= Solution
The unread measurement is the <pinching map> $\mathcal P(\rho)=\rho'=\sum_iP_i\rho P_i$, a <nonselective projective measurement>. Its output is block diagonal and commutes with each $P_i$, so $\log\rho'$ also commutes with the projectors on its support.
First the <quantum relative entropy> $D(\rho\Vert\rho')$ is finite. If $v$ lies in the kernel of $\rho'$, positivity gives
$$
0=\langle v|\rho'|v\rangle=\sum_i\|\rho^{1/2}P_iv\|^2.
$$
Every summand is zero; summing $\rho^{1/2}P_iv=0$ gives $\rho^{1/2}v=0$. Thus $\ker\rho'\subseteq\ker\rho$, equivalently $\operatorname{supp}\rho\subseteq\operatorname{supp}\rho'$.
The block-diagonal logarithm obeys $\sum_iP_i(\log\rho')P_i=\log\rho'$ on this support. Cyclicity of the trace therefore gives
$$
\operatorname{Tr}\rho\log\rho'=\operatorname{Tr}\left(\sum_iP_i\rho P_i\right)\log\rho'=\operatorname{Tr}\rho'\log\rho'.
$$
Consequently the <relative entropy of a pinched state> is exactly the entropy increase:
$$
D(\rho\Vert\rho')=-S(\rho)-\operatorname{Tr}\rho\log_2\rho'=S(\rho')-S(\rho).
$$
By the proved <nonnegativity of quantum relative entropy> and its equality condition,
$$
\boxed{S(\rho')\geq S(\rho),\qquad S(\rho')=S(\rho)\ \Longleftrightarrow\ \rho'=\rho.}
$$
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.
Back to article page
