University Physics V · Quantum Uncertainty and Commutation Relations · 8.8

Sequential Measurement & Disturbance

Two postulates do the work here: the Born rule turns a state into odds, and the projection rule says what is left behind. Chain three analysers together and you can watch the middle one destroy the answer the first one gave — then count exactly how much was lost, and where it went.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Everything measurable is packed into two rules acting on the same object. Given the spectral decomposition Â = Σₙ aₙ Π̂ₙ of a Hermitian observable, the Born rule assigns outcome aₙ the probability ⟨ψ|Π̂ₙ|ψ⟩ = ‖Π̂ₙ|ψ⟩‖², and the projection (Lüders) rule says the state to use for whatever happens next is Π̂ₙ|ψ⟩ divided by the square root of that probability. The second rule is not decoration: without it a sequence of measurements has no prediction at all, because the state entering the second apparatus is undefined.

With it, the rest is projector algebra. Π̂ₙ² = Π̂ₙ makes an immediate repeat of the same measurement certain to agree, which is what licenses the phrase 'the value of A'. But send in a second observable B̂ whose eigenbasis is not Â's, and its projector rotates the state out of Â's eigenspace, so re-measuring Â draws from a distribution instead of returning the stored answer. The cost is exactly one coherent sum: with the middle outcome recorded, the paths i → j → k contribute |⟨aₖ|bⱼ⟩|²|⟨bⱼ|aᵢ⟩|², added as probabilities; with the middle apparatus removed, the amplitudes ⟨aₖ|bⱼ⟩⟨bⱼ|aᵢ⟩ are added first and completeness collapses them to δₖᵢ.

For spin-½ with the axes at θ that is 1 against cos⁴(θ/2) + sin⁴(θ/2). What the postulate never claims is a mechanism: it is a rule for the statistics of later measurements conditioned on a recorded result.

Simple definition: A sequential measurement is a chain of measurements on one system in which each recorded outcome both fixes a probability through the Born rule and replaces the state by its normalised projection, so the next apparatus sees the projected state, never the original one.
Example: Send atoms with S_z = +ℏ/2 through an Sₓ analyser and then through a second S_z analyser: half now read −ℏ/2, because cos⁴45° + sin⁴45° = ¼ + ¼ = ½, though with the Sₓ stage removed every atom repeats +ℏ/2.

Born rule, written with projectorsP(aₙ) = ⟨ψ|Π̂ₙ|ψ⟩ = ‖Π̂ₙ|ψ⟩‖²

Covers a degenerate eigenvalue unchanged, where |⟨aₙ|ψ⟩|² would need a sum over the eigenspace — and it is the form the update rule reuses.

Π̂ₙ projects on the whole aₙ eigenspace: Π̂ₙ† = Π̂ₙ = Π̂ₙ², and Σₙ Π̂ₙ = Î. P is a pure number.

Projection (Lüders) update|ψ⟩ → Π̂ₙ|ψ⟩ / √(⟨ψ|Π̂ₙ|ψ⟩)

For a non-degenerate aₙ the output is |aₙ⟩ whatever went in: the input state survives in the odds, not in the ket.

Defined only for an outcome of non-zero probability; nonlinear in |ψ⟩, and applied to the recorded branch alone.

Repeatability is idempotenceP(aₙ again | aₙ) = ⟨aₙ|Π̂ₙ|aₙ⟩ = 1

The whole content is Π̂ₙ² = Π̂ₙ. Let the state evolve under Ĥ between the readings and the guarantee lapses at once.

Immediate repeat, no evolution in the gap, non-destructive apparatus. Probabilities, so dimensionless.

Chain rule for a recorded sequenceP(aᵢ, bⱼ, aₖ) = |⟨aₖ|bⱼ⟩|² |⟨bⱼ|aᵢ⟩|² |⟨aᵢ|ψ⟩|²

Probabilities multiply along a recorded path and add across paths: Σⱼ |⟨aₖ|bⱼ⟩⟨bⱼ|aᵢ⟩|², never |Σⱼ ⟨aₖ|bⱼ⟩⟨bⱼ|aᵢ⟩|².

Non-degenerate outcomes; each factor is a squared overlap of successive eigenvectors, all dimensionless.

Delete the middle apparatusΣⱼ ⟨aₖ|bⱼ⟩⟨bⱼ|aᵢ⟩ = ⟨aₖ|aᵢ⟩ = δₖᵢ

Unmeasured, the middle basis is invisible; measured, the same overlaps reappear squared. The gap between the two numbers is the disturbance.

