2

In studying a physics problem, I've encountered the following series of Hermite polynomials

$$ \sum_{n=0}^{\infty} \frac{H_n(x) H_n(y)}{n!2^n(n+1)^2}.$$

As of now, I am completely in a loss on whether anything can be said about this series. When looking for properties of Hermite polynomials, I've found (for example in this question) a similar expression using a Dirac delta distribution (a simple derivation of half of the formula was found here). The full version, found on Wolfram functions site, is

$$ \sum_{n=0}^{\infty} \frac{H_n(x) H_n(y)}{n!2^n}=\sqrt{\pi} \exp\left[(x^2+y^2)/2\right]\delta(x-y),$$

which seems very similar to what I'm trying to evaluate, and even shows some intuition to me that since the Hermie polynomials are orthonormal to eachother I am supposed to get a Dirac delta (with the weight function aptly treated on the right side here).

However, the series I need to evaluate is not this, but instead each term is multiplied by a factor of $(n+1)^{-2}$, and I don't see a way to handle this nicely. I've tried applying the Mehler kernel, but I can't find a choice of $\rho$ that would result in my particular series.

At this point I'm tempted to believe - based on intuition and not mathematics - that my series should evaluate to exactly the same result as without having the extra term, seeing as the Hermite polynomials are still orthonormal no matter how we weigh each of them in the sum. This seems logical to me, however this argument is ad hoc enough that it's probably false.

Does anyone have a clue on how to say anything (or even fully evaluate) about my series?

  • Look up "kernel" of an operator, Hermite functions as opposed to Hermite polynomials, and how they provide an orthonormal basis of $L^2(\mathbb{R})$ which diagonalize the Harmonic oscillator Hamiltonian $H$. The first formula without the $(n+1)^2$ and with the exponentials moved to the LHS should give the kernel of the identity operator. The same with $(n+1)^2$ should give the kernel of something like $(H+cI)^{-2}$, for some $c>0$. Also, look up "Christoffel-Darboux kernel". – Abdelmalek Abdesselam Jul 16 '21 at 16:29

1 Answers1

1

Citing Wikipedia which cites Mehler, $$\sum_{n\geq 0}\frac{H_n(x)H_n(y)}{n!}\left(\frac{u}{2}\right)^n = \frac{1}{\sqrt{1-u^2}}\,\exp\left(\frac{2u}{1+u}xy-\frac{u^2}{1-u^2}(x-y)^2\right) \tag{1}$$ follows from $$ H_n(x) = (-1)^n e^{x^2}\frac{1}{2\sqrt{\pi}}\int (is)^n e^{isx-\frac{s^2}{4}}\,ds.\tag{2} $$ As a consequence, in order to compute your series it is sufficient to (numerically) evaluate $$ \sum_{n\geq 0}\frac{H_n(x)H_n(y)}{2^n n!(n+1)^2}=\int_{0}^{1}\frac{-\log u}{\sqrt{1-u^2}}\,\exp\left(\frac{2u}{1+u}xy-\frac{u^2}{1-u^2}(x-y)^2\right)\,du. \tag{3}$$

Jack D'Aurizio
  • 353,855
  • I suppose somewhere in getting your final formula, you've used some identity for $\sum_{n=0}^\infty \frac{1}{2^n n! (n+1)^2} (-ust)^n$? Could you hint me what it was? –  Jul 19 '21 at 10:05
  • 1
    @LaBelleCroissant: I just used the material linked on Wikipedia and $$\frac{1}{(n+1)^2}=\int_{0}^{1}(-\log u)u^n,du.$$ – Jack D'Aurizio Jul 19 '21 at 13:57
  • Ok, I didn't know this formula for $\frac{1}{(n+1)^2}$, now it's all clear, thanks! –  Jul 20 '21 at 15:15
  • I know this is a somewhat old question, but I am still thinking about it, and I've just realized that I don't know how you came up with that integral for $\frac{1}{(n+1)^2}$. I mean sure, I can check that it gives the right result, but how did you find the integral in the first place? Let's say I want to do the same, but with $\frac{1}{n+1}$: how do I find a simple integral, that, evaluated from 0 to 1, gives me this fraction? –  Feb 16 '22 at 08:16
  • The integration of -u^n log(u) over (0,1) gives 1/(n+1)^2 – Jack D'Aurizio Feb 17 '22 at 06:04
  • I understand that, my question is, how did you come up with the integral of -u^n log(u) over (0,1)? I assume it is not possible to know an integral for every single series possible, so you must have an approach when writing it in an integral form –  Feb 17 '22 at 14:02