University Physics V · Atomic Spectroscopy · 13.5

Fine Structure: Relativistic, Darwin & Spin-Orbit

Dirac's equation, expanded to order α², hands the Schrödinger Hamiltonian three new operators rather than one. Learn which basis each is already diagonal in, why only L⋅S forces you into the coupled |n l j m_J⟩ basis, and how three separately l-dependent shifts add up to something that depends on l not at all.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Fine structure is not a new force; it is the Dirac equation admitting that the Schrödinger Hamiltonian was only the leading term of an expansion. Reduce Dirac to two components and to order (v/c)², and exactly three operators appear beyond p²/2m + V: the relativistic kinetic correction −p⁴/8m³c², from expanding √(p²c² + m²c⁴); a Darwin term (ħ²/8m²c²)∇²V, which for a Coulomb potential is a contact term proportional to δ³(r) and therefore reaches l = 0 only; and spin-orbit coupling ξ(r)L⋅S, whose factor of ½ is Thomas precession rather than the naive rest-frame field. Each is smaller than the gross structure by (Zα)² = 5.3 × 10⁻⁵ at Z = 1, which is what licenses first-order perturbation theory.

The cost is a change of basis. The n-manifold is 2n²-fold degenerate, so this is degenerate perturbation theory, and L⋅S is not diagonal in |n l mₗ mₛ⟩: it commutes with L², J² and J_z but not with L_z or S_z. Couple first, work in |n l j m_J⟩, and (J² − L² − S²)/2 has eigenvalue (ħ²/2)[j(j+1) − l(l+1) − ¾].

Do that, and three separately l-dependent shifts sum to (Eₙ²/2mc²)[3 − 4n/(j+½)], which depends on n and j and not on l at all — an exact degeneracy at this order, and one that quantum electrodynamics then breaks by 1058 MHz.

Simple definition: Fine structure is the order-α² splitting of an atomic level produced by three corrections to the Schrödinger Hamiltonian — relativistic kinetic energy, the Darwin contact term, and spin-orbit coupling — whose sum, for a one-electron atom, depends only on n and j.
Example: Hydrogen's n = 2 level splits into 2s₁/₂ and 2p₁/₂ together at −56.60 μeV and 2p₃/₂ at −11.32 μeV, so the fine-structure interval is 45.28 μeV, or 10 949 MHz — within 0.2% of the measured 10 969 MHz.

The three α² operatorsH_fs = −p⁴/8m³c² + (ħ²/8m²c²)∇²V + ξ(r) L⋅S

Names the three perturbations. Each is down on H₀ by (Zα/n)², which is 5.3 × 10⁻⁵ for hydrogen 1s.

ξ = (1/2m²c²)(1/r)(dV/dr) in J⁻¹ s⁻²; p in kg m s⁻¹, V in J, every term in J

Relativistic kinetic shift⟨Hᵣₑₗ⟩ = −(Eₙ²/2mc²)[4n/(l + ½) − 3]

Always negative and steepest at small l — on its own it would put hydrogen 2s 121 μeV below 2p.

Eₙ = −13.606 Z²/n² eV, mc² = 511.0 keV, l the orbital quantum number; shift in eV

Darwin contact term⟨H_D⟩ = (ħ²Ze²/8m²c²ε₀)|ψ(0)|² = +4n(Eₙ²/2mc²), l = 0

Raises s levels only — +725 μeV for hydrogen 1s — and equals what the spin-orbit formula would give as l → 0.

|ψₙ₀₀(0)|² = Z³/πn³a₀³ in m⁻³, ε₀ = 8.854 × 10⁻¹² F m⁻¹; identically zero for every l > 0

L⋅S in the coupled basis⟨L⋅S⟩ = (ħ²/2)[j(j+1) − l(l+1) − ¾], j = l ± ½

Diagonal in |n l j m_J⟩ and not in |n l mₗ mₛ⟩ — this one term dictates which basis you work in.

