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Objective:

Show that

$$ \int^{\infty}_{-\infty} x e^{-x^2} H_n(x) H_m(x) dx = \pi^{1/2} 2^{n-1} n! \delta_{m,n-1} + \pi^{1/2} 2^n (n+1)! \delta_{m,n+1} $$

My attempt at this is:


\begin{eqnarray*} \sum^\infty_{n=0} \frac{2^n s^n u^n}{n!} (s+u) \sqrt{\pi} &=& \sum^\infty_{m,n=0}\frac{s^m u^n}{m! n!} \int x H_m(x) H_n(x) e^{-x^2} dx \\ \sqrt{\pi}\bigg( \sum^\infty_{n=0} \frac{2^n s^{n+1} u^n}{n!} +\sum^\infty_{n=0} \frac{2^n s^n u^{n+1}}{n!} \bigg) &=& \sum^\infty_{m,n=0}\frac{s^m u^n}{m! n!} \int x H_m(x) H_n(x) e^{-x^2} dx \end{eqnarray*}

I'm reasonably confident up to here.

\begin{eqnarray*} \sqrt{\pi}\bigg( \sum^\infty_{n=0} \frac{2^n s^{n+1} u^n}{n!} +\sum^\infty_{n=0} \frac{2^n s^n u^{n+1}}{n!} \bigg) \delta_{m,n+1}&=& \delta_{m,n+1}\sum^\infty_{m,n=0}\frac{s^m u^n}{m! n!} \int x H_m(x) H_n(x) e^{-x^2} dx \end{eqnarray*} \begin{eqnarray*} \sqrt{\pi}\bigg( \sum^\infty_{n=0} \frac{2^n s^{n+1} u^n}{n!}\big(\delta_{m,n+1}\frac{m! n!}{s^m u^n}\big) +\sum^\infty_{n=0} \frac{2^n s^n u^{n+1}}{n!}\big(\delta_{m,n+1}\frac{m! n!}{s^m u^n}\big) \bigg) \delta_{m,n+1}&=& \int x H_m(x) H_n(x) e^{-x^2} dx \end{eqnarray*}

Then I thought maybe I could try re-indexing the n and m in the second $\big(\delta_{m,n+1}\frac{m! n!}{s^m u^n}\big)$ term.

\begin{eqnarray*} \sqrt{\pi}\bigg( \sum^\infty_{n=0} \frac{2^n s^{n+1} u^n}{n!}\big(\frac{(n+1)! n!}{s^{n+1} u^n}\big) +\sum^\infty_{n=0} \frac{2^n s^n u^{n+1}}{n!}\big(\delta_{m,n+1}\frac{n! m!}{s^n u^m}\big) \bigg) \delta_{m,n+1}&=& \int x H_m(x) H_n(x) e^{-x^2} dx \\ \bigg(\sqrt{\pi} 2^n (n+1)! + \sqrt{\pi} \sum^\infty_{n=0} \frac{2^n s^n u^{n+1}}{n!}\big(\delta_{m,n+1}\frac{n! m!}{s^n u^m}\big) \bigg) \delta_{m,n+1}&=& \int x H_m(x) H_n(x) e^{-x^2} dx \\ \sqrt{\pi} 2^n (n+1)! \delta_{m,n+1} + \sqrt{\pi} \sum^\infty_{n=0} \frac{2^n s^n u^{n+1}}{n!}\big(\frac{n! (n+1)!}{s^n u^{n+1}}\big) \delta_{m,n+1}&=& \int x H_m(x) H_n(x) e^{-x^2} dx \\ \sqrt{\pi} 2^n (n+1)! \delta_{m,n+1} + \sqrt{\pi} 2^n (n+1)! \delta_{m,n+1}&=& \int x H_m(x) H_n(x) e^{-x^2} dx \\ \sqrt{\pi} 2^{n+1} (n+1)! \delta_{m,n+1} &=& \int x H_m(x) H_n(x) e^{-x^2} dx \end{eqnarray*}


Obviously the wrong answer. I got the following hint from a friend: "You should take d^n/du^n d^m/ds^m (n-th and m-th derivatives) of the second line, answer has two different Kronecker deltas."

Can anyone specifically help me figure out how to get this $\delta_{m,n-1}$ factor? I can't figure out that one.

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  • $\begingroup$ Hermite polynomials are orthogonal to each other when using a $e^{-x^2}$ factor. So the most obvious idea would be to get rid of the additional $x$ by partial integration, and express the derivatives of $H_n(x)$ by other Hermite polynomials of lower order. $\endgroup$ Commented Oct 16, 2020 at 12:00
  • $\begingroup$ I am aware of the following: $\frac{\partial^n}{\partial u^n} g(u,x) |_{u=0} = H_n(x)$ where $g(u,x)$ is the generating function of the $H_n$ as well as the fact that $ H'_n(x) =2n H_{n-1}(x)$, but I am unclear how to "get rid of the x via integration by parts." $\endgroup$
    – Lopey Tall
    Commented Oct 16, 2020 at 12:50
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    $\begingroup$ Using the properties in here you can show that $x H_n(x) = \frac{1}{2} H_{n+1}(x) + n H_{n-1}(x)$, which together with $\int_{-\infty}^\infty \mathrm{d}x \, \mathrm{e}^{-x^2} H_m(x) H_n(x) = 2^n n! \sqrt{\pi} \delta_{m, n}$, gives you the answer. $\endgroup$
    – secavara
    Commented Oct 16, 2020 at 13:10
  • $\begingroup$ @secavara got it! :D thank you for the clear answer, please feel free to turn this into an answer! $\endgroup$
    – Lopey Tall
    Commented Oct 16, 2020 at 18:53

1 Answer 1

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@secavara's hint is, in fact, a one liner, if only you used Hermite functions, which have a flat measure, $\int_{-\infty}^\infty \psi_n(x) \psi_m(x) \,dx = \delta_{nm}$, $$\psi_n(x) = \left (2^n n! \sqrt{\pi} \right )^{-\frac12} e^{-\frac{x^2}{2}} H_n(x) = (-1)^n \left (2^n n! \sqrt{\pi} \right)^{-\frac12} e^{\frac{x^2}{2}} ~ \partial_x^n~ e^{-x^2}, \\ \sqrt{2(n+1)}~~\psi_{n+1}(x)= \left ( x- \partial_x \right ) \psi_n(x). $$

Consequently, $$ \partial_x(\psi_n\psi_m)= 2x\psi_n\psi_m -\sqrt{2(n+1)}~ \psi_{n+1}\psi_m -\sqrt{2(m+1)} ~\psi_n\psi_{m+1}. $$

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