University Quantum Mechanics II · Composite Systems, Entanglement and Density Operators · 4.3

The density operator and ⟨A⟩ = Tr(ρA)

Quantum Mechanics I only had to describe beams in which every member was prepared alike. Ovens, photocathodes and half of an entangled pair are not like that. Here you build the one operator that carries every prediction such a source makes, and learn to calculate with it before interpreting it.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Quantum Mechanics I gave a whole beam one ket, which is right only when every member was prepared alike. A real source hands out |ψₖ⟩ with classical probability pₖ, and nothing requires the |ψₖ⟩ to be orthogonal. Every measured mean is then an average over members, Σₖ pₖ⟨ψₖ|A|ψₖ⟩, and because ⟨ψ|A|ψ⟩ = Tr(|ψ⟩⟨ψ|A), linearity gathers the whole recipe into one operator, ρ = Σₖ pₖ|ψₖ⟩⟨ψₖ|, with ⟨A⟩ = Tr(ρA). The same move turns each outcome probability into Tr(ρPₙ) and, with Bayes' rule, the systems that gave aₙ into PₙρPₙ/Tr(ρPₙ).

So ρ, not the recipe, is what an experiment sees. The construction forces three properties on it: Hermitian, positive, unit trace. Conversely, by the spectral theorem any operator with those properties is an ensemble of its own orthonormal eigenvectors, weighted by eigenvalues that are generally not the pₖ you started from. Its purity Tr(ρ²) is 1 exactly for a single ket and falls towards 1/d as the ensemble spreads, and its off-diagonal elements in a basis, the coherences, are what separate a superposition from a mixture with the same populations.

The picture has two limits. One ρ can be written as many different ensembles, so it cannot say which member a given system was. And the ρ of one half of an entangled pair is not a list of kets anyone prepared; it comes from the partial trace, later in this unit.

Simple definition: The density operator ρ = Σₖ pₖ|ψₖ⟩⟨ψₖ| of an ensemble is the probability-weighted sum of the projectors onto its members' states, the one object from which every mean, Tr(ρA), and every outcome probability, Tr(ρPₙ), follows.
Example: Photons polarised at 0° with probability 0.70 and at 60° with 0.30 have ρ_HH = 0.775, ρ_VV = 0.225 and coherence ρ_HV = 0.130; a polariser at 30° passes Tr(ρP) = 0.750 of them, the same as 0.70 cos²30° + 0.30 cos²30°.

The trace rule for an ensemble⟨A⟩ = Σₖ pₖ⟨ψₖ|A|ψₖ⟩ = Tr(ρA), with ρ = Σₖ pₖ|ψₖ⟩⟨ψₖ|

Derived from ⟨ψ|A|ψ⟩ = Tr(|ψ⟩⟨ψ|A) and linearity. Every mean now needs only ρ, not the recipe; with one member it returns Quantum Mechanics I's ⟨ψ|A|ψ⟩.

pₖ ≥ 0 sum to 1; |ψₖ⟩ normalised, not necessarily orthogonal; ρ dimensionless, ⟨A⟩ in A's units (J s for a spin)

Probability of an outcomeP(aₙ) = Tr(ρPₙ)

Derived as the mean of the projector Pₙ. Since Σₙ Pₙ = I and Tr ρ = 1, the probabilities of a complete measurement add to 1.

Pₙ projects onto the whole eigenspace of the eigenvalue aₙ, degenerate or not; P(aₙ) is dimensionless

After a selective measurementρ′ = Pₙ ρ Pₙ / Tr(ρPₙ)

Derived from the projection postulate applied to each member, with Bayes' rule for the weights. Coherences inside a degenerate eigenspace survive; those between eigenspaces go.

selective measurement: only systems with outcome aₙ are kept; the denominator is P(aₙ) and restores unit trace

Every ρ is an ensemble of its own eigenvectorsρ = Σⱼ λⱼ|j⟩⟨j|, with λⱼ ≥ 0 and Σⱼ λⱼ = 1

Hermitian, positive and unit trace is all it takes. The λⱼ equal the recipe's pₖ only for orthogonal members, and the largest λⱼ is the most a one-state test can pass.

|j⟩ orthonormal eigenvectors, λⱼ dimensionless eigenvalues; the spectral theorem is quoted from linear algebra

