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

  1. Sep 30, 2012 #1
    1. The problem statement, all variables and given/known data
    Any wavepacket can be obtained by the superposition of an infinite number of plane waves using the so-called Fourier integral or Fourier transform
    [itex]f(x,t) = \frac{1}{\sqrt{2\pi}} _{-\infty}\int^\infty A(k)e^{i(kx-\omega t)} dk[/itex]

    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
    [itex]\frac{1}{\sqrt{2K}}[/itex] for -K < k < K

    2. Relevant 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:
    [itex]\left|A(p,t)^2\right| = \left|A(p)e^{\frac{ip^{2}t}{2mh-bar}}\right|^2 = \left| A(p,0) \right|^2[/itex]

    3. 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!
  2. jcsd
  3. Sep 30, 2012 #2
    You've been given [itex]A(k)[/itex], and a formula that describes [itex]f(x,t)[/itex] in terms of it. So finding [itex]f(x,0)[/itex] should just be a matter of plugging things in.

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

    Now, see if you can determine how the integral is affected by substituting in the definition of [itex]A(k)[/itex].
  4. Sep 30, 2012 #3
    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!
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