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 &lt; 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!
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.
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.