Sequential measurement and disturbance

Fix a Hermitian  = 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.

Measure Â, obtain aₙ, measure  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 is guaranteed for every outcome only when [Ĥ, Â] = 0, and holds approximately 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.

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°.

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.

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.

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 and a finite disturbance to the incompatible one, so the naive product is zero. 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.