Completeness Σⱼ |bⱼ⟩⟨bⱼ| = Î, usable only while no outcome is recorded and the branches stay coherent.

Spin-½ through three analysersP(+z again) = cos⁴(θ/2) + sin⁴(θ/2) = 1 − ½ sin²θ

Equals 1 at θ = 0° and 180°, where the observables commute, and bottoms out at ½ at θ = 90°: the worst case flips half the beam.

θ is the angle between the middle analyser's axis and z, both middle branches kept and recorded.

Two postulates, one projector

Fix a Hermitian Â = Σₙ aₙ Π̂ₙ, where Π̂ₙ projects onto the entire aₙ eigenspace. Born gives P(aₙ) = ⟨ψ|Π̂ₙ|ψ⟩ = ‖Π̂ₙ|ψ⟩‖²; Lüders replaces the state by Π̂ₙ|ψ⟩/√(⟨ψ|Π̂ₙ|ψ⟩). Take |ψ⟩ = 0.6|a₁⟩ + 0.8|a₂⟩ with a non-degenerate spectrum: P(a₁) = 0.36, P(a₂) = 0.64, and the surviving state is |a₁⟩ or |a₂⟩ — the coefficients have left the ket and live on only as odds. Now let a₂ be twofold degenerate, with |ψ⟩ = 0.6|a₁⟩ + 0.48|a₂,₁⟩ + 0.64|a₂,₂⟩. Then P(a₂) = 0.48² + 0.64² = 0.64 as before, but the projected state is (0.48|a₂,₁⟩ + 0.64|a₂,₂⟩)/0.8: the weights and relative phase inside the eigenspace are untouched. That is why the postulate is written with Π̂ₙ and not |aₙ⟩⟨aₙ|. A degenerate measurement asks a coarse question, and it damages only what it resolves.

Idempotence is the whole of repeatability

Measure Â, obtain aₙ, measure Â again at once: the second Born probability is ⟨aₙ|Π̂ₙ|aₙ⟩ = 1, because Π̂ₙ² = Π̂ₙ and the state already sits inside the eigenspace. That one line of algebra is what licenses the sentence 'the system has value aₙ'. It is a claim about what the formalism predicts for an immediate repeat, not a claim that the value was there before the first reading. Three conditions hide inside 'immediately'. The state must not evolve in the gap: switch on a field and |+z⟩ precesses, so certainty survives only for [Ĥ, Â] = 0, or for a gap short compared with ℏ/ΔE. The apparatus must be non-destructive — a photon counter that absorbs the photon leaves nothing to re-measure, and repeatability simply does not apply. And the outcome must be recorded, because the result is what selects which Π̂ₙ to apply.

A–B–A: where an amplitude sum becomes a probability sum

Prepare |+z⟩, measure Sₓ, measure S_z again. With |±x⟩ = (|+z⟩ ± |−z⟩)/√2 the chain rule gives four recorded paths, each a product of squared overlaps: +x then +z is ½ × ½ = ¼, and −x then +z is ¼ as well, so P(+z) = ½. Half the atoms now carry the opposite value of the very quantity the first magnet had just fixed. Delete the middle magnet and redo the sum with amplitudes instead: Σⱼ ⟨+z|bⱼ⟩⟨bⱼ|+z⟩ = ⟨+z|+z⟩ = 1, so P(+z) = 1. The two calculations differ by one cross term, |½ + ½|² − (¼ + ¼) = ½, and recording the middle outcome is what deletes it. At a general tilt θ the same two paths give cos⁴(θ/2) + sin⁴(θ/2) = 1 − ½ sin²θ: 0.875 at θ = 30°, 0.625 at 60°, 0.500 at 90°, back to 1 at 180°.

Compatible B is free, and some incompatible ones are too

If [Â, B̂] = 0 the two share an eigenbasis, their projectors commute, and Π̂ᴮⱼ Π̂ᴬᵢ = Π̂ᴬᵢ Π̂ᴮⱼ: the middle projection acts inside Â's eigenspace and cannot move the state out of it, so the third reading returns aᵢ with certainty and the joint distribution does not depend on the order of the two measurements. In the figure this is θ = 0° or 180°, where the middle analyser measures ±S_z again and the return probability is exactly 1. Two cautions. When aᵢ is degenerate, B̂ still changes the ket — it resolves the eigenspace — even though Â's value survives, so 'undisturbed' refers to the recorded value, not to the state. And non-commutation guarantees only that some state is disturbed, not every state: L̂ₓ and L̂_z do not commute, yet |l = 0, m = 0⟩ is a shared eigenvector, and an Lₓ–L_z–Lₓ sequence on it repeats with certainty.

