the prototypical example of a positive but not completely positive map.

The condition (\[Pcondition\]b) requires determining whether a certain $N^2 \times N^2$ matrix—the matrix ${\mathcal B}_{\beta\delta}$ given by Eq. (\[inverse\])—is positive semidefinite. In this respect it is similar to a more standard test for complete positivity, namely, to see whether the Choi operator, another $N^2 \times N^2$ matrix, is positive semidefinite [@Jamiokowski1972; @Choi1975; @wilde_2013]. In fact, it turns out that ${\mathcal B}_{\beta\delta}$ is simply the Choi operator written in a specific basis, as we now show.

From Eqs. (\[PfromE\]) and (\[inverse\]), we have $$\label{BwithE}
\begin{split}
    {\mathcal B}_{\beta\delta} &= \frac{1}{N^4}\sum_{\alpha\gamma}
    \hbox{Tr}(\hat{A}_\beta \hat{A}_\alpha \hat{A}_\delta \hat{A}_\gamma) \hbox{Tr}\left[\hat{A}_\alpha {\mathcal E}(\hat{A}_\gamma)\right]\\
    &=\frac{1}{N^3}\sum_\gamma \hbox{Tr}\left[\hat{A}_\delta \hat{A}_\gamma \hat{A}_\beta {\mathcal E}(\hat{A}_\gamma)\right].
\end{split}$$ The Choi operator is $$\label{Choidef}
    \hat{C} = \frac{1}{N}\sum_{jk} |j\rangle\langle k| \otimes {\mathcal E}(|j\rangle\langle k|).$$ Let us define the orthonormal basis $|\Psi_\alpha \rangle$ by $$\label{Psidef}
    |\Psi_\beta\rangle = (\hat{I} \otimes \hat{A}_\beta)|\Phi\rangle,$$ where $|\Phi\rangle$ is the maximally entangled state $$|\Phi\rangle = \frac{1}{\sqrt{N}} \sum_m |m\rangle \otimes |m\rangle.$$ Then we claim that $$\label{BCclaim}
    {\mathcal B}_{\beta\delta} = \langle \Psi_\beta|\hat{C}|\Psi_\delta\rangle.$$ Indeed, by plugging the definitions (\[Choidef\]) and (\[Psidef\]) into the right-hand side of Eq. (\[BCclaim\]), we find that $$\label{firstB}
\begin{split}
   \langle \Psi_\beta|\hat{C}|\Psi_\delta\rangle &= 
   \frac{1}{N^2} \sum_{jk} \hbox{Tr}\left[ |k\rangle\langle j| \hat{A}_\beta {\mathcal E}(|j\rangle\langle k|) \hat{A}_\delta \right]\\
   