ψ̄ and the sixteen bilinear covariants

Units throughout: SI, with x^μ = (ct, x), metric g = diag(1, −1, −1, −1) and ψ in m^(−3/2), so every bilinear is a density in m⁻³. Quoted from topic 4.03 in one line each: the Clifford algebra, γ^μ γ^ν + γ^ν γ^μ = 2g^μν; and the trace identities, tr 1 = 4, the trace of an odd number of gammas is zero, tr(γ^μ γ^ν) = 4g^μν and tr γ⁵ = 0. A density must be quadratic in ψ, and the first guess, ψ†Γψ, becomes ψ†(S†ΓS)ψ. For a rotation S is unitary, but a boost is a Hermitian stretch with S†S = cosh η + α⋅n̂ sinh η (topic 5.04), and S†ΓS bears no clean relation to Γ. Topic 5.03 found what replaces unitarity, γ⁰S†γ⁰ = S⁻¹. Put γ⁰ beside ψ†: ψ̄′ = ψ†S†γ⁰ = ψ†γ⁰(γ⁰S†γ⁰) = ψ̄S⁻¹, derived. Every density then obeys ψ̄′Γψ′ = ψ̄(S⁻¹ΓS)ψ, so the matrix alone decides how it transforms. Γ = 1 is a scalar at once, but not a probability: the rest spinor (1, 0, 0, 0) has ψ̄ψ = ψ†ψ, while the negative-energy rest spinor (0, 0, 1, 0) has ψ̄ψ = −ψ†ψ.

Take Γ = γ^μ. The covariance condition S⁻¹ γ^μ S = Λ^μ_ν γ^ν turns the sandwich rule into ψ̄′ γ^μ ψ′ = Λ^μ_ν ψ̄ γ^ν ψ, so j^μ = cψ̄ γ^μ ψ is a four-vector, derived. Since γ⁰γ⁰ = 1 and γ⁰γⁱ = αᵢ, its parts are j⁰ = cψ†ψ and j = cψ†αψ, Unit 4's density and current. So ψ†ψ is a time component and must grow under a boost: a plane wave at 0.60c has ψ†ψ = 1.25 ψ̄ψ, and |j|/j⁰ = 0.60 is its speed over c, so j = ρv holds in every frame, as it did non-relativistically. Now integrate. With ∂_μ j^μ = 0 (Unit 4), take the four-volume between the slice t = 0 of one frame and the slice t′ = 0 of another: the divergence integrates to zero, the sides at spatial infinity carry nothing, so equal flux crosses both slices, and because j^μ is a four-vector the flux through t′ = 0 is ∫ψ′†ψ′ d³x′. So ∫ψ†ψ d³x is the same in every frame, derived: density up by γ, length down by γ. Measured: helium and H₂ are neutral to about 10⁻²⁰ e (King, 1960), while if charge followed ∫ψ̄ψ d³x, which for a bound electron falls below 1 by about its kinetic energy over mc², helium's two electrons, with 79.0 eV of kinetic energy between them, would by themselves come about 1.5 × 10⁻⁴ e short.

Any product of gammas can be reordered with γ^μ γ^ν = −γ^ν γ^μ for μ ≠ ν and shortened with (γ^μ)² = g^μμ = ±1, no sum, until it is ±1 or ±i times a product of distinct gammas in increasing order. Such products are labelled by the subsets of the four indices 0, 1, 2, 3, so there are 2⁴ = 16: the empty product, four with one factor, six with two, four with three and one with four. Renamed, the two-factor products are −iσ^μν; the four-factor product is γ⁰γ¹γ²γ³ = −iγ⁵, and each three-factor product is γ^μ γ⁵ up to a phase, for example γ¹γ²γ³ = −iγ⁰γ⁵. Every one squares to ±1: (γ⁵)², (σ¹²)² and (γ¹γ⁵)² are 1, while (γ¹)², (σ⁰¹)² and (γ⁰γ⁵)² are −1. And every one except 1 is traceless by the quoted identities: γ^μ and γ^μ γ⁵ hold an odd number of gammas, σ^μν with μ ≠ ν has trace proportional to g^μν = 0, and tr γ⁵ = 0. Splitting a two-factor product into its symmetric and antisymmetric halves gives γ^μ γ^ν = g^μν − iσ^μν: a number plus a tensor, with the i/2 in σ^μν chosen so that σⁱʲ = εᵢⱼₖΣₖ is Hermitian with eigenvalues ±1.

