The commutator as an algebraic object

[A, B] = AB − BA is defined only where both products are, and what it returns is another operator on the same space — not a number, not a scalar measure of disagreement. It inherits units: [x̂, p̂] carries J s, the units of x times p, which is why ħ and not a pure number stands on the right. Hermiticity constrains the answer sharply. If A and B are both Hermitian, then [A, B]† = (AB)† − (BA)† = B†A† − A†B† = BA − AB = −[A, B], so the commutator of two observables is anti-Hermitian and, unless it vanishes, can never itself be an observable. Multiply by i and it becomes Hermitian: i[A, B] is measurable, and this is exactly why the canonical relation reads iħ rather than ħ. A real constant on the right would contradict Hermiticity before any physics was done. The partner object, the anticommutator {A, B} = AB + BA, is Hermitian for Hermitian A and B, and the pair split any product: AB = ½{A, B} + ½[A, B].

Bilinearity lets you expand sums slot by slot; antisymmetry gives [A, A] = 0 and lets you swap arguments at the cost of a sign; the Leibniz rule [A, BC] = [A, B]C + B[A, C] peels one factor off a product; the Jacobi identity ties three operators together. The Leibniz rule is the workhorse, and its one trap is that the spectator factor must stay on the side it started: writing [A, BC] = [A, B]C + [A, C]B is wrong unless everything in sight happens to commute. Applied to the canonical pair it gives [x̂, p̂²] = [x̂, p̂]p̂ + p̂[x̂, p̂] = iħp̂ + iħp̂ = 2iħp̂, and induction extends that to [x̂, p̂ⁿ] = iħ n p̂ⁿ⁻¹. In other words the map B ↦ [x̂, B] acts on products exactly as iħ d/dp does — it is a derivation. That single structural fact is why the classical Poisson bracket and the quantum commutator obey the same identities, and why {x, p} = 1 and [x̂, p̂] = iħ sit opposite each other in the correspondence.

Three traceless Hermitian 2×2 matrices carry the whole product table σᵢ σⱼ = δᵢⱼ I + i εᵢⱼₖ σₖ. Multiply σₓ = [[0,1],[1,0]] by σ_y = [[0,−i],[i,0]] and you get [[i,0],[0,−i]] = iσ_z; reverse the order and you get −iσ_z. So [σₓ, σ_y] = 2iσ_z and (σₓ, σ_y) = 0, the two halves of the product table read off separately. The bracket closes — a commutator of Paulis is a Pauli times a number — so these three span the Lie algebra su(2), and rescaling to Sᵢ = (ħ/2)σᵢ turns 2i εᵢⱼₖ σₖ into iħ εᵢⱼₖ Sₖ. Jacobi is worth checking here because it is cheap: [σₓ, [σ_y, σ_z]] = [σₓ, 2iσₓ] = 0, and the same for the two cyclic partners, so the sum vanishes term by term. Note also what the right-hand sides look like: 2iσ_z is traceless. Nothing in this algebra ever produces a non-zero multiple of the identity, which is the loophole the next section closes.

For finite matrices Tr(AB) = Σⱼ Σₖ Aⱼₖ Bₖⱼ = Tr(BA), because the double sum is symmetric under relabelling. Therefore Tr[A, B] = Tr(AB) − Tr(BA) = 0 for every pair of d×d matrices, with no exceptions and no dependence on d. Now take the trace of both sides of [x̂, p̂] = iħ I. The left gives 0; the right gives iħd, which is non-zero for every d ≥ 1 — for d = 1000 its modulus is 1.05 × 10⁻³¹ J s, a thousand times ħ, not a rounding error. The canonical commutation relation therefore has no finite-dimensional matrix solution: not an inaccurate one, none. Be precise about what is being forbidden. Non-commuting finite matrices are ordinary, and [σₓ, σ_y] = 2iσ_z is one. What no finite dimension permits is a commutator equal to a non-zero multiple of the identity, because the identity has trace d and commutators have trace zero.

Infinite dimensions remove the trace argument but not the obstruction. Suppose A and B are bounded operators with [A, B] = cI. The Leibniz rule and induction give [A, Bⁿ] = n c Bⁿ⁻¹, so n|c| ‖Bⁿ⁻¹‖ = ‖[A, Bⁿ]‖ ≤ 2‖A‖ ‖Bⁿ‖ ≤ 2‖A‖ ‖B‖ ‖Bⁿ⁻¹‖. If no power of B vanishes, cancel ‖Bⁿ⁻¹‖ and read off n|c| ≤ 2‖A‖‖B‖ for every n, which fails as soon as n exceeds 2‖A‖‖B‖/|c|; if some power does vanish, the smallest such n forces c = 0. That is the Wintner–Wielandt theorem, and its verdict is that x̂ and p̂ cannot both have finite norm: at least one is unbounded, and in the Schrödinger representation on L²(ℝ) both are. Unbounded Hermitian operators cannot be defined on the whole Hilbert space, only on a dense domain, so [x̂, p̂]ψ = iħψ is a statement about the vectors ψ for which both x̂p̂ψ and p̂x̂ψ exist — the Schwartz functions, say. Quoting a commutator without its domain is quoting half a theorem, and it is the half that discretised code discovers the hard way.

Almost every later result in the course is an evaluated commutator. The Heisenberg equation reads dÂ/dt = (i/ħ)[Ĥ, Â] + ∂Â/∂t, so anything that commutes with Ĥ and carries no explicit time dependence is conserved: for a free particle [Ĥ, p̂] = 0 and momentum is constant, while adding V(x̂) gives [Ĥ, p̂] = [V(x̂), p̂] = iħV′(x̂) and hence dp̂/dt = −V′(x̂), the operator form of F = −dV/dx. The Robertson bound ΔA ΔB ≥ ½|⟨[A, B]⟩| takes the commutator as its only input, so the bracket is the whole of that bound rather than a step towards it. Vanishing commutators do work too: two commuting Hermitian operators share a complete eigenbasis, which is what makes a complete set of commuting observables a usable labelling scheme for degenerate states. And Jacobi is not decoration — it makes A ↦ [A, ⋅] a derivation of the bracket itself, so that if A and B both commute with Ĥ then so does [A, B], and the conserved quantities close into an algebra.