University Quantum Mechanics II · Composite Systems, Entanglement and Density Operators · 4.7

The partial trace and reduced density operators

You rarely get to hold both halves of a system that has interacted. This lesson builds the one operator that answers every question you can put to your half, and faces its strangest consequence: a pair known exactly can have halves as uncertain as a spin-half can be.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Someone who holds only system A can measure only operators of the form A ⊗ I_B, so everything they will ever record is fixed by the numbers Tr[ρ (A ⊗ I_B)]. Asking for one operator on A's space that reproduces all of them derives it: ρ_A = Tr_B ρ = Σⱼ ⟨j_B|ρ|j_B⟩, unique, the same whichever basis of B is summed over, and itself a density operator. Applied to a pure entangled state it returns a mixed one.

Each spin of the singlet has ρ_A = I/2 exactly, unpolarised along every axis, although the pair is in a state known completely, and that mixture is not ignorance of a pure state A secretly occupies, because a pair whose half A had a definite state would be a product. In Schmidt form, |ψ⟩ = Σₖ √pₖ |uₖ⟩|vₖ⟩, both halves come out diagonal with the same eigenvalues pₖ, so −Tr(ρ_A ln ρ_A) = −Σₖ pₖ ln pₖ is one number for the pair: 0 for a product, ln 2 for the singlet. That this number measures entanglement is quoted, not derived.

The construction is pure kinematics, with no Hamiltonian, no ħ and no α in it, and its entropy stops measuring entanglement once the whole is mixed: a random mixture of |↑↓⟩ and |↓↑⟩ has the same halves, the same ln 2, and no entanglement at all.

Simple definition: The reduced density operator ρ_A = Tr_B ρ is the operator on system A's space that gives every measurement made on A alone the same statistics as the joint state ρ; it is found by summing ⟨j_B|ρ|j_B⟩ over any orthonormal basis of B.
Example: For |ψ⟩ = (√3/2)|↑↓⟩ − ½|↓↑⟩, tracing out B leaves ρ_A = diag(0.75, 0.25): spin A alone has ⟨σ_z⟩ = 0.50, purity Tr ρ_A² = 0.625 and entropy 0.562 nats, although the pair is pure.

The defining propertyTr(ρ_A A) = Tr[ρ (A ⊗ I_B)] for every observable A of system A

A ⊗ I_B is all that someone holding only A can measure, so this is the whole content of “the state of A”, and it fixes ρ_A uniquely.

ρ and ρ_A are dimensionless; both sides carry the unit of A, J for an energy or J s for a spin component. I_B is B's identity.

The partial trace, in componentsρ_A = Tr_B ρ = Σⱼ ⟨j_B|ρ|j_B⟩, so ⟨m|ρ_A|n⟩ = Σⱼ ⟨m j|ρ|n j⟩

Derived from the defining property by writing the full trace in the basis |m j⟩. A change of B's basis brings in U and U†, which cancel, so every basis gives one ρ_A.

|j_B⟩ runs over any orthonormal basis of B, |m⟩ and |n⟩ over one of A; the result is a d_A × d_A matrix of pure numbers.

From the coefficient matrixρ_A = C C†, where C has elements Cₘⱼ = ⟨m j|ψ⟩

The working formula for an explicit pure state. For two spin-halves det ρ_A = |det C|², so the eigenvalues are (1 ± √(1 − 4|det C|²))/2.

C is the d_A × d_B matrix of amplitudes, dimensionless, and C† its conjugate transpose. The other half is ρ_B = (C†C)ᵀ.

Squared Schmidt coefficients are the spectrum of each half|ψ⟩ = Σₖ √pₖ |uₖ⟩|vₖ⟩ gives ρ_A = Σₖ pₖ |uₖ⟩⟨uₖ|

Both halves share their nonzero eigenvalues even when d_A ≠ d_B; the larger space only adds zeros. So any function of the spectrum takes one value for A and B.

The Schmidt weights pₖ, squares of the Schmidt coefficients √pₖ, are dimensionless, pₖ ≥ 0 and Σₖ pₖ = 1; |uₖ⟩ orthonormal in A, |vₖ⟩ in B. Likewise ρ_B = Σₖ pₖ |vₖ⟩⟨vₖ|.