Non-selective measurement: where the coherence goes

Suppose the middle analyser fires but nobody reads the result. Then the object to propagate is the ensemble over branches, ρ̂ → Σⱼ Π̂ⱼ ρ̂ Π̂ⱼ — linear and trace-preserving, unlike the Lüders update, and the one statement of disturbance that conditions on nothing. Take ρ̂ = |+z⟩⟨+z|, which in the Sₓ basis is the matrix [[½, ½], [½, ½]], and measure Sₓ without recording: the channel keeps the diagonal and deletes the off-diagonal entries, leaving ½Î. The purity Tr ρ̂² falls from 1 to ½, every later S_z reading is 50:50, and that is precisely the number the three-analyser experiment returns. Nothing was destroyed at the level of the whole world — the missing coherence has become correlation with the apparatus — but what the spin alone can still show is exactly the diagonal that survived.

What this is not: Robertson, error–disturbance, collapse

Three separations are worth stating outright. First, the Robertson bound σ_A σ_B ≥ ½|⟨[Â, B̂]⟩| is about one prepared state: both spreads are computed in the same |ψ⟩ before any apparatus acts, no measurement appears in its proof, and it predicts no number whatever for the three-analyser experiment. Second, a genuine error–disturbance relation quantifies the apparatus, and the naive Heisenberg product ε(A)η(B) ≥ ½|⟨[Â, B̂]⟩| is false: Ozawa's universally valid form carries two extra terms, ε(A)η(B) + ε(A)σ(B) + σ(A)η(B) ≥ ½|⟨[Â, B̂]⟩|, and neutron-spin and photon experiments have measured violations of the naive one. An ideal projective measurement is the extreme case — zero error on its own observable, maximal disturbance to the incompatible one. Third, the projection rule is a recipe for later statistics; whether it names a physical event is interpretation-dependent, while the non-selective channel Σⱼ Π̂ⱼ ρ̂ Π̂ⱼ is common ground for every reading of the theory.

Change one variable at a time

Make the relationship visible.

Interactive model

Middle analyser tilt θ90 °

Final analyser tilt φ0 °

Leave φ at 0° so the last analyser repeats the first, then sweep θ: the total sags to 0.500 at 90° while the no-middle-analyser readout holds at 1.000. Now set θ = φ and watch the −n̂ path vanish — a compatible repeat costs nothing.

P(+ AT FINAL ANALYSER)0.500

RECORDED PATH +n0.250

RECORDED PATH −n0.250

NO MIDDLE ANALYSER1.000

Live interpretationP(+ AT FINAL ANALYSER): 0.500. RECORDED PATH +n: 0.250. RECORDED PATH −n: 0.250. NO MIDDLE ANALYSER: 1.000

Catch the common trap

Explain before calculating.

Misconception check

The A–B–A scrambling gets read as a clumsy apparatus knocking the spin about, with Δx Δp ≥ ℏ/2 quoted as the licence. Both halves are wrong. Robertson's inequality is a theorem about a single prepared state — σ_A and σ_B are both computed in |ψ⟩, before anything is measured — so it never mentions an apparatus and returns no number at all for the three-analyser experiment, whose answer cos⁴(θ/2) + sin⁴(θ/2) comes from projectors and overlaps alone. Nor is the magnet clumsy: an ideal projective measurement reads its own observable with zero error and still leaves the incompatible one maximally scrambled. Relations that do bound error against disturbance exist — Ozawa's and Branciard's — and they carry extra terms precisely because the naive product ε(A)η(B) ≥ ½|⟨[Â, B̂]⟩| fails, as experiments have shown. The disturbance here is the deletion of a cross term between two paths, not a jolt.

Atoms prepared with S_z = +ℏ/2 pass an Sₓ analyser that records which beam each atom took; both beams are then sent into a third analyser measuring S_z. Across all the atoms, what fraction leaves with S_z = +ℏ/2?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasySilver atoms leave a Stern–Gerlach magnet in |+z⟩. They pass a second analyser whose axis n̂ lies at θ = 60° from z in the x–z plane, with both output beams kept and their outcomes recorded, and then a third analyser along z. Find the middle-stage probabilities and the fraction reading S_z = +ℏ/2 at the end.

