Show the Fourier transformation of a Gaussian is a Gaussian.

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1. Dec 6, 2017

thomas19981

1. The problem statement, all variables and given/known data
Show, by completing the square in the exponent, that the Fourier transform of a Gaussian wavepacket $a(t)$ of width $\tau$ and centre (angular) frequency $\omega_0$:
$a(t)=a_0e^{-i\omega_0t}e^{-(t/\tau)^2}$
is a Gaussian of width $2/\tau$, centred on $\omega_0$, given by:
$a(\omega)=\frac{a_0\tau}{2\sqrt \pi}e^{-(\frac{\omega-\omega_0}{2/\tau})^2}$

2. Relevant equations
I just used the Fourier transformation:
$\frac{1}{ \sqrt{2\pi}}\int \psi(t)e^{-i\omega t} \, dt$ the limits of integration is all $\Bbb{R}$
3. The attempt at a solution
Well I subbed in $a(t)$ for $\psi(t)$ and then carried the integral through and I got:
$\frac{a_0\tau}{\sqrt2}e^{-(\frac{\omega+\omega_0}{2/\tau})^2}$. As you can see the exponent should be $\omega- \omega_0$ and there is a missing factor of $\frac{1}{\sqrt{2\pi}}$. I have looked through my work endlessly and can't find any mistakes so is that the right formula that I am using above or is there an alternative? If that is the right formula above then does that mean the solution is wrong?

So I completed the square of a(t) which gave me:
$a_0e^{-(t/\tau+(1/2)\tau i\omega_0)^2}e^{-\tau^2 \omega_0^2 /4}$
I then plugged this into the integral which gave me:
$\frac{1}{ \sqrt{2\pi}}\int a_0e^{-(t/\tau+(1/2)\tau i\omega_0)^2}e^{-\tau^2 \omega_0^2 /4}e^{-i\omega t} \, dt$

I completed the square of this which gave me:
$\frac{1}{ \sqrt{2\pi}}e^{-\tau^2 \omega_0^2 /4}\int a_0e^{-(t/\tau+(\omega+\omega_0)i\tau/2)^2}e^{-\omega^2 \tau^2 /4}e^{-\omega_0 \omega \tau^2 /2} \, dt$.
Then I simplified this down to:
$\frac{a_0\tau}{\sqrt2}e^{-\frac{\tau^2}{4}(\omega+\omega_0)^2}$.
I then simplified this down further to give me my incorrect answer at the top of the page.

Last edited by a moderator: Dec 6, 2017
2. Dec 6, 2017

Orodruin

Staff Emeritus
How are we going to find out where you have gone wrong in your work if you do not post your work?