University Physics V · The Quantum Wavefunction · 3.7

Measurement Outcomes & Projection Operators

Born's rule usually arrives as |ψ(x)|², which handles one observable in one basis. Written with projectors it handles every observable at once — degenerate levels, continuous spectra, spin — and it changes the first question you ask of an operator: not which eigenvector, but which eigenspace.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

A measurement postulate has to answer two questions, and the spectral theorem answers both with one object. Which numbers can come out? Exactly the points of the spectrum of the self-adjoint operator Â that represents the observable — nothing else, ever.

With what probability? Decompose Â as Σₙ aₙP̂ₙ, where the aₙ are the distinct eigenvalues and each P̂ₙ projects onto the whole eigenspace belonging to aₙ. Those projectors are Hermitian, idempotent, mutually orthogonal and sum to the identity, so the numbers p(aₙ) = ⟨ψ|P̂ₙ|ψ⟩ = ‖P̂ₙ|ψ⟩‖² are automatically real, automatically non-negative and automatically total one: Born's rule is not an extra assumption bolted on, it is the resolution of the identity read as a probability distribution.

Writing it as a squared length rather than a squared overlap is what buys the generality. A simple eigenvalue reduces it to |⟨n|ψ⟩|²; a gₙ-fold degenerate one adds gₙ squared moduli and never asks you to pick a basis inside the subspace; a continuous spectrum replaces the sum by a projection-valued measure and assigns probability only to intervals, because x̂ has no eigenvector in L² to overlap with. The cost is that all of this is a postulate about ensembles: it predicts frequencies over many identically prepared runs, and says nothing whatever about the next single click.

Simple definition: A measurement of an observable Â can only return a point of its spectrum, and the probability of the outcome aₙ is ⟨ψ|P̂ₙ|ψ⟩ — the squared length of the state's projection onto the entire eigenspace belonging to aₙ.
Example: For |ψ⟩ = (2|1⟩ + 3|2⟩ − |3⟩)/√14 with E₂ = E₃, the outcome E₂ owns the projector P̂ = |2⟩⟨2| + |3⟩⟨3|, so p = (3² + 1²)/14 = 10/14 = 5/7 ≈ 0.714. Both coefficients count, and they add as squared moduli.

Spectral decomposition of an observableÂ = Σₙ aₙ P̂ₙ

One object carries both halves of the postulate: the eigenvalues say what can happen, the projectors say how often.

aₙ: the distinct real eigenvalues, carrying the unit of Â; P̂ₙ projects onto the whole eigenspace of aₙ, of dimension gₙ

Born's rule, projector formp(aₙ) = ⟨ψ|P̂ₙ|ψ⟩ = ‖P̂ₙ|ψ⟩‖²

A squared length, not a squared overlap — which is why the same line survives degeneracy and a continuous spectrum unchanged.

dimensionless, for a normalised state ⟨ψ|ψ⟩ = 1; non-negative because it is a squared norm

Projector algebra: the PVM testP̂ₙ† = P̂ₙ, P̂ₙP̂ₘ = δₙₘP̂ₙ, Σₙ P̂ₙ = Î

Hermiticity makes p real, idempotence makes an immediate repeat consistent, completeness makes Σₙ p(aₙ) = 1.

Î is the identity on the whole space; the sum runs over distinct eigenvalues, never over basis vectors

Simple versus gₙ-fold degenerate eigenvalue|⟨n|ψ⟩|² → Σₖ |⟨n, k|ψ⟩|², k = 1…gₙ

Degeneracy fixes a subspace, not a basis — so add the squared moduli and never the amplitudes.

k indexes any orthonormal basis of the eigenspace; every such basis returns the same sum

Continuous spectrum: intervals, not pointsp(x ∈ Δ) = ⟨ψ|P̂_Δ|ψ⟩ = ∫_Δ |ψ(x)|² dx

The spectral measure replaces the missing eigenvector: a single point has probability zero however large the density is there.

