$\sum_{n=0}^\infty\frac{H_n(x)H_n(y)t^n}{2^nn!}$=$\frac{\exp\left[\frac{2xyt-(x^2+y^2)t^2}{1-t^2}\right]}{\sqrt{1-t^2}}$

Question

I am told to prove that :

$$\sum_{n=0}^\infty\frac{H_n(x)H_n(y)t^n}{2^nn!} = \frac{\exp\left[\frac{2xyt-(x^2+y^2)t^2}{1-t^2}\right]}{\sqrt{1-t^2}}$$

where $H_n(x)$ is Hermite polynomial.I am wondering how to prove it.please help me how to prove this. Thanks in advance!

Answer 1

Hint: Try to use the generating function for Hermite polynomials, given by $$\exp(2xt-t^2) = \sum_{n=0}^{\infty} H_n(x) \frac{t^n}{n!}$$

EDIT:

As suggested by OP, I am posting the link to the solution here.The full derivation is mentioned there with the final expression in Eq. 18 with the simple substitution $t \to \frac{t}{2}$ in that equation to obtain the result here.

Answer 2

Use the generating function given in a previous answer. Use the integral

$$ H_n(y) = \frac{2^{n/2}}{\sqrt{2\pi}} \int_{-\infty}^\infty (y\sqrt{2}+i\,u)^n \exp{(-u^2/2)} \, du $$

(This appears on the wiki for Hermite polynomials, but I've converted it to physicist notation.) Then

$$ \sum_{n=0}^\infty \frac{(t/2)^n}{n!}H_n(x)\,H_n(y) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{-u^2/2} \sum_{n=0}^\infty \frac{H_n(x)}{n!} \big(\frac{t}{\sqrt{2}}(y\sqrt{2} + iu)\big)^n = $$ $$= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-u^2/2} \exp{\big(2x(t\,y+it\,u/\sqrt{2}) - t^2/2(y\sqrt{2}+iu)^2 } \big)$$ where in the last step the generating function was used. Collect the coefficients of $u$ in the exponential argument. The constant term in front of the integral is $\exp{(2\,t\,x\,y - t^2\,y^2)}.$ Now use the well-known formula $$ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \exp\big({-a\,u^2/2 + \sqrt{2}\,i\,b\,u}\big)\,du = \exp{\big(-b^2/a\big)}$$ where $a=1-t^2$ and $b=t(x-t\,y).$ Algebra completes the proof.