We know that a map $\Phi\in\mathrm T(\mathcal X,\mathcal Y)$ being completely positive is equivalent to its Choi representation being positive: $J(\Phi)\in\operatorname{Pos}(\mathcal Y\otimes\mathcal X)$, as shown for example in Watrous' book, around pag. 82.

The proof for the completely positive case relies on writing the Choi representation $J(\Phi)$ of $\Phi$ as

$$J(\Phi)=(\Phi\otimes\mathbb1_{\mathrm L(\mathcal X)})(\operatorname{vec}(\mathbb1_{\mathcal X})\operatorname{vec}(\mathbb1_{\mathcal X})^*),$$

and noting that because $\operatorname{vec}(\mathbb1_{\mathcal X})\operatorname{vec}(\mathbb1_{\mathcal X})^*\ge0$ and $\Phi\otimes\mathbb1_{\mathrm L(\mathcal X)}$ is a positive map, then $J(\Phi)\ge0$. It is not obvious whether this sort of argument can give interesting results about $J(\Phi)$ when $\Phi\otimes\mathbb1_{\mathrm L(\mathcal X)}$ is not positive.

In other words, how does $\Phi$ being "only" positive reflect on $J(\Phi)$ (if any such simple characterisation is possible)?

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    $\begingroup$ Quick comment: Positive but not CP maps are hard to characterize, as this is essentially equivalent to characterizing separability vs. entanglement, which is a hard problem. (A fact which is puzzling at first sight: Complete positivity is much easier to characterize than positivity.) So I don't think there is a good answer to the question beyond "No.". $\endgroup$ – Norbert Schuch May 14 '19 at 23:16

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