+ħ²l/2 for j = l+½ and −ħ²(l+1)/2 for j = l−½; ħ² = 1.112 × 10⁻⁶⁸ J² s²

Spin-orbit radial factor and shift⟨1/r³⟩ = Z³/[a₀³n³l(l+½)(l+1)], ⟨H_SO⟩ = (Eₙ²/mc²)⋅n[j(j+1)−l(l+1)−¾]/[l(l+½)(l+1)]

The j = l+½ to j = l−½ interval is (Eₙ²/mc²)⋅2n/[l(l+1)] — 45.28 μeV for hydrogen 2p.

a₀ = 5.292 × 10⁻¹¹ m; l > 0 only, since ⟨1/r³⟩ diverges as l → 0

Summed shift: n and j onlyE_fs = (Eₙ²/2mc²)[3 − 4n/(j + ½)] = −(Zα)⁴mc²[n/(j+½) − ¾]/2n⁴

l has cancelled, so 2s₁/₂ and 2p₁/₂ coincide — a degeneracy only the Lamb shift removes.

α = 1/137.036, so scaling is Z⁴: He⁺ n = 2 splits 16× wider than hydrogen's, at 175 GHz

Three operators from one expansion in v/c

The Schrödinger Hamiltonian is the leading term of the Dirac Hamiltonian, and fine structure is the next one. Reduce the Dirac equation to two components — the Foldy–Wouthuysen transformation, or equivalently second-order elimination of the small spinor components — and to order (v/c)² exactly three operators appear beyond p²/2m + V(r). The kinetic one is arithmetic: √(p²c² + m²c⁴) − mc² = p²/2m − p⁴/8m³c² + …, so Hᵣₑₗ = −p⁴/8m³c². The Darwin term (ħ²/8m²c²)∇²V has no classical ancestor at all. The spin-orbit term ξ(r)L⋅S couples the two angular momenta the electron already carries. For a hydrogenic orbital ⟨v⟩/c ≈ Zα/n, so every one of them is down on H₀ by (Zα/n)²: at Z = 1 the gross structure is 13.6 eV, fine structure is a few hundred μeV, and that ratio of about 5 × 10⁻⁵ is your licence to treat all three at first order.

Degenerate perturbation theory, and the basis L⋅S demands

Hydrogen's n-manifold is 2n²-fold degenerate — n² orbital states times two spin states — so this is degenerate perturbation theory, and the rule is to work in a basis that already diagonalises the perturbation inside the manifold. Check the commutators. Hᵣₑₗ and H_D are built from p², r and V alone, so they commute with L², S², L_z and S_z, and are diagonal in either basis. L⋅S is the awkward one: [L⋅S, J²] = [L⋅S, J_z] = [L⋅S, L²] = 0, but [L⋅S, L_z] = iħ(L×S)_z ≠ 0, and similarly for S_z. So mₗ and mₛ are not good quantum numbers, while n, l, j and m_J are. Couple with Clebsch–Gordan coefficients — |j = l ± ½, m_J⟩ is a two-term combination of |mₗ = m_J − ½, ↑⟩ and |mₗ = m_J + ½, ↓⟩ — and L⋅S = (J² − L² − S²)/2 is diagonal by construction. States of different l never mix, because L² still commutes with the whole of H_fs; that is why l survives as a label even though mₗ does not.

The Darwin term is a contact term, so only s states feel it

The Darwin term is the one with no classical picture, so read it as a potential sampled over a smeared position: ⟨V(r + δr)⟩ ≈ V(r) + ⅙⟨δr²⟩∇²V, and matching the coefficient ħ²/8m²c² gives ⟨δr²⟩ = ¾(ħ/mc)², a smear of order the reduced Compton wavelength ħ/mc = 3.86 × 10⁻¹³ m. That is Zitterbewegung — interference between the positive- and negative-energy components the Dirac spinor mixes. For a Coulomb potential ∇²(1/r) = −4πδ³(r), so ∇²V = Ze²δ³(r)/ε₀ and ⟨H_D⟩ = (ħ²Ze²/8m²c²ε₀)|ψ(0)|². Only l = 0 states have ψ(0) ≠ 0, with |ψₙ₀₀(0)|² = Z³/πn³a₀³ — for hydrogen 1s that is 2.15 × 10³⁰ m⁻³, giving +725 μeV. Set against the relativistic term's −906 μeV it leaves −181 μeV, which is α²/4 of the 13.606 eV binding energy. Two large numbers cancelling down to a fifth of either is the whole character of fine structure.

