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Physics V course sequence · Interactive lesson

Fine structure: relativistic, Darwin and 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 mJ⟩ basis, and how three separately l-dependent shifts add up to something that depends on l not at all.

UPV-13.05 · Lesson 5 of 9 in this unit. Position is not a completion score.
Starting model: Fine structure: relativistic, Darwin and spin-orbitLevel diagram of hydrogen n = 2 with each fine-structure operator carried at an adjustable weight, drawn against the unperturbed level (dashed). Now: 2s₁/₂ at -56.6 μeV, 2p₁/₂ at -56.6 μeV, 2p₃/₂ at -11.3 μeV, and a 2s − 2p₁/₂ gap of 0.0 μeV.−56.6−56.6−11.32s₁/₂ (l = 0)2p₁/₂ (l = 1)2p₃/₂ (l = 1)n = 2 fine-structure shift / μeVdashed: unperturbed level+500−50−100
Start with a picture. Change the model in the lab below.
1

Build the model

Three corrections of order α² — and only n and j survive them.

The key idea

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 complete physical model and assumptions

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 Jz but not with Lz or Sz. Couple first, work in |n l j mJ⟩, 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.

1Three 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.

2Degenerate 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 HD are built from p², r and V alone, so they commute with L², S², Lz and Sz, and are diagonal in either basis. L⋅S is the awkward one: [L⋅S, J²] = [L⋅S, Jz] = [L⋅S, L²] = 0, but [L⋅S, Lz] = iħ(L×S)z ≠ 0, and similarly for Sz. So mₗ and mₛ are not good quantum numbers, while n, l, j and mJ are. Couple with Clebsch–Gordan coefficients — |j = l ± ½, mJ⟩ is a two-term combination of |mₗ = mJ − ½, ↑⟩ and |mₗ = mJ + 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 Hfs; that is why l survives as a label even though mₗ does not.

3The 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 ⟨HD⟩ = (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 the larger is the whole character of fine structure.

4The 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 μB 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 ωT ≈ 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 / 2μB = 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.

5Add 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 = +90.56 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 Efs = (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.

6Where 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 first order already misses a fifth of the 1s relativistic shift — use the exact eigenvalue, or diagonalise the Dirac Hamiltonian on a radial grid once the nucleus has finite size.

2

Key equations

Read the units, assumptions and limits before substituting.

The three α² operators

Hfs = −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

⟨HD⟩ = (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 mJ⟩ 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)], ⟨HSO⟩ = (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 only

Efs = (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
3

Interactive lab

Predict first. Explore the model. Compare the result.

Explore the model

1.00
0 to 1; steps of 0.05.
1.00
0 to 1; steps of 0.05.
1.00
0 to 1; steps of 0.05.

Try this. 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. With all three at 1 the 2s − 2p₁/₂ gap closes, and moving any one weight off 1 reopens it.

Live results

2s₁/₂ SHIFT
-56.6 μeV
2p₁/₂ SHIFT
-56.6 μeV
2p₃/₂ − 2p₁/₂
45.3 μeV
2s − 2p₁/₂ GAP
0.0 μeV

Results follow this lesson’s model and assumptions. Displayed values are rounded.

Interactive Physics V modelLevel diagram of hydrogen n = 2 with each fine-structure operator carried at an adjustable weight, drawn against the unperturbed level (dashed). Now: 2s₁/₂ at -56.6 μeV, 2p₁/₂ at -56.6 μeV, 2p₃/₂ at -11.3 μeV, and a 2s − 2p₁/₂ gap of 0.0 μeV.−56.6−56.6−11.32s₁/₂ (l = 0)2p₁/₂ (l = 1)2p₃/₂ (l = 1)n = 2 fine-structure shift / μeVdashed: unperturbed level+500−50−100
Level diagram of hydrogen n = 2 with each fine-structure operator carried at an adjustable weight, drawn against the unperturbed level (dashed). Now: 2s₁/₂ at −56.6 μeV, 2p₁/₂ at −56.6 μeV, 2p₃/₂ at −11.3 μeV, and a 2s − 2p₁/₂ gap of 0.0 μeV.

Predict → change one variable → check

Plan a change before running it. Every other input stays at the saved baseline.

Saved baseline. Relativistic term −p⁴/8m³c²: 1.00; Darwin term, l = 0 only: 1.00; Spin-orbit term ξ(r)L⋅S, l > 0: 1.00

Baseline 1.00; choose 0 to 1, in steps of 0.05.
4. Compared with the saved baseline, this result will…

This experiment is kept only while this page is open. It does not change saved course progress.

4

Catch the common trap

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 mJ = ½ its matrix is ħ²[[−½, 1/√2], [1/√2, 0]], whose eigenvalues are ħ²/2 and −ħ².

5

Practice & worked examples

3 authored examples. Make an attempt, then reveal the reasoning one step at a time.

EasyExample 1

Hydrogen'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.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

MediumExample 2

For an l = 1 electron, build the L⋅S matrix in the uncoupled basis (|mₗ = 1, ↓⟩, |mₗ = 0, ↑⟩) — the two states with mJ = 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 ⟨HSO⟩ = (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.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 6 steps revealed.

HardExample 3

Show 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.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 6 steps revealed.

6

Quick check

Select an answer and explain it before checking.

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?
7

Review & continue

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.

  • Explain the relationship without looking at the equation.
  • Check a sign, a unit and a model assumption in your working.
  • Return to the lab or worked example if a step is still unclear.

Extra challenging questions

Connect several ideas, explore supplied data, and compare your reasoning with a worked solution.