University Physics V · Quantum Uncertainty and Commutation Relations · 8.9

Quantum Spread vs Experimental Error

Two numbers get called uncertainty and only one of them is physics. This lesson shows you how to pull a state's own variance out of a histogram that a detector has smeared, a drifting preparation has widened, and a finite run has left with an error bar of its own.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Ask an experimentalist for the uncertainty and you will be handed one of three numbers. The first belongs to the state: σ_A² = Tr(ρA²) − (Tr ρA)² is a functional of the density operator and the Hermitian observable alone, computable before anyone builds an apparatus, and it is what the Robertson bound constrains. The second belongs to the run: measure A on N identically prepared copies and the sample mean carries standard error σ/√N, which falls without limit — quantum mechanics never forbids locating ⟨x⟩ to a picometre.

The third belongs to the equipment: a real detector reports a from a true a′ with response R(a|a′), so the recorded histogram is a convolution, and for an unbiased, state-independent, additive response the variances simply add, σ_obs² = σ_A² + σ_R². Preparation drift enters the same way through the law of total variance, because an ensemble whose centre wanders is a mixture ρ = Σ p_λ|ψ_λ⟩⟨ψ_λ| whose variance carries the scatter of those centres. That is the whole model, and its cost is that one histogram cannot separate the three terms: σ_A comes only from calibrating R on a state of known width, logging the drift independently, and subtracting in quadrature — and once σ_R approaches σ_obs the subtraction amplifies its own error faster than extra shots can beat it down.

Simple definition: A state's intrinsic spread is the standard deviation of the outcome distribution that ρ and the Hermitian observable A fix between them, σ_A = √[Tr(ρA²) − (Tr ρA)²]; whatever the apparatus and the finite run add sits on top of it.
Example: For a spin-½ in |+⟩ measured along z, σ² = ⟨σ_z²⟩ − ⟨σ_z⟩² = 1 − 0 = 1 exactly, so every shot returns +1 or −1 and never 0 — yet 2500 shots pin ⟨σ_z⟩ to 1/√2500 = 0.020.

Intrinsic spread of a prepared stateσ_A = √[ Tr(ρA²) − (Tr ρA)² ]

Fixed by the preparation and the choice of observable alone. No apparatus appears in it, so no better apparatus shrinks it.

ρ the density operator, A Hermitian; σ_A carries A's own unit — metres, or dimensionless for a Pauli operator.

Standard error on the ensemble meanu(⟨A⟩) = σ_obs / √N

The mean sharpens without limit while σ_A does not move: 2500 shots on |+⟩ give ⟨σ_z⟩ to ±0.020 with σ = 1 throughout.

N independent shots on freshly prepared copies, same unit as A. σ_obs is the recorded width, not the intrinsic one.

Detector response convolves the true lawp_obs(a) = ∫ R(a|a′) p(a′) da′

Says the histogram you plot is not the Born-rule distribution. Calibrate R on a near-eigenstate before quoting any width.

R the calibrated response, normalised in a for each a′; unbiased means ∫ a R(a|a′) da = a′.

Quadrature deconvolution and its priceσ_A = √(σ_obs² − σ_R²), u(σ_A) = (σ_obs/σ_A) u(σ_obs)

σ_obs = 12.4 µm with σ_R = 5.0 µm gives 11.3 µm and a prefactor of 1.09; at σ_R = 12.0 µm the same data give 3.1 µm and 3.97.

Needs an additive, state-independent response uncorrelated with the observable's own fluctuation.

Preparation drift as a classical mixtureρ = Σ p_λ |ψ_λ⟩⟨ψ_λ| → σ² = ⟨σ_λ²⟩ + Var_λ⟨A⟩_λ

A wandering centre widens the histogram exactly like a wider state. Only purity Tr ρ² or a conjugate observable separates them.

The law of total variance: mean of the within-shot variances plus the variance of the shot-to-shot centres.

What the Robertson bound actually constrainsσₓ(obs) σₚ(obs) ≥ σₓ σₚ ≥ ħ/2 = 5.27 × 10⁻³⁵ J s

Raw widths always clear the floor, so they test nothing. A claim of saturation needs deconvolved σₓ and σₚ.

The first inequality holds because response and drift only add variance; all four are widths of distributions, not resolutions.

