Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 51 / 2 / a / Solution

Write $N=|A|$. For $N>1$, the unmarked computational basis vectors in $A$ give the normalized vectorIts norm is one and $⟨a|\omega ⟩=0$, so these two vectors form an orthonormal basis for $\mathcal{V}$. Splitting the uniform superposition state into its marked and unmarked terms givesThe nonnegative coefficients select the specified angle in $[0,\pi /2]$.

There is one degenerate boundary case: if $N=1$, then $|{\psi }_A⟩=|a⟩$ and $\mathcal{V}$ is one-dimensional. No vector in $\mathcal{V}$ can complete $|a⟩$ to a two-element orthonormal basis. Thus the printed two-dimensional description needs $N>1$. For $N=1$ the search itself is already solved, with $\theta =\pi /2$ and no query.