Partial Trace

This is the canonical treatment of the partial trace, including its characterizing properties, matrix rules, examples, and computational extensions. The Reduced States and Partial Trace gateway supplies the subsystem interpretation and reading path; for the first explicit calculation here, jump to Basis calculation.

Required background. Density Operators supplies the state and trace language, and Entangled States supplies the subsystem interpretation. Familiarity with finite-dimensional product bases and operator matrices is assumed.

The partial trace removes one tensor factor from an operator while retaining exactly the information needed for the other factor. For a bipartite density operator $\rho_{AB}$,

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

What Partial Trace Does

Let

$$X_{AB} \in \mathcal L \bigl( \mathcal H_A\otimes\mathcal H_B \bigr).$$

The full trace returns a scalar:

$$\operatorname{Tr}_{AB}X_{AB} \in \mathbb C.$$

The partial trace returns an operator on the subsystem that remains:

$$\operatorname{Tr}_B X_{AB} \in \mathcal L(\mathcal H_A).$$

If $d_A$ and $d_B$ are finite, the map changes a $(d_A d_B)\times(d_A d_B)$ matrix into a $d_A\times d_A$ matrix. Tracing out $A$ instead gives a $d_B\times d_B$ matrix.

The partial trace is linear, positive, and trace-preserving:

$$\operatorname{Tr}_A \left( \operatorname{Tr}_B X_{AB} \right) = \operatorname{Tr}_{AB}X_{AB}.$$

When $X_{AB}=\rho_{AB}$ is a density operator, the result is therefore a normalized positive operator: the reduced state of subsystem $A$.

Tracing out $B$ is not the same as projecting $B$ onto one selected state. It is a sum over a complete orthonormal basis of the subsystem being ignored.

Basis Calculation

Choose an orthonormal basis $\{\lvert b\rangle_B\}$. The partial trace over $B$ can be written as a sum of contractions. Define

$$R_b \equiv I_A\otimes{}_B\langle b\rvert, \qquad R_b^\dagger = I_A\otimes\lvert b\rangle_B.$$

Then

$$\operatorname{Tr}_B X_{AB} = \sum_b R_b X_{AB}R_b^\dagger.$$

Each term inserts a bra and ket on subsystem $B$, leaving an operator on subsystem $A$.

Why the result is basis-independent

Let another orthonormal basis be related by a unitary matrix:

$$\lvert\beta\rangle = \sum_b U_{b\beta}\lvert b\rangle.$$

Define $R_\beta=I_A\otimes\langle\beta\rvert$ in the new basis.

Using

$$\sum_\beta U_{b\beta}^*U_{c\beta} = \delta_{bc},$$

the basis sum becomes

$$\begin{aligned} \sum_\beta R_\beta X_{AB}R_\beta^\dagger &= \sum_{b,c} \delta_{bc}\, R_b X_{AB}R_c^\dagger\\ &= \sum_b R_b X_{AB}R_b^\dagger. \end{aligned}$$

The basis formula is a calculation device; the resulting operator does not depend on which orthonormal basis was used.

Tensor-Product and Dyad Rules

For a simple tensor of operators,

$$\operatorname{Tr}_B \bigl( X_A\otimes Y_B \bigr) = X_A\,\operatorname{Tr}Y_B.$$

Linearity then handles every finite sum of tensor products. The most useful elementary case is the dyad rule:

$$\operatorname{Tr}_B \left( \lvert i\rangle\langle k\rvert_A \otimes \lvert j\rangle\langle\ell\rvert_B \right) = \delta_{j\ell}\, \lvert i\rangle\langle k\rvert_A.$$

This is just the ordinary trace on the $B$ factor:

$$\operatorname{Tr} \left( \lvert j\rangle\langle\ell\rvert \right) = \langle\ell\vert j\rangle = \delta_{j\ell}.$$

The cross terms in the subsystem being traced out vanish when their basis labels are different.

