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

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

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.

Predict measurements on one part

Question	Read
What does a subsystem state mean?	Reduced Density Operators
How do I compute it?	Partial Trace
How does it give probabilities?	Local Measurement Statistics
What changes if a remote outcome is known?	Conditional States
Can the same local state come from a larger pure state?	Purification
What information was lost?	Marginals and Correlations
What does local mixedness measure?	Subsystem Entropy

The local expectation-value rule

Let $M_A$ be any bounded observable on $A$; in finite dimensions every observable is bounded. 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 a trace-class operator $X_{AB}$,

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

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.

Coherence can survive a partial trace

Reduction does not mean deleting every off-diagonal matrix element. Take

$$|\psi\rangle=\frac{|00\rangle+|10\rangle+|11\rangle}{\sqrt3}.$$

The terms $|00\rangle\langle10|$ and its adjoint have the same $B$ index and survive the trace. Terms such as $|00\rangle\langle11|$ have different $B$ indices and vanish. In the $|0\rangle_A,|1\rangle_A$ basis,

$$\rho_A=\frac13\begin{pmatrix}1&1\\1&2\end{pmatrix}.$$

It has unit trace, determinant $1/9$, and purity $7/9$. Since the joint state is pure and the local state is mixed, this state is entangled. Its nonzero off-diagonal entries show that local coherence and local mixedness can coexist.

Discarding Is Not Conditioning

For a local trace-preserving measurement process on $B$, if its outcome 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

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

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 generally do not determine 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

$$C(X_A,Y_B) = \langle X_A\otimes Y_B\rangle - \langle X_A\rangle \langle Y_B\rangle.$$

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

For finite-dimensional systems, 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. In infinite dimension, the entropy subtraction can involve undefined differences of infinities; use the relative-entropy definition explained in mutual information.

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?

For a finite-dimensional reduced state, write

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

A convenient 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. Their occupied reference subspaces are related by an isometry; when the full reference dimensions agree, this can be extended to a unitary on the reference. The local state and every prediction on $A$ stay unchanged.

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. For example, the separable mixture $(|00\rangle\langle00|+|11\rangle\langle11|)/2$ and the pure Bell state $(|00\rangle+|11\rangle)/\sqrt2$ both give $S(A)=\log2$.

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.

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.

Exercises

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.

Discarded local unitaries

Prove for finite-dimensional $B$ that

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

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

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.

References

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