= Solution
Take a <Stinespring dilation> $V:B\longrightarrow B'E$ of the local <quantum channel> and define $\tau_{AB'E}=(I_A\otimes V)\rho_{AB}(I_A\otimes V^\dagger)$. Tracing out $E$ gives the prescribed output, with $A'=A$. An isometry preserves the nonzero eigenvalues of a <density operator>; therefore $S(B'E)_\tau=S(B)_\rho$, $S(AB'E)_\tau=S(AB)_\rho$, and $S(A)_\tau=S(A)_\rho$.
Consequently $I(A:B'E)_\tau=I(A:B)_\rho$. The loss of <quantum mutual information> is
$$
\begin{aligned}
I(A:B)_\rho-I(A':B')_\sigma
&=I(A:B'E)_\tau-I(A:B')_\tau\\
&=S(AB')_\tau+S(B'E)_\tau-S(B')_\tau-S(AB'E)_\tau\\
&=I(A:E\mid B')_\tau\geq0.
\end{aligned}
$$
The last inequality is <Strong subadditivity of Von Neumann entropy>, or nonnegativity of <quantum conditional mutual information>. Hence
$$
\boxed{I(A':B')_\sigma\leq I(A:B)_\rho.}
$$
This <mutual-information loss as conditional mutual information> shows exactly which correlations are discarded into the environment. No purity assumption on the original state is needed.
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