Separable Mixed States

A separable mixed state is a bipartite density operator that can be prepared as a classical random mixture of product states. It may contain ordinary correlations between subsystems, but it contains no entanglement across the specified subsystem split.

For a bipartite system

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

a mixed state is separable if it can be written as

$$\rho_{AB} = \sum_r p_r\, \rho_A^{(r)}\otimes\rho_B^{(r)}, \qquad p_r\ge0, \qquad \sum_r p_r=1.$$

If no such decomposition exists, the mixed state is entangled.

Product Mixed States

The simplest unentangled mixed state is a product density operator:

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

This represents independent subsystem states. For local observables,

$$\operatorname{Tr} \bigl[ (\rho_A\otimes\rho_B) (O_A\otimes O_B) \bigr] = \operatorname{Tr}(\rho_A O_A)\, \operatorname{Tr}(\rho_B O_B).$$

Thus product mixed states have factorized expectation values for product observables. They have no correlations between $A$ and $B$.

Product mixed states are separable, but they are not the only separable states.

Classical Mixtures of Product States

A separable state may be a mixture of different product preparations. Operationally, one can imagine a classical random variable $r$ being sampled with probability $p_r$, after which subsystem $A$ is prepared in $\rho_A^{(r)}$ and subsystem $B$ is prepared in $\rho_B^{(r)}$.

The resulting density operator is

$$\rho_{AB} = \sum_r p_r\, \rho_A^{(r)}\otimes\rho_B^{(r)}.$$

The shared label $r$ can create correlations. Those correlations are classical in the sense that they arise from common randomness in the preparation, not from nonseparability of the quantum state.

In finite-dimensional systems, one may also write separable states as mixtures of pure product projectors:

$$\rho_{AB} = \sum_r p_r\, \lvert\alpha_r\rangle\langle\alpha_r\rvert \otimes \lvert\beta_r\rangle\langle\beta_r\rvert.$$

This follows by decomposing the local density operators in each product term into pure-state ensembles.

Separable but Correlated

The standard two-qubit example is

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

This state is separable because it is already written as a convex mixture of product projectors:

$$\rho_{\mathrm{cc}} = \frac12 \bigl( \lvert0\rangle\langle0\rvert \otimes \lvert0\rangle\langle0\rvert \bigr) + \frac12 \bigl( \lvert1\rangle\langle1\rvert \otimes \lvert1\rangle\langle1\rvert \bigr).$$

It is not a product state. The reduced density operators are

$$\rho_A = \rho_B = \frac12 I.$$

If it were the product of its reduced states, it would be

$$\rho_A\otimes\rho_B = \frac14 I\otimes I,$$

which assigns probability $1/4$ to each computational-basis outcome. Instead, $\rho_{\mathrm{cc}}$ assigns probability $1/2$ to $00$, probability $1/2$ to $11$, and zero to $01$ and $10$.

Thus

$$P(0,0) = \frac12, \qquad P_A(0)P_B(0) = \frac14.$$

The state is correlated, but not entangled.

Why Not Product Is Not Enough

For pure bipartite states, the definition is simple: product means unentangled, and not product means entangled.

For mixed states, “not product” is too weak. A mixed state can fail to factor as $\rho_A\otimes\rho_B$ simply because it contains classical correlation. The state $\rho_{\mathrm{cc}}$ above is the basic example.

The correct mixed-state distinction is:

$$\text{product} \quad\Longrightarrow\quad \text{separable} \quad\Longrightarrow\quad \text{not necessarily product}.$$

Entangled mixed states are precisely the states outside the separable set.

Nonunique Ensembles

A density operator can have many different ensemble decompositions. This matters because seeing an entangled-state ensemble does not prove that the density operator is entangled.

For example, the maximally mixed two-qubit state is

$$\rho_* = \frac14 I_{AB}.$$

It can be written as a product state:

$$\rho_* = \left(\frac12 I_A\right) \otimes \left(\frac12 I_B\right).$$

So it is separable. But it can also be written as an equal mixture of Bell projectors:

$$\rho_* = \frac14 \sum_{\mu} \lvert B_\mu\rangle\langle B_\mu\rvert,$$

where $\{\lvert B_\mu\rangle\}$ is the Bell basis. The presence of entangled vectors in one ensemble decomposition does not make the density operator entangled. Entanglement of a mixed state is a property of the density operator itself, not of one chosen story about how it was prepared.

Why Mixed-State Entanglement Is Harder

Pure-state entanglement has strong tools: coefficient rank, Schmidt decomposition, and reduced-state entropy.