Coefficient and Block Rules

In product bases, write

$$\rho_{AB} = \sum_{i,k,j,\ell} \rho_{ij,k\ell} \, \lvert i\rangle\langle k\rvert_A \otimes \lvert j\rangle\langle\ell\rvert_B.$$

Then

$$\rho_A = \operatorname{Tr}_B\rho_{AB} = \sum_{i,k} \left( \sum_j \rho_{ij,kj} \right) \lvert i\rangle\langle k\rvert_A.$$

Equivalently, the matrix elements of the reduced state are

$$(\rho_A)_{ik} = \sum_j \rho_{ij,kj}.$$

The repeated $B$ label $j$ is the one being summed. The $A$ labels $i,k$ remain because the result is still an operator on $\mathcal H_A$.

Block-matrix view

With the $A$ index first and the $B$ index changing fastest, regard $X_{AB}$ as a $d_A\times d_A$ array of $d_B\times d_B$ blocks:

$$X_{AB} = \begin{pmatrix} X_{00}^{(B)}&\cdots&X_{0,d_A-1}^{(B)}\\ \vdots&\ddots&\vdots\\ X_{d_A-1,0}^{(B)}&\cdots&X_{d_A-1,d_A-1}^{(B)} \end{pmatrix}.$$

Tracing out $B$ takes the ordinary trace of every block:

$$\operatorname{Tr}_B X_{AB} = \sum_{i,k} \operatorname{Tr} \left( X_{ik}^{(B)} \right) \lvert i\rangle\langle k\rvert_A.$$

By contrast, tracing out $A$ sums the diagonal $B$-space blocks:

$$\operatorname{Tr}_A X_{AB} = \sum_i X_{ii}^{(B)}.$$

These recipes assume the declared $A$-first ordering. If a matrix uses a different flattened product basis, reorder it before applying a memorized block rule.

Pure-State Shortcut

Suppose

$$\lvert\Psi\rangle = \sum_{i,j}C_{ij}\, \lvert i\rangle_A\otimes\lvert j\rangle_B,$$

where $C$ is the $d_A\times d_B$ coefficient matrix with $A$ indexing rows. Then

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

For example,

$$\begin{aligned} (\rho_A)_{ik} &= \sum_j C_{ij}C_{kj}^*\\ &= \sum_j \rho_{ij,kj}. \end{aligned}$$

This is exactly the coefficient rule above. It avoids constructing the full $(d_A d_B)\times(d_A d_B)$ matrix $\lvert\Psi\rangle\langle\Psi\rvert$.

Local Expectation Value Property

The partial trace is defined so that local expectation values are preserved. For every operator $M_A$ on subsystem $A$,

$$\begin{aligned} &\operatorname{Tr}_{AB} \left[ \rho_{AB}(M_A\otimes I_B) \right]\\ &\quad= \operatorname{Tr}_A \left[ (\operatorname{Tr}_B\rho_{AB})M_A \right]. \end{aligned}$$

So a local observer using $\rho_A=\operatorname{Tr}_B\rho_{AB}$ gets the same predictions as an observer using the full state $\rho_{AB}$ and the embedded local observable $M_A\otimes I_B$.

This identity is often the safest way to remember which subsystem is being traced out: tracing out $B$ leaves the operator that predicts measurements on $A$. It also characterizes the partial trace uniquely, because choosing matrix units for $M_A$ fixes every matrix element of the result.

