Orthogonality of Hermite functions

[DavideGenoa]

Hi, friends! I want to show 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, i.e. that, for any $m\ne n$,

$\int_{-\infty}^\infty e^{x^2} \frac{d^m e^{-x^2}}{dx^m} \frac{d^n e^{-x^2}}{dx^n}=0$​

I have tried by integrating by parts, but I am landing nowhere...
Thank you so much for any help!

 

[mfb]

Did you try induction?

 

[dextercioby]

You can use the generating function and reduce the whole integral to a doable one.

 

[pasmith]

I note that \int_{-\infty}^\infty \phi_n(x)\phi_m(x)\,dx = \int_{-\infty}^\infty e^{-x^2}H_n(x)H_m(x)\,dx = 2\sqrt{\pi}n!\delta_{nm}. The second equality can be established by applying Sturm-Liouville theory to the Hermite equation H_n'' - 2xH_n' = -2nH_n.

 

[DavideGenoa]

Thank you all so much!
@dextercioby and pasmith: Regrettably, I do not know anything of Sturm-Liouville theory or generating functions...
@mfb: I am not sure how we could use induction with $m$ and $n$... I think that integration by parts is the key, but I am not able to manipulate the integral to get the desired result...

 

[ShayanJ]

The best way is the one suggested by pasmith because you won't do any integration.
So I suggest you read Sturm-Liouville theory chapter of Mathematical methods for physicists by Arfken.

 

[pasmith]

I note that \int_{-\infty}^\infty \phi_n(x)\phi_m(x)\,dx = \int_{-\infty}^\infty e^{-x^2}H_n(x)H_m(x)\,dx = 2\sqrt{\pi}n!\delta_{nm}. The second equality can be established by applying Sturm-Liouville theory to the Hermite equation H_n'' - 2xH_n' = -2nH_n.

Or, since \phi_n = e^{-\frac12x^2}H_n, substitution yields <br /> \phi_n&#039;&#039; + (1 - x^2)\phi_n = -2n\phi_n. Multiplying by \phi_m and integrating over the real line gives <br /> \int_{-\infty}^\infty \phi_n&#039;&#039;\phi_m + (1 - x^2)\phi_n\phi_m\,dx = -2n\int_{-\infty}^\infty \phi_n \phi_m\,dx.<br /> Integrating the first term on the left by parts twice yields <br /> \int_{-\infty}^\infty \phi_n \phi_m&#039;&#039; + (1 - x^2)\phi_n\phi_m\,dx = -2n\int_{-\infty}^\infty \phi_n \phi_m\,dx<br /> and since \phi_m&#039;&#039; + (1-x^2)\phi_m = -2m\phi_m we have <br /> -2m \int_{-\infty}^\infty \phi_n \phi_m\,dx = -2n\int_{-\infty}^\infty \phi_n \phi_m\,dx or
2(n - m)\int_{-\infty}^\infty \phi_n \phi_m\,dx = 0.

 

[mfb]

@mfb: I am not sure how we could use induction with $m$ and $n$... I think that integration by parts is the key, but I am not able to manipulate the integral to get the desired result...

It was a guess, as those things easily transform to n+-1 or m+-1 and so on via partial integration. We have much nicer solutions here now.

 

[DavideGenoa]

I thank you all very much!