Entanglement entropy of a pure joint stateS = −Tr(ρ_A ln ρ_A) = −Σₖ pₖ ln pₖ

Equal for the two halves: 0 for a product, ln 2 = 0.693 nats for the singlet. That it measures entanglement is quoted, and only for a pure whole; a mixed whole breaks it.

Dimensionless, in nats: divide by ln 2 for bits, multiply by k_B for J K⁻¹. It runs from 0 to ln min(d_A, d_B).

Purity of a halfTr ρ_A² = Σₖ pₖ²

A test with no logarithms: a pure pair is a product exactly when Tr ρ_A² = 1. For a spin-half it equals (1 + |r|²)/2, so the Bloch length of a half grades the entanglement.

Dimensionless: 1 for a product, ½ for the singlet, 1/min(d_A, d_B) for a maximally entangled pure state.

Derive ρ_A from what one half can measure

Someone holding only system A can switch on only operators of the form A ⊗ I_B, so all they can ever record is fixed by the numbers Tr[ρ (A ⊗ I_B)], one for each observable A. Look for one operator ρ_A on A's space whose Tr(ρ_A A) matches every one of them. Write the full trace in the product basis |n j⟩ = |n⟩|j⟩ and use (A ⊗ I_B)|n j⟩ = Σₘ ⟨m|A|n⟩ |m j⟩: Tr[ρ (A ⊗ I_B)] = Σ over m and n of ⟨m|A|n⟩ Σⱼ ⟨n j|ρ|m j⟩. That equals Σ over m and n of ⟨m|A|n⟩⟨n|ρ_A|m⟩ = Tr(ρ_A A) exactly when ⟨n|ρ_A|m⟩ = Σⱼ ⟨n j|ρ|m j⟩, the component form of ρ_A = Σⱼ ⟨j_B|ρ|j_B⟩. The answer is unique: two operators with equal traces against every Hermitian A agree element by element, as the choices A = |m⟩⟨n| + |n⟩⟨m| and A = i(|m⟩⟨n| − |n⟩⟨m|) show. It is independent of B's basis: a new basis |j′⟩ = Σⱼ Uⱼⱼ′ |j⟩ brings a U and a U† into the sum, and they combine to UU† = I.

Check that the partial trace hands back a state

ρ_A must itself be a density operator, and three lines show it is. It is Hermitian, because each ⟨j_B|ρ|j_B⟩ inherits ρ† = ρ. It has unit trace, because summing ⟨m j|ρ|m j⟩ over m and j is Tr ρ = 1. It is positive, because for any |φ⟩ of A, ⟨φ|ρ_A|φ⟩ = Σⱼ ⟨φ j|ρ|φ j⟩ is a sum of non-negative numbers. Two limiting cases fix its meaning. For a product ρ = ρ₁ ⊗ ρ₂ the sum gives ρ₁ Tr ρ₂ = ρ₁: an uncorrelated partner leaves A exactly as it was. And any unitary acting on B alone, ρ → (I_A ⊗ U)ρ(I_A ⊗ U†), leaves ρ_A untouched, because U and U† meet inside the trace over B and cancel by its cyclic property; the next lesson turns this into the statement that no signal crosses between the two sides. Dimensions agree: ρ and ρ_A are pure numbers, and Tr(ρ_A A) carries the unit of A. No Hamiltonian, no ħ and no α appear: the partial trace is kinematics, true of any composite at any instant.

The singlet: a known pair with maximally mixed halves

Take |ψ⁻⟩ = (|↑↓⟩ − |↓↑⟩)/√2. Its density operator has four terms: ρ = ½|↑↓⟩⟨↑↓| + ½|↓↑⟩⟨↓↑| − ½|↑↓⟩⟨↓↑| − ½|↓↑⟩⟨↑↓|. The sandwich ⟨↑_B| … |↑_B⟩ keeps only the second term and gives ½|↓⟩⟨↓|; the sandwich ⟨↓_B| … |↓_B⟩ keeps only the first and gives ½|↑⟩⟨↑|. The coherence terms carry |↓⟩⟨↑| or |↑⟩⟨↓| on B, whose diagonal elements vanish, so the trace removes them. The sum is ρ_A = I/2 exactly: Bloch vector zero, ⟨S_A⟩ = 0, and ±ħ/2 with probability ½ along every axis. The pair is pure, Tr ρ² = 1, while its half has Tr ρ_A² = ½, the least a spin-half allows; the same steps give ρ_B = I/2. Experiment agrees. The two photons from electron–positron annihilation at rest share a maximally entangled polarisation state, so each alone is unpolarised and the polarisation shows up only in coincidences: Wu and Shaknov (1950) measured a perpendicular-to-parallel coincidence ratio of 2.04 ± 0.08, against 2.00 predicted for their geometry by quantum electrodynamics.

