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Homework Help: Particle in infinite well which is suddenly expanded

  1. Apr 10, 2010 #1
    1. The problem statement, all variables and given/known data
    Well I thought this problem was easy, turned in the homework and got it wrong. My prof is hard to get help from, so hopefully someone here can help me out. A particle is in a one dimensional infinite square well with walls at x=0 and x=L. At time t=0 the well is expanded to width 2L. What is the probability the particle will be in the nth stationary state of the expanded well? I bet a lot of people have seen this before apparently it's a pretty popular problem.



    2. Relevant equations


    [tex]\left\langle\Psi_{f}\left|\Psi_{i}\right\rangle = d_{n} [/tex]


    3. The attempt at a solution

    [tex]=\frac{\sqrt{2}}{L}\int sin(\frac{n \pi x}{2 L})sin(\frac{\pi x}{L})dx[/tex]

    So I'm pretty sure I'm right so far, and the problem is just to evaluate the integral. I was sure the solution was

    [tex]d_{n}=\frac{\sqrt{2}}{2} [/tex] if n=2

    [tex]d_{n}= 0 [/tex] if n not = 2

    But apparently the answer is

    [tex] d_{n}= \frac{\sqrt{2}}{2}[/tex] if n=2
    [tex]d_{n}= \frac{4}{\pi}\frac{1}{(n^{2}-4}[/tex] if n is odd
    [tex]d_{n}= 0 [/tex] otherwise

    I can plug the integral into mathematica and of course it gives me the right answer. But I have seen integrals like this hundreds of times in solving PDE's, and they always go like I first thought. What is the difference here? This is causing me a ridiculous amount of cognitive dissonance.
     
  2. jcsd
  3. Apr 10, 2010 #2

    kuruman

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    One of the sines has a 2L in the denominator while the other has just L. That's the difference.
     
  4. Apr 10, 2010 #3
    The integral is the Fourier transform of the [tex]sin(\frac{\pi x}{L})[/tex] in the new well.
    From your result, I guess the initial wave function may not be the ground state of the of the old well ([tex]sin(\frac{\pi x}{L})[/tex])
    or your integral use a different domain (note the integration from 0 to L is different from 0 to 2L) because the wavefunction is 0 within L to 2L
    I have not tried it though.
     
  5. Apr 10, 2010 #4
    I realize that. I just don't see why it makes any difference. For n=8 there might as well be a 4 in the numerator and just L in the denominator. How the heck do I explicitly evaluate this so I can prove the answer to myself?
     
  6. Apr 10, 2010 #5

    kuruman

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    If your problem is evaluating the integral, first substitute θ = πx/L and convert the sines to exponentials. Then

    [tex]sin(n\theta/2)sin(\theta)=-\frac{(e^{i n\theta/2}-e^{-i n\theta/2})(e^{i \theta}-e^{-i \theta})}{4}[/tex]

    Multiply out the terms and you will get four integrals that look like

    [tex]\int^{\pi}_0 e^{i(n/2+1)\theta}d\theta=\left[ \frac{e^{i(n/2+1)\theta}}{i(n/2+1)}\right]^{\pi}_0[/tex]

    or, after you multiply out the terms you should be able to recognize sines and/or cosines and integrate them.

    In either case, eventually, you should be able to combine the answer to a real number and the role of the 2 in the denominator should become obvious.
     
    Last edited: Apr 10, 2010
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