Example 41.4.11 (Reduction Map).label Let $\phi: M_{n}(\complex) \to M_{n}(\complex)$ be defined by
\[\phi(x) = \text{Tr}(x)1_{M_n(\complex)}- x\]
then $\phi$ is positive, but not completely positive.
Proof. (Positive): Let $x \in M_{n}(\complex)$ with $x \ge 0$, then $\norm{x}_{M_n(\complex)}$ is the largest eigenvalue of $x$, which is bounded by the trace of $x$. As such, $\text{Tr}(x)1_{M_n(\complex)}\ge x$, and $\phi(x)$ is positive.
(Not Completely Positive): For each $1 \le i, j \le n$, $\text{Tr}(E_{ij}) = 1$ if and only if $i = j$. So the Choi matrix is given by
\begin{align*}C_{ij}&= \phi_{n}\paren{\sum_{i, j = 1}^n E_{ij} \otimes E_{ij}}= \sum_{i, j = 1}^{n} E_{ij}\otimes \phi(E_{ij}) \\&= \sum_{i, j = 1}^{n} E_{ij}\otimes (\one_{\bracsn{i = j}}\otimes 1_{M_n(\complex)}- E_{ij}) \\&= \sum_{i = 1}^{n} E_{ii}\otimes 1_{M_n(\complex)}- \sum_{i, j = 1}^{n} E_{ij}\otimes E_{ij}\\&= 1_{M_n(\complex)}\otimes 1_{M_n(\complex)}- \sum_{i, j = 1}^{n} E_{ij}\otimes E_{ij}\end{align*}
which is not positive. By Choi’s Theorem, $\phi$ is not completely positive.$\square$
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