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Normalizing Hermite Polynomials

  1. May 5, 2017 #1
    1. The problem statement, all variables and given/known data
    Evaluate the normalization integral in (22.15). Hint: Use (22.12) for one of the $H_n(x)$ factors, integrate by parts, and use (22.17a); then use your result repeatedly.


    2. Relevant equations
    (22.15) ##\int_{-\infty}^{\infty}e^{-x^2}H_n(x)H_m(x)dx = \sqrt{\pi}2^nn!## when ##n=m##
    (22.12) ##H_n(x)=(-1)^ne^{x^2}\frac{d^n}{dx^n}e^{-x^2}##
    (22.17a) ##H'_n(x)=2nH_{n-1}(x)##

    3. The attempt at a solution
    I started out by writing my integral from (22.15) as:
    ##\int_{-\infty}^{\infty}e^{-x^2}[H_n(x)]^2dx##
    Expanding one of the ##H_n(x)## terms gives:
    ##\int_{-\infty}^{\infty}e^{-x^2}H_n(x)e^{x^2}(-1)^n\frac{d^n}{dx^n}e^{-x^2}dx##
    ##=(-1)^n\int_{-\infty}^{\infty}H_n(x)\frac{d^n}{dx^n}e^{-x^2}dx##
    Integration by parts and using equation 22.17a gives:
    ##(-1)^n[H_n(x)e^{-x^2}|^{\infty}_{-\infty}-\int^{\infty}_{-\infty}2nH_{n-1}(x)\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}dx##
    The first term goes to zero and leaves:
    ##(-1)^n\int^{\infty}_{-\infty}2nH_{n-1}(x)\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}dx##
    IBP n-1 more times gives:
    ##(-1)^n2^nn!\int^{\infty}_{-\infty}H_0(x)e^{-x^2}dx=(-1)^n2^nn!\int^{\infty}_{-\infty}e^{-x^2}dx = (-1)^n2^nn!\sqrt{\pi}##

    This is correct except I have this extra factor of ##(-1)^n## that shouldn't be there, I'm not sure how to get rid of it.
     
  2. jcsd
  3. May 5, 2017 #2

    I like Serena

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    Homework Helper

    Hi rmiller,

    Let's take a closer look at the IBP:
    Shouldn't that be:
    $$(-1)^n\left[H_n(x)e^{-x^2}|^{\infty}_{-\infty}-\int^{\infty}_{-\infty}2nH_{n-1}(x)\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}dx\right] \\
    =(-1)^n \cdot \left[ -\int^{\infty}_{-\infty}2nH_{n-1}(x)\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}dx \right] \\
    =(-1)^{n-1} \int^{\infty}_{-\infty}2nH_{n-1}(x)\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}dx
    $$
    ?
     
  4. May 8, 2017 #3
    For future people, you get a factor of -2 each time you do ibp and in the end you're left with (-1)^n (-2)^n when you combine them it gives 2^n.
     
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