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)?

  • 3
    $\begingroup$ 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.". $\endgroup$ May 14, 2019 at 23:16

1 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.

  • $\begingroup$ 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? $\endgroup$
    – glS
    Nov 11, 2020 at 15:06
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    $\begingroup$ 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. $\endgroup$ Nov 11, 2020 at 15:20

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