The halves do not determine the pair

Compare the singlet with ρ_cl = ½|↑↓⟩⟨↑↓| + ½|↓↑⟩⟨↓↑|, the state of a source that emits |↑↓⟩ or |↓↑⟩ at random. Its partial traces are also I/2 on each side, and its spins are also perfectly anticorrelated along z: ⟨σ_z ⊗ σ_z⟩ = −1 for both. The difference lies entirely in the two coherences −½|↑↓⟩⟨↓↑| and −½|↓↑⟩⟨↑↓| that the partial trace throws away. σₓ ⊗ σₓ exchanges |↑↓⟩ and |↓↑⟩, so it sends |ψ⁻⟩ to −|ψ⁻⟩ and gives ⟨σₓ ⊗ σₓ⟩ = −1, while on ρ_cl it gives 0. A joint measurement separates the two pairs; no measurement on either half alone can, because every such measurement is fixed by ρ_A or ρ_B. So the map from ρ to the pair of halves loses information, and the entropies show where: S(ρ) = 0 for the singlet and ln 2 for ρ_cl, while every half of either has ln 2. The pure pair is more certain than its halves, which no classical joint distribution can be, because its Shannon entropy equals a marginal's plus a conditional entropy that is never negative.

The squared Schmidt coefficients are the spectrum of both halves

Put the pure state in Schmidt form, |ψ⟩ = Σₖ √pₖ |uₖ⟩|vₖ⟩, which the singular-value decomposition quoted earlier in this unit guarantees. Then ρ = Σ over k and k′ of √(pₖ pₖ′) |uₖ⟩⟨uₖ′| ⊗ |vₖ⟩⟨vₖ′|, and tracing B in a basis containing the |vₖ⟩ keeps only k = k′: ρ_A = Σₖ pₖ |uₖ⟩⟨uₖ|, and the same step gives ρ_B = Σₖ pₖ |vₖ⟩⟨vₖ|. The Schmidt vectors are the eigenvectors, and the squared Schmidt coefficients the nonzero eigenvalues, of both halves. If the spaces differ in size the larger half only gains zero eigenvalues, so a spin-half entangled with a spin-1 leaves the spin-1 with rank at most 2. For an explicit state there is no need to find the decomposition first: with Cₘⱼ = ⟨m j|ψ⟩, ρ_A = CC† and ρ_B = (C†C)ᵀ, and CC† and C†C share their nonzero eigenvalues, since CC†x = λx gives C†C(C†x) = λ(C†x). For two spin-halves det ρ_A = |det C|², so the eigenvalues are (1 ± √(1 − 4|det C|²))/2, and the pair is a product exactly when det C = 0.

Entropy grades pure-state entanglement, and stops there

Evaluated in ρ_A's eigenbasis, the von Neumann entropy S(ρ_A) = −Tr(ρ_A ln ρ_A) is −Σₖ pₖ ln pₖ, so for a pure pair S(ρ_A) = S(ρ_B) even when d_A ≠ d_B. It runs from 0 for a product, one pₖ equal to 1, up to ln min(d_A, d_B) for equal weights: ln 2 = 0.693 nats, one bit, for the singlet. For cos θ|↑↓⟩ − sin θ|↓↑⟩ it is −cos²θ ln cos²θ − sin²θ ln sin²θ, 0.562 nats at θ = 30°, and near a product it grows as θ²(1 − ln θ²), 0.056 nats at θ = 0.1 rad. S is a pure number, made a thermodynamic entropy by k_B = 1.381 × 10⁻²³ J K⁻¹, and local unitaries U_A ⊗ U_B leave it unchanged because they leave the pₖ unchanged. That it is the right measure of pure-state entanglement, S/ln 2 being the number of singlets per copy that many copies can be converted into and back, is quoted from quantum information theory. For a mixed pair it fails: ρ_cl has S(ρ_A) = ln 2 and no entanglement. Measures for mixed states lie outside this course.

Change one variable at a time

Make the relationship visible.

