Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 2 / 15D / b / Solution

Applying the transform twice givesThe root-of-unity filter makes the inner sum equal to $N$ exactly when $z\equiv -x(\mathrm{mod}N)$ and zero otherwise. HenceApplying modular negation twice givesTherefore ${\mathrm{QFT}}_N^2$ is an involution. Every one of its eigenvalues $\lambda$ satisfies ${\lambda }^2=1$, so its spectrum is contained in $\{1,-1\}$, as summarized by the square of the quantum Fourier transform.