# I Time evolution after "turning off" square well

1. Jun 23, 2016

### 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.

2. Jun 23, 2016

### Orodruin

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

3. Jun 23, 2016

### 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?

4. Jun 24, 2016

### stevendaryl

Staff Emeritus
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: Jun 24, 2016
5. Jun 24, 2016

thank you!

6. Jul 8, 2016