Why uncertainty is not measurement error
The spread belongs to the state, not the detector.
The uncertainty principle is often quoted as a complaint about clumsy instruments. This lesson puts numbers on why that is wrong: σₓ computed from ψ with no apparatus in sight, then the detector's resolution, the number of runs and the disturbance of a measurement added one at a time, to see which a better laboratory could remove.
Build the model
σₓ is fixed by ψ before any detector exists; a perfect detector would still record it.
1The big idea
σₓ is the standard deviation of the positions a state predicts, σₓ² = ⟨x²⟩ − ⟨x⟩², and every term in it is an integral over |ψ|². It is fixed once ψ is, before anyone chooses a detector, just as ⟨H⟩ is. Its laboratory meaning comes from an ensemble: prepare the state afresh, measure x once, repeat, and the histogram of results tends to |ψ|² with width σₓ. Once per system, because a result replaces the state (UQMI-04.04). A real detector adds an independent error of spread r, so the recorded width is √(σₓ² + r²): a better detector lowers it towards σₓ and stops there. More runs shrink the error on the mean, s/√N, without touching the width. That is why σₓσₚ ≥ ħ/2 is a statement about preparations: x is measured on one batch, p on another, and the inequality says no ψ makes both histograms narrow. Classically σₓ can be prepared to zero and every bit of recorded spread is instrumental, which is the intuition this topic replaces. The account has two edges. Removing r needs a calibration, an error independent of x, and r not much larger than σₓ. And the preparation relation says nothing about how far measuring x on one system disturbs its p, or how well one apparatus can read both at once: those are error–disturbance and joint-measurement relations, separate theorems with their own definitions and bounds.
2In one sentence
The uncertainty σₓ of a state is the standard deviation of the positions that state predicts, fixed by ψ alone and read off identically prepared systems each measured once; measurement error is what a particular detector adds on top of it.
Example. An electron in the ground state of a 1.00 nm infinite well has σₓ = 0.181 nm from ψ alone. A detector of resolution 0.10 nm records a spread of √(0.181² + 0.10²) = 0.207 nm; a perfect one would record 0.181 nm, not zero.
3Compute σₓ before naming any apparatus
Unit 3 gave the recipe in one line: σₓ² = ⟨x²⟩ − ⟨x⟩², each term an integral over |ψ|² (UQMI-03.08). Run it on the ground state of an infinite well of width a, ψ₁ = √(2/a) sin(πx/a). The density is symmetric about the middle, so ⟨x⟩ = a/2, and ⟨x²⟩ = a²(1/3 − 1/(2π²)), so σₓ = a√(1/12 − 1/(2π²)) = 0.181a. It has the dimension of a and scales with it, so a narrower box gives a narrower spread. For level n the same integrals give a√(1/12 − 1/(2n²π²)): 0.266a for n = 2, tending to a/√12 = 0.289a as n grows, the spread of a classical particle bouncing between the walls at constant speed, which spends equal time everywhere. For a = 1.00 nm the ground state has σₓ = 0.181 nm. At no point did a detector, a photon or a resolution enter the calculation. σₓ belongs to ψ in the same sense that E₁ does.
4One measurement per system, many systems
What does 0.181 nm predict? Not the jitter of one electron's readings. Prepare the state, measure x once, record the result and set that system aside; prepare again, measure again. The N results fill a histogram that approaches |ψ|², and their sample standard deviation s = √[Σ(xᵢ − x̄)²/(N − 1)] estimates the spread. Once per system, and Unit 4 has already said why: a result leaves the system in a new state concentrated near that result (UQMI-04.04), so an immediate repeat returns nearly the same value. A thousand readings of one electron sample the detector and the post-measurement state, never ψ. The same holds for momentum, and it settles what the uncertainty relation is about: σₓ and σₚ come from two separate batches drawn from one preparation, x measured on one batch and p on the other. No system in the experiment that tests σₓσₚ ≥ ħ/2 is measured twice.
5A real detector adds its own spread in quadrature
Let the detector report X = x + δ, where x is what an ideal instrument would read and δ is its own error, with mean zero, standard deviation r (the resolution), and independent of where the particle lands. Variances of independent quantities add, because the cross term 2⟨(x − ⟨x⟩)δ⟩ averages to zero, so s² = σₓ² + r². The recorded histogram is |ψ|² smeared by the detector's response: still normalised to 1, just wider and lower. Take the ground state above, σₓ = 0.181 nm. A detector with r = 0.10 nm records s = 0.207 nm; one ten times better, r = 0.010 nm, records 0.181 nm. The limit r → 0 gives s → σₓ, not zero: a perfect detector does not remove the quantum spread, it shows it undiluted. The classical case is the one students picture. A particle placed at a known point has σₓ = 0, so s = r and every bit of spread is the instrument's; quantum mechanics adds a second term to the sum that no instrument can reach.
