Show the Fourier transformation of a Gaussian is a Gaussian.

[thomas19981]

Homework Statement

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:
$a(\omega)=\frac{a_0\tau}{2\sqrt \pi}e^{-(\frac{\omega-\omega_0}{2/\tau})^2}$

Homework 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 $\Bbb{R}$

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.

 

[Orodruin]

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