In a previous post, we were concerned with channels of the form $\Phi\in\Channel(\X,\Y)$ such that $\bigl(\Phi\otimes \I_{\Lin(\Z)}\bigr)(\rho) \in \Sep(\Y:\Z)$ for every complex Euclidean space $\Z$ and every density operator $\rho\in\Density(\X\otimes\Z)$. Channels of this form have the effect of destroying entanglement that exists between the register they act on and any other registers.

There exist two channels $\Phi_0,\Phi_1\in\Channel(\X,\Y)$, both having the property described above, such that

\[

\bigtriplenorm{\Phi_0 - \Phi_1}_1

> \bignorm{\Phi_0(\rho) - \Phi_1(\rho)}_1

\]

for every $\rho\in\Density(\X)$. (Channels like this have the strange property that they destroy entanglement, and yet evaluating them on an entangled state helps to distinguish them.)

For $\lambda\in[0,1]$, consider the two channels $\Phi_0(X),\Phi_0(X)\in\Channel(X)$ defined by

\[\begin{align*}

\Phi_0(X)&=\frac{\lambda}{n+1}(\tr(X)\I_\X+X^T)+\frac{(1-\lambda)}{n}\tr(X)\I_\X \\

\Phi_1(X)&=\frac{\lambda}{n-1}(\tr(X)\I_\X-X^T)+\frac{(1-\lambda)}{n}\tr(X)\I_\X

\end{align*}\]

Then for sufficiently small $\lambda\in[0,1]$ both of the Choi representations $J(\Phi_0(X))$ and $J(\Phi_1(X))$ are in a separable neighborhood of the maximally mixed state which implies that they are both separable by some theorem. Therefore, from the results in the previous post, we have that $\Phi_0(X)$ and $\Phi_0(X)$ are entanglement destroying as described in the problem statement.

Now considering that

\[

\Phi_0(\rho) - \Phi_1(\rho)=\frac{-2\lambda}{(n+1)(n-1)}\rho^T,

\]

it follows that

\[

\bignorm{\Phi_0(\rho) - \Phi_1(\rho)}_1=\frac{2\lambda}{(n+1)(n-1)}\bignorm{\rho^T}_1=\frac{2\lambda}{(n+1)(n-1)},

\]

since $\rho\in\Density(\X)$.

Moreover, since

\[\begin{align*}

\bigtriplenorm{\Phi_0 - \Phi_1}_1&=max\{\bignorm{((\Phi_0(\rho) - \Phi_1(\rho))\otimes\I_{\Lin(\X)})(xx^\ast)}_1 \ : \ x\in S(\X\otimes\X)\} ,

\end{align*}\]

where $((\Phi_0(\rho) - \Phi_1(\rho))\otimes\I_{\Lin(\X)})(xx^\ast)$ gives the partial transpose of $xx^\ast$ (which is at most $n$ since $x\in S(\X\otimes\X)$) multiplied by the scalar quantity $\frac{2\lambda}{(n+1)(n-1)}$. Therefore $\bigtriplenorm{\Phi_0 - \Phi_1}_1=\frac{2\lambda n}{(n+1)(n-1)}$, which implies that

\[

\frac{2\lambda n}{(n+1)(n-1)}=\bigtriplenorm{\Phi_0 - \Phi_1}_1

> \bignorm{\Phi_0(\rho) - \Phi_1(\rho)}_1=\frac{2\lambda}{(n+1)(n-1)}.

\]

**Theorem:**There exist two channels $\Phi_0,\Phi_1\in\Channel(\X,\Y)$, both having the property described above, such that

\[

\bigtriplenorm{\Phi_0 - \Phi_1}_1

> \bignorm{\Phi_0(\rho) - \Phi_1(\rho)}_1

\]

for every $\rho\in\Density(\X)$. (Channels like this have the strange property that they destroy entanglement, and yet evaluating them on an entangled state helps to distinguish them.)

**Proof:**For $\lambda\in[0,1]$, consider the two channels $\Phi_0(X),\Phi_0(X)\in\Channel(X)$ defined by

\[\begin{align*}

\Phi_0(X)&=\frac{\lambda}{n+1}(\tr(X)\I_\X+X^T)+\frac{(1-\lambda)}{n}\tr(X)\I_\X \\

\Phi_1(X)&=\frac{\lambda}{n-1}(\tr(X)\I_\X-X^T)+\frac{(1-\lambda)}{n}\tr(X)\I_\X

\end{align*}\]

Then for sufficiently small $\lambda\in[0,1]$ both of the Choi representations $J(\Phi_0(X))$ and $J(\Phi_1(X))$ are in a separable neighborhood of the maximally mixed state which implies that they are both separable by some theorem. Therefore, from the results in the previous post, we have that $\Phi_0(X)$ and $\Phi_0(X)$ are entanglement destroying as described in the problem statement.

Now considering that

\[

\Phi_0(\rho) - \Phi_1(\rho)=\frac{-2\lambda}{(n+1)(n-1)}\rho^T,

\]

it follows that

\[

\bignorm{\Phi_0(\rho) - \Phi_1(\rho)}_1=\frac{2\lambda}{(n+1)(n-1)}\bignorm{\rho^T}_1=\frac{2\lambda}{(n+1)(n-1)},

\]

since $\rho\in\Density(\X)$.

Moreover, since

\[\begin{align*}

\bigtriplenorm{\Phi_0 - \Phi_1}_1&=max\{\bignorm{((\Phi_0(\rho) - \Phi_1(\rho))\otimes\I_{\Lin(\X)})(xx^\ast)}_1 \ : \ x\in S(\X\otimes\X)\} ,

\end{align*}\]

where $((\Phi_0(\rho) - \Phi_1(\rho))\otimes\I_{\Lin(\X)})(xx^\ast)$ gives the partial transpose of $xx^\ast$ (which is at most $n$ since $x\in S(\X\otimes\X)$) multiplied by the scalar quantity $\frac{2\lambda}{(n+1)(n-1)}$. Therefore $\bigtriplenorm{\Phi_0 - \Phi_1}_1=\frac{2\lambda n}{(n+1)(n-1)}$, which implies that

\[

\frac{2\lambda n}{(n+1)(n-1)}=\bigtriplenorm{\Phi_0 - \Phi_1}_1

> \bignorm{\Phi_0(\rho) - \Phi_1(\rho)}_1=\frac{2\lambda}{(n+1)(n-1)}.

\]

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