I Time evolution after "turning off" square well

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.
 

Orodruin

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The way you would expect it to, according to the Schrödinger equation for a free particle.
 
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?
 

stevendaryl

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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?
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:
thank you!
 
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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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