Lepton number and family number

The rule is additive and integer. Every charged lepton and every neutrino of family α carries L_α = +1, every antilepton of that family −1, and everything else — quarks, hadrons, photons, W, Z, Higgs — carries 0. A state's L_α is the sum over its particles, so the ledger entry for a reaction is one integer per side. Run it for μ⁻ → e⁻ + ν̄ₑ + ν_μ: L_μ reads 1 → 0 + 0 + 1 and Lₑ reads 0 → 1 + (−1) + 0, both balanced. Now run β⁻ decay with the third particle unknown, n → p + e⁻ + X. Charge and baryon number are already settled by p and e⁻, so Lₑ is the only ledger left: 0 = 0 + 1 + Lₑ(X) forces Lₑ(X) = −1, and X is the antineutrino, not the neutrino. That sign is experimental content. Reines and Cowan detected reactor ν̄ₑ through ν̄ₑ + p → e⁺ + n, which keeps Lₑ at −1 on both sides, while Davis found that the same reactor flux does not drive ν̄ₑ + ³⁷Cl → ³⁷Ar + e⁻, which would need Lₑ to go from −1 to +1.

Multiply every field of family α by one phase, ψ → e^(iφ)ψ, and leave the other two families alone. The Standard Model Lagrangian with massless neutrinos is unchanged, so Noether gives a conserved current and a Hermitian charge L̂_α = ∫ j⁰_α d³x, which in the particle basis is the number-operator combination on the formula card. The unitary family U(φ) = exp(−iφL̂_α) is that symmetry acting on states, and the Heisenberg equation d⟨L̂_α⟩/dt = (i/ℏ)⟨[H, L̂_α]⟩ reduces conservation to one statement, [H, L̂_α] = 0. It holds because every interaction term in H is family-diagonal: the W couples e to νₑ, μ to ν_μ and τ to ν_τ, never across families, and the Z, photon and Higgs couple each lepton to itself, so each term carries equal and opposite phases and is neutral. Nothing gauges this phase. Electric charge has a photon whose masslessness enforces it; L_α has no boson at all, and survives only because no renormalisable operator built from the massless-neutrino field content can break it. That is what accidental means, and why the law is exposed the moment the field content changes.

Give the neutrinos masses and the field content has changed. The states created at a W vertex, |νₑ⟩, |ν_μ⟩ and |ν_τ⟩, are eigenstates of L̂_α. The states that propagate with definite energy are eigenstates of H, |ν₁⟩, |ν₂⟩ and |ν₃⟩ with masses mᵢ, and the two bases are related by the unitary PMNS matrix, |ν_α⟩ = Σᵢ U*_αi |νᵢ⟩. If U is not diagonal the two Hermitian operators share no eigenbasis, so [H, L̂_α] ≠ 0 — the same algebra that makes S_z precess under a Hamiltonian proportional to Sₓ. Keep two flavours. In the flavour basis, dropping a multiple of the identity, H = (Δm²c⁴/4E)(−cos 2θ σ_z + sin 2θ σₓ): a flavour state is a spinor tilted by 2θ from the field, and it precesses. The probability that a ν_μ has become the other flavour after a distance L is P = sin²2θ sin²(Δm²c⁴L/4ℏcE), which in practical units is sin²2θ sin²(1.267 Δm²[eV²] L[km]/E[GeV]). For Δm² = 2.5 × 10⁻³ eV², E = 1 GeV and L = 295 km the phase is 0.934 rad, its sine squared is 0.647, and with sin²2θ = 1 only 35% of the beam still answers to L_μ.

Oscillation is a direct violation: a unit of L_μ has become a unit of L_τ or Lₑ, and no charged particle was involved. What it does not touch is the count of neutrinos. U is unitary, so Σ_β P(ν_α → ν_β) = 1 and a neutrino stays a neutrino; total L = Lₑ + L_μ + L_τ, whose generator is the sum of the three and commutes with a Dirac mass term, is untouched. The failure is also confined. Charged-lepton flavour violation such as μ → eγ must run through a virtual W and a neutrino that changes flavour on the way, so the amplitude is proportional to Σᵢ U*_μi Uₑᵢ mᵢ²/M_W². Unitarity of U kills any m-independent piece, what survives is of order (Δm²/M_W²)² ≈ 10⁻⁴⁹, and the branching ratio lands near 10⁻⁵⁴ — against a measured limit of 1.5 × 10⁻¹³ from MEG II. So in the audit keep each L_α as its own line and treat it as exact for charged leptons; strike it only for a neutrino that has travelled, where the ledger entry stops being an integer and becomes the expectation value ⟨L̂_μ⟩ = P(ν_μ → ν_μ).

Whether total L survives depends on how the mass was written. A Dirac term m ν̄_L ν_R pairs the neutrino with a distinct right-handed partner and is invariant under a common phase, so [H, L̂] = 0 still. A Majorana term ½ m ν_Lᵀ C ν_L pairs the field with itself, carries charge 2 under that phase, and turns a neutrino into an antineutrino: ΔL = ±2. The experiment that separates them is double beta decay. The two-neutrino mode (A, Z) → (A, Z + 2) + 2e⁻ + 2ν̄ₑ is observed in ¹³⁶Xe with T½ = 2.2 × 10²¹ yr and conserves Lₑ. The neutrinoless mode (A, Z) → (A, Z + 2) + 2e⁻ has ΔLₑ = 2 and needs the antineutrino emitted at one vertex to be absorbed as a neutrino at the other, which only a Majorana particle can do. Its rate is 1/T½ = G^(0ν) g_A⁴ |M^(0ν)|²(m_ββ/mₑ)² with m_ββ = |Σ Uₑᵢ² mᵢ|, so a null result is a mass bound: KamLAND-Zen's complete dataset, T½ > 3.8 × 10²⁶ yr, gives m_ββ below 28–122 meV, the spread coming from the nuclear matrix element. A signal — a line in the summed energy of the two electrons at Q = 2.458 MeV — would settle both questions at once: L is broken, and the neutrino is Majorana.

Give every particle the vector (Q, B, Lₑ, L_μ, L_τ) and sum each component across the arrow. τ⁻ → μ⁻ + ν̄_μ + ν_τ passes: L_τ 1 → 1, L_μ 0 → 1 − 1. τ⁻ → μ⁻ + γ fails two lines at once, L_τ 1 → 0 and L_μ 0 → 1, while keeping total L at 1. K⁺ → μ⁺ + ν_μ passes with L_μ 0 → −1 + 1, and it is the neutrino-or-antineutrino sign that separates ν̄ₑ + p → e⁺ + n (allowed) from νₑ + p → e⁺ + n (Lₑ: 1 → −1, forbidden). Then label the status of each line, because knowing which laws are which is the point of the audit. Q is exact and gauged. Each L_α is accidental and already known to fail, but only through neutrino mixing; for charged leptons it holds to better than one part in 10¹². Total L is accidental, unbroken by any observation so far, and tested at ΔL = 2 by 0νββ. A balanced ledger, as always, is necessary and never sufficient: it licenses a reaction without predicting its rate.