University Physics IV · Foundations of Quantum Mechanics · 5.8

Commutators & the Uncertainty Relation

Two observables that fail to commute cannot both be sharp in the same state. Here you learn to evaluate a commutator on a test function, turn it into a numerical floor under a product of standard deviations, and say exactly which claim that floor makes — and which it does not.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Operators act, and the order they act in matters. Apply position then momentum to a wave function and you do not get what momentum then position gives; the difference is not small and not state-dependent, it is the constant [x, p] = iℏ. Robertson's theorem converts that algebraic fact into a statement about statistics.

Write the shifted operators A′ = A − ⟨A⟩ and B′ = B − ⟨B⟩, note that σ_A² is the squared length of A′ψ, apply the Cauchy–Schwarz inequality to A′ψ and B′ψ, and keep only the imaginary part of their overlap; what survives is σ_A σ_B ≥ ½|⟨[A, B]⟩|. Nothing in that derivation mentions an apparatus, a disturbance, or a measurement in progress. It is a theorem about one normalised state, and each σ is the spread you would find by preparing that state many times over and measuring one observable on a fresh copy each time.

For x and p the mean commutator is the same number in every state, so the bound hardens into σₓ σₚ ≥ ℏ/2 = 5.27 × 10⁻³⁵ J s. The cost is that the inequality is usually slack: discarding the real part of the overlap throws away the position–momentum covariance, and restoring it gives the sharper Robertson–Schrödinger form. That is why an unchirped Gaussian sits exactly on the floor at the instant it is made and lifts off it as soon as it starts to spread — and why, where ⟨[A, B]⟩ happens to vanish, the bound says nothing at all.

Simple definition: The commutator [A, B] = AB − BA measures how far two observables are from being simultaneously sharp, and Robertson's theorem says that in any state the product of their standard deviations is at least half the modulus of that commutator's expectation value.
Example: Since [x, p] = iℏ, an electron prepared with σₓ = 0.10 nm must carry σₚ ≥ ℏ/(2σₓ) = 5.3 × 10⁻²⁵ kg m s⁻¹, a kinetic energy of about 0.95 eV — which is why pinning an electron inside an atom costs electronvolts.

Canonical commutator[x, p] = xp − px = iℏ

The same number in every state, which is what makes the position–momentum bound state-independent.

Test it on any differentiable ψ: x(−iℏψ′) − (−iℏ)(xψ)′ = iℏψ. ℏ = 1.055 × 10⁻³⁴ J s.

Robertson's general uncertainty relationσ_A σ_B ≥ ½ |⟨[A, B]⟩|

Turns an operator identity into a number you can test against two separately measured spreads.

σ_A = √(⟨A²⟩ − ⟨A⟩²) in the state ψ; A and B Hermitian and both variances finite.

The position–momentum floorσₓ σₚ ≥ ℏ/2 = 5.27 × 10⁻³⁵ J s

Squeeze one spread and the other must grow. No state, however cunning, gets under the line.

σₓ in m and σₚ in kg m s⁻¹, so the product carries units of action, matching ℏ.

The states that reach the floorψ ∝ exp(i p₀ x/ℏ) exp(−(x − x₀)²/(4σₓ²)) → σₓ σₚ = ℏ/2

The unique minimum-uncertainty family. Add a chirp, a phase going as x², and the product rises above ℏ/2.

Real width and no quadratic phase; any σₓ will do, since only the product is pinned.

Robertson–Schrödinger, the sharper formσ_A²σ_B² ≥ cov² + (½|⟨[A, B]⟩|)², with cov = ½⟨A′B′ + B′A′⟩

Explains why a spreading packet sits above the floor: it is busy building position–momentum correlation.

cov is the real symmetrised covariance, in units of A times B, and is zero for a real Gaussian.

A free Gaussian leaving the floorσₓ(t) σₚ = (ℏ/2)√(1 + (t/τ)²), with τ = 2mσ₀²/ℏ

The floor is touched at one instant only, and free evolution walks the product upward without bound.

σₚ is frozen at ℏ/(2σ₀); for an electron with σ₀ = 1.0 nm, τ = 17 fs.

Order matters, and by exactly iℏ

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: (xp − px)ψ = −iℏ x dψ/dx + iℏψ + iℏ x dψ/dx = iℏψ. The derivative terms cancel and 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. Two rules handle the rest of the algebra: [A, B] = −[B, A], and [A, BC] = B[A, C] + [A, B]C. From the second, [x, p²] = p[x, p] + [x, p]p = 2iℏp, and so [x, H] = iℏp/m for any H = p²/2m + V(x).

