University Physics IV · The Schrödinger Equation · 6.9

Commutators & Uncertainty

Two observables either share a basis of states or they do not, and the commutator is what decides. Learn to evaluate one properly, on a test function, then turn its expectation value into the floor beneath a product of spreads — and stay precise about what that floor is a statement about.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

A commutator is the residue left when two operations refuse to swap. Act on a test function with position and then momentum, then in the other order, and the product rule leaves exactly one term behind: (x̂p̂ − p̂x̂)f = iħf, for every f. So [x̂, p̂] = iħ times the identity — not zero, not small, and not something a careful experimenter can whittle down, since it is the same constant in every state.

Two consequences run in opposite directions. If two Hermitian operators commute, they can be diagonalised together: a complete basis of shared eigenfunctions exists, both can be sharp at once, and the pair may label states jointly, which is why hydrogen states carry n, ℓ and m. If they do not, the Cauchy–Schwarz inequality applied to the shifted vectors (Â − ⟨A⟩)ψ and (B̂ − ⟨B⟩)ψ converts the algebra into a number: σ_A σ_B ≥ ½|⟨[Â, B̂]⟩|, and for x and p that floor is ħ/2 in every state there is.

The cost is precision about what has been proved. The inequality contains no apparatus and no measurement — only two variances of one prepared state — so it constrains an ensemble of identical preparations rather than the resolution of any single reading, and it is not the source of the energy–time relation, because t is a parameter labelling the evolution and not an operator with a commutator.

Simple definition: The commutator [Â, B̂] = ÂB̂ − B̂Â measures how far two observables are from being simultaneously definite: it vanishes exactly when a complete shared eigenbasis exists, and otherwise half the modulus of its mean is a floor under the product of their standard deviations.
Example: [x̂, p̂] = iħ makes σₓ σₚ ≥ ħ/2 = 5.27 × 10⁻³⁵ J s in every state. An electron prepared with σₓ = 0.100 nm must carry σₚ ≥ 5.27 × 10⁻²⁵ kg m s⁻¹, a momentum spread of about 990 eV/c.

Commutator, and the canonical one[Â, B̂] = ÂB̂ − B̂Â · [x̂, p̂] = iħ

State-independent, so no preparation and no apparatus can make x and p compatible.

An operator equation: iħ means iħ times the identity. ħ = 1.055 × 10⁻³⁴ J s.

Evaluating [x̂, p̂] on a test functionx̂p̂f − p̂x̂f = −iħxf′ + iħ(f + xf′) = iħf

Keeping f is what stops the false cancellation that wrongly returns zero.

p̂ = −iħ d/dx, and the product rule on d(xf)/dx leaves the single term iħf.

Robertson's generalised relationσ_A σ_B ≥ ½ |⟨[Â, B̂]⟩|

Turns an algebraic identity into a floor the ensemble statistics must clear.

σ_A² = ⟨Â²⟩ − ⟨Â⟩², both operators Hermitian, every average in the same state ψ.

The x–p floor, and who sits on itσₓ σₚ ≥ ħ/2 = 5.27 × 10⁻³⁵ J s

The oscillator's nth state gives (n + ½)ħ, so only the ground state saturates it.

Equality only for a Gaussian, ψ ∝ exp[−(x − x₀)²/4σₓ²] · exp(ip₀x/ħ).

Compatibility: commuting means a shared basis[Â, B̂] = 0 ⟺ a complete common eigenbasis exists

Why quantum numbers come in compatible sets: Ĥ, L̂² and L̂_z give n, ℓ, m.

Inside a degenerate eigenvalue of Â the shared basis must be chosen, by diagonalising B̂ there.

A state-dependent bound: angular momentum[L̂ₓ, L̂_y] = iħL̂_z ⇒ σ_Lx σ_Ly ≥ ½ħ|⟨L̂_z⟩|

A vanishing bound is not compatibility — only a vanishing commutator is.

The right-hand side varies with the state, and in any m = 0 state it is zero.

