Infinitesimal transformations and the generators σ^μν

Units throughout: SI, with x⁰ = ct and metric g = diag(1, −1, −1, −1). Every proper orthochronous transformation is compounded from ones arbitrarily close to the identity, so start there: Λ^μ_ν = δ^μ_ν + ω^μ_ν. Keeping Λᵀ g Λ = g to first order gives ω_μν + ω_νμ = 0, so the antisymmetry belongs to the version with both indices down, and six components are free. Index placement matters. An active rotation through δθ about z sends x → x − δθ y and y → y + δθ x, so ω¹₂ = −δθ and ω²₁ = +δθ, antisymmetric as they stand. An active boost of rapidity δη along x has ω⁰₁ = ω¹₀ = δη, a symmetric block that becomes antisymmetric only on lowering: ω₀₁ = δη = −ω₁₀. Raising both indices gives the dictionary used below, ω^ij = εᵢⱼₖ δθₖ and ω⁰i = −δηᵢ. The sizes are honest: a 1.0 mrad rotation, or the boost to a frame moving at 3.0 × 10⁵ m s⁻¹, has a parameter of 1.0 × 10⁻³.

Try S = 1 − (i/4)σ_μνω^μν with six unknown matrices σ_μν = −σ_νμ. To first order S⁻¹ = 1 + (i/4)σ_μνω^μν, so S⁻¹γ^ρS = γ^ρ + (i/4)ω_μν[σ^μν, γ^ρ], and covariance demands γ^ρ + ω^ρ_νγ^ν for every ω. That holds if [σ^μν, γ^ρ] = 2i(g^νρ γ^μ − g^μρ γ^ν): contracting with ω_μν gives −½(ω_μ^ρ γ^μ − ω^ρ_ν γ^ν), and antisymmetry turns ω_μ^ρ into −ω^ρ_μ, leaving ω^ρ_νγ^ν. Now test σ^μν = (i/2)[γ^μ, γ^ν]. Moving γ^ρ through γ^μγν with the anticommutator twice gives [γ^μγν, γ^ρ] = 2g^νρ γ^μ − 2g^μρ γ^ν; antisymmetrising in μν doubles it, and the factor i/2 completes the required commutator. Only (γ^μ, γ^ν) = 2g^μν was used, so the result is derived for every representation. Any other solution differs by a matrix commuting with all four γ^ρ, a multiple of 1 by topic 5.02, which merely rescales S.

In the Dirac representation γ⁰ = diag(1, −1) and γⁱ carries σᵢ above the diagonal and −σᵢ below, in 2 × 2 blocks. Then γ^iγj = −σᵢσⱼ in both diagonal blocks, so σ^ij = εᵢⱼₖ Σₖ with Σₖ = diag(σₖ, σₖ), while σ⁰i = iγ⁰γⁱ = iαᵢ. With the dictionary for ω, S = 1 − (i/2)δθ⋅Σ + ½δη⋅α, which is 1 − iδθ⋅J − iδη⋅K with J = Σ/2 and K = (i/2)α. Multiplying blocks checks topic 5.01's algebra directly: [Jₓ, J_y] = iJ_z, [Jₓ, K_y] = iK_z, and [Kₓ, K_y] = −¼[αₓ, α_y] = −iJ_z, so the spinor matrices are a representation of the same group. The rotation part is two copies of Quantum Mechanics II's 1 − (i/2)δθ⋅σ, the limit any relativistic spinor must reach, and J_z has eigenvalues ±½: a spin of ħ/2 = 5.27 × 10⁻³⁵ J s, the two-beam split Stern and Gerlach measured. Nobody inserted the ½; the Clifford algebra fixed it. Boosts carry half their rapidity too, the seed of topic 5.04's half angles.

Σ is Hermitian, so a spinor rotation is unitary. α is Hermitian as well, so K = (i/2)α is anti-Hermitian and a boost S = 1 + (δη/2)α⋅n̂ is Hermitian, with S†S = 1 + δη α⋅n̂. The density therefore shifts, δ(ψ†ψ) = δη ψ†(α⋅n̂)ψ; since cψ†αψ is Unit 4's current, that is j′⁰ = j⁰ + δη n̂⋅j, the first-order law for a time component. For an electron at 0.60c, a boost of δη = 0.010 along its motion raises ψ†ψ by 0.60%; across its motion, by nothing at first order; at rest there is no current and the change is of order δη². As c → ∞ a small boost has δη = δv/c, so δρ/ρ = vδv/c² vanishes: the Galilean result. No cleverer S is unitary. Under a trace, [J_z, Kₓ] = iK_y and [Kₓ, K_y] = −iJ_z give tr(K_y²) = −tr(J_z²) in any finite-dimensional representation, derived; Hermitian J and K would make the left side non-negative and the right side non-positive, forcing both to vanish and then J = K = 0. The full theorem, that no non-trivial finite-dimensional representation of the Lorentz group is unitary, is quoted.

Unit 4's Hermiticity rule γ^μ† = γ⁰γ^μγ⁰ gives σ^μν† = (−i/2)[γ^ν†, γ^μ†] = γ⁰σ^μνγ⁰, so γ⁰σ^μν†γ⁰ = σ^μν for all six generators, derived. Hence γ⁰S†γ⁰ = 1 + (i/4)σ_μνω^μν, which is S⁻¹ at first order, and because γ⁰(S₁S₂)†γ⁰ = (γ⁰S₂†γ⁰)(γ⁰S₁†γ⁰) = S₂⁻¹S₁⁻¹ the relation survives every product, so it holds for each finite S topic 5.04 builds. Its first consequence: with ψ̄ = ψ†γ⁰, ψ̄′ψ′ = ψ†S†γ⁰Sψ = ψ̄(γ⁰S†γ⁰)Sψ = ψ̄ψ. For the 0.60c electron, ψ̄ψ = ψ†ψ/γ = 0.80 ψ†ψ and it stays there under every boost while ψ†ψ moves; the two differ at order v²/c², because the lower components are of order v/c, and agree in the non-relativistic limit. Topic 5.05 builds the full table of bilinears from this one relation.

Wigner's theorem, quoted here, demands that a symmetry connected to the identity, as every rotation and boost is, act unitarily on the Hilbert space. That space is not the four spinor components at one point. It is the space of wave functions, with ⟨ψ|ψ⟩ = ∫ψ†ψ d³x taken over a whole slice of constant time, and the operator representing Λ there sends ψ(x) to Sψ(Λ⁻¹x): it moves the argument and the slice as well as the components, so the rise of ψ†ψ by γ is paid for by the contraction of the volume it fills. The trace argument cannot be run in that space, because the generators become unbounded differential operators with no finite traces, which is why the non-trivial unitary representations the quoted theorem allows are all infinite-dimensional. That ∫ψ†ψ d³x is the same in every frame is shown in topic 5.05, from the conserved current. Measured, it is charge invariance: electrons move at different speeds in an H₂ molecule and a He atom, yet both are neutral to about 10⁻²⁰ e (King, 1960).