Density Operators, Entanglement, and Reduced States

A subsystem can be mixed even when the total state is pure. The reason is not hidden classical ignorance: local predictions discard correlations with the complementary subsystem. This page derives that statement with the partial trace, then separates three ideas that are often conflated—mixed states, classical correlations, and entanglement.

Required background. Calculus and Linear Algebra supplies finite-dimensional vector spaces, inner products, orthonormal bases, eigenvectors, and basis changes; Quantum States, Observables, and Measurement supplies state vectors, positive unit-trace operators, effects, outcome probabilities, and conditional updates.

Density operators encode predictions, not a unique preparation story

Work throughout in finite-dimensional Hilbert spaces. A density operator is a positive, unit-trace operator,

$$\rho=\rho^\dagger, \qquad \rho\ge0, \qquad \operatorname{Tr}\rho=1.$$

An ensemble that prepares $|\psi_k\rangle$ with probability $p_k$ produces

$$\rho=\sum_k p_k|\psi_k\rangle\langle\psi_k|, \qquad p_k\ge0, \qquad \sum_kp_k=1.$$

For a measurement effect $E_a$, the outcome probability is

$$p(a)=\operatorname{Tr}(\rho E_a).$$

Consequently, two ensembles with the same $\rho$ cannot be distinguished by any measurement on that system alone. For a qubit, with $|\pm\rangle=(|0\rangle\pm|1\rangle)/\sqrt2$,

$$\frac I2 =\frac12\left(|0\rangle\langle0|+|1\rangle\langle1|\right) =\frac12\left(|+\rangle\langle+|+|-\rangle\langle-|\right).$$

The two preparation lists differ, but every local probability agrees. Thus $\rho$ specifies operational predictions; it does not select a unique ignorance story. Preskill develops this ensemble ambiguity in Preskill 2015, §2.5.3, Eqs. (2.106)–(2.108), printed pp. 30–31.

The state is pure exactly when it has rank one, equivalently $\rho^2=\rho$ or $\operatorname{Tr}\rho^2=1$. Otherwise $\operatorname{Tr}\rho^2\lt1$, and the state is mixed.

The partial trace preserves every local prediction

If two systems have Hilbert spaces $\mathcal H_A$ and $\mathcal H_B$, their composite state space is $\mathcal H_{AB}=\mathcal H_A\otimes\mathcal H_B$. A product ket is $|a\rangle_A\otimes|b\rangle_B$, abbreviated $|a,b\rangle$; general joint states need not be product states. Let the joint system have state $\rho_{AB}$. The reduced state $\rho_A$ is defined by requiring

$$\operatorname{Tr}_A(\rho_AO_A) =\operatorname{Tr}_{AB}[\rho_{AB}(O_A\otimes I_B)]$$

for every operator $O_A$. In an orthonormal basis $\{|b\rangle_B\}$,

$$\rho_A=\operatorname{Tr}_B\rho_{AB} =\sum_b {}_B\langle b|\rho_{AB}|b\rangle_B.$$

This formula produces another state. For every $|a\rangle_A$,

$$\langle a|\rho_A|a\rangle =\sum_b\langle a,b|\rho_{AB}|a,b\rangle\ge0,$$

so $\rho_A$ is positive, while

$$\operatorname{Tr}_A\rho_A =\sum_{a,b}\langle a,b|\rho_{AB}|a,b\rangle =\operatorname{Tr}_{AB}\rho_{AB}=1.$$

It is also basis independent. If $|b'_\alpha\rangle=\sum_bU_{b\alpha}|b\rangle$, unitarity gives

$$\begin{aligned} \sum_\alpha{}_B\langle b'_\alpha|\rho_{AB}|b'_\alpha\rangle_B &=\sum_{b,c} \left(\sum_\alpha U^*_{b\alpha}U_{c\alpha}\right) {}_B\langle b|\rho_{AB}|c\rangle_B\\ &=\sum_b{}_B\langle b|\rho_{AB}|b\rangle_B. \end{aligned}$$

Preskill derives the reduced state and these local-expectation properties in Preskill 2015, §2.3, Eqs. (2.63)–(2.67), printed pp. 19–20.

Independent Bell-pair check in two bases

For

$$|\Phi^+\rangle =\frac{|00\rangle+|11\rangle}{\sqrt2},$$

tracing over $B$ removes the cross terms because $\langle0|1\rangle=0$:

$$\rho_A =\operatorname{Tr}_B|\Phi^+\rangle\langle\Phi^+| =\frac12\left(|0\rangle\langle0|+|1\rangle\langle1|\right) =\frac{I_A}{2}.$$

The same state is

$$|\Phi^+\rangle =\frac{|++\rangle+|--\rangle}{\sqrt2}.$$

Tracing in the $x$ basis again gives $\rho_A=(|+\rangle\langle+|+|-\rangle\langle-|)/2=I_A/2$. This explicit calculation corroborates the general basis-independence proof.