What a commutator is, and how to evaluate one

ÂB̂ means act with B̂ first, then Â. The commutator [Â, B̂] = ÂB̂ − B̂Â records how much the answer changes when you swap that order. Because these are operators and not numbers, you evaluate one by feeding it an arbitrary differentiable f(x) and simplifying until f can be cancelled at the end. With p̂ = −iħ d/dx: x̂p̂f = x(−iħf′) = −iħxf′, while p̂x̂f = −iħ d(xf)/dx = −iħ(f + xf′). Subtract, the −iħxf′ terms cancel, and iħf survives — so [x̂, p̂] = iħ Î. Drop the f and you will be tempted to cancel the whole thing to zero, which is the single most common slip in this topic. From there the algebra runs on identities rather than integrals: [Â, B̂] = −[B̂, Â] fixes the sign, and [Â, B̂Ĉ] = B̂[Â, Ĉ] + [Â, B̂]Ĉ gives [x̂, p̂²] = 2iħp̂ in one line.

The proof is Cauchy–Schwarz plus one discarded term

Put f = (Â − ⟨A⟩)ψ and g = (B̂ − ⟨B⟩)ψ. Hermiticity lets each operator move across the inner product, so ⟨f|f⟩ = σ_A² and ⟨g|g⟩ = σ_B². Cauchy–Schwarz then gives σ_A²σ_B² ≥ |⟨f|g⟩|². Write z = ⟨f|g⟩. Every complex number obeys |z|² = (Re z)² + (Im z)² ≥ (Im z)², and Im z = (z − z*)/2i works out to ⟨[Â, B̂]⟩/2i. Take the square root and σ_A σ_B ≥ ½|⟨[Â, B̂]⟩|. Notice what was thrown away in that one inequality: Re z, which is the covariance ½⟨ÂB̂ + B̂Â⟩ − ⟨A⟩⟨B⟩. Keeping it gives the stronger Robertson–Schrödinger form, and explains why a packet that has been evolving for a while sits above ħ/2 rather than on it — its position and momentum have become correlated, and the discarded term is exactly that correlation.

What the inequality is about: a prepared ensemble

Nothing in that derivation mentions an instrument. σ_A and σ_B are numbers computed from one state ψ — take ⟨Â²⟩ − ⟨Â⟩² by integration, and the same for B̂ — so the inequality is a property of the preparation and of nothing else. The experiment that tests it needs many identically prepared copies: measure position on half of them and build a histogram whose standard deviation is σₓ, measure momentum on the other half for σₚ, then multiply the two. Each individual measurement may be as sharp as the apparatus allows. Localise one electron to 10⁻¹³ m if your detector can do it, and nothing is violated, because that number is a resolution and not the spread of a state. The relation forbids exactly one thing: preparing a state whose two spreads multiply out below ħ/2.

Commuting observables share a basis of states

Suppose [Â, B̂] = 0 and Âψ = aψ. Then Â(B̂ψ) = B̂(Âψ) = a(B̂ψ), so B̂ψ is still an eigenfunction of Â with the same eigenvalue a. If that eigenvalue is non-degenerate, B̂ψ can only be a multiple of ψ, so ψ is automatically an eigenfunction of B̂ as well. If it is degenerate, B̂ maps the eigenspace into itself and you diagonalise B̂ inside that subspace to pick out a shared basis: the basis exists, but it has to be chosen rather than inherited. Hydrogen is the pay-off. Ĥ, L̂² and L̂_z commute pairwise, so a single state can be sharp in all three and carries the labels n, ℓ and m. L̂ₓ does not commute with L̂_z, which is precisely why no state is ever labelled by both at once.