The Thomas factor of ½, and where ξ(r) breaks down

Boost to the electron's rest frame, watch the proton orbit it, and the magnetic field that orbit produces has magnitude B = (1/emc²)(1/r)(dV/dr)|L|. Pair it with μ = −gₛ μ_B S/ħ and gₛ ≈ 2 and you get (1/m²c²)(1/r)(dV/dr) L⋅S — twice what experiment shows. The reason is that the electron's frame is not inertial. It accelerates, and the composition of two non-collinear boosts is a boost times a rotation, so the rest frame itself precesses at ω_T ≈ −a × v/2c². Thomas precession contributes exactly half the naive interaction with the opposite sign, leaving ξ(r) = (1/2m²c²)(1/r)(dV/dr), which for a Coulomb potential is Ze²/8πε₀m²c²r³. Hydrogen 2p's resulting 45.28 μeV interval reads as an effective internal field of 45.28 μeV / 2μ_B = 0.39 T. One caution: ⟨1/r³⟩ = Z³/[a₀³n³l(l+½)(l+1)] diverges as l → 0 while ⟨L⋅S⟩ vanishes there, so l = 0 is not a limit of this formula — it is the Darwin term's territory.

Add the three, and l cancels out

Do hydrogen n = 2 in units of u = E₂²/2mc² = (3.4014 eV)²/1.0220 MeV = 11.32 μeV. For 2s₁/₂ (l = 0): relativistic −u[4⋅2/½ − 3] = −13u = −147.17 μeV, Darwin +4ν = +8u = +90.57 μeV, spin-orbit zero, total −5u = −56.60 μeV. For 2p₁/₂ (l = 1, j = ½): relativistic −u[8/1.5 − 3] = −(7/3)u = −26.41 μeV, Darwin zero, spin-orbit −(8/3)u = −30.19 μeV, total −5u = −56.60 μeV. The same answer, from a different pair of operators. For 2p₃/₂ the spin-orbit term flips sign to +(4/3)u, giving −u = −11.32 μeV. In general E_fs = (Eₙ²/2mc²)[3 − 4n/(j + ½)] — l has cancelled. The 2p₃/₂ − 2p₁/₂ interval is 4u = 45.28 μeV, or 10 949 MHz against a measured 10 969 MHz; the 0.2% is reduced mass, higher orders in α, and QED.

Where the α² Hamiltonian stops

Three limits. First, the l-degeneracy is an artefact of stopping at α²: quantum electrodynamics adds self-energy and vacuum polarisation at order α(Zα)⁴mc² with a logarithm, raising 2s₁/₂ above 2p₁/₂ by 1057.8 MHz — 4.37 μeV, or 9.7% of the fine-structure interval, small but resolved with room to spare. Second, the nucleus here is a spinless point; hyperfine structure, smaller by roughly mₑ/mₚ, splits hydrogen 1s by 1420 MHz. Third, the expansion parameter is Zα, not α. The exact Dirac–Coulomb eigenvalue is Eₙⱼ = mc²[1 + (Zα)²/(n − j − ½ + √((j+½)² − (Zα)²))²]^(−½), and everything above is its first two terms. At Z = 1 the next term is a 10⁻⁹ correction; at Z = 82, Zα = 0.60 and the series is worthless — diagonalise the Dirac Hamiltonian on a radial grid instead.

Change one variable at a time

Make the relationship visible.

Interactive model

Relativistic term −p⁴/8m³c²1.00

Darwin term, l = 0 only1.00

Spin-orbit term ξ(r)L⋅S, l > 01.00

Pull the Darwin weight to zero: 2s₁/₂ drops to −147 μeV, far below 2p₁/₂, because the relativistic term alone favours low l. Restore it, then take the spin-orbit weight to zero and only the two p levels merge. All three at 1 is the sole setting that closes the 2s − 2p₁/₂ gap.

2s₁/₂ SHIFT-56.6 μeV

2p₁/₂ SHIFT-56.6 μeV

2p₃/₂ − 2p₁/₂45.3 μeV

2s − 2p₁/₂ GAP0.0 μeV

Live interpretation2s₁/₂ SHIFT: −56.6 μeV. 2p₁/₂ SHIFT: −56.6 μeV. 2p₃/₂ − 2p₁/₂: 45.3 μeV. 2s − 2p₁/₂ GAP: 0.0 μeV

Catch the common trap

Explain before calculating.

