Reduced Density Operators

Let a composite system $AB$ have Hilbert space

$$\mathcal H_{AB} = \mathcal H_A\otimes\mathcal H_B$$

and joint density operator $\rho_{AB}$. The reduced density operator of subsystem $A$ is

$$\rho_A = \operatorname{Tr}_B\rho_{AB}.$$

It is the unique density operator on $\mathcal H_A$ that reproduces every prediction for measurements performed only on $A$:

$$\operatorname{Tr}_A(\rho_A M_A) = \operatorname{Tr}_{AB} \left[ \rho_{AB}(M_A\otimes I_B) \right]$$

for every bounded observable $M_A$. The same identity holds for projectors, POVM effects, and arbitrary linear combinations of local operators.

Helpful background. Density operators supplies the trace rule; operators on composite systems explains the identity factor in a local measurement.

The state seen by a subsystem

A local observable on $A$ is represented on the joint Hilbert space by

$$M_A\otimes I_B.$$

This embedding says that the experimental question acts nontrivially on $A$ and does nothing to $B$. A state on $\mathcal H_A$ is operationally adequate precisely when it gives the same answers to all such questions as the joint state does.

The requirement determines $\rho_A$ uniquely. Suppose two Hermitian operators $\rho_A$ and $\tau_A$ give the same expectation value for every Hermitian $M_A$. With

$$D=\rho_A-\tau_A,$$

one has

$$\operatorname{Tr}(DM_A)=0$$

for every Hermitian $M_A$. Choosing $M_A=D$ gives

$$\operatorname{Tr}(D^2)=0.$$

Because $D^2$ is positive semidefinite, all eigenvalues of $D$ vanish and $D=0$. Thus there is exactly one local density operator containing all local statistics.

This does not mean that $\rho_A$ contains the full joint state. Operators such as $M_A\otimes N_B$ probe correlations and generally cannot be evaluated from $\rho_A$ alone.

Partial trace

In finite dimensions, for any operator $X_{AB}$; in infinite dimensions, for a trace-class operator $X_{AB}$, the partial trace over $B$ is characterized by

$$\operatorname{Tr}_A \left[ \left( \operatorname{Tr}_B X_{AB} \right)M_A \right] = \operatorname{Tr}_{AB} \left[ X_{AB}(M_A\otimes I_B) \right]$$

for every bounded $M_A$. This characterization is basis-independent and is often the cleanest definition.

To calculate it, choose an orthonormal basis $\{\lvert\mu\rangle_B\}$ of $\mathcal H_B$ and define

$$K_\mu = I_A\otimes\langle\mu\rvert.$$

Then

$$\operatorname{Tr}_B X_{AB} = \sum_\mu K_\mu X_{AB}K_\mu^\dagger.$$

Although a basis appears on the right, every orthonormal basis produces the same operator because the basis-independent characterizing identity has a unique solution.

Matrix-element rule

Let

$$(X_{AB})_{i\mu,j\nu} = \langle i,\mu\rvert X_{AB} \lvert j,\nu\rangle.$$

Then

$$\left( \operatorname{Tr}_B X_{AB} \right)_{ij} = \sum_\mu (X_{AB})_{i\mu,j\mu}.$$

The subsystem that remains supplies the free row and column labels $i,j$. The subsystem being traced out supplies one repeated label $\mu$, which is summed.

For product operators,

$$\operatorname{Tr}_B(A\otimes B) = A\,\operatorname{Tr}B.$$

In particular,

$$\operatorname{Tr}_B \left( A\otimes \lvert\mu\rangle\langle\nu\rvert \right) = \delta_{\mu\nu}A.$$

This dyad rule explains why terms with unequal $B$ labels disappear. Partial Trace page collects block-matrix, multiple-subsystem, and computational rules.

Reduction is a quantum channel

The rectangular operators $K_\mu$ introduced above map $\mathcal H_A\otimes\mathcal H_B$ to $\mathcal H_A$. The basis formula is a Kraus representation:

$$\operatorname{Tr}_B X_{AB} = \sum_\mu K_\mu X_{AB}K_\mu^\dagger.$$

Moreover,

$$\sum_\mu K_\mu^\dagger K_\mu = I_A\otimes I_B.$$

The Kraus form gives complete positivity, and the completeness relation gives trace preservation. Thus the partial trace maps joint density operators to valid subsystem density operators.

