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Particle in infinite potential well

  1. May 28, 2014 #1
    1. The problem statement, all variables and given/known data

    29xxisp.png

    Sketch the difference of probability distributions at the two times. Does the energy change with time?

    The potential well suddenly disappears, what is the form of the wavefunction?

    2. Relevant equations



    3. The attempt at a solution

    Part (a)

    At t = 0, the probability distribution is:
    [tex]|\phi|^2 = \frac{1}{a} \left[ sin^2(\frac{2\pi x}{a}) + 2 sin (\frac{2\pi x}{a})sin(\frac{4\pi x}{a}) + sin^2 (\frac{4\pi x}{a}) \right][/tex]

    At ##t = \frac{\pi}{6\omega}##:
    [tex]|\phi|^2 = \frac{1}{a} \left[ sin^2(\frac{2\pi x}{a}) + sin^2 (\frac{4\pi x}{a}) \right] [/tex]

    Thus the difference is simply ##\frac{2}{a} sin(\frac{2\pi x}{a}) sin (\frac{4\pi x}{a})##:

    33acnkn.png

    I think the energy of the particle changes with time, as simply using the TDSE ##-\frac{\hbar^2}{2m} \frac{\partial^2 \phi}{\partial x^2} = E \phi## doesn't remove the ##exp(i\omega t)## terms.


    Part (b)

    As the potential disappears, the particle becomes free. I think the general form is a linear combination of the n=2 and n=1 states.

    For a free particle, ##u_{(x)} = A e^{i\frac{px}{\hbar}} = e^{ikx}##.

    Thus, the form is:
    [tex]\phi = Ae^{i(2kx - \omega t)} + Be^{i(4kx - \omega t)}[/tex]

    Not sure if I'm right.
     
  2. jcsd
  3. May 28, 2014 #2
    Part b is not right. You have to expand the wave function using a Fourier transform. In your solution the particle jumps from a confined solution to a couple of plane waves that extend to infinity. That doesn't make sense. The Fourier transform will include all sorts of wavelengths - not just 2.
     
  4. May 28, 2014 #3
    So the solution would be:

    [tex]\phi = \sum_p a_p e^{i(\frac{xp}{\hbar} - \omega t)}[/tex]

    Does this look right?
     
  5. Jun 1, 2014 #4
    The solution seems a little short for 5 marks, in my opinion
     
  6. Jun 2, 2014 #5
    Well you should use a Fourier transform as opposed to a Fourier expansion because the particles are free at the end of the problem. That means you should have an integral instead of a sum.
     
  7. Jun 3, 2014 #6
    unscientific - I don't understand why you think the energy changes.

    That equation looks like the TISE, not the TDSE.
     
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