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Time-dependent wavefunction

  1. Nov 19, 2007 #1
    1. The problem statement, all variables and given/known data

    If given, for instance, psi(phi, 0)=[1/sqrt(2pi)](cos^2(phi/2) + isin(phi)), which is the wavefunction at t=0, how do you go about finding the wavefunction at time t, psi(phi,t)??

    2. Relevant equations

    3. The attempt at a solution

    Would it simply be psi(phi, t)=psi(phi,0)e^(-iEt/hbar)??

    The given wavefunction is for a particle on a ring.... and for that, E_n=(hbar^2)(m^2)/(2I) where I=mr^2. Is that the value of "E" I plug into the above equation??

    Is there any other constant that I have to multiply the psi(phi,t) equation by as a result of adding on the e^(-iEt/hbar) term??

    Any help would be great, I'm not entirely sure if I get this :-/
  2. jcsd
  3. Nov 19, 2007 #2


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    In general, you have to write the initial wave function as a sum of energy eigenfunctions with suitable coefficients. Then, as time evolves, each term gets multiplied by e^(-iE_n t/hbar), where E_n is the eigenenergy of that eigenfunction.
  4. Nov 19, 2007 #3
    I'm not sure how to express psi (phi, 0) as a sum of energy eigenfunctions in this case...
  5. Nov 20, 2007 #4


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    What are the energy eigenfunctions?

    If it's a free particle on a ring, they should be e^(i n phi), n=...,-1,0,+1,..., up to normalization. (The m^2 in your E_n should be n^2, no?)

    Your psi(phi,0) is some trig functions; can you express these in terms of complex exponentials?
  6. Nov 20, 2007 #5
    yes and yes. (1/sqrt(2pi))e^(i n phi) is the eigenfunction. and I really meant n^2 when I said m^2 previously...

    I just expressed my psi(phi, 0) in exponential terms. The psi I gave in the original post left out some constants, but when I do take into account those:

    psi(phi,0)=sqrt(1/4pi) (e^(i phi) + 1)

    but how does this get into a summation form..? :-/
  7. Nov 20, 2007 #6


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    It already is in a summation form. You have

    psi(phi,0) = (1/sqrt(2))( psi_1(phi) + psi_0(phi) ),

    where psi_n(phi) is the eigenfunction.
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