# 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

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.

• thanks. You are showing that $J(\Phi)$ has positive expectation value on classical states here, yes? Don't you also have to show that there is some state for which $J(\Phi)$ gives negative expectation value to call it a witness? – glS Nov 11 '20 at 15:06
• Yes. And $J(\Phi)$ must have a negative eigenvalue, otherwise $\Phi$ would be completely positive. The eigenvector corresponding to this negative eigenvalue will then be the entangled state for which it gives a negative expectation value. This is not strictly necessary, sometimes people define entanglement witness simply as operators which are positive on separable states, and non-trivial witnesses as those which can actually detect entanglement. – Mateus Araújo Nov 11 '20 at 15:20