Reduced States and Partial Trace

A joint quantum state contains more information than any one subsystem can access. The reduced state extracts exactly the information needed to predict every measurement performed locally on the retained subsystem, while discarding correlations that require joint access.

For a bipartite state $\rho_{AB}$, the reduced states are

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

This chapter organizes three distinct tasks:

$$\begin{gathered} \text{reduce a joint state} \\ \downarrow \\ \text{predict local statistics} \\ \downarrow \\ \text{separate local data from correlations}. \end{gathered}$$

The linked pages own the detailed derivations. This page supplies the shared interpretation, calculation protocol, and boundaries among discarding, conditioning, purification, and entropy.

Required background. Density Operators supplies positive trace-one operators and the trace rule; Entangled States supplies the interpretation of a pure joint state with mixed local states. Familiarity with tensor-product operator algebra is assumed for finite-product-basis expansions and subsystem-index contractions.

Chapter map

Question	Current route	Outcome
What operation produces the state available to one subsystem?	Partial Trace	basis-independent definition, index rules, block rules, and checks
What state and correlation language is required first?	Entangled States	subsystem split and joint-state classification
How does reduction become a general process?	Quantum Operations	channels, selective maps, and discarded environments

The Operational Definition

Let $M_A$ be any observable on $A$. As an observable on the composite system it is represented by $M_A\otimes I_B$. The reduced state $\rho_A$ is characterized by

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

for every $M_A$. This identity is more than a computational convenience: it says that $\rho_A$ contains all and only the information needed for measurements confined to $A$.

The same statement holds for a local measurement effect $E_a$:

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

No measurement on $A$ alone can distinguish two joint states with the same $\rho_A$. Joint measurements, correlations with $B$, or conditioning on information from $B$ may distinguish them.

What the Partial Trace Does

Choose an orthonormal basis $\{\lvert j\rangle_B\}$ for the subsystem being discarded. For an operator $X_{AB}$,

$$\begin{aligned} \operatorname{Tr}_B X_{AB} &= \sum_j (I_A\otimes\langle j\rvert_B) X_{AB} \\ &\qquad\times (I_A\otimes\lvert j\rangle_B). \end{aligned}$$

Although a basis appears in the formula, the result is basis independent. The sum contracts the bra and ket indices belonging to the same discarded factor.

In a product basis, write matrix elements as

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

Then

$$(\operatorname{Tr}_B X_{AB})_{ii'} = \sum_j (X_{AB})_{ij,i'j}.$$

The repeated $B$ index is set equal and summed; the $A$ indices remain. Tracing out $A$ instead contracts $i$ with $i'$ and leaves the $B$ indices.

Structural properties

For operators in the appropriate trace class, the partial trace is:

• linear;
• positive and completely positive;
• trace preserving from the joint operator space to the retained operator space;
• Hermiticity preserving;
• compatible with successive reductions over distinct factors.

In particular,

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

Therefore a positive, unit-trace $\rho_{AB}$ reduces to a positive, unit-trace $\rho_A$.

A Calculation Protocol

1. Declare factor and basis order

Write the Hilbert space and product basis explicitly, for example

$$\begin{aligned} \mathcal H_{AB} &= \mathcal H_A\otimes\mathcal H_B, \\ \lvert i,j\rangle &= \lvert i\rangle_A\otimes\lvert j\rangle_B. \end{aligned}$$

For two qubits, state whether the matrix order is $\lvert00\rangle,\lvert01\rangle,\lvert10\rangle,\lvert11\rangle$. Most index mistakes begin with an unstated ordering convention.

2. Identify the retained subsystem

To find $\rho_A$, trace over $B$. To find $\rho_B$, trace over $A$. Say this in words before manipulating indices.

3. Choose the most transparent representation

• Use ket-bra linearity for short sums of product projectors.
• Use the index contraction for symbolic tensors.
• Use block matrices for small finite-dimensional density matrices.
• Use a reshape-and-contract routine for numerical arrays, with dimensions and factor order supplied explicitly.

4. Run invariant checks

Verify

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

For a pure bipartite input, the nonzero spectra of $\rho_A$ and $\rho_B$ must agree. This provides a strong independent check through the Schmidt decomposition.

Three States, Two Identical Marginals

The contrast among product, classically correlated, and entangled states shows both the power and the limits of reduction.

