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Fourier transform question (pretty simple, i think)

  1. Oct 8, 2006 #1
    ok, i have a wave packet which is defined between (-pi/(2b)) and (pi/(2b)) as cos(bx), and it's zero everywhere else. here's what i've done so far:

    i normalized and solved for b, getting pi/2. so now i'm thinking i should calculate the a0, an, and bn series, and add, right? and here is where i'm stuck. how do i go about doing this? what are the limits of integration? and do i just add the results together to get the final fourier transform? any and all attempts at help is much appreciated. thanks a lot!
     
  2. jcsd
  3. Oct 8, 2006 #2

    quasar987

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    huh? I understand you want to write your wave packet as a Fourier integral, like so,

    [tex]F(x)=\int_{-\infty}^{+\infty}G(\omega)e^{i\omega x}d\omega[/tex]

    ??

    All you need to do is calculate [itex]G(\omega)[/itex] by doing the fourier transform of F:

    [tex]G(\omega)=\frac{1}{2\pi}\int_{-\infty}^{+\infty}F(x)e^{-i\omega x}dx=\frac{1}{2\pi}\int_{-\frac{\pi}{2b}}^{+\frac{\pi}{2b}}cos(bx)e^{-i\omega x}dx[/tex]
     
  4. Oct 8, 2006 #3
    i'm sorry, i guess i should've been more clear. i need to write it in k space.. as in, i need an expression for phi(k) (sorry, can't get LaTeX to display it)
     
    Last edited: Oct 8, 2006
  5. Oct 8, 2006 #4
    soo.. i don't need to normalize and solve for b?
     
  6. Oct 8, 2006 #5

    quasar987

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    Ok, from the top. This is how it goes... You have a wave packet F(x,t) propagating in the positive x direction at speed v. In the spirit of Fourier series, you're hoping to write your wave packet as a "sum" of progressive waves of carefully chosen amplitudes: [itex]\phi(k)e^{k(x-vt)}[/itex] ([itex]\phi(k)[/itex] represents the amplitude of the wave of wave number k. It is the analogue in Fourier integrals of the coefficients [itex]c_n[/itex] in Fourier series):

    [tex]F(x,t)=\int_{-\infty}^{+\infty}\phi(k)e^{ik(x-vt)}dk[/tex]

    But what is [itex]\phi(k)[/itex]? You know that at t=0, the wave has the form [itex]F(x,0)=\cos(bx)[/itex] for [itex]-\pi/2b<x<\pi/2b[/itex] and 0 elsewhere. In this case, your wave packet becomes

    [tex]F(x,0)=F(x)=\int_{-\infty}^{+\infty}\phi(k)e^{ikx}dk[/tex]

    This is a form that is suitable for a Fourier transform. According to Fourier's inversion theorem then,

    [tex]\phi(k)=\frac{1}{2\pi}\int_{-\infty}^{+\infty}F(x)e^{-ikx}dx[/tex]

    [tex]\phi(k)=\frac{1}{2\pi}\int_{-\frac{\pi}{2b}}^{+\frac{\pi}{2b}}cos(bx)e^{-ikx}dx[/tex]
     
  7. Oct 8, 2006 #6

    quasar987

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    In post #4, you're talking about normalization, this suggest that F(x,t) is intended as a wave function in the quantum mechanical sense? In that case, yes, b cannot have just about any value. For F to be a wave function satisfying the probabilistic interpretation of QM, [itex]FF^*[/itex] needs to have the properties of a probability density function. One of these is that the "sum" of the probabilities be 1. In this case,

    [tex]1=\int_{-\infty}^{+\infty}FF^*dx[/tex]

    [tex]1=\int_{-\frac{\pi}{2b}}^{+\frac{\pi}{2b}}cos^2(bx)dx[/tex]

    What condition does that set on b?
     
    Last edited: Oct 8, 2006
  8. Oct 8, 2006 #7
    Think I got it :) I got b as [tex]\frac{1}{\pi}[/tex] and my final equation in k space as [tex]\phi(k) = \frac {\pi}{\frac{\pi^2}{4}+k^2}e^{2ik}[/tex] .


    Thanks a lot for the help!
     
  9. Oct 8, 2006 #8

    quasar987

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    I got b=pi/2, but I could be wrong. Still it's worth double checking your work I think.
     
  10. Oct 8, 2006 #9
    er, you're right. That's what I got too. I was looking at something else. Thanks so much for the help, it's truly appreciated.
     
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