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

Question

We know that a map $Phiinmathrm T(mathcal X,mathcal Y)$ being completely positive is equivalent to its Choi representation being positive: $J(Phi)inoperatorname{Pos}(mathcal Yotimesmathcal X)$, 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

$$J(Phi)=(Phiotimesmathbb1_{mathrm L(mathcal X)})(operatorname{vec}(mathbb1_{mathcal X})operatorname{vec}(mathbb1_{mathcal X})^*),$$

and noting that because $operatorname{vec}(mathbb1_{mathcal X})operatorname{vec}(mathbb1_{mathcal X})^*ge0$ and $Phiotimesmathbb1_{mathrm L(mathcal X)}$ 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 $Phiotimesmathbb1_{mathrm L(mathcal X)}$ is not positive.

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

Mateus Araújo · Accepted Answer

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 $|Omegarangle := sum_i |iirangle$, such that $J(Phi) = I otimes Phi ( |OmegaranglelangleOmega|)$, and let $A,B$ be positive semidefinite operators of the appropriate dimensions. Then
$$operatorname{tr}big[I otimes Phi ( |OmegaranglelangleOmega|) Aotimes Bbig] = operatorname{tr}big[|OmegaranglelangleOmega| Aotimes Phi^dagger (B)big] ge 0,$$
as the adjoint of a positive map is always positive.

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

One Answer

Add your own answers!

Ask a Question