Using given Fourier transform to find the equation for the wave packet.

[Jennifer_ea]

Homework Statement

Any wavepacket can be obtained by the superposition of an infinite number of plane waves using the so-called Fourier integral or Fourier transform
$f(x,t) = \frac{1}{\sqrt{2\pi}} _{-\infty}\int^\infty A(k)e^{i(kx-\omega t)} dk$

Find at t=0 the representation of the wavepacket f(x) associated with the flat distribution given by:

A(k) =
0 for k<-K and k>K
$\frac{1}{\sqrt{2K}}$ for -K < k < K

Homework Equations

The textbook I found that isn't leaving me entirely confused has replaced k with p (momentum), but that doesn't seem to be overly relevant to my lack of understanding. The one I've found that seem to be in the ball park is:
$\left|A(p,t)^2\right| = \left|A(p)e^{\frac{ip^{2}t}{2mh-bar}}\right|^2 = \left| A(p,0) \right|^2$

The Attempt at a Solution

I have figured out that A(k) is the Fourier transform, but after that I run into a brick wall. I can't even seem to get far enough to be able to make useful searches. I'm getting the impression that I need to do some more manipulation so that I can use the above equation, but right now it'd just be blind hammering without understanding why.

I feel like there may have a linking concept I'm not getting, even just a nudge in the right direction would be extremely helpful!

 

[Chopin]

You've been given $A(k)$, and a formula that describes $f(x,t)$ in terms of it. So finding $f(x,0)$ should just be a matter of plugging things in.

Since $t=0$, you can drop the $\omega t$ term from the integral. So right off the bat, we have:
$$f(x, 0) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} A(k)e^{ikx} dk$$

Now, see if you can determine how the integral is affected by substituting in the definition of $A(k)$.

 

[Jennifer_ea]

Wow, I was so focused on trying to figure out what the transform did that I completely overlooked the basics. Did not cross my mind once about the omega disappearing.

Thanks a bunch! You're right, it's just simple substitution now. Lesson learned!