Microcausality: [φ(x), φ†(y)] = 0 at spacelike separation

Natural units ħ = c = 1 throughout, with ħc = 197.327 MeV fm turning masses into inverse lengths, and the metric (+, −, −, −), so x² = t² − r² is negative outside the light cone. Topic 9.03's field, quoted in one line and taken from its box of volume V to infinite volume (Σₖ → V∫ d³p/(2π)³, √V aₖ → aₚ, √V bₖ → bₚ), is φ(x) = ∫ d³p (aₚ e^(−ip⋅x) + bₚ† e^(ip⋅x)) / ((2π)³ √(2Eₚ)), with [aₚ, a_q†] = [bₚ, b_q†] = (2π)³ δ³(p − q). Acting on the vacuum, φ†(y) can only create a particle at y, because its b-term annihilates |0⟩, and φ(x) can only absorb it. So ⟨0|φ(x)φ†(y)|0⟩ = ∫ d³p e^{(−ip⋅(x−y))} / ((2π)³ 2Eₚ) ≡ D(x − y). Each piece is Lorentz invariant: p⋅(x − y) is a scalar, and d³p/2Eₚ = d⁴p δ(p² − m²) θ(p⁰), where θ(p⁰) survives because a proper orthochronous transformation never flips the sign of a timelike p⁰. D therefore depends only on (x − y)² and, inside the cone, on whether x is later or earlier than y.

At equal times and separation r the angular integral gives D(0, r) = (1/(4π²r)) ∫₀^∞ p sin(pr) / Eₚ dp, which needs a convergence factor e^(−εp). Extend the even integrand to the whole real line, replace sin(pr) by e^(ipr)/i, which the symmetry allows, and close the contour in the upper half plane: the only obstruction is the branch cut of Eₚ = √(p² + m²) rising from p = im. Wrapping it with p = iρ gives D(0, r) = (1/(4π²r)) ∫ from m to ∞ of ρ e^(−ρr) / √(ρ² − m²) dρ, every piece of which decays at least as e^(−mr); a standard representation of K₁, quoted, makes this mK₁(mr)/(4π²r). Checks: the dimension is mass over length, mass² here; as m → 0, K₁(z) → 1/z returns the massless 1/(4π²r²); for mr ≫ 1 the endpoint ρ ≈ m dominates and gives √(πm/2) e^(−mr)/(4π² r³⁄²). For the charged pion, measured at mc² = 139.570 MeV, ħ/mc = 1.414 fm, and at r = 3.0 fm the amplitude is 0.25 of its massless value: positive, real, and not zero.

Now commute the fields instead of sandwiching them between vacua. The a-terms give [aₚ, a_q†] = (2π)³ δ³(p − q) times e^{(−ip⋅x + iq⋅y)}, which integrates to D(x − y). The b-terms give [bₚ†, b_q] = −(2π)³ δ³(p − q) times e^{(ip⋅x − iq⋅y)}, which integrates to −D(y − x). Cross terms vanish because a and b commute. So [φ(x), φ†(y)] = D(x − y) − D(y − x), a c-number with the same value in every state, while [φ(x), φ(y)] = 0 identically. Read it as two processes: a particle created at y and absorbed at x, minus an antiparticle created at x and absorbed at y. Since D(y − x) = D(x − y)*, the difference is 2i Im D(x − y) = iΔ(x − y), with Δ the real Pauli-Jordan function. One more check comes free: differentiating in x⁰ at equal times gives −i∫ d³p e^(ip⋅r)/(2π)³ = −iδ³(r), half from each term: topic 9.03's canonical [φ(x), π(y)] = iδ³(x − y), read with the derivative on x rather than y.

Take x − y spacelike, (t, r) along one axis with |t| < r. A boost at velocity t/r makes the events simultaneous at separation s = √(r² − t²); a rotation by π about a perpendicular axis turns (0, s) into (0, −s); the inverse boost then lands on (−t, −r). All three are proper and orthochronous, so D, which depends only on (x − y)² and the time order inside the cone, takes the same value at x − y and y − x, and the commutator vanishes. At equal times the step is immediate, since D(0, r) depends on |r| alone. Inside the cone x − y and y − x lie in the future and past halves, and no orthochronous transformation exchanges them. Continuing s → iλ gives, quoted, D = m[Y₁(mλ) + iJ₁(mλ)]/(8πλ) for x⁰ > 0, so Δ = mJ₁(mλ)/(4πλ) inside, with −δ(x²)/(2π) on the cone. It first vanishes at mλ = 3.832, at λ = 5.42 fm for the pion. As m → 0 only the δ survives: the massless commutator lives on the cone itself.

The cancellation needs both terms, and the second exists only because φ contains b†. A field of positive frequencies alone, φ₊ = ∫ d³p aₚ e^(−ip⋅x) / ((2π)³ √(2Eₚ)), has [φ₊(x), φ₊†(y)] = D(x − y), which for a pion-mass field 3.0 fm apart at equal times is 7.1 × 10⁻⁴ fm⁻², real and positive. Its time derivative at equal times is −(i/2)δ³(r), half the canonical value, and the antiparticle supplies the other half. No reshuffling of positive-frequency modes helps: such an amplitude is the boundary value of a function analytic in complex time, and, quoted, it cannot vanish across the spacelike region without vanishing everywhere. The real field of topic 9.02 makes the same subtraction with aₚ† playing both roles. Once the elementary commutators vanish, Leibniz's rule [AB, C] = A[B, C] + [A, C]B extends the result to every polynomial in φ, φ† and their derivatives at x against any at y: the charge density, the energy density, φ†φ. An operation exp(iεO(y)) then leaves every ⟨O′(x)⟩ at spacelike separation unchanged, to all orders in ε.

D itself does not vanish, and it should not: it is a correlation of the vacuum across spacelike separation, the field cousin of an entangled pair, and like one it carries no signal. Its range is set by a measured mass. For charged pions it is ħ/mc = 1.41 fm, the length that sets the e^{(−r/1.41 fm)} tail of charged-pion exchange between nucleons, quoted; the neutral pion, at 134.977 MeV, gives 1.46 fm. For an electron-mass field it is 386 fm, and across an atomic 0.1 nm the factor e^(−mr) is e⁽⁻²⁵⁹⁾ ≈ 3 × 10⁻¹¹³, which is why atomic physics never meets it; as c → ∞ the range ħ/mc goes to zero altogether. The same two terms will return: ⟨0|T φ(x)φ†(y)|0⟩ = θ(x⁰ − y⁰)D(x − y) + θ(y⁰ − x⁰)D(y − x), the Feynman propagator, is only named here, and its frame-dependent time ordering is harmless precisely because the two pieces agree at spacelike separation. Everything here assumes free fields quantised with commutators; topic 9.08 shows what anticommutators do to this cancellation.