# Are there separable $\rho$ that cannot be decomposed with less than $\operatorname{rank}(\rho)^2$ pure product states?

In What separable $\rho$ only admit separable pure decompositions with more than $\mathrm{rank}(\rho)$ terms?, examples were given of separable states $$\rho$$ with separable decompositions requiring more than $$\operatorname{rank}(\rho)$$ components. In particular, symmetric Werner states provided one such class of examples, having rank $$\binom{n+1}{2}$$, but requiring no less than $$n^2$$ rank-one separable states to be decomposed.

On the other hand, the standard proof with Carathéodory's theorem sets $$\operatorname{rank}(\rho)^2$$ as the general upper bound for the sufficient number of elements in a separable pure decomposition of $$\rho$$, and thus examples such as symmetric Werner states clearly do not saturate this bound, as they require $$n^2 < \binom{n+1}{2}^2$$ terms. For example, for $$n=3$$ these states have $$\operatorname{rank}=6$$, but can be decomposed with $$n^2=9<6^2$$ terms.

Is there any example of a state that saturates this upper bound? In other words, a separable state $$\rho$$ that cannot be decomposed with less than $$\operatorname{rank}(\rho)^2$$ pure product states?