Mixed-state entanglement is harder because one must rule out every possible separable decomposition:

$$\rho_{AB} \ne \sum_r p_r\, \rho_A^{(r)}\otimes\rho_B^{(r)}$$

for all choices of probabilities and local states.

Useful criteria include partial-transpose tests, entanglement witnesses, semidefinite programs, and special formulas for low-dimensional systems. These belong to a later mixed-state entanglement toolkit. The definition, however, is already fixed: separable means convex mixture of product states.

Examples

A product mixed state is

$$\rho_A\otimes\rho_B.$$

It is separable and uncorrelated.

A diagonal classical two-qubit state

$$\rho = \sum_{i,j=0}^{1} p_{ij} \lvert ij\rangle\langle ij\rvert, \qquad p_{ij}\ge0, \qquad \sum_{ij}p_{ij}=1,$$

is separable because it is a mixture of computational-basis product projectors. It can still be classically correlated if $p_{ij}$ does not factor as $p_i^A p_j^B$.

A Bell pure state such as

$$\lvert\Phi^+\rangle = \frac{1}{\sqrt2} \bigl( \lvert00\rangle+\lvert11\rangle \bigr)$$

is not separable, because for pure states separability reduces to productness.

Common Mistakes

• Calling every nonproduct mixed state entangled.
• Calling every correlated mixed state entangled.
• Treating one ensemble decomposition as the unique meaning of a density operator.
• Thinking that a mixture of entangled kets is automatically an entangled mixed state.
• Forgetting to specify the subsystem split $A|B$.
• Assuming that separable means uncorrelated; separable states may have classical correlations.

Cross-Links

• Product States
• Classical Correlation versus Entanglement
• Entangled States
• Bell States
• Local Unitary Equivalence
• Entanglement Depends on a Decomposition
• Reduced Density Operators
• Partial Trace
• Schmidt Decomposition
• Schmidt Rank
• Concurrence for Two Qubits
• Negativity and PPT Criterion
• Entanglement Witnesses
• LOCC Preview
• Multipartite Separability
• Formula Sheet
• Density Operators
• Pure Versus Mixed States

References

• R. F. Werner, “Quantum States with Einstein-Podolsky-Rosen Correlations Admitting a Hidden-Variable Model,” Physical Review A 40, 4277-4281, 1989.
• A. Peres, Quantum Theory: Concepts and Methods, Kluwer, 1995.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, 2010.
• R. Horodecki, P. Horodecki, M. Horodecki, and K. Horodecki, “Quantum Entanglement,” Reviews of Modern Physics 81, 865-942, 2009.
• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014.

Exercises

1. Show that

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

is separable but not product.

Solution

It is separable because it is a convex mixture of product projectors. Its reduced states are

$$\rho_A=\rho_B=\frac12 I.$$

The product of the reduced states is

$$\rho_A\otimes\rho_B = \frac14 I\otimes I,$$

which is not equal to $\rho_{\mathrm{cc}}$. Therefore the state is separable but not product.

2. Compute the computational-basis probabilities for $\rho_{\mathrm{cc}}$ and show that the outcomes are correlated.

Solution

The nonzero joint probabilities are

$$P(0,0)=\frac12, \qquad P(1,1)=\frac12.$$

The marginal probabilities are

$$P_A(0)=P_B(0)=\frac12.$$

If the outcomes were independent, $P(0,0)$ would equal $P_A(0)P_B(0)=1/4$. Instead $P(0,0)=1/2$, so the outcomes are correlated.

3. Explain why an equal mixture of Bell projectors can still be separable.

Solution

The equal mixture of the four Bell projectors is the maximally mixed state:

$$\rho_* = \frac14 I_{AB}.$$

But this also factors as

$$\rho_* = \left(\frac12 I_A\right) \otimes \left(\frac12 I_B\right).$$

Since it is a product density operator, it is separable. A density operator can have many ensemble decompositions, so one decomposition using entangled kets does not prove mixed-state entanglement.

4. For

$$\rho_p = p\lvert00\rangle\langle00\rvert + (1-p)\lvert11\rangle\langle11\rvert, \qquad 0\le p\le1,$$

for which $p$ is the state product?

Solution

The state is separable for every $p$ because it is a convex mixture of product projectors. Its reduced states are

$$\rho_A=\rho_B = p\lvert0\rangle\langle0\rvert + (1-p)\lvert1\rangle\langle1\rvert.$$

For $0<p<1$, the product $\rho_A\otimes\rho_B$ contains nonzero probabilities for $\lvert01\rangle$ and $\lvert10\rangle$, while $\rho_p$ does not. Thus $\rho_p$ is product only at $p=0$ or $p=1$.