Infinite sum of Hermite polynomials with same order, but different argument

Question

I am looking for any possible simplification of the following sum for positive reals $\alpha,\beta$ and positive integer $n$: $$ \sum_{t=-\infty}^{\infty}e^{-\beta(t+\alpha)^{2}}H_{n}(t+\alpha) $$ I've looked around and all I can find is sums of Hermites of different orders. This one, however, keeps the order and changes argument. Thanks!

Answer 1

When $n=0$ this sum can be expressed in terms of theta function. Since $H_n(t+\alpha)$ is a polynomial in $t+\alpha$, and as a consequence in $t$, this sum can be represented as a sum of derivatives of theta functions.

For $\beta=1$ a simple expression can be obtained. Consider the case $n\ge 1$. Using the integral representation of Hermite polynomials $$ e^{-(t+\alpha)^2}H_n(t+\alpha)=\frac{2^{n+1}}{\sqrt{\pi}}\int_0^\infty e^{-x^2}x^n\cos\left(2xt+2x\alpha-\frac{\pi n}{2}\right)dx $$ one can obtain when $\beta=1$ \begin{align} \sum_{t=-\infty}^{\infty}e^{-(t+\alpha)^{2}}H_{n}(t+\alpha)&=\frac{2^{n+1}}{\sqrt{\pi}}\sum_{t=-\infty}^{\infty}\int_0^\infty e^{-x^2}x^n\cos\left(2xt+2x\alpha-\frac{\pi n}{2}\right)dx\\ &=\frac{2^{n+1}}{\sqrt{\pi}}\int_0^\infty e^{-x^2}x^n\sum_{t=-\infty}^{\infty}\cos 2xt\cos\left(2x\alpha-\frac{\pi n}{2}\right)dx. \end{align} Dirac comb can help to simplify the integral: \begin{align} \sum_{t=-\infty}^{\infty}e^{-\beta(t+\alpha)^{2}}H_{n}(t+\alpha)&=\frac{2^{n+1}}{\sqrt{\pi}}\int_0^\infty e^{-x^2}x^n\sum_{t=-\infty}^{\infty}\delta\left(\frac{x}{\pi}-t\right)\cos\left(2x\alpha-\frac{\pi n}{2}\right)dx\\ &=\pi^{n+1/2}{2^{n+1}}\sum_{t=1}^{\infty}e^{-\pi^2t^2}t^n\cos\left(2\pi \alpha t-\frac{\pi n}{2}\right). \end{align} One can see that this sum can be represented as $n$-th derivative of theta function.