6More runs locate the mean, not the particle
N runs yield two different numbers, and they behave oppositely as N grows. The recorded spread s belongs to the preparation plus the detector. More runs estimate it better, to roughly s/√(2(N − 1)) for Gaussian-shaped data, but s itself does not move. The error on the mean, u(x̄) = s/√N, says how well ⟨x⟩ has been located, and it shrinks without limit. With s = 0.207 nm, 100 runs locate ⟨x⟩ to 0.021 nm and 10 000 runs to 0.0021 nm, a hundredth of the width of the distribution, while every one of those runs still scatters by 0.207 nm. So “measure more and the uncertainty goes down” is true of ⟨x⟩ and false of σₓ. Reporting ⟨x⟩ = 0.500 ± 0.002 nm for the 1.00 nm box violates nothing: the ± describes an average, and no uncertainty relation limits how well an average can be known.
7Calibrate the detector, and know when subtraction fails
To turn s into σₓ you need r from somewhere else: a calibration run on a source known to be far narrower than r, or a scan across a sharp edge. Then σₓ = √(s² − r²), and the propagation rules from University Physics I carry the uncertainties through. The method works when r is comparable to σₓ or smaller: s = 0.50 ± 0.01 nm and r = 0.30 ± 0.01 nm give σₓ = 0.400 ± 0.015 nm. It fails when the detector dominates: s = 1.02 ± 0.01 nm and r = 1.00 ± 0.01 nm give s² − r² = 0.040 ± 0.029 nm², so σₓ = 0.20 ± 0.07 nm, a 35% uncertainty built from two 1% measurements, because a small difference of large squares inherits their errors whole. The subtraction also assumes δ is unbiased and independent of x. A detector whose response changes across the screen, or which shifts every reading, has to be modelled, not subtracted.
8Disturbance and joint measurement are different theorems
σₓσₚ ≥ ħ/2 was proved in UQMI-04.07 and UQMI-04.08 from ψ and the Schwarz inequality. No apparatus appears in the proof, so it cannot be a statement about apparatus. Heisenberg's 1927 microscope asked a different question: if one instrument measures x on a system with error ε(x), how far does it disturb that same system's p, by η(p)? That question has its own theorems. Ozawa's relation, ε(x)η(p) + ε(x)σₚ + σₓη(p) ≥ ħ/2, mixes the errors with the state's own spreads and lets the product ε(x)η(p) alone fall below ħ/2; the matching product for neutron spin components was measured below its naive bound in 2012, and other definitions of error give other bounds. A third question, reading x and p together on every system, gives readouts whose spreads obey σ(X)σ(P) ≥ ħ, twice the preparation floor. Three questions, three theorems, and only the first is the uncertainty relation of this unit.
Key equations
What each relation says, and the units it is written in.
Spread of the prepared state
σₓ² = ⟨x²⟩ − ⟨x⟩²
Every symbol is fixed once ψ is. No resolution, no detector and no N appear: this is the number that σₓσₚ ≥ ħ/2 constrains.
⟨xⁿ⟩ = ∫ xⁿ |ψ(x)|² dx; σₓ in m, ψ in m(−1/2); the recipe of UQMI-03.08
Recorded spread, detector included
s² = σₓ² + r²
Let r → 0 and s → σₓ, not zero. A perfect detector shows the quantum spread undiluted; it cannot remove it.
s: spread of the recorded positions; r: detector resolution; both in m. The detector's error is unbiased and independent of x.
Error on the mean
u(x̄) = s/√N
Locates ⟨x⟩ ever more finely as N grows, and says nothing about how sharply the state is located: 100 times the runs divide u by 10 and leave s alone.
u(x̄) and s in m; N systems, each prepared afresh and measured once
How well the spread itself is known
u(s) ≈ s/√(2(N − 1))
More runs pin s down without shrinking it: 400 runs fix s to about 3.5%, and the value they fix is still s.
u(s) and s in m; for Gaussian-shaped data
The preparation relation, read operationally
σₓ σₚ ≥ ħ/2
σₓ comes from one batch and σₚ from another batch of the same preparation. No system is measured twice, so nothing in it is disturbed.