Schmidt decomposition answers the central question

Every finite-dimensional bipartite pure state has a Schmidt decomposition

$$|\Psi\rangle_{AB} =\sum_{i=1}^{r}\sqrt{\lambda_i}\, |i\rangle_A|\widetilde i\rangle_B, \qquad \lambda_i\gt0, \qquad \sum_i\lambda_i=1.$$

Here $\{|i\rangle_A\}_{i=1}^{r}$ and $\{|\widetilde i\rangle_B\}_{i=1}^{r}$ are orthonormal families, $r$ is the Schmidt rank, and the numbers $\sqrt{\lambda_i}$ are the Schmidt coefficients. Taking the two partial traces gives

$$\rho_A=\sum_i\lambda_i|i\rangle\langle i|, \qquad \rho_B=\sum_i\lambda_i|\widetilde i\rangle \langle\widetilde i|.$$

Thus $\rho_A$ and $\rho_B$ have the same nonzero eigenvalues $\lambda_i$. For a globally pure state, the following statements are equivalent: the Schmidt rank is greater than one, the state is entangled, and either marginal is mixed. This equivalence does not extend to a globally mixed state.

The von Neumann entropy is

$$S(\rho)=-\operatorname{Tr}(\rho\ln\rho).$$

Here $\ln$ is the natural logarithm, entropy is measured in nats, and $0\ln0$ is defined as zero. A bipartite pure state therefore obeys

$$S(\rho_A)=S(\rho_B) =-\sum_i\lambda_i\ln\lambda_i.$$

Preskill proves the Schmidt decomposition and common reduced spectrum in Preskill 2015, §2.4, Eqs. (2.79)–(2.90), printed pp. 23–24.

Classical correlation and entanglement are different

A bipartite mixed state is separable when it can be written as

$$\rho_{AB}^{\rm sep} =\sum_k p_k\,\rho_A^{(k)}\otimes\rho_B^{(k)}.$$

Here $p_k\ge0$, $\sum_kp_k=1$, and every $\rho_A^{(k)}$ and $\rho_B^{(k)}$ is itself a density operator. If no such convex decomposition exists, the state is entangled. Compare the Bell state with a classically correlated mixture,

$$\rho_\Phi=|\Phi^+\rangle\langle\Phi^+|, \qquad \rho_{\rm cc} =\frac12\left(|00\rangle\langle00|+|11\rangle\langle11|\right).$$

Test	Bell state $\rho_\Phi$	Classical mixture $\rho_{\rm cc}$
Global status	Pure and entangled	Mixed and separable
Marginals	$\rho_A=\rho_B=I/2$	$\rho_A=\rho_B=I/2$
Global purity	$\operatorname{Tr}\rho_\Phi^2=1$	$\operatorname{Tr}\rho_{\rm cc}^2=1/2$
$z$ correlation	$\langle\sigma_z\otimes\sigma_z\rangle=1$	$\langle\sigma_z\otimes\sigma_z\rangle=1$
$x$ correlation	$\langle\sigma_x\otimes\sigma_x\rangle=1$	$\langle\sigma_x\otimes\sigma_x\rangle=0$

The $z$ measurement sees matching bits in both states. The $x$ operator, however, exchanges $|00\rangle$ and $|11\rangle$, so it detects the coherent off-diagonal terms of $\rho_\Phi$ that are absent from $\rho_{\rm cc}$. Neither a mixed marginal nor one classical-looking correlation is by itself an entanglement test.

Purification restores a pure state on a larger space

Let the spectral decomposition of a mixed state be

$$\rho_A=\sum_{i=1}^{r}\lambda_i|i\rangle\langle i|.$$

Choose a reference space $R$ of dimension at least $r=\operatorname{rank}\rho_A$ and define

$$|\Psi_\rho\rangle_{AR} =\sum_{i=1}^{r}\sqrt{\lambda_i}\,|i\rangle_A|i\rangle_R.$$

Then

$$\operatorname{Tr}_R|\Psi_\rho\rangle\langle\Psi_\rho| =\sum_i\lambda_i|i\rangle\langle i| =\rho_A.$$

Thus every finite-dimensional mixed state is the reduction of a pure state, and a reference system of dimension $r$ suffices. Purification is not unique: a fixed minimal purification maps into a larger reference space by an isometry; two minimal purifications are related by a unitary on the reference, as are purifications using reference spaces of the same dimension. Preskill derives this construction in Preskill 2015, §2.5.5, Eqs. (2.111)–(2.121), printed pp. 34–35.

