University Physics V · Quantum Uncertainty and Commutation Relations · 8.3

The Robertson Bound from Cauchy-Schwarz

Most courses assert Δx Δp ≥ ħ/2 and move on. Here you prove the general theorem it belongs to — four lines of Hilbert-space geometry — and then learn to read what the proof throws away, because the floor it hands you is often well below the truth and sometimes exactly zero.

Open the lab ↓Practice & worked examples ↓

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Connect the measurement to the mechanism.

Robertson's theorem is Cauchy-Schwarz wearing physics notation. Fix a normalised |ψ⟩ and two Hermitian observables Â and B̂, then centre them: with ΔÂ = Â − ⟨Â⟩Î, the variance σ_A² = ⟨ψ|ΔÂ²|ψ⟩ is literally the squared norm of the vector ΔÂ|ψ⟩. So σ_A σ_B is a product of two lengths, and Cauchy-Schwarz says it is at least the modulus of one complex number, z = ⟨ψ|ΔÂ ΔB̂|ψ⟩.

Splitting that product into its Hermitian and anti-Hermitian halves, ΔÂ ΔB̂ = ½(ΔÂ, ΔB̂) + ½[Â, B̂], sorts z into a real part — the symmetrised covariance — and an imaginary part, half the mean commutator, because a Hermitian operator has real expectation and an anti-Hermitian one has imaginary expectation. Keep both squares and you have Schrödinger's relation; discard the covariance square and you have Robertson's, σ_A σ_B ≥ ½|⟨[Â, B̂]⟩|. That discard is where the price is paid.

The right-hand side is not a constant of nature but an expectation value in the very state whose spreads sit on the left, so it moves as the state moves and can vanish outright: for L̂x and L̂y in |l, 0⟩ it reads zero while the true product is ħ²l(l+1)/2. Robertson is a floor, never an estimate.

Simple definition: The Robertson relation says that for any two Hermitian observables and any normalised state, the product of their standard deviations is at least half the modulus of the expectation of their commutator, taken in that same state.
Example: For x̂ and p̂ the commutator is the constant iħÎ, so the floor is the state-independent ħ/2 = 5.27 × 10⁻³⁵ J s; the infinite well's ground state returns σₓ σₚ = 0.5679 ħ, which clears that floor by 13.6%.

Centred operators and their spreadsΔÂ = Â − ⟨Â⟩Î, σ_A² = ⟨ψ|ΔÂ²|ψ⟩ = ‖ΔÂ|ψ⟩‖²

Rewriting a variance as a squared norm is what lets a Hilbert-space inequality say anything about measured spreads.

Â Hermitian, so ⟨Â⟩ is real and ΔÂ is Hermitian; σ_A carries Â's own unit — m for x̂, kg m s⁻¹ for p̂.

Cauchy-Schwarz on the centred vectors⟨f|f⟩⟨g|g⟩ ≥ |⟨f|g⟩|², with |f⟩ = ΔÂ|ψ⟩ and |g⟩ = ΔB̂|ψ⟩

This is the only inequality in the derivation. Every later line is an identity, so slack begins here or at the one deliberate discard.

Pure geometry, no physics; equality holds exactly when |g⟩ = λ|f⟩ for some complex λ, or one vector is zero.

Hermitian and anti-Hermitian splitΔÂ ΔB̂ = ½(ΔÂ, ΔB̂) + ½[Â, B̂]

It sorts z = ⟨ΔÂ ΔB̂⟩ into a real and an imaginary part, and |z|² = (Re z)² + (Im z)² does the rest.

Anticommutator Hermitian, so real mean; commutator anti-Hermitian, so imaginary mean; and [ΔÂ, ΔB̂] = [Â, B̂] exactly.

Schrödinger's relation, before anything is droppedσ_A² σ_B² ≥ ¼|⟨(ΔÂ, ΔB̂)⟩|² + ¼|⟨[Â, B̂]⟩|²

Strictly stronger than Robertson, and the term that keeps a bound alive when the mean commutator vanishes.

