Gaussian distribution integral?

[21joanna12]

when considering the quantum harmonic oscillator, you get that the wave function takes the form

$psi=ae^{-\frac{m\omega}{2\hbar}x^2}$

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

Thank you in advance!

 

[DrClaude]

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
$$
\frac{1}{\sqrt{c}} \int_{-\infty}^{\infty} e^{- \tilde{x}^2} d\tilde{x}
$$

 

[ShayanJ]

Use the substitution $y=\sqrt{\frac{m\omega}{2\hbar}} x$ and check the proof here!

 

[statdad]

when considering the quantum harmonic oscillator, you get that the wave function takes the form

$\psi=ae^{-\frac{m\omega}{2\hbar}x^2}$

I have been trying to integrate $\psi ^2$ to find the constant a so that the wave function is normalised, and I know the trick with converting to polar coordinates to integrate $e^{-x^2}$, but I cannot figure out how to integrate the more complicated version above. I know that the constant should have the value $\left(\frac{m\omega}{\pi \hbar}\right)^{\frac{1}{4}}$ 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 $x^2$ in your expression for $\psi$? If not then the comments above should serve you well: if you do, are you really trying to integrate something
$$\propto \int_{-\infty}^\infty e^{-cx^4} \, dx$$