1
$\begingroup$

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!

$\endgroup$
2
1
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ At the first shift I get $\int^b_a x^k\frac{d^n}{dx^n}e^{-x^2}dx=x^k\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}|^b_a-\int_a^b kx^{k-1}\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}dx$, at the $j$-th step I have the "left" term outside the integral in the form $x^{k-j+1}\frac{d^{n-j}}{dx^{n-j}}e^{-x^2}|^b_a$, but I'm not able to prove that it approaches 0 as $b\to\infty$... Thank you so much again! $\endgroup$ – Self-teaching worker Nov 13 '14 at 16:41
  • 1
    $\begingroup$ That expression is just of the type $P(x)\exp(-x^2)$ with some polynomial $P$, which vanishes at $\pm\infty$ because $\exp(-x^2)$ decays faster than any polynomial. This can be proven using elementary properties of the exponential function. $\endgroup$ – Daniel Nov 13 '14 at 19:09
  • $\begingroup$ Oh, sorry, now I understand: thank you so much!!! $\endgroup$ – Self-teaching worker Nov 13 '14 at 19:30
  • 1
    $\begingroup$ You're welcome :) $\endgroup$ – Daniel Nov 13 '14 at 19:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.