Eigenbasis expansion and the probability of an eigenvalue
Squared overlaps give the odds of each eigenvalue.
This is where a wavefunction starts making predictions you can check with a counter. Decide what you will measure, expand ψ on that operator's eigenfunctions, and the squared coefficients say which readings can occur and in what proportion over many identical runs, before any apparatus has been built.
Build the model
A measurement returns one eigenvalue; the prepared state fixes only the odds.
1The big idea
Quantum mechanics does not forecast the reading of one measurement; it forecasts how often each allowed reading turns up across many identically prepared systems, each measured once. The model is short. The allowed readings of Q are the eigenvalues qₙ of Q̂, and the eigenfunction ψₙ is the state in which qₙ is certain (UQMI-04.02). Completeness (UQMI-03.07) lets any prepared ψ be written as Σ cₙψₙ, the overlap cₙ = ∫ ψₙ*ψ dx measures how much of each certain state ψ holds, and the postulate is that |cₙ|² is the probability of reading qₙ. The rest follows: the probabilities add to one because ψ is normalised, their weighted mean is the ⟨Q⟩ an integral also gives, eigenfunctions sharing an eigenvalue pool their probabilities, and a continuous spectrum smears the list into a density. The list is built for one question and forgets every phase, so it is not the state: two states with identical energy odds can put a particle on opposite sides of a box. The model has three edges. Completeness is asserted, not proved. The rule describes an ideal detector that resolves every eigenvalue; a real one reports a band, and you add the probabilities inside it. And it is silent on which outcome a given run produces and on the state left behind, which is where UQMI-04.04 takes over.
2In one sentence
The probability that a measurement of Q returns the eigenvalue qₙ is |cₙ|², where cₙ = ∫ ψₙ* ψ dx is the overlap of the prepared state ψ with the normalised eigenfunction ψₙ of Q̂.
Example. An electron in a 1.00 nm box prepared as ψ = (√3 ψ₁ + ψ₂)/2 gives E₁ = 0.376 eV in 75% of energy measurements and E₂ = 1.504 eV in 25%. Their mean, 0.658 eV, is never read.
3Choose the observable, then expand on its eigenfunctions
The measurement postulate says a measurement of Q returns one of the eigenvalues qₙ of Q̂, which UQMI-03.05 guarantees are real, and UQMI-04.02 showed that the eigenfunction ψₙ is the state in which qₙ is certain. So the question an experiment puts to a prepared ψ is how much of each certain state it contains. Completeness (UQMI-03.07) says the answer exists, ψ = Σ cₙψₙ, and orthonormality returns each coefficient as an overlap, cₙ = ∫ ψₙ*ψ dx. The expansion belongs to the question, not to the state. On a ring of circumference L, ψ = √(2/L) cos(2πx/L) is an eigenfunction of Ĥ with E = h²/(2mL²), so an energy measurement is certain. Expand the same ψ on the momentum eigenfunctions e(2πinx/L)/√L and it has c₁ = c₋₁ = 1/√2 and every other cₙ zero: a momentum measurement returns +h/L or −h/L, and never 0. Change the observable and you change the basis, the coefficients, and the list of possible readings.
4Square the overlap, and let the sum check you
The postulate turns each coefficient into odds: P(qₙ) = |cₙ|². Because the eigenfunctions are complete and orthonormal, Σ|cₙ|² = ∫|ψ|² dx = 1, the physical statement that every run returns some eigenvalue. Use it as the first check on any expansion. Prepare ψ = (3ψ₁ + 4iψ₃)/5 in a box: energy runs return E₁ with probability 9/25 = 0.36 and 9E₁ with 16/25 = 0.64, total 1.00. They never return 4E₁, an allowed eigenvalue with zero overlap. Symmetry often shows a zero without an integral: a state even about the centre of the box has zero overlap with every ψₙ of even n, since those are odd about the centre, so 4E₁, 16E₁ and 36E₁ are absent however many runs you make. If your |cₙ|² add to 0.97, the missing 0.03 is not an inefficient detector; it is a lost term or an unnormalised ψ.
5What the squared modulus throws away
|cₙ|² discards the phase of cₙ, so states that differ only in the relative phases of their coefficients give identical statistics for Q. They are still different states, and another observable tells them apart. In a box of width a take ψ = (ψ₁ + e(iα)ψ₂)/√2. Energy runs return E₁ and 4E₁ half the time each, whatever α is. The chance of finding the particle in the left half is the integral of |ψ|² from 0 to a/2, which is ½ + (4/(3π)) cos α, because the cross term ψ₁ψ₂ integrates to 4/(3π) = 0.424 over that half. At α = 0 the particle turns up on the left in 92.4% of runs, at α = π in 7.6%, at α = π/2 in 50%. So no energy histogram, however long, reconstructs ψ, and the |cₙ|² for one observable are not the state. Only a phase multiplying every cₙ at once changes nothing measurable.
