Assume no previously shared <entanglement>. Alice's encoded ensemble $\{p_x,\rho_x\}$ lives in a <Hilbert space> of dimension $2^n$. Its <Holevo quantity> satisfies
$$
\chi_{\rm in}=S\left(\sum_xp_x\rho_x\right)-\sum_xp_xS(\rho_x)
\leq\log_2(2^n)=n,
$$
using nonnegativity of <Von Neumann entropy> and the <maximum entropy of a quantum state>. The <data-processing inequality for quantum relative entropy> implies $\chi_{\rm out}\leq\chi_{\rm in}$ for the transmission <quantum channel>. Finally, the <Holevo bound> applied to Bob's <quantum measurement> gives
$$
\boxed{I(X:Y)\leq\chi_{\rm out}\leq n\text{ bits}}.
$$
This argument also works when the output system has larger dimension than the input. Prior shared <entanglement> changes the communication resource: <superdense coding> can transmit two classical bits per sent qubit, so the absence of that resource is part of this bound's setting.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 323 / 3 / i / Solution

{title2=$S(\rho)\leq\log_2d$}

On a $d$-dimensional <Hilbert space>, the <Von Neumann entropy> is maximized uniquely by $I/d$. Indeed, <nonnegativity of quantum relative entropy> gives
$$
0\leq D(\rho\|I/d)=\log_2d-S(\rho),
$$
with equality exactly when $\rho=I/d$.

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 Incoming links: Maximum entropy of a quantum state