The Orthogonality of Hermite functions 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!
 A: Forget about doing the integral. It can be done but it is not the most straight forward way.
Consider Hermite's equation;
$ \frac{d^2y}{dx^2} -2x\frac{dy}{dx} = -\lambda y $
Now if we consider the "Hermite differential operator":
$L_H \equiv \frac{d^2}{dx^2} - 2x\frac{d}{dx}$
and we can see that: $ L_H[y] = -\lambda y $
We can put it in Sturm-Liouville form (if you don't know what that is , you should look up Sturm Liouville theory) :
$ L_H = e^{x^2}[e^{-x^2} \frac{d^2}{dx^2} - 2xe^{-x^2}\frac{d}{dx}]$
So now we know that the Hermite polynomials solve the DE. Consider two different Hermite polynomials $H_n,H_m$  where:
$\frac{d^2H_n}{dx^2} -2x\frac{dH_n}{dx} = -\lambda_n H_n  \, \, \, (1)$
$\frac{d^2H_m}{dx^2} -2x\frac{dH_m}{dx} = -\lambda_m H_m \, \, \, (2) $
Now multiply $(1)$ by $H_m$ and $(2)$ by $H_n$:
$H_m\frac{d^2H_n}{dx^2} -2xH_m\frac{dH_n}{dx} = -\lambda_n H_nH_m  \, \, \, (3)$
$H_n\frac{d^2H_m}{dx^2} -2xH_n\frac{dH_m}{dx} = -\lambda_m H_mH_n \, \, \, (4) $
Now subtract $(4)$ from $(3)$ then divide by $e^{x^2}$ on the both sides and on the LHS you use the reverse product rule to get a derivative (I leave the details for you to workout):
$\frac{d}{dx}[e^{-x^2}(H_m\frac{dH_n}{dx} - H_n\frac{dH_m}{dx})] = (\lambda_m - \lambda_n)e^{-x^2}H_nH_m $
Then integrate both sides from $-\infty$ to $+\infty$
Using properties of the Hermite polynomials the LHS will be zero and you are left with:
$(\lambda_m - \lambda_n)\int_{-\infty}^{\infty}e^{-x^2} H_mH_n dx = 0$
and since $m \neq n$ then $\lambda_m - \lambda_n \neq 0$ so the integral must be equal zero.
