# How do I compute the eigenfunctions of the Fourier Transform? [closed]

I read today (ref) that the Continuous Fourier Transform has four eigenvalues: +1, +i, -1, and -i. Associated with each eigenvalue is a space of eigenfunctions: functions which retain their form after undergoing the Fourier transform. Perhaps the best known example is the Gaussian: the Fourier transform of a Gaussian is again Gaussian.

A more general example is the Hermite-Gauss functions (Gaussian multiplied by Hermite polynomial). These are also eigenfunctions of the Fourier transform, which is why TEMxx modes (with Hermite-Gauss transverse profiles) are stable modes of laser beam propagation.

This leads me to wonder a number of things:

• How do I show that the Fourier transform has just these four eigenvalues?
• Given an arbitrary function, how do I find the projections onto the four eigenspaces?
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## closed as off topic by David Z♦, J. M., kennytm, ptomato, Pratik DeoghareNov 9 '10 at 10:43

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Wouldn't math.stackexchange be a better place to ask this? – Gerard Nov 8 '10 at 23:08
The fractionally iterated Fourier transform is useful in optics, for describing near-field diffraction. – DarenW Dec 4 '10 at 7:27
For eigenvalues, just apply Fourier four times to any function. You get the original back - eigenvalue 1. Now find the fourth roots of that value. – DarenW Dec 4 '10 at 7:38

## 1 Answer

You'd have better luck on math.stackexchange.com, but here's a short outline of how to attack your problem.

• By definition, to calculate the eigenvalues of the Fourier transform, solve the equation $$\mathcal{F}{f(x)} = \lambda f(x)$$ or $$\int_{-\infty}^\infty f(x) e^{-2\pi i X x} dx - \lambda f(x) = 0$$
• To decompose a function into Hermite-Gaussian functions $HG_n(x)$, i.e. find $c_n$ in $$f(x) = \sum_n c_n HG_n(x),$$ take the inner product of $f(x)$ with each Hermite-Gauss function: $$c_n = \int_{-\infty}^\infty f(x)\overline{HG_n(x)} dx$$ where $\bar{x}$ denotes the complex conjugate of $x$.

Disclaimer: this is all from memory, so please inform me of the inevitable mistakes, and I'll be glad to correct them.

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