Quantum Computing Asked on April 12, 2021
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)?
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.
Correct answer by Mateus Araújo on April 12, 2021
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