Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 51 / 1 / a / Solution

The answer qubit enters the Boolean quantum oracle in $|-⟩=(|0⟩-|1⟩)/\sqrt{2}$, because the Pauli X gate followed by the Hadamard gate maps $|0⟩$ to $|-⟩$. Since $X|-⟩=-|-⟩$, quantum phase kickback multiplies the $|xy⟩$ component by $(-1{)}^{f(x,y)}$ and leaves the answer qubit unchanged. The two input qubits enter in the uniform quantum superposition $|++⟩$. ThusEquivalently, using the Controlled-Z gate and the Pauli Z gate,Here the exponent in the phase may be evaluated as an ordinary integer, since its parity agrees with the exclusive or expression. In particular, $d$ contributes only a global phase.