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For a bipartite quantum state $\rho_{AB}$, if its reduced state on party A, $\rho_A$, is maximally mixed, then $\rho_{AB}$ is a convex combination of maximally entangled states and product state $I/d\otimes \rho_B$? It seems true. But how to prove it?

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    $\begingroup$ Also another conter-example: $\frac{1}{2}\rho _1\otimes |0\rangle \langle 0|+\frac{1}{2}\rho _2\otimes |1\rangle \langle 1|$. $\endgroup$
    – narip
    Commented Jun 6, 2023 at 6:34

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I don't think this is true. Let $$ |\psi\rangle=\alpha|00\rangle+\beta|11\rangle, $$ and $$ \rho=\frac12|\psi\rangle\langle\psi|+\frac{1}{2}(|\beta|^2|00\rangle\langle 00|+|\alpha|^2|11\rangle\langle 11|). $$ $\rho$ is maximally mixed on either qubit individually, but is not a mixture of a maximally entangled state and the maximally mixed state.

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