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Time Evolution of the Gaussian packet

  1. Jul 14, 2007 #1
    1. The problem statement, all variables and given/known data
    Can someone give me a general idea (I do not need and probably will not understand the rigorous mathematics) of how you can use the propagator equation

    [tex] \psi(x,t) = \int{U(x,t;x',t')\psi(x',t')dx'}[/tex],

    the equation for a Gaussian wave packet,

    [tex]\psi(x',0)= e^{i p_0 x'/\hbar}\frac{e^-x'^2/2\Delta^2}{(\pi\Delta^2)^{1/4}}[/tex],

    to arrive at the equation [tex]\psi(x,t) = [\pi^{1/2}(\Delta+\frac{i +\hbar t}{m\Delta})]^{-1/2}

    exp{\frac{-(x-p_0 t/m)^2}{2\Delta^2(1+i \hbar t/m \Delta^2}}

    X exp[\frac{i p_0}{\hbar}(x - \frac{p_0 t}{2m}) ] [/tex]

    Is this just a really complicated integration method? Is that a cross product or multiplcation? If it is a cross product how does come in? See Shankar 154 for more context.

    2. Relevant equations

    3. The attempt at a solution
    I really do not know of any integration methods that give a cross product so I don't know. Just a general explanation would be helpfull.
    Last edited: Jul 15, 2007
  2. jcsd
  3. Jul 14, 2007 #2


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    Homework Helper

    If the propagator "U" is the free propagator then you can actually just write it down explicity and then you just have to do a gaussian integral to get the result. If "U" is the propagator for almost any other system the problem gets hella harder.

    The "X" is not a "cross product." how could it be a cross product!?... there are no vectors!
  4. Jul 15, 2007 #3
    You're right about the cross-product. Then I have no idea why the author used an X instead of a dot for multiplication. It is actually a free propagator. So that is just the result of the Gaussian integral with the propagator and the initial psi...I see.
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