How does a map being "only" positive reflect on its Choi representation?

Question

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)?

Answer 1

If $\Phi$ is positive but not completely positive, then it gives an operator that has positive trace with separable quantum states, that is, an entanglement witness.

To see that, let $|\Omega\rangle := \sum_i |ii\rangle$, such that $J(\Phi) = I \otimes \Phi ( |\Omega\rangle\langle\Omega|)$, and let $A,B$ be positive semidefinite operators of the appropriate dimensions. Then $$\operatorname{tr}\big[I \otimes \Phi ( |\Omega\rangle\langle\Omega|) A\otimes B\big] = \operatorname{tr}\big[|\Omega\rangle\langle\Omega| A\otimes \Phi^\dagger (B)\big] \ge 0,$$ as the adjoint of a positive map is always positive.