γ⁵, chirality and the massless Weyl equations

Units are SI, with x⁰ = ct, ∂_μ = ∂/∂x^μ and metric g = diag(1, −1, −1, −1). Topic 5.05 defined γ⁵ = iγ⁰γ¹γ²γ³, and its properties follow from γ^μ γ^ν + γ^ν γ^μ = 2g^μν by counting swaps. Carrying any γ^μ through the product passes three gammas that anticommute with it and one equal to it, so γ⁵γ^μ = −γ^μ γ⁵. Bringing equal factors together in (γ⁰γ¹γ²γ³)² takes 3 + 2 + 1 = 6 swaps, so it equals (γ⁰)²(γ¹)²(γ²)²(γ³)² = −1 and (γ⁵)² = 1. Reversing the order for the adjoint also takes six swaps, and with γ⁰ Hermitian and each γⁱ anti-Hermitian that gives γ⁵† = γ⁵. Because γ⁵ = −γ⁰γ⁵γ⁰, its trace equals minus itself and vanishes, so the eigenvalues +1 and −1 occur twice each, and P_R = (1 + γ⁵)/2 and P_L = (1 − γ⁵)/2 are complementary projectors. All derived. In the Dirac representation γ⁵ has off-diagonal blocks of 1: it swaps large and small components, so a slow electron's ⟨γ⁵⟩ is of order v/c, and Pauli theory never sees chirality.

Since γ⁵ anticommutes with every γ^μ it commutes with any product of two, hence with σ^μν = (i/2)(γ^μ γ^ν − γ^ν γ^μ) and with every S(Λ) = exp(−(i/4)σ_μν ω^μν) of topics 5.03 and 5.04. A spinor with P_R ψ = 0 in one frame has it in every frame, derived: chirality is a label no observer can change. In Unit 4's Weyl representation γ⁵ = diag(−1, 1) in 2 × 2 blocks, so ψ = (ψ_L, ψ_R), and the generators fall into the same blocks: J = Σ/2 = diag(σ/2, σ/2), but K = (i/2)α = diag(−iσ/2, +iσ/2). Exponentiating, a rotation multiplies both halves by exp(−(i/2)φσ⋅n̂), while a boost gives exp(−(η/2)σ⋅n̂) on ψ_L and exp(+(η/2)σ⋅n̂) on ψ_R, derived. Both halves carry spin ½ but stretch oppositely: boosting an electron at rest to 0.800c along its spin, rapidity ln 3, multiplies ψ_R by √3 = 1.732 and ψ_L by 1/√3 = 0.577, so their weights go from 50:50 to 90:10. They are the two inequivalent two-component representations of the Lorentz group, labelled (½, 0) and (0, ½) in a classification quoted here.

Unit 4's Hamiltonian in Weyl blocks is H = [[−cσ⋅p, mc²], [mc², cσ⋅p]], so a plane wave of energy E obeys (E + cσ⋅p)ψ_L = mc²ψ_R and (E − cσ⋅p)ψ_R = mc²ψ_L; covariantly iħσ̄^μ ∂_μ ψ_L = mcψ_R and iħσ^μ ∂_μ ψ_R = mcψ_L, with σ^μ = (1, σ) and σ̄^μ = (1, −σ). Only the mass crosses between halves. So chirality is not conserved: γ⁵ commutes with each αᵢ but anticommutes with β, giving [H, γ⁵] = 2mc²βγ⁵, or ∂_μ(ψ̄γ^μ γ⁵ψ) = (2mc/ħ)iψ̄γ⁵ψ, both sides in m⁻⁴, derived. For a helicity state, σ⋅p̂ξₕ = hξₕ, the equations fix the mixture: with u†u = 2E/c, |u_R|² = (E + hpc)/c and |u_L|² = (E − hpc)/c, so ⟨γ⁵⟩ = hβ. At rest the halves are equal, √(mc)(ξ, ξ); at 10.0 MeV of kinetic energy the wrong chirality carries (1 − β)/2 = 5.91 × 10⁻⁴ of the density, matching (mc²/2E)². The product ūu = 2 Re(u_L†u_R) = 2mc is the same in every frame.

