University Physics IV · Atomic Physics · 11.7

Spin-Orbit Coupling & the Zeeman Effect

Every alkali line you have ever resolved is a doublet, and every one of them splits again in a field. This topic hands you the two energies responsible — ζ L⋅S from the atom's own internals, and g_J μ_B B m_J from the magnet — and, more usefully, teaches you to decide which of them is in charge before choosing a formula.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Sit in the electron's rest frame and the nucleus becomes a current loop: a charge +Ze circling at radius r makes a magnetic field along L, about 0.7 T at hydrogen's 2p electron, and the electron's spin moment has an energy in it. That is the origin of the term — an internal magnetic interaction, needing no magnet — and because both factors are angular momenta the energy is written ζ L⋅S, whose value is fixed by the total J = L + S. Doing the frame change honestly costs a factor of two: the electron's frame rotates as well as moves, and Thomas precession halves the naive result.

Two corrections of the same order in α arrive alongside it, the relativistic kinetic term and the Darwin contact term that acts only where ψ(0) ≠ 0, so fine structure is a sum of three effects whose total depends on n and j alone. All of it scales as α², one part in 19 000 of the gross level spacing, but as Z⁴/n³ across the periodic table, which turns a 45 µeV curiosity in hydrogen into sodium's 2.1 meV doublet and caesium's 554 cm⁻¹ chasm. Apply a field and the same bookkeeping runs again: while μ_B B stays below the fine-structure gap, J survives and levels move by g_J μ_B B m_J; push past it and the field pulls L and S apart, mₗ and mₛ take over, and the anomalous pattern collapses back towards a normal triplet.

Simple definition: Spin-orbit coupling is the magnetic energy of an electron's spin moment sitting in the field its own orbital motion produces, written ζ L⋅S, and it splits every l > 0 level into two components labelled by the total angular momentum j = l ± ½.
Example: In sodium's 3p level ζ ≈ 1.43 meV, so the j = 3/2 and j = 1/2 components sit (l + ½)ζ = 2.14 meV apart — the 0.6 nm gap between the D₂ line at 589.0 nm and the D₁ line at 589.6 nm.

Spin-orbit HamiltonianH_SO = (1/2m²c²)(1/r)(dV/dr) L⋅S

Names the interaction as internal and magnetic — the field comes from the electron's own orbit, so no applied field is needed.

m is the electron mass and V(r) the central potential in J; the leading ½ is Thomas precession.

L⋅S from the total angular momentum⟨L⋅S⟩ = (ħ²/2)[j(j+1) − l(l+1) − s(s+1)]

Squaring J = L + S turns an operator product you cannot evaluate into three quantum numbers you already know.

Bracket is a pure number: for l = 1, s = ½ it is +1 at j = 3/2 and −2 at j = 1/2.

Hydrogenic splitting and its Z⁴ scalingΔE_fs = Z⁴α²Eₕ / [2n³ l(l+1)]

Hydrogen 2p: 4.53 × 10⁻⁵ eV, or 0.365 cm⁻¹. For a screened valence electron, replace Z by an effective charge.

Eₕ = 27.211 eV, α² = 5.325 × 10⁻⁵; ΔE_fs is the j = l+½ to j = l−½ gap, in eV.

Fine-structure energy of hydrogenEₙⱼ = −(13.6 eV/n²)[1 + (α²/n²)(n/(j+½) − ¾)]

Sums the spin-orbit, relativistic and Darwin terms; the 1058 MHz Lamb shift that separates 2s₁⁄₂ from 2p₁⁄₂ is QED beyond it.

Depends on n and j only, so 2s₁⁄₂ and 2p₁⁄₂ come out degenerate.

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

Projects μ = −(μ_B/ħ)(L + 2S) onto J, which is why the shifts are not plain multiples of μ_B B.

Pure number: 2 for ²S₁⁄₂, 2/3 for ²P₁⁄₂, 4/3 for ²P₃⁄₂, and exactly 1 when s = 0.

Weak-field Zeeman shift, and where it failsΔE = g_J μ_B B m_J, valid while μ_B B ≪ ΔE_fs

Sodium's 2.14 meV doublet keeps this honest to about 10 T; hydrogen's 2p doublet is already breaking up near 0.8 T.

μ_B = 5.788 × 10⁻⁵ eV T⁻¹, B in tesla, m_J running −j to +j in unit steps.

