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Quantum core · institution-dependent · Interactive lesson

Measurement, eigenstates and 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.

UPIV-05.06 · Lesson 6 of 9 in this unit. Position is not a completion score.
Starting model: Measurement, eigenstates and state updateThree Stern–Gerlach analysers in a row. Atoms enter as |↑z⟩; analyser 2 measures along an axis 90° from z and passes fraction 0.500; analyser 3 measures at 0° and its − branch delivers 0.250 of the original beam. Bars: surviving fraction after each stage.each analyser rewrites the state it readsanalyser 2 leaves |+θ⟩ behind, not |↑z⟩|↑z⟩ in → θ = 90° → φ = 0° → − branchbeam 1.000 · after 2 0.500 · detector 0.250dashed = whole beam
Start with a picture. Change the model in the lab below.

Quantum core · institution-dependent. Departments differ in how much quantum mechanics they include in Physics IV or teach in a dedicated course. Follow your university’s required mathematical depth and assessment; these numbered units are not automatically optional.

1

Build the model

The state you had is not the state you are left with.

The key idea

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 = √(1/5) ψ₁ + √(4/5) ψ₂ in an infinite well, an energy reading returns E₁ with probability 1/5 or 4E₁ with probability 4/5 — never the mean 3.4E₁, which is on no rung of the ladder — and whichever it returns, the state left behind is that ψₙ alone.

The complete physical model and assumptions

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.

1Hermiticity is what supplies the menu

An observable is represented by a Hermitian operator Â, and two of the three properties this lesson needs follow straight from that. 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. The third property, completeness, does not follow from Hermiticity alone; it is the spectral theorem for a self-adjoint operator that hands you Σₙ|aₙ⟩⟨aₙ| = 1̂, and every expansion written below quietly leans on it. Granted that, the spectrum — the whole list of aₙ — fixes in advance every number the apparatus is allowed to display, and nothing else can ever appear on the dial, in any state whatever. For a particle in a box of width L the energy menu is Eₙ = n²π²ħ²/2mL², isolated numbers with nothing in between, so 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.

2Project, 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 ∫ uₙ*(x) ψ(x) dx over the eigenfunctions uₙ. 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 = (ψ₁ + √2 ψ₂ + ψ₃)/2 in the infinite well: the squared moduli are 1/4, 1/2 and 1/4, which do sum to 1, and ⟨E⟩ = (1/4)E₁ + (1/2)(4E₁) + (1/4)(9E₁) = 4.5E₁. Notice what that mean is not. It falls between E₂ = 4E₁ and E₃ = 9E₁ and equals neither, 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. Put a phase on the ψ₂ term and the position density is rearranged — ψ₂ is antisymmetric about the midpoint, so even ⟨x⟩ slides off the centre of the well — while all three energy probabilities stay exactly where they were.

3The 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⟩). That square root is nothing but √P(aₙ), so the division only restores unit norm — yet 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.

4Continuous 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 come from ∫|ψ|²dx across the window rather than from |ψ|² at a point. 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.

5Two 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. The word carrying the weight there is recorded. Prepare Sz = +ħ/2, send the beam through an Sₓ analyser, keep its + output and read Sz again: the Medium example counts the atoms, and the count is not zero, because the middle reading installed |+x⟩ and |+x⟩ holds no memory of z. Now rebuild the same hardware so the two x beams rejoin with nothing written down about which one an atom took. Not one magnet has moved, but the amplitudes now add before they are squared, they reassemble |↑z⟩ exactly, and the −z channel goes dark. Two apparatuses that look alike, two different answers: the postulate is applied where a value becomes a fact, and applying it one step too early predicts the wrong count.

6What 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.

2

Key equations

Read the units, assumptions and limits before substituting.

Eigenvalue equation and spectrum

Â|aₙ⟩ = aₙ|aₙ⟩, with aₙ real

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

 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⟩ = ∫ uₙ*(x) ψ(x) dx

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

uₙ(x) = ⟨x|aₙ⟩ is the eigenfunction, a different object from the number aₙ; ⟨aₘ|aₙ⟩ = δₘₙ, Σₙ|aₙ⟩⟨aₙ| = 1̂, and the cₙ are dimensionless

Born's rule for outcomes

P(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†, and r runs over the degeneracy of aₙ — a single term when aₙ is non-degenerate

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 row

P(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
3

Interactive lab

Predict first. Explore the model. Compare the result.

Explore the model

90 °
0 to 180 °; steps of 5.
0 °
0 to 180 °; steps of 5.

Try this. Leave φ at 0 and sweep θ: at θ = 0 the detector reads zero, because a repeated z reading never flips, and by θ = 90° a quarter of the beam is arriving with Sz = −ħ/2. Then set φ = 90° as well and the detector empties again — analyser 3 is now asking the question analyser 2 has just answered.

Live results

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

Results follow this lesson’s model and assumptions. Displayed values are rounded.

Interactive Physics IV modelThree Stern–Gerlach analysers in a row. Atoms enter as |↑z⟩; analyser 2 measures along an axis 90° from z and passes fraction 0.500; analyser 3 measures at 0° and its − branch delivers 0.250 of the original beam. Bars: surviving fraction after each stage.each analyser rewrites the state it readsanalyser 2 leaves |+θ⟩ behind, not |↑z⟩|↑z⟩ in → θ = 90° → φ = 0° → − branchbeam 1.000 · after 2 0.500 · detector 0.250dashed = whole beam
Three Stern–Gerlach analysers in a row. Atoms enter as |↑z⟩; analyser 2 measures along an axis 90° from z and passes fraction 0.500; analyser 3 measures at 0° and its − branch delivers 0.250 of the original beam. Bars: surviving fraction after each stage.

Predict → change one variable → check

Plan a change before running it. Every other input stays at the saved baseline.

Saved baseline. Analyser 2 axis θ from z: 90 °; Analyser 3 axis φ from z: 0 °

Baseline 90; choose 0 to 180, in steps of 5.
4. Compared with the saved baseline, this result will…

This experiment is kept only while this page is open. It does not change saved course progress.

4

Catch the common trap

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 Sz 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 Sz reading — and only after — the ignorance description becomes the right one. That is exactly what the projection encodes.

5

Practice & worked examples

3 authored examples. Make an attempt, then reveal the reasoning one step at a time.

EasyExample 1

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

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 4 steps revealed.

MediumExample 2

A beam of 8.0 × 10⁵ silver atoms per second leaves a Stern–Gerlach magnet in the Sz = +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 Sz = −ħ/2 channel. Then find that rate if the two x paths are instead recombined coherently, with no record of which was taken.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 4 steps revealed.

HardExample 3

An 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 of  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.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

6

Quick check

Select an answer and explain it before checking.

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?
7

Review & continue

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.

  • Explain the relationship without looking at the equation.
  • Check a sign, a unit and a model assumption in your working.
  • Return to the lab or worked example if a step is still unclear.

Extra challenging questions

Connect several ideas, explore supplied data, and compare your reasoning with a worked solution.