Noether currents and conservation laws

Take ℒ = ψ̄(iγ^μ∂_μ − m)ψ and the global phase ψ → e^(iα)ψ, so δψ = iψ per unit α and δψ̄ = −iψ̄. The only derivative in ℒ acts on ψ, with ∂ℒ/∂(∂_μψ) = iψ̄γ^μ, so the Noether current is (iψ̄γ^μ)(iψ) = −ψ̄γ^μψ; the overall sign is conventional and one writes j^μ = ψ̄γ^μψ. Its divergence is (∂_μψ̄)γ^μψ + ψ̄γ^μ∂_μψ, and the Dirac equation with its adjoint supply γ^μ∂_μψ = −imψ and ∂_μψ̄γ^μ = +imψ̄, so the two terms cancel: ∂_μ j^μ = 0, but only on-shell. The time component j⁰ = ψ†ψ integrates to Q = ∫ψ†ψ d³x, and once ψ is expanded in creation and annihilation operators, Q = Σ(b†b − d†d) = N − N̄, particles minus antiparticles. Nothing used the mass: the current is conserved at m = 0 too, and the same construction on each of the fifteen Weyl fields per generation is where every additive quantum number in the Standard Model comes from.

Two of those currents couple to gauge fields, and that changes their status. Electric charge is the generator Q = T₃ + Y/2 that the Higgs vacuum leaves unbroken, so its current couples to a massless photon and dQ/dt = 0 is protected by gauge invariance itself: a photon mass and a decaying electron would both break U(1)_EM, and the limit on e⁻ → νγ exceeds 6 × 10²⁸ yr. Colour is subtler. The quark current J^μa = ψ̄γ^μTaψ obeys only D_μJ^μa = 0, because gluons carry colour and exchange it with quarks; the genuinely conserved Noether current adds the gluon field's own term, f^abc A^b_ν F^cνμ. In practice colour never enters a ledger at all: confinement makes every asymptotic state a colour singlet with Q^a = 0 on both sides, so the law is satisfied identically rather than checked. Spontaneous breaking does not repeal Noether either — the currents of the three broken electroweak generators are still conserved, but their charges fail to annihilate the vacuum, so they label no states. Energy, momentum and angular momentum come from the Poincaré generators: exact, never accidental.

List every gauge-invariant, renormalisable operator you can build from the Standard Model fields — dimension four at most — and something falls out unasked. Each term has equal numbers of quark and antiquark fields, and of lepton and antilepton fields, so the Lagrangian is invariant under the global phases q → e^(iα/3)q and ℓ → e^(iβ)ℓ that nobody wrote down. Noether then hands over B = ⅓∫Σ_q q†q d³x and L = ∫Σ_ℓ ℓ†ℓ d³x, plus separate Lₑ, L_μ and L_τ, since with no right-handed neutrino nothing can rotate one lepton doublet into another. These are accidental symmetries: consequences of the field content and of renormalisability, not principles. The ⅓ in B is a bookkeeping choice that makes the proton carry B = 1; the physics is the phase. Proton stability to 10³⁴ yr is therefore enforced by no gauge field — it is the statement that the operators which could end it happen to have dimension six, suppressed by two powers of whatever scale lies beyond.

Noether's proof is classical. In the quantum theory the fermion measure is not invariant under a chiral rotation, and the B and L currents — which SU(2)_L sees only through left-handed doublets — pick up the same anomaly: ∂_μ j^μ_B = ∂_μ j^μ_L = (N_g/32π²)(g² W⋅W̃ − g′² B⋅B̃). Two consequences are immediate. Subtract, and B − L is exactly conserved, an accidental symmetry that survives quantisation. Add, and B + L changes whenever the gauge field's Chern–Simons number does: each unit of ΔN_CS shifts B and L by N_g = 3, one per generation, through a twelve-fermion vertex such as uud ccs ttb → e⁺μ⁺τ⁺. At zero temperature that needs tunnelling between gauge vacua, suppressed by exp(−4π/α_W) ≈ 4 × 10⁻¹⁶², with α_W = α/sin²θ_W from α = 1/128 and sin²θ_W = 0.231, so no laboratory will see it. Above the electroweak temperature of about 100 GeV the barrier is crossed thermally by sphaleron transitions with no such factor, and B + L is simply not conserved in the early universe — which is why baryogenesis is a question the model must answer rather than a bookkeeping triviality.

Parity, charge conjugation and time reversal are not continuous: there is no one-parameter family U(α) = exp(−iαG), no generator, no current, nothing to integrate. P and C act as multiplicative labels, ±1 on their eigenstates; T is antiunitary, so it supplies no eigenvalue label at all, only a relation between a process and its time-reversed partner. Each is a separate empirical question. The Standard Model answers all three the same way. The projector P_L = (1 − γ₅)/2 in every charged-current vertex breaks P and C maximally; the CKM phase δ breaks CP in the quark sector at the level of the Jarlskog invariant J ≈ 3 × 10⁻⁵; and T is broken by exactly as much as CP, because CPT was never the model's to choose. The Lüders–Pauli–Jost theorem says any Lorentz-invariant, local field theory with a Hermitian Hamiltonian and the usual spin–statistics connection is CPT-symmetric, so every particle and its antiparticle share mass and lifetime; the neutral kaons test that at |m_K⁰ − m_K̄⁰|/m_K < 6 × 10⁻¹⁹. A CPT violation would not be a new parameter in the model. It would be the end of the framework.

Rank the model's conserved quantities by what protects them. Exact from spacetime symmetry: energy, momentum, angular momentum. Exact from an unbroken gauge symmetry: electric charge and colour. Exact by accident and surviving the anomaly: B − L. Accidental but anomalous: B, L and B + L, safe at zero temperature, violated at 100 GeV. Accidental and already dead: Lₑ, L_μ and L_τ, because neutrino oscillation rotates one family into another, and the model as written — no ν_R, no dimension-5 operator — cannot even describe it. Two things stand outside. Neutrino mass needs either a Dirac partner ν_R, which keeps total L, or the Weinberg operator (LH)²/Λ, which breaks it by two units and makes neutrinoless double beta decay the deciding experiment. And gravity: it has no generator here, and a black hole that swallows a baryon and evaporates remembers its electric charge but not its baryon number — the expectation that quantum gravity respects gauge charges and no global ones is the deepest reason to call B and L accidental.