For Hermitian $X$, $\|X\|_1=\max_{-I\leq H\leq I}\operatorname{Tr}(HX)$. The adjoint of a trace-preserving completely positive map is unital and positive, so $-I\leq H\leq I$ implies $-I\leq\Lambda^*(H)\leq I$. Therefore
$$
\begin{aligned}
\|\Lambda(\rho)-\Lambda(\sigma)\|_1
&=\max_{-I\leq H\leq I}\operatorname{Tr}[\Lambda^*(H)(\rho-\sigma)]\\
&\leq\|\rho-\sigma\|_1.
\end{aligned}
$$
Hence <trace distance> is contractive under quantum channels.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 323 / 4 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-323/4/a/solution