advanced32 min studyLesson 366 in Physics

Partial traces and reduced states

Build a local density operator from a composite state and distinguish discarding a subsystem from conditioning on its measurement result.

Before you begin

On this page

The question for today

What you’ll understand

Identify tensor-product coordinates and operators for two subsystems.
Compute a partial trace in a product basis.
Explain why local statistics can remain unchanged while joint correlations differ.

A subsystem does not generally have its own pure-state vector even when the combined system does. The reduced density operator gives exactly the probabilities accessible to measurements on that subsystem alone. It is obtained by a partial trace, which contracts the inaccessible subsystem's indices. This operation summarizes local predictions without choosing a measurement outcome on the discarded subsystem and without asserting that the global correlations have physically disappeared.

The tensor product fixes which observations are local

For two distinguishable subsystems, the joint Hilbert space is the tensor product of their individual spaces. A product basis has pairs of indices, and a two-qubit state has four amplitudes rather than two separately chosen single-qubit vectors. An observable acting only on A has the form A's operator tensored with B's identity. The specified tensor factorization matters: calling a four-component column entangled makes sense only relative to a chosen division into subsystems.

A product state has coefficients that factor into an A vector times a B vector. Most joint vectors do not factor. To predict an A measurement, sum over the unobserved B alternatives at the probability-operator level, retaining any A coherences associated with the same B index. This contraction gives the unique local operator reproducing every A-only expectation. Tracing the entire matrix instead would give only the scalar normalization and discard all local-state information.

\mathcal H_{AB}=\mathcal H_A\otimes\mathcal H_B,\qquad (\rho_A)_{ij}=\sum_\mu(\rho_{AB})_{i\mu,j\mu},\qquad \operatorname{tr}[\rho_{AB}(O_A\otimes I_B)]=\operatorname{tr}(\rho_AO_A)

ρA=trBρAB is dimensionless and acts only on HA. Indices i,j label an orthonormal A basis and μ an orthonormal B basis. The equality defines the partial trace operationally; OA and its expectation carry the local observable's units.↔ Scroll wide equations horizontally.

Partial trace preserves some coherence and removes other terms

A matrix unit built from product kets contracts by the overlap of its B factors. Orthogonal B records eliminate the corresponding A cross term, while identical B records leave it intact. The result is independent of the orthonormal basis used to perform the B contraction. This basis independence separates a physical local description from a computational choice of index labels. The reduced operator remains positive with unit trace.

For a pure joint state, arrange coefficients into a matrix C with A indexing rows and B indexing columns. Then rho A equals C times C dagger. This gives a quick route to local populations and coherences without writing a full four-by-four projector. A mixed reduced state is compatible with a perfectly pure global preparation. Its mixedness reflects inaccessible correlations, not necessarily a classical random choice made by the preparation device.

\operatorname{tr}_B\left(|i\rangle\langle j|\otimes|\mu\rangle\langle\nu|\right)=\delta_{\mu\nu}|i\rangle\langle j|,\qquad |\Psi\rangle=\sum_{i,\mu}C_{i\mu}|i\mu\rangle\ \Longrightarrow\ \rho_A=CC^\dagger

All kets and coefficients refer to normalized orthonormal product bases. C has squared Frobenius norm one for a normalized pure state. The Kronecker delta is the trace of the B matrix unit; it is not a physical measurement selection.↔ Scroll wide equations horizontally.

Discarding is different from selecting a reported outcome

A trace-preserving operation on B alone cannot change A's reduced state when the B outcome is unreported. This no-signalling statement concerns the averaged local description. If B is measured and a particular classical outcome is reported, the conditional state of A can differ from its earlier reduced state. Selection on the reported record changes which subensemble is being described; obtaining that record still requires ordinary communication.

Two different joint states can have the same reduced operator and therefore remain indistinguishable by A-only measurements. Correlation measurements involving both subsystems can nevertheless tell them apart. The reduced state is consequently sufficient for local questions but not a complete reconstruction of the composite preparation. This is why replacing an entangled pure state by a product of its reduced operators usually gives incorrect joint probabilities even though it preserves separate marginal probabilities.

Work it through

A reduced state that retains off-diagonal terms

For $|\Psi\rangle=(|00\rangle+|01\rangle+|11\rangle)/\sqrt3$ , compute ρA, p(A=0), the expectation of XA, and the local purity. The first label belongs to A.

Arrange the normalized coefficient matrix as $C=\tfrac1{\sqrt3}\begin{pmatrix}1&1\\0&1\end{pmatrix}$ . Its rows share a B=1 contribution, so an off-diagonal A term survives the partial trace.
Multiply to get $\rho_A=CC^\dagger=\tfrac13\begin{pmatrix}2&1\\1&1\end{pmatrix}$ . Thus p(A=0)=2/3 and $\langle X_A\rangle=(\rho_A)_{01}+(\rho_A)_{10}=2/3$ . A diagonal-only reduction would miss the latter prediction.
The purity is $(4+1+2)/9=7/9$ . It is below one even though the full ket was normalized and pure. Its determinant 1/9 is positive, confirming that both local eigenvalues are nonzero.

The result

The reduced matrix is one third of [[2,1],[1,1]], with outcome-zero probability 2/3, X expectation 2/3, and purity 7/9. Tracing B removed information without removing every A coherence.

The living notebook

Explore the model

Interactive lab

Interactive lab · probability

Repeated local outcomes

For independent copies of a Bell pair, a local computational outcome has probability 1/2. Reset and set both controls to Success probability p=0.5 and Trials n=10; predict count mean 5 and variance 2.5. For a fresh comparison, Reset and set both controls to p=0.5 and n=20; predict mean 10 and variance 5. The bars are exact binomial probabilities, not simulated measurements.

Exact probabilityHighlighted outcomeMean np

Mean · np: 5
Variance · np(1−p): 2.5
Total probability: 1

Success probability p0.5Trials n10

t = 0 s

These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forRepeated local statistics can be ordinary binomial even when a pair's global state is entangled. This lab has no tensor-product state, partner outcome, or joint-correlation model.

Pause & reason

Check your understanding

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Read the checkpoint explanation

The partial trace summarizes unconditional local predictions and can retain off-diagonal terms. Conditioning requires an instrument and a specific nonzero-probability outcome. A reduced state may be mixed even when the composite state is pure.

Your turn

Try it yourself

Compare $|\Phi_\pm\rangle=(|00\rangle\pm|11\rangle)/\sqrt2$ . Find each reduced state on A and the expectation of $X_A\otimes X_B$ . Can A alone distinguish the sign?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The B overlap kills the |00⟩⟨11| term in a partial trace. The joint X operator swaps |00⟩ and |11⟩.

Compare with the solution

Both reduced states are $\rho_A=I/2$ , since only the two equally weighted diagonal A terms remain. The joint operator maps each Bell state to itself times its sign, giving expectations +1 for Φ+ and −1 for Φ−. Thus an A-only apparatus cannot distinguish these two preparations, although joint correlation measurements can. Replacing either joint state with $(I/2)\otimes(I/2)$ would predict zero X correlation and lose the distinguishing information.

Keep the idea

Take it with you

Tensor-product structure identifies local operators and subsystem questions.
A partial trace reproduces unconditional local statistics and may retain coherence.
Identical reduced states can coexist with different measurable joint correlations.

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