University Physics V · Introduction to Quantum Information · 20.3
Measurement, the Born Rule & 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.
Build the model
Connect the measurement to the mechanism.
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 Hermitian 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.
- Simple definition
- 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
- For |ψ⟩ = cos 30°|0⟩ + sin 30°|1⟩ a σz measurement gives p₀ = ⟨ψ|0⟩⟨0|ψ⟩ = cos² 30° = 0.750 and p₁ = 0.250; the survivor is |0⟩ or |1⟩, so an immediate repeat returns the same outcome with probability 1.
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]
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 was
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 eigenspaces
One dot product replaces a matrix trace; at r = 0 every axis gives 1/2, the signature of the maximally mixed state.
r is the Bloch vector, |r| ≤ 1; n is the unit axis measured; both dimensionless
Buys what projectors cannot: error-free discrimination of non-orthogonal states, and honest models of lossy detectors.
Eₖ Hermitian with non-negative eigenvalues; need not be projectors, need not commute, may outnumber the dimension
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 ⊗ Hanc
A 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. 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.
Take 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. On one qubit the whole calculation is a dot product. With ρ = (I + r⋅σ)/2 and P_± = (I ± n⋅σ)/2, the identities Tr σᵢ = 0 and Tr(σᵢ σⱼ) = 2δᵢⱼ give p_± = (1 ± r⋅n)/2 and ⟨σₙ⟩ = p+ − p_− = r⋅n. Take the pure state at polar angle 60° on the Bloch sphere, r = (sin 60°, 0, cos 60°), and measure σz, n = (0, 0, 1): p+ = (1 + 0.5)/2 = 0.750. Measure σₓ instead: p+ = (1 + 0.866)/2 = 0.933. Shrink r to zero and every axis returns 0.500, which is why the maximally mixed state carries no information about any preparation. The projection of r onto n is the probability, and the figure below draws exactly that.
What 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.
When a projector is the wrong tool: a POVM in three lines
Suppose a source sends |0⟩ or |+⟩ with equal probability and you must never guess wrong, though you may answer 'don't know'. Two-outcome projective measurements on C² have the form (|v⟩⟨v|, I − |v⟩⟨v|), and no choice of v is orthogonal to |0⟩ while its complement is orthogonal to |+⟩, so errors are forced. A third outcome needs three positive operators summing to I in a two-dimensional space, which no set of orthogonal projectors can provide. Take E₁ = a|1⟩⟨1|, which annihilates |0⟩ and so fires only for |+⟩; E₂ = a|−⟩⟨−|, which fires only for |0⟩; and E_? = I − E₁ − E₂. The operator |1⟩⟨1| + |−⟩⟨−| has matrix [[½, −½], [−½, 3/2]] with eigenvalues 1 ± 1/√2, so E_? ≥ 0 holds up to a = 1/(1 + 1/√2) = 2 − √2 = 0.586. At that value each state is identified with probability a/2 = 0.293 and the rest is 'don't know', with zero errors. That 1 − |⟨0|+⟩| is the Ivanovic–Dieks–Peres bound, and no projective scheme reaches it.
Every 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.
What 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.
Change one variable at a time
Make the relationship visible.
Set n to 90° with r at 60° and length 1: p(+n) rises from the 0.75 of σz to 0.93 for σₓ. Then align n with r: the perpendicular vanishes, p(+n) reaches 1, and the unread-measurement purity climbs back to (1 + r²)/2, because a state already diagonal in the measured basis has no coherence to lose.
p(+n) = Tr(ρP₊)0.750
p(−n) = Tr(ρP₋)0.250
⟨σₙ⟩ = r⋅n0.500
PURITY AFTER UNREAD0.625
Live interpretationp(+n) = Tr(ρP₊): 0.750. p(−n) = Tr(ρP₋): 0.250. ⟨σₙ⟩ = r⋅n: 0.500. PURITY AFTER UNREAD: 0.625
Catch the common trap
Explain before calculating.
A qubit in |ψ⟩ = (|0⟩ + |1⟩)/√2 is measured in the σz basis and outcome 0 is obtained. What state does it hold afterwards, and with what probability does an immediate σₓ measurement then return +1?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA qubit is prepared in |ψ⟩ = (√3|0⟩ + |1⟩)/2. Write the projectors for a σₓ measurement, find both outcome probabilities, check them against ⟨σₓ⟩ = Tr(ρ σₓ), and state the post-measurement state for each outcome.
- Operator and basis: σₓ has eigenvectors |±⟩ = (|0⟩ ± |1⟩)/√2, so P_± = |±⟩⟨±| = (I ± σₓ)/2 = ½[[1, ±1], [±1, 1]], and P+ + P_− = I.
- Amplitudes: ⟨+|ψ⟩ = (√3 + 1)/(2√2) and ⟨−|ψ⟩ = (√3 − 1)/(2√2). Born rule: p+ = |⟨+|ψ⟩|² = (4 + 2√3)/8 = (2 + √3)/4 = 0.933 and p_− = (2 − √3)/4 = 0.067; the sum is 1.
- Cross-check: ⟨σₓ⟩ = p+ − p_− = √3/2 = 0.866. Directly, ρ = |ψ⟩⟨ψ| has off-diagonal element √3/4, so Tr(ρ σₓ) = 2 × √3/4 = 0.866; the state sits at polar angle 60° on the Bloch sphere, and rₓ = sin 60° agrees.
- Update: outcome + leaves P+|ψ⟩/√p+ = |+⟩ and outcome − leaves |−⟩. Either is a σₓ eigenstate, so repeating the measurement returns the same sign with probability 1.
Answerp+ = (2 + √3)/4 ≈ 0.933, p_− = (2 − √3)/4 ≈ 0.067, ⟨σₓ⟩ = √3/2 ≈ 0.866; the qubit is left in |+⟩ or |−⟩.
MediumA 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?
- 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.
Answerρ = [[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.
HardAlice sends Bob either |0⟩ or |+⟩, each with probability ½. Bob must never misidentify a state but may answer 'inconclusive'. Show that no projective measurement achieves this, construct the three-element POVM that does, find the largest allowed scale factor, and give the success and inconclusive probabilities.
- ⟨0|+⟩ = 1/√2 ≠ 0. A projective measurement on C² is (|v⟩⟨v|, I − |v⟩⟨v|). Error-free identification needs the '|+⟩' outcome to annihilate |0⟩, forcing |v⟩ = |1⟩, and the '|0⟩' outcome I − |1⟩⟨1| = |0⟩⟨0| then fires on |+⟩ half the time. 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|+⟩|. 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.
Answeraₘₐₓ = 2 − √2 ≈ 0.586, with E₁ = a|1⟩⟨1|, E₂ = a|−⟩⟨−|, E_? = I − E₁ − E₂. Each state is identified with probability 0.293, the answer is inconclusive with probability 0.707, and the error rate is exactly zero.