Can the reduced state of a mixed entangled state be pure?

Question

For an entangled pure state, the Schmidt decomposition is such that there are at least two non-zero Schmidt coefficients. Tracing out one subsystem implies that the other subsystem is mixed.

Explicitly, we have

$$\psi = \sum_k\sqrt{\lambda_k}\vert k\rangle_A \vert k\rangle_B$$

$$\rho = \vert\psi\rangle\langle\psi\vert = \sum_{k,k'}\sqrt{\lambda_k}\sqrt\lambda_{k'}\vert k\rangle_A \vert k\rangle_B\langle k\vert_A\langle k'\vert_B$$

Taking the partial trace gives us a sum of the form below with at least two nonzero terms. $$\rho_A = \sum_k \lambda_k \vert k\rangle\langle k\vert$$

This is a diagonal matrix with rank 2 or more and is hence a mixed state.

Is there a similar argument one can make for the case where $\rho$ is a mixed entangled state? Alternatively, if this is not true, can one provide a counter example for which the reduced state is pure but the state $\rho$ is still entangled?

Answer 1

A mixed entangled state is a mixture of pure entangled states, $$ \rho = \sum p_i |\psi_i\rangle\langle \psi_i| $$ with (at least some) $|\psi_i\rangle$ entangled. We then have $$ \rho_A = \mathrm{tr}_B \rho = \sum p_i \rho^A_i\ , $$ with $\rho^A_i = \mathrm{tr}_B |\psi_i\rangle\langle \psi_i|$. Since at least one $|\psi_i\rangle$ is entangled, at least one $\rho^A_i$ has $\mathrm{rank}(\rho^A_i)\ge 2$. Since all $\rho^A_i\ge0$, we have that $$ \mathrm{rank}(\sum p_i\rho^A_i)\ge \mathrm{max}(\mathrm{rank}(\rho^A_i))\ge 2\ . $$

Thus, for any entangled mixed state, the reduced state $\rho_A$ is mixed as well. (Note, however, that the same is true for almost all separable mixed states as well.)