Two-Qubit Matrix Rule

For two qubits in the ordered product basis

$$\lvert00\rangle,\; \lvert01\rangle,\; \lvert10\rangle,\; \lvert11\rangle,$$

the reduced density matrix of qubit $A$ has entries

$$\rho_A = \begin{pmatrix} \rho_{00,00}+\rho_{01,01} & \rho_{00,10}+\rho_{01,11} \\ \rho_{10,00}+\rho_{11,01} & \rho_{10,10}+\rho_{11,11} \end{pmatrix}.$$

The convention is

$$\rho_{ij,k\ell} = \langle i,j\rvert \rho_{AB} \lvert k,\ell\rangle.$$

The pair $(i,j)$ labels the matrix row and $(k,\ell)$ labels the matrix column. For example,

$$\rho_{01,11} = \langle0,1\rvert \rho_{AB} \lvert1,1\rangle.$$

This rule says: keep the $A$ labels and sum over matching $B$ labels.

Bell State Example

For

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

the density matrix in the same ordered basis is

$$\rho_{AB} = \frac{1}{2} \begin{pmatrix} 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 \end{pmatrix}.$$

Using the two-qubit rule,

$$\begin{aligned} (\rho_A)_{00} &= \rho_{00,00}+\rho_{01,01} = \frac{1}{2}, \\ (\rho_A)_{11} &= \rho_{10,10}+\rho_{11,11} = \frac{1}{2}, \\ (\rho_A)_{01} &= \rho_{00,10}+\rho_{01,11} = 0, \\ (\rho_A)_{10} &= \rho_{10,00}+\rho_{11,01} = 0. \end{aligned}$$

Therefore

$$\rho_A = \frac{1}{2} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \frac{I_A}{2}.$$

The same calculation gives $\rho_B=I_B/2$.

Same marginal from a different joint state

The separable, classically correlated state

$$\rho_{\mathrm{cc}} = \frac12 \lvert00\rangle\langle00\rvert + \frac12 \lvert11\rangle\langle11\rvert$$

has matrix

$$\rho_{\mathrm{cc}} = \frac12 \begin{pmatrix} 1&0&0&0\\ 0&0&0&0\\ 0&0&0&0\\ 0&0&0&1 \end{pmatrix}.$$

The same block trace gives

$$\operatorname{Tr}_B\rho_{\mathrm{cc}} = \frac12 I_A.$$

The Bell state and $\rho_{\mathrm{cc}}$ have the same reduced state but different joint coherence and correlations. This illustrates directly that the partial trace is many-to-one.

Calculation workflow and checks

1. Declare the tensor-factor order and the subsystem being discarded.
2. Choose a product, dyad, coefficient-matrix, or block representation.
3. Contract only the input and output indices of the discarded subsystem.
4. Reassemble the operator on the retained Hilbert space.
5. Check dimension, trace, Hermiticity, positivity, and the product rule.
6. Verify one local expectation value when the bookkeeping is uncertain.

For a density operator, these checks require a $d_A\times d_A$ positive matrix of trace one. A wrong dimension usually means that the wrong factor was retained; a wrong trace usually signals a missing diagonal term or normalization factor.

Diagrammatic Preview

In tensor-network notation, an operator on $AB$ has an input and output leg for $A$ and an input and output leg for $B$. Taking $\operatorname{Tr}_B$ connects the $B$ output leg back to the $B$ input leg and sums over that internal label. The $A$ input and output legs remain open, so the result is an operator on $A$.

This picture is often the cleanest way to remember the operation: a trace closes a pair of matching legs. A partial trace closes only the legs belonging to the subsystem being discarded.

Multiple Subsystems

For three subsystems,

$$\mathcal H_{ABC} = \mathcal H_A\otimes\mathcal H_B\otimes\mathcal H_C,$$

one may trace over more than one subsystem:

$$\rho_A = \operatorname{Tr}_{BC}\rho_{ABC} = \operatorname{Tr}_B \bigl( \operatorname{Tr}_C\rho_{ABC} \bigr).$$

Partial traces over distinct factors commute:

$$\operatorname{Tr}_B\operatorname{Tr}_C\rho_{ABC} = \operatorname{Tr}_C\operatorname{Tr}_B\rho_{ABC}.$$

The order of tensor factors still matters for index bookkeeping. If the register order is $A,B,C$, then tracing out the middle subsystem means summing over the middle input-output pair, not simply deleting adjacent rows and columns of a matrix.

