
Quantum information · institution-dependent · Interactive lesson
Measurement, the Born rule and POVMs
Every protocol in this unit ends the same way: a state meets a set of operators that sum to the identity, and one of them fires. Learn to compute that firing probability as a trace, to write down what survives, and to recognise when a projector is too blunt an instrument and a POVM is the honest tool.
UPV-20.03 · Lesson 3 of 9 in this unit. Position is not a completion score.Quantum information · institution-dependent. Your university determines the required depth of these quantum topics, including the use of operators, differential equations and computation. Institution-dependent does not automatically mean optional.
Build the model
Probability enters quantum mechanics in exactly one place.
The key idea
A measurement is a set of positive operators Eₖ on the state space that sum to the identity; outcome k occurs with probability Tr(ρ Eₖ), and when the Eₖ are orthogonal projectors Pₖ the state left behind is Pₖ ρ Pₖ divided by that probability.
Example. A detector that fires only on |1⟩, and then with efficiency 0.80, has click element E₁ = 0.80|1⟩⟨1| and no-click element E₀ = |0⟩⟨0| + 0.20|1⟩⟨1|: the two sum to I, yet neither is a projector. On |+⟩ it clicks with probability Tr(ρ E₁) = 0.80 × ½ = 0.400 and stays silent with probability 0.600.
The complete physical model and assumptions
Unitary evolution is deterministic and reversible; nothing in the Schrödinger equation ever produces a probability. Probability enters through one postulate. A measurement is a set of positive operators that resolve the identity — orthogonal projectors Pₖ = |k⟩⟨k| with Σ Pₖ = I in the projective case — and outcome k occurs with probability pₖ = Tr(ρ Pₖ), the state then becoming Pₖ ρ Pₖ / pₖ. The form is not arbitrary: Gleason's theorem shows that any assignment of probabilities to the projectors of a space of dimension three or more, additive over orthogonal sets, must be Tr(ρP) for some density operator ρ. What the postulate buys is the entire predictive content of the theory — expectation values, uncertainties, Bell correlations, key rates — and a rule strong enough to generalise: relax the projectors to positive operators Eₖ ≥ 0 with Σ Eₖ = I and you have a POVM, which describes noisy detectors and measurements with more outcomes than dimensions, and which Naimark's theorem reads as an ordinary projective measurement on a larger space. What it costs is a break in the formalism. The update ρ → Pₖ ρ Pₖ / pₖ is not unitary, cannot be derived from the Hamiltonian, and says nothing about how one outcome comes to be selected. The rule gives statistics, and the statistics are exactly right; a mechanism is something it never claimed to supply.
1A measurement is a resolution of the identity
Start from the observable, not the outcome. A Hermitian A on Cⁿ has real eigenvalues aₖ and orthogonal eigenspaces, so the spectral theorem writes A = Σ aₖ Pₖ with Pₖ the projector onto the k-th eigenspace, Pⱼ Pₖ = δⱼₖ Pₖ and Σ Pₖ = I. The measurement is the set {Pₖ}; the numbers aₖ are labels. For σz the projectors are P₀ = (I + σz)/2 = |0⟩⟨0| and P₁ = (I − σz)/2 = |1⟩⟨1|, and for σₙ = n⋅σ along any unit axis n they are P_± = (I ± n⋅σ)/2, which the spin-measurement topic UPV-10.07 builds and turns into the probabilities (1 ± r⋅n)/2 for a qubit with Bloch vector r. Degeneracy makes a projector rank higher than one: σz ⊗ I on two qubits has eigenvalue +1 on the whole plane spanned by |00⟩ and |01⟩, so P+ = |0⟩⟨0| ⊗ I has rank 2 and asks nothing about the second qubit. Name the operator, the basis it is diagonal in, and the rank of each projector before touching a state; that is where most errors are decided.
2Take the trace against the state
The Born rule is pₖ = Tr(ρ Pₖ). For a pure state ρ = |ψ⟩⟨ψ| the trace collapses to ⟨ψ|Pₖ|ψ⟩, which is |⟨k|ψ⟩|² when Pₖ has rank one; for a mixture it weights each member's probability by its ensemble weight, which is exactly what the linearity of the trace does for you. That linearity has a sharp consequence: pₖ depends on ρ and on nothing else about how the state was made. An equal mix of |0⟩ and |1⟩ and an equal mix of |+⟩ and |−⟩ are both I/2, so no measurement, projective or POVM, can ever tell the two recipes apart; the density operator is not shorthand for an ensemble, it is everything any ensemble behind it can show. The maximally mixed state I/d answers each question in proportion to the size of the subspace asked about, pₖ = Tr Pₖ / d = rank(Pₖ)/d: on two qubits, P = |0⟩⟨0| ⊗ I fires with probability 2/4 = 0.500 and P = |00⟩⟨00| with 1/4 = 0.250. The same trace gives a POVM's statistics unchanged, pₖ = Tr(ρ Eₖ), so everything in this section holds for non-projective measurements too.
