University Physics IV · Foundations of Quantum Mechanics · 5.6

Measurement, Eigenstates & State Update

Two questions get answered here, and they are different questions. What values can come out of an apparatus, and with what probability — and then, what the system is in once the pointer has settled. The first is an eigenvalue problem. The second is a postulate, fitted to experiment, with no equation of motion behind it.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Two postulates sit on top of the state and the operators. The first says an observable's Hermitian operator carries a spectrum, and a single measurement returns one member of it and nothing else: expand the state in that operator's orthonormal eigenbasis, |ψ⟩ = Σ cₙ|aₙ⟩ with cₙ = ⟨aₙ|ψ⟩, and Born's rule assigns |cₙ|² to the outcome aₙ, with completeness making the list sum to one. That much is a recipe for statistics over identically prepared copies.

The second postulate answers a different question — what you are holding afterwards — and its whole content is repeatability: the reading is projected out, P̂ₙ|ψ⟩ divided by √P(aₙ), so an immediate repeat returns the same value with certainty and every other component of the state is gone. Where does it come from? Not from the Schrödinger equation, which is linear, unitary and deterministic, and therefore cannot turn one superposition into one random outcome.

It comes from the bench: sequential Stern–Gerlach magnets, repeated spectroscopy, quantum non-demolition readings. The cost is a theory carrying two incompatible rules of evolution, and no statement anywhere in the formalism of where the first stops and the second begins.

Simple definition: A measurement of an observable returns one of the eigenvalues of its Hermitian operator, with probability equal to the squared modulus of the state's component along the matching eigenvector, and leaves the system in that eigenvector.
Example: For ψ = √(1/3) ψ₁ + √(2/3) ψ₃ in an infinite well, an energy reading returns E₁ with probability 1/3 or 9E₁ with probability 2/3 — never 6.33E₁, the mean — and whichever it returns, the state left behind is that ψₙ alone.

Eigenvalue equation and spectrumÂ|aₙ⟩ = aₙ|aₙ⟩, with aₙ real

The spectrum is the complete menu of readings; a value off it has probability zero in every state.

Â Hermitian; aₙ carries the unit of the observable — J for energy, J s for spin, m for position

Expansion in the eigenbasis|ψ⟩ = Σₙ cₙ|aₙ⟩, cₙ = ⟨aₙ|ψ⟩ = ∫ aₙ*(x) ψ(x) dx

One inner product per outcome turns the state into a probability list — no dynamics required.

⟨aₘ|aₙ⟩ = δₘₙ and Σₙ|aₙ⟩⟨aₙ| = 1̂; the cₙ are dimensionless complex numbers

Born's rule for outcomesP(aₙ) = |cₙ|² = |⟨aₙ|ψ⟩|², Σₙ |cₙ|² = 1

Squared moduli, so a global phase drops out and relative phase never enters this observable's odds.

For a degenerate aₙ, add the squared moduli over the whole subspace: P(aₙ) = ⟨ψ|P̂ₙ|ψ⟩

Projection postulate, Lüders form|ψ⟩ → P̂ₙ|ψ⟩ / √(⟨ψ|P̂ₙ|ψ⟩), P̂ₙ = Σᵣ |aₙ, r⟩⟨aₙ, r|

Fixes what you hold afterwards and makes an immediate repeat certain. It is a postulate, not a solution.

P̂ₙ² = P̂ₙ = P̂ₙ†; the denominator is just √P(aₙ), so dividing by it is renormalisation, not physics

Expectation value as an outcome average⟨Â⟩ = ⟨ψ|Â|ψ⟩ = Σₙ aₙ |cₙ|²

Ties the sandwich integral to a laboratory mean — and warns that the mean is often not a possible reading.

Unit of Â; needs many identically prepared copies, and ⟨Â⟩ need not lie in the spectrum at all

Two readings in a rowP(aₙ then bₘ) = |⟨bₘ|aₙ⟩|² · |⟨aₙ|ψ⟩|²

Turns a Stern–Gerlach chain into arithmetic: ½ × ½ = ¼ for the z → x → z sequence.

