Baryon number: an accidental symmetry

In second quantisation every quark flavour f and colour c carries a number operator N̂_q = Σ b†b and an antiquark counterpart N̂_q̄ = Σ d†d. Define B̂ = (1/3) Σ_{f, c} (N̂_q − N̂_q̄), equivalently (1/3) Σ_f ∫ d³x :q_f† q_f:, the normal ordering being what makes an antiquark count −1/3. B̂ is Hermitian, its eigenvalue is an integer on every colour-singlet state, and on a product state |ψ₁⟩⊗|ψ₂⟩ it acts as B̂⊗1 + 1⊗B̂, so eigenvalues add — that is what makes a ledger legitimate. Read the entries off the quark content: p, n, Λ, Σ, Ω⁻ and every qqq state carry +1, their antiparticles −1, a meson qq̄ carries 1/3 − 1/3 = 0, leptons, photons, W, Z, gluons and the Higgs carry 0, and a nucleus carries its mass number A. Then p + p → p + p + p + p̄ reads 2 → 1 + 1 + 1 − 1 = 2 and is allowed; the invariant s = 2mₚ² + 2E mₚ must reach (4mₚ)², so E = 7mₚ and the beam needs kinetic energy 6mₚ c² = 5.63 GeV. That number is why the Bevatron was built to 6.2 GeV.

A conserved additive number is the generator of a continuous symmetry, so ask which. U(α) = exp(−iαB̂) rotates every quark field by q → e^(−iα/3) q and leaves leptons and bosons alone; α is one constant for all of spacetime, so this is a global U(1), and Noether's theorem gives the current j_B^μ = (1/3) Σ_f q̄_f γ^μ q_f with B̂ = ∫ j_B⁰ d³x. Conservation reduces to [H, B̂] = 0, and here the commutator can be checked by counting fields: every term of the Standard Model Hamiltonian — gauge couplings q̄γ^μD_μq, Yukawas q̄φq, and the pure-boson terms — contains quark and antiquark fields in pairs, so each term changes N̂_q − N̂_q̄ by zero. Compare electric charge. Its phase is local, α(x), and making the Lagrangian tolerate that forces the photon into existence. A massless vector coupled to B would exist the same way, and would pull on iron and aluminium by different amounts per unit mass, because their baryon-to-mass ratios differ. Eötvös-type balances see no such force, so B is not gauged: it is a symmetry the theory happens to have, not one it was built on.

B is conserved only because no allowed operator can break it, and allowed has three parts: Lorentz invariant, invariant under SU(3)_c × SU(2)_L × U(1)_Y, and mass dimension at most four. A quark field has dimension 3/2. A colour singlet from a quark and an antiquark uses 3 ⊗ 3̄ ∋ 1, but that is q̄q with B = 0. Two quarks give 3 ⊗ 3 = 6 ⊕ 3̄, no singlet at all. Three quarks give 3 ⊗ 3 ⊗ 3 ∋ 1 through ε_abc, with B = 1 — but that object has dimension 9/2 and is a fermion, so it cannot stand alone; it needs a fourth fermion, a lepton, to make a Lorentz scalar, and then the dimension is 6. Renormalisability throws it out. So B conservation is a theorem about the field content, and it evaporates the moment dimension six is admitted. The operator that survives every other check is O₆ = ε_abc (u_R^a d_R^b)(u_R^c e_R)/Λ²: hypercharges 4/3 − 2/3 + 4/3 − 2 = 0, colour singlet, Lorentz scalar, and coefficient 1/Λ² because the total must come out at dimension four. Acting on a proton it removes three quarks and creates a positron: ΔB = ΔL = −1.

Once a 1/Λ² operator exists, dimensional analysis prices it. If a heavy boson of mass M_X and coupling g mediates the four-fermion contact, the amplitude for p → e⁺π⁰ carries g²/M_X², the rate carries α_X²/M_X⁴, and the only other scale in the problem is the proton mass: Γₚ ≈ α_X² mₚ⁵/M_X⁴ in natural units. Put in M_X = 10¹⁶ GeV and α_X = 1/40. With mₚ⁵ = 0.727 GeV⁵, Γ = 6.25 × 10⁻⁴ × 0.727/10⁶⁴ = 4.5 × 10⁻⁶⁸ GeV, and τ = ħ/Γ = 6.58 × 10⁻²⁵ GeV s / 4.5 × 10⁻⁶⁸ GeV = 1.4 × 10⁴³ s = 4.6 × 10³⁵ years. The fourth power is the lesson: every factor of 10 in M_X moves the lifetime by 10⁴, while the coupling shifts it by one decade across any plausible range. Run it backwards. Super-Kamiokande's τ/BR(p → e⁺π⁰) > 2.4 × 10³⁴ years gives M_X⁴ > α_X² mₚ⁵ τ/ħ, that is M_X > 4.8 × 10¹⁵ GeV. Minimal SU(5) unification put M_X near 10¹⁵ GeV and predicted about 10³¹ years; the accident, measured, killed it.

The classical statement ∂_μ j_B^μ = 0 does not survive quantisation. Because the weak interaction is chiral, the triangle diagram with two W bosons gives ∂_μ j_B^μ = ∂_μ j_L^μ = (n_g g²/32π²) W^a_μν W̃^(aμν), with n_g = 3 and the same coefficient for B and L because each generation has three colours of quark doublet at 1/3 and one lepton doublet at 1. So j_B − j_L is exactly conserved and j_B + j_L is not. The right-hand side is a total derivative whose spacetime integral is an integer, the change in Chern–Simons number, so one transition changes B and L by 3 ΔN_CS each: the 't Hooft vertex turns nine quarks into three antileptons, ΔB = ΔL = −3. At zero temperature this is tunnelling through a barrier about 9 TeV high, the sphaleron, and the rate carries e^{(−4π/α_W)} ≈ e⁽⁻³⁷⁰⁾ ≈ 10⁻¹⁶⁰: unobservable. Above about 130 GeV the plasma hops the barrier freely and B + L is washed out. So the observed asymmetry n_B/n_γ ≈ 6 × 10⁻¹⁰ is a problem the accident forces on you: Sakharov's first condition is B violation, the model has some, and any primordial B + L is erased while B − L stays.

Two classes of search test the accident, and the ledger tells them apart. p → e⁺π⁰ and p → K⁺ν̄ have ΔB = −1 and ΔL = −1, so Δ(B − L) = 0; they are what dimension six and grand unification predict, and a water Cherenkov detector looks for a back-to-back positron and π⁰ with invariant mass 938 MeV and near-zero net momentum, Fermi motion in oxygen smearing both. Neutron–antineutron oscillation has ΔB = 2 with ΔL = 0, so Δ(B − L) = 2; it needs six quark fields, dimension nine, coefficient 1/Λ⁵, and a mixing δm ≈ Λ_QCD⁶/Λ⁵ that makes a free neutron oscillate on the time scale ħ/δm. The ILL free-neutron limit, τₙₙ̄ > 8.6 × 10⁷ s, reaches Λ near 4 × 10⁵ GeV — ten orders below what proton decay reaches, because δm falls as Λ⁻⁵ from a hadronic scale of 0.2 GeV and an oscillation time of 10⁸ s is a far weaker constraint than a lifetime of 10⁴² s. The pairing is the point: an observed ΔB = 1 decay with Δ(B − L) = 0 points at a unification scale; an observed ΔB = 2 points at a few hundred TeV and, like a Majorana neutrino, breaks B − L itself. And a limit is a lower bound on Λ, never a proof of the law.