Cauchy–Schwarz gives Robertson in four lines

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.

The ℏ/2 floor, and the states that touch 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 must itself vanish, which forces c to be purely imaginary and so kills 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.

Where the commutator vanishes, there is no floor

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 be zero in the state at hand: [Lₓ, L_y] = iℏL_z, yet in the l = 0 ground state of hydrogen ⟨L_z⟩ = 0, so the bound reads σ(Lₓ) σ(L_y) ≥ 0 — and indeed both spreads are zero even though the operators do not commute. Robertson's inequality is necessary, never sufficient; a vacuous bound is not evidence of compatibility.

Schrödinger's covariance term and the spreading packet

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 an electron as a real Gaussian of width σ₀ = 1.0 nm: at t = 0 the covariance is zero and the product sits exactly on the floor. Free evolution never touches the momentum distribution, so σₚ stays frozen at ℏ/(2σ₀) = 5.3 × 10⁻²⁶ kg m s⁻¹, while σₓ grows as σ₀√(1 + (t/τ)²) with τ = 2mσ₀²/ℏ = 17 fs. The covariance climbs as σₚ²t/m, and the two sides of Schrödinger's inequality stay equal at every instant: the Gaussian saturates the sharper bound always, and the weaker Robertson bound only once. By t = 100 fs the product has reached 5.9 times ℏ/2.

What the inequality is a statement about

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, histogram the results; prepare the same state again from scratch, measure momentum, histogram those. Multiply the two widths. The claim is that the product of those two numbers never falls below ℏ/2, however good either instrument is. It does not say a single position measurement is fuzzy — one electron can be located to whatever resolution your detector supports, and no inequality is threatened. Nor does it say that measuring position kicks the momentum; that is a separate claim about measurement disturbance, with its own operators and its own inequalities. Read every σ in this unit as the spread over repeated preparations.

Change one variable at a time

Make the relationship visible.

Interactive model

Initial width σ₀1.0 nm

Elapsed time t / τ0.0 τ

Leave the time at zero and note that the marker sits on the curve: that is a minimum-uncertainty Gaussian. Then run time forward and watch it slide horizontally — free evolution never touches σₚ, so the product can only climb.

POSITION SPREAD σₓ1.00 nm

MOMENTUM SPREAD σₚ98.7 eV/c

PRODUCT IN UNITS OF ℏ/21.00

ELAPSED TIME t0.0 fs

Live interpretationPOSITION SPREAD σₓ: 1.00 nm. MOMENTUM SPREAD σₚ: 98.7 eV/c. PRODUCT IN UNITS OF ℏ/2: 1.00. ELAPSED TIME t: 0.0 fs

Catch the common trap

Explain before calculating.

Misconception check

The usual gloss — you cannot measure position and momentum at the same time, because measuring one disturbs the other — is not what the inequality says. σₓ and σₚ are numbers computed from ψ alone, fixed the moment the state is prepared and before any instrument is switched on, and they are the widths of two histograms gathered in two separate runs on identically prepared copies. A single electron's position can be recorded as precisely as your detector allows, and that measurement violates nothing. Disturbance is a genuine and separate topic with its own error–disturbance relations, which look different from Robertson's and which experiments have used to show the naive Heisenberg version failing where Ozawa's corrected version holds. The floor is a fact about which states exist, not about how clumsy instruments are.

An electron source prepares every electron in the same state. One run measures position and returns a spread σₓ = 50 pm; a second run, on freshly prepared electrons, measures momentum and returns σₚ = 3.0 × 10⁻²⁴ kg m s⁻¹. What do the two numbers tell you? Take ℏ/2 = 5.27 × 10⁻³⁵ J s.

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyVerify the canonical commutator by acting with x and p = −iℏ d/dx in both orders on an arbitrary differentiable ψ(x), then use the product rule for commutators to evaluate [x, p²].

