Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 3 / 10C / b / i / Solution

Let $|-⟩=(|0⟩-|1⟩)/\sqrt{2}$. Applying the oracle with this answer qubit produces phase kickback:Inside $A$, the $X$ gates on both search qubits conjugate the phase oracle for $11$. Hencewhere $I_0=I-2|00⟩⟨00|$.

The first oracle similarly acts on the search register as $I_{x_0}=I-2|x_0⟩⟨x_0|$. Reading the remaining gates from right to left, the search-register operation after its initial Hadamards is thereforewhile the answer qubit remains in $|-⟩$.