Collapse, repeatability, and the state after measurement
A measurement result is also a preparation.
The Born rule says what a measurement will read. This lesson is about what it leaves behind, because that state is the initial condition for everything after it: you will write it down, renormalise it, hand it back to the Schrödinger equation, and learn which parts of the recipe experiments actually test.
Build the model
A result is also a preparation: what you read decides what happens next.
1The big idea
A measurement does two things, and the Born rule covers only the first: it produces a result, qₙ with probability |cₙ|². The second is that it prepares a state. The evidence is repeatability, since an ideal measurement repeated at once returns the same value, and together with the Born rule that fact fixes what is left behind: a state in which qₙ has probability 1, which is the eigenfunction ψₙ up to a phase no experiment can see. When the result cannot tell several eigenfunctions apart, because they share an eigenvalue or the instrument is coarse, the rule keeps all of them with their relative phases, removes only what the result excludes, and divides by √P to restore normalisation. From that instant the Schrödinger equation is back in charge, so the result stays sharp for ever when the state left behind is stationary, and in general only for times short against ħ over the energy differences it contains. The model stops in three places. It is a rule for computing the statistics of later measurements, not a mechanism anyone has watched, and it says nothing about when or how the change happens. It cannot be applied literally to a continuous spectrum, because δ(x − x₀) is not a normalisable state, so a real detector projects onto a band. And a sharp-edged band is itself an idealisation that would hand the particle infinite mean energy; physical detectors have soft edges.
2In one sentence
The projection rule: a measurement that returns qₙ leaves the system in the matching eigenfunction ψₙ, or, when the result cannot tell several eigenfunctions apart, in the part of ψ it cannot exclude, renormalised. That state is where every later calculation starts.
Example. An electron in a well in ψ = 0.60ψ₁ + 0.80ψ₂ gives E₂ in 64% of runs. In each of those runs the state afterwards is ψ₂, and a second energy measurement returns E₂ again with probability 1, whether it comes a nanosecond or an hour later.
3Repeatability is what fixes the state left behind
The previous topic gave the odds: measuring Q̂ on ψ = Σ cₙψₙ returns qₙ with probability |cₙ|². It said nothing about the state afterwards, and experiment supplies the missing fact. An ideal measurement repeated at once returns the same result: a photon that has just passed a vertical polariser passes an identical one straight after, every time. Now borrow one line from Unit 3: the eigenfunctions of an observable Q̂ can be chosen orthonormal and form a complete set, so whatever state ψ′ the first measurement leaves can be written ψ′ = Σ dₘψₘ with Σ |dₘ|² = 1. The repeat returns qₙ with probability |dₙ|², so certainty needs |dₙ|² = 1, and normalisation then forces every other dₘ to zero. The state left behind is e(iα)ψₙ, and a global phase changes no prediction. Collapse is not extra drama bolted on: given the Born rule, it is what repeatability means. Not every real measurement is ideal, since a photodiode absorbs the photon it counts, so the rule describes measurements built to be repeatable.
4Renormalise, and keep what the result cannot tell apart
The projected piece cₙψₙ is not normalised: ∫ |cₙψₙ|² dx = |cₙ|². Dividing by |cₙ|, the square root of the probability of what actually happened, is the conditional in conditional probability; from now on you compute only among the runs that gave qₙ. When the result cannot separate several eigenfunctions, because they share the eigenvalue or because the instrument cannot resolve them, keep every one of them with its amplitude and relative phase, and remove only what the result rules out. Take c = 0.60, 0.64 and 0.48 on ψ₁, ψ₂ and ψ₃, with ψ₂ and ψ₃ sharing the eigenvalue q. The outcome q has probability 0.64² + 0.48² = 0.64, and the state left behind is (0.64ψ₂ + 0.48ψ₃)/0.80 = 0.80ψ₂ + 0.60ψ₃. Check it three ways: 0.80² + 0.60² = 1, the amplitude ratio is still 4 : 3, and ψ/√P keeps the units m⁻¹⁄² because P is a pure number.
