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Fourier transform for a localized free particle

  1. Feb 18, 2009 #1
    1. The problem statement, all variables and given/known data
    This is from Griffiths Introduction to Quantum Mechanics, Problem 2.21.

    Suppose a free particle, which is initially localized in the range -a<x<a, is released at time t=0:
    [tex] \Psi(x,0) = \begin{cases}
    \frac{1}{\sqrt{2a}}, & \text{if } -a<x<a,\\
    0, & \text{otherwise},\end{cases} [/tex]
    Determine the Fourier transform [tex]\phi(k)[/tex] of [tex]\Psi(x,0)[/tex] and comment on the behavior of [tex]\phi(k)[/tex] for very small and very large values of a. How does this relate to the uncertainty principle?

    2. Relevant equations
    [tex] \phi(k) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} \Psi(x,0) e^{-ikx}\, dx = \frac{1}{2\sqrt{\pi a}} \int_{-a}^{+a} e^{-ikx}\, dx = -\frac{1}{2ik\sqrt{\pi a}} \left(e^{-ika}-e^{ika}\right) = \frac{\sin (ka)}{k\sqrt{\pi a}}[/tex]


    3. The attempt at a solution
    The one Fourier transform with a greater [tex]a[/tex] has more rapid oscillations and a higher peak at k=0, when compared to the one with a smaller [tex]a[/tex]. A big value of [tex]a[/tex] corresponds to a widely spread particle, while a small [tex]a[/tex] stands for a particle well localized at x=0.

    My problem is that I don't see the connection between my Fourier transform and uncertainty. As far as I understand, rapid oscillations serve to pick up some specific values of energy, and the resulting wave function is close to a discrete linear combination, while slow oscillations pick up a wide spectrum of energies. But what does this have to do with uncertainty? Both seem quite uncertain to me...
     
    Last edited: Feb 18, 2009
  2. jcsd
  3. Feb 18, 2009 #2

    lanedance

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    Homework Helper

    In terms of uncertainty look at your initial wave function, what does it represent? It represnts a probability amplitude to be found at position x, if a measurement is made. The bigger a is, the more uncertainty in the initial position of the particle when a measurement is made

    Now have a think about what the fourier transfrom is 'transforming' to...ie it is shifting from description in position space to...

    Then look at your solution effectively sinka/k, with a normalisation factor. Sounds like you've plotted it up but always a good place to start. Amplitude decays like 1/k and oscillates as sinka.

    I wouldn't get too tied up with the oscillations at this stage they are more an artifact of the orginal position distribution shape, which is rectangular. Concentrate on the main peak for now

    You're close with you comments of wide v narrow energy distribution, but why make the jump to energy? A k state for a free particle is an energy eigenstate, but work in k space for the moment. Think about the width of the peak and how this affects the range of outcomes for a measurement of k. Then consider how this relates to the uncertainty in the in the original position distribution...
     
    Last edited: Feb 18, 2009
  4. Feb 19, 2009 #3
    Yes, I actually have a graph:
    http://img410.imageshack.us/my.php?image=221eo7.pdf"
    I only don't know if you can see it, it's a PDF.

    Anyway, OK, I concentrate on the main peak. For a big [tex]a[/tex] the particle can be in a wide range of positions, so it has a high position uncertainty, but the Fourier transform in such a case gives a narrow and high peak at the center, which serves to pick up a small range of energies (roughly acts like a delta function):

    [tex] \Psi(x,t) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} \phi(k) e^{ikx} e^{-iE(k)t/\hbar}\, dk[/tex]

    which is the opposite of what I expect. An initially uncertain particle gets to transformed into something with a well defined energy and position (center).
     
    Last edited by a moderator: Apr 24, 2017
  5. Feb 19, 2009 #4

    lanedance

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    Homework Helper

    getting close, but you need to understand the the fourier transfrom is taking you to momentum (k) space, so instead of a description in position (x)you now have a description in momentum (k) (ie the bottom axis on graph changes from x to k after the transform)

    So once we have transformed to k space, the particle is no longer described in terms of position (the position information is inhernet in the wave function but not expressed directly)

    Your observations about the realative peak widths were right, though interpretation of the k distributtion was a bit misguided. Try thinking again about these peaks in term of x & k uncertainties...

    have a go thinking it through first but if you need a further hint:
    google heisenberg
     
    Last edited: Feb 19, 2009
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