Past exam of the laser science and quantum information processing course of the University of Oxford / 2021 / 4 / d / Solution

The extra Pauli gate on the control is removed by $Z_{a}G=C_Z$. Conjugating the target by a Hadamard gate then givesThus, from left to right in the network, apply $H_b$, then $G$, then $Z_a$ and $H_b$; the last two gates commute.

For a direct proof, the control-zero block of $G$ is $I_b$ and its control-one block is $-Z_b$. Multiplication by $Z_a$ changes the latter into $Z_b$. Since $H_b{I}_b{H}_b=I_b$ and $H_b{Z}_b{H}_b=X_b$, the result iswhich is exactly the CNOT gate on all inputs, including quantum superpositions.