Misconception check

Fine structure is not the spin-orbit interaction. Spin-orbit is one of three terms, and it is not always the largest — at l = 0 it is exactly zero, yet hydrogen 1s still shifts by −181 μeV, carried entirely by the relativistic term at −906 μeV and the Darwin term at +725 μeV. Even where L⋅S survives it does not dominate: 2p₁/₂ takes −26.41 μeV from the relativistic term and −30.19 μeV from spin-orbit, and needs both to reach −56.60 μeV. Two smaller versions of the same error are worth naming. A rest-frame magnetic-field derivation gives twice the right coupling unless Thomas precession supplies the ½. And ⟨L⋅S⟩ is not ħ²mₗ mₛ, because in the uncoupled basis the operator is not diagonal — for l = 1 and m_J = ½ its matrix is ħ²[[−½, 1/√2], [1/√2, 0]], whose eigenvalues are ħ²/2 and −ħ².

In hydrogen at order α², the 2s₁/₂ and 2p₁/₂ levels come out at exactly the same energy, −56.60 μeV below the unperturbed n = 2 level. Which statement accounts for that degeneracy?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyHydrogen's 1s level has l = 0, so the spin-orbit term contributes nothing. Evaluate the relativistic and Darwin shifts separately, add them, and express the total as a fraction of the 13.606 eV binding energy. Take E₁ = −13.606 eV and mc² = 511.0 keV.

Common factor: E₁²/2mc² = (13.606 eV)²/(1.0220 × 10⁶ eV) = 185.12/1.0220 × 10⁶ = 1.8113 × 10⁻⁴ eV = 181.13 μeV.
Relativistic: ⟨Hᵣₑₗ⟩ = −(E₁²/2mc²)[4n/(l+½) − 3]. With n = 1 and l = 0, the bracket is 4/(½) − 3 = 8 − 3 = 5, so ⟨Hᵣₑₗ⟩ = −5 × 181.13 = −905.65 μeV.
Darwin: ⟨H_D⟩ = +4n(E₁²/2mc²) = 4 × 181.13 = +724.52 μeV. Spin-orbit: ⟨L⋅S⟩ = (ħ²/2)[¾ − 0 − ¾] = 0, so it contributes nothing at l = 0.
Sum: −905.65 + 724.52 = −181.13 μeV. Two shifts of order 10⁻³ eV cancel down to a fifth of either.
As a fraction: 181.13 μeV / 13.606 eV = 1.331 × 10⁻⁵, which is α²/4 = 5.325 × 10⁻⁵/4. The closed formula agrees: (E₁²/2mc²)[3 − 4⋅1/1] = 181.13 × (−1) = −181.13 μeV.

Answer⟨Hᵣₑₗ⟩ = −905.65 μeV, ⟨H_D⟩ = +724.52 μeV, ⟨H_SO⟩ = 0, total −181.13 μeV — a fraction α²/4 = 1.33 × 10⁻⁵ of the binding energy.

MediumFor an l = 1 electron, build the L⋅S matrix in the uncoupled basis (|mₗ = 1, ↓⟩, |mₗ = 0, ↑⟩) — the two states with m_J = ½ — diagonalise it, and identify the eigenvalues with j = 3/2 and j = 1/2. Then get the hydrogen 2p spin-orbit interval in gigahertz from ⟨H_SO⟩ = (Eₙ²/mc²)⋅n[j(j+1) − l(l+1) − ¾]/[l(l+½)(l+1)], with E₂ = −3.4014 eV and mc² = 511.0 keV.