6Mean and spread come from the same histogram
Weight each eigenvalue by its probability and you have the mean, ⟨Q⟩ = Σ qₙ|cₙ|², which must agree with the integral ∫ψ*Q̂ψ dx of UQMI-03.08; two routes to one number is the second check. Do the same with qₙ² and σQ² = Σ qₙ²|cₙ|² − ⟨Q⟩²; with only two outcomes this reduces to σQ = |q₁ − q₂|√(P₁P₂). For ψ = (3ψ₁ + 4iψ₃)/5, ⟨E⟩ = 0.36E₁ + 0.64 × 9E₁ = 6.12E₁ and σE = 8E₁ × √(0.36 × 0.64) = 3.84E₁. Neither is a reading. An experiment prepares N copies and measures each once, and the number of E₁ results is binomial, NP ± √(NP(1 − P)): 144 ± 10 in 400 runs. The observed fractions close on |cₙ|² as 1/√N, and the sample mean on ⟨E⟩ with standard error σE/√N, but σE itself does not shrink, because it belongs to the prepared state rather than to the counting.
7A degenerate eigenvalue: add probabilities, not amplitudes
When several orthonormal eigenfunctions share one eigenvalue, the instrument reports the eigenvalue and cannot say which eigenfunction it came from, so P(q) = Σₖ|cₖ|² over the set. On a ring, E₁ = h²/(2mL²) belongs to both e(±2πix/L)/√L. Prepare ψ = [√0.2 + 1.2 cos(2πx/L) + 0.4i sin(2πx/L)]/√L. On the momentum eigenfunctions the coefficients are c₀ = √0.2, c₁ = 0.8 and c₋₁ = 0.4, so momentum runs give +h/L with probability 0.64 and −h/L with 0.16, while energy runs give E₁ with 0.64 + 0.16 = 0.80 and zero energy with 0.20. Expand instead on √(2/L) cos and √(2/L) sin, an equally good orthonormal pair for E₁: the shares become 0.72 and 0.08, and the sum is still 0.80. The split depends on the basis; the probability of the eigenvalue does not. Adding amplitudes before squaring gives (0.8 + 0.4)² = 1.44, a probability above one, which is how that mistake gives itself away.
8A continuous spectrum: from sticks to a density
On a ring of circumference L the momentum eigenvalues pₙ = nh/L are spaced Δp = h/L apart. For a ψ that is zero outside a stretch shorter than L, cₙ = √(Δp) φ(pₙ), where φ(p) = ∫ e(−ipx/ħ)ψ dx/√(2πħ) is in (kg m s⁻¹)⁻¹⁄², so the units check. Then |cₙ|² = |φ(pₙ)|²Δp: enlarge the ring and each eigenvalue's probability falls as 1/L while the number of eigenvalues in any window grows as L, so the sum over the window tends to ∫|φ|² dp. For ψ = e(ip₀x/ħ)/√w on an arc of width w, |cₙ|² = (w/L) sin²θₙ/θₙ² with θₙ = π(pₙ − p₀)w/h. With p₀ = 0, P(p = 0) is 0.125 on a ring eight times the arc and 0.010 on one a hundred times, yet the share inside |p − p₀| < h/w stays at 0.903. On the open line no single p has any probability, and |φ(p)|² is probability per kg m s⁻¹, just as the Born density |ψ(x)|² of UQMI-01.07 is this rule for x̂.
Key equations
What each relation says, and the units it is written in.
Overlap with an eigenfunction
cₙ = ∫ ψₙ*(x) ψ(x) dx
The conjugate sits on the eigenfunction. Orthonormality (UQMI-03.06) removes every other term of ψ = Σ cₘψₘ, leaving how much of the certain state ψₙ the prepared state holds.
ψₙ is a normalised eigenfunction, Q̂ψₙ = qₙψₙ. Both functions are in m⁻¹⁄², so cₙ is a dimensionless complex number.
Probability of an eigenvalue
P(qₙ) = |cₙ|²
Only eigenvalues are ever read. One with cₙ = 0 never appears in any number of runs, and no value between two eigenvalues appears at all.
qₙ carries the units of Q: J for Ĥ, kg m s⁻¹ for p̂. P is a pure number from 0 to 1.
The probabilities exhaust the outcomes
Σₙ |cₙ|² = ∫ |ψ|² dx = 1
A built-in check. A total of 0.97 means a missing eigenfunction or an unnormalised ψ, never a detector that sometimes fails to fire.
Needs a complete set of ψₙ (UQMI-03.07, asserted rather than proved) and a normalised ψ.
Mean of the outcome histogram
⟨Q⟩ = Σₙ qₙ |cₙ|²
An ensemble average, not a reading: ψ = (ψ₁ + ψ₂)/√2 in a box has ⟨E⟩ = 2.5E₁, which no single run returns.
