Solution

Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 58 / 3 / ii / Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 10D / a / Solution

{c}

For $|\beta_{xz}\rangle=(Z^zX^x\otimes I)|\beta_{00}\rangle$, an arbitrary qubit obeys
$$
|\alpha\rangle_C|\beta_{00}\rangle_{AB}
=\frac12\sum_{x,z\in\{0,1\}}
|\beta_{xz}\rangle_{CA}X^xZ^z|\alpha\rangle_B.
$$
A Bell-basis measurement reveals $(x,z)$, after which the receiver applies the inverse Pauli correction.


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 Incoming links: Bell-basis teleportation identity