Three numbers all called uncertainty

Fix a density operator ρ and a Hermitian observable A. The Born rule gives outcome aₖ the probability Tr(ρPₖ) with Pₖ the spectral projector, and that distribution has variance σ_A² = Tr(ρA²) − (Tr ρA)², written down with no reference to a laboratory. It is an ensemble statement: one shot returns one eigenvalue and has no width at all, so σ_A means something only over many identically prepared copies. Now run the experiment. The dataset carries a second number, the standard error on the mean u(⟨A⟩) = σ_obs/√N, which says how well this run located ⟨A⟩ and nothing about how wide the state is. The apparatus carries a third, its resolution σ_R, the width the response R(a|a′) would produce if fed a perfectly sharp value. Three numbers, three owners: the state, the run, the machine. Almost every confusion in this topic is one of them quoted where another was meant.

More shots sharpen the mean, never the state

Take a spin-½ in |+⟩ = (|0⟩ + |1⟩)/√2 measured along z with eigenvalues ±1. Then ⟨σ_z⟩ = 0, and because σ_z² = I we get ⟨σ_z²⟩ = 1, so the variance is 1 − 0 = 1 in any state whatever the mean. Every single shot returns +1 or −1, never something between. Run 2500 shots and the sample mean has standard error 1/√2500 = 0.020; run 250 000 and it is 0.0020. The mean is being located a hundred times better than the width of the distribution it is drawn from, and nothing objects, because the theorem constrains σ_z and not u(⟨σ_z⟩). This is why an atomic clock quotes a fractional instability falling as 1/√N while each atom keeps its full projection noise: the standard quantum limit says σ = 1 per shot, so a phase estimate improves as N^(−1/2) unless the atoms are entangled. Note what the argument needs — N independent, freshly prepared copies. Measurement projects the state, so these are repetitions of a preparation, never repeated looks at one system.

Calibrate the detector, then subtract variances

A real instrument reports a when the truth is a′ with conditional probability R(a|a′), so the histogram you plot is the convolution of R with the Born-rule density. If R is additive, unbiased, and independent of the value being measured, its effect on the second moment is simple: variances add, σ_obs² = σ_A² + σ_R². Two consequences follow. First you must calibrate — send in a state whose own width is negligible, a tightly localised reference ion, a pinhole, a resolved single emitter, and the recorded width is σ_R. Second you subtract in quadrature, not linearly: σ_obs = 12.4 µm with σ_R = 5.0 µm gives √(153.76 − 25.0) = 11.3 µm, where naive subtraction would have claimed 7.4 µm. The subtraction has a price the propagation makes explicit, u(σ_A) = (σ_obs/σ_A) u(σ_obs), a prefactor of 1.09 here but 3.97 had σ_R been 12.0 µm. Once the detector is as wide as the signal, more shots tighten u(σ_obs) as N^(−1/2) while the amplification sits where it is; you need a better camera, not a longer run.

Preparation drift is a mixture, and total variance hides it

Suppose the preparation is not perfectly repeatable: the trap centre wanders, a detuning drifts, the field breathes. Shot λ prepares |ψ_λ⟩ with probability p_λ, so the ensemble is the mixture ρ = Σ p_λ|ψ_λ⟩⟨ψ_λ|, and the law of total variance splits its spread into the mean within-shot variance plus the variance of the shot-to-shot centres, σ² = ⟨σ_λ²⟩ + Var_λ(⟨A⟩_λ). A Gaussian packet of true width 30.0 nm whose centre jitters with rms 16.0 nm records √(900 + 256) = 34.0 nm, arithmetically identical to a genuinely wider pure state. No histogram of A alone can tell them apart. Two things can. The conjugate observable: a rigid displacement acts as exp(−i d p/ħ), which commutes with p and leaves the momentum distribution untouched, so the drifted ensemble reports σₓ σₚ = (34.0/30.0)(ħ/2) = 1.13 ħ/2 while every member of it saturates the bound. Or purity: Tr ρ² < 1 for the mixture and exactly 1 for a pure state, reachable by a swap test or by tomography.

