= Solution
Introduce an orthonormal classical-label basis $|i\rangle$ and the <classical-quantum state>
$$
\Gamma_{XA}=\sum_ip_i|i\rangle\langle i|\otimes\rho_i.
$$
Its <reduced density operators> are $\Gamma_X=\sum_ip_i|i\rangle\langle i|$ and $\Gamma_A=\overline\rho=\sum_ip_i\rho_i$. If $\lambda_{ij}$ are the eigenvalues of $\rho_i$, the eigenvalues of $\Gamma_{XA}$ are $p_i\lambda_{ij}$. Thus the <entropy of an orthogonal quantum mixture> is
$$
S(\Gamma_{XA})=H(p)+\sum_ip_iS(\rho_i),\qquad S(\Gamma_X)=H(p),
$$
where $H(p)=-\sum_ip_i\log_2p_i$ is the <Shannon entropy>. Apply the proved <Subadditivity of Von Neumann entropy> to $X$ and $A$ and cancel $H(p)$. This gives <Concavity of Von Neumann entropy>:
$$
\boxed{S\left(\sum_ip_i\rho_i\right)\geq\sum_ip_iS(\rho_i).}
$$
Zero-probability terms may simply be omitted. The equality condition from the preceding solution implies that equality holds precisely when all the positive-probability states are the same: the block equations $p_i\rho_i=p_i\overline\rho$ follow from $\Gamma_{XA}=\Gamma_X\otimes\Gamma_A$.
Back to article page
