If a pure state, $\rho_{AB}$, has subsystems described by mixed density matrices, the overall state is entangled (as far as I understand).

Can you conclude the same with an initially mixed bipartite density matrix i.e. if $\rho_{AB}$ is a mixed density matrix, it is entangled if its subsystems are described by mixed density matrices $\rho_A$ and $\rho_B$? Is there a counterexample; a separable, but mixed, $\rho_{AB}$ that has mixed subsystems?


1 Answer 1


Let $\rho_1$ and $\rho_2$ be mixed density matrices. Then $\rho=\rho_1\otimes\rho_2$ is mixed and separable.

In response to the comment: Let us study the case $$\rho_1=\rho_2 = \frac{1}{2} \,\mathbb I_2 \quad ,$$ where $\mathbb I_2$ is the identity matrix on $H\cong \mathbb C^2$, i.e. these matrices are the maximally mixed density matrices of a qubit system.

Then these are the reduced density matrices of both the maximally entangled two-qubit Bell state

$$\sigma = |\psi^-\rangle \langle \psi^-| $$ and the separable mixed state $$\rho=\rho_1 \otimes \rho_2 \quad . $$ Additionally, consider a Werner state of the form $$\omega = \alpha \,\sigma + (1-\alpha)\, \rho \quad, $$ with some restriction on $\alpha$ such that it is an entangled mixed density matrix. We again find that $\rho_1$ and $\rho_2$ are its reduced density matrices.

In conclusion, we see that the two (mixed) reduced density matrices $\rho_1$ and $\rho_2$ could arise from a pure entangled, mixed separable or mixed entangled state. However, they cannot arise from a pure separable (product) state, since every pure state with those reduced density matrices necessarily has a Schmidt rank greater than one and is thus entangled.

  • $\begingroup$ Is there anything we can conclude about the separability of the $\rho$ if we know $\rho_1$ and $\rho_2$ are mixed? You've shown that $\rho$ can be separable, but can it also be entangled or can we only find out by using something like the Concurrence? $\endgroup$
    – Angus
    Jan 13 at 22:17
  • 1
    $\begingroup$ @Angus Does this help (especially point 3. in the answer)? $\endgroup$ Jan 13 at 22:27
  • $\begingroup$ @Angus I've edited the answer and tried to give an example. Please double check everything. Does this help? $\endgroup$ Jan 13 at 23:14
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    $\begingroup$ It does, thanks for the great answer. $\endgroup$
    – Angus
    Jan 13 at 23:31

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