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Expansion coefficients of a wave packet

  1. Dec 6, 2014 #1
    1. The problem statement, all variables and given/known data
    What are the expansion coefficients of a wavepacket [tex]\Psi (x) = \sqrt{\frac{2}{L}}sin \frac{\pi x}{L}[/tex] in the basis Ψn(x) of a particle in a periodic box of size L?

    2. Relevant equations
    [tex]\Psi (r,t) = {\sum_{n}^{}} a_{n}(t) \Psi _{n}(r)[/tex]

    3. The attempt at a solution
    [tex]\left \langle \Psi _{m}| \Psi \right \rangle= {\sum_{n}^{}} a_{n}(t)\left \langle \Psi _{m}| \Psi_{n} \right \rangle[/tex]

    all zero except for m=n therefore

    [tex]a_{n}(t)=\left \langle \Psi _{n}| \Psi \right \rangle[/tex]

    so I have a term for the coefficients but how do I apply it to the specific wavefunction? espcially as it has no n in it.

    Any help as always very much appreciated. Thanks.
     
  2. jcsd
  3. Dec 6, 2014 #2

    Orodruin

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    What are the basis functions for your potential?
     
  4. Dec 6, 2014 #3
    hmm do you mean the basis wavefunctions? if it's a periodic box then

    [tex]\Psi _{n}=\sqrt{\frac{1}{L}} exp(\frac{i2\pi nx }{L})[/tex]

    so i substitute this into the above an(t) equation?

    thanks for the reply!

    edit i expanded the exponential in terms of cos and sin but it gets a bit complicated, also not sure what to do with the Σan thats on the right hand side.
     
    Last edited: Dec 6, 2014
  5. Dec 6, 2014 #4

    Orodruin

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    It will be more fruitful to expand the sine in terms of exponentials ...
     
  6. Jan 4, 2015 #5
    Apologies for the late reply. I'm getting really lost with this one, I'm really not sure my method is correct as there is no time dependance in either wavefunction, so why am I using an(t).

    If I use:
    [tex]a_{n}(t)=\left \langle \Psi _{n}| \Psi \right \rangle[/tex]
    with

    [tex]\Psi _{n}=\sqrt{\frac{1}{L}} exp(\frac{i2\pi nx }{L})[/tex]
    [tex]\Psi (x) = \sqrt{\frac{2}{L}}sin \frac{\pi x}{L}[/tex]

    but then expand

    [tex]\Psi (x) = \sqrt{\frac{2}{L}}sin \frac{\pi x}{L}[/tex]

    [tex]\Psi (x) = \sqrt{\frac{2}{L}}sin \frac{\pi x}{L} = \sqrt{\frac{2}{L}}\frac{1}{2i}(exp\frac{in\pi}{L}- exp \frac{-in\pi}{L})[/tex]

    so I get

    [tex]a_{n}(t) = \int_{0}^{L} \sqrt{\frac{1}{L}} exp(\frac{i2\pi nx }{L})\sqrt{\frac{2}{L}}\frac{1}{2i}(exp\frac{in\pi}{L}- exp \frac{-in\pi}{L}) dx[/tex]

    this cant be correct?

    any further help on this would be appreciated, I still dont really understand what the question is asking to be honest.
     
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