Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 2 / 15D / a / iii / Solution

Replacing the resource by $(I\otimes U)|{\varphi }^{+}⟩$ applies $U$ to Bob's uncorrected output. If $P\in \{I,Z,X,XZ\}$ labels the Bell outcome, the uncorrected state is $UP|\alpha ⟩$, and the unchanged correction producesThus the four outputs are $U|\alpha ⟩$, $ZUZ|\alpha ⟩$, $XUX|\alpha ⟩$, and $(XZ{)}^{-1}U(XZ)|\alpha ⟩$, up to global phases.