Finite spinor rotations and boosts

Units are SI with c kept, and the metric is g = diag(1, −1, −1, −1). A rotation through φ about a fixed axis n̂ is N rotations through φ/N, each 1 − (i/2)(φ/N)Σ⋅n̂ by topic 5.03. Steps about one axis commute, so as N → ∞ their product tends to S_R = exp(−(i/2)φΣ⋅n̂), and likewise S_B = exp(½η α⋅n̂) for a boost of rapidity η = artanh(v/c). The series close because the Clifford algebra makes both matrices square to one: (α⋅n̂)² = ½nᵢnⱼ(αᵢαⱼ + αⱼαᵢ) = nᵢnⱼδᵢⱼ = n̂⋅n̂ = 1, and the same for Σ⋅n̂. Even powers are 1 and odd powers the matrix itself, so the cosine and sine series, or the hyperbolic ones, can be read off: S_R = cos(φ/2) − iΣ⋅n̂ sin(φ/2) and S_B = cosh(η/2) + α⋅n̂ sinh(η/2), both derived. Each is dimensionless, as φ and η are, and each falls back to 1 plus topic 5.03's first-order term as φ or η → 0. Parameters about one axis add: S_B(η₁)S_B(η₂) = S_B(η₁ + η₂), so rapidities add although velocities do not.

Test the closed form against topic 5.02's covariance condition at a finite angle. For n̂ = ẑ the generator is Σ_z = iγ¹γ², which anticommutes with γ¹ and γ² and commutes with γ⁰ and γ³. Carrying γ¹ through S_R therefore reverses the sign of its exponent: S_R⁻¹γ¹S_R = e^(iφΣ_z)γ¹ = γ¹ cos φ − γ² sin φ, while γ⁰ and γ³ pass through untouched. That is Λ^μ_νγ^ν for an active rotation through the whole of φ, so the finite matrix satisfies covariance exactly (derived). The half angle sits in S, which acts once on ψ; a bilinear ψ̄Γψ carries S on both sides and turns through φ. Rotate an electron at rest with spin along +x through 90°: its two upper components gain phases of −45° and +45°, their relative phase becomes 90°, and ψ†Σψ points along +y. S_R is block-diagonal in every frame, two copies of Quantum Mechanics II's e^{(−iφσ⋅n̂/2)}, so on a rest spinor it is exactly that two-component rotation, the limit a relativistic spinor must reach. The ½ is the eigenvalue of J = Σ/2 from topic 5.03, not a choice made here.

At φ = 2π the closed form gives S_R = cos π = −1 about every axis, while Λ is the identity. Covariance cannot object, since S and −S give the same S⁻¹γ^μS. Then why not declare S = +1 for every full turn? Because the composition law S(Λ₁)S(Λ₂) = S(Λ₁Λ₂) forbids it. A half-turn has S_R(π) = −iΣ⋅n̂, and whichever sign it is given its square is −(Σ⋅n̂)² = −1, although two half-turns make the identity. So the identity Λ is reached with both +1 and −1, and the map from Λ to S is two-valued (derived); the matrices S form SL(2, C), the double cover of the proper orthochronous Lorentz group (quoted). In the parameter φ the spinor closes its loop only at 4π, where S_R = cos 2π = +1. The topological reason, quoted here rather than proved, is that in SO(3) the loop of one full turn cannot be shrunk to a point while the loop of two turns can. A spinor field is therefore defined only up to this sign, which no bilinear can see, as the previous section showed, but which matters wherever two parts of one state are compared.

Divide a spin-½ beam into two equal halves, rotate one through φ, and recombine. The count rate goes as |ψ + S_Rψ|², so with ⟨ψ|ψ⟩ = 1 it is I/Iₘₐₓ = ½[1 + Re⟨ψ|S_Rψ⟩], and ⟨ψ|S_Rψ⟩ = cos(φ/2) − i⟨Σ⋅n̂⟩ sin(φ/2). The spin enters only the imaginary part, so I/Iₘₐₓ = ½[1 + cos(φ/2)] for every polarisation, and for an unpolarised beam too (derived). The period is 4π: at φ = 2π the spin is back where it started, yet the recombined beam is dark. Neutrons make the test practical, since a field B precesses their spin through φ = γₙBt with γₙ = 1.832 × 10⁸ rad s⁻¹ T⁻¹. The neutron is no Dirac point particle, but S_R depends only on the representation, not on the dynamics. Werner and co-workers found 704 ± 38° in 1975, and Rauch and co-workers 716.8 ± 3.8° in 1978 (both quoted): 0.8 standard deviations from the 720° derived here, and 94 from the 360° a vector would need.

α⋅n̂ is Hermitian, so S_B = exp(½η α⋅n̂) is Hermitian too, with real eigenvalues e^(η/2) and e^(−η/2), each twice. Their product is 1, so det S_B = 1, as for S_R, but no eigenvalue has modulus 1 unless η = 0, so S_B is not unitary: S_B†S_B = cosh η + α⋅n̂ sinh η. What survives is topic 5.03's pseudo-unitarity, S_B⁻¹ = γ⁰S_B†γ⁰ = cosh(η/2) − α⋅n̂ sinh(η/2), because γ⁰ anticommutes with α. The covariance check runs as for rotations: for n̂ = x̂, αₓ = γ⁰γ¹ anticommutes with γ⁰ and γ¹, so S_B⁻¹γ⁰S_B = γ⁰e^(ηαₓ) = γ⁰ cosh η + γ¹ sinh η, the time row of an active boost (derived). A passive boost, in which the observer moves, flips the sign of α⋅n̂. In the Weyl representation, where α = diag(−σ, σ), S_B = diag(e^{(−ησ⋅n̂/2)}, e^{(ησ⋅n̂/2)}): one Weyl half is stretched where the other is squeezed, whereas a rotation multiplies both halves by the same unitary e^{(−iφσ⋅n̂/2)}. Angles come round again; rapidities never do.

Unit 4's rest spinors are u(0, s) = √(2mc)(χₛ ; 0) and v(0, s) = √(2mc)(0 ; ζₛ), where ζₛ is Unit 4's ηₛ renamed to free η for the rapidity. In the Dirac representation α⋅n̂ swaps the blocks and multiplies by σ⋅n̂, so S_B u(0, s) = √(2mc)(cosh(η/2)χₛ ; sinh(η/2)σ⋅n̂ χₛ). With E = mc² cosh η and p = mc sinh η n̂, √(2mc) cosh(η/2) = √((E + mc²)/c) and tanh(η/2) = pc/(E + mc²), so this is Unit 4's u(p, s) to the last factor, and S_B v(0, s) is v(p, s) the same way (derived). Then u†u = 2mc[cosh²(η/2) + sinh²(η/2)] = 2E/c and ūu = 2mc[cosh²(η/2) − sinh²(η/2)] = 2mc. Lower over upper is about v/2c at low speed, Unit 6's small component, and tends to 1 as v → c. The γ = cosh η is that of every moving clock: muons stored at γ = 29.33 at CERN lived 64.419 ± 0.058 μs, against 29.33 × 2.1970 μs = 64.44 μs (quoted).