Let $\Delta=\rho-\sigma=\Delta_+-\Delta_-$ be the <positive-negative decomposition of a Hermitian operator>. Since $\operatorname{Tr}\Delta=0$,
$$
\operatorname{Tr}\Delta_+
=\operatorname{Tr}\Delta_-
=\frac12\|\Delta\|_1.
$$
For $0\leq P\leq I$,
$$
\operatorname{Tr}(P\Delta)
\leq\operatorname{Tr}(P\Delta_+)
\leq\operatorname{Tr}\Delta_+.
$$
Equality is attained by the projector onto the positive eigenspace of $\Delta$. This proves the <variational characterization of trace distance>
$$
\boxed{D(\rho,\sigma)
=\max_{0\leq P\leq I}\operatorname{Tr}[P(\rho-\sigma)]}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 323 / 4 / a / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-323/4/a/ii/solution