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Determining the Wave Function

  1. Aug 16, 2014 #1
    This is Problem comes from Griffiths Quantum Mechanics textbook; specifically, it is problem 2.5 (b).

    A particle in an infinite square well has its initial wave function an even mixture of the first two stationary states:

    [itex]\displaystyle \Psi(x,0) = A[\psi_1(x) + \psi_2(x)][/itex]

    Here is the part of the problem that I am having a little trouble with:

    (b) Find [itex]\displaystyle \Psi(x,t)[/itex] and [itex]\displaystyle |\Psi(x,t)|^2[/itex]|. Express the latter as a sinusoidal function of time, as in Example 2.1. To simplify the result, let [itex]\omega \equiv \frac{\pi^2 \hbar}{2ma^2}[/itex]

    According to the answer key, even after t=0, the wave function continues to be a mixture of the first two stationary states. Why is that? I am having a little difficulty understanding this. Why can't it be a new 'mixture?'

    As far as I understand, to calculate the wavefunction for all future times, we use the equations

    [itex]\displaystyle \Psi (x,t) = \sum_{n=1}^{\infty} c_n \sqrt{\frac{2}{a}} \sin \left( \frac{n \pi}{a} x \right) e^{-i(n^2 \pi^2 \hbar/2ma^2)t}[/itex]

    and

    [itex]\displaystyle c_n = \sqrt{\frac{2}{a}} \int_0^a \sin \left( \frac{n \pi}{a} x \right) \Psi (x,0) dx [/itex]
     
    Last edited: Aug 16, 2014
  2. jcsd
  3. Aug 16, 2014 #2

    vela

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    I'm not sure I understand your question. ##\Psi## will evolve with time, but it will remain a superposition of ##\psi_1## and ##\psi_2##.
     
  4. Aug 16, 2014 #3
    That's precisely my question. Why will it remain as a superposition of the two states, even after t=0?
     
  5. Aug 16, 2014 #4

    vela

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    At t=0, ##c_n = 0## for ##n>2##, right? So the coefficient of ##\psi_n## for ##n>2## will be 0 as well for ##t>0##.
     
  6. Aug 16, 2014 #5
    Oh, so the coefficients do not change with time?
     
  7. Aug 16, 2014 #6

    vela

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    The coefficients of ##\psi_1## and ##\psi_2## change with time as they pick up a phase factor. If you look at the formula for ##\Psi(x,t)##, you should be able to see that the coefficient of ##\psi_1##, for instance, is ##c_1 e^{-i\omega t}##. But the ##c_n##'s are defined in terms of ##\Psi## when ##t=0##, so they're just constants.
     
  8. Aug 16, 2014 #7

    Fredrik

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    Are you familiar with the time evolution operator in the form ##U(t)=e^{-iHt/\hbar}##? See if you can find it in your book. Once you understand it, you should find it easy to use it to answer your question.
     
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