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For a bipartite separable quantum state $\rho$ acting on Hilbert space $H\otimes H'$ with $\dim H=D$ and $\dim H'=D'$, what is the minimum number of separable state needed for a decomposition? That is, what is the smallest $r$ such that

$$\rho=\sum_{i=1}^r A_i\otimes B_i.$$

The problem was widely studied, if you require the decomposition to be pure, i.e., $A_i$ and $B_i$ are pure. A general upper bound for that is $D^2D'^{2}$ (Though for qubit-qubit state, it is 4. )

I am wondering, what if one removes the requirement of $A_i$ and $B_i$ being pure, how does the upper bound change?

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    $\begingroup$ related: quantumcomputing.stackexchange.com/q/13031/55 $\endgroup$
    – glS ♦
    Commented Jul 8 at 17:18
  • $\begingroup$ Thanks for including the URL, this post is about decomposition into pure product states, and my main question is to understand a general decomposition without pure-state-constrain. Or if there is a way to see that they are equivelaent. $\endgroup$
    – Yujie Zhang
    Commented Jul 8 at 19:21
  • $\begingroup$ They're not equivalent. Suppose $\rho$ is the maximally mixed state of two qubits. Then by simply considering the rank of the operators we find $r=4$ with the pure state constraint and $r=1$ without it. $\endgroup$
    – Adam Zalcman
    Commented Jul 9 at 4:35

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