Conditioning changes ensembles; ignoring outcomes cannot signal

Let a local instrument on $B$ have Kraus operators $K_\alpha$ satisfying $\sum_\alpha K_\alpha^\dagger K_\alpha=I_B$. The outcome-$\alpha$ subnormalized state at $A$ is

$$p_\alpha\rho_{A|\alpha} =\operatorname{Tr}_B\!\left[ (I\otimes K_\alpha)\rho_{AB}(I\otimes K_\alpha^\dagger) \right].$$

The probability $p_\alpha$ is the trace of this positive operator; when $p_\alpha\gt0$, division by it defines the normalized conditional state $\rho_{A|\alpha}$. Individual conditional states can differ. If the outcome is not communicated, however, the state is the sum over outcomes:

$$\begin{aligned} \sum_\alpha p_\alpha\rho_{A|\alpha} &=\operatorname{Tr}_B\sum_\alpha (I\otimes K_\alpha)\rho_{AB}(I\otimes K_\alpha^\dagger)\\ &=\operatorname{Tr}_B\!\left[ \rho_{AB}\left(I\otimes\sum_\alpha K_\alpha^\dagger K_\alpha\right) \right]\\ &=\operatorname{Tr}_B\rho_{AB}=\rho_A. \end{aligned}$$

The second line uses cyclicity only for operators acting on the subsystem that is traced out; it is not unrestricted cyclicity of an operator-valued partial trace. Hence a trace-preserving local operation at $B$ cannot change unconditional statistics at $A$. Learning which conditional ensemble was selected still requires a classical record. Preskill discusses the ensemble and no-signalling distinction in Preskill 2015, §2.5.3, Eqs. (2.106)–(2.108) and surrounding discussion, printed pp. 30–31.

Common pitfalls

Treating a density operator as one hidden ensemble. A single $\rho$ can have many pure-state decompositions. Only additional preparation information chooses among them.

Calling every mixed marginal entangled. For a globally pure bipartite state, mixedness of a marginal diagnoses entanglement. For a globally mixed state, separability must be tested separately.

Calling every von Neumann entropy thermodynamic. The same functional appears in several settings, but its physical meaning depends on the state, subsystem, regulator, and operational question.

The finite-dimensional boundary toward gravity

Every derivation above assumed an explicit finite tensor product $\mathcal H_A\otimes\mathcal H_B$. It does not establish a sharp spatial factorization in continuum quantum field theory, gauge theory, or gravity; ultraviolet structure and constraints require a more careful algebraic treatment Donnelly and Giddings 2016, Introduction, printed pp. 2–3.

In perturbation theory about flat spacetime in spacetime dimension $D\gt3$, Donnelly and Giddings show that a compactly supported operator carrying nonzero Poincaré charges requires long-range gravitational dressing. This obstructs an automatic exact local subsystem, but it does not rule out useful approximate subsystems in every regime Donnelly and Giddings 2016, §III, printed pp. 6–10. Nothing on this page derives Hawking radiation, black-hole entropy, generalized entropy, or an information paradox.

The next page turns from discarded subsystems to open spatial channels: Scattering, Bound States, and Tunneling.

Exercises

Pure entanglement versus a classical mixture

For $0\le q\le1$, compare

$$|\Psi_q\rangle =\sqrt q\,|00\rangle+\sqrt{1-q}\,|11\rangle$$

with

$$\rho_{\rm cc}(q) =q|00\rangle\langle00|+(1-q)|11\rangle\langle11|.$$

Show that both have the same reduced state—and hence the same reduced-state purity and entropy—and the same $zz$ correlation. Then determine when each joint state is entangled, calculate its $xx$ correlation, and compare the two global purities and entropies.

Solution

In both cases the partial trace over $B$ removes any cross term and gives

$$\rho_A =q|0\rangle\langle0|+(1-q)|1\rangle\langle1| =\operatorname{diag}(q,1-q).$$

Therefore

$$\operatorname{Tr}\rho_A^2=q^2+(1-q)^2, \qquad S(\rho_A)=-q\ln q-(1-q)\ln(1-q).$$

The pure state is already in Schmidt form. It is entangled exactly when $0\lt q\lt1$; at either endpoint it is a product state. The mixed state is separable for every $q$ because its displayed expression is already a convex sum of product states.

Globally, the two states differ:

$$\operatorname{Tr}\rho_{\Psi_q}^2=1, \qquad S(\rho_{\Psi_q})=0,$$

whereas

$$\operatorname{Tr}\rho_{\rm cc}^2=q^2+(1-q)^2, \qquad S(\rho_{\rm cc})=-q\ln q-(1-q)\ln(1-q).$$

Both states contain equal computational-basis bits, so

$$\langle\sigma_z\otimes\sigma_z\rangle=1.$$

Since $\sigma_x\otimes\sigma_x$ exchanges $|00\rangle$ and $|11\rangle$,

$$\langle\sigma_x\otimes\sigma_x\rangle_{\Psi_q} =2\sqrt{q(1-q)}, \qquad \langle\sigma_x\otimes\sigma_x\rangle_{\rho_{\rm cc}}=0.$$

The difference is coherence between the two branches, not the local marginal or the $zz$ correlation.

References

• Donnelly, William, and Steven B. Giddings. “Observables, Gravitational Dressing, and Obstructions to Locality and Subsystems.” Physical Review D 94 (2016): 104038. doi:10.1103/PhysRevD.94.104038. Open PDF.
• Preskill, John. Quantum Information, Chapter 2, revised July 2015, §§2.3–2.5, printed pp. 16–35. California Institute of Technology lecture notes. Official PDF.