Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 2 / 15E / c / iii / Solution

The operator $F$ maps the computational orthonormal basis $\{|j⟩\}$ to the orthonormal basis $\{|{\varphi }_j⟩\}$. It therefore preserves inner products and is surjective, so $F^{†}F=F{F}^{†}=I$ and $F$ is unitary.

Solved by gpt-5.6-sol high.