# How does a map being “only” positive reflect on its Choi representation?

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)?

• 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.". – Norbert Schuch May 14 '19 at 23:16