# Time Evolution After Sudden Potential Change

1. Feb 27, 2017

### BOAS

1. The problem statement, all variables and given/known data
A particle in an infinite potential well $V(x) = 0, -\frac{a}{2} \leq x \leq \frac{a}{2}$, and infinite elsewhere is in it's ground state. Subsequently, the potential is removed and the particle is free to move.

How does the probability distribution in x and p change immediately after the walls are removed?

2. Relevant equations

3. The attempt at a solution

I have found and normalised my wavefunction in position space $\psi (x, t=0) = \sqrt{\frac{2}{a}} \sin(\frac{\pi x}{a} + \frac{\pi}{2})$

I have taken the fourier transform of this wavefunction to find the wavefunction in momentum space $\psi (p, t=0) = \frac{2 \hbar^{\frac{3}{2}} \sqrt{\pi}}{\sqrt{\frac{1}{a}} (\hbar^2 \pi^2 - a^2 p^2)} \cos(\frac{pa}{2 \hbar})$

with these, i can calculate the probability density in x and p at t=0, but i am unsure of how to proceed once the potential has been removed.

I think I might need to make use of this relation: $|\psi(t)\rangle = e^{-iHt / \hbar} |\psi(0)\rangle$, where I use the hamiltonian of the free particle to compute how my initial wavefunction evolves.

I am not familiar with Dirac notation however - Is that equivalent to the hamiltonian acting on the wavefunction in the argument of the exponential?

something like $\psi (x,t) = e^{-iH \psi(x, t=0) t / \hbar}$

Last edited: Feb 27, 2017
2. Feb 27, 2017

### BOAS

I am really confused.

Do I just need to make use of $\hat{H} \psi (x,0) = E \psi (x,0)$, and $\psi (x,t) = e^{-iEt/\hbar} \psi (x,0)$?

Using the hamiltonian for a free particle