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Gaussian distribution integral?

  1. Dec 10, 2014 #1
    when considering the quantum harmonic oscillator, you get that the wave function takes the form

    [itex]psi=ae^{-\frac{m\omega}{2\hbar}x^2}[/itex]

    I have been trying to integrate [itex]\psi ^2[/itex] to find the constant a so that the wave function is normalised, and I know the trick with converting to polar coordinates to integrate [itex]e^{-x^2}[/itex], but I cannot figure out how to integrate the more complicated version above. I know that the constant should have the value [itex]\left(\frac{m\omega}{\pi \hbar}\right)^{\frac{1}{4}}[/itex] if the wavefunction is to be normalised, but I can't figure out how to do this?

    Thank you in advance!
     
  2. jcsd
  3. Dec 10, 2014 #2

    DrClaude

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    Staff: Mentor

    You have
    $$
    \int_{-\infty}^{\infty} e^{- c x^2} dx
    $$
    Make the substitution ##\tilde{x} = \sqrt{c} x##, ##d\tilde{x} = \sqrt{c} dx##, and you get
    $$
    \frac{1}{\sqrt{c}} \int_{-\infty}^{\infty} e^{- \tilde{x}^2} d\tilde{x}
    $$
     
  4. Dec 10, 2014 #3

    ShayanJ

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    Gold Member

    Use the substitution [itex] y=\sqrt{\frac{m\omega}{2\hbar}} x [/itex] and check the proof here!
     
  5. Dec 12, 2014 #4

    statdad

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    Homework Helper

    Do you want the [itex] x^2 [/itex] in your expression for [itex] \psi [/itex]? If not then the comments above should serve you well: if you do, are you really trying to integrate something
    [tex]
    \propto \int_{-\infty}^\infty e^{-cx^4} \, dx
    [/tex]
     
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