PurityTr(ρ²) = Σⱼ λⱼ² = Σₖₗ pₖpₗ|⟨ψₖ|ψₗ⟩|²

Equal to 1 exactly when ρ = |ψ⟩⟨ψ|. Overlapping members push it above Σpₖ²: 0.685 against 0.580 for photons at 0° and 60° weighted 0.70 and 0.30.

dimensionless and the same in every basis; 1/d ≤ Tr(ρ²) ≤ 1 in a d-dimensional state space

Coherences are capped by populations|ρₘₙ|² ≤ ρₘₘ ρₙₙ

From positivity of every 2 × 2 principal block of ρ. a|1⟩ + b|2⟩ reaches the cap with ρ₁₂ = ab*; the mixture with populations |a|² and |b|² has ρ₁₂ = 0.

ρₙₙ = ⟨n|ρ|n⟩ populations and ρₘₙ = ⟨m|ρ|n⟩ = ρₙₘ* coherences in a chosen basis; all dimensionless

Average the members, then gather the average into ρ

A source prepares |ψₖ⟩ with probability pₖ: photons polarised at 0° with p₁ = 0.70 and at 60° with p₂ = 0.30, say. The pₖ are ordinary probabilities about the preparation, and the states overlap, ⟨0°|60°⟩ = cos 60° = 0.50. Quantum Mechanics I supplies each member's mean, so the ensemble's is ⟨A⟩ = Σₖ pₖ⟨ψₖ|A|ψₖ⟩. Insert the completeness relation: Tr(|ψ⟩⟨ψ|A) = Σₙ⟨n|ψ⟩⟨ψ|A|n⟩ = ⟨ψ|A|ψ⟩. The trace is linear, so the sum becomes ⟨A⟩ = Tr(ρA) with ρ = Σₖ pₖ|ψₖ⟩⟨ψₖ|: derived, not postulated. In the (H, V) basis |θ⟩⟨θ| has entries cos²θ, sin²θ and sin θ cos θ, giving ρ_HH = 0.775, ρ_VV = 0.225 and ρ_HV = 0.130. Checks: ρ is dimensionless, so Tr(ρA) carries the units of A, and one member with p₁ = 1 returns ⟨ψ|A|ψ⟩. What you must not do is add the kets. Any combination of |0°⟩ and |60°⟩ is one pure polarisation, which some polariser passes every time; no polariser passes more than 80.4 per cent of this ensemble.

Probabilities, and what a measurement leaves, are traces too

An outcome probability is the mean of a projector. With Pₙ projecting onto the eigenspace of aₙ, P(aₙ) = Σₖ pₖ⟨ψₖ|Pₙ|ψₖ⟩ = Tr(ρPₙ). A polariser at angle χ is |χ⟩⟨χ|, so the example passes ρ_HH cos²χ + ρ_VV sin²χ + 2ρ_HV sin χ cos χ, which is 0.750 at χ = 30°. Now keep only the systems that gave aₙ. Quantum Mechanics I's projection postulate sends member k to Pₙ|ψₖ⟩/√⟨ψₖ|Pₙ|ψₖ⟩, and Bayes' rule says a survivor was member k with probability pₖ⟨ψₖ|Pₙ|ψₖ⟩/P(aₙ). Multiply, and the norms cancel: ρ′ = PₙρPₙ/Tr(ρPₙ), derived rather than postulated anew. If every system is kept and the results merely recorded, weighting each branch by P(aₙ) gives Σₙ PₙρPₙ: the populations in the measured basis are unchanged and the coherences between different eigenvalues are gone. Measuring H or V on the example this way leaves diag(0.775, 0.225), and the 30° polariser then passes 0.638 instead of 0.750.

What every density operator must be, and why that is enough

The construction forces three properties. ρ is Hermitian, a real-weighted sum of projectors. It is positive: for any |φ⟩, ⟨φ|ρ|φ⟩ = Σₖ pₖ|⟨φ|ψₖ⟩|² ≥ 0, so no outcome gets a negative probability. And Tr ρ = Σₖ pₖ = 1. The converse uses the spectral theorem, quoted from linear algebra: a Hermitian ρ equals Σⱼ λⱼ|j⟩⟨j| with orthonormal |j⟩, positivity makes each λⱼ ≥ 0 and unit trace makes them sum to 1. So every such operator is an ensemble of its own eigenvectors, and its largest eigenvalue is the most any test projecting on one state can pass. The photon example has λ = 0.804 and 0.196, with eigenvectors at 12.6° and 102.6°: a rotating polariser peaks at 0.804 near 12.6°, not at 0°, the likelier member's direction. The weights 0.70 and 0.30 are not the eigenvalues; they would be only for orthogonal members, at 90°. That one ρ has many recipes is the subject of the next topic.