A vanishing bound is not the same as compatibility

Robertson's right-hand side depends on the state, so it can collapse even where the operators flatly refuse to commute. Take [L̂ₓ, L̂_y] = iħL̂_z in the state |ℓ = 1, m = 0⟩. There ⟨L̂_z⟩ = 0, so the bound reads σ_Lx σ_Ly ≥ 0 and says nothing whatever. The true spreads are far from zero: ⟨L̂ₓ²⟩ + ⟨L̂_y²⟩ = ⟨L̂²⟩ − ⟨L̂_z²⟩ = 2ħ² − 0, and symmetry splits that evenly, so σ_Lx = σ_Ly = ħ and the product is ħ² = 1.11 × 10⁻⁶⁸ J² s². The inequality is necessary, never sufficient. Compatibility is settled by the operator identity [Â, B̂] = 0 holding across the whole domain, not by the size of a bound in one chosen state. The x–p case is unusual precisely because iħ is state-independent, so its floor can never go soft.

Energy and time are not a commutator pair

ΔE Δt ≥ ħ/2 looks like the same theorem, but it cannot be, because there is no self-adjoint time operator to put inside the commutator. Pauli's argument: if T̂ satisfied [T̂, Ĥ] = iħ, then exp(−iεT̂/ħ) applied to an energy eigenstate would shift its eigenvalue by any ε you like, forcing the spectrum of Ĥ to fill the whole real line — impossible for a Hamiltonian bounded below. In quantum mechanics t is a parameter labelling the evolution, not an observable of the system. Two honest statements survive. A state that decays with lifetime τ has an energy width Γ ≈ ħ/τ: the 2p level of hydrogen, τ = 1.6 ns, gives Γ = 6.6 × 10⁻²⁶ J = 0.41 μeV. And the Mandelstam–Tamm bound, with Δt = σ_A/|d⟨Â⟩/dt|, says no expectation value can shift appreciably in a time shorter than ħ/2σ_E.

Change one variable at a time

Make the relationship visible.

Interactive model

Commutator size |⟨[A, B]⟩|1.00 ħ

Spread σA of the prepared state1.00 a

Spread σB of the prepared state1.00 ħ/a

Hold the commutator at 1 ħ and shrink σA: the curve never moves, so the open dot climbs and drags the least allowed σB up with it. Then set the commutator to 0 — the floor collapses onto the axes and every state is legal, which is what a compatible pair looks like.

PRODUCT σA σB1.00 ħ

FLOOR ½|⟨[A, B]⟩|0.50 ħ

MARGIN ABOVE THE FLOOR0.50 ħ

LEAST σB AT THIS σA0.50 ħ/a

Live interpretationPRODUCT σA σB: 1.00 ħ. FLOOR ½|⟨[A, B]⟩|: 0.50 ħ. MARGIN ABOVE THE FLOOR: 0.50 ħ. LEAST σB AT THIS σA: 0.50 ħ/a

Catch the common trap

Explain before calculating.

An electron is prepared many times over in the same state, which has σₓ = 2.0 × 10⁻¹¹ m. On one copy a detector then locates the electron to within 1.0 × 10⁻¹³ m, two hundred times finer than that spread. Which statement is correct?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyAct with x̂p̂ − p̂x̂ on an arbitrary differentiable f(x), taking p̂ = −iħ d/dx, to establish the canonical commutator. Then use the identity [Â, B̂Ĉ] = B̂[Â, Ĉ] + [Â, B̂]Ĉ to evaluate [x̂, p̂²].

Feed the commutator a test function. Acting with p̂ first and then x̂: x̂p̂f = x(−iħf′) = −iħxf′.
In the other order the derivative sees a product: p̂x̂f = −iħ d(xf)/dx = −iħ(f + xf′).
Subtract. The −iħxf′ terms cancel and one term survives: (x̂p̂ − p̂x̂)f = −iħxf′ + iħf + iħxf′ = iħf.
f was arbitrary, so the result is an operator identity, [x̂, p̂] = iħ Î — a constant in every state, not a number that depends on ψ.
For the square, take B̂ = Ĉ = p̂: [x̂, p̂²] = p̂[x̂, p̂] + [x̂, p̂]p̂ = p̂(iħ) + (iħ)p̂ = 2iħp̂.

Answer[x̂, p̂] = iħ Î and [x̂, p̂²] = 2iħp̂. The second is the engine of Ehrenfest's first result: d⟨x⟩/dt = (i/ħ)⟨[Ĥ, x̂]⟩ = ⟨p̂⟩/m.

