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Wave function

  1. Oct 17, 2006 #1
    Consider a particle of mass m in the normal ground sate of an infinite square well potential of width a/2. Its normalized wave function at time t=0 is given by

    [tex] \Psi(x,0) = \frac{2}{\sqrt{a}} \sin \frac{2 \pi x}{a} [/tex] for 0 <x <a/2
    0 elsewhere

    At this time the well suddenly changes to an infinite square well with width a without affectring the wave function.

    By writing Psi(x,t) as a linear superposition of the energy eigenfunctions of the new potentaail find the probability taht a subsequency measurement of the enrgy will yield th result
    [tex] E_{1} = \frac{\hbar^2 \pi^2}{2ma^2} [/tex]
    {Hint: A linear superposition of square well eigenfunctions is a Fourier sine series and teh coefficients of teh series are given by simple integrals)

    Now we know that
    [tex] \Psi_{1}(x,0) = \sum_{n=0}^{\infty} c_{n} \psi_{n}(x) [/tex]

    and the wave function didnt change
    from teh hint
    if i find the coefficients all i do is find an approximation to teh wave function. the new potentail would be

    V(x) = 0 for 0<x<a
    infinity elsewhere

    and we know the wave function for that potential it is
    [tex] \Psi_{2}(x,0) = \sqrt{\frac{2}{a}} \sin\frac{\pi x}{a} [/tex]
    for the first exceited state

    so am i to find Psi 1 in terms of Psi 2??
     
    Last edited: Oct 17, 2006
  2. jcsd
  3. Oct 17, 2006 #2

    Galileo

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    You have to find the probability a measurement of the energy yields E_1. That probability is equal to [itex]|c_1|^2[/itex].

    You can get the term c1 from
    [tex] \Psi_{1}(x,0) = \sum_{n=0}^{\infty} c_{n} \psi_{n}(x) [/tex]
    by taking the inner product with [itex]\psi_1[/itex] on both sides, since the eigenstates are orthonormal.
     
  4. Oct 17, 2006 #3
    so with the change in dimensions of the well, the fact taht the wave funcion didnt change allows us to compute Cn using the wave function given to us??

    so
    [tex] c_{1} = \frac{2}{a/2} \int_{0}^{a/2} \sin \frac{\pi}{a} x \Psi(x,0) dx [/tex]

    wher [tex] \Psi(x,0) = \frac{2}{\sqrt{a}} \sin \frac{2 \pi x}{a} [/tex]

    right?
     
    Last edited: Oct 17, 2006
  5. Oct 17, 2006 #4

    Galileo

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    You can ALWAYS do that. If you expand your wave function on an orthonormal basis:

    [tex]|\Psi\rangle = \sum_n c_n|\phi_n\rangle[/tex]
    then we can find the coefficients by taking the inner product:
    [tex]c_n = \langle \phi_n|\Psi\rangle[/tex]

    (Maybe the notation is new to you, but I hope you get what it means).

    It could possibly be a factor of [itex]\frac{2\sqrt{2}}{a}[/itex] in front of there, but otherwise it looks good.
     
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