Quantum spread vs experimental error

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.

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 in any state whatever, so here the variance is 1 − 0² = 1; in general it is 1 − ⟨σ_z⟩², reaching 1 only when the mean vanishes. 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 then being located five hundred times more finely 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: each atom still returns σ = 1 per shot, so a phase estimate improves only as N^(−1/2), the standard quantum limit, 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.

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.

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.

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.

Write the budget before the run. The recorded variance is a sum of terms with different owners, σ_obs² = ⟨σ_λ²⟩ + 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.