Spin-statistics: what is shown and what is assumed

Units throughout are ħ = c = 1 with metric (+, −, −, −), so D carries length⁻² and distances are counted in ħ/mc, 386.16 fm for the electron. Write [A, B]_η = AB − ηBA, with η = +1 for a commutator and −1 for an anticommutator. Give each particle mode [b, b†]_η = 1 and leave the antiparticle open, [d, d†]_η = σ = ±1, so σ is the sign of ⟨0|d d†|0⟩. For the complex scalar the mode sum gives [φ(x), φ†(y)]_η = D(x − y) + [d†, d]_η D(y − x), and [d†, d]_η = −ησ. For the Dirac field Unit 4's spin sums, quoted, read Σ u ū = p̸ + m and Σ v v̄ = p̸ − m here, and since (i∂̸ₓ + m) acting on e^{(ip⋅(x−y))} returns −(p̸ − m), the same bookkeeping gives (i∂̸ₓ + m)[D(x − y) + ησD(y − x)]. Both read D(x − y) − λD(y − x) with λ = ησ(−1)²ˢ. The extra sign for spin ½ is no choice of ours: (p̸ + m)v = 0, with the positive normalisation v†v = 2Eₚ, fixes Σ v v̄, which equals (−1)²ˢ times the particle's spin sum taken at −p. Derived.

D(x − y) = ∫ d³p/((2π)³2Eₚ) e^{(−ip⋅(x−y))} is Lorentz invariant because d³p/2Eₚ is. When x − y is spacelike a proper, orthochronous Lorentz transformation carries it to −(x − y); at equal times it is a rotation by π. So outside the light cone D(x − y) = D(y − x) = D(r), and the spacelike value is (1 − λ)D(r), acted on by (i∂̸ + m) for spin ½: zero when λ = +1, twice the amplitude when λ = −1, with no third case. Inside the cone no such transformation exists, and the commutator that carries signals survives. Topic 9.04's D(r) = mK₁(mr)/(4π²r), quoted, never vanishes: 4π²r²D is 0.602 at r = ħ/mc, 0.0202 at 5ħ/mc and 1.86 × 10⁻⁴ at 10ħ/mc. Checks: m/r has dimensions length⁻², and as m → 0 it becomes the massless 1/(4π²r²). A wrong sign therefore fails at every separation. Exponentially small is not zero, and microcausality is an equality, not an estimate.

Topic 9.06's Hamiltonian, quoted, is H = ∫ d³p/(2π)³ Σₛ Eₚ(b†b − d d†) whatever the algebra. Writing d d† = ηd†d + σ, the state d†|0⟩ lies −ησEₚ above the vacuum, and for spin ½ that is λEₚ. So the sign that decides causality also decides the positron's energy: for the Dirac field they are one condition, λ = +1, and two algebras meet it. Anticommutators with σ = +1 give positive norm, positive energy and a causal field. Commutators with σ = −1, topic 9.06's [d, d†] = −1, give positrons of energy +Eₚ and a commutator that is term for term the causal anticommutator; they fail only because ⟨0|d d†|0⟩ = −1. The last choice, commutators with [d, d†] = +1, keeps the norm positive and fails twice: 1000 positrons at rest lie 511 MeV below the vacuum, with no floor, and the spacelike commutator is (i∂̸ + m)2D(r). All derived. Causality alone cannot exclude commutators for spin ½; positive norm and positive energy together do.

For the scalar (−1)²ˢ = +1 and λ = ησ, so causality again admits two algebras: commutators with σ = +1 and anticommutators with σ = −1. Positive norm picks commutators. Topic 9.08 showed the failure of the alternative directly, quoted here: with (d, d†) = +1 the spacelike anticommutator is D(x − y) + D(y − x) = 2D(r). For a field of the charged-pion mass, ħ/mc = 1.414 fm, that is 3.71 × 10⁻² fm⁻² at r = 1.0 fm and 1.43 × 10⁻³ fm⁻² at 3.0 fm, the scale of nucleon spacings. A real scalar is its own antiparticle and has no σ to choose: its anticommutator is 2D(r) outside the cone with no way out. The two halves of the result therefore lean on different assumptions, as in Pauli's 1940 paper: microcausality excludes Fermi statistics for integer spin, positive energy excludes Bose statistics for half-integer spin, and positive norm is assumed throughout. One sign, (−1)²ˢ, sits between them.

Quantum Mechanics II postulated that states of identical spin-½ particles are antisymmetric. The algebra that survived now delivers it. Because (b†ₚ, b†ₖ) = 0, b†ₖ b†ₚ|0⟩ = −b†ₚ b†ₖ|0⟩, and k = p gives (b†ₚ)² = 0: exclusion. Moving each ψ⁺ through the creators with (b, b†) = 1 turns ⟨0|ψ⁺(x₁)ψ⁺(x₂) b†ₖ b†ₚ|0⟩ into fₚ(x₁)fₖ(x₂) − fₖ(x₁)fₚ(x₂), and N particles give an N × N Slater determinant with N! terms, 6 for N = 3; the one-particle limit N = 1 is the bare wavefunction. The labels include spin, so the antisymmetry is of the whole space-and-spin state. Derived, for free fields. Measured checks: helium's lowest spin-triplet level is 1s2s ³S₁, 19.82 eV above the ground state, because the 1s² triplet a symmetric state would allow is absent; and Rasetti's 1929 Raman lines of ¹⁴N₂ alternate 2 : 1 in intensity, even J stronger, as spin-1 nuclei obeying Bose statistics require.

Six assumptions carried the argument. Lorentz invariance made D(x − y) = D(y − x) outside the cone and fixed the spin sums. Energy bounded below removed [d, d†] = +1. Positive norm removed the causal, negative-norm algebra of each spin. Microcausality removed the positive-norm anticommuting scalar. Free fields supplied the mode expansions every step ran on. Three space dimensions made spin an integer or half-integer, so (−1)²ˢ = ±1, and made a double exchange contractible, so the exchange phase is ±1. Quoted, not derived: Pauli's 1940 proof for free fields of every spin, building on Fierz in 1939; the 1958 theorems of Lüders and Zumino and of Burgoyne, which reach interacting fields from Wightman's axioms on vacuum expectation values; and the escape in two dimensions, where a double exchange winds around the partner and e^(iθ) is free, seen as θ = π/3 for ν = 1/3 quantum Hall quasiparticles in 2020. No proof from non-relativistic quantum mechanics alone is generally accepted; there, spin and statistics remain separate postulates.