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Time Evolution After Sudden Potential Change

  1. Feb 27, 2017 #1
    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.

    From reading the wikipedia page here: https://en.wikipedia.org/wiki/Hamiltonian_(quantum_mechanics)#Free_particle

    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}##

    Thanks for any help you can give!
     
    Last edited: Feb 27, 2017
  2. jcsd
  3. Feb 27, 2017 #2
    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
     
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