Valid only once aₙ has been recorded; alternatives left unrecorded add amplitudes instead

Hermiticity is what supplies the menu

An observable is represented by a Hermitian operator Â, and Hermiticity does three jobs at once. Its eigenvalues are real, so a reading can be a number on a dial. Eigenvectors belonging to different eigenvalues are orthogonal, so distinct outcomes are perfectly distinguishable — no reading is ever ambiguous between two spectrum values. And the eigenvectors span the space, giving Σₙ|aₙ⟩⟨aₙ| = 1̂, so every state has an expansion in them. The spectrum, the list (aₙ), is therefore the complete menu of what one measurement can return; anything off it has probability zero whatever the state. For a particle in a box of width L the energy menu is Eₙ = n²π²ħ²/2mL², isolated numbers with nothing in between, and a reading of 3.5E₁ is not merely unlikely but impossible. For x̂ the menu is the whole real line, which needs the continuous version of everything that follows. One detail matters later: a degenerate eigenvalue owns a subspace, not a single vector.

Project, square, and check the total is one

Given |ψ⟩ and the eigenbasis, one inner product per outcome does all the work: cₙ = ⟨aₙ|ψ⟩, which in one dimension is the overlap integral ∫ aₙ*(x) ψ(x) dx. Born's rule is then P(aₙ) = |cₙ|², and completeness guarantees Σ|cₙ|² = ⟨ψ|ψ⟩ = 1, so the check costs nothing. Take ψ = √(1/3) ψ₁ + √(2/3) ψ₃ in the infinite well: P(E₁) = 1/3, P(9E₁) = 2/3, and ⟨E⟩ = (1/3)E₁ + (2/3)(9E₁) = 19E₁/3 ≈ 6.33E₁. Notice what that mean is not. 6.33E₁ is nowhere in the spectrum, so no single electron ever registers it; it is the average of many readings on identically prepared electrons, and it agrees term by term with the integral ⟨ψ|Ĥ|ψ⟩. Notice also that only moduli entered. Change the relative phase between ψ₁ and ψ₃ and you change ψ(x) and every position statistic, without moving one energy probability.

The state after the reading, and why repeats agree

Born's rule gives the odds and says nothing about what you now hold. The projection postulate does: with P̂ₙ the projector onto the aₙ eigenspace, |ψ⟩ → P̂ₙ|ψ⟩ / √(⟨ψ|P̂ₙ|ψ⟩). The denominator is √P(aₙ), and dividing by it is renormalisation rather than physics — but forgetting it is the commonest slip in the algebra. The reason to believe the rule at all is repeatability. Measure again immediately and P̂ₙ|aₙ⟩ = |aₙ⟩, so the same value returns with probability 1. Feed the +ħ/2 output of a Stern–Gerlach magnet into a second identical magnet and the − channel stays empty. When aₙ is degenerate the correct rule is Lüders': project onto the whole subspace and keep the relative phases inside it. The cruder "collapse picks one eigenvector" is a genuinely different state, and the Hard example below separates the two predictions by 0.90 against 0.50.

Continuous spectra: a real detector projects onto a window

Position has no normalisable eigenstates — δ(x − x₀) has infinite norm and infinite kinetic energy — so the postulate cannot literally leave you in one. What a real detector implements is a projection onto an interval: ψ(x) → ψ(x)⋅χ(x) / ‖ψχ‖, where χ is 1 inside the accepted window and 0 outside. That state is normalisable, and probabilities are ∫|ψ|²dx over the window rather than |ψ|² itself. It is also where the cost of a sharp reading shows up. Chop a wave function down to Δx = 1.0 nm and the momentum spread is at least ħ/2Δx = 1.055×10⁻³⁴ ÷ 2.0×10⁻⁹ = 5.3×10⁻²⁶ kg m s⁻¹, which for an electron is a velocity spread of 5.3×10⁻²⁶ ÷ 9.11×10⁻³¹ = 5.8×10⁴ m s⁻¹. The projection did not reveal a momentum the electron had; it rebuilt the momentum content by cutting the function.

Two readings in a row, and the phase that does not survive

Once an intermediate value has been recorded, probabilities multiply along the recorded path: P(aₙ then bₘ) = |⟨bₘ|aₙ⟩|² · |⟨aₙ|ψ⟩|². Prepare S_z = +ħ/2, pass the beam through an Sₓ analyser and keep its + output, then measure S_z again. Half survives the x analyser, and half of that emerges at S_z = −ħ/2 — one quarter of the beam, in a channel the first magnet had already emptied. The middle apparatus did not filter the beam; it replaced its state, and |+x⟩ carries no memory of z. Now take the same hardware and recombine the two x paths coherently, with no record of which was taken. The amplitudes add instead of the probabilities, they rebuild |↑z⟩ exactly, and the −z channel goes empty again. The projection postulate applies at the point where a value becomes a fact — apply it earlier and you predict the wrong count.

