Since $|\phi\rangle\langle\phi|\geq0$ and a channel is completely positive,
$$
\boxed{C=(T\otimes\operatorname{id})(|\phi\rangle\langle\phi|)\geq0}.
$$
Writing
$$
C=\frac1d\sum_{j,k}T(|j\rangle\langle k|)
\otimes|j\rangle\langle k|
$$
and tracing over the output system gives
$$
\operatorname{Tr}_1C
=\frac1d\sum_{j,k}\operatorname{Tr}T(|j\rangle\langle k|)
|j\rangle\langle k|
=\frac1d\sum_j|j\rangle\langle j|.
$$
Thus
$$
\boxed{\operatorname{Tr}_1C=I_d/d},
$$
where the factor $1/d$ follows from the normalized maximally entangled state.

Solved by gpt-5.6-sol high.


Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 323 / 1 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-323/1/b/solution