For example, positivity can be checked directly. Let $\lvert\alpha\rangle\in\mathcal H_A$. Then

$$\begin{aligned} \langle\alpha\rvert \rho_A \lvert\alpha\rangle &= \sum_\mu \langle\alpha,\mu\rvert \rho_{AB} \lvert\alpha,\mu\rangle \\ &\geq0. \end{aligned}$$

Thus

$$\rho_A^\dagger=\rho_A, \qquad \rho_A\geq0, \qquad \operatorname{Tr}_A\rho_A=1.$$

Calling the partial trace a channel does not imply that someone measured $B$. It may describe a physical discard operation, but it also serves as a mathematical restriction from all joint observables to the local observable algebra of $A$.

Local predictions

If $\{E_a\}$ is a POVM performed on $A$, its joint-space effects are $E_a\otimes I_B$. The outcome probabilities satisfy

$$\begin{aligned} p(a) &= \operatorname{Tr}_{AB} \left[ \rho_{AB}(E_a\otimes I_B) \right] \\ &= \operatorname{Tr}_A(\rho_AE_a). \end{aligned}$$

Therefore complete tomography using only measurements on $A$ can reconstruct $\rho_A$, but not $\rho_{AB}$.

For an observable $M_A$,

$$\langle M_A\rangle_{\rho_A} = \left\langle M_A\otimes I_B \right\rangle_{\rho_{AB}}.$$

For a correlation observable,

$$\left\langle M_A\otimes N_B \right\rangle_{\rho_{AB}},$$

both factors matter. No function of $\rho_A$ alone can reproduce every such value for every compatible joint state.

Reduction preserves every statistic of $A$-local effects $E_A\otimes I_B$. It does not retain enough information to reconstruct $B$ or the correlations between the two subsystems.

Product states and correlations

For a product state,

$$\rho_{AB} = \rho_A\otimes\rho_B,$$

one obtains

$$\operatorname{Tr}_B\rho_{AB} = \rho_A\operatorname{Tr}\rho_B = \rho_A.$$

The same local density matrix can arise from very different joint states. Consider

$$\begin{aligned} \rho_{\mathrm{unc}} &= \frac{I_{AB}}{4}, \\ \rho_{\mathrm{cc}} &= \frac12 \left( \lvert00\rangle\langle00\rvert + \lvert11\rangle\langle11\rvert \right), \\ \rho_{\mathrm{ent}} &= \lvert\Phi^+\rangle\langle\Phi^+\rvert. \end{aligned}$$

Here

$$\lvert\Phi^+\rangle = \frac{ \lvert00\rangle+\lvert11\rangle }{\sqrt2}.$$

All three have

$$\rho_A=\rho_B=\frac I2.$$

Yet they have different global structure:

• $\rho_{\mathrm{unc}}$ is uncorrelated;
• $\rho_{\mathrm{cc}}$ is separable but classically correlated;
• $\rho_{\mathrm{ent}}$ is a pure entangled state.

For example,

$$\begin{aligned} \langle Z\otimes Z\rangle_{\mathrm{unc}} &=0, & \langle X\otimes X\rangle_{\mathrm{unc}} &=0, \\ \langle Z\otimes Z\rangle_{\mathrm{cc}} &=1, & \langle X\otimes X\rangle_{\mathrm{cc}} &=0, \\ \langle Z\otimes Z\rangle_{\mathrm{ent}} &=1, & \langle X\otimes X\rangle_{\mathrm{ent}} &=1. \end{aligned}$$

The first correlation separates the product state from the two correlated states; the second separates the classical correlation from the Bell state. Identical marginals do not imply identical joint physics.

Pure joint states and shared spectra

Write a normalized pure state in product bases as

$$\begin{aligned} \lvert\Psi\rangle &= \sum_{i,\mu} C_{i\mu} \lvert i\rangle_A \lvert\mu\rangle_B, \\ 1 &= \sum_{i,\mu} \lvert C_{i\mu}\rvert^2. \end{aligned}$$

The reduced-state matrix elements are

$$\begin{aligned} (\rho_A)_{ij} &= \sum_\mu C_{i\mu}C_{j\mu}^*, \\ (\rho_B)_{\mu\nu} &= \sum_i C_{i\mu}C_{i\nu}^*. \end{aligned}$$

If $C$ denotes the coefficient array with row index $i$ and column index $\mu$, then

$$\rho_A=CC^\dagger, \qquad \rho_B=C^{\mathsf T}C^*.$$

The transpose in the second expression follows from this explicit coefficient convention. Equivalently, $\rho_B$ is the transpose of $C^\dagger C$. Both reduced states have the squared singular values of $C$ as their nonzero eigenvalues.