Purity tells a single ket from a mixture

Tr(ρ²) = Σⱼ λⱼ², and because the λⱼ are non-negative and sum to 1, Σⱼ λⱼ² ≤ (Σⱼ λⱼ)² = 1, with equality only when one eigenvalue is 1: ρ = |ψ⟩⟨ψ| and ρ² = ρ. The floor is 1/d, reached by ρ = I/d: 0.500 for a spin-½, 0.333 for a spin-1. Through the recipe, Tr(ρ²) = Σₖₗ pₖpₗ|⟨ψₖ|ψₗ⟩|², which exceeds Σₖ pₖ² whenever members overlap: 0.685 for the photons against 0.49 + 0.09 = 0.580. Both results are derived; purity is basis-independent, and the unitary evolution of the von Neumann equation, later in this unit, leaves it fixed. A measured check: an electron beam polarised along its axis, with P = Tr(ρσ_z), has ρ = diag((1 + P)/2, (1 − P)/2), so Tr(ρ²) = (1 + P²)/2. Bulk GaAs photocathodes, whose 3:1 excitation ratio is quoted from the optical selection rules, cap P at 0.50 and purity at 0.625; strained-layer cathodes at accelerator sources are measured near P = 0.85, purity 0.86. Still a mixture.

Coherences: what the off-diagonal elements record

In an orthonormal basis |n⟩ the diagonal elements ρₙₙ = P(n) are populations and the off-diagonal ρₘₙ are coherences. The ket a|1⟩ + b|2⟩ has ρ₁₂ = ab*; the mixture of |1⟩ and |2⟩ with the same populations has ρ₁₂ = 0. The trace rule shows when that matters: Tr(ρA) contains 2 Re(ρ₁₂A₂₁), Quantum Mechanics I's cross term, so an observable diagonal in this basis cannot tell the two apart and one with A₂₁ ≠ 0 can. Positivity caps the size. Every 2 × 2 principal block of ρ has a non-negative determinant, so |ρₘₙ|² ≤ ρₘₘρₙₙ, and for a two-level system equality means pure. The photon example has ρ_HV = 0.130 against a cap of √(0.775 × 0.225) = 0.418. Coherence belongs to a state and a basis together: diag(0.8, 0.2) has none in (H, V), yet in the ±45° basis its elements are ρ₊₊ = ρ₋₋ = 0.5 and ρ₊₋ = 0.3. Only ρ = I/d lacks coherences in every basis.

Change one variable at a time

Make the relationship visible.

Interactive model

Weight p of the 0° member0.70

Polarisation α of the other member60 °

Polariser angle χ30 °

Open α to 90°: the members become orthogonal, ρ_HV vanishes, the curves merge and λ₊, λ₋ become the larger and the smaller of p and 1 − p. Then close α towards 0° at fixed p: the weights never change, yet λ₋ falls to 0 and Tr(ρ²) climbs to 1. The peak always sits between the two members.

PASSED AT χ, Tr(ρP_χ)0.750

LARGER EIGENVALUE λ₊0.804

SMALLER EIGENVALUE λ₋0.196

PURITY Tr(ρ²)0.685

Live interpretationPASSED AT χ, Tr(ρP_χ): 0.750. LARGER EIGENVALUE λ₊: 0.804. SMALLER EIGENVALUE λ₋: 0.196. PURITY Tr(ρ²): 0.685

Catch the common trap

Explain before calculating.

