# Fixed points of the Fourier transform

This previous post looked at the hyperbolic secant distribution. This distribution has density

and characteristic function sech(t). It’s curious that the density and characteristic function are so similar.

The characteristic function is essentially the Fourier transform of the density function, so this says that the hyperbolic secant function, properly scaled, is a fixed point of the Fourier transform. I’ve long known that the normal density is its own Fourier transform, but only recently learned that the same is true of the hyperbolic secant.

## Hermite functions

The Hermite functions are also fixed points of the Fourier transform, or rather eigenfuctions of the Fourier transform. The eigenvalues are 1, i, -1, and i. When the eigenvalues are 1, we have fixed points.

There are two conventions for defining the Hermite functions, and multiple conventions for defining the Fourier transform, so the truth of the preceding paragraph depends on the conventions used.

For this post, we will define the Fourier transform of a function f to be

Then the Fourier transform of exp(-x²/2) is the same function. Since the Fourier transform is linear, this means the same holds for the density of the standard normal distribution.

We will define the Hermite polynomials by

using the so-called physics convention. Hn is an nth degree polynomial.

The Hermite functions ψn(x) are the Hermite polynomials multiplied by exp(-x²/2). That is,

With these definitions, the Fourier transform of ψn(x) equals (-i)n ψn(x). So when n is a multiple of 4, the Fourier transform of ψn(x) is ψn(x).

[The definition Hermite functions above omits a complicated constant term that depends on n but not on x. So our Hermite functions are proportional to the standard Hermite functions. But proportionality constants don’t matter when you’re looking for eigenfunctions or fixed points.]

## Hyperbolic secant

Using the definition of Fourier transform above, the function sech(√(π/2) x) is its own Fourier transform.

This is surprising because the Hermite functions form a basis for L²(ℝ), and all have tails on the order of exp(-x²), but the hyperbolic secant has tails like exp(-x). Each Hermite function eventually decays like exp(-x²), but this happens later as n increases, so an infinite sum of Hermite functions can have thicker tails than any particular Hermite function.