What the postulate costs

Everything else in this unit runs on one linear equation, iħ ∂ψ/∂t = Ĥψ: deterministic, unitary, norm-preserving, and reversible. The update rule is none of those. It is discontinuous; it is non-unitary, since projection throws probability away before rescaling; and because of that rescaling it is not even linear in ψ. Applied to a superposition it yields one outcome, and a linear deterministic equation applied to the same superposition provably cannot. The two rules also disagree about jurisdiction. Nothing in the formalism marks the boundary between "system" and "apparatus", and you may slide that cut a long way — into the magnet, the detector, the experimenter — without changing any prediction anyone has tested. Decoherence explains part of the puzzle: coupling to an environment suppresses the interference terms extraordinarily fast for anything macroscopic. But it converts a superposition into a list of possibilities, not into one actuality. Use the postulate; it has never failed a measurement. Do not mistake it for a derivation.

Change one variable at a time

Make the relationship visible.

Interactive model

Analyser 2 axis θ from z90 °

Analyser 3 axis φ from z0 °

Set θ = 0 and the detector count vanishes — a repeated z reading never flips. Push θ to 90° and a quarter of the beam arrives with S_z = −ħ/2, a value the first analyser had already removed. The middle reading did not filter the beam; it replaced its state.

P(+) AT ANALYSER 20.500

P(−) AT ANALYSER 30.500

FRACTION AT DETECTOR0.250

⟨Sz⟩ AFTER ANALYSER 20.000 ħ

Live interpretationP(+) AT ANALYSER 2: 0.500. P(−) AT ANALYSER 3: 0.500. FRACTION AT DETECTOR: 0.250. ⟨Sz⟩ AFTER ANALYSER 2: 0.000 ħ

Catch the common trap

Explain before calculating.

Misconception check

The tempting reading of |cₙ|² is ignorance: the electron was already in one eigenstate and the measurement merely revealed which. That is a physically different state, and the difference is testable. Take |ψ⟩ = (|↑z⟩ + |↓z⟩)/√2. Measure S_z and both stories predict ±ħ/2 at fifty-fifty, so on that observable they agree exactly. But that ψ is precisely |↑x⟩: measure Sₓ on it and you get +ħ/2 every single time, whereas a genuine fifty-fifty mixture of |↑z⟩ and |↓z⟩ gives ±ħ/2 at fifty-fifty. The superposition carries a relative phase; the mixture has none, and any observable that is sensitive to that phase separates them. What is true is that after the S_z reading — and only after — the ignorance description becomes the right one. That is exactly what the projection encodes.

An electron in an infinite square well of ground-state energy E₁ is prepared in ψ = √(1/3) ψ₁ + √(2/3) ψ₃, where the ψₙ are the normalised energy eigenfunctions. A single energy measurement is made on one such electron. Which statement is correct?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyAn electron is prepared in |ψ⟩ = (3|↑z⟩ + 4i|↓z⟩)/5. Find the probability of each S_z outcome, the expectation value ⟨S_z⟩, and the state immediately after a reading of −ħ/2.

Check normalisation: |3/5|² + |4i/5|² = 9/25 + 16/25 = 1 ✓. The coefficients are already the components along the S_z eigenvectors, so no overlap integral is needed.
Born's rule squares the moduli: P(+ħ/2) = |3/5|² = 9/25 = 0.36 and P(−ħ/2) = |4i/5|² = 16/25 = 0.64. The modulus discards the i, so |4i/5|² = (4/5)².
Expectation value: ⟨S_z⟩ = 0.36(+ħ/2) + 0.64(−ħ/2) = (0.36 − 0.64)(ħ/2) = −0.14 ħ = −0.14 × 1.055×10⁻³⁴ = −1.5×10⁻³⁵ J s. It lies between the two eigenvalues and equals neither.
Projection: keep only the −ħ/2 term and renormalise. (4i/5)|↓z⟩ ÷ √0.64 = (4i/5)|↓z⟩ ÷ (4/5) = i|↓z⟩, which differs from |↓z⟩ only by a global phase and is therefore the same physical state.

