= Solution
Use two results: the isometric <Stinespring dilation> of a <quantum channel>, and <Strong subadditivity of Von Neumann entropy>. Let $V:B\to B'E$ dilate the given operation, and define
$$
\omega_{AB'E}=(I_A\otimes V)\rho_{AB}(I_A\otimes V^\dagger),\qquad\sigma_{AB'}=\operatorname{Tr}_E\omega.
$$
An <linear isometry of Hilbert spaces> preserves the nonzero <eigenvalues>, so $S(B'E)_\omega=S(B)_\rho$ and $S(AB'E)_\omega=S(AB)_\rho$. Subtracting the two <coherent information> expressions gives
$$
\begin{aligned}
I(A\rangle B)_\rho-I(A\rangle B')_\sigma
&=S(B'E)_\omega-S(AB'E)_\omega-S(B')_\omega+S(AB')_\omega\\
&=I(A:E|B')_\omega\geq0.
\end{aligned}
$$
The last inequality is <Strong subadditivity of Von Neumann entropy>, in the form $S(AB')+S(B'E)\geq S(B')+S(AB'E)$. Thus the <data-processing inequality for coherent information> is
$$
\boxed{I(A\rangle B)_\rho\geq I(A\rangle B')_\sigma.}
$$
The lost <coherent information> is precisely the <quantum conditional mutual information> between the reference $A$ and discarded environment $E$, conditional on the retained output $B'$.
Back to article page
