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By making use of the generating function $$\sum_{m=0}^\infty \frac{H_m(x)}{m!} t^m=e^{-t^2 + 2xt} $$ for the real Hermite polynomials $H_m$, we get easily that $$(*)\quad \sum_{m,n=0}^\infty \frac{u^m}{m!} \frac{v^n}{n!} H_m(x+\frac{y}{2}) H_n(x-\frac{y}{2}) =e^{-(u^2+v^2) + 2x(u+v)+y(u-v)} .$$

I am looking for a reference which I find the analog of $(*)$ for the complex Hermite polynomials, i.e., a formula as follows $$\sum_{??}\, ?? \, H_{m,n}(z,\bar z) H_{k,l}(w,\bar w) = ?? .$$

Thank you in advance

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2 Answers 2

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The formula (*) is trivial since it breaks up into a product of a sum over $m$ with a sum over $n$. The same reasoning works for the complex analog as a sum over $m,n,k,l$, using $$ \sum_{m,n\geq 0}H_{m,n}(z,\bar{z})\frac{u^m}{m!}\frac{v^n}{n!} = e^{uz+v\bar{z}-uv}. $$ More interesting would be to find the complex analogue of the formula $$ \sum_{n\geq 0}\frac{H_n(x)H_n(y)}{n!}\left(\frac u2\right)^n = $$ $$ \frac{1}{\sqrt{1-u^2}} \exp\left( \frac{2u}{1+u}xy-\frac{u^2}{1-u^2} (x-y)^2\right). $$ I suspect that this should be possible.

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  • $\begingroup$ This is a nice problem! $\endgroup$ Dec 31, 2016 at 5:57
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The OP's question has been addressed By Richard Stanley.

So, we make attempt at Stanley's question on the complex counterpart to $$\sum_{n\geq0}\frac{H_n(x)H_n(y)}{n!}\left(\frac{u}2\right)^2 =\frac1{\sqrt{1-u^2}}\cdot\exp\left(\frac{2u}{1+u}xy-\frac{u^2}{1-u^2}(x-y)^2\right).$$ Mimicking the RHS and after some (tedious) infinite series manipulation with the generating function $$\sum_{m,n\geq0}H_{m,n}(z,\bar{z})\frac{u^mv^n}{m!n!}=e^{uz+v\bar{z}-uv},$$ we arrive at the desired double-indexed formula $$\sum_{m,n\geq0}H_{m,n}(z,\bar{z})H_{n,m}(w,\bar{w})\frac{u^mv^m}{m!n!} =\frac1{1-uv}\cdot \exp\left(\frac{-uv(\vert z\vert^2+\vert w\vert^2)+uz\bar{w}+v\bar{z}w}{1-uv}\right).$$ Notice the "annoying" index-swap in $H_{n,m}(w,\bar{w})$. We may use $H_{n,m}(w,\bar{w})=\overline{H_{m,n}(w,\bar{w})}$.

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