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 finite-dimensional 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.

For infinite-dimensional normal states, separability means the trace-norm closure of finite convex mixtures of product states. A separable state need not then admit an exact countable product ensemble. The rigorous extension is developed in Separable and Entangled Mixed States.

Helpful background. Product states explains factorization; density operators explains why different ensembles can describe the same state.

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 bounded 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 without entanglement. The label is classical, but the conditional local states need not commute or be mutually distinguishable. Therefore “separable” does not mean that every kind of quantum correlation vanishes; see the nonorthogonal example below.

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.

Product, separable, and entangled are different conditions

For pure states, failure to factor means entanglement. For mixed states, a failure to factor can instead result from shared randomness, as in $\rho_{\mathrm{cc}}$.

The sets are nested: every product state is separable, and every separable state is a density operator. Some separable states are product and some are correlated. Entangled states lie 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.

The partial-transpose test is complete for two qubits and qubit–qutrit systems; in higher dimensions it can miss entanglement. An entanglement witness can certify entanglement through an expectation value. Failure to find a separable ensemble by guessing is not itself a criterion.

Nonorthogonal preparations need not be fully classical

Consider

$$\rho_{AB}=\frac12|0\rangle\langle0|\otimes|0\rangle\langle0| +\frac12|1\rangle\langle1|\otimes|+\rangle\langle+|, \qquad |+\rangle=\frac{|0\rangle+|1\rangle}{\sqrt2}.$$

This is separable by its displayed preparation. Measuring $A$ in the $Z$ basis reads the classical label without changing the state when outcomes are ignored. On $B$, however, the two conditional projectors do not commute:

$$[|0\rangle\langle0|,|+\rangle\langle+|] =\frac12\begin{pmatrix}0&1\\-1&0\end{pmatrix}\ne0.$$

No basis makes both conditional states diagonal, so they cannot be read as two orthogonal classical labels on $B$. This is the distinction behind quantum discord: separable states can retain quantum correlations in a sense broader than entanglement. The comparison of classical correlation and entanglement develops this caution without requiring discord as an entanglement test.

Common pitfalls

One product ensemble proves separability. One entangled ensemble does not prove entanglement; another ensemble may consist entirely of product states.

A mixed marginal does not prove mixed-state entanglement. It can reflect local noise or shared classical randomness. The reduced-purity criterion applies when the joint state is known to be pure.

Shared randomness does not make local quantum states orthogonal. Separability allows arbitrary local density operators, including noncommuting ones.

References

• Ryszard Horodecki, Paweł Horodecki, Michał Horodecki, and Karol Horodecki, “Quantum entanglement,” Reviews of Modern Physics 81, 865–942 (2009). DOI.
• Michael A. Nielsen and Isaac L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press, 2010, section 2.5.
• H. Ollivier and W. H. Zurek, “Quantum Discord: A Measure of the Quantumness of Correlations,” Physical Review Letters 88, 017901 (2002 issue; published online 14 December 2001). DOI; preprint.
• John Watrous, The Theory of Quantum Information, Cambridge University Press, 2018, chapter 6. Author-hosted text.

Exercises

1. Prove that a convex mixture $q\rho+(1-q)\sigma$ of two separable states is separable for $0\le q\le1$.

Solution

If $\rho=\sum_i p_i\rho_A^{(i)}\otimes\rho_B^{(i)}$ and $\sigma=\sum_j r_j\sigma_A^{(j)}\otimes\sigma_B^{(j)}$, concatenate the two ensembles with weights $qp_i$ and $(1-q)r_j$. These weights are nonnegative and sum to $q+(1-q)=1$. Thus the separable set is convex. The set of product states is not convex, as $\rho_{\mathrm{cc}}$ demonstrates.

2. For the nonorthogonal example above, compute $\rho_A$ and $\rho_B$ and explain why it is not a product state.

Solution

$$\rho_A=I/2,\qquad \rho_B=\frac12\bigl(|0\rangle\langle0|+|+\rangle\langle+|\bigr) =\begin{pmatrix}3/4&1/4\\1/4&1/4\end{pmatrix}.$$

In a product state, conditioning on either $Z_A$ outcome would leave $B$ in the same state $\rho_B$. Here the two conditional states are $|0\rangle\langle0|$ and $|+\rangle\langle+|$, so the systems are correlated even though the preparation is separable.

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 $p=1$.