The bound constrains the state's widths, not your resolution

Robertson reads σ_A σ_B ≥ |⟨[A, B]⟩|/2, and every symbol in it refers to the distribution the state assigns. Feed it recorded widths and you get something true but empty, because response and drift only add variance: σₓ(obs) σₚ(obs) ≥ σₓ σₚ ≥ ħ/2. An experiment quoting a raw product of 3ħ/2 has not tested the bound, it has tested its own camera. The interesting measurements run the other way — showing a squeezed or ground-state packet sitting at the floor — and those are precisely the ones that cannot be made without a calibrated deconvolution, since the claim is that nothing is left once the instrument is removed. Two boundaries deserve naming. The bound is not a statement about resolution: a camera with 1 µm pixels imaging a 50 µm packet is entirely legal, because σₓ describes the ensemble histogram while the pixel describes one reading. And error–disturbance relations, the Ozawa and Branciard inequalities, are separate theorems about a measurement's own error and the kick it delivers, with different hypotheses and different proofs.

Build the error budget, then vary one term at a time

Write the budget before the run. The recorded variance is a sum of terms with different owners, σ_obs² = σ_A² + Var_λ(⟨A⟩_λ) + σ_R², while the reported mean carries u(⟨A⟩) = σ_obs/√N plus any calibration offset, which does not fall with N at all. Then test each term by moving only what it depends on. Increase N: the error on the mean should fall as N^(−1/2) and nothing else should move, and if σ_obs grows too then the preparation is drifting on the timescale of the run. Change the preparation — retune the trap frequency, recool, change the squeezing — and σ_A moves while σ_R does not. Defocus deliberately, rebin the pixels, attenuate the signal: σ_R moves while σ_A does not. Interleave the calibration state through the run rather than measuring R once at the start, because drift in R itself is the one failure this budget cannot see. What survives all three checks is a number about ρ; what does not is a number about your morning.

Change one variable at a time

Make the relationship visible.

Interactive model

State spread σ_ψ12 nm

Detector response σ_R8 nm

Shots averaged N25 shots

Hold the state at 12 nm and push the detector to 24 nm: the solid curve inflates to 26.8 nm, overstating the state 2.24-fold. Then drive N from 1 to 400 and the bar on the mean shrinks twentyfold while neither curve moves at all.

STATE σ12.0 nm

OBSERVED σ14.4 nm

ERROR ON MEAN2.88 nm

WIDTH INFLATION1.20 times

Live interpretationSTATE σ: 12.0 nm. OBSERVED σ: 14.4 nm. ERROR ON MEAN: 2.88 nm. WIDTH INFLATION: 1.20 times

Catch the common trap

Explain before calculating.

Misconception check

The usual error is to read Δx as the precision of a position measurement, and conclude either that a camera resolving 1 µm beats the uncertainty principle on a 50 µm packet, or that averaging 10 000 shots has narrowed the state to 0.5 µm. Neither is true and neither is forbidden. Δx is the standard deviation of the outcome distribution ρ assigns to the position operator, fixed by the preparation before any apparatus is switched on. Resolution belongs to the detector and may be far finer; the standard error on the mean belongs to the dataset and falls as 1/√N without limit. What the theorem forbids is a state whose position and momentum histograms are both narrow, not an apparatus that reads either one sharply. And disturbance is a separate subject: Robertson proves the bound by Cauchy-Schwarz on two mean-shifted operators, with no disturbance in it anywhere.

An ion's axial position is imaged on 10 000 identically prepared shots, and the histogram has standard deviation 12.4 µm. Imaging a tightly localised reference source gives a point-spread standard deviation of 5.0 µm. What should be quoted for the state's intrinsic spread σₓ, and for the uncertainty on the mean position?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA spin-½ is prepared in |+⟩ = (|0⟩ + |1⟩)/√2 and measured in the σ_z basis with eigenvalues ±1. Find the intrinsic spread of a single measurement, the standard error on ⟨σ_z⟩ after 2500 shots, and the number of shots needed to pin ⟨σ_z⟩ down to 0.0020.

Mean first: ⟨σ_z⟩ = ⟨+|σ_z|+⟩ = ½(⟨0| + ⟨1|)(|0⟩ − |1⟩) = ½(1 − 1) = 0.
Second moment: σ_z² = I, so ⟨σ_z²⟩ = 1 in any state at all. The variance is 1 − 0² = 1, so σ = 1 exactly — every shot returns +1 or −1 and never a value in between.
Standard error after N shots: u = σ/√N = 1/√2500 = 1/50 = 0.020.
Invert for the target: N = (σ/u)² = (1/0.0020)² = 2.5 × 10⁵ shots. σ stays at 1 the whole way, being a property of |+⟩ and σ_z that no amount of data touches.

