In Andy's answer to the question "What are fixed points of the Fourier Transform" on Math Overflow, he shows that the Fourier Transform has eigenvalues $\{+1, +i, -1, -i \}$ and that the projections of any function onto the corresponding four eigenspaces may be found through some simple linear algebra.

I would like to get a better feeling for these four eigenspaces of the fourier transform.

  1. How can I find some interesting members of each of these eigenspaces?
  2. How can I show that Hermite-Gaussians are in one (or more?) of the eigenspaces?
  3. How can one define usable projection operators onto these eigenspaces?
  4. The wikipedia article on the Fourier Transform mentions that Wiener defined the Fourier Transform via these projections. What exactly was Wiener's approach?

1 Answer 1


One method is to apply the Bargmann transform:

$$B_f(z) = \int_{-\infty}^{\infty} f(x) \exp \left(2 \pi x z - \pi x^2 - \frac{\pi}{2}z^2 \right) dy$$

to the eigenvalue equation:

$$f(x) = \lambda \int_{-\infty}^{\infty}f(y) \exp(-2 \pi ix y) dy$$

to obtain the following relation for the Bargmann transformed eigenfunctions:

$$B_f(z) = \lambda B_f(-iz) $$

whose solutions are the monomials:

$B_f(z) = z^n$, corresponding to the eigenvalues $i^{n \bmod{4}}$.

Now, the inverse Bargmann transform of the monomials are just the Hermite functions.

Remark: The application of the Bargmann transform to the eigenvalue equation entails a change of the integration order which is possible because the Hermite functions are bounded.

The projection operators onto the $n$th subspace in the Bargamann representation are given by the kernels

$$ P(v,\bar{z})=\frac{(v\bar{z})^n}{n!} ,$$

which act on the Bargmann transformed functions according to:

$$ Pf(v) = \int_{\mathbb{C}}P(v,\bar{z}) f(z) \exp(-z\bar{z})dzd\bar{z} .$$

In the time domain the projection kernels are just $P(t, \tau) = H_n(\tau) H_n(t)$ due to the orthonormality of the Hermite functions.

  • $\begingroup$ Why would this give all possible eigenfunctions? $\endgroup$
    – JT_NL
    Nov 18, 2010 at 13:32
  • $\begingroup$ The sum of the projection kernels $\sum_{n=1}^{\infty}P(v,\bar{z})=exp(v\bar{z})$ is just the reproducing kernel of the Bargmann space, thus the eigenfunctions constitute of a complete basis of the Bargmann space $\endgroup$ Nov 18, 2010 at 14:07

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