15

What are the eigenfunctions and the eigenvalues of the exponential kernel?

The exponential kernel is defined as $$k(x,x')=\sigma^2\exp\left(\frac{||x-x'||}{l}\right)$$ where both $\sigma>0$ and $l>0$.

Mercers theorem tell us that for every kernel function $k(x,x')$ there exists a decomposition in eigenfunctions $\phi_i(x)$ and corresponding eigenvalues $\lambda_i$ such that

$$k(x,x')=\sum_{i=1}^\infty \lambda_i \phi_i(x)\phi_i(x)$$

The Fourier transform $$\mathcal{F}(k)(\omega)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} k(r) e^{i\omega r}dr$$ of the function $$k(r)=\sigma^2 \exp\left(\frac{||r||}{l}\right)$$ with $r=x-x'$ is $$\mathcal{F}(k)(\omega)=\frac{\sqrt{\frac{2}{\pi }} \sigma^2 l}{l^2 \omega ^2+1}.$$ How to proceed from here?

Julian Karch
  • 1,890
  • 1
  • 18
  • 29

2 Answers2

1

First of all, your question is not quite well-posed. The reason is that Mercer's theorem only applies for the case of a kernel defined on a finite measure space. Practically, this means that in order to apply the theorem, the eigenfunctions $\phi_i$ are in fact taken with respect to the operator $$K_{\mu}(f)= \left( x\mapsto \int_{\mathbb{R}} K(x,y)f(y)\mu(dy)\right)$$ where $\mu(dy)=p(y)dy$ is a probability measure. The $\phi_i$ are then orthonormal wrt to the inner product defined by $<f,g>=\int f(x)g(x)\mu(dx)$.

It is simple to see that the condition $\mu(\mathbb{R})<\infty$ is necessary for Mercer's theorem to hold. Consider the identity:

$$\int_{\mathbb{R}} e^{-(x-y)^2}xdx=\sqrt{\pi}y$$

This shows that the function $f(x)=x$ is an eigfunction of the operator $\int K(x,y)f(y)dy$. But evidently $\int_{\mathbb{R}} f(x)^2dx=\infty$, which shows that it is not possible to construct an orthonormal basis of $K$, without introducing a weighting function $p(y)$.

Secondly, I am assuming there should be a minus sign in the definition of the kernel $e^{-|x-y|}$, otherwise the resulting kernel fails to be positive definite.

Simon Segert
  • 1,974
  • Aren't you quoting the solution to a different problem? The question concerns an exponential kernel, not a Gaussian one. – whuber Jan 17 '22 at 18:45
  • 1
    @whuber, oops, you are right, I misread the question. but the first point still stands – Simon Segert Jan 17 '22 at 18:48
0

Assuming the Hamiltonian of your system is the |x> operator, what you are really trying to find is the reciprocal space of x. The easiest way to do this is to take the fourier transform of k(x,x') which by definition is a linear combination of of the k(x,x') states. You can read more here: https://en.wikipedia.org/wiki/Fourier_transform

I also recommend you switch variables to r = x-x' to simplify the math.

Greg Petersen
  • 361
  • 1
  • 6
  • Hey thanks for your comment. I added your suggestion to my original question. How to proceed from there? – Julian Karch Jan 14 '16 at 12:29
  • 1
    Actually I should take a step back. What operator are you investigating? To find eigenvalues and eigenfunctions you need to define this first. I realized I was assuming it was the Fourier Transform operator. – Greg Petersen Jan 15 '16 at 19:11