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Homework Help: Wavepacket expansion coefficients

  1. Jan 2, 2008 #1
    I'm trying to get my head around the idea of expansion coefficients when describing a wavefunction as

    [tex]\Psi(\textbf{r}, t) = \sum a_{n}(t)\psi_{n}(\textbf{r})[/tex]

    As I understand it, the expansion coefficients are the [tex]a_{n}[/tex] s which include a time dependence and also dictate the probability of obtaining an eigenvalue whereby [tex]\sum |a_{n}(t)|^{2} = 1[/tex]. I also understand that the expectation values of operators can be given as function of the [tex]a_{n}(t)[/tex] coefficients given the orthonormality in the eigenfunctions, whereby [tex]<H> = \sum |a_{n}(t)|^{2} E_{n}[/tex].

    If I'm looking at the wavepacket:

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

    How would I determine the expansion coefficients of the wavepacket in the basis states [tex]\psi_{n}(x)[/tex] for the particle in the periodic box, length L? I'm completely confused about the terminology here.

    Any help/explanation would be massively appreciated
  2. jcsd
  3. Jan 2, 2008 #2


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    you fist find the corresponding eigenfunctions [itex] \phi (x) [/itex]for the particle in the periodic box, length L? Then you do this:

    [tex] a_n = \int \psi ^*(x) \phi _n(x) dx [/tex]


    [tex] \psi (x) = \sum a_n \phi _n(x) [/tex]

    wave functions are normalised here.

    So now find the eigenfunction for a box with lenght L, and do the integral.
    Last edited: Jan 2, 2008
  4. Jan 2, 2008 #3
    I thought that the eigenfunctions [tex]\psi(x)[/tex]were already specific by
    [tex]\psi(x) = \sqrt{\frac{2}{L}}sin(\frac{\pi x}{L})[/tex]?
  5. Jan 2, 2008 #4


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    ok your post was not clear.

    You state that your wave packed was:
    [tex]\psi(x) = \sqrt{\frac{2}{L}}sin(\frac{\pi x}{L})[/tex]

    But that is the wave fucntion for the groud state for the box with lenght L.

    The eigenfunctions are altough:

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

    So IF your wave function was [tex]\psi(x) = \sqrt{\frac{2}{L}}sin(\frac{\pi x}{L})[/tex], then it is trivial to find the expansion coefficients in the basis [itex] \phi _n(x) [/itex]
  6. Jan 2, 2008 #5
    Sorry, I should have probably just transcribed the question as it's written here:

    I thought that maybe I'd need to use this relation:

    [tex]a_{n}(t) = \int \psi^{*}_{m}(r) \Psi(r,t) dV[/tex]

    But that gives me a sinĀ² integral which seems very involved for the question...
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