1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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 sin² integral which seems very involved for the question...
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?