This is the matrix form of the Schmidt decomposition. For a pure bipartite state:

• $\rho_A$ and $\rho_B$ have the same nonzero spectrum;
• their ranks equal the Schmidt rank;
• $\operatorname{Tr}(\rho_A^2)=\operatorname{Tr}(\rho_B^2)$;
• one reduced state is pure exactly when the joint state is a product state;
• a mixed reduced state is equivalent to entanglement across the $A\mid B$ bipartition.

The final statement requires the joint state to be pure. If $\rho_{AB}$ is mixed, a mixed marginal by itself does not certify entanglement.

Partially entangled pair

Consider

$$\begin{aligned} \lvert\Psi\rangle &= \sqrt p\,\lvert00\rangle + e^{i\phi} \sqrt{1-p}\,\lvert11\rangle, \\ 0&\leq p\leq1. \end{aligned}$$

The reductions are

$$\rho_A = \rho_B = p\lvert0\rangle\langle0\rvert + (1-p)\lvert1\rangle\langle1\rvert.$$

The real phase $\phi$ is present in the joint coherence but absent from either local state. For example, $\langle X\otimes X\rangle=2\sqrt{p(1-p)}\cos\phi$ changes with the phase even though the marginals do not. Their purity is

$$\operatorname{Tr}(\rho_A^2) = p^2+(1-p)^2.$$

The state is a product at $p=0$ or $p=1$, entangled for $0<p<1$, and maximally entangled at $p=1/2$.

Two-qubit matrix example

In the ordered basis

$$\left( \lvert00\rangle, \lvert01\rangle, \lvert10\rangle, \lvert11\rangle \right),$$

consider the physical state

$$\rho_{AB} = \begin{pmatrix} a&0&0&w\\ 0&b&z&0\\ 0&z^*&c&0\\ w^*&0&0&d \end{pmatrix},$$

where

$$\begin{aligned} a,b,c,d&\geq0, & a+b+c+d&=1, \\ \lvert w\rvert^2&\leq ad, & \lvert z\rvert^2&\leq bc. \end{aligned}$$

The reductions are

$$\begin{aligned} \rho_A &= \begin{pmatrix} a+b&0\\ 0&c+d \end{pmatrix}, \\[6pt] \rho_B &= \begin{pmatrix} a+c&0\\ 0&b+d \end{pmatrix}. \end{aligned}$$

The off-diagonal joint entries $w$ and $z$ connect basis states with different labels on the subsystem being traced out, so they vanish from both marginals. They may still affect joint observables and cannot be declared physically irrelevant.

Marginal is not conditional

The reduced state is an unconditioned marginal: it predicts measurements on $A$ without selecting an outcome on $B$.

Let outcome $b$ on $B$ have Kraus operators $L_{b\alpha}$ and effect $F_b=\sum_\alpha L_{b\alpha}^\dagger L_{b\alpha}$. The unnormalized conditional state of the untouched subsystem $A$ is

$$\begin{aligned} \widetilde\rho_{A|b} &=\sum_\alpha\operatorname{Tr}_B\bigl[(I\otimes L_{b\alpha})\rho_{AB}(I\otimes L_{b\alpha}^\dagger)\bigr]\\ &=\operatorname{Tr}_B\bigl[\rho_{AB}(I\otimes F_b)\bigr]. \end{aligned}$$

Its trace is $p(b)$; divide by $p(b)>0$ to obtain $\rho_{A|b}$. The second equality uses partial-trace cyclicity on $B$. Although the right-hand product need not look Hermitian, the first expression proves positivity.

For a fixed joint state, this remote conditional state depends only on the effect $F_b$, not on which instrument realizes that effect on $B$. The measured subsystem’s state and the remaining joint state generally do depend on the instrument. See conditional states for the distinction.