Computational Implementation

For finite-dimensional systems, the safest implementation is to reshape the matrix into a tensor with one row index and one column index for each subsystem.

For a bipartite operator with dimensions $d_A$ and $d_B$, use components

$$X_{a b, a' b'}.$$

Then

$$(\operatorname{Tr}_B X)_{a a'} = \sum_{b=0}^{d_B-1} X_{a b, a' b}.$$

In array language, this is a contraction over the $B$ row index and the $B$ column index. A good implementation should make the subsystem dimensions and basis ordering explicit rather than inferring them from the total matrix size.

Apply the calculation checks above to the numerical result, allowing only the stated floating-point tolerance when testing Hermiticity, trace, and nonnegative eigenvalues.

Common Mistakes

• Tracing out the subsystem you meant to keep.
• Projecting onto one state of $B$ instead of summing over a complete basis of $B$.
• Forgetting that only terms with matching traced-out indices survive.
• Applying an $A$-first block rule to a matrix stored in a different product-basis order.
• Confusing “take the trace of every block” with “sum the diagonal blocks”; which recipe applies depends on the factor being traced out.
• Thinking the partial trace destroys normalization; for a normalized density operator, the reduced density operator still has trace one.
• Expecting the reduced state to preserve joint correlations with the subsystem that was traced out.
• Treating the basis formula as basis-dependent in its result. The calculation uses a basis, but the operator it returns is basis-independent.
• Building the full pure-state density matrix when the smaller coefficient-matrix shortcut would suffice.

Cross-Links

• Reduced States and Partial Trace
• Density Operators
• Entangled States
• Quantum Operations
• No-Broadcasting Theorem

References

• J. von Neumann, Mathematical Foundations of Quantum Mechanics, Princeton University Press, 1955.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, 2010.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• A. Peres, Quantum Theory: Concepts and Methods, Kluwer Academic Publishers, 1995.
• J. Watrous, The Theory of Quantum Information, Cambridge University Press, 2018.
• H.-P. Breuer and F. Petruccione, The Theory of Open Quantum Systems, Oxford University Press, 2002.

Exercises

1. Compute

$$\operatorname{Tr}_B \left( |0\rangle\langle1|_A \otimes |1\rangle\langle1|_B \right).$$

Solution

The trace of $|1\rangle\langle1|_B$ is $1$, so

$$\operatorname{Tr}_B \left( |0\rangle\langle1|_A \otimes |1\rangle\langle1|_B \right) = |0\rangle\langle1|_A.$$

2. Compute

$$\operatorname{Tr}_B \left( |0\rangle\langle1|_A \otimes |0\rangle\langle1|_B \right).$$

Solution

The trace of $|0\rangle\langle1|_B$ is $\langle1|0\rangle=0$, so the result is

$$0.$$

3. Trace out qubit $B$ from the product state $|01\rangle\langle01|$.

Solution

Write

$$|01\rangle\langle01| = |0\rangle\langle0|_A \otimes |1\rangle\langle1|_B.$$

Taking the trace over $B$ gives

$$\operatorname{Tr}_B(|01\rangle\langle01|) = |0\rangle\langle0|_A.$$

4. For the Bell state $|\Phi^+\rangle$, compute the reduced state of qubit $B$.

Solution

By symmetry, the reduced state of qubit $B$ is the same as that of qubit $A$:

$$\rho_B=\frac{I_B}{2}.$$

Directly tracing out $A$ gives the same result because the terms $|00\rangle\langle11|$ and $|11\rangle\langle00|$ have zero overlap on subsystem $A$.

5. If $\rho_A=\operatorname{Tr}_B\rho_{AB}$, what operator on $AB$ corresponds to measuring an observable $M_A$ only on subsystem $A$?