3What survives: Lüders' rule and the loss of coherence
After outcome k the state is Pₖ ρ Pₖ / pₖ, or Pₖ|ψ⟩/√pₖ for a pure state. Two consequences matter. First, a rank-one projector produces a pure state whatever went in: measure σz on any qubit, read 1, and you hold |1⟩ exactly, the earlier amplitudes gone rather than hidden. Second, a higher-rank projector preserves structure inside its eigenspace. Measuring σz on qubit A of (|00⟩ + |01⟩ + |10⟩)/√3 and reading 0 leaves (|00⟩ + |01⟩)/√2: the relative amplitudes of the surviving branch are untouched, which is Lüders' refinement of von Neumann's rule and the reason later measurements on B still make sense. If the outcome is taken but not read, the state is the average Σ pₖ (Pₖ ρ Pₖ / pₖ) = Σ Pₖ ρ Pₖ, which keeps the diagonal of ρ in the measured basis and deletes the off-diagonal terms. For |+⟩ under an unread σz measurement, purity Tr ρ² falls from 1 to 1/2. No unitary does that: unitaries preserve Tr ρ² and are deterministic, so the update is a different kind of map.
4When a projector is the wrong tool: a POVM in three lines
Suppose a source sends |ψ₁⟩ or |ψ₂⟩ with equal probability, their overlap s = |⟨ψ₁|ψ₂⟩| strictly between 0 and 1, and you must never guess wrong, though you may answer 'don't know'. A projective measurement on C² has two outcomes, (|v⟩⟨v|, I − |v⟩⟨v|). An outcome that names ψ₂ must annihilate ψ₁, which fixes |v⟩ = |ψ₁⊥⟩, the state orthogonal to ψ₁; its partner |ψ₁⟩⟨ψ₁| then fires on ψ₂ with probability s², so it can only mean 'don't know'. The best projective scheme therefore names one state and never the other, an average success of (1 − s²)/2. Naming either state needs three outcomes in two dimensions, and only a POVM supplies them. Take E₁ = a|ψ₁⊥⟩⟨ψ₁⊥|, blind to ψ₁ and so a certain sign of ψ₂; E₂ = a|ψ₂⊥⟩⟨ψ₂⊥|, a certain sign of ψ₁; and E_? = I − E₁ − E₂. The sum rule holds for every a, so positivity is the binding constraint. The two rank-one operators overlap by |⟨ψ₁⊥|ψ₂⊥⟩| = s, their sum has eigenvalues 1 ± s, and E_? ≥ 0 only while a ≤ 1/(1 + s). At that value each state is identified with probability a(1 − s²) = 1 − s, the Ivanovic–Dieks–Peres bound, which beats (1 − s²)/2 for every s between 0 and 1. The figure draws E₁ and E₂ as spokes pointing away from the states they must ignore, because orthogonal states sit at opposite ends of the Bloch sphere.
5Every POVM is a projector on a bigger space
A POVM element Eₖ ≥ 0 factorises as Eₖ = Mₖ† Mₖ, and the natural update is ρ → Mₖ ρ Mₖ† / pₖ. The factorisation is not unique: Mₖ = Uₖ √Eₖ works for any unitary Uₖ, and every such choice gives identical statistics, so a POVM fixes what you will see and not what you will hold afterwards. Naimark's theorem removes any suspicion that POVMs are a second postulate. Attach an ancilla in |0⟩ₐ, apply a unitary U on system plus ancilla, and measure the ancilla projectively; the effective operator on the system is Eₖ = V†(I ⊗ Pₖ)V with V|ψ⟩ = U(|ψ⟩ ⊗ |0⟩ₐ) an isometry. A qubit with three outcomes is the standard example: the trine Eₖ = (2/3)|φₖ⟩⟨φₖ|, with the three Bloch vectors nₖ 120° apart on the equator, sums to I because the three vectors sum to zero: Σ Eₖ = (1/3)(3I + Σ nₖ⋅σ) = I. In NumPy this is three 2 × 2 matrices and one assertion, and the Naimark isometry is the 6 × 2 stack of the √Eₖ, completed to a 6 × 6 unitary with scipy.linalg.qr.
