Spin-orbit coupling and the Zeeman effect

The Coulomb Hamiltonian p²/2m − Ze²/4πε₀r is the first term of an expansion in v/c, and since v/c ≈ Zα/n for a bound electron the next terms are smaller by (Zα)². Three arrive together. Spin-orbit coupling is the spin moment in the magnetic field of the electron's own orbit, Thomas factor included; that operator is built in Spin-orbit coupling, fine structure, and the Landé g factor (UPIV-09.09), and this lesson takes one line from it: in a state of definite j, ⟨L⋅S⟩ = (ħ²/2)[j(j+1) − l(l+1) − s(s+1)]. The relativistic kinetic correction comes from expanding √(p²c² + m²c⁴). The Darwin term comes from the Dirac equation and acts only at the nucleus. All three perturb a degenerate n-shell, and degenerate perturbation theory is safe only in a basis that already diagonalises the perturbation. The coupled states |n l j m_J⟩ do that for all three at once: p⁴ and the Darwin term are spherically symmetric and ignore spin, so they commute with L², S², J² and J_z, and L⋅S is diagonal once j is fixed. First-order theory is then three expectation values, added.

For a Coulomb potential (1/r)(dV/dr) = Ze²/4πε₀r³, so the first-order spin-orbit shift is (Ze²/8πε₀m²c²)⟨1/r³⟩⟨L⋅S⟩, and the hydrogenic integral ⟨1/r³⟩ = Z³/[a₀³n³ l(l+½)(l+1)] carries the dependence on the state. One Z from the potential and three from the shrinking orbit give Z⁴/n³, and the result is E_SO = Z⁴α²Eₕ[j(j+1) − l(l+1) − ¾]/[4n³ l(l+½)(l+1)]. Writing the interaction as (ζ/ħ²)L⋅S, with ζ an energy, the doublet j = l ± ½ is split by (l + ½)ζ, which for a hydrogenic level is Z⁴α²Eₕ/[2n³ l(l+1)]: 45.3 µeV for hydrogen 2p, or 10.9 GHz. Sodium's 3p doublet, 2.13 meV, is 47 times larger, yet the formula with the bare Z = 11 overshoots it by a factor of 92; inverting it gives Z_eff ≈ 3.6, because the 3p orbital penetrates the ten-electron core. The same fourth power is why caesium's 6p doublet reaches 554 cm⁻¹. At l = 0 the term gives nothing at all, since L⋅S = 0, and every s level is left to the other two corrections.

Expand the kinetic energy: √(p²c² + m²c⁴) − mc² = p²/2m − p⁴/8m³c² + …. The second term is the perturbation, and its expectation value needs no new integral. Since p²/2m = Eₙ − V, ⟨p⁴⟩ = 4m²[Eₙ² − 2Eₙ⟨V⟩ + ⟨V²⟩], and with ⟨1/r⟩ = Z/(n²a₀) and ⟨1/r²⟩ = Z²/[n³(l + ½)a₀²] the result is Eᵣₑₗ = −(Eₙ²/2mc²)[4n/(l + ½) − 3]. It is negative for every state and largest for s states, whose electrons spend the most time close to the nucleus, where they move fastest. The Darwin term is H_D = (ħ²/8m²c²)∇²V. For the Coulomb potential ∇²(1/r) = −4πδ³(r), so H_D = (πħ²/2m²c²)(Ze²/4πε₀)δ³(r), a contact interaction that samples only |ψ(0)|². Only s states have ψ(0) ≠ 0, and with |ψ(0)|² = Z³/(πn³a₀³) the shift is +2nEₙ²/mc². It is often pictured as the electron's position being smeared over its reduced Compton wavelength ħ/mc, so that it feels the Coulomb potential averaged over that distance; the operator itself is what the Dirac equation delivers at this order.

Collect the pieces in units of Eₙ²/2mc². For j = l + ½, Eᵣₑₗ + E_SO = 3 − 4n/(l + 1); for j = l − ½ it is 3 − 4n/l; and for an s state, where the Darwin term stands in for spin-orbit, Eᵣₑₗ + E_D = 3 − 4n. Every case is 3 − 4n/(j + ½), so Eₙⱼ = −(13.6 eV/n²)[1 + (α²/n²)(n/(j + ½) − ¾)]: the energy remembers n and j and has forgotten l. Three checks close it. Dimensions: Eₙ²/mc² is an energy. Limits: α → 0, which is c → ∞, returns the Bohr levels, and the correction is always negative, largest for small j. Correspondence with the exact theory: the Dirac equation, solved for hydrogen, expands to this same expression at order α⁴mc², so the three first-order pieces are the whole of the α² physics. What survives is a degeneracy. 2s₁⁄₂ and 2p₁⁄₂ reach the same energy by different routes, the first lowered by the relativistic term and raised by Darwin, the second lowered by both the relativistic and spin-orbit terms. Only the 1058 MHz Lamb shift, vacuum polarisation and electron self-energy from QED, one order of α further on, separates them.

A field adds (μ_B/ħ)(L + 2S)⋅B. While μ_B B is well below the fine-structure gap, j survives, and the result to import from UPIV-09.09 is one line: each level shifts by g_J μ_B B m_J with g_J = 1 + [j(j+1) + s(s+1) − l(l+1)]/[2j(j+1)], which is 2 for ²S₁⁄₂, 2/3 for ²P₁⁄₂ and 4/3 for ²P₃⁄₂. A spectrometer sees differences. A line's components sit at (g_J′m_J′ − g_J m_J)μ_B B for Δm_J = 0, ±1, primes marking the upper level. Sodium D₁, ²P₁⁄₂ → ²S₁⁄₂, gives four: π components (Δm_J = 0) at ±2/3 and σ components (Δm_J = ±1) at ±4/3, in units of μ_B B. D₂, ²P₃⁄₂ → ²S₁⁄₂, gives six: π at ±1/3 and σ at ±1 and ±5/3. Viewed across the field, π light is linearly polarised along B and σ light across it; viewed along B, the π components are missing and the two σ groups are circularly polarised in opposite senses. The shifts are small: at 1 T, μ_B B = 57.9 µeV, about three parts in 10⁵ of the 2.10 eV photon. The normal Lorentz triplet at 0 and ±μ_B B needs g_J = 1 in both levels, which singlet lines, S = 0, guarantee; the other pattern is called anomalous only because it was seen before spin was known.

The Landé result is first-order perturbation theory in the field on top of fine structure, so it holds only while μ_B B ≪ ΔE_fs. The crossing field ΔE_fs/μ_B is 37 T for sodium's 2.13 meV doublet, beyond any bench magnet, but 0.78 T for hydrogen's 45.3 µeV 2p doublet. Past it the field outweighs spin-orbit coupling, mₗ and mₛ become the good labels, and a p level has energies μ_B B(mₗ + 2mₛ) + ζmₗ mₛ. The electric-dipole operator acts on position and not on spin, so Δmₛ = 0, and a p → s line then has components at μ_B B Δmₗ with Δmₗ = 0, ±1: the normal Lorentz triplet, each outer member split into a pair ζ apart by the leftover spin-orbit term. The ten weak-field components do not simply slide into three. Only m_J stays good at every field, so upper states of equal m_J are mixed by the field and repel, and line strength drains out of the four components that would need a spin flip until they vanish, while the six survivors gather into the triplet. The figure draws exactly that: every component at its exact position, as tall as its emission strength.