σₚ in kg m s⁻¹; ħ = 1.055 × 10⁻³⁴ J s; proved in UQMI-04.07 and UQMI-04.08
Reading x and p together on every system
σ(X) σ(P) ≥ ħ
A separate theorem with its own floor, twice ħ/2, because the meter's own noise adds to each spread. It limits that apparatus, not the state.
X, P: the two readings of one apparatus that measures both on each system, unbiased on every state; σ(X) in m, σ(P) in kg m s⁻¹
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. Drag r to 0: the solid curve narrows onto the dashed one and stops at σₓ, however perfect the detector. Then drag N from 1 to 100: the bar shrinks tenfold while neither curve moves, and the σₚ floor ignores r and N alike.
Live results
- RECORDED SPREAD s
- 50.0 nm
- ERROR ON MEAN s/√N
- 10.0 nm
- STATE SHARE σₓ²/s²
- 64 %
- σₚ FLOOR ħ/(2σₓ)
- 1.32 10⁻²⁷ kg m s⁻¹
Catch the common trap
The mistake this lesson is built to prevent.
The usual mistake is to read σₓσₚ ≥ ħ/2 as “measuring position disturbs the particle, so position can never be measured accurately.” Three things are wrong with it. The inequality is proved from ψ alone (UQMI-04.07), with no detector in the proof, so it cannot be about how accurately detectors read. It places no limit on a single position measurement: r can be as small as the technology allows, and the recorded spread then approaches σₓ from above, not zero, because σₓ was never the detector's. And σₓ and σₚ are measured on different systems of the same preparation, so no system is disturbed by the other measurement at all. What the relation forbids is a preparation: no ψ has both |ψ(x)|² and the momentum distribution narrow. Disturbance is real, but it is governed by separate error–disturbance relations with their own bounds.
Test it: what changes?
- Recorded spread s narrows onto σₓ, not zero
- State share σₓ²/s² rises to 100%
- Error on the mean falls to σₓ/√N
- σₚ floor ħ/(2σₓ) does not move
- Error on the mean s/√N shrinks tenfold
- Recorded spread s unchanged
- State share σₓ²/s² unchanged
- σₚ floor ħ/(2σₓ) unchanged
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
A source prepares every electron in the same state, for which |ψ(x)|² across a screen is a Gaussian with σₓ = 45 nm. The position detector's calibrated resolution is r = 28 nm, with an error independent of where each electron lands. Predict the standard deviation of the recorded positions, and what it becomes with a perfect detector.
View solution →
- The detector reports X = x + δ, with x distributed as |ψ|² and δ its own error. Independent spreads add in quadrature: s² = σₓ² + r².
- s² = (45 nm)² + (28 nm)² = 2025 nm² + 784 nm² = 2809 nm², so s = 53 nm.
- The state's share of the recorded variance is 2025/2809 = 0.72: 72% comes from ψ and 28% from the detector.
- A perfect detector has r → 0, so s → σₓ = 45 nm. The recorded spread shrinks by 8 nm and stops; it cannot reach zero, because 45 nm was never the detector's.
- Checks: every term is a length squared, so the dimensions agree. In the classical limit a particle placed at one point has σₓ = 0, and then s = r = 28 nm, all of it instrumental.
Answer. s = 53 nm with this detector; a perfect detector would record 45 nm, and no detector can record less.
MediumExample 2
A single rubidium-87 atom (mass 1.443 × 10⁻²⁵ kg) is loaded into an optical tweezer, released and imaged once; the cycle is repeated 400 times. The 400 positions have standard deviation s = 0.85 μm, and imaging a point-like bead calibrates the resolution at r = 0.40 μm, taken as exact. Find σₓ with its uncertainty, the error on the mean position, and the smallest velocity spread the state allows. What changes with 1600 cycles?
View solution →
- Each cycle prepares a fresh atom and measures it once, so the 400 results sample |ψ|² blurred by the imaging. Remove the resolution in quadrature: σₓ = √(s² − r²) = √(0.7225 − 0.1600) μm = √0.5625 μm = 0.75 μm.
- How well s is known: u(s) ≈ s/√(2(N − 1)) = 0.85 μm/√798 = 0.030 μm. Through the subtraction, u(σₓ) = (s/σₓ) u(s) = (0.85/0.75)(0.030 μm) = 0.034 μm, so σₓ = 0.750 ± 0.034 μm.
