Discarding the outcome of the complete projective measurement gives
$$
\boxed{\sigma=\sum_iP_i\rho P_i}.
$$
For $m$ projectors, let $\omega=e^{2\pi i/m}$ and $U=\sum_j\omega^jP_j$. Then the pinching identity is
$$
\sigma=\frac1m\sum_{k=0}^{m-1}U^k\rho U^{-k}.
$$
The <Concavity of Von Neumann entropy> and its invariance under <unitary operators> imply
$$
\boxed{S(\sigma)\geq\frac1m\sum_kS(U^k\rho U^{-k})
=S(\rho)}.
$$
Equality holds exactly when every conjugate in the average is the same, equivalently
$$
\boxed{[\rho,P_i]=0\quad\hbox{for every }i}.
$$
Thus equality holds when the input already has no coherence between distinct measurement subspaces.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 323 / 5 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-323/5/c/solution