Product, Separable, and Entangled States

The words product, separable, and entangled answer different questions about a composite state. The distinctions are elementary for pure bipartite states and substantially harder for mixed states. They also make sense only after the physical subsystem decomposition has been specified.

Required background. Density Operators supplies the distinction between pure and mixed states and the trace rule. Familiarity with tensor-product linear algebra is assumed for expansions in product bases.

Helpful background. Correlations and Covariance supplies the distinction between marginal statistics and joint correlations.

Compare three kinds of joint state

Question	Read
When do local statistics factor?	Product States
How can shared randomness correlate two systems?	Separable Mixed States
What makes a state entangled?	Entangled States
Which correlations establish entanglement?	Classical Correlation versus Entanglement
What are the standard two-qubit examples?	Bell States and Singlet and Triplet States
What changes when we change coordinates or subsystems?	Local Unitary Equivalence and Entanglement Depends on a Decomposition

Product and separable density operators

First take finite-dimensional factors of dimension at least two, with a specified physical split

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

A density operator $\rho_{AB}$ is a product state if

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

for density operators $\rho_A$ and $\rho_B$. Every pair of local measurements has independent outcome statistics in a product state.

A density operator is separable if it admits at least one decomposition

$$\begin{aligned} \rho_{AB} &= \sum_k p_k\, \rho_A^{(k)}\otimes\rho_B^{(k)}, \\ p_k&\geq0, \qquad \sum_k p_k=1. \end{aligned}$$

The label $k$ may carry shared randomness: sample $k$, then prepare the indicated local states. Consequently a separable state can be correlated. This definition does not imply that it has zero quantum discord, which is a different way to characterize correlations.

A density operator is entangled if it is not separable: no convex decomposition of the preceding form exists. Thus

$$\begin{aligned} \{\text{product states}\} &\subsetneq \{\text{separable states}\}, \\ \{\text{separable states}\} &\subsetneq \{\text{all states}\}. \end{aligned}$$

For a pure state $\rho_{AB}=\lvert\Psi\rangle\langle\Psi\rvert$, separability reduces to factorization:

$$\lvert\Psi\rangle = \lvert\psi\rangle_A\otimes\lvert\phi\rangle_B.$$

There is no separate class of correlated but separable pure states. A bipartite pure state is either product or entangled.

Pure-State Decision Test

Expand a normalized pure state in a product basis:

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

For finite-dimensional bipartite systems, the following conditions are equivalent:

$$\begin{aligned} \lvert\Psi\rangle\text{ is product} &\iff \operatorname{rank}C=1, \\ &\iff \operatorname{rank}\rho_A=1, \\ &\iff \operatorname{Tr}(\rho_A^2)=1, \\ &\iff S(\rho_A)=0, \end{aligned}$$

where

$$\rho_A = \operatorname{Tr}_B \lvert\Psi\rangle\langle\Psi\rvert$$

and $S(\rho)=-\operatorname{Tr}(\rho\log\rho)$. If any one of these tests fails, the pure state is entangled across the $A|B$ split.

For the singular-value construction behind these equivalences, see Schmidt Decomposition.

Mixed-state tests need different information

For a mixed state, a mixed marginal does not diagnose entanglement. The marginal can be mixed because of classical mixing, entanglement, or both.

A useful sequence is:

1. Verify that $\rho_{AB}$ is positive semidefinite and has unit trace.
2. Compute $\rho_A=\operatorname{Tr}_B\rho_{AB}$ and $\rho_B=\operatorname{Tr}_A\rho_{AB}$.
3. Test whether $\rho_{AB}=\rho_A\otimes\rho_B$. Equality proves the state is product.
4. If the state is not product, look for an explicit separable decomposition. Finding one proves separability.
5. If no decomposition is apparent, use a criterion valid for the state’s dimensions and assumptions. Failure to find a decomposition is not itself proof of entanglement.

Mixed-state separability is a convex-membership problem. No single elementary scalar test solves it in arbitrary dimensions. Several common tools have deliberately limited scopes:

Tool	What a positive result establishes	Scope or limitation
explicit product decomposition	separability	sufficient in every finite dimension, but may be hard to find
partial transpose with a negative eigenvalue	entanglement	sufficient in every finite bipartite dimension
positive partial transpose	separability only in special dimensions	necessary in general; sufficient for $2\times2$ and $2\times3$ systems
entanglement witness with negative expectation	entanglement	detects only states seen by that witness
concurrence	two-qubit entanglement and an associated formation measure	not a general high-dimensional criterion

Use each tool only within the scope stated in the table. In particular, a positive partial transpose does not prove separability beyond $2\times2$ and $2\times3$, and failure of one witness to detect a state is inconclusive.