With m = 0 the halves decouple into Weyl's two-component equations, Eψ_R = cσ⋅pψ_R and Eψ_L = −cσ⋅pψ_L, each covariant on its own because S(Λ) is block-diagonal. Since (σ⋅p)² = p², E = ±pc, the m → 0 limit of E² = p²c² + m²c⁴: every solution moves at c, and no observer can overtake it. On the positive-energy solution σ⋅p̂ψ_R = +ψ_R and σ⋅p̂ψ_L = −ψ_L, so a right-chiral massless particle has helicity +1 and a left-chiral one −1, derived. The result does not need the Weyl blocks. Multiplying them out shows γ⁵α = Σ, a matrix identity and therefore true in every representation, so with H = cα⋅p, Σ⋅p̂ψ = (E/pc)γ⁵ψ on every massless plane wave: helicity is chirality times the sign of the energy, and on negative-energy solutions the two are opposite, which Unit 8 reinterprets. A Weyl field holds one helicity of particle, half a Dirac field's states; the massive equation needs both halves only because m couples them. The neutrino, observed only with helicity −1, is the case in point, named here only.

Topic 5.06's S_P = γ⁰ anticommutes with γ⁵, so γ⁰P_L = P_R γ⁰: the mirror image of a left-chiral spinor is right-chiral, derived. In the Weyl representation γ⁰ has off-diagonal blocks of 1 and does it literally, ψ′_L(t, x) = η_P ψ_R(t, −x) and ψ′_R(t, x) = η_P ψ_L(t, −x). Check it on Weyl's equations: inversion reverses p but not the axial spin σ, so Eψ = −cσ⋅pψ becomes Eψ = +cσ⋅pψ, the other equation, and helicity −1 becomes +1. A single Weyl field therefore cannot be parity invariant, because its mirror image obeys an equation it does not. The massive Dirac equation is invariant precisely because the mass couples ψ_L to ψ_R and ψ_R to ψ_L with the same mc², so the swap carries the pair of equations into itself. The comparison with topic 5.06 is close but not exact. A scalar and a pseudoscalar respond identically to every proper transformation, so only P tells them apart; ψ_L and ψ_R already respond oppositely to boosts, by exp(∓(η/2)σ⋅n̂), and what no proper transformation can do is carry one into the other. P does exactly that. A coupling that sees ψ_L alone is maximally parity-violating, which is what the weak interaction's V − A form, quoted in topic 5.06, does.

Write ψ_L = P_L ψ, whose adjoint is ψ̄_L = (P_L ψ)†γ⁰ = ψ̄ P_R, since γ⁵ is Hermitian and anticommutes with γ⁰. Then ψ̄_L Γ ψ_R = ψ̄ P_R Γ P_R ψ. Pushing P_R through an odd number of gammas turns it into P_L, and P_L P_R = 0; an even number leaves it P_R. So ψ̄γ^μ ψ = ψ̄_L γ^μ ψ_L + ψ̄_R γ^μ ψ_R, while ψ̄ψ = ψ̄_L ψ_R + ψ̄_R ψ_L, derived; in Weyl blocks ψ̄γ^μ ψ = ψ_L†σ̄^μ ψ_L + ψ_R†σ^μ ψ_R and ψ̄ψ = 2 Re(ψ_L†ψ_R). Vector and axial couplings never connect opposite chiralities; scalar, pseudoscalar and tensor couplings connect only opposite ones. The photon couples to the vector current, so it cannot flip chirality, and at high energy a helicity flip costs an amplitude of order mc²/E, the rule topic 5.09 uses to switch off backward Coulomb scattering. A coupling to ψ_L alone, as in β decay, with nothing else picking a spin direction, emits an electron of helicity h with probability (1 − hβ)/2, so ⟨σ⋅p̂⟩ = −v/c: the polarisation measured in β decay since 1957.