The first term is the squared symmetrised covariance; both terms carry the units of A²B².

The Robertson boundσ_A σ_B ≥ ½|⟨ψ|[Â, B̂]|ψ⟩|

A floor, not a prediction: it can be met exactly, exceeded by a factor of five, or read zero and say nothing at all.

Both sides are evaluated in the same |ψ⟩; the units are those of A×B. For x̂ and p̂ the right side is the constant ħ/2.

When the bound is saturatedΔB̂|ψ⟩ = λ ΔÂ|ψ⟩ with λ purely imaginary

For x̂ and p̂ this is the first-order equation (p̂ − ⟨p̂⟩)ψ = λ(x̂ − ⟨x̂⟩)ψ, whose normalisable solutions are Gaussians.

Proportionality saturates Cauchy-Schwarz and gives z = λ σ_A², so Re λ = 0 is exactly the vanishing of the covariance.

A variance is the squared length of a centred vector

Fix a normalised |ψ⟩ and a Hermitian Â. Because Â is Hermitian, ⟨Â⟩ = ⟨ψ|Â|ψ⟩ is real, so ΔÂ = Â − ⟨Â⟩Î is Hermitian too, and σ_A² = ⟨ψ|ΔÂ²|ψ⟩ = ⟨ΔÂψ|ΔÂψ⟩ = ‖ΔÂ|ψ⟩‖². That one rewriting is what makes the theorem available: a variance, a statistical claim about repeated measurements on identically prepared systems, has become the squared length of a definite vector in Hilbert space, and everything after it is geometry. Two consequences arrive immediately. σ_A = 0 if and only if ΔÂ|ψ⟩ = 0, that is, |ψ⟩ is an eigenvector of Â — a sharp value is an eigenvector statement, not the limit of careful preparation. And ΔÂ|ψ⟩ is orthogonal to |ψ⟩, since ⟨ψ|ΔÂ|ψ⟩ = 0 by construction; for a qubit that pins it to a single ray, which is why every pure spin-½ state will turn out to saturate the stronger Schrödinger relation exactly.

Cauchy-Schwarz is the only inequality in the proof

Set |f⟩ = ΔÂ|ψ⟩ and |g⟩ = ΔB̂|ψ⟩. Cauchy-Schwarz on any inner-product space gives ⟨f|f⟩⟨g|g⟩ ≥ |⟨f|g⟩|², which reads σ_A² σ_B² ≥ |⟨ψ|ΔÂ ΔB̂|ψ⟩|² once Hermiticity is used to move ΔÂ across to the bra. It is worth recalling why it is true, because the equality case is the part you will use later: expand the non-negative quantity ‖|g⟩ − (⟨f|g⟩/‖f‖²)|f⟩‖², the squared length of the piece of |g⟩ orthogonal to |f⟩, and the inequality falls out, with equality exactly when that piece is zero — that is, when |g⟩ is a multiple of |f⟩. Nothing else in the derivation is an inequality. Every remaining line is an identity, so any looseness in the final bound was created either here, by |f⟩ and |g⟩ failing to be parallel, or by the single deliberate discard that comes next.

Sorting the complex number into real and imaginary parts

The product of two Hermitian operators is not Hermitian, so z = ⟨ΔÂ ΔB̂⟩ is complex, and the trick is to split it by symmetry: ΔÂ ΔB̂ = ½(ΔÂ, ΔB̂) + ½[ΔÂ, ΔB̂]. The anticommutator is Hermitian, so its expectation is real; the commutator is anti-Hermitian, so its expectation is purely imaginary. And because a multiple of the identity commutes with everything, [ΔÂ, ΔB̂] = [Â, B̂] exactly — the centring shifts drop out of the commutator, which is why the final answer never mentions ⟨Â⟩ or ⟨B̂⟩. So Re z = ½⟨(ΔÂ, ΔB̂)⟩, the symmetrised covariance, and Im z = ½⟨[Â, B̂]⟩/i. Now |z|² = (Re z)² + (Im z)². Keep both squares and you have Schrödinger's relation, σ_A² σ_B² ≥ ¼|⟨(ΔÂ, ΔB̂)⟩|² + ¼|⟨[Â, B̂]⟩|². Drop the covariance square — legal, since it is never negative — and what remains is Robertson. The famous relation is deliberately the weaker of the two.

