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Orthogonality of Hermite functions

  1. Nov 12, 2014 #1
    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!
     
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  3. Nov 12, 2014 #2

    mfb

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    Did you try induction?
     
  4. Nov 12, 2014 #3

    dextercioby

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    You can use the generating function and reduce the whole integral to a doable one.
     
  5. Nov 12, 2014 #4

    pasmith

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    I note that [tex]\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}.[/tex] The second equality can be established by applying Sturm-Liouville theory to the Hermite equation [tex]H_n'' - 2xH_n' = -2nH_n.[/tex]
     
  6. Nov 13, 2014 #5
    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...
     
  7. Nov 13, 2014 #6

    ShayanJ

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    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.
     
  8. Nov 13, 2014 #7

    pasmith

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    Or, since [itex]\phi_n = e^{-\frac12x^2}H_n[/itex], substitution yields [tex]
    \phi_n'' + (1 - x^2)\phi_n = -2n\phi_n.[/tex] Multiplying by [itex]\phi_m[/itex] and integrating over the real line gives [tex]
    \int_{-\infty}^\infty \phi_n''\phi_m + (1 - x^2)\phi_n\phi_m\,dx = -2n\int_{-\infty}^\infty \phi_n \phi_m\,dx.
    [/tex] Integrating the first term on the left by parts twice yields [tex]
    \int_{-\infty}^\infty \phi_n \phi_m'' + (1 - x^2)\phi_n\phi_m\,dx = -2n\int_{-\infty}^\infty \phi_n \phi_m\,dx
    [/tex] and since [itex]\phi_m'' + (1-x^2)\phi_m = -2m\phi_m[/itex] we have [tex]
    -2m \int_{-\infty}^\infty \phi_n \phi_m\,dx = -2n\int_{-\infty}^\infty \phi_n \phi_m\,dx[/tex] or
    [tex]2(n - m)\int_{-\infty}^\infty \phi_n \phi_m\,dx = 0.[/tex]
     
  9. Nov 13, 2014 #8

    mfb

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    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.
     
  10. Nov 13, 2014 #9
    I thank you all very much!!!!!
     
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