Product state

For

$$\rho_{AB} = \lvert0\rangle\langle0\rvert_A \otimes \lvert1\rangle\langle1\rvert_B,$$

the marginals are

$$\rho_A=\lvert0\rangle\langle0\rvert, \qquad \rho_B=\lvert1\rangle\langle1\rvert.$$

Both are pure because the global pure state factorizes.

Classically correlated state

Consider

$$\rho_{\mathrm{cc}} = \frac12 \left( \lvert00\rangle\langle00\rvert + \lvert11\rangle\langle11\rvert \right).$$

Tracing over either qubit gives

$$\rho_A=\rho_B=\frac{I}{2}.$$

The local states are maximally mixed, but the joint state is separable and carries classical correlation.

Bell state

For

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

one again finds

$$\rho_A=\rho_B=\frac{I}{2}.$$

This time the global state is pure and entangled. The same local marginals therefore arise from two globally different situations:

Joint state	Local states	Joint information omitted by reduction
product projector	pure	none between the factors
diagonal separable mixture	maximally mixed	classical correlation
Bell projector	maximally mixed	entanglement and joint coherence

The example demonstrates a general rule: marginals do not determine the joint state.

Discarding Is Not Conditioning

If the outcome of a measurement on $B$ is ignored, the state available for local predictions on $A$ is the reduced state $\rho_A$. No outcome label appears.

If a projective measurement $\{Q_b\}$ is performed on $B$ and outcome $b$ is learned, define the unnormalized conditional operator

$$\begin{aligned} \widetilde\rho_{A|b} &= \operatorname{Tr}_B \left[ (I_A\otimes Q_b) \rho_{AB} \right. \\ &\qquad\left. \times (I_A\otimes Q_b) \right]. \end{aligned}$$

Its trace is the outcome probability,

$$p_b = \operatorname{Tr}_A\widetilde\rho_{A|b},$$

and, when $p_b>0$, the normalized conditional state is

$$\rho_{A|b} = \frac{\widetilde\rho_{A|b}}{p_b}.$$

The distinction is informational:

Situation	State assigned to $A$	Required information
no measurement on $B$, or outcome ignored	$\rho_A$	none from $B$
outcome $b$ learned	$\rho_{A\vert b}$	classical record $b$
all outcomes averaged	$\sum_b p_b\rho_{A\vert b}=\rho_A$	record discarded

Conditioning can change the state assigned by an observer who learns $b$. It does not permit faster-than-light signaling because an observer at $A$ who lacks the record still uses the unchanged average $\rho_A$. Quantum Operations supplies the instrument formalism; here the essential composite-system distinction is between an outcome-conditioned state and the unchanged nonselective marginal.

Marginals Do Not Contain Correlations

The pair $(\rho_A,\rho_B)$ fixes all separate local statistics but generally not the expectation of a product observable $X_A\otimes Y_B$. Define the connected correlation

$$\begin{aligned} C(X_A,Y_B) &= \langle X_A\otimes Y_B\rangle \\ &\quad- \langle X_A\rangle \langle Y_B\rangle. \end{aligned}$$

Every connected correlation vanishes in a product state, but checking one pair of observables is not enough to prove product structure.

The quantum mutual information

$$I(A{:}B) = S(\rho_A)+S(\rho_B)-S(\rho_{AB})$$

measures total correlation. It vanishes exactly for product states in finite dimensions, but it does not by itself separate classical correlation from entanglement. That separation requires the joint-state classification developed in the state-classification chapter, not the marginals alone.

Purification Reverses the Viewpoint

Reduction maps a larger state to a subsystem state. Purification asks the inverse existence question: can a mixed state be represented as the reduction of a larger pure state?

If

$$\rho_A = \sum_r \lambda_r \lvert r\rangle\langle r\rvert_A,$$

then a canonical purification on $A\otimes R$ is

$$\lvert\Psi\rangle_{AR} = \sum_r\sqrt{\lambda_r}\, \lvert r\rangle_A\lvert r\rangle_R.$$

Direct reduction gives

$$\operatorname{Tr}_R \lvert\Psi\rangle\langle\Psi\rvert_{AR} = \rho_A.$$

Purifications are not unique. Once the reference space is large enough, purifications of the same $\rho_A$ are related by an isometry on the reference. That isometry changes the auxiliary representation without changing the reduced state or any prediction made from $\rho_A$.