- Error on the mean: u(x̄) = s/√N = 0.85 μm/√400 = 0.0425 μm. The mean position is located twenty times more finely than any single image.
- Velocity floor from UQMI-04.08: σₚ ≥ ħ/(2σₓ) = 1.055 × 10⁻³⁴ J s/(2 × 0.75 × 10⁻⁶ m) = 7.03 × 10⁻²⁹ kg m s⁻¹, so σᵥ = σₚ/m ≥ (7.03 × 10⁻²⁹ kg m s⁻¹)/(1.443 × 10⁻²⁵ kg) = 4.87 × 10⁻⁴ m s⁻¹. The imaging resolution appears nowhere in this bound.
- With 1600 cycles, u(x̄) halves to 0.021 μm and u(s) halves to 0.015 μm, but s stays near 0.85 μm, σₓ near 0.75 μm and the velocity floor at 0.49 mm s⁻¹. Four times the data sharpens what is known about the state; it does not sharpen the state.
Answer. σₓ = 0.750 ± 0.034 μm; ⟨x⟩ located to 0.0425 μm; σᵥ ≥ 4.9 × 10⁻⁴ m s⁻¹. At 1600 cycles both errors halve, while σₓ and the velocity floor do not change.
HardExample 3
A proposed apparatus reads both x and p on every electron, adding independent noise of standard deviation a to its position reading X and b to its momentum reading P. Its meter is itself a quantum system, so ab ≥ ħ/2. The electrons arrive in a minimum-uncertainty Gaussian state with σₓ = 0.50 nm. Find σₚ, then σ(X), σ(P) and their product for a = 0.50 nm with b at its floor, and show that no choice of a and b brings σ(X)σ(P) below ħ.
View solution →
- The state saturates the bound of UQMI-04.08, so σₚ = ħ/(2σₓ) = 1.055 × 10⁻³⁴ J s/(1.00 × 10⁻⁹ m) = 1.05 × 10⁻²⁵ kg m s⁻¹.
- Meter at its floor with a = 0.50 nm: b = ħ/(2a) = 1.05 × 10⁻²⁵ kg m s⁻¹, equal to σₚ.
- Independent noise adds in quadrature: σ(X) = √(0.50² + 0.50²) nm = 0.71 nm and σ(P) = √2 × 1.055 × 10⁻²⁵ kg m s⁻¹ = 1.49 × 10⁻²⁵ kg m s⁻¹. Their product is 0.707 × 10⁻⁹ m × 1.491 × 10⁻²⁵ kg m s⁻¹ = 1.05 × 10⁻³⁴ J s, which is ħ, twice the preparation floor.
- In general, σ(X)²σ(P)² = (σₓ² + a²)(σₚ² + b²) = σₓ²σₚ² + a²b² + σₓ²b² + a²σₚ². The first two terms are each at least ħ²/4. Because y² + z² ≥ 2yz for any real y and z, the last two sum to at least 2σₓσₚab ≥ 2(ħ/2)(ħ/2) = ħ²/2. The total is at least ħ², so σ(X)σ(P) ≥ ħ.
- A mismatched meter costs more: a = 1.0 nm with b = ħ/(2a) gives σ(X) = 1.12 nm and σ(P) = 1.18 × 10⁻²⁵ kg m s⁻¹, a product of 1.25ħ. Equality needs a = σₓ and b = σₚ with every factor at its floor.
- Limiting case: drop the momentum reading and a can go to zero, leaving σ(X) → σₓ, so a position measurement alone has no floor of its own. The ħ appears only because one apparatus answers both questions, a separate theorem about that apparatus rather than the preparation relation again.
Answer. σₚ = 1.05 × 10⁻²⁵ kg m s⁻¹; σ(X) = 0.71 nm and σ(P) = 1.49 × 10⁻²⁵ kg m s⁻¹, with product ħ = 1.05 × 10⁻³⁴ J s. In general σ(X)σ(P) ≥ ħ, twice the ħ/2 that limits the state.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
Recorded spread is √(σₓ² + r²), never below σₓ.
Compute σₓ from ψ alone, then add the detector's independent resolution in quadrature, s² = σₓ² + r². More runs shrink the error on the mean, s/√N, but not s. Subtract r in quadrature only after calibration and when r is not much larger than σₓ. σₓσₚ ≥ ħ/2 limits preparations, not detectors.