I am trying to prove an identity involving Hermite polynomials using other identities from Wikipedia, but I can't find the way. I have checked the identity in Mathematica for many values of $n$ and it holds for all values of $n$ I have tried. The identity is
$$ \text{He}_k(x) \text{He}_l(x) = \sum_{q=0}^l \frac{k! \, l!}{(l-q)! \, q! \, (k-l+q)!} \, \text{He}_{2q+k-l} (x) \, ,$$
where $k \geq l$ (this is important: the sum only runs until the smallest of $\{k, l\}$) and $\text{He}_n(x)$ is the probabilists' Hermite polynomial
$$\text{He}_n(x) = (-1)^n e^{\frac{x^2}{2}} \frac{d^n}{dx^n} e^{-\frac{x^2}{2}} \, .$$
Any ideas? At first I was trying to prove the particular case $k=l$: I asked the question here and obtained and answer, using Laguerre polynomials and Mehler's formula. But now I think the proof doesn't work for this more general case, mainly because Mehler's formula is only valid for the the product of two Hermite polynomials of the same order (for the case $k=l$).
Best Answer
In the work N. Nielsen, Recherches sur les polynomes d'Hermite, Det. kgl. Danske Videnskabernes Selskabs, Mathematisk-fysyske Meddelelser, I, 6, pp. 1-78 the following formula is proved in page 33:
$$H_n(x,a) H_p(x,a) = \sum_{s=0}^{\text{max}(n,p)} \frac{(2a)^s}{s!} {n+p-2s \choose n-s} H_{n+p-2s} (x,a) \, ,$$
where the convention is $n! H_n(x,\frac{1}{2}) = \text{He}_n(x)$. Therefore, for $k \geq l$
$$\text{He}_k(x) \, \text{He}_l(x) = \sum_{s=0}^{l} \frac{k! \, l!}{s! \, (k+l-2s)!} {k+l-2s \choose l-s} \text{He}_{k+l-2s} (x) \, \\ = \sum_{s=0}^{l} \frac{k! \, l!}{s! \, (l-s)! \, (k-s)!} \text{He}_{k+l-2s} (x) \, \\ = \sum_{q=0}^{l} \frac{k! \, l!}{(l-q)! \, q! \, (k-l+q)!} \text{He}_{k+2q-l} (x) \, , $$
where $q = l-s$.