P̂_Δ = ∫_Δ |x⟩⟨x| dx; |ψ(x)|² has units L⁻¹, so only an interval yields a pure number

Expectation as a weighted spectrum⟨Â⟩ = Σₙ aₙ p(aₙ) = ⟨ψ|Â|ψ⟩

The mean is a consequence of the outcome probabilities, so it is a check on them rather than a shortcut past them.

carries the unit of Â; lies in the spectrum only when |ψ⟩ is an eigenstate of Â

The spectrum is the whole menu of results

Fix the observable first: it is a self-adjoint operator Â on the state space, and the numbers a measurement of it can return are exactly the spectrum σ(Â) — the eigenvalues where those exist, plus the continuous part where they do not. Nothing else is on the menu, and the state has no vote in what the menu contains; |ψ⟩ decides only how often each item is served. An electron in a 1.00 nm infinite well has Ĥ with Eₙ = n²h²/8mL² = n² × 0.376 eV, so 0.376, 1.504 and 3.384 eV are possible readings while 1.00 eV is not — not improbable, impossible. Position is the other extreme: x̂ on L²(ℝ) has a purely continuous spectrum, every real number belongs to it, and yet no single value carries positive probability. So the opening question of any measurement problem is not what is the probability but what is the spectrum, because the answer partitions the identity into the outcomes you then have to weigh.

Build the projectors before you build the probabilities

The spectral theorem writes Â = Σₙ aₙP̂ₙ over the distinct eigenvalues, and the recipe is mechanical. Diagonalise; collect the eigenvectors sharing one eigenvalue; orthonormalise inside each collection; set P̂ₙ = Σₖ|n, k⟩⟨n, k|, a sum over an orthonormal basis of that eigenspace. Then run the three tests: P̂ₙ† = P̂ₙ, P̂ₙP̂ₘ = δₙₘP̂ₙ, Σₙ P̂ₙ = Î. Take Â = diag(1, 3, 3) eV. It has two distinct eigenvalues, so two projectors and not three: P̂₁ = diag(1, 0, 0) and P̂₂ = diag(0, 1, 1). The second is where the content sits. Rotate inside its eigenspace to |±⟩ = (|2⟩ ± |3⟩)/√2 and rebuild: |+⟩⟨+| + |−⟩⟨−| is ½(|2⟩⟨2| + |2⟩⟨3| + |3⟩⟨2| + |3⟩⟨3|) plus ½(|2⟩⟨2| − |2⟩⟨3| − |3⟩⟨2| + |3⟩⟨3|), and the cross terms cancel to leave |2⟩⟨2| + |3⟩⟨3| exactly. P̂₂ is an attribute of Â; the basis is an attribute of whoever diagonalised it.

Born's rule is a squared length, not a squared overlap

With the projectors in hand the postulate is one line: p(aₙ) = ⟨ψ|P̂ₙ|ψ⟩. Use P̂† = P̂ and P̂² = P̂ and it becomes ⟨ψ|P̂ₙ†P̂ₙ|ψ⟩ = ‖P̂ₙ|ψ⟩‖², a squared norm — automatically real and non-negative, bounded above by ‖ψ‖² = 1, and totalling one because Σₙ P̂ₙ = Î. A global phase cancels between bra and ket, as it must. Work the example: |ψ⟩ = (2|1⟩ + 3|2⟩ − |3⟩)/√14 measured by the Â of the last section. P̂₂|ψ⟩ = (3|2⟩ − |3⟩)/√14, whose squared norm is (9 + 1)/14 = 5/7 = 0.714; p(a₁) = 4/14 = 2/7 = 0.286; together they make 1. Now repeat in the rotated basis: ⟨+|ψ⟩ = (3 − 1)/√28 and ⟨−|ψ⟩ = (3 + 1)/√28, so the sum of squared moduli is (4 + 16)/28 = 5/7 again. That invariance is the payoff of writing the rule with a projector rather than a bra — there is no basis inside a degenerate subspace left to justify.

