Infinite sum of Hermite polynomials with same order, but different argument 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!
 A: 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.
