.79 × 10⁻⁵) ≈ 0.39 T — a field you would need a solid laboratory magnet to match, supplied by the atom itself.
(b) The regimes cross when the two energy scales meet: B = ΔE/μ_B = 4.5 × 10⁻⁵/5.79 × 10⁻⁵ = 0.78 T. Below it use j, mⱼ and g_J; well above it use mₗ and mₛ.
(c) B = 3.0 T is nearly four times the crossover field, so 2p is in the Paschen-Back regime: shifts are (mₗ + 2mₛ)μ_B B with μ_B B = 5.79 × 10⁻⁵ × 3.0 = 1.74 × 10⁻⁴ eV.
mₗ + 2mₛ over mₗ = −1, 0, +1 and mₛ = ±½ takes the values −2, −1, 0, 0, +1, +2: five equally spaced level positions 1.74 × 10⁻⁴ eV apart, with the middle one doubly degenerate.
The residual spin-orbit term ξħ²mₗ mₛ, of order 10⁻⁵ eV here, nudges the (mₗ = ±1, mₛ = ∓½) pair off exact coincidence — Paschen-Back is a limit, not an exact symmetry.

AnswerBᵢₙₜ ≈ 0.39 T; crossover at ≈ 0.78 T; at 3.0 T the pattern is five nearly equally spaced levels 1.74 × 10⁻⁴ eV apart, the central one doubly degenerate up to the small spin-orbit remnant.