The generalised uncertainty relation from Cauchy–Schwarz

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Statement

For any two observables, the product of their standard deviations in a given state is bounded below by half the modulus of the expectation of their commutator.

Why it matters

The position–momentum relation is usually introduced through a hand-waving microscope argument that suggests measurement disturbance. That story is wrong, or at best a different statement. Uncertainty is a property of the state, present before any measurement, and this derivation shows it follows from nothing but the geometry of Hilbert space.

Assumptions

Derivation

Result

Limiting cases and special structures

• Equality holds when |f⟩ ∝ |g⟩ and the anticommutator term vanishes. Working through the resulting differential equation gives a Gaussian — the minimum-uncertainty state, and the reason coherent states matter.
• Angular momentum components obey [L̂_x,L̂_y] = iℏL̂_z, so the bound depends on the state and vanishes when ⟨L̂_z⟩ = 0 — two components can be simultaneously sharp in such a state.
• The zero-point energy of the oscillator follows independently: minimising ⟨E⟩ = p²/2m + ½mω²x² subject to xp ≥ ℏ/2 gives ½ℏω, matching D-010 exactly.

Breaks when

Failure modes

• Reading Δx as an experimental resolution. It is the standard deviation of the distribution the state predicts.
• Believing the relation says a particle "has" a position and momentum we cannot know. It says no state assigns both sharply.

Worked number — position and momentum

Take  = x̂, B̂ = p̂ with [x̂,p̂] = iℏ. The commutator expectation is the constant iℏ, so ½|⟨[x̂,p̂]⟩| = ½|iℏ| = ℏ/2. The result reads ΔxΔp ≥ ℏ/2 ≈ 5.27×10⁻³⁵ J·s — the tightest bound any state can reach, saturated only by the Gaussian.

Run the check → commutator-to-bound verifier for x̂, p̂ and L̂ components

Discussion

The whole relation is a geometric fact wearing physical clothes. In step 3 the deviation states |f⟩ = Â′|ψ⟩ and |g⟩ = B̂′|ψ⟩ are just two vectors in Hilbert space, and Cauchy–Schwarz says |⟨f|g⟩|² ≤ ⟨f|f⟩⟨g|g⟩, which is nothing more than |cos θ| ≤ 1 for the "angle" between them. The uncertainty product ΔA ΔB measures the lengths of these two vectors; the bound is saturated only when they are collinear. So the deep content is: two observables fail to be simultaneously sharp precisely to the extent that acting with one operator on the state points you in a different Hilbert-space direction than acting with the other. There is no measurement, no disturbance, no observer — only the shape of the state.

Why does the commutator set the floor rather than the full inner product ⟨f|g⟩? Because ⟨f|g⟩ = ⟨Â′B̂′⟩ is generally complex, and step 4 resolves it into a real part (the anticommutator, symmetric) and an imaginary part (the commutator, antisymmetric). For Hermitian operators these are cleanly separated: ⟨{Â′,B̂′}⟩ is real and ⟨[Â′,B̂′]⟩ is pure imaginary. Discarding the anticommutator in step 5 keeps only the imaginary axis, and that is exactly the piece the algebra forces to be non-zero when the operators do not commute. This is also why the bound is state-dependent: ⟨[Â,B̂]⟩ is an expectation value, so it can shrink or even vanish in special states — unless, as for x̂ and p̂, the commutator is a non-zero constant and no escape exists.

The relation as derived here is Robertson's (1929); throwing away the anticommutator makes it strictly weaker than Schrödinger's (1930), which keeps that term: (ΔA)²(ΔB)² ≥ (½⟨{Â′,B̂′}⟩)² + (½i⟨[Â,B̂]⟩)². The extra term is the quantum covariance and matters for correlated (e.g. squeezed) states. Structurally the ½|⟨[Â,B̂]⟩| floor is a shadow of the Lie algebra of observables: the commutator's structure constants dictate which pairs are jointly measurable. For canonical pairs [x̂,p̂] = iħ the constant is ħ itself, which is why ħ sets the area of the irreducible phase-space cell — the same ħ that appears in the Fourier bandwidth theorem, since x and p representations are Fourier conjugates and Cauchy–Schwarz there is the bandwidth–duration bound. Note too that Robertson's bound is not the last word: entropic uncertainty relations (Maassen–Uffink) give state-independent floors that survive even where ⟨[Â,B̂]⟩ = 0, curing a known weakness of the variance form.

Common misconceptions. (i) This is not the Heisenberg microscope. Measurement–disturbance is a separate relation with a separate derivation and a different bound; the quantity here, ΔA, is the standard deviation of the distribution the state already predicts, computable before any apparatus is switched on. (ii) It does not say a particle secretly "has" a sharp x and p we are too clumsy to read; it says no state assigns both sharply. (iii) The energy–time relation is not an instance of this theorem, because time is a parameter, not a Hermitian operator, so step 4 has nothing to split.

Worked examples

Example 1 — the atomic-scale kinetic energy floor. An electron is prepared so that its position spread is Δx = 0.10 nm, roughly the size of an atom. With ⟨p̂⟩ = 0 (a bound state at rest on average), find the smallest momentum spread and the resulting floor on the mean kinetic energy.

Example 2 — spin-½ and the state-dependence of the bound. For spin the components obey [Ŝ_x,Ŝ_y] = iħŜ_z. Evaluate the bound and the actual uncertainty product for the state |↑_z⟩, then repeat for |→_x⟩, to show the floor moves with the state.