Write L⋅S = L_zS_z + ½(L₊S₋ + L₋S₊). Diagonal elements: ⟨1,↓|L_zS_z|1,↓⟩ = (ħ)(−ħ/2) = −ħ²/2, and ⟨0,↑|L_zS_z|0,↑⟩ = 0.
Off-diagonal: S₋|↑⟩ = ħ|↓⟩ and L₊|1,0⟩ = ħ√(1⋅2 − 0⋅1)|1,1⟩ = √2 ħ|1,1⟩, so ½L₊S₋|0,↑⟩ = (ħ²/√2)|1,↓⟩, giving ⟨1,↓|L⋅S|0,↑⟩ = ħ²/√2. The block is ħ²[[−½, 1/√2], [1/√2, 0]].
Characteristic equation in units of ħ²: λ² + ½λ − ½ = 0, so λ = (−½ ± √(¼ + 2))/2 = (−0.5 ± 1.5)/2, giving λ = +½ and λ = −1, i.e. +ħ²/2 and −ħ².
Match to (ħ²/2)[j(j+1) − 2 − ¾]: j = 3/2 gives (ħ²/2)(3.75 − 2.75) = +ħ²/2, and j = ½ gives (ħ²/2)(0.75 − 2.75) = −ħ². The coupled basis was already the eigenbasis; the uncoupled one was not.
Energies: E₂²/mc² = (3.4014)²/(5.110 × 10⁵) eV = 11.570/5.110 × 10⁵ = 2.2641 × 10⁻⁵ eV = 22.64 μeV. With l = 1, l(l+½)(l+1) = 3 and n = 2: ⟨H_SO⟩ = 22.64 × 2 × (+1)/3 = +15.09 μeV for j = 3/2, and 22.64 × 2 × (−2)/3 = −30.19 μeV for j = ½.
Interval: 15.09 − (−30.19) = 45.28 μeV. In frequency, 45.28 × 10⁻⁶ eV × 2.4180 × 10¹⁴ Hz eV⁻¹ = 1.0949 × 10¹⁰ Hz = 10.95 GHz.

AnswerIn the m_J = ½ block L⋅S has eigenvalues +ħ²/2 (j = 3/2) and −ħ² (j = ½); hydrogen's 2p spin-orbit interval is 45.28 μeV = 10.95 GHz.

HardShow that hydrogen's 2s₁/₂ and 2p₁/₂ are degenerate at order α² by evaluating all three fine-structure terms for each, then find the 2p₃/₂ − 2p₁/₂ interval, scale it to He⁺, and compare the degeneracy with the measured 1057.8 MHz Lamb shift. Use E₂ = −3.4014 eV and mc² = 511.0 keV.

Set u = E₂²/2mc² = 11.570/(1.0220 × 10⁶) eV = 1.1321 × 10⁻⁵ eV = 11.32 μeV. Since Eₙ²/mc² = 2u, the spin-orbit shift is 2u⋅n[j(j+1) − l(l+1) − ¾]/[l(l+½)(l+1)].
2s₁/₂ (n = 2, l = 0, j = ½): relativistic −u[4⋅2/½ − 3] = −13u = −147.17 μeV; Darwin +4ν = +8u = +90.57 μeV; spin-orbit 0. Total −5u = −56.60 μeV.
2p₁/₂ (l = 1, j = ½): relativistic −u[8/1.5 − 3] = −(7/3)u = −26.41 μeV; Darwin 0; spin-orbit 2u⋅2(0.75 − 2 − 0.75)/3 = −(8/3)u = −30.19 μeV. Total −5u = −56.60 μeV — degenerate with 2s₁/₂, from a different pair of operators.
2p₃/₂: relativistic −(7/3)u again; spin-orbit 2u⋅2(3.75 − 2.75)/3 = +(4/3)u = +15.09 μeV; total −u = −11.32 μeV. Interval 2p₃/₂ − 2p₁/₂ = 4u = 45.28 μeV = 10 949 MHz, against 10 969 MHz measured.
He⁺: Eₙ ∝ Z² so E_fs ∝ Eₙ² ∝ Z⁴. With Z = 2 the interval is 16 × 10.949 = 175.2 GHz.
The Lamb shift is 1057.8 MHz = 4.37 μeV, raising 2s₁/₂ above 2p₁/₂. That is 1057.8/10949 = 9.7% of the fine-structure interval — outside this Hamiltonian entirely, and the reason a degeneracy that looks exact here is not.

AnswerBoth j = ½ levels sit at −5u = −56.60 μeV, degenerate at order α². The 2p₃/₂ − 2p₁/₂ interval is 45.28 μeV = 10.95 GHz, and 175.2 GHz in He⁺; the Lamb shift lifts the degeneracy by 9.7% of it.