Auditing a proposed reaction

Start from the operator statement behind every conservation law in this unit: a Hermitian generator G with [H, G] = 0. Take eigenstates G|i⟩ = gᵢ|i⟩ and G|f⟩ = g_f|f⟩ and sandwich the commutator: ⟨f|HG − GH|i⟩ = gᵢ⟨f|H|i⟩ − g_f⟨f|H|i⟩, using the Hermiticity of G to let it act leftward on ⟨f|. So (gᵢ − g_f)⟨f|H|i⟩ = 0, and one of the two factors must vanish. If the eigenvalues differ the amplitude is identically zero — that is a selection rule, and it holds to every order in the interaction, not only the first. On a multi-particle state G is the sum of one-particle copies, G⊗1⊗1 + 1⊗G⊗1 + …, so a product state's eigenvalue is the total of its constituents'. Those two facts are the whole audit: add the eigenvalues on each side, subtract, and any non-zero row kills the amplitude. Read the same line the other way and you have the limit of the method. When gᵢ = g_f the equation is satisfied trivially and ⟨f|H|i⟩ is unconstrained: it may be large, small, or zero for some reason the ledger knows nothing about.

Each particle gets v = (Q, B, Lₑ, L_μ, L_τ, S), with its spin j kept to one side. Charge in units of e. Baryon number (n_q − n_q̄)/3: 1 for every baryon, 0 for every meson and lepton. Lepton-family number +1 for e⁻ and νₑ, −1 for e⁺ and ν̄ₑ, and likewise for the μ and τ families — the row that puts an antineutrino, not a neutrino, beside the electron in beta decay. Strangeness S = −(nₛ − nₛ̄), so the sign is the trap: K⁺ = us̄ has S = +1, Λ⁰ = uds has S = −1, Ξ⁻ = dss has S = −2 and Ω⁻ = sss has S = −3. An antiparticle negates every entry; a photon or a Z is the zero vector and a W± carries only Q = ±1. So p = (1, 1, 0, 0, 0, 0), n = (0, 1, 0, 0, 0, 0), π⁻ = (−1, 0, 0, 0, 0, 0), K⁰ = ds̄ = (0, 0, 0, 0, 0, +1), Σ⁺ = uus = (1, 1, 0, 0, 0, −1), e⁻ = (−1, 0, 1, 0, 0, 0), ν̄_μ = (0, 0, 0, −1, 0, 0). Spins: 0 for the pseudoscalar mesons, ½ for the octet baryons and the leptons, 3/2 for Δ and Ω⁻, 1 for the photon. A wrong entry here makes every later step wrong with it, so build the table before doing any arithmetic.

π⁻ + p → K⁻ + Σ⁺ looks like ordinary strange-particle production. Initial: (−1, 0, 0, 0, 0, 0) + (1, 1, 0, 0, 0, 0) = (0, 1, 0, 0, 0, 0). Final: K⁻ = sū is (−1, 0, 0, 0, 0, −1) and Σ⁺ = uus is (1, 1, 0, 0, 0, −1), summing to (0, 1, 0, 0, 0, −2). Residual (0, 0, 0, 0, 0, −2): charge and baryon number pass, strangeness fails by two units. Swap the kaon for K⁺ = us̄ and the σ for Σ⁻ = dds: the final vector becomes (0, 1, 0, 0, 0, 0) and every row is zero — associated production, in which a strange and an antistrange quark are created together. Now a decay: p → e⁺ + π⁰. Initial (1, 1, 0, 0, 0, 0); final (1, 0, −1, 0, 0, 0). Residual (0, −1, −1, 0, 0, 0): B and Lₑ fail by one unit each, in the same direction, so the combination B − L is unchanged. That survivor is not an accident of the example. It is why grand unified theories, which violate B and L together, predict this exact channel, and why it is the one Super-Kamiokande searches for.

H = Hₛ + H_EM + H_w, and the three pieces do not commute with the same generators. Q, B and each L_ℓ commute with all three, so a failure there closes every door, except that for L_ℓ the neutrino-mass terms leave it ajar, at a μ → eγ branching ratio near 10⁻⁵⁴. S commutes with Hₛ and H_EM but not with H_w: at a charged-current vertex an s quark becomes a u quark, changing S by exactly one unit, so |ΔS| = 1 means weak or nothing, and |ΔS| = 2 needs two W exchanges — Ξ⁻ → n + π⁻ sits below 2 × 10⁻⁵. Semileptonic decays obey the sharper rule ΔS = ΔQ_had, which is what forbids Σ⁺ → n + e⁺ + νₑ at first order. If every row is zero the process takes the strongest interaction available: hadrons in and out with no photon means strong, near 10⁻²³ s, so a Δ resonance of width 117 MeV lives 5.6 × 10⁻²⁴ s; a photon, or a decay into photons, means electromagnetic, π⁰ → γγ at 8.4 × 10⁻¹⁷ s; a neutrino anywhere, or ΔS ≠ 0, means weak, Λ⁰ at 2.6 × 10⁻¹⁰ s and K⁺ at 1.2 × 10⁻⁸ s. Parity, being multiplicative, is a label rather than a row: conserved by Hₛ and H_EM, ignored by H_w.

J is not additive, so it cannot be a row of the vector. The products' spins couple through Clebsch–Gordan coefficients to any total from |j₁ − j₂| to j₁ + j₂, that total couples to the orbital L between them, and the audit asks only whether the parent's J lies in the resulting range. Λ⁰ → p + π⁻: ½ ⊗ 0 = ½, then L = 0 gives J = ½ and L = 1 gives ½ or 3/2, so the parent's ½ is reachable both ways. Parity picks between them: the product parity (+)(−)(−1)^L is + for L = 1 and − for L = 0, against the Λ⁰'s +, so a parity-conserving interaction would allow only the p-wave. The weak decay uses both, and the measured asymmetry α = 0.75 is their interference. What is strictly additive is the fermion count: half-integer J on one side demands an odd number of fermions on the other. Two-body n → p + e⁻ has one fermion becoming two, so its final J is an integer for every L, and the proposal fails on angular momentum before the lepton row is even reached.

K⁺ → μ⁺ν_μ, K⁺ → π⁺π⁰, K⁺ → π⁺π⁺π⁻ and K⁺ → e⁺νₑ pass the same audit: Q +1 → +1, B 0 → 0, the lepton family balanced, ΔS = −1 at one weak vertex. Measured, they take 63.6%, 20.7%, 5.6% and 1.6 × 10⁻⁵ of the decays, and nothing in the ledger ranks them. The rate is Fermi's golden rule, Γ = (2π/ħ)|⟨f|H_w|i⟩|²ρ(E), and a branching ratio divides one such Γ by the sum of all of them, so it needs every amplitude and every phase space. The electron mode is the sharpest lesson. Its amplitude is proportional to m_ℓ, because a spin-0 kaon forces the ℓ⁺ and the ν into the same helicity, and (mₑ/m_μ)² × 1.10 = 2.6 × 10⁻⁵ against 2.49 × 10⁻⁵ measured — a factor of about 40,000 the ledger cannot see. The audit also inherits the status of its symmetries. Q is exact and gauge-protected; B and L are accidental, so "forbidden" means only "not yet seen", for p → e⁺π⁰ at lifetimes below 10³⁴ years; lepton-family number is already known to fail through oscillation.