Auditing neutron beta decay

Energy is conserved because H generates time translation, U(t) = exp(−iHt/ħ), and the weak Hamiltonian carries no explicit time dependence. For a decay at rest that statement collapses to a subtraction of tabulated rest masses. With mₙ = 939.565 MeV, mₚ = 938.272 MeV and mₑ = 0.511 MeV, Q = mₙ − mₚ − mₑ = 0.782 MeV — 782.3 keV with the full-precision masses, since mₙ − mₚ = 1293.3 keV. The antineutrino mass, below 1 eV, changes nothing at this level. Q > 0 is the verdict: a free neutron can decay. Run the same subtraction for the mirror process p → n + e⁺ + νₑ and Q = −1.293 − 0.511 = −1.804 MeV, so a free proton cannot; electron capture, p + e⁻ → n + νₑ, is closed too, with Q = −0.782 MeV, which is why hydrogen is stable while free neutrons are not. Inside a nucleus the binding-energy difference between parent and daughter joins the ledger and can flip the sign either way — that is how β⁺ emitters exist — but for the bare particles the masses alone decide.

Each additive quantum number is the eigenvalue of a Hermitian generator that commutes with H: electric charge from the gauged U(1), baryon number and lepton-family number from accidental global U(1) phases, the last only approximately, since neutrino oscillation breaks Lₑ in flight though never at a decay vertex. Because the generators commute with each other a state carries all of them at once, and because they commute with H each sum is fixed. Assign them particle by particle. Charge: n 0, p +1, e⁻ −1, ν̄ₑ 0, so 0 → +1 − 1 + 0 = 0. Baryon number: n 1 (udd), p 1 (uud), leptons 0, so 1 → 1. Electron-family number: e⁻ +1, ν̄ₑ −1, hadrons 0, so 0 → +1 − 1 = 0, while L_μ and L_τ are zero on both sides. Strangeness is 0 → 0, consistent with a ΔS = 0 charged-current vertex. The ledger is also what fixes the antineutrino: n → p + e⁻ + νₑ would give Lₑ = 0 → +2 and is forbidden, though charge, B and Q are untouched. The audit does not care that ν̄ₑ is invisible in a detector — it demands the particle by bookkeeping, which is exactly how Pauli argued for it in 1930.

Rotational invariance gives [H, Jᵢ] = 0, so total J is a ledger too, with the twist that angular momenta combine as a direct sum, not a number. The neutron has j = ½. The final spins live in C² ⊗ C² ⊗ C², dimension 8, and Clebsch–Gordan reduces it: ½ ⊗ ½ = 0 ⊕ 1, then (0 ⊕ 1) ⊗ ½ = ½ ⊕ ½ ⊕ 3/2, with dimensions 2 + 2 + 4 = 8. For an allowed transition the leptons leave in the L = 0 s-wave, so J_final = S and only the two S = ½ multiplets can match the neutron; S = 3/2 is shut. The two open multiplets are physically distinct. In one the e⁻ν̄ₑ pair is in the spin singlet (Fermi, coupling g_V), in the other the spin triplet (Gamow–Teller, coupling g_A), and with λ = g_A/g_V = −1.275 the rate splits as 1 : 3λ², about 17 % Fermi to 83 % Gamow–Teller. The same ledger kills the two-body candidate n → p + e⁻: two spin-½ particles plus an integer orbital L can only reach integer J, never the neutron's ½ — a half-integer total needs an odd number of fermions on each side.

P̂ = −iħ∇ commutes with H, so in the neutron's rest frame pₚ + pₑ + p_ν̄ = 0. With three bodies that leaves two continuous parameters — the electron energy and the electron–antineutrino angle — instead of the single fixed momentum a two-body decay would give. Weighting every configuration equally, the density of final states for an electron of kinetic energy Tₑ is dN/dTₑ ∝ F(Z, Eₑ) pₑ Eₑ (Q − Tₑ)², where pₑ and Eₑ are the electron momentum and total energy, (Q − Tₑ)² is the antineutrino's phase space with the proton's sub-keV recoil neglected, and the Fermi function F is within a few per cent of 1 for Z = 1 except at the lowest electron energies. The curve rises from zero, peaks near 250 keV and falls to zero at the endpoint. The endpoint is where the antineutrino leaves with no momentum, so the proton must recoil with pₚ = −pₑ. The electron there is relativistic, γ = 2.53, so pₑ²c² = Tₑ² + 2mₑc²Tₑ = 1.41 MeV², and Tₚ = pₑ²/(2mₚ) = 0.75 keV. The endpoint is therefore Q − 0.75 keV = 781.6 keV, and a Kurie plot of √(N/(pₑ Eₑ F)) against Tₑ is a straight line through it.

A closed ledger says the decay may happen; the interaction says how. At quark level n(udd) → p(uud) turns one d into a u, and a flavour change is something neither the gluon nor the photon can do — both couple diagonally in flavour. Only the charged-current weak vertex d → u + W⁻, followed by W⁻ → e⁻ + ν̄ₑ, does it, with ΔS = 0 and a Cabibbo factor V_ud = 0.974 at the hadronic end. That single fact sets the scale of the lifetime: a strong decay of a 940 MeV hadron takes about 10⁻²³ s and an electromagnetic one 10⁻¹⁶ to 10⁻²⁰ s, while the neutron survives 878 s because the amplitude carries G_F ≈ 1.17 × 10⁻⁵ GeV⁻² and the phase space is squeezed by mₑ⁵. It also explains why g_A ≠ g_V: the vector current is protected by CVC, so g_V = 1, but the axial current is renormalised by the strong interaction inside the nucleon to λ = g_A/g_V = −1.275 — a number measured through decay asymmetries and derivable from no ledger.

Fermi's golden rule, Γ = (2π/ħ)|M|²ρ(E), separates what the audit gave from what it did not. The audit gave the kinematics, Q and the shape of the spectrum in the previous section, but not the number ρ: integrating that shape with the Coulomb-dependent Fermi function gives f = 1.689 in units of mₑ⁵, which no ledger computes. Everything else is the matrix element: G_F from muon decay, |V_ud| = 0.974 from the CKM matrix, the factor (1 + 3λ²) from summing the Fermi and Gamow–Teller channels, and a 3.9 % radiative correction. Assembled, 1/τ = G_F²|V_ud|²(1 + 3λ²) mₑ⁵ f (1 + Δ_R)/(2π³ħ) gives 880 s against the measured 878.4 ± 0.5 s. Turn it round and the neutron becomes an instrument: with τ and λ measured, the formula returns |V_ud| and tests CKM unitarity. What is still open is experimental — bottle experiments that count surviving neutrons give 878 s, beam experiments that count emitted protons give about 888 s, and the 9 s gap is roughly four standard deviations. Nothing in the conservation audit even knows there is a lifetime to disagree about.