Rank-one dephasing ($\Delta (\rho )={\sum }_i{P}_i\rho P_i,\mathrm{rank}{P}_i=1$)

In an orthonormal basis, rank-one dephasing is the quantum channel $\Delta (\rho )={\sum }_i|{\psi }_i⟩⟨{\psi }_i|\rho |{\psi }_i⟩⟨{\psi }_i|$. It removes off-diagonal entries while preserving the basis probabilities. Its output Von Neumann entropy is their Shannon entropy, and the relative-entropy identity for rank-one dephasing quantifies the entropy increase.

Because $\log \Delta (\rho )$ is diagonal, $\mathrm{Tr}(\rho \log \Delta (\rho ))=\mathrm{Tr}(\Delta (\rho )\log \Delta (\rho ))$. Hence $D_{\mathrm{rel}}(\rho \Vert \Delta (\rho ))=S(\Delta (\rho ))-S(\rho )$. Support inclusion under rank-one dephasing handles zero probabilities, and Klein's inequality makes the difference nonnegative. Equality holds exactly when the state was already diagonal.

If $p_i=⟨{\psi }_i|\rho |{\psi }_i⟩=0$, positivity gives $\sqrt{\rho }|{\psi }_i⟩=0$, so the corresponding basis vector is in $\ker \rho$. Therefore $\ker \Delta (\rho )\subseteq \ker \rho$, and $\mathrm{supp}\rho \subseteq \mathrm{supp}\Delta (\rho )$. This makes the quantum relative entropy of $\rho$ relative to its rank-one dephasing finite.