Write the middle eigenvectors in the z basis: |+n̂⟩ = cos(θ/2)|+z⟩ + sin(θ/2)|−z⟩ and |−n̂⟩ = −sin(θ/2)|+z⟩ + cos(θ/2)|−z⟩. With θ = 60°, the half-angle is 30°, cos 30° = 0.8660 and sin 30° = 0.5000.
Born rule at the middle stage: P(+n̂) = |⟨+n̂|+z⟩|² = cos²30° = 0.750, and P(−n̂) = sin²30° = 0.250.
Lüders update: the recorded branches are exactly |+n̂⟩ and |−n̂⟩, so their S_z statistics are P(+z | +n̂) = cos²30° = 0.750 and P(+z | −n̂) = sin²30° = 0.250.
Chain and add over the two recorded paths: P(+z) = 0.750 × 0.750 + 0.250 × 0.250 = 0.5625 + 0.0625 = 0.625, which matches 1 − ½ sin²60° = 1 − 0.375 = 0.625.
With the middle analyser removed the amplitudes would instead add to ⟨+z|+z⟩ = 1, so recording the middle outcome has cost 0.375 of the beam.

AnswerP(+n̂) = 0.750 and P(−n̂) = 0.250; P(S_z = +ℏ/2) = 0.625, so 37.5% of the atoms have had their z value flipped by an apparatus that never measured S_z.

MediumA 1.00 mW beam is polarised vertically. (a) Send it straight into a horizontal polariser. (b) Put a polariser at +45° in front of that horizontal one. (c) Replace the 45° polariser by a calcite analyser at 45° that splits the light into +45° and −45° channels and detects which channel each photon took; pass each channel through its own horizontal polariser and add the transmitted powers. Find the power delivered in each case, ignoring absorption losses.

Choose the basis |H⟩, |V⟩ and write |±45⟩ = (±|H⟩ + |V⟩)/√2, so ⟨+45|V⟩ = 1/√2, ⟨H|+45⟩ = 1/√2, ⟨−45|V⟩ = 1/√2 and ⟨H|−45⟩ = −1/√2.
(a) One projection: P = |⟨H|V⟩|² = 0, so the transmitted power is 0 mW.
(b) A polariser is a projective measurement that keeps one branch and absorbs the other, so a single path survives: P = |⟨+45|V⟩|² × |⟨H|+45⟩|² = ½ × ½ = ¼, giving 0.250 mW.
(c) Both branches are recorded, so their probabilities add: P = |⟨H|+45⟩⟨+45|V⟩|² + |⟨H|−45⟩⟨−45|V⟩|² = ¼ + ¼ = ½, giving 0.500 mW.
Compare with the coherent sum that case (a) performs over the same two paths: ⟨H|+45⟩⟨+45|V⟩ + ⟨H|−45⟩⟨−45|V⟩ = ½ − ½ = 0 = ⟨H|V⟩. Recording which channel deletes the cross term 2 × (½) × (−½) = −½, and that is what turns 0 into ½.

Answer(a) 0 mW, (b) 0.250 mW, (c) 0.500 mW. The middle element does not rotate the light; it deletes the interference term −½ that had made H and V orthogonal.

HardA spin-½ prepared in |+z⟩ passes N analysers in series; the k-th has its axis in the x–z plane at angle kΘ/N from z, with Θ = 90°, and only the '+' beam continues to the next stage. (a) Give the probability of surviving all N stages. (b) Evaluate it for N = 1, 3 and 10. (c) Find the large-N behaviour. (d) State the survivors' final state, and say what has been done to the spin.

Each stage projects the previous stage's '+' eigenvector onto an axis rotated by Θ/N, so every overlap is the same: |⟨+n̂ₖ|+n̂ₖ₋₁⟩|² = cos²(Θ/2N).
The stages lie along one recorded path, so probabilities multiply: P(N) = [cos²(Θ/2N)]^N, which for Θ = 90° is [cos²(45°/N)]^N.
N = 1: cos²45° = 0.500. N = 3: cos²15° = 0.93301, cubed, gives 0.8122. N = 10: cos²4.5° = 0.993844, to the tenth power, gives 0.9401.
Large N: cos²(Θ/2N) ≈ 1 − Θ²/4N², so P ≈ (1 − Θ²/4N²)^N ≈ 1 − Θ²/4N = 1 − π²/16N = 1 − 0.617/N, which tends to 1. At N = 10 the estimate reads 0.938 against the exact 0.940.
The survivor was last projected onto the final axis, so it is exactly |+x⟩. Measurement alone has carried the spin through 90° with a fidelity approaching 1, whereas the single 90° measurement of the Easy example transports only half the atoms.

AnswerP(N) = [cos²(45°/N)]^N, equal to 0.500, 0.812 and 0.940 for N = 1, 3, 10 and rising as 1 − π²/(16N) → 1; the survivors are in |+x⟩. Frequent projection steers a state rather than scrambling it — the quantum Zeno dragging effect.