Orthogonality of Hermite functions

Question

I would like to prove to myself that Hermite functions, defined by $\varphi_n(x)=(-1)^n e^{x^2/2}\frac{d^n e^{-x^2}}{dx^n}$, $n\in\mathbb{N}$ are an orthogonal system in $L^2(\mathbb{R})$, i.e. that, for any $m\ne n$, $$\int_{-\infty}^\infty e^{x^2} \frac{d^m e^{-x^2}}{dx^m} \frac{d^n e^{-x^2}}{dx^n}dx=0 $$ I have tried by integrating by parts as the book where I find such a statement, Kolmogorov-Fomin's Элементы теории функций и функционального анализа, says that it can be proved, but I am landing nowhere since I have no experience in manipulating such integrals... Thank you so much for any answer!

Answer 1

It is actually easier to prove the orthogonality of the Hermite polynomials $$ H_n=(-1)^n \exp(x^2)\frac{d^n}{dx^n}\exp(-x^2) $$ with respect to the weight $d\omega=\exp(-x^2)dx$. It should be obvious that both claims are the same.

First notice, that each $H_n$ is obviously a polynomial of degree $n$ (this is a simple proof by induction). Let $k$ be any nonnegative integer smaller than $n$, then there holds $$ \int_\mathbb{R}x^kH_n(x)d\omega=\int_\mathbb{R}x^kH_n(x)\exp(-x^2)dx=\int_\mathbb{R}x^k\frac{d^n}{dx^n}\exp(-x^2)dx=0 $$From the last expression, you can actually see that this integral is zero, as you can partially integrate and shift the derivation to $x^k$ until the term vanishes (remember $k<n$). The boundary terms obviously vanish as $\exp(-x^2)$ decays rapidly. As $H_m$ is only a linear combination of $x^k$, this directly implies, that for any $m<n$,$$ \int_\mathbb{R}H_m(x)H_n(x)d\omega=\int_\mathbb{R}\varphi_m(x)\varphi_n(x)dx=0 $$ which proves the claim.