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Homework Help: Normalizing wavefunction of (x^2)e^(-x^2)?

  1. Sep 19, 2013 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations


    3. The attempt at a solution


    I think I multiplied ψ*ψ correctly... I'm surprised a little that there is an x^2 with this problem. We have not discussed these integrals in class. I watched a video on how to integrate e[itex]^{-x^2}[/itex], but don't know what to do with the product x^2 and e^(-2cx^2)
  2. jcsd
  3. Sep 19, 2013 #2


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    Did you take statistical mechanics ? You learn about integrating gaussians there. If not, then at least a course in mathematical methods of physics should have explained you this.

    Anyways, think I(c) = int_R exp(-cx2) dx. Compute it then differentiate the result wrt c. (c is a parameter).
  4. Sep 19, 2013 #3
    I'll give you a hint.
    I presume you know that: [itex]\int_{- \infty}^{+\infty} e^{-\lambda x^2}dx = \sqrt{\pi/ \lambda}[/itex]. Consider the function:
    [tex]I(\lambda )=\int_{- \infty}^{+\infty}e^{-\lambda x^2}dx[/tex]
    [tex]\frac{d}{d\lambda }I(\lambda )=-\int_{- \infty}^{+\infty}x^2e^{-\lambda x^2}dx[/tex]
  5. Sep 19, 2013 #4


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    You could also integrate by parts.
  6. Sep 19, 2013 #5
    Thanks for the help, guys! Didn't take statistical mechanics, unfortunately...

    Does this mean that the derivative of [itex]\sqrt{\frac{\pi}{\lambda}}[/itex] is my answer?
  7. Sep 19, 2013 #6
    vela is correct, but kostas230's trick is one of the most beautiful in mathematics and one of the most important in physics. The technique is referred to as using a "generating functional."
  8. Sep 19, 2013 #7
    I don't know if you ever had to evaluate the Gaussian integral, but here's an elegant way to do it. Consider the integral:

    [tex]I= \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} e^{-(x^2+y^2)}dxdy[/tex]

    By changing from cartesian coordinates to polar coordinates, the integral above becomes:
    [tex]I= \int_{0}^{+\infty}\int_{0}^{2\pi}re^{-r^2}drd\theta[/tex]

    I trust that you are able to evaluate the integral. Can you find the gaussian integral [itex]I'=\int_{-\infty}^{+\infty}e^{-x^2}dx[/itex] using the above results? ;)
  9. Sep 19, 2013 #8
    Thanks kostas230, appreciate your help:smile:... The result should be in the form of a Hermite polynomial, is that correct? I think the derivative of re[itex]^{-r^2}[/itex] is


    Not quite sure what to do next
  10. Sep 20, 2013 #9
    No, I said: compute the integral, not find the derivative of the integrated function
  11. Sep 20, 2013 #10
    If you are asking me to find the integral of re[itex]^{-r^2}[/itex]drd[itex]\theta[/itex] from 0[itex]\leq[/itex]r[itex]\leq[/itex][itex]\infty[/itex] and 0[itex]\leq\theta[/itex][itex]\leq[/itex]2[itex]\pi[/itex] I believe it is [itex]\pi[/itex]. And then the integral with a coefficient λ in the power would be [itex]\frac{\pi}{\lambda}[/itex]
  12. Sep 20, 2013 #11


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    That is right, and it allows you to find the value of the original Gaussian integral, if you find the relation between that and the formula kostas230 used.
  13. Sep 20, 2013 #12
    Thank you, mfb.

    So is it that I[itex]^{2}[/itex]= [itex]\frac{\pi}{\lambda}[/itex]

    So the derivative of [itex]\frac{\pi}{\lambda}[/itex] is -[itex]\frac{\pi}{\lambda^2}[/itex]?

    Since the derivative would be 2I for I[itex]^{2}[/itex], should I divide -[itex]\frac{\pi}{\lambda^2}[/itex] by 2, so that I=-[itex]\frac{\pi}{2\lambda^2}[/itex] ?

    Although A is a positive value.. would the negative sign in front of kostas' integral cancel this one, so that I =[itex]\frac{\pi}{2\lambda^2}[/itex] ?
    Last edited: Sep 20, 2013
  14. Sep 20, 2013 #13


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    I think you are mixing two different approaches here.
  15. Sep 20, 2013 #14
    My first post was on an algorithmic process in finding integrals of the form: [itex]I_n=\int_{-\infty}^{+\infty}x^n e^{-\lambda x^2}dx[/itex] presuming that you know the integral [itex]I_0=\int_{-\infty}^{+\infty}e^{-\lambda x^2}dx=\sqrt{\pi}[/itex]. My other posts where to show you a way to compute the integral [itex]\int_{-\infty}^{+\infty}e^{-\lambda x^2}dx=\sqrt{\pi}[/itex].
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