Helicity, energy projectors and spin sums

Units are SI with c and ħ kept, and the metric is diag(1, −1, −1, −1). At each momentum the free equation has two positive-energy plane waves of the same E, so a second label is needed. At rest Σ_z served (topic 4.05), but topic 4.07 derived [H, Σ] = 2ic α×p, so [H, Σ_z] = 2ic(αₓ p_y − α_y pₓ): spin along a fixed axis commutes with H only if p is zero or lies along that axis. Dot the same commutator into p instead. The momentum commutes with the free H and with α, so [H, Σ⋅p] = 2ic(α×p)⋅p, a triple product containing p twice, which is zero. J gives the same answer: (r×p)⋅p vanishes identically, so (ħ/2)Σ⋅p = J⋅p, built from two conserved operators. On a plane wave |p| is a number, so the helicity Σ⋅p̂ commutes with H and with p, and because (Σ⋅p̂)² = 1 its eigenvalues are +1 and −1. All of this is derived.

Topic 4.06 made the lower pair of u(p, s) equal to cσ⋅p/(E + mc²) acting on the upper pair χ. Choose χ = χₕ, the eigenspinor of σ⋅p̂ with eigenvalue h, and the lower pair becomes hκχₕ with κ = |p|c/(E + mc²) = γβ/(γ + 1). So u(p, h) = √((E + mc²)/c)(χₕ, hκχₕ), and Σ⋅p̂ returns h from both pairs: an eigenvector, derived. At 0.80c along z, E = 5mc²/3 = 0.852 MeV, |p|c = 4mc²/3 = 0.681 MeV and κ = 0.5, so in the Dirac representation u(p, +) = √(8mc/3)(1, 0, ½, 0), with u†u = (8mc/3)(5/4) = 2E/c and ū u = (8mc/3)(3/4) = 2mc, topic 4.06's normalisation. Limits check it: as v → 0, κ ≈ v/2c → 0 and topic 4.05's rest spinors return; as v → c, κ → 1 and the pairs become equal in size, the massless form topic 5.08 ties to chirality. Taking η as eigenspinors of σ⋅p̂ makes v(p, s) helicity eigenvectors too; which physical helicity they carry is Unit 8's convention.

Build the plane wave on χ = spin up along z while p points at angle θ from z. Its lower pair is κ(σ⋅p̂)χ, and the identity (σ⋅p̂)σ_z(σ⋅p̂) = 2p̂_z σ⋅p̂ − σ_z, with χ†(σ⋅p̂)χ = cos θ, gives u†Σ_z u ∝ 1 + κ² cos 2θ. Divide by u†u ∝ 1 + κ² and use κ² = (E − mc²)/(E + mc²): ⟨Σ_z⟩ = cos²θ + (mc²/E) sin²θ, derived. The spin along p is untouched and the spin across p shrinks by 1/γ. At 0.80c with p along x, θ = 90° and ⟨Σ_z⟩ = 0.60, so Σ_z returns −1 one time in five although χ was spin up; the flipped spin sits in the lower pair, κσₓχ, which is κ times spin down. The same state has ⟨Σ⋅p̂⟩ = cos θ at every speed, an even mixture of h = ±1 at 90°, and those weights never change because helicity is conserved. Only at θ = 0°, at θ = 180° or at rest is the state an eigenvector of Σ_z.

Let an electron move along +z at 0.50c with h = +1. For a plane wave along z, L_z = 0, so J_z = (ħ/2)Σ_z = +ħ/2. An observer moving along +z at 0.60c sees the electron at (0.50 − 0.60)/(1 − 0.30) c = −0.143c. A boost along z commutes with rotations about z, so it cannot change J_z (quoted; Unit 5 builds the boost on spinors), but the momentum has reversed, so this observer measures h = −1. Any massive particle can be overtaken, so its helicity is conserved yet frame-dependent. A massless one cannot: (c − V)/(1 − V/c) = c for every V below c, so its helicity is the same for all observers. Experiment draws the same line, quoted: Goldhaber, Grodzins and Sunyar (1958) found the neutrino's helicity negative, consistent with −1, its mass being far too small for any real observer to overtake it, while β⁻ electrons are measured, from 1957 on, to emerge with longitudinal polarisation −v/c, approaching −1 only as v → c.

With p̸ = γ⁰E/c − γ⋅p, the algebra γ^μγν + γ^νγμ = 2g^μν gives p̸p̸ = p_μp^μ = E²/c² − p² = m²c² on the mass shell. So Λ±(p) = (±p̸ + mc)/(2mc) obey Λ±² = (2m²c² ± 2mc p̸)/(4m²c²) = Λ±, Λ₊Λ₋ = (m²c² − p̸p̸)/(4m²c²) = 0 and Λ₊ + Λ₋ = 1, and since Tr p̸ = 0, Tr Λ± = 2, room for two spins each; all derived. Topic 4.06's (p̸ − mc)u = 0 and (p̸ + mc)v = 0 then give Λ₊u = u, Λ₋u = 0, Λ₋v = v and Λ₊v = 0. Read the third with care: v(p, s) multiplies exp(+ip_μx^μ/ħ), a wave of energy −E and momentum −p, so Λ₋(p) does not select the negative-energy waves of momentum p, which are v(−p, s). Nor is Λ± Hermitian, γⁱ being anti-Hermitian: at 0.80c along z, u†(p, ↑)v(p, ↑) = 2|p| = 8mc/3 while ū v = 0. The projectors are orthogonal in ψ̄ψ, not in ψ†ψ. At rest Λ₊ = (1 + γ⁰)/2 = diag(1, 1, 0, 0), as topic 4.05 requires.

The four spinors u(p, s) and v(p, s) satisfy ū u = 2mc and v̄ v = −2mc for equal spins, zero between different spins, and ū v = v̄ u = 0 for every pair, so Σₛ[u ū − v v̄]/(2mc) returns every u and every v unchanged: it is the identity. Its first half sends u to u and v to zero, exactly as Λ₊ does, so Σₛ u ū = 2mcΛ₊ = p̸ + mc; likewise Σₛ v v̄ = −2mcΛ₋ = p̸ − mc. Both are derived, and neither cares whether s labels helicity or spin along z, since a unitary mixing of the two spin states leaves the sum alone. Checks: Tr(p̸ + mc) = 4mc = 2 × 2mc, and at rest p̸ + mc = diag(2mc, 2mc, 0, 0). An unobserved spin now costs nothing: |ū′Γu|² = (ū′Γu)(ūΓ̄u′) with Γ̄ = γ⁰Γ†γ⁰, and summing over s and s′ gives Tr[Γ(p̸ + mc)Γ̄(p̸′ + mc)]. For Γ = 1 that is 4(p_μp′^μ + m²c²), which at p′ = p returns 8m²c², two spins times (2mc)². Topic 5.09 runs it with Γ = γ⁰ to reach Mott's factor, which electron-scattering data confirm.