The floor moves, because it is itself an expectation value

Read the right-hand side carefully: ½|⟨ψ|[Â, B̂]|ψ⟩| is evaluated in the same |ψ⟩ whose spreads stand on the left. Position and momentum are the exception that hides this, because [x̂, p̂] = iħÎ is a multiple of the identity, its expectation is iħ in every state, and the floor is the constant ħ/2 = 5.27 × 10⁻³⁵ J s. Almost nothing else behaves that way. For angular momentum [L̂x, L̂y] = iħL̂z gives the floor (ħ/2)|⟨L̂z⟩|, a function of the state. In |l, m⟩ the exact product is σ(L̂x) σ(L̂y) = ½ħ²(l(l+1) − m²), since ⟨L̂x²⟩ = ⟨L̂y²⟩ = ½(⟨L̂²⟩ − ⟨L̂z²⟩) and ⟨L̂x⟩ = ⟨L̂y⟩ = 0, while the floor is ½ħ²|m|. At m = ±l the two agree exactly. At l = 2, m = 1 the product is 2.5ħ² against a floor of 0.5ħ², a factor of five. At m = 0 the floor is zero and the theorem says nothing, while the product is 3ħ².

Which step lost the inequality, and how to tell

Two conditions must hold for equality, and they are independent. Cauchy-Schwarz needs ΔB̂|ψ⟩ = λ ΔÂ|ψ⟩; then z = λσ_A², and the discard costs nothing only if Re λ = 0. So a state saturates Robertson exactly when the two centred vectors are parallel with a purely imaginary constant, which for x̂ and p̂ is a first-order ODE solved only by Gaussians. Diagnosis matters more than the verdict. The infinite well's ground state gives σₓ σₚ = 0.5679 ħ, 13.6% clear of the floor, and its covariance is exactly zero because the eigenfunction is real — so Schrödinger's relation is no stronger here, and the whole loss is Cauchy-Schwarz: no Gaussian can vanish at both walls. The state |l, 0⟩ fails differently. There ΔL̂x|ψ⟩ and ΔL̂y|ψ⟩ are exactly orthogonal, so z = 0 and both parts of Schrödinger's relation vanish together. Same empty bound, different culprit.

Checking Robertson in NumPy, and the truncation trap

Numerically the theorem is three lines: build Hermitian arrays A and B, centre them with dA = A - (v.conj() @ A @ v) * np.eye(N), read each spread as √((v.conj() @ dA @ dA @ v).real), and compare the product with 0.5 * abs(v.conj() @ (A @ B - B @ A) @ v). Assert it for a hundred random normalised v and it will hold every time — which proves less than it appears to. On any N × N truncation the trace of a commutator is zero, so no finite matrices satisfy [x̂, p̂] = iħÎ; on the oscillator's Fock truncation the exact result is iħ(Î − N|N−1⟩⟨N−1|), with the entire error dumped on the top rung. Truncate at N = 60 and the n = 0 state returns σₓ σₚ = 0.5000 ħ against a floor of 0.5000 ħ, both correct. The n = 59 state returns 29.5 ħ for both — the inequality passes, saturated, and both numbers are wrong, since the true product is 59.5 ħ. A satisfied bound is not a converged calculation.

Change one variable at a time

Make the relationship visible.

Interactive model

Polar angle θ from +z40 °

Azimuth φ in the xy-plane35 °

Set θ = 90° with φ = 45°: the vertical part collapses, so Robertson reads ≥ 0 and says nothing while the circle is still half a unit wide. Then set φ = 0 at any θ — the arrow stands straight up, nothing is discarded, and the bound is exact.

