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Infinite square well, momentum space

  1. Mar 13, 2007 #1
    Problem:
    Find the momentum-space wave function [itex] \Phi_n(p,t)[/itex] for the [itex]n[/itex]th stationary state of the infinite square well.

    Equations:

    [tex] \Psi_n(x,t) = \psi_n(x) \phi_n(t) [/tex]

    [tex] \psi_n(x) = \sqrt{\frac{2}{a}}\sin(\frac{n\pi}{a}x) [/tex]

    [tex] \phi_n(t) = e^{-iE_n t/\hbar} [/tex]

    [tex] \Phi_n(p,t) = \frac{1}{\sqrt{2\pi\hbar}} \int^{\infty}_{-\infty} e^{-ipx/\hbar} \Psi_n(x,t) dx [/tex]

    Attempt:

    [tex] \Phi_n(p,t) = \frac{\phi_n(t)}{\sqrt{a\pi\hbar}} \int^{\infty}_{-\infty} e^{-ipx/\hbar} \sin(\frac{n\pi}{a}x) dx [/tex]

    [tex] = \frac{\phi_n(t)}{\sqrt{a\pi\hbar}} \frac{1}{2i} \int^{\infty}_{-\infty}\Bigg(e^{i(\frac{n\pi}{a} - \frac{p}{\hbar})x} - e^{i(\frac{-p}{\hbar} - \frac{n\pi}{a})x}\Bigg) dx [/tex]

    [tex] = \frac{\phi_n(t)}{\sqrt{a\pi\hbar}} \frac{1}{2i} 2\pi \Bigg(\delta(\frac{n\pi}{a} - \frac{p}{\hbar}) - \delta(\frac{p}{\hbar} + \frac{n\pi}{a})\Bigg) [/tex]

    This doesn't seem right to me. Do I have this right, or am I missing something somewhere?
     
    Last edited: Mar 13, 2007
  2. jcsd
  3. Mar 13, 2007 #2

    Dick

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    OOOOPS. Except you DON'T want to integrate x from -infinity to +infinity. The wavefunction only lives in the box.
     
  4. Mar 13, 2007 #3
    Oh, right... thanks
     
  5. Mar 13, 2007 #4

    Dick

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    I corrected my previous flip answer.
     
  6. Mar 13, 2007 #5
    I know, I saw. Thanks.
     
  7. Mar 14, 2007 #6

    dextercioby

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    He SHOULD integrate from - to + infinity, it's just that outside the box the wavefunction is strictly 0...
     
  8. Oct 21, 2010 #7
    What happens when you only integrate from 0 to a? When I try I get a seemingly uselessly messy answer, is there some trick to use delta functions or something I am missing?
     
  9. Sep 30, 2011 #8
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