# 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.

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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
Staff Emeritus
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:

$\psi(x,t=0) = \int dk e^{ikx} \tilde{\psi}(k)$

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

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

$\psi(x,t) = \int dk e^{ikx - \omega t} \tilde{\psi}(k)$

We can write this in another way:

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

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

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

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

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

$\psi(x,t) = \int dx' G(x', x, t) \psi(x',t=0)$

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

The amplitude (density) for finding the particle at $x$ at time $t$ ($\psi(x,t)$ is the sum over all points $x'$ of the amplitude for finding the particle at $x'$ at time $t$ ($\psi(x, t=0)$) times the amplitude for the particle to go from $x'$ at time $t=0$ to $x$ in time $t$ ($G(x', x, t)$.

Last edited:
Mentz114
thank you!

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.