Interactive model

Schmidt angle θ30 °

Coherence kept w1.0

Hold w = 1 and sweep θ from 0° to 45°: the dots meet at the centre, the filled marker climbs to ln 2 and the open one stays at 0, a pure pair with maximally mixed halves. Then lower w to 0: neither half moves, yet the whole's entropy rises to meet them and ⟨σₓ ⊗ σₓ⟩ falls to 0.

EIGENVALUE p = cos²θ0.750

S(ρ_A) = S(ρ_B), EACH SPIN0.562 nats

S(ρ), THE WHOLE PAIR0.000 nats

⟨σₓ ⊗ σₓ⟩-0.87

Live interpretationEIGENVALUE p = cos²θ: 0.750. S(ρ_A) = S(ρ_B), EACH SPIN: 0.562 nats. S(ρ), THE WHOLE PAIR: 0.000 nats. ⟨σₓ ⊗ σₓ⟩: −0.87

Catch the common trap

Explain before calculating.

Misconception check

The error is to read a reduced state as ignorance. The singlet's ρ_A = I/2 is the same operator as an unpolarised beam's, so it is tempting to say each spin left the source up or down and the mixture only records which one we have not looked at. The pair contradicts this. It is in a known pure state, and if spin A were in any definite pure state |φ⟩ the pair would have to be the product |φ⟩|χ⟩, with one Schmidt coefficient instead of two. The ignorance story also makes a different, testable prediction: pairs made as |↑↓⟩ or |↓↑⟩ at random have the same halves, I/2 each, and give ⟨σₓ ⊗ σₓ⟩ = 0 where the singlet gives −1. ρ_A is a complete account of what A alone will show, not a list of hidden alternatives; the correlations that decide between the stories live only in the pair's ρ.

Two spins are in |ψ⟩ = (2|↑↑⟩ + |↓↑⟩ + |↓↓⟩)/√6. What is the von Neumann entropy of spin A's reduced state, and why?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyTwo spin-halves are prepared in |ψ⟩ = (3|↑↓⟩ + 4i|↓↑⟩)/5. Find the reduced density matrix of spin A, the expectation values ⟨S_z⟩ and ⟨Sₓ⟩ of spin A alone, and the entanglement entropy.

In the coefficient matrix C, rows for A and columns for B, the only nonzero entries are 3/5 for |↑↓⟩ and 4i/5 for |↓↑⟩. Normalisation: 9/25 + 16/25 = 1.
ρ_A = CC†. Its diagonal is |3/5|² = 0.36 and |4i/5|² = 0.64. Its off-diagonal element ⟨↑|ρ_A|↓⟩ sums, over B's two states, the amplitude of |↑j⟩ times the conjugate amplitude of |↓j⟩, and each product has a zero factor. So ρ_A = diag(0.36, 0.64). The phase i has dropped out: it lived in a coherence of the pair, which the trace over B discards.
⟨S_z⟩ = (ħ/2)(0.36 − 0.64) = −0.14ħ = −1.48 × 10⁻³⁵ J s. ⟨Sₓ⟩ = ħ Re⟨↑|ρ_A|↓⟩ = 0. The Bloch vector is (0, 0, −0.28), well inside the ball.
S = −0.36 ln 0.36 − 0.64 ln 0.64 = 0.3678 + 0.2856 = 0.653 nats, or 0.943 bits, just below the singlet's ln 2 = 0.693 nats.
Checks: tracing out A instead gives ρ_B = diag(0.64, 0.36), with the same eigenvalues and entropy, and ⟨S_z⟩ of A plus that of B is 0, as it must be for a state built only from |↑↓⟩ and |↓↑⟩.

Answerρ_A = diag(0.36, 0.64); ⟨S_z⟩ = −0.14ħ = −1.48 × 10⁻³⁵ J s and ⟨Sₓ⟩ = 0; S = 0.653 nats, the same for spin B.

MediumA pair of spins is left in |ψ⟩ = (|↑↑⟩ + 2|↑↓⟩ + 2|↓↓⟩)/3. Find both reduced density matrices and their Bloch vectors, the eigenvalues they share and the entanglement entropy. Is the pair entangled?