Ignoring the outcome restores the original marginal:

$$\sum_b\widetilde\rho_{A|b} =\operatorname{Tr}_B\bigl[\rho_{AB}(I\otimes\sum_bF_b)\bigr] =\rho_A.$$

Thus a local trace-preserving operation on $B$ cannot change the unconditioned state of $A$.

What reduction preserves and discards

Reduction preserves:

• normalization, Hermiticity, and positivity;
• every probability distribution for a measurement on $A$ alone;
• every expectation value of $M_A\otimes I_B$;
• the nonzero Schmidt spectrum when the joint state is pure.

Reduction does not preserve enough information to recover:

• the state of $B$;
• correlations between $A$ and $B$;
• whether a mixed marginal arose from classical correlation, entanglement, or an uncorrelated mixed preparation;
• relative phases that appear only in joint coherences;
• the original joint state.

The map $\rho_{AB}\mapsto\rho_A$ is many-to-one and has no inverse on the set of joint states. A purification constructs a larger pure state with a chosen marginal, but that construction is highly nonunique.

When $B$ is an uncontrolled environment, $\rho_A(t)$ is the state used to predict accessible observations. Even if $\rho_{AB}(t)$ evolves unitarily, the reduced evolution of $A$ need not be unitary because information and correlations can flow into $B$.

Infinite-dimensional qualification

The formulas above are automatic for finite-dimensional spaces. In infinite dimensions, a density operator is trace class. If $\rho_{AB}$ is trace class, there is a unique trace-class $\rho_A=\operatorname{Tr}_B\rho_{AB}$ satisfying

$$\operatorname{Tr}_A(\rho_A M_A) = \operatorname{Tr}_{AB} \left[ \rho_{AB}(M_A\otimes I_B) \right]$$

for every bounded $M_A$. The basis sum defining the partial trace converges in trace norm. These hypotheses matter: one should not treat arbitrary infinite matrices as though finite-dimensional trace manipulations automatically apply. See Trace-Class and Hilbert–Schmidt Operators for the operator-class background.

Interpreting local mixedness

A mixed local state can arise from classical uncertainty about a preparation, entanglement with another system, or both. Local statistics alone do not determine which joint preparation produced it.

If the joint state is known to be pure, mixedness of its marginal implies entanglement. This is a statement about the joint state, not a need to assign a hidden pure vector to the subsystem. Proper and improper mixtures develops this interpretive distinction.

Reduced-State Dynamics

If only $A$ undergoes a unitary $U_A$, then

$$\rho_A \longmapsto U_A\rho_A U_A^\dagger.$$

If only $B$ undergoes a trace-preserving operation and its outcome is not selected, $\rho_A$ does not change, as shown above.

A joint unitary generally can change the reduced state:

$$\rho_A(t) = \operatorname{Tr}_B \left[ U_{AB}(t)\, \rho_{AB}(0)\, U_{AB}^\dagger(t) \right].$$

Interactions can transfer purity and build correlations, so the reduced evolution need not be unitary. If initial system-environment correlations are present, $\rho_A(0)$ alone may not determine $\rho_A(t)$; the initial joint state can matter.

If the initial joint state is $\rho_A\otimes\tau_B$ with a fixed environment state $\tau_B$, this construction defines a quantum channel on arbitrary inputs $\rho_A$. Without that fixed preparation assumption, do not infer a single reduced map from one trajectory.

Common pitfalls

A partial trace discards a subsystem, not selected matrix entries. Match the discarded row and column labels and sum over them; keep both free indices of the subsystem that remains.

A marginal is not a conditional state. Tracing out $B$ requires no measurement. Selecting a known result on $B$ is a different operation.

Complex conjugations matter. With coefficients $C_{i\mu}$, the correct formulas are $\rho_A=CC^\dagger$ and $\rho_B=C^{\mathsf T}C^*$. Using $C^\dagger C$ for the latter generally conjugates the wrong local coherences.

References

• Heinz-Peter Breuer and Francesco Petruccione, The Theory of Open Quantum Systems, Oxford University Press, 2002, sections on reduced dynamics.
• Michael A. Nielsen and Isaac L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press, 2010, sections 2.4–2.5.
• John Watrous, The Theory of Quantum Information, Cambridge University Press, 2018, chapters 1–2. Author-hosted text.

