The anticommuting scalar field violates microcausality

Units are ħ = c = 1 with metric (+, −, −, −), so a negative (x − y)² means spacelike, and lengths are quoted in fm through ħc = 197.327 MeV fm. Topic 9.03's expansion stays as it was: φ(x) = ∫ d³p (2π)⁻³ (2Eₚ)^(−1/2) [aₚ e^(−ip⋅x) + bₚ† e^(ip⋅x)], with p⁰ = Eₚ = √(p² + m²). It solves the Klein-Gordon equation whatever the operators obey, because the equation is linear and acts only on the plane waves. What changes is the algebra. Write [A, B]₊ = AB + BA and impose [aₚ, aₖ†]₊ = [bₚ, bₖ†]₊ = (2π)³δ³(p − k), with every other pair, a with b included, anticommuting to zero. Two consequences follow at once. [bₚ†, bₚ†]₊ = 0 means (bₚ†)² = 0, so no mode holds two quanta: these would be spin-0 fermions. And because a and b anticommute with each other, the cross terms aₚbₖ + bₖaₚ in φφ† + φ†φ cancel, so [φ(x), φ†(y)]₊ is a c-number waiting to be computed.

Two pairs survive in [φ(x), φ†(y)]₊. The particle pair gives [aₚ, aₖ†]₊ e^{(−ip⋅x + ik⋅y)}, and the k integral returns ∫ d³p (2π)⁻³ (2Eₚ)⁻¹ e^{(−ip⋅(x−y))} = D(x − y), topic 9.04's amplitude. The antiparticle pair gives [bₚ†, bₖ]₊ e^{(ip⋅x − ik⋅y)}, and the whole topic sits in its sign: an anticommutator is symmetric, [bₚ†, bₖ]₊ = +[bₖ, bₚ†]₊, where a commutator gives [bₚ†, bₖ] = −[bₖ, bₚ†]. So [φ(x), φ†(y)]₊ = D(x − y) + D(y − x), derived here, against the commutator's D(x − y) − D(y − x). Topic 9.04 showed that D(x − y) = D(y − x) for spacelike x − y, because a continuous, proper orthochronous Lorentz transformation, under which D is invariant, carries x − y into −(x − y) there; for timelike x − y no such transformation exists. The equality that made the commutator vanish now doubles the anticommutator: 2D = mK₁(mσ)/(2π²σ), with σ = √(−(x − y)²) and the closed form quoted from 9.04. K₁ is positive, so this is positive at every spacelike σ, and the same holds for a real scalar quantised this way.

Could a different normalisation rescue the rule? Set [bₚ, bₖ†]₊ = w(2π)³δ³(p − k) with w real. The antiparticle term enters as +wD(y − x), so the spacelike anticommutator is (1 + w)D, zero only at w = −1. Now read w. If bₚ annihilates the vacuum, the state bₖ†|0⟩ has ⟨0|bₚbₖ†|0⟩ = w(2π)³δ³(p − k); if bₚ† annihilates it instead, the state bₖ|0⟩ has ⟨0|bₚ†bₖ|0⟩ = w(2π)³δ³(p − k) again. Either way w is the antiparticle's norm, and the only causal value is negative. With commutators the same bookkeeping gives coefficient −w, so w = +1 cancels with a positive norm when bₚ|0⟩ = 0. The canonical side tells the same story. With π = ∂ₜφ†, differentiating gives [φ(t, x), π(t, y)]₊ = (i/2)(1 − w)δ³(x − y): zero for w = 1, and the canonical iδ³ only for w = −1. With positive norms the anticommuting scalar is not even a canonical theory.

Fields are not observables, and the Dirac field shows that fields failing to commute at spacelike separation do no harm, provided the observables built from them commute. So test the observables. Expanding the right side of [AB, CF] = A[B, C]₊F − AC[B, F]₊ + [A, C]₊FB − C[A, F]₊B gives eight terms, six of which cancel, leaving ABCF − CFAB. Put A = φ†(x), B = φ(x), C = φ†(y), F = φ(y). [φ(x), φ(y)]₊ and [φ†(x), φ†(y)]₊ vanish identically, and the other two both equal S = [φ(x), φ†(y)]₊, so [φ†φ(x), φ†φ(y)] = S[φ†(x)φ(y) − φ†(y)φ(x)], derived. The bracket is no zero operator: in the state |k⟩ = aₖ†|0⟩, with the vacuum chosen in the next section and a box of volume V, its expectation is i sin(k⋅(x − y))/(EₖV). At spacelike separation S = 2D ≠ 0, so the density at x fails to commute with the density at y, and a measurement at one event can change the statistics of one outside its light cone. Under commutators the parallel identity puts [B, C], [B, F], [A, C] and [A, F] in those places, and all four vanish there.

Now the energies. With φ† written to the left of φ, as H = ∫ d³x (∂ₜφ†∂ₜφ + ∇φ†⋅∇φ + m²φ†φ) reads, topic 9.03's mode sum is H = ∫ d³p (2π)⁻³ Eₚ(aₚ†aₚ + bₚbₚ†). If bₚ annihilates the vacuum, bₚbₚ† = −bₚ†bₚ + (2π)³δ³(0) gives each b-quantum energy −Eₚ, and with infinitely many modes the energy has no lower bound. But the symmetric anticommutator lets dₚ ≡ bₚ† take over: [dₚ, dₖ†]₊ = (2π)³δ³(p − k) is the same algebra, dₚ|0⟩ = 0 defines the vacuum, and H = ∫ d³p (2π)⁻³ Eₚ(aₚ†aₚ + dₚ†dₚ), every quantum adding +Eₚ with positive norm. Topic 9.06 found no such escape for the Dirac field, because [b†, b] = −[b, b†] turns the same relabelling into a negative norm. Side by side, with the spin-½ entries quoted from 9.06 and 9.07: spin 0 with commutators and spin ½ with anticommutators pass both tests, spin ½ with commutators fails positivity, and spin 0 with anticommutators passes positivity and fails causality. Only spin 0 and spin ½ are derived here; Pauli's 1940 theorem for free fields of every spin, quoted in topic 9.09, extends the pattern: integer spin with Fermi statistics breaks causality, and half-integer spin with Bose statistics breaks positivity.

Dimensions: with ħ = c = 1 the action ∫ d⁴x ∂φ†⋅∂φ is a pure number, so φ carries mass and D carries mass², and mK₁(mσ)/(2π²σ) is a mass over a length, mass² again; 1 fm⁻² = 0.03894 GeV². Massless limit: K₁(z) → 1/z, so 2D → 1/(2π²σ²), a power law reaching every distance. Large σ: 2D → √m e^(−mσ)/(2πσ)³⁄², shrinking over ħ/mc but never reaching zero. For a scalar of the charged pion's measured mass, 139.57 MeV, ħ/mc = 1.414 fm, and the equal-time anticommutator is 0.0371 fm⁻² at 1 fm, 0.00563 fm⁻² at 2 fm and 1.53 × 10⁻⁴ fm⁻² at 5 fm, where the asymptotic form already lies within 10%. The commutator rule gives zero at all three. Nature takes the side the argument predicts: charged pions, measured to have spin 0, show an excess of like-sign pairs at small relative momentum, an effect first identified in antiproton annihilation by Goldhaber, Goldhaber, Lee and Pais in 1960. That is Bose enhancement; an anticommuting pion field would have produced a Pauli suppression instead.