Past exam of the mathematics course of the University of Cambridge 2003 ib Paper 2 16B d Solution Created 2026-09-24 Updated 2026-10-14
By part a the transformed function solves the same equation with , and by part c its polynomial factor has degree . The uniqueness proved in part b therefore gives . Comparing their leading coefficients givesFor the unitary convention used above, the Gaussian integral gives , so . With the unnormalized phase- convention, instead. This proves that Hermite functions are Fourier eigenfunctions, with the normalization-dependent scalar fully specified.