Misconception check

The tempting shortcut reads the recipe's weights as the eigenvalues of ρ, so that Tr(ρ²) = Σₖ pₖ² and an equal-weight source is as mixed as a two-level system can be. That holds only for orthogonal members. Prepare spin-½ particles in |+z⟩ or |+x⟩ with probability 0.50 each. The shortcut says ρ = I/2, purity 0.50, and every Stern-Gerlach axis splits the beam evenly. In fact ρ = [[0.75, 0.25], [0.25, 0.25]] in the S_z basis, its eigenvalues are (1 ± 1/√2)/2 = 0.854 and 0.146, Tr(ρ²) = 0.75, and an analyser along the bisector of z and x finds +ħ/2 with probability 0.854. The weights describe how the source was run; the eigenvalues describe the state, and only they bound what a measurement can find. The same slip in the other direction, adding √pₖ|ψₖ⟩ into one ket, overstates the coherences: it gives ρ₁₂ = 0.354 in place of 0.25 and a pure state in place of a mixture.

For each photon a source picks, with equal probability, a polarisation of 0° or 60°. A polariser is turned until it passes as many photons as possible. What is the largest fraction it can pass, and at what angle?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA source emits spin-½ particles in |+z⟩ with probability 0.60 and in |+x⟩ with probability 0.40. Write ρ in the S_z basis, use the trace rule for ⟨S_z⟩ and ⟨Sₓ⟩ in J s, check both member by member, and find the probability of Sₓ = −ħ/2.

Projectors in the (|+z⟩, |−z⟩) basis: |+z⟩⟨+z| = [[1, 0], [0, 0]] and |+x⟩⟨+x| = ½[[1, 1], [1, 1]]. So ρ = 0.60[[1, 0], [0, 0]] + 0.20[[1, 1], [1, 1]] = [[0.80, 0.20], [0.20, 0.20]], Hermitian with trace 1.00.
⟨S_z⟩ = (ħ/2)Tr(ρσ_z) = (ħ/2)(0.80 − 0.20) = 0.30ħ = 0.30 × 1.0546 × 10⁻³⁴ J s = 3.16 × 10⁻³⁵ J s. ⟨Sₓ⟩ = (ħ/2)Tr(ρσₓ) = (ħ/2)(ρ₁₂ + ρ₂₁) = 0.20ħ = 2.11 × 10⁻³⁵ J s.
Member by member: |+z⟩ gives ⟨S_z⟩ = ħ/2 and ⟨Sₓ⟩ = 0, |+x⟩ gives 0 and ħ/2, so the averages are 0.60 × ħ/2 = 0.30ħ and 0.40 × ħ/2 = 0.20ħ. Both agree, and both carry the units of ħ, as a spin component must.
With P₋ₓ = |−x⟩⟨−x| = ½[[1, −1], [−1, 1]], P(Sₓ = −ħ/2) = Tr(ρP₋ₓ) = ½(0.80 − 0.20 − 0.20 + 0.20) = 0.30. Check: 0.60 × ½ + 0.40 × 0 = 0.30.

Answerρ = [[0.80, 0.20], [0.20, 0.20]]; ⟨S_z⟩ = 0.30ħ = 3.16 × 10⁻³⁵ J s, ⟨Sₓ⟩ = 0.20ħ = 2.11 × 10⁻³⁵ J s, and P(Sₓ = −ħ/2) = 0.30.

MediumAn electron in an infinite well of width 0.60 nm is prepared in ψ₁ with probability 0.30 and in (ψ₁ + ψ₂)/√2 with probability 0.70. Find ρ in the energy basis, ⟨E⟩, ⟨x⟩, the probability of the left half and Tr(ρ²), and compare each with the mixture of ψ₁ and ψ₂ that has the same populations.