AnswerP(+ħ/2) = 0.36, P(−ħ/2) = 0.64, ⟨S_z⟩ = −0.14 ħ ≈ −1.5×10⁻³⁵ J s, and the state left behind is |↓z⟩ up to a global phase.

MediumA beam of 8.0 × 10⁵ silver atoms per second leaves a Stern–Gerlach magnet in the S_z = +ħ/2 channel. It passes an Sₓ magnet whose +ħ/2 output is kept, then a third magnet aligned along z. Find the rate in the final S_z = −ħ/2 channel. Then find that rate if the two x paths are instead recombined coherently, with no record of which was taken.

Rewrite the input in the basis actually being measured: |↑z⟩ = (|↑x⟩ + |↓x⟩)/√2, so P(+ħ/2 along x) = |1/√2|² = 1/2. Rate past the second magnet: 8.0×10⁵ × 0.5 = 4.0×10⁵ s⁻¹.
The projection postulate replaces the state with |↑x⟩ — the z preparation is gone. Re-expand: |↑x⟩ = (|↑z⟩ + |↓z⟩)/√2, so P(−ħ/2 along z) = |1/√2|² = 1/2.
Because the x value was recorded, probabilities multiply along the path: 4.0×10⁵ × 0.5 = 2.0×10⁵ s⁻¹. That is a quarter of the entering beam, arriving in a channel the first magnet had emptied.
Recombining the x paths coherently records no x value, so the amplitudes add instead: (|↑x⟩ + |↓x⟩)/√2 rebuilds |↑z⟩ exactly, and the −ħ/2 channel of the third magnet receives 0 atoms per second.

Answer2.0 × 10⁵ atoms s⁻¹ when the x value is recorded; 0 atoms s⁻¹ when the two x paths are recombined coherently.

HardAn observable has eigenvalue a = 2 with eigenvector |1⟩, and eigenvalue a = 5 twofold degenerate with orthonormal eigenvectors |2⟩ and |3⟩. A system is prepared in |ψ⟩ = (|1⟩ + 2|2⟩ + i|3⟩)/√6 and a reading gives a = 5. A second observable B̂ has eigenvectors |±⟩ = (|2⟩ ± i|3⟩)/√2 inside that subspace. Find P(a = 5), the state left behind, and P(+). Then compare with the naive rule that collapse picks a single eigenvector.

Normalisation: (1 + 4 + 1)/6 = 1 ✓. The a = 5 eigenspace is spanned by |2⟩ and |3⟩, so P̂₅ = |2⟩⟨2| + |3⟩⟨3| and P(5) = ⟨ψ|P̂₅|ψ⟩ = (4 + 1)/6 = 5/6 ≈ 0.833. The other 1/6 belongs to a = 2.
Lüders' rule projects onto the whole subspace: P̂₅|ψ⟩ = (2|2⟩ + i|3⟩)/√6, whose norm is √(5/6). Dividing gives |ψ′⟩ = (2|2⟩ + i|3⟩)/√5. The relative phase between |2⟩ and |3⟩ has survived, because the reading never resolved it.
Project onto |+⟩ = (|2⟩ + i|3⟩)/√2, whose bra is (⟨2| − i⟨3|)/√2. Then ⟨+|ψ′⟩ = (1/√10)(2 − i⋅i) = 3/√10, so P(+) = 9/10 = 0.90 and P(−) = 1/10 = 0.10, summing to 1 ✓.
The naive rule instead collapses to |2⟩ with probability 4/5 or |3⟩ with probability 1/5. Since |⟨+|2⟩|² = |⟨+|3⟩|² = 1/2, it predicts P(+) = (4/5)(1/2) + (1/5)(1/2) = 1/2 = 0.50.
0.90 against 0.50 is nearly a factor of two in a countable rate, so the degenerate case is no matter of convention. Experiment picks Lüders, and the reason is structural: the projection must destroy only the coherence the reading actually distinguished.

AnswerP(a = 5) = 5/6 ≈ 0.833; the state left behind is (2|2⟩ + i|3⟩)/√5; P(+) = 0.90. The naive single-eigenvector rule predicts 0.50 and is ruled out.