What a degenerate outcome leaves unasked

Degeneracy costs information, and it is worth naming exactly which. Measuring Â on |ψ⟩ = (2|1⟩ + 3|2⟩ − |3⟩)/√14 returns a₂ with probability 5/7, and that one number is everything Â can say about the plane spanned by |2⟩ and |3⟩: the 3 : −1 split between them, and any relative phase between them, are invisible to it. They are not destroyed. Introduce a second observable B̂ that commutes with Â and takes distinct values on |2⟩ and |3⟩ — the two together forming a complete set of commuting observables — and the joint projectors separate into |2⟩⟨2| and |3⟩⟨3|. The single outcome of probability 10/14 splits in two, 9/14 = 0.643 and 1/14 = 0.071, which add back to 10/14 exactly as they must. That is the operational content of degeneracy: an eigenvalue with gₙ greater than 1 is a coarse-grained outcome, and the fine grain is recoverable only by an observable that resolves it. It is also why the postulate names the projector rather than the eigenvector.

A continuous spectrum has intervals, not eigenvectors

For x̂ the eigenvalue equation x̂|x⟩ = x|x⟩ has no solution inside L²(ℝ), so there is no |a⟩ to overlap with and no |⟨a|ψ⟩|² to square. The spectral theorem still supplies projectors, one for each interval: P̂_Δ = ∫_Δ |x⟩⟨x| dx, obeying P̂_Δ P̂_Δ′ = P̂ over the intersection, and P̂ over all of ℝ equal to Î. The rule then reads unchanged, p(x ∈ Δ) = ⟨ψ|P̂_Δ|ψ⟩ = ∫_Δ |ψ(x)|² dx. Two consequences follow at once. First, |ψ(x)|² is a density carrying inverse length, not a probability: for the n = 1 state of a 1.00 nm infinite well, |ψ₁(L/2)|² = 2/L = 2.00 nm⁻¹, so a detector window 0.010 nm wide at the centre registers 2.00 × 0.010 = 0.020. Second, a single point is a set of zero length, so the outcome x = L/2 exactly has probability zero however large the density is there. The finite window is not an experimental blemish to be idealised away; it is what makes the question well posed.

Frequencies, not verdicts — and how to check them in NumPy

The postulate predicts a distribution, and a distribution shows only in repetition. Prepare N identical copies, measure each once, and the fraction returning aₙ approaches p(aₙ) with standard deviation √(p(1 − p)/N): at p = 5/7 and N = 10 000 that is 0.0045, so ten thousand runs pin the third decimal and no further. A single click is never wrong and never a test. The same discipline transfers to code. Diagonalise with vals, vecs = np.linalg.eigh(A); group the eigenvalues by a stated tolerance rather than exact equality, because a degenerate pair comes back as 2.9999999998 and 3.0000000001; stack that group's columns as V and set P = V @ V.conj().T. Check before trusting: np.allclose(P @ P, P) and np.allclose(sum(Ps), np.eye(n)). With the state as a column v, the probability is np.vdot(v, P @ v).real, and the imaginary part you discarded should sit near 1e-16 — if it does not, the matrix is not Hermitian. One warning earns the whole section: eigh returns an arbitrary orthonormal basis inside a degenerate block, so never read a single column's overlap. Sum the block.

Change one variable at a time

Make the relationship visible.

Interactive model

Angle α to the a₂ plane55 °

Angle β inside the a₂ plane30 °

Relative phase φ of c₃0 °

Fix α and β, sweep φ: the four filled bars never move, because ⟨ψ|P̂₂|ψ⟩ adds squared moduli and ⟨2|3⟩ = 0 kills the cross term — yet the open bar swings. At α = 90°, β = 45°, φ = 0° it reaches 2.00, twice the total probability. Slide β alone: the narrow bars trade height, their sum does not.

p(a₁) = ⟨ψ|P̂₁|ψ⟩0.329

p(a₂) = ‖P̂₂|ψ⟩‖²0.671

TOTAL p(a₁) + p(a₂)1.000

WRONG RULE |c₂ + c₃|²1.252

Live interpretationp(a₁) = ⟨ψ|P̂₁|ψ⟩: 0.329. p(a₂) = ‖P̂₂|ψ⟩‖²: 0.671. TOTAL p(a₁) + p(a₂): 1.000. WRONG RULE |c₂ + c₃|²: 1.252

Catch the common trap

Explain before calculating.

