= Solution
Use <Klein's inequality>, or equivalently <nonnegativity of quantum relative entropy>. First check the support needed for the logarithm. If $q_y=0$, positivity gives $|\rho_{yz}|^2\leq\rho_{yy}\rho_{zz}=0$ for every $z$; the corresponding row and column vanish. Thus <support inclusion under rank-one dephasing> gives $\operatorname{supp}\rho\subseteq\operatorname{supp}\sigma$, and the logarithms may be evaluated on this support.
Since $\log\sigma$ is diagonal in the dephasing basis,
$$
\operatorname{Tr}\rho\log_2\sigma=\sum_{y:q_y>0}q_y\log_2q_y=\operatorname{Tr}\sigma\log_2\sigma.
$$
The <relative-entropy identity for rank-one dephasing> follows:
$$
D(\rho\|\sigma)=\operatorname{Tr}\rho(\log_2\rho-\log_2\sigma)=S(\sigma)-S(\rho).
$$
<Klein's inequality> gives $D(\rho\|\sigma)\geq(\operatorname{Tr}\rho-\operatorname{Tr}\sigma)/\ln2=0$. Therefore
$$
\boxed{S(\Lambda(\rho))\geq S(\rho).}
$$
Equality holds precisely when $\rho=\sigma$, meaning that the input was already diagonal in the chosen basis. This quantifies why <rank-one dephasing> removes coherence without reducing the entropy.
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