In some physics related problem, I found out the curious identity $$\sum\limits_{n_1+n_2+n_3=n}\frac{n!}{n_1!\,n_2!\,n_3!}\,H_{2n_1}(x)\,H_{2n_2}(y)\,H_{2n_3}(z)=\frac{H_{2n+1}(r)}{2r},$$ where $H_n(x)=(-1)^ne^{x^2}\frac{d^n}{dx^n}e^{-x^2}$ are Hermite polynomials and $r=\sqrt{x^2+y^2+z^2}$. Is this identity known?
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asked Nov 29, 2018 at 5:31
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1 Answer
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If we define the generating functions $F(x,t)=\sum_{n=0}^{\infty}H_{2n}(x)\frac{t^n}{n!}$ and $G(x,t)=\sum_{n=0}^{\infty}H_{2n+1}(x)\frac{t^n}{n!}$ then your identity is equivalent to $$F(x,t)F(y,t)F(z,t)=\frac{G\left(\sqrt{x^2+y^2+z^2},t\right)}{2\sqrt{x^2+y^2+z^2}}.$$ This is in turn an immediate corollary to the fact that we have $$F(x,t)=\frac{1}{(1+4t)^{1/2}}\exp\left(\frac{4tx^2}{1+4t}\right)$$ $$G(x,t)=\frac{2x}{(1+4t)^{3/2}}\exp\left(\frac{4tx^2}{1+4t}\right)$$ and these have appeared in the literature in various contexts. Here is a paper that has a physics style proof, and here is one that derives it from the exponential generating function of $H_n(x)H_n(y)$.
answered Nov 29, 2018 at 7:03
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1$\begingroup$ Thanks for the references! I proved the identity not by using this new generating function (which was unknown for me) but by using $H_n(x)=e^{-\frac{1}{4}\,\frac{d}{dx}}\,(2x)^n$, analogous formula for scalar Hermite polynomials (with $\frac{d}{dx}$ replaced by Laplacian) and relations to the Laguerre polynomials. Of course this specific generating function makes the proof almost trivial. $\endgroup$ Commented Nov 29, 2018 at 7:48
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1$\begingroup$ There is a typo in the above comment: the correct formula is $H_n(x)=e^{-\frac{1}{4}\,\frac{d^2}{dx^2}}\,(2x)^n$. $\endgroup$ Commented Nov 29, 2018 at 8:24