The field in the L⋅S term is the atom's own

Change to the frame in which the electron is instantaneously at rest and the nucleus, charge +Ze, circles it. A circulating charge is a current loop, so the electron sits in a field Bᵢₙₜ = (Ze/4πε₀mc²r³)L pointing along the orbital angular momentum, and its spin moment μₛ = −(e/m)S has energy −μₛ⋅Bᵢₙₜ there. Both factors carry an angular momentum, so the energy goes as L⋅S. Two things separate this from an applied field. The size: hydrogen 2p numbers, with ⟨1/r³⟩ = 1/(24a₀³), give Bᵢₙₜ ≈ 0.7 T, so an unaided atom generates what a laboratory electromagnet has to work for — which is why fine structure exists in every spectrum ever taken. And the factor of one-half. The electron's rest frame accelerates, so it is not inertial, and a chain of boosts around a curved path compounds into a rotation. That Thomas precession removes exactly half of the naive interaction. It is not a fudge: the Dirac equation produces the same ½ without ever mentioning frames.

Turn L⋅S into quantum numbers by squaring J

You cannot evaluate L⋅S in the |l, mₗ⟩|s, mₛ⟩ basis, because the operator mixes those states. Square the total instead. J = L + S gives J² = L² + 2L⋅S + S², so L⋅S = ½(J² − L² − S²), and in a state of definite j, l and s that is (ħ²/2)[j(j+1) − l(l+1) − s(s+1)]. Everything after this is arithmetic. For a p electron, l = 1 and s = ½: j = 3/2 gives (ħ²/2)(3.75 − 2 − 0.75) = +ħ²/2, and j = 1/2 gives (ħ²/2)(0.75 − 2 − 0.75) = −ħ². The upper component sits (3/2)ζ above the lower, and in general the two components of a one-electron level are split by (l + ½)ζ, with j = l + ½ on top for a less-than-half-filled shell. The same algebra applied to a many-electron term with total L and S gives the Landé interval rule: the gap between adjacent J levels is ζJ for the upper J. A measured ladder of intervals not in that ratio is telling you LS coupling has broken down.

Z⁴/n³ — why hydrogen whispers and caesium shouts

The size of ζ comes from ⟨1/r³⟩, which for a hydrogenic orbital is Z³/[a₀³n³l(l+½)(l+1)]. Multiply by the Z already in the potential and ζ carries Z⁴/n³, so the gap between the two j components is ΔE_fs = Z⁴α²Eₕ/[2n³l(l+1)]. Hydrogen 2p, Z = 1, n = 2, l = 1: (5.325 × 10⁻⁵ × 27.211)/(2 × 8 × 2) = 4.53 × 10⁻⁵ eV — 0.365 cm⁻¹, or 10.9 GHz. Now sodium's 3p. Feed in the bare Z = 11 and the formula returns 0.196 eV, while the measured D-line splitting is 2.14 × 10⁻³ eV: the bare charge overshoots by a factor of 92. Ten inner electrons screen the nucleus, and inverting the formula gives Z_eff⁴ = 160, so Z_eff ≈ 3.6 — not 1, because the 3p orbital penetrates the core far enough to feel several times the net +1 seen from outside. The fourth power is unforgiving in both directions: it is why fine structure is a footnote in hydrogen and why caesium's 6p doublet spans 554 cm⁻¹, wide enough to read as two different colours.

Fine structure is three terms of order α², not one

Spin-orbit is not the only α² correction. Expanding √(p²c² + m²c⁴) past ½mv² adds −p⁴/8m³c², whose expectation value is −(Eₙ²/2mc²)[4n/(l+½) − 3] — comparable in size to the spin-orbit term and present for every l. The Dirac equation also supplies a Darwin term proportional to ∇²V, and since ∇²(1/r) = −4πδ³(r) it is a contact interaction: it shifts only states with ψ(0) ≠ 0, that is l = 0, which is precisely where L⋅S vanishes and spin-orbit has nothing to say. All three scale as α² ≈ 5.3 × 10⁻⁵ against the gross structure; that ratio is what the name 'fine structure constant' records. Their sum is unexpectedly tidy: Eₙⱼ = −(13.6 eV/n²)[1 + (α²/n²)(n/(j+½) − ¾)], a function of n and j but not of l. So 2s₁⁄₂ and 2p₁⁄₂ are predicted degenerate. They are not — they differ by 1058 MHz, the Lamb shift — and that residue is vacuum polarisation and electron self-energy, one order in α beyond anything here.