The product of two different members is labelled by the symmetric difference of two different subsets, which is never empty, so Γ_A Γ_B is, whenever A ≠ B, a phase times a member other than 1, with zero trace. With Γ^A = (Γ_A)⁻¹ = ±Γ_A this gives tr(Γ^A Γ_B) = 4δ^A_B, derived. If Σ c_A Γ_A = 0, multiplying by Γ^B and tracing gives 4c_B = 0, so the sixteen are independent, and sixteen independent matrices in the sixteen-dimensional space of 4 × 4 matrices are a basis: M = ¼Σ tr(M Γ^A) Γ_A for every M, derived. Apply it to M = ψψ̄, whose trace against Γ^A is ψ̄ Γ^A ψ: ψψ̄ = ¼[ψ̄ψ + (ψ̄γ⁵ψ)γ⁵ + V_μ γ^μ − A_μ γ^μ γ⁵ + ½T_μν σ^μν]. The sixteen bilinears are the coordinates of one matrix. For (1, 0, 0, 0), ψψ̄ = diag(1, 0, 0, 0) = ¼(1 + γ⁰ + γ³γ⁵ + σ¹²), so ψ̄ψ = V⁰ = A³ = T¹² = 1 and the other twelve vanish. And a matrix commuting with all four γ^μ commutes with all sixteen, so with every matrix: it is a multiple of 1, the uniqueness topics 5.02 and 5.06 lean on.

Insert SS⁻¹ between neighbouring gammas: S⁻¹ γ^μ γ^ν S = (S⁻¹ γ^μ S)(S⁻¹ γ^ν S) = Λ^μ_α Λ^ν_β γ^α γ^β. A product of gammas thus transforms with one Λ per index, and σ^μν is an antisymmetric tensor, derived. Write γ⁵ = −(i/4!)ε_μνρσ γ^μ γ^ν γ^ρ γ^σ with ε₀₁₂₃ = −1: four factors of Λ contracted into ε give det Λ times ε, the definition of a determinant, so S⁻¹γ⁵S = (det Λ)γ⁵ for any Λ that has an S, derived. For rotations and boosts det Λ = 1, as a second route confirms: γ⁵ anticommutes with every γ^μ and so commutes with every σ^μν and every S. Hence iψ̄γ⁵ψ is invariant, ψ̄ γ^μ γ⁵ ψ a four-vector and ψ̄ σ^μν ψ a tensor. Quoting topic 5.06's S_P = γ⁰ for Λ_P = diag(1, −1, −1, −1), the same law makes iψ̄γ⁵ψ odd and sends the axial vector to (−A⁰, A) while the vector goes to (V⁰, −V). Reality fixes the last factors: (ψ̄Γψ)* = ψ̄(γ⁰Γ†γ⁰)ψ, and γ⁰Γ†γ⁰ = Γ for fifteen of the sixteen but −γ⁵ for γ⁵, so ψ̄γ⁵ψ alone is imaginary: hence the i in iψ̄γ⁵ψ.

For a free spinor with ūu = 2mc, all sixteen follow from p^μ and the spin four-vector s^μ: iūγ⁵u = 0; ū γ^μ u = 2p^μ; ū γ^μ γ⁵ u = 2mc s^μ; and ū σ^μν u = −2ε^μνρσ p_ρ s_σ, each derived by checking it at rest, because two tensors equal in one frame are equal in all. Put the spin along z and boost along z to 0.60c, so γ = 1.25 and γβ = 0.75. In units of 2mc, V goes from (1, 0, 0, 0) to (1.25, 0, 0, 0.75); A goes from (0, 0, 0, 1) to (0.75, 0, 0, 1.25), V's mirror image in the light cone, its time part γβ now marking positive helicity; and the tensor, at rest only the spin density T¹² = 1, is untouched, as a boost along z leaves B_z untouched. A transverse spin fares the other way, keeping A and growing a σ⁰ⁱ part, as the second worked example shows. The invariants survive: V_μV^μ = 1.25² − 0.75² = 1, A_μA^μ = 0.75² − 1.25² = −1, V_μA^μ = 1.25 × 0.75 − 0.75 × 1.25 = 0 and ½T_μν T^μν = (T¹²)² = 1. For any spinor the Fierz identities, quoted, give V_μV^μ = (ψ̄ψ)² + (iψ̄γ⁵ψ)², A_μA^μ = −V_μV^μ and V_μA^μ = 0. What the laws cannot do is tell 1 from γ⁵, or γ^μ from γ^μ γ⁵: each pair transforms identically under every rotation and boost. Only a transformation with det Λ = −1 can, and parity, topic 5.06, is the one that does.