The Foldy-Wouthuysen transformation to order (v/c)²

SI units throughout; where a four-vector appears the metric is (+, −, −, −). Unit 4's Hamiltonian is H = cα⋅π + βmc² + qφ with π = p − qA. In the Dirac representation β = diag(1, 1, −1, −1) and each αᵢ has only off-diagonal blocks σᵢ, so sort the terms by how they treat β. ℰ = qφ is even, βℰ = ℰβ, and acts within the upper and within the lower pair; O = cα⋅π is odd, βO = −Oβ, and is the only term that couples them. The aim is a representation in which H commutes with β. Count before computing: for a bound electron O ~ cp ~ mc²(v/c) and ℰ ~ mc²(v/c)², and [O, X] is no larger than OX, because the gradient it leaves falls on X, which varies over the same length ħ/p as ψ, and so brings down about p/ħ; so each factor O/mc² costs one power of v/c. In hydrogen v/c ≈ α, so O ≈ 3.7 keV and ℰ ≈ 27 eV against mc² = 511 keV. Working to order (v/c)² means keeping even terms through mc²(v/c)⁴, about α⁴mc² = 1.4 meV, the scale of the fine structure.

Quote the rule for a change of representation ψ' = e^(iS)ψ: H' = H + i[S, H] − ½[S, [S, H]] − (i/6)[S, [S, [S, H]]] + (1/24)[S, [S, [S, [S, H]]]] − ħṠ − (iħ/2)[S, Ṡ] + …. With S = −iβO/2mc², β² = 1 and βO = −Oβ give i[S, βmc²] = −O, so the odd term cancels at leading order. Its partners follow: i[S, O] = βO²/mc², while −½[S, [S, βmc²]] = −βO²/2mc², so the net kinetic term is βO²/2mc². Stop at i[S, H] and you double the kinetic energy. Collecting every term through mc²(v/c)⁴, derived, the even part is ℰ' = ℰ + β(O²/2mc² − O⁴/8m³c⁶) − (1/8m²c⁴)[O, [O, ℰ]] − (iħ/8m²c⁴)[O, Ȯ], and the new odd part is O' = (β/2mc²)([O, ℰ] + iħȮ) − O³/3m²c⁴. Check it numerically with random 8 × 8 matrices: halve O, with ℰ and Ȯ scaled to match, and the even residual falls by 64 = 2⁶, as a first neglected term of order mc²(v/c)⁶ requires.

O contains A(r, t), so S depends on time: the rule's −ħṠ = iħβȮ/2mc² joins the odd part, and −(iħ/2)[S, Ṡ] gives the even term −(iħ/8m²c⁴)[O, Ȯ]. Both arrive paired with ℰ, in the combination [O, ℰ] + iħȮ. Evaluate it. [cα⋅π, qφ] = −iħqc α⋅∇φ, since only p fails to commute with φ, and iħȮ = −iħqc α⋅∂A/∂t. The sum is iħqc α⋅(−∇φ − ∂A/∂t) = iħqc α⋅E, derived: the potentials have gone and the gauge-invariant field remains. The outer commutator follows from αᵢαⱼ = δᵢⱼ + iεᵢⱼₖΣₖ and [πᵢ, Eᵢ] = −iħ∂ᵢEᵢ: [O, iħqc α⋅E] = qħ²c²∇⋅E + qħc²Σ⋅(E × π − π × E). Divided by −8m²c⁴, the first piece is the Darwin term and the second the spin-orbit term; topics 6.07 and 6.08 take each apart. Dimensions: qħEπ/m²c² is C × J s × V m⁻¹ × kg m s⁻¹ over kg² m² s⁻², which reduces to J, as an energy must.

O' is of order mc²(v/c)³: [O, ℰ]/mc² and O³/m²c⁴ are each that size. Repeat the step with S' = −iβO'/2mc²: it cancels O' and leaves O'' = (β/2mc²)([O', ℰ'] + iħȮ'), of order mc²(v/c)⁵, and S'' = −iβO''/2mc² removes that in turn, leaving odd coupling at mc²(v/c)⁷. None of this alters the even part at the order kept, because an odd term X shifts energies only at second order, by about X²/2mc²; for O' that is mc²(v/c)⁶, near 8 × 10⁻⁸ eV in hydrogen against α⁴mc² = 1.4 meV. So H_FW = βmc² + ℰ' to order (v/c)², derived. Its upper block follows from topic 6.02's (σ⋅π)² = π² − qħσ⋅B, which turns βO²/2mc² into π²/2m − (qħ/2m)σ⋅B, while −O⁴/8m³c⁶ becomes −π⁴/8m³c² once its field-dependent pieces, (v/c)² corrections to the Zeeman energy, are dropped. Every term is Hermitian, because each e^(iS) is unitary, and none contains the energy eigenvalue.

With φ = A = 0 the series can be tested against an exact answer. Let K = βα⋅p̂, p̂ = p/|p|. Then K² = −1, so e^(Kθ) = cos θ + K sin θ, and K anticommutes with both α⋅p̂ and β, hence with H. So e^(Kθ)He^(−Kθ) = e^(2Kθ)H = α⋅p̂(c|p| cos 2θ − mc² sin 2θ) + β(mc² cos 2θ + c|p| sin 2θ), derived. The odd part vanishes for tan 2θ = |p|/mc, leaving H' = β√(p²c² + m²c⁴): one rotation separates the two energy branches exactly. The perturbative steps are this rotation expanded. The first S has angle p/2mc; the second, built from −O³/3m²c⁴, adds −(p/mc)³/6, the next term of ½ arctan(p/mc). Expanding the root gives mc² + p²/2m − p⁴/8m³c² + p⁶/16m⁵c⁴ − …, so H₊ with no fields is right through p⁴. The series converges only for |p| below mc: at p = mc, kinetic energy 211.7 keV, the order-(v/c)² terms give 191.6 keV, 9.5% low.

Collected, H₊ = mc² + π²/2m + qφ − π⁴/8m³c² − (qħ/2m)σ⋅B − (qħ/8m²c²)σ⋅(E × π − π × E) − (qħ²/8m²c²)∇⋅E. The order E × π − π × E is the Hermitian one, since (E × π)† = −π × E. For a static central potential, topics 6.07 and 6.08 reduce the last two terms to Quantum Mechanics II's spin-orbit and Darwin operators, which with its expectation values give its Eₙⱼ. At n = 2 that puts 2P₃/₂ above 2P₁/₂ by α⁴mc²/32h = 10.95 GHz, against 10.969 GHz measured. The 0.18% gap is the anomalous moment quoted in topic 6.04, which multiplies the spin-orbit splitting by g − 1 = 1 + 2a and so raises it 0.23%, less the 0.05% the reduced mass removes. Three limits bound the result. Momenta must stay below mc, which for a 1s electron means Zα well below 1. Fields must vary little over ħ/mc = 3.86 × 10⁻¹³ m, which hydrogen's a₀ exceeds 137-fold. And fields must stay far below the critical field m²c³/eħ = 1.3 × 10¹⁸ V m⁻¹, quoted, at which the vacuum creates pairs and no rotation can keep the electron's block closed.