Coulomb scattering of a Dirac electron: the Mott factor

Units are SI, with x⁰ = ct and metric diag(1, −1, −1, −1); four-vector products are written p_μp′^μ, and p⋅p′ always means the three-vector product. Treat the nucleus as a fixed point charge. Its potential V(r) = −Zαħc/r multiplies the 4 × 4 identity, because it is the time component of topic 5.08's vector coupling. Box-normalise the free solutions as ψ = u(p, s)e^(ip⋅r/ħ)/√(2EL³/c), so Unit 4's u†u = 2E/c puts one electron in the box. Then ⟨p′s′|V|ps⟩ = Ṽ(q)u†(p′, s′)u(p, s)c/(2EL³), with q = p′ − p. Quantum Mechanics II's golden rule needs the final-state density per steradian, L³p²/((2πħ)³v), because dE/dp = v at any speed, and the incident flux v/L³. The box cancels and dσ/dΩ = (p/(2πħ²v))²|Ṽ(q)|²|u′†u|²/(2E/c)², derived. As v → 0, p/v → m and the overlap ratio becomes χ′†χ: the familiar Born formula, with the spin carried through untouched.

For V = −Zαħc/r, Ṽ(q) = −4πZαħ³c/q², derived by screening with e^(−r/a), transforming, and letting a grow without limit. A fixed source scatters elastically, so |p′| = |p| and q = 2p sin(θ/2), and the spin-free part of the cross-section is (Zαħc)²/(4p²v² sin⁴(θ/2)). That is the complete first-order answer for a spinless particle: to first order in V the Klein–Gordon equation reads p̂²φ/(2m) + (E/mc²)Vφ = [(E² − m²c⁴)/(2mc²)]φ, a Schrödinger problem with V scaled by γ, and (γm)² = (p/v)². Dimensions: (J m)² over J² is m², per steradian. As v → 0, pv → 2T and Geiger and Marsden's measured sin⁻⁴(θ/2) law returns. At 1.00 MeV, pv = 1.34 MeV against 2T = 2.00 MeV, so kinematics alone raise the cross-section 2.23-fold over the non-relativistic formula. Whatever spin does must come on top, through the overlap ratio.

An unpolarised beam weights each initial spin by 1/2, and a detector blind to spin adds the final ones, so the factor to compute is (1/2)Σ|u′†u|²/(2E/c)², summed over s and s′. Write u′†u = ū′γ⁰u; since γ⁰ is Hermitian its complex conjugate is ūγ⁰u′. The product ū′γ⁰u ūγ⁰u′ is a closed chain, and Unit 4's spin sum Σₛ u ū = p̸ + mc, used twice, turns it into (1/2)Tr[γ⁰(p̸ + mc)γ⁰(p̸′ + mc)] without writing a spinor. Traces of an odd number of γ matrices vanish, Tr(γ⁰γ⁰) = 4, and the trace identity for four γ matrices gives Tr(γ⁰p̸γ⁰p̸′) = 4(2p⁰p′⁰ − p_μp′^μ) = 4(E²/c² + p⋅p′). The full trace is 4(E²/c² + p² cos θ + m²c²) = (8E²/c²)[1 − β² sin²(θ/2)], with β = pc/E. Halve it, divide by (2E/c)², and the Mott factor appears, derived: 0.595 at β = 0.90 and 90°. Only the Clifford algebra was used here; the Born approximation entered once, through the overlap.

The trace says how much scatters, not which spins. Explicit spinors say which. In the Dirac representation a helicity state is u(p, h) = √((E + mc²)/c) (χₕ, hrχₕ), with σ⋅p̂χₕ = hχₕ, h = ±1 and r = pc/(E + mc²). Put p along z and p′ at θ in the xz-plane; then u†(p′, h′)u(p, h) = [(E + mc²)/c](1 + hh′r²)χ′†χ. Because r² = (E − mc²)/(E + mc²), the bracket is 2E/c when helicity is kept and 2mc when it flips, while the two-component overlaps are cos(θ/2) and ±sin(θ/2). Relative to 2E/c the amplitudes are cos(θ/2) and (mc²/E) sin(θ/2), derived, the same for either starting helicity, and cos²(θ/2) + (1 − β²) sin²(θ/2) rebuilds the trace term by term, an independent check. As β → 0 the split is cos² against sin²: the spin stays put while p turns, so a flipped helicity only means the reference axis moved. As β → 1 the lower components lock spin to momentum, and flips die as 1/γ².

At θ = π each amplitude vanishes for its own reason. A point charge is symmetric under rotations about the beam, so J_z is conserved, and a plane wave along ±z has L_z = 0, so J_z is the spin projection on z. An h = +1 electron arriving along +z has J_z = +ħ/2; leaving along −z with h = +1 needs J_z = −ħ/2, which a potential symmetric about the beam cannot supply at any energy: that is the zero of cos(θ/2). A flip conserves J_z, but the coupling ψ̄γ⁰ψ, like any vector coupling, connects only like chiralities (topic 5.08), and chirality equals helicity up to terms of order mc²/E. So backward scattering is Rutherford's times 1/γ², 2.4 × 10⁻³ at 10 MeV. A contrast confirms the reason: a Lorentz-scalar potential would bring in ū′u, which flips chirality, and its spin average, 1 − β² cos²(θ/2) as the Medium example derives, is largest straight back. The zero is measured and exploited: electron scattering at 180° picks out a nucleus's magnetic response, because charge scattering is switched off there.

First Born needs Zα ≪ 1, and Zα is 0.095 for aluminium but 0.58 for gold. Mott's 1929 exact partial-wave solution, quoted here, sums the Coulomb series to all orders; expanded for light nuclei it adds McKinley and Feshbach's πZαβ sin(θ/2)[1 − sin(θ/2)] to the factor, also quoted. That term changes sign with the projectile's charge, so electrons and positrons, identical in first Born, differ; for highly relativistic electrons the fractional difference is largest at backward angles, where the Mott factor is small, although the term itself vanishes at exactly 180°. A transversely polarised beam scatters left and right differently only beyond first Born, where the two spin amplitudes acquire a relative phase; Mott polarimeters use the tens-of-per-cent asymmetry on gold foils that first Born puts at zero. A point nucleus needs qR ≪ ħ: a charge cloud multiplies the cross-section by |F(q)|² ≈ 1 − q²⟨r²⟩/(3ħ²), and Hofstadter's Stanford experiments read nuclear radii off exactly this departure from the Mott cross-section. A fixed nucleus needs E ≪ Mc², 25 GeV for aluminium.