For a bipartite input,
$$
(T\otimes\operatorname{id})(\rho_{AB})
=\sum_k\sigma_k^A\otimes
\operatorname{Tr}_A[(M_k\otimes I)\rho_{AB}].
$$
Each second factor is positive and has trace equal to the probability of measurement outcome $k$. Dividing nonzero factors by their traces therefore writes the output as a convex combination of product states. Hence every measure-and-prepare channel is an <entanglement-breaking channel>.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 323 / 2 / c / i / Solution

{wiki}

A channel is entanglement breaking when applying it to one side of every bipartite state always produces a separable state. Equivalently, its Choi matrix is separable, and equivalently it has a measure-and-prepare representation.


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