 Hermite functions are Fourier eigenfunctions

Use the unitary Fourier transform with phase $e^{-i\lambda x}$. It commutes with $-d^2/d{x}^2+x^2$, because differentiation transforms to multiplication by $i\lambda$ and multiplication by $x$ transforms to $id/d\lambda$. The Hermite differential equation has a unique polynomial solution of each degree up to scale, and multiplying by the Gaussian gives a Hermite function. Transforming a polynomial times this Gaussian again gives a polynomial times the Gaussian of the same degree; comparison of leading coefficients gives the eigenvalue $(-i{)}^n$.