Units of Q. The integral ∫ ψ* Q̂ ψ dx of UQMI-03.08 must return the same number.
A degenerate eigenvalue
P(q) = Σₖ |cₖ|², summed over orthonormal ψₖ that all satisfy Q̂ψₖ = qψₖ
The detector reports q, not which ψₖ. Each |cₖ|² depends on the pair you chose inside the degenerate set; their sum does not.
cₖ = ∫ ψₖ* ψ dx, dimensionless. On a ring, E₁ = h²/(2mL²) belongs to both e(2πix/L)/√L and e(−2πix/L)/√L.
A continuous spectrum gives a density
P(p₁ ≤ p ≤ p₂) = ∫ |φ(p)|² dp, from p₁ to p₂
One exact value has probability zero; only a window has odds. On a ring, |cₙ|² = |φ(pₙ)|² h/L, and the sum becomes this integral as L → ∞.
φ(p) = ∫ e(−ipx/ħ) ψ(x) dx / √(2πħ), so |φ|² is in (kg m s⁻¹)⁻¹; ħ = 1.055 × 10⁻³⁴ J s.
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. Hold p₀ at 0 and step L from 2w to 8w: the sticks crowd together and each shrinks as w/L, while the bars close onto the fixed dashed curve. Then shift p₀ by 0.25 h/w: the sticks stay put and only their heights change. The last readout pools the degenerate pair n = ±1 into P(E₁).
Live results
- SPACING Δp = h/L
- 0.250 h/w
- |cₙ|² NEAREST p₀
- 0.250
- SAME STICK ÷ Δp
- 1.000 w/h
- P(E₁) = |c₁|² + |c₋₁|²
- 0.351
Catch the common trap
The mistake this lesson is built to prevent.
Reading |cₙ|² as the probability that the system already was in ψₙ, or already had the value qₙ, before anyone looked. If that were so, the prepared ensemble would be a mixture, a fraction |cₙ|² of systems sitting in each ψₙ, and every other observable would show the weighted average of the separate ψₙ statistics, with no cross terms. Test it on the ring state ψ = √(2/L) cos(2πx/L), which returns momentum +h/L or −h/L with probability ½ each. Each momentum eigenfunction has the uniform density |ψ|² = 1/L, so a 50:50 mixture of them would find the particle equally often everywhere on the ring. The superposition has |ψ|² = (2/L) cos²(2πx/L) instead, with nodes at x = L/4 and 3L/4 where no run ever finds it. The |cₙ|² are odds for a measurement not yet made, not a census of values the system was carrying.
Test it: what changes?
- Stick spacing Δp = h/L shrinks fourfold
- Stick at p = 0 falls as w/L, to 0.125
- |cₙ|² ÷ Δp traces the fixed density |φ(p)|²
- Sticks stay at the same pₙ = nh/L
- Stick heights change, recentred on the new p₀
- Σ|cₙ|² still equals 1
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
An electron in an infinite well of width 0.50 nm is prepared in ψ = N[2ψ₁ + (1 − i)ψ₃], where the ψₙ are the normalised energy eigenfunctions. Find N, list the energies an energy measurement can return with their probabilities, and find ⟨E⟩.
View solution →
- Normalise with the sum rule: |2|² + |1 − i|² = 4 + 2 = 6, so 6N² = 1 and N = 1/√6. The phase of 1 − i enters only through its modulus.
- Energies: Eₙ = n²π²ħ²/(2ma²), and (J s)²/(kg m²) = J as it should. With a = 0.50 × 10⁻⁹ m and m = 9.11 × 10⁻³¹ kg, E₁ = 2.41 × 10⁻¹⁹ J = 1.504 eV, so E₃ = 9E₁ = 13.54 eV.
- Probabilities: P(E₁) = 4/6 = 0.667 and P(E₃) = 2/6 = 0.333, adding to 1. P(E₂) = 0: 6.02 eV is an eigenvalue of the well, but this state holds no ψ₂.
- Mean: ⟨E⟩ = (2/3)E₁ + (1/3)(9E₁) = (11/3)E₁ = 5.52 eV. It lies between the two readings, as a weighted mean must, and is neither of them.
- Limiting case: remove the ψ₃ term and ⟨E⟩ → E₁ with probability 1, the determinate ground state.
Answer. N = 1/√6; readings of 1.50 eV with probability 2/3 and 13.5 eV with 1/3, never 6.02 eV; ⟨E⟩ = 11E₁/3 = 5.52 eV, a value no single run returns.