In the product xp the right-hand operator acts first: (xp)ψ = x(−iℏ dψ/dx) = −iℏ x dψ/dx.
In px the position acts first, so the derivative meets a product: (px)ψ = −iℏ d(xψ)/dx = −iℏ(ψ + x dψ/dx).
Subtract: (xp − px)ψ = −iℏ x dψ/dx + iℏψ + iℏ x dψ/dx = iℏψ. The derivative terms cancel, and the result holds for every ψ, so [x, p] = iℏ as an operator identity rather than a statement about one state.
For [x, p²] use [A, BC] = B[A, C] + [A, B]C with B = C = p: [x, p²] = p[x, p] + [x, p]p = iℏp + iℏp = 2iℏp.

Answer[x, p] = iℏ times the identity operator, and [x, p²] = 2iℏp.

MediumA neutron is bound inside a nucleus, so its position spread is roughly σₓ = 2.0 fm. Estimate the smallest momentum spread compatible with that, express it both as σₚ c in MeV and in SI units, and convert it into a kinetic-energy scale. Take ℏc = 197.3 MeV fm, mc² = 939.6 MeV for the neutron, and 1 MeV = 1.602 × 10⁻¹³ J.

Robertson with [x, p] = iℏ gives σₚ ≥ ℏ/(2σₓ).
Work in ℏc units to keep the arithmetic clean: σₚ c ≥ ℏc/(2σₓ) = 197.3 MeV fm ÷ (2 × 2.0 fm) = 49.3 MeV.
In SI: σₚ = 49.3 MeV/c = (49.3 × 1.602 × 10⁻¹³ J) ÷ (3.00 × 10⁸ m s⁻¹) = 2.6 × 10⁻²⁰ kg m s⁻¹.
Kinetic scale: E ≈ (σₚ c)²/(2mc²) = (49.3)²/(2 × 939.6) = 2430/1879 = 1.3 MeV.
Check the assumption: σₚ c = 49 MeV is far below mc² = 940 MeV, so the non-relativistic form was fair. The same bound with an electron and 0.10 nm gave about 1 eV — one relation, six orders of magnitude apart, because the length and the mass changed.

Answerσₚ ≥ 2.6 × 10⁻²⁰ kg m s⁻¹, that is σₚ c ≥ 49 MeV, giving a kinetic-energy scale of about 1.3 MeV.

HardAn electron is prepared at t = 0 as a real Gaussian of width σ₀ = 1.0 nm with zero mean momentum, then evolves freely. Find the time at which its width has grown by a factor √2, evaluate the covariance ½⟨xp + px⟩ at that instant, and check both the Robertson and the Robertson–Schrödinger inequalities there. Take m = 9.11 × 10⁻³¹ kg and ℏ = 1.055 × 10⁻³⁴ J s.

Free evolution leaves the momentum distribution untouched, so σₚ is frozen at ℏ/(2σ₀) = (1.055 × 10⁻³⁴)/(2.0 × 10⁻⁹) = 5.27 × 10⁻²⁶ kg m s⁻¹, and at t = 0 the product is exactly ℏ/2.
The width obeys σₓ(t) = σ₀√(1 + (t/τ)²) with τ = 2mσ₀²/ℏ, so a growth factor of √2 means t = τ. Numerically τ = 2(9.11 × 10⁻³¹)(1.0 × 10⁻⁹)²/(1.055 × 10⁻³⁴) = 1.73 × 10⁻¹⁴ s = 17 fs.
At t = τ: σₓ = 1.41 nm while σₚ is unchanged, so σₓ σₚ = (1.41 × 10⁻⁹)(5.27 × 10⁻²⁶) = 7.46 × 10⁻³⁵ J s, which is √2 × ℏ/2.
The covariance grows as cov = σₚ²t/m = (5.27 × 10⁻²⁶)²(1.73 × 10⁻¹⁴)/(9.11 × 10⁻³¹) = 5.27 × 10⁻³⁵ J s, exactly ℏ/2 at this instant.
Robertson–Schrödinger: √(cov² + (ℏ/2)²) = √2 × ℏ/2 = 7.46 × 10⁻³⁵ J s, matching the product term for term — the Gaussian saturates the sharper bound at every time. Robertson alone demands only ≥ 5.27 × 10⁻³⁵ J s, so it is now loose by a factor √2.
Reading: the packet leaves the ℏ/2 floor the moment position and momentum become correlated, and correlation is exactly what free spreading builds.

Answerτ = 17 fs; cov = ℏ/2 = 5.27 × 10⁻³⁵ J s; σₓ σₚ = 7.46 × 10⁻³⁵ J s = √2 ℏ/2, saturating Robertson–Schrödinger while sitting √2 above the Robertson floor.