TITLE: Fourier transform as exponential of Hermitian operator
QUESTION [12 upvotes]: The Fourier transform
$$F: L^2\rightarrow L^2 \qquad \hat{f}(\omega) \equiv (Ff)(\omega) \equiv \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)e^{-i\omega t}dt$$
can be viewed as a unitary operator with inverse (and hence, adjoint) 
$$(F^{-1}Ff)(t) = (F^\dagger F f)(t) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty}\hat{f}(\omega)e^{i\omega t}d\omega \, .$$
All unitary operators $U$ can be written as the exponential of a Hermitian operator $H$, i.e. $U = \exp(iH)$.
What is the Hermitian operator $H$ that satisfies $F = \exp(iH)$, and does it have any physical significance or uses?

REPLY [5 votes]: This hermitian generator is simply the Hamiltonian of harmonic oscillator $H=\frac{1}{2}(x^2+p^2)$. This is simple to understand. In Heisenberg picture evolution with this Hamiltonian is just rotations in $x,p$ plane. Fourier transform is transformation which maps $x \to p$ and $p \to -x$ so it is a particular rotation with angle $\frac{\pi}{2}$. Hence $\mathcal F = \exp (-i \frac{\pi H}{2})$.