5From the instant after, the Schrödinger equation is in charge
Collapse acts at the measurement time t₀ and nowhere else. After it, ψafter is simply the initial condition for the Schrödinger equation: expand it as dₘ = ∫ φₘ* ψafter dx over the energy eigenfunctions and attach e(−iEₘτ/ħ), with τ = t − t₀. If the measured observable was the energy, ψafter is a single φₙ, only its phase turns, and a repeat returns Eₙ a second or a year later. When ψafter mixes several energies, the result is repeatable only for times short against ħ divided by the differences between them. For an electron in a 1.0 nm well, E₁ = h²/(8mL²) = 0.376 eV, and a position band over the middle third leaves mostly φ₁ and φ₃, which beat with period h/(E₃ − E₁) = 1.37 fs. Evolving the full expansion, the same detector fires on a repeat 0.20 fs later with probability 0.66, and 0.50 fs later with only 0.28. “Immediately” in the repeatability rule means short on that scale, not quick by the experimenter's standards.
6A real detector projects onto a band
Position has a continuous spectrum, and its would-be eigenfunction δ(x − x₀) cannot be normalised, so no measurement can leave a particle exactly at x₀. What a detector reports is a band, a ≤ x ≤ b, and the rule carries over with an integral for the sum: P = ∫ |ψ|² dx over the band, and ψafter = ψ/√P inside it, zero outside. The old density survives inside, scaled up by 1/P. For the ground state of a 1.0 nm well and a detector over the middle half, P = 1/2 + 1/π = 0.818, and the density inside rises by a factor of 1.22. Two limits check the rule. A band covering the whole well gives P = 1 and changes nothing: a result that excludes nothing disturbs nothing. A narrow band of width δ around x₀ gives P ≈ |ψ(x₀)|² δ, so the peak left behind grows as 1/δ. The sharp band is itself an idealisation. Its edges make ψafter jump, the weights |dₙ|² then fall only as 1/n² on average, and ⟨E⟩ = Σ Eₙ |dₙ|² diverges; a physical detector has soft edges, which keeps the energy it hands the particle finite.
7A rule for later statistics, not a film of a process
No experiment has watched a wavefunction jump. What experiments record are sequences of results, and the projection rule is the recipe for their joint probabilities: P(qₙ, then rₘ) = |cₙ|² × |∫ χₘ* ψₙ dx|², with χₘ the eigenfunctions of the second observable. Everything testable about collapse lives in numbers like that, and the rule passes its consistency checks. A coarse result followed by a fine one reproduces the fine statistics: in the example above, q occurs in 0.64 of runs, ψ₂ is then found in 0.80² = 0.64 of those, and the product 0.4096 is exactly |c₂|², the chance of resolving ψ₂ at the start. Sequences also expose what cannot be sharp together: a click over the middle half of the well sharpens x but leaves E₁ only 0.818 likely. A detector that stays silent collapses the state too, onto ψ/√(1 − P) outside its band: learning where the particle is not is still learning. And the classical limit is built in. A macroscopic pointer's state is far narrower than any band we can read, so P ≈ 1, ψafter ≈ ψ, and measurement looks like pure revelation.
Key equations
What each relation says, and the units it is written in.
Collapse onto a non-degenerate eigenvalue
ψ(x) → (cₙ/|cₙ|) ψₙ(x), when the result is qₙ
The prefactor is a pure phase, so the state left behind is ψₙ and Q is now sharp. How often this branch happens is |cₙ|², from the previous topic.
cₙ = ∫ ψₙ* ψ dx is dimensionless. ψ and ψₙ in m⁻¹⁄². qₙ carries the units of the observable Q̂.
Coarse or degenerate result
ψafter = (Σ over qₙ in B of cₙψₙ)/√PB, with PB = Σ over qₙ in B of |cₙ|²
Keep every component the result cannot exclude, with its amplitude and relative phase, and divide by √PB so that ∫ |ψafter|² dx = 1 again.
B is the set of eigenvalues the result cannot tell apart. ψafter in m⁻¹⁄². PB is a pure number.
Position detector covering a ≤ x ≤ b
ψafter(x) = ψ(x)/√P for a ≤ x ≤ b, and 0 elsewhere, where P = ∫ |ψ(x)|² dx from a to b
Inside the band the old density survives, scaled up by 1/P. A band over the whole region gives P = 1 and changes nothing.
a and b in m. |ψ|² in m⁻¹, so P is a pure number. The band stands in for the non-normalisable δ(x − x₀).
Evolution restarts from the new state
Ψ(x, t) = Σₘ dₘ φₘ(x) e(−iEₘτ/ħ), with dₘ = ∫ φₘ* ψafter dx and τ = t − t₀
The |dₘ|² are frozen after t₀, so an energy result repeats for ever. A state that mixes energies repeats only for times short against ħ/(Eₘ − Eₖ).
φₘ in m⁻¹⁄² and Eₘ in J are the energy eigenfunctions and energies. t₀ is the measurement time. ħ = 1.055 × 10⁻³⁴ J s.
