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

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?

• What do you mean by "similar argument"? Mar 27, 2019 at 8:42
• @Norbert Schuch I mean an argument that proves that the reduced state of a mixed entangled state is always mixed. Mar 27, 2019 at 9:30
• But this doesn't teach us anything: The reduced state of a non-entangled mixed states will also be mixed (with few exceptions). Mar 27, 2019 at 9:44
• @NorbertSchuch yes, that's true. But I was wondering if either one can prove the statement that all entangled mixed states have reduced states that are mixed or find a counterexample of an entangled mixed state with a reduced state that is pure. Mar 27, 2019 at 10:50
• Then, PLEASE, ask that CLEARLY in your question! -- Other than that, the statement is true. Mar 27, 2019 at 12:57

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.)