Fourier transform as exponential of Hermitian operator

Question

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?

Answer 1

I found that question quite interesting and found a very nice paper on the arXiv: Classical Structures in Quantum Mechanics and Applications. It deals with the Fourier transform in the framework of quantum mechanics (QM). In QM the Fourier transform is used frequently to transform from position to momentum space: $$\hat F|q(x)\rangle=|p(x)\rangle,$$ where $|q(x)\rangle$ and $|p(x)\rangle$ are quantum eigenkets of the position and momentum observables and $\hat F$ is the operator of the Fourier transform. $\hat F$ can be defined elegantly as $$\hat F=\int^\infty_{-\infty}dx|p(x)\rangle\langle q(x)|.$$

$\hat F$ is unitary and therefore can be expressed via the exponential of a hermitan operator. For the Fourier transform in this setting this hermitian generator is the number operator defined with the ladder operators: $ \hat N=\hat a^\dagger \hat a$:$$\hat F=\exp(i(\pi/2)\hat N).$$ This implies a nice bridge between second quantisation: ladder operators, number operator,.. and the canonical represantaion in position or momentum states.

For a more detailed explanation of this, the consequences and possible generalisations I highly recommend reading that paper. Some points coming up in that paper related are "Weyl-Wigner Formalism" and "Quantum Symplectic Group".

Answer 2

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 $\mathcal F = \exp (-i \frac{\pi H}{2})$.