Purification is a representation theorem, not a claim that every mixed state has one uniquely identifiable hidden environment. Whether the reference is physical, hypothetical, or computational depends on the problem.

Interpreting Subsystem Entropy

The von Neumann entropy of a reduced state is

$$S(A) = -\operatorname{Tr}(\rho_A\log\rho_A).$$

Its interpretation depends on the global state.

Pure global state

If $\rho_{AB}$ is pure, then

$$S(A)=S(B),$$

and this common value is the bipartite entanglement entropy. It vanishes exactly for a product pure state.

Mixed global state

If $\rho_{AB}$ is mixed, $S(A)$ can reflect local statistical mixing, classical correlation, quantum correlation, or combinations of them. It is not by itself an entanglement measure. The classically correlated state and Bell state above both give $S(A)=\log2$ despite different entanglement classifications.

The two cases above are the essential interpretation rule: subsystem entropy is an entanglement measure for a bipartite pure global state, but not for an arbitrary mixed global state.

Multiple Subsystems

For a tripartite state $\rho_{ABC}$, reductions can be nested:

$$\begin{aligned} \rho_{AB} &= \operatorname{Tr}_C\rho_{ABC}, \\ \rho_A &= \operatorname{Tr}_B\rho_{AB} \\ &= \operatorname{Tr}_{BC}\rho_{ABC}. \end{aligned}$$

Traces over distinct factors commute:

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

What must not change silently is the labeling and ordering of factors in the chosen matrix or tensor representation.

For a local unitary acting only on the discarded subsystem,

$$\begin{aligned} &\operatorname{Tr}_B \left[ (I_A\otimes U_B) X_{AB} \right. \\ &\qquad\left. \times (I_A\otimes U_B^\dagger) \right] = \operatorname{Tr}_B X_{AB}. \end{aligned}$$

The retained state is invariant because the partial trace is basis independent and $U_B$ merely changes the discarded basis.

Continuous Variables and Infinite Dimensions

In a position basis for two particles, a density kernel may be written

$$\rho(x_A,x_B;x_A',x_B').$$

Tracing out $B$ contracts its two arguments:

$$\rho_A(x_A;x_A') = \int dx_B\, \rho(x_A,x_B;x_A',x_B).$$

The notation is the continuous counterpart of the finite index sum. Functional-analytic care is required: the kernel must represent a trace-class operator, generalized position kets are distributions, and the diagonal contraction must exist in the appropriate operator sense rather than only as a formal integral.

Common Mistakes

• Tracing over the subsystem to be kept. Name the output before contracting indices.
• Using inconsistent basis order. A correct block formula with the wrong product-basis order gives the wrong marginal.
• Taking an elementwise trace. The partial trace contracts matched tensor indices; it is not deletion of arbitrary rows and columns.
• Assuming reduction preserves purity. A pure entangled state has mixed reduced states.
• Inferring the global state from marginals. Different joint states can share every one-body marginal.
• Confusing discarding with conditioning. A postselected state requires an outcome record and normalization.
• Calling local entropy entanglement in a mixed joint state. That identification is valid for pure bipartite states, not generally.
• Treating a basis formula as basis dependent. The calculation uses a basis; the map does not.
• Dropping domain assumptions in infinite dimensions. Trace-class conditions matter.

Reading paths

Core calculation. Read Density Operators → Entangled States → Partial Trace.

Processes and information. Continue to Quantum Operations and then No-Broadcasting Theorem.

References

• J. von Neumann, Mathematical Foundations of Quantum Mechanics, Princeton University Press, 1955.
• A. Peres, Quantum Theory: Concepts and Methods, Kluwer, 1995.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press, 2010.
• J. Preskill, Lecture Notes for Physics 229: Quantum Information and Computation, California Institute of Technology.
• J. Watrous, The Theory of Quantum Information, Cambridge University Press, 2018.
• M. M. Wilde, Quantum Information Theory, 2nd ed., Cambridge University Press, 2017.
• I. Bengtsson and K. Życzkowski, Geometry of Quantum States, 2nd ed., Cambridge University Press, 2017.