Misconception check

The trap is to add amplitudes inside a degenerate eigenspace. Told that E₂ = E₃, students reason that the two components lead to the same outcome, so they must interfere, and write p = |c₂ + c₃|² where the rule says |c₂|² + |c₃|². Take |ψ⟩ = (|1⟩ + |2⟩ + |3⟩)/√3: the projector gives ⟨ψ|P̂₂|ψ⟩ = 1/3 + 1/3 = 2/3, while the amplitude sum gives |1 + 1|²/3 = 4/3 — a probability greater than one, which is the wrong rule announcing itself. Amplitudes interfere when they are components of one vector along one direction; |2⟩ and |3⟩ are orthogonal directions, so the cross terms in ⟨ψ|P̂₂|ψ⟩ carry a factor ⟨2|3⟩ = 0 and are simply absent. The same error wears a second costume: reading one eigenvector out of a diagonaliser and calling its overlap the answer — 1/3 here. The projector belongs to the eigenspace, not to whichever vector the algorithm handed back.

A three-level system has Â with a₁ = 1.0 eV on |1⟩ and a₂ = 3.0 eV twofold degenerate on span(|2⟩, |3⟩). The state is |ψ⟩ = (2|1⟩ + 2|2⟩ − |3⟩)/3. What is the probability that a measurement of Â returns 3.0 eV?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA spin-half is prepared as |ψ⟩ = (√3|↑⟩ + e^(iπ/3)|↓⟩)/2, where |↑⟩ and |↓⟩ are the Ŝ_z eigenkets with eigenvalues ±ℏ/2. Name the projectors, give both outcome probabilities and ⟨Ŝ_z⟩, and say what the phase e^(iπ/3) does.

Name the operator and its spectrum. Ŝ_z is self-adjoint on C² with two nondegenerate eigenvalues, +ℏ/2 and −ℏ/2, so those are the only two numbers the apparatus can ever record.
Each eigenvalue is simple, so its projector is one outer product: P̂₊ = |↑⟩⟨↑| and P̂₋ = |↓⟩⟨↓|. Check the resolution of the identity, P̂₊ + P̂₋ = Î, and orthogonality, P̂₊P̂₋ = 0, which holds because ⟨↑|↓⟩ = 0.
p(+ℏ/2) = ⟨ψ|P̂₊|ψ⟩ = |⟨↑|ψ⟩|² = |√3/2|² = 3/4 = 0.750, and p(−ℏ/2) = |e^(iπ/3)/2|² = 1/4 = 0.250. They sum to 1, as Σₙ P̂ₙ = Î guarantees.
The modulus kills the phase: |e^(iπ/3)| = 1, so it changes neither probability. It is a relative phase, not a global one, and it is invisible only to this observable — a measurement of Ŝₓ is sensitive to it.
⟨Ŝ_z⟩ = Σₙ aₙp(aₙ) = 0.750(+ℏ/2) + 0.250(−ℏ/2) = ℏ/4 = 2.64 × 10⁻³⁵ J s. No single run returns this number, because it is not in the spectrum.

Answerp(+ℏ/2) = 3/4 and p(−ℏ/2) = 1/4; ⟨Ŝ_z⟩ = ℏ/4 = 2.64 × 10⁻³⁵ J s. The phase e^(iπ/3) is invisible to Ŝ_z because Born's rule squares a modulus, but a measurement of Ŝₓ would see it.

