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 decomposition? That is

$\rho=\sum_{i=1}^r A_i\otimes B_i$, find the smallest $r$

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 remove the requirement of $A_i$ and $B_i$ being pure? is there a better upper bound?

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?