Correlation Is Not the Same as Entanglement

Let $X_A$ and $Y_B$ be local observables. In a product state,

$$\langle X_A\otimes Y_B\rangle = \langle X_A\rangle \langle Y_B\rangle,$$

so every connected correlation vanishes:

$$\operatorname{Cov}(X_A,Y_B) = \langle X_A\otimes Y_B\rangle - \langle X_A\rangle\langle Y_B\rangle =0.$$

The converse is false for one chosen pair of observables. Vanishing covariance for $X_A$ and $Y_B$ does not prove that the state is product; other observables may reveal correlations.

For finite-dimensional systems, the quantum mutual information

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

vanishes exactly for product states. It measures total correlation, not entanglement alone. A separable state can have $I(A{:}B)>0$.

A decisive two-qubit comparison

Consider the classically correlated state

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

and the Bell state density operator

$$\rho_{\Phi^+} = \lvert\Phi^+\rangle\langle\Phi^+\rvert, \qquad \lvert\Phi^+\rangle = \frac{\lvert00\rangle+\lvert11\rangle}{\sqrt2}.$$

Both states give perfectly matched outcomes in the computational basis, and both have maximally mixed one-qubit marginals:

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

Their difference lies in joint coherence. Expanding the Bell projector gives

$$\rho_{\Phi^+} =\frac12\bigl( \lvert00\rangle\langle00\rvert + \lvert00\rangle\langle11\rvert + \lvert11\rangle\langle00\rvert + \lvert11\rangle\langle11\rvert \bigr).$$

The off-diagonal terms are absent from $\rho_{\mathrm{cc}}$. A measurement in another local basis exposes the difference; for Pauli observables,

$$\begin{aligned} \operatorname{Tr} \left( \rho_{\mathrm{cc}}\, \sigma_x\otimes\sigma_x \right)&=0, \\ \operatorname{Tr} \left( \rho_{\Phi^+}\, \sigma_x\otimes\sigma_x \right)&=1. \end{aligned}$$

This example blocks two common shortcuts: identical marginals do not determine the joint state, and strong correlation in one basis does not establish entanglement.

Local Changes and Entangling Dynamics

A local unitary has the form $U_A\otimes U_B$. It transforms a state as

$$\rho_{AB}' = (U_A\otimes U_B) \rho_{AB} (U_A^\dagger\otimes U_B^\dagger).$$

Local unitaries preserve product, separable, and entangled classes. For a separable state,

$$\rho_{AB}' = \sum_k p_k \left( U_A\rho_A^{(k)}U_A^\dagger \right) \otimes \left( U_B\rho_B^{(k)}U_B^\dagger \right),$$

which remains an explicit separable decomposition. Applying the inverse local unitary proves the converse.

A genuinely joint unitary need not preserve the classes. For example,

$$\operatorname{CNOT} \left(H\otimes I\right) \lvert00\rangle = \lvert\Phi^+\rangle.$$

The input is product and the output is entangled. A product of local unitaries cannot make this change because it preserves factorization; the interaction term can because its evolution is not of the form $U_A\otimes U_B$.

Entanglement Is Relative to a Subsystem Split

The statement “this state is entangled” is incomplete unless the tensor-product decomposition is understood. A state may be product with respect to one physically meaningful partition and entangled with respect to another.

Three points must be kept separate:

• Changing local bases does not change entanglement. This is a local-unitary transformation within a fixed partition.
• Changing the subsystem decomposition can change entanglement. Particle, mode, spatial-region, center-of-mass, and relative-coordinate factorizations need not classify a vector identically.
• A mathematical refactorization is not automatically physical. The partition should correspond to preparation, control, measurement, locality, conserved structure, or an accessible operator algebra.

This dependence is not a defect in the definition. Entanglement describes a relation among subsystems, so the physically chosen tensor-factor or operator-algebra split is part of the question. A mathematical refactorization alone does not create a new laboratory partition.

Identical particles require additional care. Formal particle labels introduced before symmetrization or antisymmetrization are not automatically operational subsystems. Mode or region partitions often provide the physically meaningful alternative, and exchange symmetry by itself must not be counted as operationally accessible particle entanglement.

Bell States and Coupled Spins

The four Bell states form an orthonormal basis of two-qubit space. Each has Schmidt coefficients $(1/\sqrt2,1/\sqrt2)$ and is maximally entangled for the two-qubit partition.