Quote from Quantum Mechanics I: Eₙ = n²h²/(8mₑL²), x₁₁ = x₂₂ = L/2, x₁₂ = −16L/(9π²) = −0.1081 nm, and the left-half integral of ψ₁ψ₂ is 4/(3π) = 0.4244.
ρ = 0.30[[1, 0], [0, 0]] + 0.35[[1, 1], [1, 1]] = [[0.65, 0.35], [0.35, 0.35]], trace 1. The same-population mixture is diag(0.65, 0.35), with no coherence.
E₁ = (6.626 × 10⁻³⁴ J s)² / (8 × 9.109 × 10⁻³¹ kg × (0.60 × 10⁻⁹ m)²) = 1.674 × 10⁻¹⁹ J = 1.045 eV. H is diagonal, so ⟨E⟩ = 0.65E₁ + 0.35 × 4E₁ = 2.05E₁ = 2.14 eV for both states.
⟨x⟩ = Tr(ρx) = (ρ₁₁ + ρ₂₂)L/2 + 2ρ₁₂x₁₂ = 0.300 nm − 0.70 × 0.1081 nm = 0.224 nm, against 0.300 nm for the mixture; every term is a length.
P(left) = Tr(ρP_L) = ½ + 2ρ₁₂ × 0.4244 = 0.797, against 0.500 for the mixture. Member check: 0.30 × 0.500 + 0.70 × 0.924 = 0.797.
Tr(ρ²) = 0.65² + 0.35² + 2 × 0.35² = 0.790, the mixture's is 0.545, and the weights' Σpₖ² is only 0.580. Dimension check: ⟨x|ρ|x⟩ = Σₖ pₖ|ψₖ(x)|² is a density per length, 2 × 0.65/(0.60 nm) = 2.17 nm⁻¹ at x = L/2.

Answer⟨E⟩ = 2.14 eV for both; ⟨x⟩ = 0.224 nm and P(left) = 0.797 for the ensemble against 0.300 nm and 0.500 for the mixture; Tr(ρ²) = 0.790 against 0.545.

HardA spin-1 source emits (|+1⟩ + |0⟩ + |−1⟩)/√3 with probability 0.60 and |0⟩ with probability 0.40. An ideal S_z² measurement, which tells m = 0 from m = ±1 but not +1 from −1, keeps only systems giving ħ². Find the fraction kept, their ρ and purity, and P(m_y = 0) in a later S_y measurement, set against a rule deleting all coherences. What if all systems are kept, results unread?

In (|+1⟩, |0⟩, |−1⟩), ρ = 0.60 × ⅓(all-ones matrix) + 0.40|0⟩⟨0| = [[0.20, 0.20, 0.20], [0.20, 0.60, 0.20], [0.20, 0.20, 0.20]]. Tr(ρ²), the sum of squared elements, is 8 × 0.04 + 0.36 = 0.68, matching Σₖₗ pₖpₗ|⟨ψₖ|ψₗ⟩|² = 0.36 + 0.16 + 2 × 0.24 × ⅓ = 0.68.
S_z² = ħ² diag(1, 0, 1), so ħ² is doubly degenerate, with P = |+1⟩⟨+1| + |−1⟩⟨−1|. Kept fraction: Tr(ρP) = 0.20 + 0.20 = 0.40; member check 0.60 × ⅔ + 0.40 × 0 = 0.40.
PρP keeps rows and columns +1 and −1: [[0.20, 0, 0.20], [0, 0, 0], [0.20, 0, 0.20]]. Divided by 0.40 it is |ξ⟩⟨ξ| with |ξ⟩ = (|+1⟩ + |−1⟩)/√2, so Tr(ρ′²) = 1.00. The coherence inside the degenerate eigenspace survived, and selection raised the purity from 0.68 to 1. Limits: a pure-|0⟩ source passes nothing, a pure-superposition source passes ⅔, into the same |ξ⟩.
With S_y = (S₊ − S₋)/(2i), S₊|−1⟩ = √2ħ|0⟩ and S₋|+1⟩ = √2ħ|0⟩ cancel, so S_y|ξ⟩ = 0: |ξ⟩ is the m_y = 0 state and P(m_y = 0) = ⟨ξ|ρ′|ξ⟩ = 1.00. The delete-every-coherence rule gives diag(0.5, 0, 0.5) and ⟨ξ|ρ|ξ⟩ = ½(0.5 + 0.5) = 0.50.
Keeping everything unread gives PρP + QρQ with Q = |0⟩⟨0|: [[0.20, 0, 0.20], [0, 0.60, 0], [0.20, 0, 0.20]]. Tr(ρ″²) = 4 × 0.04 + 0.36 = 0.52, below 0.68: coherences between m = 0 and m = ±1 are gone, the one between +1 and −1 remains.

Answer0.40 is kept, in the pure state (|+1⟩ + |−1⟩)/√2 with Tr(ρ²) = 1.00, so S_y then gives m_y = 0 with probability 1.00, not the 0.50 of the delete-all rule; kept unread, the ensemble has Tr(ρ²) = 0.52.