MediumAn electron is prepared with a position spread σₓ = 0.150 nm, about an atomic diameter. Find the smallest momentum spread the state may have, and the kinetic-energy scale σₚ²/2m that goes with it. Then repeat for a proton confined to σₓ = 2.0 fm, a nuclear diameter, and say what the comparison shows.

Rearranging σₓ σₚ ≥ ħ/2 gives σₚ ≥ ħ/(2σₓ). No potential enters: this is the floor any state of that width must clear.
Electron: σₚ ≥ (1.055 × 10⁻³⁴ J s)/(2 × 1.50 × 10⁻¹⁰ m) = 3.52 × 10⁻²⁵ kg m s⁻¹.
Its energy scale: σₚ²/(2mₑ) = (3.52 × 10⁻²⁵)²/(2 × 9.109 × 10⁻³¹) = 6.78 × 10⁻²⁰ J = 0.42 eV.
Proton: σₚ ≥ (1.055 × 10⁻³⁴)/(2 × 2.0 × 10⁻¹⁵) = 2.64 × 10⁻²⁰ kg m s⁻¹, and σₚ²/(2mₚ) = (2.64 × 10⁻²⁰)²/(2 × 1.673 × 10⁻²⁷) = 2.08 × 10⁻¹³ J = 1.30 MeV.
One inequality, two wildly different answers: the confinement scale sets the energy scale. Fractions of an electronvolt for an electron in an atom, megaelectronvolts for a nucleon in a nucleus — a factor of three million, and the reason chemistry and nuclear physics live where they do.

Answerσₚ ≥ 3.52 × 10⁻²⁵ kg m s⁻¹ and about 0.42 eV for the electron; σₚ ≥ 2.64 × 10⁻²⁰ kg m s⁻¹ and about 1.30 MeV for the proton. The eV-versus-MeV split follows from ħ/(2σₓ) alone.

HardAn electron is in the angular state |ℓ = 1, m = 0⟩. Using [L̂ₓ, L̂_y] = iħL̂_z, write down what Robertson's relation bounds σ_Lx σ_Ly by. Then compute the actual product from ⟨L̂²⟩ and ⟨L̂_z²⟩, and say what the comparison proves about the relation.

Robertson gives σ_Lx σ_Ly ≥ ½|⟨[L̂ₓ, L̂_y]⟩| = ½ħ|⟨L̂_z⟩|. In |1, 0⟩ we have L̂_z|1, 0⟩ = 0, so ⟨L̂_z⟩ = 0 and the bound reads σ_Lx σ_Ly ≥ 0.
The means themselves vanish. L̂ₓ and L̂_y are built from L̂₊ and L̂₋, which shift m by ±1, so their diagonal elements in |ℓ, m⟩ are zero: ⟨L̂ₓ⟩ = ⟨L̂_y⟩ = 0, and hence σ_Lx² = ⟨L̂ₓ²⟩.
Use L̂² = L̂ₓ² + L̂_y² + L̂_z²: ⟨L̂ₓ²⟩ + ⟨L̂_y²⟩ = ℓ(ℓ + 1)ħ² − m²ħ² = 2ħ² − 0 = 2ħ².
Nothing distinguishes x from y in an m = 0 state, so each takes half: ⟨L̂ₓ²⟩ = ⟨L̂_y²⟩ = ħ², giving σ_Lx = σ_Ly = ħ and σ_Lx σ_Ly = ħ² = 1.11 × 10⁻⁶⁸ J² s².
The bound of zero is satisfied but empty, while the true product is as large as this ℓ = 1 state can make it. Incompatibility is settled by [L̂ₓ, L̂_y] ≠ 0 as an operator identity; the numerical bound can go soft in particular states and never certifies that two observables may be sharp together.

AnswerRobertson gives only σ_Lx σ_Ly ≥ 0, while the actual product is ħ² = 1.11 × 10⁻⁶⁸ J² s². A vanishing bound proves nothing: only [Â, B̂] = 0 does.