Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 2 / 15C / b / Solution

The Hadamard gate conjugates the Pauli X gate into the Pauli Z gate:When the control qubit is $|0⟩$, all three circuits act as the identity on the target. When it is $|1⟩$, the right-hand circuit acts on the target as $HXH=Z$. It therefore agrees with the Controlled-Z gate on every computational basis state, provingConjugating both sides by $I\otimes H$ and using $H^2=I$ immediately gives the reverse identity