Commutators and the uncertainty relation

An operator is an instruction, and instructions compose in an order. Position multiplies, sending ψ to xψ; momentum differentiates, sending ψ to −iℏ dψ/dx. Act with both, both ways round, on the same differentiable ψ. In the product xp the right-hand operator goes first, so (xp)ψ = x(−iℏ dψ/dx) = −iℏ x dψ/dx. In px the position acts first, and the product rule bites: (px)ψ = −iℏ d(xψ)/dx = −iℏ(ψ + x dψ/dx). Subtract, and the derivative terms cancel: (xp − px)ψ = iℏψ. What is left is iℏ times ψ itself, for every ψ — which is what entitles you to write the operator identity [x, p] = iℏ rather than a claim about one particular state. That identity also tells you how large the state space has to be. Traces obey Tr(AB) = Tr(BA), so any commutator of finite matrices is traceless, while iℏ times an N-dimensional identity has trace iℏN ≠ 0; no finite list of matrices can carry x and p, and the pair demands an infinite-dimensional space to live on. Two rules then handle the rest of the algebra without any further test functions: [A, B] = −[B, A], and [A, BC] = B[A, C] + [A, B]C.

Let A and B be Hermitian, with means ⟨A⟩ and ⟨B⟩ in a normalised state ψ, and define A′ = A − ⟨A⟩ and B′ = B − ⟨B⟩. Hermiticity is what lets you move a factor across the inner product, so the variance is a squared length: σ_A² = ⟨ψ|A′²|ψ⟩ = ‖A′ψ‖². Now apply Cauchy–Schwarz to the two vectors A′ψ and B′ψ: ‖A′ψ‖²‖B′ψ‖² ≥ |⟨A′ψ, B′ψ⟩|², that is σ_A²σ_B² ≥ |⟨A′B′⟩|². Split the complex number ⟨A′B′⟩ into its real part ½⟨A′B′ + B′A′⟩, the covariance, and its imaginary part ⟨[A, B]⟩/2i, using [A′, B′] = [A, B] since the shifts are only numbers. Drop the real part, keep the imaginary one, take a square root: σ_A σ_B ≥ ½|⟨[A, B]⟩|. Every step is algebra on a single state vector, which is why no apparatus appears anywhere in it.

Set A = x and B = p. The commutator is a constant, so ⟨[x, p]⟩ = iℏ in every state and the bound loses all state-dependence: σₓ σₚ ≥ ℏ/2 = 5.27 × 10⁻³⁵ J s. Equality needs two conditions at once. Cauchy–Schwarz is tight only when B′ψ = cA′ψ for some complex c, and integrating that first-order equation gives a Gaussian; then the discarded real part, which is c σₓ², must itself vanish, forcing c to be purely imaginary and so killing any quadratic phase. The minimum-uncertainty states are therefore exactly the real Gaussians carrying a plane-wave factor, and nothing else reaches the line. Most states sit comfortably above it. For the hydrogen ground state ⟨x²⟩ = a₀² and ⟨pₓ²⟩ = ℏ²/(3a₀²), so σₓ σ(pₓ) = a₀ × ℏ/(√3 a₀) = ℏ/√3, which is 1.15 times the floor: bound, but not minimal.

The bound is only as strong as ⟨[A, B]⟩ in the state you are holding, and that number can vanish two ways. If the operators commute outright — x with p_y, or energy with total angular momentum and one of its components in a central potential — the right-hand side is zero for every state, a common eigenbasis exists, and both observables can be sharp together. That is exactly what a complete set of commuting observables buys you. The subtler case is a non-commuting pair whose mean commutator happens to vanish in the state at hand, and there the bound goes quiet without anything becoming compatible. Take [Lₓ, L_y] = iℏL_z in the l = 1, m = 0 state: ⟨L_z⟩ = 0, so Robertson demands only σ(Lₓ) σ(L_y) ≥ 0. Yet ⟨Lₓ²⟩ = ⟨L_y²⟩ = (⟨L²⟩ − ⟨L_z²⟩)/2 = (2ℏ² − 0)/2 = ℏ², and both means are zero, so each spread is ℏ and their product is ℏ² — as far from zero as the bound is empty. Robertson's inequality is necessary, never sufficient, and a bound that has collapsed to zero is no evidence that two observables can be sharp together.

Robertson threw away the real part of ⟨A′B′⟩. Put it back and you get the Robertson–Schrödinger inequality, σ_A²σ_B² ≥ cov² + (½|⟨[A, B]⟩|)². For x and p, that covariance is precisely what free evolution manufactures. Start from a real Gaussian of width σ₀: position and momentum are uncorrelated, cov = 0, and the product sits on the floor. Free evolution leaves the momentum distribution alone, because p commutes with the free Hamiltonian, so σₚ never moves from ℏ/(2σ₀); meanwhile the fixed spread of speeds σₚ/m pulls the fast components ahead of the slow ones and builds correlation, cov = σₚ²t/m. Measure both in units of ℏ/2 and the bookkeeping is clean: cov is t/τ and the product is √(1 + (t/τ)²), with τ = 2mσ₀²/ℏ, so the two sides of Schrödinger's inequality stay equal at every instant while Robertson's is matched only at t = 0. One τ puts the packet √2 above the floor; six τ puts it 6.1 times above. Nothing pushes the packet open, and nothing lets it return to the line.

σₓ is a property of ψ, computable before anything is measured: σₓ² = ⟨x²⟩ − ⟨x⟩². Testing the bound therefore takes two ensembles, not one clever apparatus. Prepare the state, measure position on each copy, histogram the results and record the width; prepare the same state again from scratch, measure momentum on each copy, histogram those. Multiply the two widths and see whether the product falls below ℏ/2, which the theorem says it never can. Two practical conditions ride along. Each detector's resolution must sit well inside the σ it is reporting, or the histogram you record is the instrument's width rather than the state's. And the source must be reproducible, since drift between preparations broadens both histograms and inflates the product for a reason that has nothing to do with quantum mechanics. Single-slit diffraction is the familiar realisation, the slit width setting σₓ and the far-field angular spread reporting σₚ; it is an order-of-magnitude check rather than a saturation test, because a hard-edged slit is not a Gaussian and never sits on the line.