 Solution

Put $p_i=⟨{\psi }_i|\rho |{\psi }_i⟩$. The rank-one dephasing is ${\rho }^{\prime }={\sum }_i{p}_i|{\psi }_i⟩⟨{\psi }_i|$. If $p_i=0$, positivity givesso $\rho |{\psi }_i⟩=0$. The kernel of ${\rho }^{\prime }$ is exactly the span of these zero-probability basis vectors and is therefore contained in the kernel of $\rho$. Taking orthogonal complements proves support inclusion under rank-one dephasing:Here the support of a positive operator is the orthogonal complement of its kernel.

On that support, ${\log }_2{\rho }^{\prime }$ is diagonal in the measurement basis, givingConsequently the relative-entropy identity for rank-one dephasing isKlein's inequality gives $D_{\mathrm{rel}}(\rho \Vert {\rho }^{\prime })\ge 0$, since both density operators have trace one. For singular $\rho$, first restrict to $\mathrm{supp}{\rho }^{\prime }$ where ${\rho }^{\prime }$ is positive definite, replace $\rho$ by $(\rho +\eta I)/(1+\eta \dim \mathrm{supp}{\rho }^{\prime })$, and let $\eta \downarrow 0$. The support inclusion ensures that the limit is finite. ThusThis proves entropy increase under nonselective projective measurement using Klein's inequality. Equality holds exactly when $\rho ={\rho }^{\prime }$, so the original density operator was already diagonal in the chosen basis.