MediumExample 2
An electron is released in ψ(x) = √(2b³/π)/(x² + b²) with b = 0.20 nm; its momentum wavefunction is φ(p) = √(b/ħ) e(−b|p|/ħ). Find the probability that a momentum measurement gives |p| < ħ/b, the probability that it gives exactly p = 0, and the probability that a detector accepting a window 1.0 × 10⁻²⁶ kg m s⁻¹ wide, centred on p = 0, fires.
View solution →
- Density: |φ|² = (b/ħ) e(−2b|p|/ħ). Units: b/ħ is m/(J s) = (kg m s⁻¹)⁻¹, a probability per unit momentum. Normalisation: 2(b/ħ)(ħ/2b) = 1, as it must.
- Window: P(|p| < ħ/b) = 2∫ (b/ħ) e(−2bp/ħ) dp from 0 to ħ/b = 1 − e⁻² = 0.865, where ħ/b = 5.27 × 10⁻²⁵ kg m s⁻¹. Widen the window without limit and P → 1.
- A single value: the interval from 0 to 0 has zero width, so P(p = 0) = 0 exactly, even though the density there is at its largest, b/ħ = 1.90 × 10²⁴ (kg m s⁻¹)⁻¹.
- Detector: P = 1 − e(−bδp/ħ) = 1 − e(−0.0190) = 0.0188 for δp = 1.0 × 10⁻²⁶ kg m s⁻¹. The narrow-window limit, density × width = 0.0190, agrees to 1%.
- Discrete check: on a ring of circumference 50 nm, p = 0 would be an eigenvalue with |c₀|² ≈ |φ(0)|² h/L = 2πb/L = 0.025, and the 79 eigenvalues inside |p| < ħ/b would hold 0.863, already close to 0.865.
Answer. P(|p| < ħ/b) = 1 − e⁻² = 0.865; P(p = 0 exactly) = 0; the 1.0 × 10⁻²⁶ kg m s⁻¹ window fires with probability 0.019.
HardExample 3
An electron in an infinite well of width a = 1.00 nm is prepared in ψ(x) = A sin²(πx/a). Normalise it, find the probabilities of E₁ and E₃ by projection, say which energies never occur, get ⟨E⟩ from the integral, and compare the two routes to ⟨E⟩ to learn what the remaining outcomes contribute. In 2000 runs, how many readings of each energy do you expect?
View solution →
- Normalise: ∫ sin⁴(πx/a) dx from 0 to a = 3a/8, so A = √(8/(3a)) = 1.63 nm⁻¹⁄². Symmetry: ψ is even about x = a/2 and every ψₙ of even n is odd about it, so c₂ = c₄ = … = 0, and 4E₁, 16E₁, … never occur.
- Project with ψₙ = √(2/a) sin(nπx/a) and u = πx/a: c₁ = A√(2/a)(a/π)∫ sin³u du from 0 to π = (4/(√3 a))(a/π)(4/3) = 16/(3√3π) = 0.980, so P(E₁) = 256/(27π²) = 0.9607.
- For n = 3, ∫ sin 3u sin²u du from 0 to π = −4/15, so c₃ = −16/(15√3π) = −0.196 and P(E₃) = 0.0384; the minus sign is a phase and drops out. Sum rule: 0.9607 + 0.0384 = 0.9991, leaving 0.0009 for n ≥ 5.
- Integral route: ψ vanishes at both walls, so ⟨E⟩ = (ħ²/2m)∫|ψ′|² dx = (ħ²/2m)A²(π/a)²(a/2) = (4/3)E₁. With E₁ = 0.376 eV, ⟨E⟩ = 0.501 eV, which is not an eigenvalue: no run returns it.
- Spectral route: E₁(0.9607 + 9 × 0.0384) = 1.3065E₁, short of 1.3333E₁ by 0.027E₁. That remainder comes from readings of 25E₁ and above, which carry only 0.09% of the probability: rare high readings pull the mean.
- Counts in N = 2000 runs, NP ± √(NP(1 − P)): 1921 ± 9 readings of 0.376 eV, 77 ± 9 of E₃ = 3.38 eV, and about 2 at 25E₁ = 9.40 eV or higher.
Answer. A = 1.63 nm⁻¹⁄²; P(E₁) = 0.961, P(E₃) = 0.038, no even-n energies, about 0.001 above E₃; ⟨E⟩ = 4E₁/3 = 0.501 eV, never read; in 2000 runs about 1921 ± 9 at 0.376 eV and 77 ± 9 at 3.38 eV.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
Expand on Q̂'s eigenfunctions, then P(qₙ) = |cₙ|².
Compute cₙ = ∫ ψₙ*ψ dx with the conjugate on the eigenfunction, check that Σ|cₙ|² = 1, and get ⟨Q⟩ = Σ qₙ|cₙ|². Pool |cₖ|² over a degenerate eigenvalue, integrate |φ(p)|² over a window for a continuous spectrum, and remember the list forgets every phase.
Next: Collapse, repeatability, and the state after measurement →