Weak field: J survives and the Landé factor does the work

Switch on B. The atom's moment is μ = −(μ_B/ħ)(L + 2S), and the 2 on the spin against the 1 on the orbit is what wrecks the classical picture: unless L or S is zero, μ is not parallel to J, so its energy in a field is not simply proportional to J_z. While μ_B B stays well below the fine-structure splitting, spin-orbit still rules: J stays a good quantum number, L and S precess fast around J while J precesses slowly around B, and only the component of μ along J survives the averaging. That projection is exactly what g_J = 1 + [j(j+1) + s(s+1) − l(l+1)]/[2j(j+1)] computes, giving ΔE = g_J μ_B B m_J. Different terms then take different g: 2 for ²S₁⁄₂, 2/3 for ²P₁⁄₂, 4/3 for ²P₃⁄₂. Sodium's D₁ line, ²P₁⁄₂ → ²S₁⁄₂, therefore splits into four components at ±2/3 and ±4/3 of μ_B B, and D₂, ²P₃⁄₂ → ²S₁⁄₂, into six at ±1/3, ±1 and ±5/3. That is the anomalous Zeeman effect; 'anomalous' only records that it was found before spin was.

Strong field: Paschen-Back pulls L and S apart

The weak-field result is an expansion, and it has a stopping point. Compare the two energies directly: μ_B B = 5.788 × 10⁻⁵ eV per tesla, against ΔE_fs. Sodium's 3p doublet, 2.14 × 10⁻³ eV, matches it at 37 T, so a bench electromagnet at 1 T is safely weak-field. Hydrogen's 2p doublet, 4.53 × 10⁻⁵ eV, matches it at 0.78 T — the anomalous pattern in hydrogen breaks up in a field an undergraduate laboratory can reach. Past that crossing the external field couples to L and S more strongly than they couple to each other; J stops being a good quantum number and mₗ and mₛ take over, with energies μ_B B(mₗ + 2mₛ) plus a residual ζ mₗ mₛ. The six states of a ²P term then collect onto slopes +2, +1, 0, 0, −1, −2, and because a photon cannot flip the spin the many anomalous components collapse back towards the normal Lorentz triplet. In between, only mⱼ labels anything, and the levels must be found by diagonalising the 2 × 2 blocks it labels — which is why levels of equal mⱼ repel instead of crossing.

Change one variable at a time

Make the relationship visible.

Interactive model

Spin-orbit constant ζ1.4 meV

Applied magnetic field B12 T

Hold B below 5 T and the six levels fan out as straight lines with Landé slopes 4/3 and 2/3 — the anomalous Zeeman effect. Drag past the dashed μ_B B = ζ mark and the fan bends: by 50 T the slopes have become the Paschen-Back integers mₗ + 2mₛ = +2, +1, 0, 0, −1, −2.

ZERO-FIELD GAP 1.5ζ2.100 meV

ZEEMAN ENERGY μ_B B0.695 meV

RATIO μ_B B / ζ0.50

m_J = +½ LEVEL GAP2.422 meV

Live interpretationZERO-FIELD GAP 1.5ζ: 2.100 meV. ZEEMAN ENERGY μ_B B: 0.695 meV. RATIO μ_B B / ζ: 0.50. m_J = +½ LEVEL GAP: 2.422 meV

Catch the common trap

Explain before calculating.

Misconception check

Two errors travel together. The first is treating the anomalous Zeeman effect as the exception: students write ΔE = μ_B B m_J, get the normal triplet, and reach for g_J only when a question says 'anomalous'. It is the other way round. Because μ = −(μ_B/ħ)(L + 2S) carries a 2 on the spin and a 1 on the orbit, μ is not parallel to J whenever both L and S are non-zero, so g_J μ_B B m_J is the general answer and the normal triplet is the special case S = 0, g_J = 1. The second is calling fine structure 'the spin-orbit splitting'. Spin-orbit is one of three corrections of order α²; the relativistic kinetic term is comparable in size, and the Darwin term is what rescues l = 0, which spin-orbit cannot touch because L⋅S vanishes there. Quote a hydrogen fine-structure level from ζ⟨L⋅S⟩ alone and you are wrong by a factor of order one, and you falsely predict that s states shift not at all.

Sodium's 3p level is split into ²P₃⁄₂ and ²P₁⁄₂ by 2.14 meV. A sodium lamp is placed in a 0.50 T field, where μ_B B = 0.029 meV. Into how many components does the D₂ line, ²P₃⁄₂ → ²S₁⁄₂, split, and at what shifts?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA single electron occupies a 3d orbital, so l = 2 and s = ½. Evaluate ⟨L⋅S⟩ in units of ħ² for both fine-structure components, and confirm that the ²D₅⁄₂–²D₃⁄₂ gap is (l + ½)ζ.

j = l ± ½ gives j = 5/2 and j = 3/2. For both, l(l+1) = 6 and s(s+1) = 3/4.
⟨L⋅S⟩/ħ² = ½[j(j+1) − l(l+1) − s(s+1)]. For j = 5/2: ½[8.75 − 6 − 0.75] = ½(2) = +1.
For j = 3/2: ½[3.75 − 6 − 0.75] = ½(−3) = −1.5. The j = l + ½ component lies higher, as it must for ζ > 0.
The gap is ζ[1 − (−1.5)] = 2.5ζ, and (l + ½)ζ = 2.5ζ. ✓ At equal ζ a d electron splits 2.5/1.5 = 1.67 times as widely as a p electron.