6What the rule does not say
Gleason's theorem tells you the form of the probabilities is forced once you accept that a probability is assigned to each projector and adds over orthogonal sets, in any dimension three or more. It does not tell you why one outcome happens. Unitary evolution maps |+⟩ to a single definite vector; the postulate maps it to |0⟩ half the time and |1⟩ half the time, and no linear operator does that, so the update is added by hand. Decoherence, treated in the last topic of this unit, explains why the off-diagonal terms vanish in a pointer basis when the environment is traced out; it produces the mixture Σ Pₖ ρ Pₖ and stops there, one outcome short. Two things the rule does guarantee: repeated projective measurements agree, and measuring one half of an entangled pair leaves the other half's reduced state exactly as it was, Σₖ TrA[(Pₖ ⊗ I) ρ (Pₖ ⊗ I)] = TrA ρ = ρB, so no choice of measurement can send a signal. The postulate is a statistical contract, precise to every decimal place ever tested, and silent about mechanism by construction.
Key equations
Read the units, assumptions and limits before substituting.
Born rule, projective case
pₖ = Tr(ρ Pₖ) = ⟨ψ|Pₖ|ψ⟩ for pure ρ = |ψ⟩⟨ψ|
Σ pₖ = Tr(ρ Σ Pₖ) = Tr ρ = 1: normalisation is the resolution of the identity, not a separate condition.
Pₖ = |k⟩⟨k| with Pₖ² = Pₖ = Pₖ† and Σ Pₖ = I; pₖ is dimensionless, in [0, 1]Lüders update after outcome k
ρ → Pₖ ρ Pₖ / pₖ|ψ⟩ → Pₖ|ψ⟩ / √pₖ
Pₖ² = Pₖ makes the measurement repeatable: measure again at once and outcome k returns with probability 1.
defined only when pₖ > 0; for a rank-one Pₖ the result is |k⟩⟨k| whatever ρ wasExpectation value without diagonalising
⟨A⟩ = Σ aₖ pₖ = Tr(ρ A), with A = Σ aₖ Pₖ
The mean is a trace against A itself, so you diagonalise only when you need the full distribution.
aₖ are the eigenvalues of A in its own units; the Pₖ project onto its eigenspacesUnread measurement: the non-selective update
ρ → ρ' = Σₖ Pₖ ρ Pₖ, Tr ρ'² ≤ Tr ρ²
Measuring and not reading is not the same as not measuring: the coherence between outcomes is gone, and since unitaries preserve Tr ρ², no unitary on the system alone can restore it.
the pₖ-weighted average of the Lüders states; keeps the blocks of ρ inside each eigenspace and deletes those between them; Tr ρ' = 1POVM: positive operators summing to I
Eₖ ≥ 0, Σ Eₖ = I, pₖ = Tr(ρ Eₖ)
Buys what projectors cannot: error-free identification of either of two non-orthogonal states, at the Ivanovic–Dieks–Peres rate 1 − |⟨ψ₁|ψ₂⟩|, and honest models of lossy detectors.
Eₖ Hermitian with non-negative eigenvalues; need not be projectors, need not commute, may outnumber the dimensionKraus form and Naimark dilation
Eₖ = Mₖ† Mₖρ → Mₖ ρ Mₖ† / pₖEₖ = V†(I ⊗ Pₖ)V with V†V = I
Every POVM is a projective measurement on system plus ancilla, so no new postulate is needed, only more dimensions.
Mₖ = Uₖ √Eₖ for any unitary Uₖ, so Eₖ fixes statistics, not the update; V is an isometry into H ⊗ HancInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Start at α = 90°, the |0⟩ and |+⟩ of the Hard example, and raise a: the left bar passes the projective line at a = 0.50, and at a = 0.59 it pokes above the IDP line just as the right bar drops below zero, because that E_? would give some state a negative probability. Then close α: as the states merge, 1 − s falls towards 0 and the largest valid a = 1/(1 + s) towards ½.
Live results
- OVERLAP s = |⟨ψ₁|ψ₂⟩|
- 0.707
- LARGEST VALID a = 1/(1 + s)
- 0.586
- LOWEST EIGENVALUE OF E_?