PRODUCT σ(Sx) σ(Sy)0.790 ħ²/4

ROBERTSON FLOOR0.766 ħ²/4

COVARIANCE, Re z-0.194 ħ²/4

PRODUCT / FLOOR1.03 ×

Live interpretationPRODUCT σ(Sx) σ(Sy): 0.790 ħ²/4. ROBERTSON FLOOR: 0.766 ħ²/4. COVARIANCE, Re z: −0.194 ħ²/4. PRODUCT / FLOOR: 1.03 ×

Catch the common trap

Explain before calculating.

Misconception check

The bound is remembered as a property of the pair Â, B̂ when it is a property of the pair and the state together. x̂ and p̂ are the exception that teaches the wrong habit: [x̂, p̂] = iħÎ is a multiple of the identity, so the floor is ħ/2 in every state and never moves. Almost nothing else does that. For L̂x and L̂y the floor is (ħ/2)|⟨L̂z⟩|, and in |l = 2, m = 0⟩ it is exactly zero — from which students conclude either that both spreads vanish or that the state is illegal. Neither: σ(L̂x) = σ(L̂y) = √3 ħ, a product of 3ħ². A zero floor does not say the observables are compatible, because [L̂x, L̂y] = iħL̂z is not the zero operator, only its mean is; it says the theorem has declined to speak. Nor is the repair always Schrödinger's stronger relation, since in any L̂z eigenstate the covariance vanishes too.

In the state |l = 1, m = 0⟩ the Robertson bound for L̂x and L̂y reads σ(L̂x) σ(L̂y) ≥ (ħ/2)|⟨L̂z⟩| = 0. What have you learned about the two spreads?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA spin-½ particle is prepared in |+z⟩. Evaluate σ(Ŝx) and σ(Ŝy), compute the Robertson bound for the pair Ŝx, Ŝy, and decide whether the state saturates it.

Ŝx|+z⟩ = (ħ/2)|−z⟩ and Ŝy|+z⟩ = (iħ/2)|−z⟩. Both results are orthogonal to |+z⟩, so ⟨Ŝx⟩ = ⟨Ŝy⟩ = 0 and the operators are already centred: ΔŜx = Ŝx and ΔŜy = Ŝy.
Spreads: σ(Ŝx)² = ⟨Ŝx²⟩ = (ħ/2)²⟨σ̂x²⟩ = ħ²/4, because σ̂x² = Î, and the same for Ŝy. So σ(Ŝx) = σ(Ŝy) = ħ/2 and the product is ħ²/4.
Floor: [Ŝx, Ŝy] = iħŜz, and ⟨Ŝz⟩ = +ħ/2 in |+z⟩, so ½|⟨[Ŝx, Ŝy]⟩| = ½ħ(ħ/2) = ħ²/4.
The two sides agree, and the equality test says why: ΔŜy|+z⟩ = (iħ/2)|−z⟩ = i ΔŜx|+z⟩, so λ = i is purely imaginary and the discarded covariance, 2σ(Ŝx)² Re λ, is zero.

Answerσ(Ŝx) = σ(Ŝy) = ħ/2, product ħ²/4, floor ħ²/4 — saturated exactly. Every Ŝz eigenstate does this, since ΔŜy|ψ⟩ = ±i ΔŜx|ψ⟩ there.

MediumAn electron sits in the ground state of an infinite square well of width L = 1.00 nm. Compute σₓ and σₚ, test the product against the Robertson floor for x̂ and p̂, and say which step of the derivation is responsible for the shortfall from saturation.