Exercises

Exercise 1: Trace a product operator

Let $X_{AB}=A\otimes B$, where $A$ and $B$ are finite-dimensional operators. Show that

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

Solution

Choose an orthonormal basis $\{\lvert j\rangle_B\}$. Then

$$\begin{aligned} \operatorname{Tr}_B(A\otimes B) &= \sum_j (I_A\otimes\langle j\rvert) (A\otimes B) \\ &\quad\times (I_A\otimes\lvert j\rangle) \\ &= \sum_j A\langle j\rvert B\lvert j\rangle \\ &= A\operatorname{Tr}B. \end{aligned}$$

For density operators $\rho_A\otimes\rho_B$, this gives $\operatorname{Tr}_B(\rho_A\otimes\rho_B)=\rho_A$ because $\operatorname{Tr}\rho_B=1$.

Exercise 2: A block-matrix partial trace

In the ordered basis $\lvert00\rangle,\lvert01\rangle,\lvert10\rangle,\lvert11\rangle$, write a two-qubit operator as a $2\times2$ array of $2\times2$ blocks,

$$X_{AB} = \begin{pmatrix} B_{00} & B_{01} \\ B_{10} & B_{11} \end{pmatrix},$$

where each block acts on $B$. Express $\operatorname{Tr}_B X_{AB}$.

Solution

The outer block labels are the retained $A$ indices, while the trace contracts the two $B$ indices inside each block. Therefore

$$\operatorname{Tr}_B X_{AB} = \begin{pmatrix} \operatorname{Tr}B_{00} & \operatorname{Tr}B_{01} \\ \operatorname{Tr}B_{10} & \operatorname{Tr}B_{11} \end{pmatrix}.$$

This rule depends on the declared factor and basis order. With a different ordering, the same numerical matrix must be regrouped differently before tracing.

Exercise 3: Discarded local unitaries

Prove for finite-dimensional $B$ that

$$\begin{aligned} &\operatorname{Tr}_B \left[ (I_A\otimes U_B) \rho_{AB} \right. \\ &\qquad\left. \times (I_A\otimes U_B^\dagger) \right] = \operatorname{Tr}_B\rho_{AB}. \end{aligned}$$

Solution

Evaluate the partial trace in a basis $\{\lvert j\rangle_B\}$:

$$\begin{aligned} \rho_A' &= \sum_j (I_A\otimes\langle j\rvert U_B) \rho_{AB} (I_A\otimes U_B^\dagger\lvert j\rangle). \end{aligned}$$

The vectors $\lvert\widetilde j\rangle=U_B^\dagger\lvert j\rangle$ form another orthonormal basis. Hence the sum is simply the basis-independent definition of $\operatorname{Tr}_B\rho_{AB}$.

Exercise 4: Conditioning versus averaging

For the Bell state $\lvert\Phi^+\rangle$, qubit $B$ is measured in the computational basis. Find the two conditional states of $A$ and show that their probability-weighted average equals the original reduced state.

Solution

Outcome $0$ occurs with probability $1/2$ and leaves $A$ in $\lvert0\rangle\langle0\rvert$. Outcome $1$ also occurs with probability $1/2$ and leaves $A$ in $\lvert1\rangle\langle1\rvert$. The averaged state is

$$\begin{aligned} \sum_b p_b\rho_{A|b} &= \frac12\lvert0\rangle\langle0\rvert + \frac12\lvert1\rangle\langle1\rvert \\ &= \frac{I}{2} = \rho_A. \end{aligned}$$

Someone who learns the outcome uses a pure conditional state; someone without that record uses the unchanged maximally mixed reduced state.

Exercise 5: Nested reductions

Let $\rho_{ABC}$ be a tripartite density operator. Show from the product-basis index formula that tracing out $B$ and then $C$ gives the same state on $A$ as tracing out $BC$ at once.

Solution

Write the matrix elements as $\rho_{ijk,i'j'k'}$. Tracing over $B$ and $C$ successively gives

$$\begin{aligned} (\rho_A)_{ii'} &= \sum_k\sum_j \rho_{ijk,i'jk} \\ &= \sum_{j,k} \rho_{ijk,i'jk}. \end{aligned}$$

The second line is exactly the contraction obtained by treating $(j,k)$ as a joint basis index for $BC$. Finite sums commute, so the order of tracing distinct factors does not matter.