Problems

1. A proton is localised to Δx = 2.0 fm (a nuclear scale) with ⟨p̂⟩ = 0. Find the minimum momentum spread and comment on whether a non-relativistic treatment is safe. (m_p = 1.67×10⁻²⁷ kg.)
   Solution

   Δp ≥ ħ/(2Δx) = 1.055×10⁻³⁴/(2×2.0×10⁻¹⁵) = 2.6×10⁻²⁰ kg m s⁻¹. The associated speed spread is Δv = Δp/m_p = 2.6×10⁻²⁰/1.67×10⁻²⁷ = 1.6×10⁶ m s⁻¹, about 0.5% of c, so a non-relativistic treatment is marginally acceptable. (For an electron in the same 2 fm the momentum would give pc ≈ 49 MeV ≫ m_ec² = 0.511 MeV, so electrons cannot be confined to nuclei — a classic uncertainty argument against nuclear electrons.)
2. A spin-1 system has [L̂_x,L̂_y] = iħL̂_z. In the state |m_z = +1⟩, ⟨L̂_z⟩ = ħ. Give the bound on ΔL_x ΔL_y, and state one m_z value for which the bound vanishes.
   Solution

   ΔL_x ΔL_y ≥ ½ħ|⟨L̂_z⟩| = ½ħ·ħ = ħ²/2 = 5.6×10⁻&sup6;⁹ J²s². The bound vanishes for m_z = 0, since then ⟨L̂_z⟩ = 0 — and indeed in the m_z = 0 state both ⟨L̂_x⟩ and ⟨L̂_y⟩ can be arranged consistently with a small joint spread, though not exactly zero because L_x, L_y still fail to share a full eigenbasis with L_z.
3. Estimate the ground-state energy of a 1D harmonic oscillator of frequency ω by minimising ⟨E⟩ = ⟨p̂²⟩/2m + ½mω²⟨x̂²⟩ over states with ⟨x̂⟩ = ⟨p̂⟩ = 0, using the saturated relation ⟨x̂²⟩⟨p̂²⟩ = ħ²/4.
   Solution

   With ⟨x̂⟩ = ⟨p̂⟩ = 0 we have ⟨x̂²⟩ = (Δx)², ⟨p̂²⟩ = (Δp)². Saturating, ⟨p̂²⟩ = ħ²/(4⟨x̂²⟩). Write u = ⟨x̂²⟩: ⟨E⟩(u) = ħ²/(8mu) + ½mω²u. Set d⟨E⟩/du = −ħ²/(8mu²) + ½mω² = 0 ⇒ u = ħ/(2mω). Back-substitute: ⟨E⟩_min = ħ²/(8m)·(2mω/ħ) + ½mω²·ħ/(2mω) = ħω/4 + ħω/4 = ½ħω. This matches the exact zero-point energy (D-010) exactly, because the oscillator ground state is the Gaussian minimum-uncertainty state that saturates the bound — a coincidence of exactness, not an estimate.
4. Starting from step 5 of the derivation but keeping the anticommutator term, obtain the Schrödinger uncertainty relation, and evaluate it for  = x̂, B̂ = p̂ in a state with covariance C ≡ ½⟨{x̂′,p̂′}⟩ = ħ/2.
   Solution

   |⟨f|g⟩|² is the full modulus-squared of the complex number in step 4, so retaining both parts: (ΔA)²(ΔB)² ≥ (½⟨{Â′,B̂′}⟩)² + (½i⟨[Â,B̂]⟩)² = C² + (½ħ)², where C is the symmetrised covariance. With C = ħ/2: (Δx)²(Δp)² ≥ (ħ/2)² + (ħ/2)² = ħ²/2, i.e. Δx Δp ≥ ħ/√2 ≈ 0.707ħ. The Robertson floor ħ/2 is beaten: a state with x–p correlation (a rotated/squeezed Gaussian) cannot reach ħ/2 — the covariance term stiffens the bound. This is the extra content the standard relation throws away in step 5.
5. Prove that equality in the position–momentum relation forces a Gaussian wavefunction. (Use the two conditions for saturation: Cauchy–Schwarz equality and vanishing anticommutator term.)
   Solution

   Cauchy–Schwarz is an equality iff |g⟩ = λ|f⟩ for some λ ∈ ℂ, i.e. p̂′ψ = λ x̂′ψ. The anticommutator (real-part) term must also vanish: ⟨{x̂′,p̂′}⟩ = 0. Now ⟨{x̂′,p̂′}⟩ = ⟨f|(λ+λ*)|f⟩… more directly, ⟨{x̂′,p̂′}⟩ = 2•Re[λ*]⟨x̂′²⟩·(…) forces Re(λ) ∝ the covariance to be zero, so λ is pure imaginary, λ = iκ with κ real. Writing p̂′ = −iħ d/dx − ⟨p⟩ and x̂′ = x − ⟨x⟩, the eigen-equation (−iħ d/dx − ⟨p⟩)ψ = iκ(x − ⟨x⟩)ψ gives dψ/ψ = [i⟨p⟩/ħ − (κ/ħ)(x−⟨x⟩)] dx. Integrating: ψ(x) = N exp[i⟨p⟩x/ħ] exp[−κ(x−⟨x⟩)²/2ħ]. Normalisability needs κ > 0, giving a Gaussian of width (Δx)² = ħ/2κ modulated by a plane wave of mean momentum ⟨p⟩. Thus the minimum-uncertainty states are exactly the Gaussian wave packets — the coherent and squeezed states of the oscillator.