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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?

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