I Time evolution after "turning off" square well

[Swamp Thing]

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]

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

 

[Swamp Thing]

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]

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).

 

[Swamp Thing]

thank you!

 

[AdaggerA]

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.