C has rows (1, 2) for A up and (0, 2) for A down, all divided by 3. det C = (1 × 2 − 2 × 0)/9 = 2/9 ≠ 0, so the pair is entangled, which counting its three terms could not have decided.
ρ_A = CC† has populations 5/9 and 4/9 and coherence ⟨↑|ρ_A|↓⟩ = 4/9. ρ_B = (C†C)ᵀ has populations 1/9 and 8/9 and coherence 2/9. Both traces are 1.
With r = (2 Re⟨↑|ρ|↓⟩, −2 Im⟨↑|ρ|↓⟩, ⟨↑|ρ|↑⟩ − ⟨↓|ρ|↓⟩) for each half: r_A = (8/9, 0, 1/9) and r_B = (4/9, 0, −7/9). They point in different directions but have one length, √65/9 = 0.896.
Eigenvalues: λ(1 − λ) = det ρ_A = |det C|² = 4/81, so λ = (1 ± √65/9)/2 = 0.948 and 0.052, which is (1 ± |r|)/2 for either spin.
S = −0.9479 ln 0.9479 − 0.0521 ln 0.0521 = 0.0507 + 0.1539 = 0.205 nats. Limiting check: drop the |↑↑⟩ term and the state becomes (|↑⟩ + |↓⟩)|↓⟩/√2, a product with det C = 0 and S = 0, so the small |↑↑⟩ amplitude is what entangles the pair.

Answerρ_A: populations 5/9, 4/9, coherence 4/9; ρ_B: 1/9, 8/9, coherence 2/9. Shared eigenvalues 0.948 and 0.052, Bloch lengths 0.896, S = 0.205 nats: entangled, weakly.

HardIn the np₁/₂, mⱼ = +½ level of a one-valence-electron atom, orbit and spin are entangled: |ψ⟩ = √(2/3)|mₗ = +1⟩|↓⟩ − √(1/3)|mₗ = 0⟩|↑⟩, quoted here and derived in Unit 5. Find the spin's reduced state ρ_S, ⟨S_z⟩, ⟨L_z⟩ and the entanglement entropy, then the g-factor defined by ⟨μ_z⟩ = −g_J μ_B mⱼ, taking the electron's gₛ = 2.00232.

The state lives on orbit (l = 1, dimension 3) ⊗ spin (dimension 2) and is already in Schmidt form, since |mₗ = +1⟩ ⊥ |mₗ = 0⟩ and |↓⟩ ⊥ |↑⟩. The Schmidt weights are p = 2/3 and 1/3.
Trace out the orbit: ρ_S = (1/3)|↑⟩⟨↑| + (2/3)|↓⟩⟨↓|, Bloch vector (0, 0, −1/3). Trace out the spin: ρ_L = (2/3)|+1⟩⟨+1| + (1/3)|0⟩⟨0|, with a third eigenvalue 0 on |mₗ = −1⟩. Both give S = −(2/3) ln(2/3) − (1/3) ln(1/3) = 0.637 nats.
From the reduced states: ⟨S_z⟩ = (ħ/2)(1/3 − 2/3) = −ħ/6 = −1.76 × 10⁻³⁵ J s, and ⟨L_z⟩ = ħ(2/3 × 1 + 1/3 × 0) = 2ħ/3. Check: ⟨J_z⟩ = 2ħ/3 − ħ/6 = ħ/2 = mⱼ ħ.
The moment μ = −(μ_B/ħ)(L + gₛ S) is a sum of one-part operators, so each term needs only its own reduced state: ⟨μ_z⟩ = −μ_B(2/3 − gₛ/6) = −μ_B(4 − gₛ)/6. Setting this equal to −g_J μ_B × ½ gives g_J = (4 − gₛ)/3; both sides are in J T⁻¹, with μ_B = 9.274 × 10⁻²⁴ J T⁻¹.
With gₛ = 2, g_J = 2/3, the Landé value for a ²P₁/₂ level. With gₛ − 2 = 0.00232, close to α/π and quoted from quantum electrodynamics, g_J = 0.66589: a first-order shift in α of about −α/(3π) = −7.7 × 10⁻⁴. The measured g_J of caesium's 6²P₁/₂ level is 0.66590 ± 0.00009.

Answerρ_S = diag(1/3, 2/3); ⟨S_z⟩ = −ħ/6 = −1.76 × 10⁻³⁵ J s, ⟨L_z⟩ = 2ħ/3; S = 0.637 nats; g_J = (4 − gₛ)/3 = 0.6659, against 0.66590 ± 0.00009 measured for Cs 6²P₁/₂.

Derive ρA from what one half can measure