- 0.146
- P(IDENTIFIED) = a(1 − s²)
- 0.250
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Bloch angle α between ψ₁ and ψ₂: 90 °; POVM scale a: 0.50
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The trap is to read pₖ = Tr(ρ Pₖ) as ignorance: the qubit 'really was' in |0⟩ or |1⟩ and the measurement looked. That reading fits a mixture diagonal in the measured basis and fails for a superposition. |+⟩ is an eigenstate of σₓ with the definite value +1, yet σz on it gives 0 or 1 with probability 1/2 each. If those σz values had been carried all along, a σₓ measurement afterwards would still find +1; it finds +1 only half the time, because the σz measurement replaced the state. The pure |+⟩ and the mixture I/2 give identical σz statistics and different σₓ statistics, so the probabilities are not a hidden label being revealed. The CHSH violation two topics on rules out the last refuge of that picture, local pre-existing values for every axis at once.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
Two qubits are prepared in |+⟩|+⟩ and the parity σz ⊗ σz is measured. Write the two projectors and their ranks, find the outcome probabilities and ⟨σz ⊗ σz⟩ = Tr[ρ (σz ⊗ σz)], give the state after the outcome +1, and compare it with what is left by reading σz on each qubit and keeping only the parity.
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.
- Operator and basis: σz ⊗ σz is diagonal in (|00⟩, |01⟩, |10⟩, |11⟩) with eigenvalues +1, −1, −1, +1. The measurement is P+ = |00⟩⟨00| + |11⟩⟨11| for +1 and P− = |01⟩⟨01| + |10⟩⟨10| for −1; each has rank 2, P+ P− = 0 and P+ + P− = I.
- State: |+⟩|+⟩ = (|00⟩ + |01⟩ + |10⟩ + |11⟩)/2. Born rule: p+ = ⟨ψ|P+|ψ⟩ = ¼ + ¼ = 0.500 and p− = 0.500; the sum is 1.
- Mean: ⟨σz ⊗ σz⟩ = (+1)(0.500) + (−1)(0.500) = 0. Cross-check on the product state: ⟨σz⟩ = 0 on each |+⟩, so the mean of the product is 0 × 0 = 0.
- Lüders update for +1: P+|ψ⟩ = (|00⟩ + |11⟩)/2 has squared norm ½ = p+, so the state becomes (|00⟩ + |11⟩)/√2. The relative amplitude of |00⟩ and |11⟩ inside the eigenspace survives, and a repeated parity measurement returns +1 with probability 1.
- Reading each qubit's σz instead applies the four rank-one projectors |00⟩⟨00|, |01⟩⟨01|, |10⟩⟨10| and |11⟩⟨11|; keeping only the parity leaves, for outcome +1, the mixture ½(|00⟩⟨00| + |11⟩⟨11|), purity ½. σₓ ⊗ σₓ tells the two apart: it swaps |00⟩ and |11⟩, so (|00⟩ + |11⟩)/√2 returns +1 with certainty, while the mixture gives ⟨σₓ ⊗ σₓ⟩ = 0, a 50/50 split.
Answer. P+ = |00⟩⟨00| + |11⟩⟨11| and P− = |01⟩⟨01| + |10⟩⟨10|, rank 2 each; p+ = p− = 0.500 and ⟨σz ⊗ σz⟩ = 0; outcome +1 leaves (|00⟩ + |11⟩)/√2, an eigenstate of σₓ ⊗ σₓ, whereas two single-qubit readings leave a mixture of purity 0.5 whose σₓ ⊗ σₓ result is 50/50.
MediumExample 2
A source emits |0⟩ with probability 0.7 and |+⟩ with probability 0.3. Write ρ as a matrix, find its Bloch vector and purity, then the probability that σₓ reads +1. If σz is measured and reads 1, what is the state? If σz is measured but the result is discarded, what purity is left?
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.
- ρ = 0.7|0⟩⟨0| + 0.3|+⟩⟨+| = 0.7[[1, 0], [0, 0]] + 0.15[[1, 1], [1, 1]] = [[0.85, 0.15], [0.15, 0.15]], with Tr ρ = 1.
- Bloch components rᵢ = Tr(ρ σᵢ): rₓ = 2 × 0.15 = 0.30, ry = 0, rz = 0.85 − 0.15 = 0.70, so |r|² = 0.58 and purity Tr ρ² = (1 + 0.58)/2 = 0.79. Direct check: 0.85² + 2(0.15²) + 0.15² = 0.7225 + 0.045 + 0.0225 = 0.79.
- Born rule for σₓ = +1: p+ = Tr(ρ P+) = (1 + rₓ)/2 = 0.65. As a trace, P+ = ½[[1, 1], [1, 1]] gives ½(0.85 + 0.15 + 0.15 + 0.15) = 0.65.
