 Solution

For the quantum state ensemble $\mathcal{E}=\{p_i,{\rho }_i\}$ with average $\overline{\rho }={\sum }_i{p}_i{\rho }_i$, the Holevo quantity isThe second equality follows by expanding the trace definition: ${\sum }_i{p}_i\mathrm{Tr}{\rho }_i\log \overline{\rho }=\mathrm{Tr}\overline{\rho }\log \overline{\rho }$. All positive-weight states are supported within $\overline{\rho }$, so the expression is finite.

For a CPTP map $\Lambda$, the output average is $\Lambda (\overline{\rho })$, with unchanged probabilities $p_i$. The data-processing inequality for quantum relative entropy givesThus the Holevo quantity cannot increase under a quantum channel.

The underlying monotonicity can also be seen directly from Strong subadditivity of Von Neumann entropy. Attach the classical label $X$, and realize $\Lambda$ by a Stinespring representation of a completely positive map with output $B$ and discarded environment $E$. The isometry preserves $I(X:BE)$, whose input value is $\chi (\mathcal{E})$. The difference from the output value iswhich is precisely strong subadditivity. This explains why discarding an environment loses accessible ensemble information.

Here “operation” means the unconditional channel, including the classical outcome when a measurement is retained. Conditioning on a selected outcome can increase the Holevo quantity. For instance, take equally likely vectors $\sqrt{a}|0⟩\pm \sqrt{1-a}|1⟩$, with $1/2<a<1$. Their input Holevo quantity is $h_2(a)<1$. The filter $K=\mathrm{diag}(\sqrt{(1-a)/a},1)$ succeeds on either state with probability $2(1-a)$ and produces the orthogonal states $|+⟩,|-⟩$ after normalization. The success-conditioned quantity is $1$. This does not violate the unconditional statement.