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Using generating function to normalize wave function

  1. Feb 9, 2017 #1
    1. The problem statement, all variables and given/known data
    Prove that ##\psi_n## in Eq. 2.85 is properly normalized by substituting generating functions in place of the Hermite polynomials that appear in the normalization integral, then equating the resulting Taylor series that you obtain on the two sides of your equation. As a by-product, you should also find that ## \psi_m ## and ## \psi_n ## are orthogonal when m ≠ n.


    2. Relevant equations
    Eq 2.85: $$ \psi_n(x) = \bigg(\frac{m\omega}{\pi\hbar}\bigg)^{1/4} \frac{1}{\sqrt{2^nn!}}H_n(\xi) e^{-\xi^2/2} $$
    where $$\xi \equiv \sqrt{\frac{m\omega}{\hbar}}x $$
    and ## H_n(\xi) ## are the Hermite polynomials.
    3. The attempt at a solution
    Ok so the main thing throwing me off here is where it says to equate the two taylor series obtained because I don't really see how I'm supposed to obtain a series on both sides.

    First off, when I go to normalize I use
    $$ \int_{-\infty}^{\infty} |\psi_n|^2d\xi = 1$$
    and equating all the constants in front of the ## H_n(\xi) ## term to ## A## the integral becomes
    $$ \int_{-\infty}^{\infty} A^2 H_n^2 e^{-\xi^2} d\xi = 1 $$
    Then, substituting the generating function (using the form ## e^{2\xi z-z^2} ##) in, this becomes
    $$ \int_{-\infty}^{\infty} A^2 e^{-(\xi-z)^2} = 1 $$
    But this is a simple gaussian integral right? So then its just
    $$ \int_{-\infty}^{\infty} A^2 e^{-(\xi-z)^2} = \sqrt{\pi}A^2 $$

    But I feel like all this is wrong because I don't see how a Taylor series comes into play. Am I doing the normalization all wrong or is there just something I'm not seeing?

    Any advice is much appreciated!!
     
  2. jcsd
  3. Feb 9, 2017 #2

    vela

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    I haven't worked out the problem, but I did notice a mistake in your attempt. The integral involves the Hermite polynomial squared, so you need to square the generating function to get a term that looks like that. Squaring the Taylor series the generating function is equal to will also produce the cross terms that you expect to vanish.
     
  4. Feb 9, 2017 #3
    Oh thank you I hadn't caught that mistake!
    So it should be ##\int_{-\infty}^{\infty} A^2 e^{4\xi z-z^2}e^{-\xi^2}d\xi## ? And should it be an integral of ##d\xi## or ##dx##?
    But I'm still not sure how ## e^{4\xi z-2z^2} = \sum \frac{z^{2n}}{(n!)^2}H_n^2## gives me cross terms that vanish? What do you mean by that?
     
  5. Feb 9, 2017 #4

    vela

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    You need to be a little more careful here. Remember ##(a+b)^2 \ne a^2 + b^2##.
    $$(e^{2\xi z- z^2})^2 = \left[\sum_{n=0}^\infty H_n(\xi) \frac{z^n}{n!}\right]^2 = \left[\sum_{m=0}^\infty H_m(\xi) \frac{z^m}{m!}\right]\left[\sum_{n=0}^\infty H_n(\xi) \frac{z^n}{n!}\right]$$
     
  6. Feb 9, 2017 #5

    TSny

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    This step is not valid. ##H_n(\xi)## does not equal the generating function. So, you can't replace ##H_n(\xi)## by the generating function and still expect the integral to equal 1.

    In normalizing the ## \psi_n(x)## and checking their orthonormality, you essentially need to find the integrals $$ I_{nm} = \int_{-\infty}^{\infty} H_n(\xi) H_m(\xi) e^{-\xi^2} d\xi $$

    The suggestion to obtain ##I_{nm}## is to consider the corresponding quantities $$G_{nm} = \int_{-\infty}^{\infty} g(\xi, y) g(\xi, z) e^{-\xi^2} d\xi$$ where the ##g##'s are the generating functions ##g(\xi, y) = e^{2\xi y-y^2} ## and ##g(\xi, z) = e^{2\xi z-z^2} ##.

    See if you can evaluate ##G_{nm}## in two different ways:

    (1) Using the expressions above for the generating functions

    (2) Using the expansions of the generating functions, where the expansions are in terms of the Hermite polynomials.
     
    Last edited: Feb 9, 2017
  7. Feb 9, 2017 #6
    Oh yeah I see. I should've noticed that after seeing everyones comments about mod squared.

    Yeah I was questioning that as I was typing it since ##e^{2\xi-z^2} = \sum \frac{z^n}{n!}H_n## and not just ##H_n## but I wasn't sure if it was possible to rearrange the summation.

    So combining your suggestion with vela's I need to evaluate something like ## \int_{-\infty}^{\infty} A^2 \bigg[ \sum\frac{z^m}{m!}H_m\bigg]\bigg[\sum \frac{z^n}{n!}H_n\bigg] e^{-\xi^2} d\xi ## ? Any tips on how to evaluate an integral like this?
     
  8. Feb 9, 2017 #7

    TSny

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    You will not need to evaluate this expression directly. (However, I think that one of the sums should involve the variable ##y## instead of ##z## if you are going to follow the suggestion I gave in post #5. This yields the "second way" of evaluating ##G_{nm}## that I listed in post #5.)

    The idea is to compare this result with the "first way" of evaluating ##G_{nm}## that I listed in post #5.

    This "first way" should lead to a simple result for ##G_{nm}## in terms of ##y## and ##z##. You can then expand this result in a Taylor series expansion in powers of ##yz## and compare term-by-term with your double sum expression above (after writing one of your sums in terms of ##y## instead of ##z##).
     
  9. Feb 10, 2017 #8
    Ok I think I see what you are saying. So I evaluate ## \int e^{2\xi y-y^2}e^{2\xi z-z^2} e^{-\xi^2} d\xi ## and compare it to ##\int \bigg[\sum \frac{y^m}{m!}H_m\bigg]\bigg[\sum \frac{z^n}{n!}H_n\bigg] e^{-\xi^2}d\xi ## something like that?
     
  10. Feb 10, 2017 #9

    TSny

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    Yes, that should work.
     
  11. Feb 10, 2017 #10
    Ok I got $$ \int e^{2\xi y-y^2}e^{2\xi z-z^2}e^{-\xi^2}d\xi = \sqrt{\pi}e^{2yz} = \sqrt{\pi} \sum \frac{y^lz^l}{l!} $$
    Is this the correct series? Cause in comparison to $$ \int \frac{y^m}{m!}H_m \frac{z^n}{n!}H_n e^{-\xi^2} d\xi $$ the series and gaussian would cancel for m=n=l leaving just ##\int H_n H_m## which equals 1 since they're orthogonal. Is that right?
     
  12. Feb 10, 2017 #11

    TSny

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    Looks good except the right hand side is missing 2 raised to some power.
    You want to use this analysis to prove that you have orthogonality and to deduce the value of ## \int H_n^2 e^{-\xi^2} d\xi ## . Since ##y## and ##z## are arbitrary numbers, terms of the same power in ##y## and ##z## in the comparison must match.
     
  13. Feb 10, 2017 #12
    Ok I think I understand it all now. Thank you so much for y'all's help!!
     
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