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Square shape wave packet spreading

  1. May 29, 2012 #1
    for a 1D free particle with initial wave function [itex]\phi(x')[/itex] square shaped(e.g. [itex]\phi(x')=1,x'\in [a,b][/itex],otherwise it vanishes),
    my question is: how does it evolve with time [itex]t[/itex]?


    if we deal with it in [itex]P[/itex] basis, it is easily solved, using the propagator [itex]U(t)=∫|p'><p'|e^{-\frac{ip'^2 t}{2m\hbar}}dp'[/itex];

    but if we directly solve SE in [itex]X[/itex] basis, where [itex]P[/itex] must be written as [itex]-i\frac{∂}{∂x'}[/itex], the initial wavefunction is not continous, so the equation becomes improper at the ends of the interval[itex][a,b][/itex],


    so why dose the SE equation seems so distinct in these 2 representations? what goes wrong in [itex]X[/itex] representation?
     
  2. jcsd
  3. May 29, 2012 #2
    my question comes thus:

    suppose we set up a device to detect a particle, it can detect the particle when the particle occur with in the region [itex][a,b][/itex],
    so when the device really detected the particle, the wave function of the particle must collaspe to one vanishes without the region, in a special case, to a square shaped one.

    I want to know its latter evolution, but I got confused when I tried to solve the SE directly in the Schrodinger representation .

    Thanks for any tip.
     
    Last edited: May 29, 2012
  4. May 29, 2012 #3

    vanhees71

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    You can simply use the solution in momentum space and transform it to position space. That's also nicely done numerically if you can't find an analytic expression. In fact you have

    [tex]\psi(t,x)=\langle x|\psi(t) \rangle=\int \mathrm{d} p \langle x | p \rangle \langle p|\psi> = \int \mathrm{d} p \frac{\exp(\mathrm{i} p x)}{\sqrt{2 \pi}} \tilde{\psi}(t,p).[/tex]

    Now you have

    [tex]\tilde{\psi}(t,p)=\langle p |\exp[-\mathrm{i} \hat{p}^2/(2m)] \psi(t=0) \rangle = \exp[-\mathrm{i} p^2/(2m)] \tilde{\psi}_0(p).[/tex]

    The wave function at [itex]t=0[/itex] in the momentum-space representation is of course

    [tex]\tilde{\psi}_0(p)=\int \mathrm{d} x \frac{\exp(-\mathrm{i} p x)}{(2 \pi)^{1/2}} \psi_0(x).[/tex]

    Now you have just to plug everything together.
     
  5. May 29, 2012 #4
    vanhees71, thank you for your reply.

    Your method is exactly what I mean by solving it using the propagator [itex]U(t)[/itex] in terms of [itex]P[/itex]'s eigenbras and eigenkets.

    What I wonder is why can't I solve it in coordinate space.

    For example, after infinitesimal interval [itex]\Delta t[/itex], by [itex]-i\hbar\frac{\partial}{\partial t}|\psi\rangle=H|\psi\rangle[/itex], we have
    [itex]\Delta|\psi\rangle=\frac{\Delta t}{i\hbar}H|\psi\rangle[/itex].

    if we work in [itex]P[/itex] basis, that's no problem, since then
    [itex]\Delta\psi(p)=\frac{p^2\Delta t}{i 2m\hbar}\psi(p)[/itex], and [itex]\psi(p)[/itex] behaves very well.

    however, if we work in [itex]X[/itex] basis,
    [itex]\Delta\psi(x)=i\hbar\Delta t\frac{\partial^2}{\partial t^2}\psi(x)[/itex], but look at how [itex]\frac{\partial^2}{\partial t^2}\psi(x)[/itex] behaves here, it just becomes improper.

    with this perspective, how could these two methods lead to the same result?
     
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