= Solution
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 gives
$$
0=\langle v|\sigma|v\rangle=\sum_i\langle P_iv|\rho|P_iv\rangle
=\sum_i\|\rho^{1/2}P_iv\|^2.
$$
Every summand is zero. Since $\sum_iP_i=I$, 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> gives
$$
\operatorname{Tr}(\sigma\log\sigma)
=\sum_i\operatorname{Tr}(P_i\rho P_i\log\sigma)
=\operatorname{Tr}\left(\rho\sum_iP_i\log\sigma P_i\right)
=\operatorname{Tr}(\rho\log\sigma).
$$
Expand the <quantum relative entropy>:
$$
D(\rho\|\sigma)=\operatorname{Tr}(\rho\log\rho)-\operatorname{Tr}(\rho\log\sigma)
=S(\sigma)-S(\rho).
$$
Its nonnegativity now proves the <entropy increase under nonselective projective measurement>:
$$
\boxed{S(\sigma)\geq S(\rho).}
$$
The equality case in <nonnegativity of quantum relative entropy> says equality holds exactly when $\rho=\sigma$. Equivalently,
$$
\boxed{P_i\rho P_j=0\text{ for }i\ne j,\quad\text{or, equivalently, }[\rho,P_i]=0\text{ for every }i.}
$$
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.
Back to article page
