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I was stuck in proving the orthogonality of Hermite functions. Assume our Hermite functions is defined as $H_{n} = e^{-x^{2}/2}\frac{d^{n}}{dx^{n}}e^{-x^{2}/2}$. What I want to show is $$\int_{-\infty}^{\infty} H_{n}H_{m}e^{-x^{2}}dx = 0-(1).$$

If we modify the weight $e^{-x^{2}}$ in (1.) to be $e^{-x^{2}/2}$, in this case, we have $$\int_{-\infty}^{\infty} H_{n}H_{m}e^{-x^{2}/2} dx = \int_{-\infty}^{\infty} H_{m} (\frac{d^{n}}{dx^{n}}e^{-x^{2}/2}) dx -(2)$$ To prove $\int_{-\infty}^{\infty} H_{m} (\frac{d^{n}}{dx^{n}}e^{-x^{2}/2}) dx=0$, it is equivalent to show that $\int_{-\infty}^{\infty} x^{m} (\frac{d^{n}}{dx^{n}}e^{-x^{2}/2}) dx=0$ for any m < n , which can be proved by using integration by parts m-times. However, this argument can not be applied to (1.).

Any hint or reference on how to prove (1.) would be greatly appreciated. Thanks!

Roddick Yu
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    Use the generating function $f(x,t) = e^{-x^2/2 +2xt-t^2}$ then $g(ut)=\int_{-\infty}^\infty f(x,t)f(x,u)dx = \sum_{m,n} t^m u^n <H_n,H_m>$ where $g$ has a closed-form – reuns Oct 27 '19 at 20:07
  • This is wrong! Set $f(x) = e^{-x^2/2}$. Then $f'' = (x^2-1)f$ and so $\int H_2H_0e^{-x^2},dx = \int (ff'')f^2f^2,dx = \int f^5f'',dx = \int (x^2-1)e^{-3x^2},dx$, which is not zero. – amsmath Oct 27 '19 at 20:51
  • Hi, $H_{n}$ is defined as $e^{-x^{2}/2}\frac{d^{n}}{dx^{n}}e^{\frac{-x^{2}}{2}}$, so it is a pure polynomial, $\int H_{2}H_{0}e^{-x^{2}}$ should be $\int (x^{2}-1)e^{-x^{2}}dx$ – Roddick Yu Oct 30 '19 at 12:57

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