What survives of the state before
P(φ found again) = |∫ φ* ψafter dx|² = PB, when φ was the state before
A click that was 82% likely leaves 82% of the ground state behind: the less likely the result, the more the measurement disturbs.
φ is normalised and an eigenfunction of the next observable. PB is the probability of the result just obtained.
Probability of a sequence of results
P(qₙ, then rₘ) = |cₙ|² × |∫ χₘ* ψₙ dx|²
This is what the projection rule is for, and the only way it is ever tested. Order matters whenever the χₘ differ from the ψₙ.
χₘ are the eigenfunctions of the second observable R̂, with eigenvalues rₘ. R̂ is measured before ψₙ has time to evolve.
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. Start at the middle half: the click is 0.818 likely, and a later energy test finds E₁ with the same 0.818. Squeeze the band to 0.40–0.60 nm and the density inside jumps by 2.58; open it wall to wall and the solid curve lands on the dashed one, since a result that excludes nothing changes nothing.
Live results
- P(CLICK)
- 0.818
- DENSITY RESCALE 1/P
- 1.22
- ⟨x⟩ AFTER THE CLICK
- 0.500 nm
- P(E₁) IF ENERGY IS MEASURED
- 0.818
Catch the common trap
The mistake this lesson is built to prevent.
“Once measured, the value sticks.” Collapse is easily read as setting the particle to qₙ for good, so that a position found in a band stays found and every later measurement is a foregone conclusion. The rule claims much less. It fixes the state at the instant of measurement, and from then on the Schrödinger equation evolves ψafter like any other initial condition. A repeat stays certain for all time when ψafter is stationary, as after an energy measurement, and most other results do not leave a stationary state. After a click over the middle third of a 1.0 nm well, a second identical detector fires with probability 0.66 at 0.20 fs and only 0.28 at 0.50 fs, because the state has spread out of the band. The mirror-image error, that the particle now sits at a point, fails too: the result was a band, and inside it the state keeps its old shape.
Test it: what changes?
- Click probability falls below 0.818
- Density rescale 1/P rises from 1.22 to 2.58
- Later P(E₁) still equals P(click), now lower
- Click becomes certain: P = 1
- State after the click is ψ₁, unchanged
- Later energy test gives E₁ with certainty
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
An electron in an infinite well of width 0.80 nm is prepared in ψ = (ψ₁ − 2ψ₂)/√5, where ψₙ are the well's energy eigenfunctions. An energy measurement returns E₂. How likely was that result, what is the state straight afterwards, and what does a second energy measurement 2.0 ms later return?
View solution →
- Normalisation first: (1/√5)² + (2/√5)² = 1/5 + 4/5 = 1, so the coefficients can be read as they stand. P(E₂) = |−2/√5|² = 4/5 = 0.80.
- The levels: E₁ = h²/(8mL²) = (6.626 × 10⁻³⁴ J s)²/(8 × 9.109 × 10⁻³¹ kg × (0.80 × 10⁻⁹ m)²) = 9.41 × 10⁻²⁰ J = 0.588 eV, and E₂ = 4E₁ = 2.35 eV.
- Collapse keeps only the E₂ component and divides by |c₂|: (−2/√5)ψ₂ ÷ (2/√5) = −ψ₂. The minus sign is a global phase and changes no prediction, so the state is ψ₂.
- From the measurement on, Ψ(x, t) = −ψ₂(x) e(−iE₂τ/ħ), with τ = t − t₀. Only the phase turns, the weight on ψ₂ stays 1, and the repeat returns 2.35 eV with certainty, 2.0 ms later or at any other time. Before the first measurement that probability was 0.80, at every time.
- Check: averaged over all runs, 20% left in ψ₁ at 0.588 eV and 80% left in ψ₂ at 2.35 eV give 0.2 × 0.588 + 0.8 × 2.35 = 2.00 eV, which is ⟨E⟩ of the prepared state. The energy measurement sorted the runs without changing their mean.
Answer. P(E₂) = 0.80. The state afterwards is ψ₂ (the sign is an unobservable phase), and a second energy measurement returns E₂ = 2.35 eV with certainty, whenever it is made.
MediumExample 2
An electron in a 0.60 nm infinite well is in ψ = (2ψ₁ + 2iψ₂ + ψ₃)/3, with ψₙ the energy eigenfunctions. A detector fires only for energies above 3.0 eV and cannot tell one such level from another. It fires. Find the probability of that, the state left behind, what a detector that resolves every level would then report, and ⟨E⟩ before and after.
