$\mu\mathbf{I}$.

Suppose that signal mode $a$ is subject to a pure-loss channel. This means that Eve is using a beam splitter of transmissivity $\tau$ mixing the signal mode with a vacuum mode $e$. At the output of the beam splitter, mode $B$ is detected by Bob, while mode $E$ is stored in a quantum memory coherently detected by Eve (this is a collective entangling cloner attack with $\omega=1 $).

Since the average state of mode $a$ is thermal and mode $e$ is in the vacuum, no entanglement can be present between the two output ports $B$ and $E$ of the beam splitter. This implies that their entanglement of formation must be zero $E_{f}(B,E)=0$ and, therefore, the optimal backward rate $K(\blacktriangleleft )$ must be equal to the discord $D(B|A)$ of the Gaussian state $\rho_{AB}$. Since this output state has CM$$\mathbf{V}_{AB}=\left(
\begin{array}{cc}
\mu\mathbf{I} & \sqrt{\tau(\mu^{2}-1)}\mathbf{Z} \\
\sqrt{\tau(\mu^{2}-1)}\mathbf{Z} & (\tau\mu+1-\tau)\mathbf{I}\end{array}
\right) ,$$ its discord is easy to compute and is equal to [@Bconj] $$D(B|A)=h(\mu)-h[\tau+(1-\tau)\mu].$$ For large modulation ($\mu\rightarrow+\infty$), we have the asymptotic expression $$K(\blacktriangleleft)=D(B|A)=\log_{2}\left( \frac{1}{1-\tau}\right) ,$$ which is positive for any $0<\tau<1$. One can check that this rate can be achieved by heterodyne detections at Bob’s side (and coherent detection at Alice’s side).

