Midterm consolidation: symmetry, scattering, Dirac theory

Units are SI with x⁰ = ct, the metric is (+, −, −, −), the γ matrices are in the Dirac representation with γ⁰ = β = diag(1, 1, −1, −1), and spinors carry Unit 4's u†u = 2E/c. Every answer then gets three checks and one tag. Dimensions: a cross-section in m² sr⁻¹, a spinor in √(kg m s⁻¹), every energy a multiple of mc² = 0.511 MeV. A limit: β → 0 must return Rutherford, the leading order in Zα Bohr's Z² × 13.6 eV, and v/c → 0 the Pauli equation of Quantum Mechanics II. A measured number. The tag says where the result lives: derived in the one-particle theory (Θ² = −1, both Mott formulas, g = 2, the Foldy-Wouthuysen operators, the n- and j-only energies, ψ^c); read off the filled sea (the positron's charge +e, energy +E, momentum and spin); or quoted (Bargmann's proof, Pauli's theorem, Mott's exact solution, Schwinger's α/2π, the Z ≈ 172 finite-nucleus limit, every measured value). A quoted result presented as derived is wrong even when its number is right.

Units 1 and 2 reach their results without solving a wave equation. Wigner's theorem allows only unitary and antiunitary symmetries; time reversal must be antiunitary, and for spin ½ Θ = −iσ_yK gives Θ² = −1, a value no phase choice alters, so an odd-electron level in zero magnetic field is at least doubly degenerate. Rotation and parity then cut a spin-½ amplitude on a spinless target down to g(θ) + ih(θ)σ⋅n; an unpolarised beam leaves with P = 2 Im(gh*)/(|g|² + |h|²) along n, and time reversal makes the analysing power equal to it. Exchange fixes the identical-particle weights, ¼|f + f′|² + ¾|f − f′|² for spin-½ fermions and |f + f′|² for spin-0 bosons. With the exact Coulomb phase the cross term at 90° does not depend on η, so the rate there is half the classical sum for protons and twice it for α particles, wherever the nuclear force can be neglected. The limit check: make the particles distinguishable and the cross term goes, leaving Rutherford.

Unit 5 puts spin into the dynamics. In the first Born approximation a static Coulomb field couples through γ⁰, so the amplitude is proportional to ū(p′)γ⁰u(p). Summing final and averaging initial spins with Σu ū = p̸ + mc turns |M|² into ½Tr[(p̸′ + mc)γ⁰(p̸ + mc)γ⁰], and only Tr(γ^μγν) = 4g^μν and the vanishing of odd traces are needed: the trace is 4(E²/c² + p⋅p′ + m²c²), and with p⋅p′ = p² cos θ it is 8(E/c)²(1 − β² sin²(θ/2)). Dividing by the spinless 4(E/c)² leaves the factor. Checks: (Zαħc)²/(pc)² is an area; as β → 0 the factor goes to 1 and pv to 2T, giving Rutherford. At 180° the factor is 1 − β², 0.114 for a 1.00 MeV electron, and it vanishes as β → 1: a vector coupling conserves helicity, and reversing the momentum with J_z fixed along the axis would flip it. This is a property of one Dirac particle's current, not of exchange.

The spinor at momentum p is the rest spinor √(2mc)(χ ; 0) boosted by S = cosh(ζ/2) + α⋅p̂ sinh(ζ/2), with cosh ζ = E/mc². Its lower pair is tanh(ζ/2) = pc/(E + mc²) of the upper, about v/2c when slow: roughly α/2 = 0.0036 in hydrogen's ground state. S is Hermitian, not unitary, so u†u rises from 2mc to 2E/c while ū u stays 2mc. That small ratio is Unit 6's expansion parameter. Write H = βmc² + O + V with the odd operator O = cα⋅π, where π = p − qA is the minimal-coupling momentum. The unitary e^{(βO/(2mc²))}, its exponent anti-Hermitian because β and O anticommute, removes O at first order, and the even term it leaves, βO²/(2mc²) = β(σ⋅π)²/(2m), already holds −(qħ/2m)βΣ⋅B: g = 2 with nothing added. Carried to order (v/c)², it leaves −βp⁴/(8m³c²) and −[O, [O, V]]/(8m²c⁴), whose two parts are the spin-orbit term with the Thomas half built in and the Darwin term (ħ²/(8m²c²))∇²V. In hydrogen's 2S level the Darwin term alone is mc²α⁴/16 = 9.06 × 10⁻⁵ eV, or 21.9 GHz.

Unit 7 turns two first-order radial equations into a series with exponent √(κ² − (Zα)²); making it end leaves n − |κ| = 0, 1, 2, … (0 only for negative κ) and E = mc²[1 + (Zα)²/(n − |κ| + √(κ² − (Zα)²))²]^(−1/2). Three checks follow. Expanded, it is mc² − mc²(Zα)²/(2n²) − (mc²(Zα)⁴/(2n⁴))(n/(j + ½) − ¾), exactly the Foldy-Wouthuysen answer. It knows only |κ| = j + ½, so 2S₁/₂ (κ = −1) and 2P₁/₂ (κ = +1) coincide at every order, and the measured 1057.8 MHz between them, the Lamb shift, lies outside this theory. And it ends: the 1s energy mc²√(1 − (Zα)²) reaches zero at Zα = 1, where Klein-Gordon s states had already failed at Zα = ½. Against data, hydrogen's 2P₃/₂ − 2P₁/₂ comes out 10.944 GHz with the reduced mass, and 10.969 GHz is measured: the 25 MHz left over is radiative. In U⁹¹⁺ the point-nucleus 1s binding, 132.28 keV, exceeds the measured value by 0.46 keV, nuclear size plus radiative effects.

Unit 8 changes no equation; it rereads one. In the Dirac representation ψ^c = iγ²ψ* = (ψ₄*, −ψ₃*, −ψ₂*, ψ₁*) solves the Dirac equation with q → −q and carries a sea wave of energy −E, momentum −p and spin −s into a positive-energy wave with +E, +p and +s: the quantum numbers derived for the hole by subtracting one state from the filled sea, with helicity unchanged. The same reading turns Klein's anomalous step current into pair creation, at the scale E_c = m²c³/(eħ) = 1.32 × 10¹⁸ V m⁻¹ with suppression exp(−πE_c/E). S_P = γ⁰ gives fermion and antifermion opposite intrinsic parity, so positronium has P = (−1)^(L+1): parapositronium is 0⁻, and its two annihilation photons are measured polarised at right angles. On the paper, hole theory is examinable as this bookkeeping and its failures: an infinite, unobservable sea charge, no sea at all for bosons, and no description of particle number changing in time, which is what Unit 9's quantised fields supply.