Answerσ = 1 exactly; u(⟨σ_z⟩) = 0.020 after 2500 shots; 2.5 × 10⁵ shots to reach 0.0020. Only the estimate of the mean improves — the width never does.

MediumA minimum-uncertainty Gaussian packet with σₓ = 30.0 nm is released from a trap whose centre jitters shot to shot with rms 16.0 nm, independently of the state. Find the recorded position width, the momentum width, and the uncertainty product an experimenter would report from the raw histograms. Take ħ = 1.0546 × 10⁻³⁴ J s.

The ensemble is a mixture of displaced copies, ρ = ∫ p(d) |ψ_d⟩⟨ψ_d| dd. The law of total variance gives σ_obs² = σₓ² + Var(d) = 30.0² + 16.0² = 900 + 256 = 1156 nm², so σ_obs = 34.0 nm.
The displacement is generated by momentum: |ψ_d⟩ = exp(−i d p/ħ)|ψ⟩, and that operator commutes with p, so every displaced copy carries the identical momentum distribution. The jitter adds nothing to σₚ.
Minimum uncertainty for the undisplaced packet: σₚ = ħ/(2σₓ) = 1.0546 × 10⁻³⁴ / (6.00 × 10⁻⁸) = 1.758 × 10⁻²⁷ kg m s⁻¹.
Raw product: σ_obs σₚ = (3.40 × 10⁻⁸)(1.758 × 10⁻²⁷) = 5.98 × 10⁻³⁵ J s, against ħ/2 = 5.27 × 10⁻³⁵ J s — a ratio of exactly 34.0/30.0 = 1.13.

Answerσ_obs = 34.0 nm, σₚ = 1.76 × 10⁻²⁷ kg m s⁻¹, product 5.98 × 10⁻³⁵ J s = 1.13 ħ/2. Every member saturates the bound; the classically drifting ensemble does not.

HardA Ca-40 ion in a 1.20 MHz axial trap is imaged over N = 1600 shots, giving a histogram of width σ_obs = 25.0 nm. A localised calibration source gives point-spread σ_R = 15.0 ± 0.5 nm, and independently logged shot-to-shot centres drift with rms 12.0 ± 0.4 nm. Extract the state's own width with its uncertainty and convert it to a mean phonon number. Take m = 39.96 u and ħ = 1.0546 × 10⁻³⁴ J s.

Strip the detector first: σ² = σ_obs² − σ_R² = 625 − 225 = 400 nm², i.e. 20.0 nm. That still contains the drift.
Strip the drift by the law of total variance: σₛₜₐₜₑ² = 400 − 12.0² = 400 − 144 = 256 nm², so σₛₜₐₜₑ = 16.0 nm.
Uncertainty on a Gaussian sample width: u(σ_obs) = σ_obs/√(2(N−1)) = 25.0/√3198 = 0.442 nm. Propagating σₛₜₐₜₑ² = σ_obs² − σ_R² − σ_d² gives contributions (25/16)(0.442) = 0.691, (15/16)(0.5) = 0.469 and (12/16)(0.4) = 0.300 nm, so in quadrature u(σₛₜₐₜₑ) = 0.89 nm.
Ground-state scale of the trap: σ₀ = √(ħ/2mω) with m = 39.96 × 1.6605 × 10⁻²⁷ = 6.636 × 10⁻²⁶ kg and ω = 2π × 1.20 × 10⁶ = 7.540 × 10⁶ s⁻¹, giving σ₀ = 10.27 nm.
A thermal oscillator state has σₓ² = σ₀²(2n̄ + 1), so 2n̄ + 1 = (16.0/10.27)² = 2.43 and n̄ = 0.71, with u(n̄) = ½(2n̄ + 1)(2u(σₛₜₐₜₑ)/σₛₜₐₜₑ) = ½(2.43)(0.111) = 0.13.
Had the camera never been calibrated, the raw 25.0 nm would have given 2n̄ + 1 = (25.0/10.27)² = 5.93 and n̄ = 2.5 — roughly three quarters of the apparent heating belonging to the detector rather than to the ion.

Answerσₛₜₐₜₑ = 16.0 ± 0.9 nm, so n̄ = 0.71 ± 0.13: close to the motional ground state but not in it. Skipping the calibration would have reported n̄ ≈ 2.5.