For two spin-$1/2$ systems, the coupled basis reorganizes the same four-dimensional space into a spin-one triplet and a spin-zero singlet:

Coupled state	Product or entangled?	Bell-state relation
$\lvert1,1\rangle=\lvert\uparrow\uparrow\rangle$	product	none
$\lvert1,0\rangle$	entangled	symmetric Bell state
$\lvert1,-1\rangle=\lvert\downarrow\downarrow\rangle$	product	none
$\lvert0,0\rangle$	entangled	antisymmetric Bell state

Thus “triplet” does not mean “entangled”: two members are product in the chosen spin partition and one is entangled. Total-spin labels classify angular-momentum transformation properties; product versus entangled classifies factorization across the chosen two-spin split.

Exercises

Classify four states

Classify each two-qubit state as product, separable but nonproduct, or entangled:

$$\begin{aligned} \rho_1&=\lvert01\rangle\langle01\rvert, \\ \rho_2&=\frac12\lvert00\rangle\langle00\rvert +\frac12\lvert11\rangle\langle11\rvert, \\ \rho_3&=\lvert\Phi^+\rangle\langle\Phi^+\rvert, \\ \rho_4&=\frac{I_A}{2}\otimes\lvert0\rangle\langle0\rvert_B. \end{aligned}$$

Solution

$\rho_1$ is a pure product state because $\lvert01\rangle=\lvert0\rangle_A\otimes\lvert1\rangle_B$.

$\rho_2$ is separable because its displayed expression is a convex mixture of product projectors. It is not product: its two outcomes are perfectly correlated, whereas its marginals are both $I/2$.

$\rho_3$ is entangled. The pure vector $\lvert\Phi^+\rangle$ has Schmidt rank two, and either reduced state has purity $1/2$.

$\rho_4$ is product by inspection, even though the $A$ factor is mixed.

Same marginals, different joint states

Show that $\rho_{\mathrm{cc}}$ and $\rho_{\Phi^+}$ have the same one-qubit marginals. Explain why this does not make them equally entangled.

Solution

Taking the partial trace over $B$ gives

$$\operatorname{Tr}_B\rho_{\mathrm{cc}} = \frac12 \left( \lvert0\rangle\langle0\rvert + \lvert1\rangle\langle1\rvert \right) = \frac{I}{2}.$$

The diagonal terms of $\rho_{\Phi^+}$ give the same result, while its cross terms vanish under the partial trace because $\langle0|1\rangle=0$. Symmetry gives $\rho_B=I/2$ for both states.

The joint operators differ. $\rho_{\mathrm{cc}}$ is explicitly a separable mixture, whereas the pure state $\rho_{\Phi^+}$ has Schmidt rank two and is entangled. Marginals do not uniquely determine a joint state.

An entangling circuit

Starting from $\lvert00\rangle$, apply a Hadamard gate to qubit $A$ and then a controlled-NOT with $A$ as control. Verify that the result is entangled.

Solution

The Hadamard gate gives

$$(H\otimes I)\lvert00\rangle = \frac{1}{\sqrt2} \left( \lvert00\rangle+\lvert10\rangle \right).$$

The controlled-NOT maps $\lvert00\rangle$ to itself and $\lvert10\rangle$ to $\lvert11\rangle$, so

$$\operatorname{CNOT}(H\otimes I)\lvert00\rangle = \frac{\lvert00\rangle+\lvert11\rangle}{\sqrt2}.$$

Its coefficient matrix is diagonal with two nonzero entries and therefore has rank two. Equivalently, either reduced state is $I/2$. The output is entangled across the two-qubit partition.

References

• I. Bengtsson and K. Życzkowski, Geometry of Quantum States, 2nd ed., Cambridge University Press, 2017.
• M. Horodecki, P. Horodecki, and R. Horodecki, “Separability of Mixed States: Necessary and Sufficient Conditions,” Physics Letters A 223, 1–8, 1996.
• R. Horodecki, P. Horodecki, M. Horodecki, and K. Horodecki, “Quantum Entanglement,” Reviews of Modern Physics 81, 865–942, 2009.
• 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. von Neumann, Mathematical Foundations of Quantum Mechanics, Princeton University Press, 1955.
• R. F. Werner, “Quantum States with Einstein–Podolsky–Rosen Correlations Admitting a Hidden-Variable Model,” Physical Review A 40, 4277–4281, 1989.