Solution

The corresponding composite-system observable is

$$M_A\otimes I_B.$$

The defining expectation-value property is

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

6. Verify trace compatibility for a product operator:

$$\operatorname{Tr}_A \left[ \operatorname{Tr}_B \bigl( X_A\otimes Y_B \bigr) \right] = \operatorname{Tr}_{AB} \bigl( X_A\otimes Y_B \bigr).$$

Why does the result extend to finite sums of product operators?

Solution

First trace over $B$:

$$\operatorname{Tr}_B \bigl( X_A\otimes Y_B \bigr) = X_A\,\operatorname{Tr}Y_B.$$

Then

$$\begin{aligned} \operatorname{Tr}_A \left[ X_A\,\operatorname{Tr}Y_B \right] &= \operatorname{Tr}X_A\, \operatorname{Tr}Y_B\\ &= \operatorname{Tr}_{AB} \bigl( X_A\otimes Y_B \bigr). \end{aligned}$$

Both the full trace and partial trace are linear, so the identity extends term by term to finite sums.

7. Use the coefficient-matrix shortcut for

$$\lvert\Psi\rangle = \frac{ 2\lvert00\rangle +\lvert01\rangle +\lvert11\rangle }{\sqrt6}.$$

Compute both reduced states and check their traces.

Solution

The coefficient matrix is

$$C = \frac{1}{\sqrt6} \begin{pmatrix} 2&1\\ 0&1 \end{pmatrix}.$$

Therefore

$$\rho_A = CC^\dagger = \frac16 \begin{pmatrix} 5&1\\ 1&1 \end{pmatrix},$$

and

$$\rho_B = C^{\mathsf T}C^* = \frac16 \begin{pmatrix} 4&2\\ 2&2 \end{pmatrix}.$$

Both have unit trace:

$$\operatorname{Tr}\rho_A = \operatorname{Tr}\rho_B =1.$$

8. For the Bell state $\lvert\Phi^+\rangle$, compare (a) tracing out $B$, (b) projecting onto the outcome $B=0$, and (c) performing a complete computational-basis measurement on $B$ and discarding its outcome. Identify the output object in each case and explain why (a) and (c) give the same local state on $A$ even though they are different operations on the joint system.

The expressions for (a) and the unnormalized branch in (b) are

$$\operatorname{Tr}_B \left( \lvert\Phi^+\rangle\langle\Phi^+\rvert \right)$$

and

$$\bigl( I_A\otimes{}_B\langle0\rvert \bigr) \lvert\Phi^+\rangle\langle\Phi^+\rvert \bigl( I_A\otimes\lvert0\rangle_B \bigr).$$

Solution

The partial trace sums over a complete basis of $B$ and gives

$$\operatorname{Tr}_B \left( \lvert\Phi^+\rangle\langle\Phi^+\rvert \right) = \frac12 I_A.$$

The second expression selects only the $B=0$ component:

$$\begin{aligned} &\bigl( I_A\otimes{}_B\langle0\rvert \bigr) \lvert\Phi^+\rangle\langle\Phi^+\rvert \bigl( I_A\otimes\lvert0\rangle_B \bigr)\\ &\qquad= \frac12 \lvert0\rangle\langle0\rvert_A. \end{aligned}$$

It is an unnormalized conditional operator with trace $1/2$, the probability of the selected outcome. After normalization it becomes $\lvert0\rangle\langle0\rvert_A$.

For (c), the nonselective measurement produces the joint state

$$\rho'_{AB} = \frac12\lvert00\rangle\langle00\rvert + \frac12\lvert11\rangle\langle11\rvert.$$

This operation removes joint coherence and therefore differs from merely computing a reduced state. Nevertheless, $\operatorname{Tr}_B\rho'_{AB}=I_A/2$, the same local state obtained in (a). Projection selects one branch, nonselective measurement changes the joint state and averages all branches, and partial trace extracts the local state without specifying a measurement.