The floor first. [x̂, p̂] = iħÎ is a multiple of the identity, so ⟨[x̂, p̂]⟩ = iħ in every state, and here no boundary term intrudes: ⟨x̂p̂⟩ − ⟨p̂x̂⟩ = iħ∫|ψ|² dx directly. The floor is the constant ħ/2 = 5.273 × 10⁻³⁵ J s.
Position: for ψ₁ = √(2/L) sin(πx/L), ⟨x̂⟩ = L/2 and ⟨x̂²⟩ = L²(1/3 − 1/(2π²)), so σₓ² = L²(1/12 − 1/(2π²)) = L²(0.083333 − 0.050661) = 0.032673 L², giving σₓ = 0.18076 L = 1.8076 × 10⁻¹⁰ m.
Momentum: ⟨p̂⟩ = 0 by symmetry, and ⟨p̂²⟩ = 2mE₁ = π²ħ²/L², so σₚ = πħ/L = π × 1.0546 × 10⁻³⁴ ÷ 1.00 × 10⁻⁹ = 3.3131 × 10⁻²⁵ kg m s⁻¹.
Product: σₓ σₚ = 1.8076 × 10⁻¹⁰ × 3.3131 × 10⁻²⁵ = 5.989 × 10⁻³⁵ J s = 0.5679 ħ. The bound holds, with 13.6% to spare.
Locating the loss: ψ₁ is real, so ⟨(Δx̂, Δp̂)⟩ = 0 and Schrödinger's relation gives the same floor. The slack is all in Cauchy-Schwarz — saturation would need Δp̂|ψ⟩ = λΔx̂|ψ⟩ with λ imaginary, whose solutions are Gaussians, and no Gaussian vanishes at both walls.

Answerσₓ = 0.1808 nm, σₚ = 3.313 × 10⁻²⁵ kg m s⁻¹, so σₓ σₚ = 0.5679 ħ against a floor of 0.5 ħ — 13.6% above it. The covariance is exactly zero, so all the slack is Cauchy-Schwarz: the eigenstate is a sine, not a Gaussian.

HardFor orbital angular momentum with l = 1, compare the states |1, 0⟩ and |1, 1⟩. Compute σ(L̂x) σ(L̂y) and the Robertson floor in each, and identify which step of the derivation loses the inequality where it is loose.

In the ordered basis (|1, 1⟩, |1, 0⟩, |1, −1⟩), L̂x = (ħ/√2)[[0,1,0],[1,0,1],[0,1,0]] and L̂y = (ħ/√2)[[0,−i,0],[i,0,−i],[0, i,0]], with L̂z = ħ diag(1, 0, −1).
Take |ψ⟩ = |1, 0⟩ = (0, 1, 0)ᵀ. Then L̂x|ψ⟩ = (ħ/√2)(1, 0, 1)ᵀ and L̂y|ψ⟩ = (ħ/√2)(−i, 0, i)ᵀ, so ⟨L̂x⟩ = ⟨L̂y⟩ = 0 and each squared norm is ħ². Hence σ(L̂x) = σ(L̂y) = ħ and the product is ħ².
The floor is ½ħ|⟨L̂z⟩| = 0, so the theorem reads ħ² ≥ 0. The loss is at Cauchy-Schwarz: ⟨f|g⟩ = (ħ²/2)[(1)(−i) + (1)(i)] = 0, so the two centred vectors are orthogonal, z = 0, and Schrödinger's covariance term is empty as well.
Now take |ψ⟩ = |1, 1⟩ = (1, 0, 0)ᵀ. Then L̂x|ψ⟩ = (ħ/√2)(0, 1, 0)ᵀ and L̂y|ψ⟩ = (ħ/√2)(0, i, 0)ᵀ, so σ(L̂x) = σ(L̂y) = ħ/√2 and the product is ħ²/2.
Its floor is ½ħ|⟨L̂z⟩| = ½ħ(ħ) = ħ²/2 — saturated. Both equality conditions hold: |g⟩ = i|f⟩ is proportional with λ = i purely imaginary, so the covariance 2σ(L̂x)² Re λ vanishes.
The rule behind both cases: in |l, m⟩, σ(L̂x) σ(L̂y) = ½ħ²(l(l+1) − m²) while the floor is ½ħ²|m|, so the ratio is (l(l+1) − m²)/|m|. It is 1 at m = ±l and diverges as m → 0.