- σz reads 1 with p₁ = ρ₁₁ = 0.15, and P₁ ρ P₁ / p₁ = (0.15|1⟩⟨1|)/0.15 = |1⟩⟨1|: a rank-one projector leaves a pure state however mixed the input, and the 0.7/0.3 history is erased.
- Unread σz: ρ' = P₀ ρ P₀ + P₁ ρ P₁ = diag(0.85, 0.15). Purity 0.85² + 0.15² = 0.7225 + 0.0225 = 0.745, down from 0.79; the Bloch vector has shrunk from (0.30, 0, 0.70) to (0, 0, 0.70) because the off-diagonal 0.15 was deleted.
Answer. ρ = [[0.85, 0.15], [0.15, 0.15]], r = (0.30, 0, 0.70), purity 0.79; p(σₓ = +1) = 0.65; after σz = 1 the state is |1⟩; an unread σz leaves diag(0.85, 0.15) with purity 0.745.
HardExample 3
Alice sends Bob either |0⟩ or |+⟩, each with probability ½. Bob must never misidentify a state but may answer 'inconclusive'. Find the best a projective measurement on the qubit can do, construct the three-element POVM that does better, find the largest allowed scale factor, and give the success and inconclusive probabilities.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- ⟨0|+⟩ = 1/√2 ≠ 0. A projective measurement on C² is (|v⟩⟨v|, I − |v⟩⟨v|). An outcome read as '|+⟩' must annihilate |0⟩, forcing |v⟩ = |1⟩, and its partner I − |1⟩⟨1| = |0⟩⟨0| then fires on |+⟩ half the time, so it can only be read as 'inconclusive'. The best projective scheme identifies |+⟩ half the time and |0⟩ never, an average success of ½ × ½ = 0.250; naming either state needs three outcomes, and three outcomes in two dimensions require a POVM.
- Let E₁ = a|1⟩⟨1|, which gives zero on |0⟩ and so can only be triggered by |+⟩; E₂ = a|−⟩⟨−|, zero on |+⟩ and triggered only by |0⟩; and E_? = I − E₁ − E₂. Positivity of E₁ and E₂ needs a ≥ 0; the real constraint is E_? ≥ 0.
- |1⟩⟨1| + |−⟩⟨−| = [[0, 0], [0, 1]] + ½[[1, −1], [−1, 1]] = [[½, −½], [−½, 3/2]], with trace 2 and determinant ½, so eigenvalues 1 ± 1/√2. E_? ≥ 0 requires a(1 + 1/√2) ≤ 1, so aₘₐₓ = 1/(1 + 1/√2) = 2 − √2 = 0.5858.
- At a = 2 − √2, E_? = [[0.7071, 0.2929], [0.2929, 0.1213]]: its determinant is 0.7071 × 0.1213 − 0.2929² = 0 and its trace is 2√2 − 2 = 0.8284, so E_? is a rank-one positive operator, as tight as positivity allows.
- Success: given |0⟩, p(E₂) = a|⟨−|0⟩|² = a/2 = 0.2929 and p(E₁) = a|⟨1|0⟩|² = 0; given |+⟩, p(E₁) = a|⟨1|+⟩|² = a/2 = 0.2929 and p(E₂) = 0. The average success is (2 − √2)/2 = 1 − 1/√2 = 0.2929, inconclusive 1/√2 = 0.7071, error 0.
- This is the Ivanovic–Dieks–Peres bound 1 − |⟨0|+⟩|, above the 0.250 of the best projective scheme. Accepting errors instead, the Helstrom projective measurement identifies correctly with probability ½(1 + √(1 − |⟨0|+⟩|²)) = ½(1 + 1/√2) = 0.854 but is wrong 14.6 % of the time; which is better depends on what a wrong answer costs.
Answer. aₘₐₓ = 2 − √2 ≈ 0.586, with E₁ = a|1⟩⟨1|, E₂ = a|−⟩⟨−|, E_? = I − E₁ − E₂. Each state is identified with probability 0.293, against an average 0.250 for the best projective scheme; the answer is inconclusive with probability 0.707, and the error rate is exactly zero.
Quick check
Select an answer and explain it before checking.
Review & continue
A measurement is a set of positive operators Eₖ on the state space that sum to the identity; outcome k occurs with probability Tr(ρ Eₖ), and when the Eₖ are orthogonal projectors Pₖ the state left behind is Pₖ ρ Pₖ divided by that probability.
- 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.