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I Time evolution after "turning off" square well

  1. Jun 23, 2016 #1
    How how can we calculate the future evolution of a particle after the infinite square well potential is (somehow) turned off, releasing it into a free state? Assuming that it was in the ground state before.
     
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  3. Jun 23, 2016 #2

    Orodruin

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    The way you would expect it to, according to the Schrödinger equation for a free particle.
     
  4. Jun 23, 2016 #3
    Apart from brute force / numerical, is there a nice way to solve this (or to get useful qualitative properties of the solution)? For example, in the absence of ##V(x)##, can we transform the ##\psi(x,t=0)## into ##\psi_p(p,t=0)## and find the future ##\psi_p## in a simple way?
     
  5. Jun 24, 2016 #4

    stevendaryl

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    After turning off the potential, the wave function will evolve as a superposition of plane waves.

    Write:

    [itex]\psi(x,t=0) = \int dk e^{ikx} \tilde{\psi}(k)[/itex]

    where [itex]\tilde{\psi}(k) = \frac{1}{2\pi} \int dx e^{-ikx} \psi(x,t=0)[/itex]

    After the potential turns off, [itex]e^{ikx}[/itex] evolves into [itex]e^{i (kx - \omega t)}[/itex], where [itex]\omega = \frac{E_k}{\hbar} = \frac{\hbar k^2}{2m}[/itex]. So for [itex]t > 0[/itex],

    [itex]\psi(x,t) = \int dk e^{ikx - \omega t} \tilde{\psi}(k)[/itex]

    We can write this in another way:

    [itex]\psi(x,t) = \int dx' \int dk e^{ikx - \omega t} \int dx' e^{-i k x'} \psi(x',t=0)[/itex]

    [edit: added factor of [itex]\frac{1}{2\pi}[/itex]]

    Now under the questionable assumption that we can swap the order of integration, we can also write:

    [itex]\psi(x,t) = \int dx' \psi(x', t=0) \frac{1}{2\pi} \int dk e^{ik(x - x') - \omega t}[/itex]

    If we define [itex]G(x', x, t) = \frac{1}{2\pi} \int dk e^{ik(x-x') - \omega t}[/itex], then we can write:

    [itex]\psi(x,t) = \int dx' G(x', x, t) \psi(x',t=0)[/itex]

    where [itex]G[/itex] is the "Green function" for the Schrodinger equation. This can be interpreted in terms of amplitudes:

    The amplitude (density) for finding the particle at [itex]x[/itex] at time [itex]t[/itex] ([itex]\psi(x,t)[/itex] is the sum over all points [itex]x'[/itex] of the amplitude for finding the particle at [itex]x'[/itex] at time [itex]t[/itex] ([itex]\psi(x, t=0)[/itex]) times the amplitude for the particle to go from [itex]x'[/itex] at time [itex]t=0[/itex] to [itex]x[/itex] in time [itex]t[/itex] ([itex]G(x', x, t)[/itex].
     
    Last edited: Jun 24, 2016
  6. Jun 24, 2016 #5
    thank you!
     
  7. Jul 8, 2016 #6
    So, it's essentially just multiplying the wavefunction you have just before turning off the potential with the free particle propagator and integrating over all space as stevendaryl put it.
     
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