Exercises

1. Prove that

$$\operatorname{Tr}_B(\rho_A\otimes\rho_B) = \rho_A$$

for normalized density operators.

Solution

The product rule gives

$$\operatorname{Tr}_B(\rho_A\otimes\rho_B) = \rho_A\operatorname{Tr}\rho_B.$$

Because $\rho_B$ is normalized,

$$\operatorname{Tr}\rho_B=1,$$

and the result follows.

2. Starting from the basis definition of the partial trace, derive

$$\left( \operatorname{Tr}_B X \right)_{ij} = \sum_\mu X_{i\mu,j\mu}.$$

Solution

Insert the basis definition between $\langle i\rvert$ and $\lvert j\rangle$:

$$\begin{aligned} \langle i\rvert \operatorname{Tr}_B(X) \lvert j\rangle &= \sum_\mu \langle i\rvert \left( I_A\otimes\langle\mu\rvert \right) X \left( I_A\otimes\lvert\mu\rangle \right) \lvert j\rangle \\ &= \sum_\mu \langle i,\mu\rvert X \lvert j,\mu\rangle \\ &= \sum_\mu X_{i\mu,j\mu}. \end{aligned}$$

3. Show directly that $\rho_A=\operatorname{Tr}_B\rho_{AB}$ is positive whenever $\rho_{AB}$ is positive.

Solution

For any $\lvert\alpha\rangle\in\mathcal H_A$ and any orthonormal basis $\{\lvert\mu\rangle\}$ of $B$,

$$\begin{aligned} \langle\alpha\rvert \rho_A \lvert\alpha\rangle &= \sum_\mu \langle\alpha,\mu\rvert \rho_{AB} \lvert\alpha,\mu\rangle. \end{aligned}$$

Every term is nonnegative because $\rho_{AB}\geq0$. Hence the sum is nonnegative for every $\lvert\alpha\rangle$, so $\rho_A\geq0$.

4. The states $\rho_{\mathrm{unc}}=I_{AB}/4$ and $\rho_{\mathrm{cc}}=(|00\rangle\langle00|+|11\rangle\langle11|)/2$ have the same initial marginals. Apply the joint unitary $U|a,b\rangle=|a\mathbin{\oplus}b,b\rangle$, where $\oplus$ is addition modulo two. Find the final state of $A$ for each preparation. Can a single map of the initial $\rho_A$ describe both experiments?

Solution

The maximally mixed joint state is invariant under every unitary, so it leaves $\rho_A'=I_A/2$. The correlated state instead transforms as

$$U\rho_{\mathrm{cc}}U^\dagger =\frac12\bigl(|00\rangle\langle00|+|01\rangle\langle01|\bigr) =|0\rangle\langle0|_A\otimes\frac{I_B}{2}.$$

Thus its final marginal is $\rho_A'=|0\rangle\langle0|$. Both experiments start with $\rho_A=\rho_B=I/2$, yet the same interaction produces different final states of $A$. Their initial correlations distinguish the preparations. No function of $\rho_A$ alone can assign both outputs to the same input; specifying a fixed uncorrelated environment is an additional assumption behind the usual reduced channel construction.

5. For

$$\lvert\Psi\rangle = \sqrt p\,\lvert00\rangle + e^{i\phi} \sqrt{1-p}\,\lvert11\rangle,$$

with $0\le p\le1$ and real $\phi$, compute both reduced states, their purity, and the values of $p$ for which the joint state is entangled.

Solution

The joint density operator contains diagonal terms

$$p\lvert00\rangle\langle00\rvert + (1-p)\lvert11\rangle\langle11\rvert$$

and two cross terms with different $B$ labels. The cross terms vanish under $\operatorname{Tr}_B$, giving

$$\rho_A = p\lvert0\rangle\langle0\rvert + (1-p)\lvert1\rangle\langle1\rvert.$$

By symmetry, $\rho_B=\rho_A$. Their purity is

$$\operatorname{Tr}(\rho_A^2) = p^2+(1-p)^2.$$

The reduction is pure only for $p=0$ or $p=1$. Because the joint state is pure, it is entangled exactly when the reduction is mixed:

$$0<p<1.$$

6. Verify that $\rho_{\mathrm{unc}}$, $\rho_{\mathrm{cc}}$, and $\rho_{\mathrm{ent}}$ from the text all have marginal $I/2$ on $A$. Then explain how $Z\otimes Z$ and $X\otimes X$ distinguish them.

