The Darwin term from the double commutator [O, [O, qΦ]]

Units are SI throughout, with Unit 5's metric signature (+, −, −, −) wherever four-vectors appear. Topic 6.06 split the Dirac Hamiltonian into βmc², an even potential energy V = qΦ and the odd O = cα⋅π, and removed O with the rotation e^(iS), S = −iβO/(2mc²). Write T = iS = βO/(2mc²). The series e^T H e^(−T) = H + [T, H] + ½[T, [T, H]] + … acting on V gives at second order ½[T, [T, V]] = −(1/(8m²c⁴))[O, [O, V]], because β anticommutes with O and with the odd [O, V], and β² = 1; that line is quoted from 6.06. Count its size before computing it. O carries one power of cp, and each commutator with V trades a p for −iħ∇ acting on V, so the term is of order (ħ/mc)²∇²V, or V(ħ/(mcL))² for a potential that changes over a length L. In hydrogen L ≈ a₀ and ħ/(mca₀) = α = 7.30 × 10⁻³: the term is α² of the Coulomb energy, the same order as the kinetic correction −p⁴/(8m³c²).

Take static fields first. V commutes with A, so [O, V] = cα⋅[p, V] = −iħcα⋅∇V, and the second commutator is −iħc²[α⋅π, α⋅∇V] = −iħc² αᵢαⱼ(πᵢ∂ⱼV − ∂ᵢV πⱼ). The Dirac α's obey αᵢαⱼ = δᵢⱼ + iεᵢⱼₖΣₖ, with Σ the 4 × 4 spin matrix, and that identity splits the bracket in two. The symmetric part is [πᵢ, ∂ᵢV] = [pᵢ, ∂ᵢV] = −iħ∇²V, from which A has dropped out. The antisymmetric part is iΣ⋅(π × ∇V − ∇V × π). So [O, [O, V]] = −ħ²c²∇²V + ħc²Σ⋅(π × ∇V − ∇V × π), and multiplying by −1/(8m²c⁴) gives H_D = (ħ²/(8m²c²))∇²V, derived, beside the spin term topic 6.07 reduced to S⋅L. The Darwin term is a commutator: it survives only because p and ∇V do not commute, which is why it carries ħ² and has no classical counterpart. A spinless Klein-Gordon particle, whose odd operator is quadratic in p rather than linear, gets no such term at this order (quoted), so the term belongs to the Dirac structure even though it contains no spin.

If the fields depend on time, so does S, and topic 6.06 added −(iħ/(8m²c⁴))[O, Ȯ] beside the double commutator. With Ȯ = −qcα⋅∂A/∂t the two combine, because [O, V] + iħȮ = −iħcqα⋅(∇Φ + ∂A/∂t) = iħcqα⋅E. The whole second-order term is then −(1/(8m²c⁴))[O, iħcqα⋅E], and the same Pauli split gives H_D = −(qħ²/(8m²c²))∇⋅E, derived, beside the spin-orbit term in its E × π form. Only the field appears, so the result is gauge invariant: a gauge that puts everything into A recovers the Darwin term from [O, Ȯ] alone. Gauss's law, ∇⋅E = ρ/ε₀, then says what the term measures: H_D = −(qħ²/(8m²c²ε₀))ρ(r), the density of source charge at the particle's own position. Three checks follow. As c → ∞ the term vanishes as 1/c², leaving Schrödinger's equation. A free particle, or one in a radiation field in empty space, has ρ = 0 and no Darwin term at all. For an electron, q = −e, near positive charge ρ is positive and so is H_D: the term pushes energies up wherever the electron overlaps the nucleus.

For a hydrogen-like ion V = −Ze²/(4πε₀r), and ∇²(1/r) = −4πδ³(r) gives ∇²V = (Ze²/ε₀)δ³(r). So H_D = (πħ²/(2m²c²))(Ze²/(4πε₀))δ³(r), derived: exactly the operator Quantum Mechanics II quoted. Dimensions: (ħ/mc)² = 1.491 × 10⁻²⁵ m² is an area, Ze²/(4πε₀) is in J m and δ³(r) in m⁻³, leaving joules. Its expectation value is the coefficient times |ψₙₗₘ(0)|², and Quantum Mechanics II's hydrogenic values, reused, give |ψₙ₀₀(0)|² = Z³/(πa₀³n³) and zero for l ≥ 1, since ψ grows as rˡ. With a₀ = ħ/(mcα) and e²/(4πε₀) = αħc, ΔE_D = Z⁴α⁴mc²/(2n³) for s states only, derived. In hydrogen 1S rises 0.7245 meV (175.2 GHz) and 2S 90.57 μeV (21.90 GHz); as a fraction of the binding energy the shift is Z²α²/n, 5.3 × 10⁻⁵ for 1S. A real proton spreads its charge over about 1 fm, but the shift integrates |ψ|² against the source density, and across R/a₀ ≈ 2 × 10⁻⁵ the 1s density is flat: the term sees the enclosed charge, not its shape.

Three operators of order (v/c)² now stand on derived ground: the kinetic correction −p⁴/(8m³c²) from O⁴ (topic 6.06), spin-orbit with the Thomas half built in (6.07), and the Darwin term. Quantum Mechanics II's expectation values, reused as first-order perturbation theory, give, in units of Eₙ²/(2mc²) with Eₙ = −Z²α²mc²/(2n²): kinetic −(4n/(l + ½) − 3); spin-orbit, for l ≥ 1, 2n[j(j + 1) − l(l + 1) − ¾]/(l(l + ½)(l + 1)); Darwin, for l = 0, +4n. For an s state, j = ½, the sum is −(8n − 3) + 4n = 3 − 4n, which is 3 − 4n/(j + ½). For l ≥ 1 kinetic and spin-orbit already combine to that form for both j = l ± ½. So Eₙⱼ − Eₙ = (Eₙ²/(2mc²))(3 − 4n/(j + ½)) for every state, derived. Hydrogen's n = 2 makes it concrete, with u = α⁴mc²/128 = 11.32 μeV: 2S₁/₂ is −13u + 8u = −5u, 2P₁/₂ is −(7/3)u − (8/3)u = −5u, and 2P₃/₂ is −(7/3)u + (4/3)u = −u. The Darwin term is precisely what lifts 2S₁/₂ onto 2P₁/₂.

The interval 2P₃/₂ − 2P₁/₂ = 4u = α⁴mc²/32 = 10.95 GHz stands against a measured 10.969 GHz; the 0.18% remainder is the anomalous moment, +0.23% (topic 6.04), and nuclear recoil, −0.05%, while order α⁶ adds under 0.01%. The Darwin term's own test is the 2S level. Without it 2S₁/₂ would sit 8u = 21.90 GHz below 2P₁/₂; measured, it sits 1.0578 GHz above, the Lamb shift, quoted here and owed to the quantised field of Unit 13. So the measured 2S position confirms the derived contact term to 4.8% of its size, and the remainder is radiative. The expansion behind it runs in ħ/(mcL), and near a point nucleus the Coulomb field has no smallest L. At this order that costs nothing, because ⟨δ³(r)⟩ = |ψ(0)|² is finite, but the next order brings more derivatives of the singular potential, whose s-state expectation values diverge, so the α⁶ energies come from Unit 7's exact solution rather than from more terms. The picture of a position jittering over ħ/mc, which reproduces the coefficient, is Unit 8's.