The Klein-Gordon current and its indefinite density

SI units and the metric diag(1, −1, −1, −1) hold throughout, with x^μ = (ct, r) and ∂^μ = ((1/c)∂/∂t, −∇) as fixed in topic 3.01, whose equation (∂_μ∂^μ + κ²)φ = 0, κ = mc/ħ, is used here as given. Multiply it by φ*, multiply its complex conjugate by φ, and subtract. The κ² terms cancel because m is real, and φ*∂_μ∂^μφ − φ∂_μ∂^μφ* = ∂_μ(φ*∂^μφ − φ∂^μφ*), since the terms ∂_μφ*∂^μφ cancel in pairs. So j^μ = (iħ/2m)(φ*∂^μφ − φ∂^μφ*) obeys ∂_μ j^μ = 0, which with j^μ = (cρ, j) reads ∂ρ/∂t + ∇⋅j = 0. The prefactor is chosen so that j = (ħ/2mi)(φ*∇φ − φ∇φ*), Schrödinger's current unchanged: the minus sign in ∂ⁱ = −∂/∂xⁱ combines with the i in front, and −i = 1/i. Dimensions: ħ/m is m² s⁻¹, 4.24 × 10⁻⁷ m² s⁻¹ for a pion; ∂^μ adds m⁻¹ and |φ|² is m⁻³, so j is m⁻² s⁻¹ and ρ = j⁰/c is m⁻³, as a flux and a density must be.

Schrödinger's |ψ|² works because iħ∂ψ/∂t = Hψ turns ∂|ψ|²/∂t into an expression in ψ alone. The Klein-Gordon equation fixes ∂²φ/∂t², not ∂φ/∂t, so ∂|φ|²/∂t = 2 Re(φ*∂φ/∂t) is not determined by φ at all, and nothing keeps ∫|φ|² d³x constant. Covariance gives a second, independent reason. A conserved density must be the time component of a four-vector, so that a boost which shrinks a volume by 1/γ raises the density by γ and leaves the count alone; |φ|² is a Lorentz scalar and cannot do that. The simplest four-vector built from φ and φ* with nothing else to point along carries one ∂^μ, and of the two combinations the symmetric one, ∂^μ|φ|², is not conserved, which leaves the antisymmetric one. The non-relativistic limit checks the result. Write φ = e^{(−imc²t/ħ)}ψ with ψ varying on the scale of the kinetic energy: ∂φ/∂t ≈ −(imc²/ħ)φ, and ρ = |ψ|²[1 + O(Eₖᵢₙ/mc²)]. For a pion with 1 MeV of kinetic energy the correction is 0.72%.

Put φ = N e^{(−i(Et − p⋅r)/ħ)} with E = ±√(p²c² + m²c⁴). Then ∂^μφ = −(i/ħ)p^μφ and the current is j^μ = (p^μ/m)|N|²: ρ = (E/mc²)|N|² and j = (p/m)|N|². Their ratio j/ρ = pc²/E is the group velocity of topic 3.01, and for positive E the factor E/mc² is exactly the γ that length contraction demands of a density. A pion of momentum 250 MeV/c has E = 286.3 MeV, so ρ = 2.051|N|²; setting |N|² equal to the number density, Schrödinger-style, would overcount by that factor. Now take the negative root with the same |N|. Nothing in the algebra objects: ρ = −2.051|N|² while j = (p/m)|N|² still points along p, a negative density travelling at j/ρ = −0.873c, against its own momentum. A density that is negative on a solution of the free equation cannot be a probability. Complex conjugation makes the pairing exact: φ* solves the same equation, carries energy −E and momentum −p, and j^μ[φ*] = −j^μ[φ] term by term.

Throwing away the negative energies does not rescue ρ. Superpose φ = a₁e^(−iθ₁) + a₂e^(−iθ₂) with θₖ = (Eₖt − pₖ⋅r)/ħ, both energies positive and the amplitudes real. The cross terms of φ*∂φ/∂t bring down E₁ and E₂ in turn, so ρ = [E₁a₁² + E₂a₂² + (E₁ + E₂)a₁a₂ cos Δ]/mc² with Δ = θ₁ − θ₂. As a quadratic form in a₁ and a₂ this is positive at every phase only if (E₁ + E₂)²/4 ≤ E₁E₂, which fails whenever E₁ ≠ E₂: the arithmetic mean beats the geometric mean. Where cos Δ = −1, ρₘᵢₙ = (a₁ − a₂)(E₁a₁ − E₂a₂)/mc², negative for a₂/a₁ between E₁/E₂ and 1. A pion at rest plus one with p₂c = 2.4mc² = 335 MeV, so E₂ = 2.6mc², and a₂ = 0.70a₁ gives ρ from 4.79a₁² down to −0.246a₁², negative over 14% of each 3.70 fm beat, while |φ|² never drops below 0.09a₁². The mean stays positive, 2.27a₁², because the cross term averages away. As E₂ → E₁ the window closes, so the pockets need momentum differences of order mc and are always narrower than ħ/mc = 1.41 fm.

Schrödinger's equation is first order in time, so ψ(r, 0) alone fixes the future and ∂ψ/∂t is computed, never chosen. The Klein-Gordon equation needs φ(r, 0) and ∂φ/∂t(r, 0) separately, just as a plucked string needs both its shape and its velocity. Pick any φ(r, 0) and set ∂φ/∂t = −iω(r)φ with any real ω(r). Then ρ(r, 0) = (ħω(r)/mc²)|φ(r, 0)|²: positive where ω is positive, negative where it is negative, zero where you choose ω = 0, with the same |φ|² in every case. Fourier analysis shows what the choice does. For each momentum p the two data fix one positive-frequency amplitude a₊(p) and one negative-frequency amplitude a₋(p), and in a box of volume V the conserved total is Q = V Σ (Eₚ/mc²)(|a₊(p)|² − |a₋(p)|²), with Eₚ = +√(p²c² + m²c⁴). Only data with every a₋(p) = 0 guarantee a positive Q, and even those leave negative pockets, as the previous section showed. So ρ is not a probability density, and no choice of normalisation repairs it.

The integral Q = ∫ρ d³x is still worth keeping. It is constant, because ∂_μ j^μ = 0 turns dQ/dt into a surface flux that vanishes for a localised field, and it is Lorentz invariant, because j^μ is a conserved four-vector. Two derived facts say what kind of number it is. A real field has ρ = 0 identically, so a particle described by a real field carries Q = 0. Conjugation φ → φ* is again a solution and reverses j^μ, so every solution has a partner of opposite Q with the same mass. Pauli and Weisskopf in 1934 read eQ as electric charge, with ρ an indefinite charge density rather than a probability; that reading is quoted here and derived in Unit 9, where the field is quantised. The measured spinless mesons fit it: π⁺ and π⁻ carry charges +e and −e with the common measured mass 139.570 MeV, as a complex field and its conjugate should, while the π⁰, 134.977 MeV, is neutral and its own antiparticle, as a real field must be. A quantity that can be negative fits a theory of particles and antiparticles, not of one particle.