Symmetry, generators and conservation

A symmetry is a transformation that leaves every measurable prediction alone. Wigner's theorem turns that into algebra: any such map on state space is a unitary (or antiunitary) operator. For a continuous family the antiunitary option drops out, because U(a) = U(a/2)U(a/2) and the product of two antiunitary operators is unitary; Stone's theorem then writes the family as exp(−iaG/ħ) with a single Hermitian G. Translations show the mechanism. Shifting a wavefunction by a means U(a)ψ(x) = ψ(x − a), and Taylor's theorem writes ψ(x − a) = Σ (−a)ⁿ/n! dⁿψ/dxⁿ = exp(−a d/dx)ψ. Matching exponents, −iG/ħ = −d/dx, so G = −iħ d/dx = p̂: the generator of translations is the momentum operator, Hermitian because integrating by parts moves the derivative across the inner product with a sign that cancels the i. The same reading gives Ĵ for rotations, H for time translation and ħQ̂/e for a global phase, and in every case the product aG has the dimensions of ħ, so aG/ħ is a pure number: a length pairs with a momentum, an angle with an angular momentum, a time with an energy.

The dynamics respect the symmetry when transforming a state before or after evolving it makes no difference, which is U†(a) H U(a) = H for every a — apparently infinitely many conditions. Expand to first order, U(a) = 1 − iaG/ħ + O(a²): U†HU = H + (ia/ħ)(GH − HG) + O(a²) = H − (ia/ħ)[H, G], so the whole family is respected if and only if [H, G] = 0; the converse holds because [H, G] = 0 implies [H, Gⁿ] = 0 for every power in the exponential. One commutator is a finite computation. Take a spin-½ with H = (ħω/2)(cos θ σ_z + sin θ σₓ) and ask whether G = σ_z = 2S_z/ħ, the generator of rotations about z in units of ħ/2, is conserved. Only the σₓ piece fails to commute: [H, σ_z] = (ħω/2) sin θ [σₓ, σ_z] = −iħω sin θ σ_y. The commutator vanishes exactly when sin θ = 0, that is when the field defining H lies along the axis σ_z generates rotations about. At θ = 30° with ω = 1.0 rad ns⁻¹ the operator norm is ħω sin θ = 0.50 ħ ns⁻¹, and that number is about to set the rate at which σ_z fails to be conserved.

Differentiate ⟨ψ(t)|G|ψ(t)⟩ using iħ d|ψ⟩/dt = H|ψ⟩ and its conjugate: d⟨G⟩/dt = (i/ħ)⟨[H, G]⟩ + ⟨∂G/∂t⟩. For a generator with no explicit clock the last term is absent, and the rate of change of ⟨G⟩ in any state is the expectation of the commutator in that state. Run the spin example from |+z⟩, the σ_z = +1 eigenstate. The Bloch vector precesses about n = (sin θ, 0, cos θ) at angular rate ω, so ⟨σ_z⟩(t) = cos²θ + sin²θ cos ωt and ⟨σ_y⟩(t) = −sin θ sin ωt. Check the equation directly: (i/ħ)⟨[H, σ_z]⟩ = (i/ħ)(−iħω sin θ)⟨σ_y⟩ = −ω sin²θ sin ωt, and differentiating ⟨σ_z⟩(t) gives the same. At θ = 30° the expectation falls from 1 to cos 2θ = 0.50 in half a period, at ωt = π, so a system prepared with a definite σ_z has a 25% chance of showing the opposite value by then; at θ = 90° it swings all the way to −1; at θ = 0 nothing moves. Note the subtlety at t = 0: the rate is zero for every θ, because the expectation of any commutator [H, G] in an eigenstate of G is g⟨H⟩ − g⟨H⟩ = 0. Conservation is a claim about all times, and only the operator statement [H, G] = 0 delivers it.

Commuting Hermitian operators can be diagonalised together, so [H, G] = 0 lets you label energy eigenstates |E, g⟩ by the eigenvalue of G as well. Expand a state of definite g in that basis: every component carries the same g, and time evolution exp(−iHt/ħ) only multiplies each component by a phase exp(−iEt/ħ). The state stays inside the g sector, so the probability of measuring any other value g′ is zero at every later time — not small, zero. This is stronger than a frozen mean. Since [H, G] = 0 implies [H, Gⁿ] = 0, every moment of the G distribution is fixed, and with it the entire distribution of outcomes. The joint basis is unique only where H is non-degenerate: in the hydrogen n = 2 level four states share one energy, and choosing L_z rather than Lₓ to split them is a choice you make, not one nature makes. That freedom is what the reaction audit later in this unit uses when it labels a particle state by charge, baryon number and J_z alongside its mass.

Parity sends x → −x and has no small version: there is no continuous a to differentiate, so no Hermitian generator, and nothing additive comes out. What survives is P itself, unitary with P² = 1, hence eigenvalues η = ±1. If [H, P] = 0 those eigenvalues are conserved, but composing systems multiplies them rather than adding: a two-particle state has η = η₁η₂(−1)^L, the orbital factor coming from Y_Lm(−r̂) = (−1)^L Y_Lm(r̂). Two pions, each of intrinsic parity −1, have total parity +1 in an s-wave and −1 in a p-wave. Charge conjugation works the same way: the π⁰ has C = +1 and each photon C = −1, so π⁰ → 3γ would need (−1)³ = −1 and is forbidden, with its branching fraction measured to be below 3.1 × 10⁻⁸. Contrast the additive ledger: two particles of charge +1 make +2, never (+1)(+1). The reaction audit in this unit carries both kinds of column, and confusing them — adding parities, multiplying charges — is the commonest bookkeeping error.

The theorem is only as good as the H you feed it. A gauged symmetry such as the electromagnetic U(1) is built into every allowed term, so [H, Q̂] = 0 is exact. Baryon and lepton number are different: [H_SM, B̂] = 0 holds for every renormalisable term one can write, but nothing forces it, and a single higher-dimension operator such as qqqℓ/Λ² would break it — which is why they are called accidental and why proton-decay searches matter. Isospin is broken explicitly, by m_d − mᵤ and by electromagnetism, so [H, Î] is small rather than zero and ⟨Î⟩ drifts at a rate set by the size of the breaking term: an approximate symmetry, and still a usable column in the ledger. Finally, H is itself the generator of time translation, with dH/dt = ∂H/∂t; a Hamiltonian with an explicit clock, such as an atom in a laser field, conserves nothing called energy until you enlarge the system to include the field. The symmetry argument never tells you which symmetries nature has — it tells you what follows, exactly, once you have named one.