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.

This chapter is a classification guide. Its central workflow is

$$\begin{gathered} \text{declare the partition} \\ \downarrow \\ \text{identify the state type} \\ \downarrow \\ \text{apply the suitable criterion} \\ \downarrow \\ \text{interpret the result}. \end{gathered}$$

The detailed constructions and proofs live in the linked pages. Here the emphasis is on choosing the right definition, test, and conclusion.

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.

Chapter map

Question	Current route	Outcome
What does entanglement mean for pure and mixed states?	Entangled States	product, separable, and entangled classifications across a stated split
How are local states extracted?	Reduced States and Partial Trace	subsystem interpretation and reduction workflow
How do these distinctions enter Bell tests?	Bell Locality and Quantum Correlations	entanglement, Bell nonlocality, and signaling kept distinct

State Classes

Fix a bipartite Hilbert space

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

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

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

for density operators $\rho_A$ and $\rho_B$. It contains no correlations between the chosen subsystems.

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 ordinary shared randomness. Consequently, a separable state can have correlations even though it has no entanglement.

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.

The Schmidt decomposition unifies the criteria. In Schmidt form,

$$\begin{aligned} \lvert\Psi\rangle &= \sum_{r=1}^{R} \sqrt{\lambda_r}\, \lvert u_r\rangle_A \lvert v_r\rangle_B, \\ \lambda_r&>0, \qquad \sum_r\lambda_r=1. \end{aligned}$$

The state is product exactly when the Schmidt rank is $R=1$. Its reduced states have the same nonzero eigenvalues $\{\lambda_r\}$, so either marginal diagnoses the pure-state entanglement.

Mixed-State Decision Test

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:

$$\begin{aligned} \operatorname{Cov}(X_A,Y_B) &= \langle X_A\otimes Y_B\rangle \\ &\quad- \langle X_A\rangle\langle Y_B\rangle =0. \end{aligned}$$

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

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

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,

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

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.

A Reliable Classification Protocol

When presented with a composite state, record the following before assigning a label.

1. State the partition

Write the ordered factors and their physical meaning:

$$\mathcal H = \mathcal H_A\otimes\mathcal H_B.$$

If the factors are modes, regions, particles, or degrees of freedom, say so explicitly.

2. Identify the state representation

Determine whether the object is a normalized vector, a rank-one projector, or a genuinely mixed density operator. Do not apply a pure-state coefficient-rank test to an arbitrary density matrix.

3. Use the strongest exact test available

For a pure bipartite state, use Schmidt rank, coefficient-matrix rank, or reduced-state purity. For a mixed state, first test product structure, then seek a separable decomposition or a dimension-appropriate entanglement criterion.

4. Separate diagnosis from quantification

Showing that a state is entangled is a yes-or-no classification. Assigning an amount of entanglement requires a specified measure and operational setting. Entanglement entropy is canonical for bipartite pure states but is not a universal mixed-state measure.

5. State the scope of the conclusion

Name the partition, dimensions, assumptions, and criterion. A good conclusion reads, for example: “The two-qubit state is entangled across $A|B$ because its partial transpose has a negative eigenvalue.”

Common Mistakes

• Reading entanglement from notation. A sum of product-basis vectors may still factor after algebraic simplification.
• Calling every nonproduct mixed state entangled. Separable mixtures are generally nonproduct and may be correlated.
• Using mixed marginals as a mixed-state test. Pure global states allow this shortcut; general mixed states do not.
• Equating correlation with entanglement. Correlation in one basis can arise from shared classical randomness.
• Treating a failed search as a proof. Not finding a separable decomposition does not establish nonexistence.
• Ignoring local basis invariance. Local-unitary changes can alter coefficients while preserving entanglement.
• Leaving the partition implicit. Entanglement is defined relative to specified subsystems.
• Confusing exchange symmetry with usable entanglement. Identical-particle states require operationally meaningful modes or algebras.

Reading paths

Classification. Read Density Operators and then Entangled States.

Subsystem consequences. Continue to the Reduced States and Partial Trace gateway and Partial Trace.

Foundations. Use Bell Locality and Quantum Correlations only after the state classification is secure.

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.
• R. F. Werner, “Quantum States with Einstein–Podolsky–Rosen Correlations Admitting a Hidden-Variable Model,” Physical Review A 40, 4277–4281, 1989.
• 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.
• I. Bengtsson and K. Życzkowski, Geometry of Quantum States, 2nd ed., Cambridge University Press, 2017.

Exercises

Exercise 1: 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.

Exercise 2: 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.

Exercise 3: Local-unitary invariance

Prove that if $\rho_{AB}$ is separable, then

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

is separable. Why does this also show that a local unitary cannot turn an entangled state into a separable one?

Solution

Write a separable decomposition

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

Then

$$\begin{aligned} \rho_{AB}' &= \sum_k p_k\, \left(U_A\rho_A^{(k)}U_A^\dagger\right) \\ &\quad\otimes \left(U_B\rho_B^{(k)}U_B^\dagger\right). \end{aligned}$$

Every transformed factor is a density operator, so this is a separable decomposition. If a local unitary could map an entangled state to a separable state, applying $U_A^\dagger\otimes U_B^\dagger$ would map that separable state back to the original entangled state, contradicting the result just proved.

Exercise 4: 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.

Exercise 5: What must be specified?

A vector is reported to be “entangled,” but no subsystem decomposition or experimental access model is given. What information is missing, and what change of description would leave the classification invariant?

Solution

One must specify a factorization such as $\mathcal H=\mathcal H_A\otimes\mathcal H_B$ and explain what $A$ and $B$ physically represent. For identical particles or field modes, the accessible observables or mode partition may also need to be stated.

Independent basis changes within $A$ and $B$ are local unitaries and leave the product, separable, or entangled classification invariant. Replacing the physical subsystem split by a genuinely different factorization can change the classification.