The covariance condition S⁻¹ γ^μ S = Λ^μ_ν γ^ν

Observer O′ labels events by x′^μ = Λ^μ_ν x^ν with Λᵀ g Λ = g, as topic 5.01 set up; this lesson takes the passive view, one event and two labellings of it. Relativity asks O′ to write (iħ γ^μ ∂′_μ − mc)ψ′ = 0: the same m, ħ and c, and the same numerical γ^μ. That last demand is the one people skip. The γ's are not fields with values to be measured at a point, and they are not components of a vector; they are a bookkeeping of the spinor index, chosen once for every frame. So all of the frame dependence must sit in ψ. Take ψ′(x′) = S(Λ)ψ(x). It is linear, because a superposition for O is a superposition for O′; it compares ψ′ and ψ at the same event; and S does not depend on x, because Λ is the same everywhere, while an x-dependent S would let ∂′ act on S and add a term (∂′S)ψ that the Dirac equation does not contain. Quantum Mechanics II did the same for rotations: a two-component spinor at the rotated point picks up D¹⁄²(R).

Every step here is derived. The chain rule gives ∂_μ = (∂x′^ν/∂x^μ)∂′_ν = Λ^ν_μ ∂′_ν, with constant coefficients. Put this, and ψ(x) = S⁻¹ψ′(x′), into O's equation: (iħ γ^μ Λ^ν_μ ∂′_ν − mc)S⁻¹ψ′ = 0. S is constant, so it passes through ∂′; multiplying on the left by S gives (iħ S γ^μ S⁻¹ Λ^ν_μ ∂′_ν − mc)ψ′ = 0. This is O′'s Dirac equation when the matrix in front of each ∂′_ν is γ^ν (that is, when S γ^μ S⁻¹ Λ^ν_μ = γ^ν), and conjugating by S turns this into S⁻¹ γ^ν S = Λ^ν_μ γ^μ. The mass term came through untouched, because mc S S⁻¹ = mc: m is a Lorentz scalar, and every constraint on S comes from the derivative term. Dimensions check: ħ γ^μ ∂_μ ψ and mcψ both carry kg m s⁻¹ times the units of ψ, while S, Λ and γ^μ are pure numbers, so both sides of the condition are dimensionless. The count is sobering: four 4 × 4 equations, 64 complex conditions on the 16 entries of S.

Call the right-hand sides Γ^μ = Λ^μ_ν γ^ν. Their anticommutators are Λ^μ_ρ Λ^ν_σ (γ^ρ γ^σ + γ^σ γ^ρ) = 2Λ^μ_ρ Λ^ν_σ g^ρσ = 2g^μν; the last step is the definition of a Lorentz transformation. That is derived, and it makes the Γ's a second set of Dirac matrices. For a frame moving at β = 0.6 along x, Γ⁰ = 1.25γ⁰ − 0.75γ¹, and (Γ⁰)² = 1.5625 − 0.5625 = 1 because γ⁰ and γ¹ anticommute. Pauli's fundamental theorem, quoted and not proved here, says that any two sets of 4 × 4 matrices with the same anticommutators are similar, so an invertible S with S⁻¹ γ^μ S = Γ^μ exists for every Λ. Its proof is an existence argument built from the sixteen products of gammas and hides S's structure, which is why topics 5.03 and 5.04 construct S instead. One consequence is immediate: Γ⁰ is not Hermitian, since (γ¹)† = −γ¹, whereas S⁻¹ γ⁰ S would be Hermitian if S were unitary. A boost cannot act on Dirac spinors through a unitary S.

Suppose S₁ and S₂ both work. Then S₁⁻¹ γ^μ S₁ = S₂⁻¹ γ^μ S₂, so M = S₂S₁⁻¹ commutes with every γ^μ. In the Dirac representation, commuting with γ⁰ = diag(I₂, −I₂) makes M block diagonal, M = diag(A, D); commuting with each γⁱ (off-diagonal blocks σᵢ and −σᵢ) gives Aσᵢ = σᵢ D and Dσᵢ = σᵢ A, hence Aσᵢσⱼ = σᵢσⱼ A. Since σᵢσⱼ = δᵢⱼ + iεᵢⱼₖ σₖ, A commutes with all three Pauli matrices, so A is a multiple of I₂, and then D = A. So M = κI and S is fixed only up to a factor: derived. The condition is blind to κ because S enters next to S⁻¹. Setting det S = 1 trims κ to a fourth root of 1, and building S from generators, as topic 5.04 does, leaves ±S, quoted. The sign is physical. For rotations S reduces to the two-component rotation, whose −1 at 2π neutron interferometry has measured: Werner and co-workers (1975) needed a spin precession of 704° ± 38° to bring the interference pattern back, consistent with 720° and not with 360°.

S carries a spinor index running from 1 to 4, Λ a spacetime index running from 0 to 3, and the covariance condition is the one place they meet. That ψ has four components is a fact about the Clifford algebra, not about spacetime: the smallest matrices with four mutually anticommuting members are 4 × 4 (quoted), and the match with four dimensions is a coincidence, not a rule, since in 2 + 1 dimensions two components suffice and in 5 + 1 eight are needed. Watch both act on the same column. For a frame moving at β = 0.6 along x, Λ takes (1, 0, 0, 0) to (1.25, −0.75, 0, 0), mixing time with x through cosh η and sinh η. The S of topic 5.04 takes the spin-up rest spinor (1, 0, 0, 0) to (1.061, 0, 0, −0.354), mixing an upper component with a lower one through cosh(η/2) and sinh(η/2). After a 2π rotation Λ is the identity while S, as topic 5.04 shows, is −I. The four-vector you were looking for is the bilinear ψ̄γ^μ ψ, whose law topic 5.05 derives from this condition.

Two limits test the condition before any S is built, and both are derived. With Λ = I it reads S⁻¹ γ^μ S = γ^μ for every μ, so S commutes with every γ^μ and, by the uniqueness argument, S = κI: no change of frame, no change of ψ beyond a constant that the normalisation sets to 1. For a pure rotation Λ⁰_ν = δ⁰_ν, so the μ = 0 equation is S⁻¹ γ⁰ S = γ⁰ and S commutes with γ⁰ = diag(I₂, −I₂). S is then block diagonal in the Dirac representation: a rotation never mixes upper components with lower ones. That is the non-relativistic picture, in which the upper pair is the Pauli spinor of Quantum Mechanics II, and the γⁱ equations then tie each block to its two-component rotation, which topic 5.04 solves. A boost has Λ⁰ᵢ ≠ 0, so S⁻¹ γ⁰ S must acquire γⁱ terms and S must mix upper with lower components, as the β = 0.6 example did. Applied to inversion, Λ = diag(1, −1, −1, −1), the same condition is solved in topic 5.06.