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Time Evolution of Gaussian Wave packets

  1. Nov 4, 2008 #1
    Hi, I'm trying to derive a wave equation for a gaussian wavepacket for both the position (x) and the momentum (k), for a wave packet of width sigma, at some initial position x0 and with an initial momentum k0.

    Now I have worked out the initial wavepacket equation to be:

    psi(x) = (sigma*sqrt(pi))^1/2 * exp(-(x-x0)^2 / 2*sigma^2) * exp(ik0(x-x0)

    and I've fourier transformed the result to get the initial momentum wavepacket:

    phi(k) = (sigma/sqrt(pi))^1/2 * exp(-(sigma^2 * (k-k0)^2) / 2) * exp(-ik0x)

    Now i'm not sure how to progess in order to achieve a time dependent version of the equation.... Ive tried a few methods but i'm not too sure.... Any help would be great!
  2. jcsd
  3. Nov 4, 2008 #2


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    is that last "x" supposed to be an "x0"? there shouldn't be any x dependence in phi(k)...

    So, to get the time-dependence you need to have a hamiltonian... are we just dealing with the free-particle hamiltonian?
  4. Nov 4, 2008 #3
    woops yeah it is meant to be x0..... :)
    yes it's just the free particle hamiltonian, no potential. I tried using a separation of variables technique:
    psi(x,t) = X(x)T(t), and then sticking that in schrodingers equation using psi(x) = X(x), but I couldn't get that to work....
  5. Nov 4, 2008 #4
    Wow, that's quite a coincidence. We asked pretty much the same question at nearly exactly the same time... my thread is the one immediately below yours in the thread listing.

    The time dependence of the evolving wave packet is given there as a fourier transform... but I couldn't do the inverse transform.


  6. Nov 4, 2008 #5


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    yup, pretty similar questions.

    thegaussian should have a look at schieghovens thread to see how he stuck in the time dependence.
  7. Nov 4, 2008 #6


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    the difference is that since thegaussian has a free particle hamiltonian (so that omega=k^2/2m rather than omega=|k|) the integral he ends up having to do is just another gaussian intergral...
  8. Nov 10, 2008 #7

    Hans de Vries

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    There's a treatement on the spreading of the wave function in the
    openings chapter on the Klein Gordon equation in my book here:


    See section 9.9. You may want to have a look at 9.8 first.

    Regards, Hans
  9. Nov 13, 2008 #8
    I found an answer in closed form, and posted it on the other thread


    It's for the massless relativistic case -- I'm not sure whether it's the case you wanted. (The answer there is a bit overcomplicated, since I subsequently realised you could just take the real part of the solution I gave, and get a simpler answer.)

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