View solution →
- Levels: E₁ = h²/(8mL²) = 1.0445 eV for L = 0.60 nm, so E₂ = 4E₁ = 4.18 eV and E₃ = 9E₁ = 9.40 eV. Only E₂ and E₃ lie above 3.0 eV. Normalisation: 4/9 + 4/9 + 1/9 = 1.
- P(fire) = |2i/3|² + |1/3|² = 4/9 + 1/9 = 5/9 = 0.556.
- Remove the ψ₁ term, which the result excludes, keep the other two with their phases, and divide by √(5/9): ψafter = (2iψ₂ + ψ₃)/√5 = 0.894iψ₂ + 0.447ψ₃. Check: 0.80 + 0.20 = 1.
- The resolving detector now finds E₂ with probability 0.80 and E₃ with 0.20. Consistency: (5/9) × 0.80 = 4/9, exactly |c₂|² for the original state, so a coarse result followed by a fine one reproduces the fine statistics.
- In units of E₁: ⟨E⟩ before = (4/9)(1) + (4/9)(4) + (1/9)(9) = 29/9, and 29/9 × 1.0445 eV = 3.37 eV. After = 0.80 × 4 + 0.20 × 9 = 5.00, and 5.00 × 1.0445 eV = 5.22 eV. Because two energies survive, |ψafter|² still sloshes, with period h/(E₃ − E₂) = 0.79 fs; the kept relative phase i fixes the timing of the slosh, not whether it happens, and a detector that had named the level would have left a stationary density.
Answer. P = 5/9 ≈ 0.556 and ψafter = (2iψ₂ + ψ₃)/√5. A resolving detector then reads E₂ = 4.18 eV with probability 0.80 or E₃ = 9.40 eV with 0.20, and ⟨E⟩ rises from 3.37 eV to 5.22 eV.
HardExample 3
An electron is in the ground state ψ₁ of a 1.0 nm infinite well. A position detector covering the middle third, L/3 ≤ x ≤ 2L/3, clicks. Find the probability of the click, the state and peak density left behind, the probabilities that an energy measurement now returns E₁, E₂ and E₃, and the time scale on which the click stops being repeatable.
View solution →
- P = ∫ (2/L) sin²(πx/L) dx from L/3 to 2L/3 = [x/L − sin(2πx/L)/(2π)] between those limits = 1/3 + √3/(2π) = 0.333 + 0.276 = 0.609. Limit check: the same integral over the whole well gives 1.
- ψafter = ψ₁/√0.609 = 1.28ψ₁ inside the band and 0 outside. Its peak density, at x = L/2, is (2/L)/0.609 = 3.28 nm⁻¹, an inverse length as |ψ|² must be, and ∫ |ψafter|² dx = 0.609/0.609 = 1.
- Overlap with ψ₁: (∫ ψ₁² dx over the band)/√P = P/√P = √P, so P(E₁) = 0.609. The band and ψ₁ are symmetric about L/2 while ψ₂ is antisymmetric, so that overlap vanishes and P(E₂) = 0.
- Overlap with ψ₃: (1/√P) ∫ (2/L) sin(3πx/L) sin(πx/L) dx over the band = (1/√P)[sin(2πx/L)/(2π) − sin(4πx/L)/(4π)] between the limits = −3√3/(4π√P) = −0.4135/0.7804 = −0.530, so P(E₃) = 0.281. The last 0.110 is spread over E₅, E₇ and above.
- Timing: E₁ = h²/(8mL²) = 0.376 eV and E₃ − E₁ = 8E₁ = 3.01 eV, so the dominant ψ₁–ψ₃ beat has period h/(E₃ − E₁) = 1.37 fs. The click is repeatable only for times well under that, whereas an energy result, once obtained, would repeat for ever.
Answer. P(click) = 0.609, leaving 1.28ψ₁ in the band with peak density 3.28 nm⁻¹. Then P(E₁) = 0.609, P(E₂) = 0, P(E₃) = 0.281, with 0.110 higher up, and the click stays repeatable only for times short against 1.4 fs.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
Project, renormalise, then let the Schrödinger equation take over.
After result qₙ the state is ψₙ; for a coarse or degenerate result keep every component the result cannot exclude, with its phase, and divide by √PB. Then evolve it: a repeat stays certain for ever when ψafter is stationary, but in general only for times short against ħ/(Eₘ − Eₖ).
Next: Commutators and the canonical relation [x̂, p̂] = iħ →