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The unread measurement is the pinching map $\mathcal{P}(\rho )={\rho }^{\prime }={\sum }_i{P}_i\rho P_i$, a nonselective projective measurement. Its output is block diagonal and commutes with each $P_i$, so $\log {\rho }^{\prime }$ also commutes with the projectors on its support.

First the quantum relative entropy $D(\rho \Vert {\rho }^{\prime })$ is finite. If $v$ lies in the kernel of ${\rho }^{\prime }$, positivity givesEvery summand is zero; summing ${\rho }^{1/2}{P}_{i}v=0$ gives ${\rho }^{1/2}v=0$. Thus $\ker {\rho }^{\prime }\subseteq \ker \rho$, equivalently $\mathrm{supp}\rho \subseteq \mathrm{supp}{\rho }^{\prime }$.

The block-diagonal logarithm obeys ${\sum }_i{P}_i(\log {\rho }^{\prime })P_i=\log {\rho }^{\prime }$ on this support. Cyclicity of the trace therefore givesConsequently the relative entropy of a pinched state is exactly the entropy increase:By the proved nonnegativity of quantum relative entropy and its equality condition,Equivalently, equality holds exactly when all off-diagonal blocks $P_i\rho P_j$ for $i\ne j$ already vanish, or when $\rho$ commutes with every measurement projector. This argument proves the equality condition even when the projectors are degenerate and the states singular.

Let $\mathcal{P}$ be the pinching map determined by the projectors, and put $\sigma =\mathcal{P}(\rho )$. This is a nonselective projective measurement; its projectors are its Kraus operators. We will prove the exact relative entropy of a pinched state identity.

First the support of $\rho$ is contained in that of $\sigma$. Indeed, if $v$ lies in the kernel of $\sigma$, positivity givesEvery summand is zero. Since ${\sum }_i{P}_i=I$, also ${\rho }^{1/2}v=0$, and thus $v$ is in the kernel of $\rho$. Therefore the logarithm of $\sigma$ may be used on its support without an infinite relative-entropy term.

The operator $\sigma$ is block diagonal, so each $P_i$ commutes with $\log \sigma$ on that support. Cyclicity of the trace givesExpand the quantum relative entropy:Its nonnegativity now proves the entropy increase under nonselective projective measurement:The equality case in nonnegativity of quantum relative entropy says equality holds exactly when $\rho =\sigma$. Equivalently,Thus equality means the state already has no coherence between distinct measurement blocks. It need not be diagonal inside a block of rank greater than one.

Take the rank-one orthogonal projections $P_i=|{\varphi }_i⟩⟨{\varphi }_i|$. Their pinching map givesThis has exactly the specified diagonal entries and no off-diagonal entries. Applying the entropy increase under nonselective projective measurement proved in part (c) therefore givesWriting $q_i=⟨{\varphi }_i|\rho |{\varphi }_i⟩$, the output Von Neumann entropy is the Shannon entropy $S({\rho }_d)=-{\sum }_i{q}_i\log q_i$. Equality holds precisely when $\rho$ was already diagonal in this basis.

The pinching map is self-adjoint for the trace pairing, and $\log \mathcal{P}(\rho )$ is block diagonal. Thus $\mathrm{Tr}\rho \log \mathcal{P}(\rho )=\mathrm{Tr}\mathcal{P}(\rho )\log \mathcal{P}(\rho )$. The kernel of $\mathcal{P}(\rho )$ is contained in that of $\rho$, so the relative entropy is finite on its support. Expanding the quantum relative entropy gives the displayed identity. Its nonnegativity proves entropy increase, with equality exactly when the original state is already block diagonal.