Gaussian distribution integral?

  • #1
126
2
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!
 

Answers and Replies

  • #2
DrClaude
Mentor
7,551
3,893
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
$$
\frac{1}{\sqrt{c}} \int_{-\infty}^{\infty} e^{- \tilde{x}^2} d\tilde{x}
$$
 
  • #3
2,792
594
Use the substitution [itex] y=\sqrt{\frac{m\omega}{2\hbar}} x [/itex] and check the proof here!
 
  • #4
statdad
Homework Helper
1,495
35
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!
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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