Answer⟨L⋅S⟩ = +ħ² for ²D₅⁄₂ and −1.5ħ² for ²D₃⁄₂, so the gap is 2.5ζ — exactly (l + ½)ζ with l = 2.

MediumThe sodium D lines lie at 589.0 nm and 589.6 nm. Find the 3p spin-orbit splitting in meV, then use ΔE_fs = Z_eff⁴α²Eₕ/[2n³l(l+1)], with α² = 5.325 × 10⁻⁵ and Eₕ = 27.211 eV, to find the effective nuclear charge the 3p electron feels. Compare it with the bare Z = 11.

Work in wavenumbers: 10⁷/589.0 = 16977.9 cm⁻¹ and 10⁷/589.6 = 16960.7 cm⁻¹, a difference of 17.3 cm⁻¹.
With 1 cm⁻¹ = 1.2398 × 10⁻⁴ eV, ΔE_fs = 17.3 × 1.2398 × 10⁻⁴ = 2.14 × 10⁻³ eV = 2.14 meV.
Rearrange: Z_eff⁴ = 2 ΔE_fs n³ l(l+1)/(α²Eₕ) = 2 × 2.142 × 10⁻³ × 27 × 2 / (1.449 × 10⁻³) = 0.2313 / 1.449 × 10⁻³ = 160.
So Z_eff = 160¹⁄⁴ = 3.6, roughly a third of the bare charge. Putting Z = 11 in instead would give 11⁴α²Eₕ/108 = 0.196 eV, which is 92 times the observed splitting.

AnswerΔE_fs = 2.14 meV (17.3 cm⁻¹) and Z_eff ≈ 3.6. Because ζ goes as Z⁴, a threefold error in the effective charge becomes a factor-of-92 error in the energy.

HardSodium vapour emitting the D₂ line (²P₃⁄₂ → ²S₁⁄₂, 589.0 nm) sits in a 0.80 T field. Show that the weak-field treatment applies, list the six components in units of μ_B B, then find the wavelength gap between adjacent components and the resolving power needed to see it. Take μ_B = 5.788 × 10⁻⁵ eV T⁻¹, the 3p splitting as 2.14 meV, and hc = 1239.84 eV nm.

Regime check: μ_B B = 5.788 × 10⁻⁵ × 0.80 = 4.63 × 10⁻⁵ eV = 0.0463 meV, which is 2.14/0.0463 = 46 times below the spin-orbit splitting. J is a good quantum number, so use g_J μ_B B m_J.
Landé factors: ²P₃⁄₂ gives g = 1 + (3.75 + 0.75 − 2)/(2 × 3.75) = 4/3, so its m_J = ±3/2, ±1/2 levels shift by ±2 and ±2/3. ²S₁⁄₂ gives g = 1 + (0.75 + 0.75 − 0)/(2 × 0.75) = 2, so its levels shift by ±1. Units of μ_B B throughout.
Apply Δm_J = 0, ±1 and subtract lower from upper: +2 − 1 = +1; +2/3 − 1 = −1/3; +2/3 + 1 = +5/3; −2/3 − 1 = −5/3; −2/3 + 1 = +1/3; −2 + 1 = −1. Six components at ±1/3, ±1, ±5/3.
Adjacent components are all 2/3 apart, so ΔE = (2/3)(4.63 × 10⁻⁵) = 3.09 × 10⁻⁵ eV.
Convert to wavelength: Δλ = λ²ΔE/hc = (589.0)² × 3.09 × 10⁻⁵ / 1239.84 = 10.71/1239.84 = 8.6 × 10⁻³ nm.
Resolving power R = λ/Δλ = 589.0/0.0086 = 6.8 × 10⁴ — a good grating, not a hand spectroscope.

Answerμ_B B is 46 times below the fine-structure gap, so the weak-field pattern holds: six lines at ±1/3, ±1 and ±5/3 of μ_B B, adjacent pairs 8.6 pm apart at 589 nm, needing R ≈ 6.8 × 10⁴.