The conserved current and its positive density

Units are SI with ħ and c kept, x^μ = (ct, x) and the metric diag(1, −1, −1, −1), as in Unit 3. Topics 4.01 and 4.02 supply all that is needed: iħ∂ₜψ = −iħc αₖ∂ₖψ + βmc²ψ, summed over k, with αₖ and β Hermitian 4 × 4 matrices. Multiply by ψ† on the left. Take the Hermitian conjugate, −iħ∂ₜψ† = iħc(∂ₖψ†)αₖ + mc²ψ†β, which uses αₖ† = αₖ and β† = β, and multiply by ψ on the right. Subtracting the second from the first removes both mc²ψ†βψ terms and leaves iħ[ψ†∂ₜψ + (∂ₜψ†)ψ] = −iħc[ψ†αₖ∂ₖψ + (∂ₖψ†)αₖψ]. Divide by iħ and both brackets are total derivatives: ∂ₜ(ψ†ψ) + ∂ₖ(cψ†αₖψ) = 0. Hermiticity is the whole proof, since a non-Hermitian β would leave a mass term behind as a source or sink of probability. Dimensions close: with ∫ψ†ψ d³x = 1, ψ carries m^(−3/2), ρ is in m⁻³, j in m⁻² s⁻¹, and both terms of the equation in m⁻³ s⁻¹.

Unit 3's conserved Klein–Gordon density, ρ_KG = (iħ/2mc²)(φ*∂ₜφ − φ∂ₜφ*), contains a time derivative because that equation is second order, and a second-order equation takes φ and ∂ₜφ as independent initial data, so ρ_KG can be made negative anywhere at t = 0; on a plane wave it is (E/mc²)|φ|², negative on the lower branch. Dirac's equation is first order, its initial data are ψ alone, and its density holds no derivative: ρ = Σₐ|ψₐ|², zero only where all four components vanish. With fields that fall off at infinity, d/dt ∫ρ d³x = −∮j⋅dS = 0, so a state normalised once stays normalised. Times −e the same integral is the electron's charge, and Borexino's 2015 bound τ ≥ 6.6 × 10²⁸ yr on e⁻ → νₑγ says nature conserves it too. The non-relativistic limit is the check: on the positive branch the lower pair is smaller than the upper by about v/2c, so ρ → |ψ₁|² + |ψ₂|², the Pauli density, with a correction of order v²/4c². For a 1 keV electron the lower pair holds 0.098% of ρ.

In the Dirac representation αₖ carries σₖ in its off-diagonal blocks, so with ψ = (φ, χ) split into upper and lower two-spinors, ρ = φ†φ + χ†χ while j = 2c Re(φ†σχ). The current is an interference between the halves, and a spinor with χ = 0 carries none. A bound comes free from the algebra of topic 4.01: for a unit vector n̂, (n̂⋅α)² = Σ nᵢnₖαᵢαₖ = 1, because the αs square to 1 and anticommute in pairs, so the Hermitian n̂⋅α has eigenvalues ±1 and −ψ†ψ ≤ ψ†(n̂⋅α)ψ ≤ ψ†ψ. Point n̂ along j: |j| ≤ cρ at every point of every solution. Check it on a plane wave with p along z and spin up. The equation reduces to the 2 × 2 block [[mc², cp], [cp, −mc²]] acting on (ψ₁, ψ₃), whose positive-energy eigenvector has r = ψ₃/ψ₁ = cp/(E + mc²). Then j_z/ρ = 2cr/(1 + r²) = pc²/E, the group velocity dE/dp, always below c and tending to Schrödinger's p/m as p → 0. At p = mc it is c/√2 = 0.71c, where p/m would claim c itself.

Multiply the equation by β/c and name γ⁰ = β and γᵏ = βαₖ, the matrices of topic 4.03: (iħγ^μ ∂_μ − mc)ψ = 0. Its Hermitian conjugate reads −iħ ∂_μ ψ† γ^μ† − mcψ† = 0. Because γ⁰ is Hermitian and each γᵏ anti-Hermitian, γ⁰γ^μ†γ⁰ = γ^μ for every μ, so multiplying on the right by γ⁰ gives the adjoint equation iħ ∂_μ ψ̄ γ^μ + mc ψ̄ = 0 for ψ̄ = ψ†γ⁰, derived rather than assumed. Then ∂_μ(ψ̄ γ^μ ψ) = (∂_μ ψ̄)γ^μ ψ + ψ̄ γ^μ ∂_μ ψ = (imc/ħ)ψ̄ψ − (imc/ħ)ψ̄ψ = 0. Since γ⁰γ⁰ = 1 and γ⁰γᵏ = αₖ, the current j^μ = c ψ̄ γ^μ ψ has j⁰ = cρ and spatial part j: the same law, written so that Unit 5 can prove j^μ a four-vector. Read with that result, ρ is a time component, not a scalar. The scalar is ψ̄ψ = φ†φ − χ†χ, which can be negative, and on a positive-energy plane wave it equals (mc²/E)ρ. And |j| ≤ cρ makes j^μ timelike or null and future-pointing, so ρ is positive in every frame.

The same 2 × 2 block has a second eigenvector, at energy −E, with ψ₁/ψ₃ = −cp/(E + mc²). At p = 0.75 mc it is proportional to (1, 0, −3, 0): ρ = 10|N|², positive, while j_z = 2c(1)(−3)|N|² = −6c|N|², so probability flows at −0.60c, against the momentum, which is the group velocity dE/dp = −pc²/|E| of the lower branch. Klein–Gordon gives its negative-energy plane waves a negative density; Dirac gives this one a positive density, so ρ no longer marks the states that sank Klein–Gordon's probability reading, and the spectrum is still unbounded below. Mix the branches at one momentum with weights cos²θ and sin²θ and the contrast sharpens. The cross terms cancel in ρ, because eigenvectors of a Hermitian matrix with different eigenvalues are orthogonal, so Dirac's ρ counts cos²θ + sin²θ = 1, while Klein–Gordon's counts cos²θ − sin²θ. The current, unlike ρ, keeps a cross term, oscillating at 2E/ħ ≥ 2mc²/ħ = 1.55 × 10²¹ s⁻¹, the Zitterbewegung that topic 4.09 quotes and Unit 8 derives. Positivity makes a probability reading possible; whether it is one electron's probability is for Unit 8 to settle.