MediumAn observable Â on C³ has a₁ = 4 on |1⟩ and a₂ = −2 twofold degenerate on span(|2⟩, |3⟩). The state is |ψ⟩ = (|1⟩ + 3|2⟩ − 2|3⟩)/√14. (a) Find p(−2) using the projector. (b) Recompute it in the rotated eigenspace basis |±⟩ = (|2⟩ ± |3⟩)/√2. (c) Give ⟨Â⟩.

Check the norm first: (1² + 3² + 2²)/14 = 14/14 = 1, so ⟨ψ|ψ⟩ = 1 and no rescaling is needed.
Group by distinct eigenvalue rather than by basis vector: P̂₂ = |2⟩⟨2| + |3⟩⟨3|, so P̂₂|ψ⟩ = (3|2⟩ − 2|3⟩)/√14.
p(−2) = ‖P̂₂|ψ⟩‖² = (9 + 4)/14 = 13/14 = 0.9286, and p(4) = 1/14 = 0.0714. The two total 1, as completeness requires.
In the rotated basis, ⟨+|ψ⟩ = (3 − 2)/(√2⋅√14) = 1/√28 and ⟨−|ψ⟩ = (3 + 2)/(√2⋅√14) = 5/√28, so the sum of squared moduli is (1 + 25)/28 = 26/28 = 13/14 — identical, because |+⟩⟨+| + |−⟩⟨−| = |2⟩⟨2| + |3⟩⟨3|.
⟨Â⟩ = 4(1/14) + (−2)(13/14) = (4 − 26)/14 = −11/7 = −1.571, a number outside the spectrum (4, −2) that no single run can produce.

Answerp(−2) = 13/14 ≈ 0.929 and p(4) = 1/14 ≈ 0.071, identical in either eigenspace basis; ⟨Â⟩ = −11/7 ≈ −1.571, which is not itself a possible outcome.

HardAn electron occupies the n = 1 state of an infinite well of width L = 1.00 nm, ψ₁(x) = √(2/L) sin(πx/L). (a) Write the projector for the central third and evaluate the probability of finding the electron there. (b) Repeat on N = 2000 identical electrons: give the expected count and its standard deviation. (c) Why does x = L/2 exactly carry probability zero although |ψ₁|² is largest there, and what would a 0.010 nm window register?

x̂ has a purely continuous spectrum, so there is no eigenket in L²(ℝ) to overlap with. The spectral theorem supplies an interval projector instead: P̂_Δ = ∫_Δ |x⟩⟨x| dx with Δ = [L/3, 2L/3], and p = ⟨ψ₁|P̂_Δ|ψ₁⟩ = ∫_Δ |ψ₁(x)|² dx.
Use sin²θ = (1 − cos 2θ)/2, so |ψ₁|² = (1/L)[1 − cos(2πx/L)], whose antiderivative is x/L − sin(2πx/L)/(2π).
Evaluate the bracket: at 2L/3 it is 2/3 − sin(4π/3)/2π = 0.6667 + 0.1378 = 0.8045, and at L/3 it is 1/3 − sin(2π/3)/2π = 0.3333 − 0.1378 = 0.1955. The difference is p = 1/3 + √3/(2π) = 0.609.
Counting is binomial: the expected count is Np = 2000 × 0.609 = 1218, with standard deviation √(Np(1 − p)) = √(2000 × 0.609 × 0.391) = √476 = 21.8, so 1218 ± 22 electrons, a fraction 0.609 ± 0.011.
A single point is a set of zero length, so the integral over it vanishes and p at x = L/2 is 0. What is large there is the density, |ψ₁(L/2)|² = 2/L = 2.00 nm⁻¹ — a density, not a probability. A real detector integrates its window: 2.00 nm⁻¹ × 0.010 nm = 0.020, or 2.0%.

Answerp = 1/3 + √3/(2π) = 0.609, giving 1218 ± 22 counts from 2000 runs. The point x = L/2 carries probability zero because probability comes from ∫|ψ|²dx over an interval; a 0.010 nm window there registers 0.020.