Solution

For $I_{AB}/4$,

$$\operatorname{Tr}_B \left( \frac{I_A\otimes I_B}{4} \right) = \frac{I_A}{2}.$$

For $\rho_{\mathrm{cc}}$, tracing either $\lvert00\rangle\langle00\rvert$ or $\lvert11\rangle\langle11\rvert$ over $B$ leaves the corresponding $A$ projector, so the equal mixture gives $I_A/2$. The Bell-state cross terms vanish under the partial trace, giving the same result.

Their correlations are

$$\begin{aligned} \langle Z\otimes Z\rangle &= 0,\ 1,\ 1, \\ \langle X\otimes X\rangle &= 0,\ 0,\ 1, \end{aligned}$$

in the order product, classically correlated, entangled. Thus local marginals agree while joint observables distinguish the states.

7. Reduce the two-qubit matrix

$$\rho_{AB} = \begin{pmatrix} a&0&0&w\\ 0&b&z&0\\ 0&z^*&c&0\\ w^*&0&0&d \end{pmatrix}$$

over $B$ and over $A$.

Solution

For $\rho_A$, sum entries with matching $B$ labels:

$$\begin{aligned} (\rho_A)_{00} &= \rho_{00,00}+\rho_{01,01} = a+b, \\ (\rho_A)_{11} &= \rho_{10,10}+\rho_{11,11} = c+d. \end{aligned}$$

The off-diagonal entries vanish, so

$$\rho_A = \begin{pmatrix} a+b&0\\ 0&c+d \end{pmatrix}.$$

Similarly,

$$\rho_B = \begin{pmatrix} a+c&0\\ 0&b+d \end{pmatrix}.$$

The entries $w$ and $z$ connect unequal labels on either traced subsystem and do not appear in the marginals.

8. The Bell state $\lvert\Phi^+\rangle$ is measured in the $Z$ basis on subsystem $B$. Find the conditional states of $A$ for both outcomes and their unconditioned average.

Solution

Outcome $0$ occurs with probability $1/2$ and leaves

$$\rho_{A\mid0} = \lvert0\rangle\langle0\rvert.$$

Outcome $1$ also occurs with probability $1/2$ and leaves

$$\rho_{A\mid1} = \lvert1\rangle\langle1\rvert.$$

If the outcome is not retained, the average state is

$$\begin{aligned} \rho_A &= \frac12\rho_{A\mid0} + \frac12\rho_{A\mid1} \\ &= \frac12 \left( \lvert0\rangle\langle0\rvert + \lvert1\rangle\langle1\rvert \right) \\ &= \frac I2. \end{aligned}$$

Conditioning changes the state assigned using a known outcome; ignoring the outcome returns the original reduced state.

9. A qubit–qutrit pure state is

$$\lvert\Omega\rangle = \frac{ \lvert0,0\rangle +i\lvert1,1\rangle +\lvert1,2\rangle }{\sqrt3}.$$

Find its coefficient matrix, $\rho_A$, and $\rho_B$. Verify that the two reduced states have the same nonzero eigenvalues.

Solution

With $A$ indexing rows and $B$ indexing columns,

$$C = \frac{1}{\sqrt3} \begin{pmatrix} 1&0&0\\ 0&i&1 \end{pmatrix}.$$

Therefore

$$\rho_A = CC^\dagger = \frac13 \begin{pmatrix} 1&0\\ 0&2 \end{pmatrix},$$

and

$$\rho_B = C^{\mathsf T}C^* = \frac13 \begin{pmatrix} 1&0&0\\ 0&1&i\\ 0&-i&1 \end{pmatrix}.$$

The eigenvalues of $\rho_A$ are $1/3$ and $2/3$. The qutrit state has the same two nonzero eigenvalues plus one zero eigenvalue:

$$\operatorname{spec}(\rho_B) = \left\{ \frac23,\frac13,0 \right\}.$$

The qutrit off-diagonal entries are $i/3$ and $-i/3$. Replacing $C^{\mathsf T}C^*$ by $C^\dagger C$ would reverse these phases while leaving the eigenvalues unchanged.