Minimal coupling and gauge invariance of the Dirac equation

Units are SI, the metric is (+, −, −, −) and A^μ = (φ/c, A). The rule is quoted from UAQM-03.04: replace ∂_μ by D_μ = ∂_μ + (iq/ħ)A_μ, which is p_μ → p_μ − qA_μ. Dirac's equation becomes (iħγ^μ D_μ − mc)ψ = 0. Multiply by cγ⁰, use γ⁰γ⁰ = 1 and γ⁰γⁱ = αⁱ, and iħ∂ψ/∂t = Hψ with H = cα⋅π + βmc² + qφ, where π = p − qA is the kinetic momentum; for the electron q = −e and π = p + eA. Units: qA is C × T m = kg m s⁻¹, a momentum, so qcα⋅A is in J. Free-field limit: A^μ → 0 gives back this unit's free equation. Two features matter later. H is linear in π, so the coupling −qcα⋅A + qφ is exact, with no A² term to keep or drop. And in 2 × 2 blocks qφ sits on the diagonal while cσ⋅π sits off it, so the vector potential acts by mixing the large and small components. In hydrogen qφ = −e²/(4πε₀r), −27.2 eV at r = a₀, and Unit 7 solves that H exactly.

Transform the potentials by A_μ → A_μ − ∂_μ Λ, which is A → A + ∇Λ and φ → φ − ∂Λ/∂t, and the spinor by ψ → e^(iqΛ/ħ)ψ. Differentiating the phase adds (iq/ħ)∂_μ Λ ψ, and the new −(iq/ħ)∂_μ Λ in D_μ removes it, so D′_μ ψ′ = e^(iqΛ/ħ)D_μ ψ. The step that is new for spinors comes next: the phase must pass through every γ^μ matrix, and it does, because e^(iqΛ/ħ) multiplies all four components by the same number and is a multiple of the 4 × 4 identity. So (iħγ^μ D′_μ − mc)ψ′ = e^(iqΛ/ħ)(iħγ^μ D_μ − mc)ψ = 0, derived. Two consequences follow. All four components, including those that carry the negative-energy solutions, couple with the same charge q, a fact Unit 8 has to reinterpret. And gauge and Lorentz transformations act on different things: the first changes only the phase, while a boost (Unit 5) mixes the components through a 4 × 4 matrix. Check: Λ is in T m² = V s, so qΛ/ħ is dimensionless.

From iħ∂ψ/∂t = Hψ and its adjoint, ∂(ψ†ψ)/∂t = (1/iħ)[ψ†(Hψ) − (Hψ)†ψ]. Each term of H without a derivative, βmc², qφ and −qcα⋅A, is a Hermitian matrix times a real function, so it contributes ψ†Mψ − ψ†M†ψ = 0 at every point. Only cα⋅p survives, giving ∂(ψ†ψ)/∂t + ∇⋅(cψ†αψ) = 0, derived, the free form unchanged. Compare the Schrödinger current (ħ/m)Im(ψ*∇ψ) − (q/m)A|ψ|²: it holds a gradient, the gauge phase adds (q/m)∇Λ|ψ|² to it, and the A term is there to cancel that. The Dirac current holds no derivative, so e^(−iqΛ/ħ) and e^(iqΛ/ħ) cancel inside ψ†αψ and ρ and j are gauge invariant on sight. The A term is hidden, not lost: Unit 5's Gordon decomposition, quoted here, splits j into a convective part built from π, which contains it, and a spin part. Units: ψ in m^(−3/2), ρ in m⁻³, j in m⁻² s⁻¹. Free-field limit: A → 0 gives the current of this unit's fourth topic.

In the Heisenberg picture v = (i/ħ)[H, r]. Only cα⋅p fails to commute with x, and [pₓ, x] = −iħ, so vₓ = cαₓ: the velocity is cα in any field, where Schrödinger's was π/m. The field shows up in how π moves instead. Using [πₓ, π_y] = iqħB_z, the gradient of qφ and ∂π/∂t = −q∂A/∂t, dπ/dt = q(ℰ + cα × B), derived, with ℰ = −∇φ − ∂A/∂t the electric field: the Lorentz force with v replaced by cα. In a positive-energy plane wave this unit's spinor u = N(χ, cσ⋅πχ/(E + mc²)) gives u†αu/u†u = cπ/E, so ⟨cα⟩ = c²π/E = ∂E/∂π, the group velocity. Limits: for π ≪ mc it is π/m and the force is Newton's; as π grows it approaches c and never passes it. At π = mc it is 0.707c, and an electron with 1.00 MeV of kinetic energy has E = 1.511 MeV and moves at 0.941c. Time-of-flight measurements on MeV electrons, Bertozzi's among them, find speeds crowding just below c, as this predicts.

Since αₓ² = 1 and αₓ is traceless, each component of cα has eigenvalues +c and −c, twice each. The components do not commute, αₓα_y − α_yαₓ = 2iΣ_z, so vₓ and v_y cannot be measured together, and c²α⋅α = 3c² as an operator, as though every speed were √3 c. Yet no packet outruns light: Ehrenfest's d⟨r⟩/dt = ⟨cα⟩ moves a free positive-energy packet at the average of c²p/E over its momenta, below c. The tension is real. cα does not commute with H, so a plane wave of definite energy is not a velocity eigenstate, and an eigenstate of α_z must mix both energy signs: among free positive-energy states of momentum p, α acts as cp/E, smaller than 1 in size, so none returns +1. A packet that does mix the signs trembles at angular frequency 2E/ħ, at least 2mc²/ħ = 1.55 × 10²¹ s⁻¹, with an amplitude near ħ/2mc = 1.93 × 10⁻¹³ m, Zitterbewegung, quoted here and derived in Unit 8, where the negative-energy states get their meaning. This lesson derives the operator and its expectation; the resolution waits.

H shows neither the A² of (p − qA)²/2m nor any spin–field term, and both are inside it. The kinetic momenta fail to commute: [πₓ, π_y] = −q([pₓ, A_y] + [Aₓ, p_y]) = iqħB_z. With αᵢαⱼ = δᵢⱼ + iεᵢⱼₖΣₖ from this unit, (α⋅π)² = π² + (i/2)εᵢⱼₖΣₖ[πᵢ, πⱼ] = π² − qħΣ⋅B, derived. For φ = 0 the cross terms of H² cancel because β anticommutes with every αᵢ, so H² = c²π² + m²c⁴ − qħc²Σ⋅B. Klein–Gordon with the same coupling has c²π² + m²c⁴ and nothing more; the spinor brings a moment whose size no parameter sets, which Unit 6 reduces to −(q/m)S⋅B, g = 2. Minimal means adding nothing gauge invariance does not demand, and that is a hypothesis: a Pauli term κσ^μν F_μν ψ is just as gauge invariant and would move g freely. Measurement decides. The electron's g = 2.0023 is 2 up to a radiative 0.1%; the proton's 5.59, and a neutron moment with q = 0, show minimal coupling of a point particle failing for them.