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Hermite polynomals

  1. May 17, 2010 #1
    1. The problem statement, all variables and given/known data

    Prove that <x^2> for an harmonic oscillator = (1+2n)/2 (all other constants were taken to be = to 1) using the Hermite's generating function.

    2. Relevant equations

    <x^2> = an integral of exp(-x^2)*x^2*[Hn(x)]^2 = (2n+1)/2 (all constants are taken to be equal to 1)


    3. The attempt at a solution

    I tried the standard way of multiplying the generating function by x^2 and exp(-x^2). Then, I took the integral. I had: integral of x^2*exp(-x^2 - 2z^2 + 4xz)dx = sum of z^(2n)/(n!)^2 * integral of x^2*exp(-x^2)*[Hn(x)]^2dx. So the only thing that remained to do was to calculate the integral of x^2*exp(-x^2 - 2z^2 + 4xz)dx from -infinity to +infinity. This integral is easily done by multiplying it by exp(2z^2) and writing it as integral of x^2*exp(-(x-2z)^2). By changing variables from x-2z = u, three integrals remain: integral of u^2*exp(-u^2)du + 4z^2*integral of exp(-u^2)du and the other one will vanish because the integrand is odd. By doing the integrations I obtain Exp(2z^2)*((Pi^0.5)*0.5 + 4z^2*Pi^0.5). Expanding the exponential and making this solution equal to sum of z^(2n)/(n!)^2 * integral of x^2*exp(-x^2)*[Hn(x)]^2dx, and taking the coefficients of the same power in z, I obtain as a final results (4n+1)/2 instead of (2n+1)/2 (which I am able to obtain using recursion relations). I have already tried to look for mistakes in the development of the solution, but I really cannot find. Everything seems completely allright to me, except for the result. The 2n in 2n +1/2 is a consequence of the 4z1^2 in the integral of exp(-x^2) which comes from completing the square in the initial integration, so I cannot see where this is wrong, although i am 100% sure there is something wrong.

    I would be extremely thankful if someone could help me.

    Thanks
     
  2. jcsd
  3. May 17, 2010 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Careful,

    [tex]\begin{aligned}\left(\sum_{n=0}^{\infty}\frac{z^n}{n!}H_n(x)\right)^2 &= \sum_{n,m=0}^\infty \frac{z^{n+m}}{n!m!}H_n(x)H_m(x) \\ &= \frac{z^{2n}}{(n!)^2}\left[H_n(x)\right]^2+\sum_{\stackrel{n,m=0}{n\neq m}}^\infty \frac{z^{n+m}}{n!m!}H_n(x)H_m(x) \\ &\neq \frac{z^{2n}}{(n!)^2}\left[H_n(x)\right]^2\end{aligned}[/tex]
     
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