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Particle-in-a-box and the uncertainty principle

  1. Oct 17, 2004 #1
    For a particle-in-a-box it can be shown that the possible energies are given by

    [tex] E_n = \frac{n^2h^2}{8mL^2} [/tex]

    where L is the length of the box. The corresponding momentum are given by:

    [tex] p_n = \frac{nh}{2L} [/tex]

    I don't think it's a problem that the energy has a definite value ([tex] \Delta E = 0 [/tex]) since it is a stationary state ([tex] \Delta t = \infty [/tex]).

    But how is it possible for the momentum to be definite ([tex] \Delta p = 0 [/tex]) and, at the same time, the particle to be confined within the box ([tex] \Delta x < \infty [/tex]). Doesn't this violate the uncertainty principle [tex]

    \Delta x \Delta p_x \geq \frac{h}{2\pi} [/tex].
     
  2. jcsd
  3. Oct 17, 2004 #2

    Dr Transport

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    Do the calculation and find out.......it just takes a few integrals to find the answer.
     
  4. Oct 17, 2004 #3
    By my understanding it does. We have [tex] \Delta x \approx L/2 [/tex] and [tex] \Delta p_x = 0 [/tex], and thus:

    [tex] \Delta x \Delta p_x = (L/2) \cdot 0 = 0 \leq \frac{h}{2\pi} [/tex]

    A violation! What is wrong with this??
     
  5. Oct 17, 2004 #4

    vanesch

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    What you write as p_x is in fact |p_x|.
    Indeed, given the energy, the momentum can be to the left or to the right, so there is an uncertainty on p_x (even if there isn't on |p_x|).

    cheers,
    Patrick.
     
  6. Oct 17, 2004 #5
    Oh, you're right.. So we have [tex] \Delta p_x \approx \frac{h}{2L} [/tex]?
     
  7. Oct 17, 2004 #6

    vanesch

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    Without plunging into the calculation, I'd guess so. If (for n = 1), we can say that |p| = Sqrt[ 2 m E_1] = h/2L. This is of course a bit naive, because the particle is not really free, it undergoes an interaction at the walls. But it is "most of the time" free. So we'd think then that the corresponding momentum density function can be represented by two dirac pulses, one at -h/2L and one at h/2L. The average momentum being 0, this means that the standard deviation is h/2L for p_x.
    But as I said, we've forgotten the role of the potential energy here. Indeed, the position representation of such a momentum double dirac function would be a sine or cosine, with period 2L. Well, this is exactly the ground state !
    It is sin(x / (pi L) ) ! But wait... it is this function within the interval 0-L, but the wave function is 0 outside the interval, while the Fourier transform of our 2 dirac peaks would indicate sin(x/(pi L) ) function everywhere. So we have multiplied in the position space with a square window function (1 in the interval 0-L, 0 outside). This translates into a convolution with a sin(p)/p function in the momentum representation, which will smoothen out the dirac peaks and will probably change a bit the value of Delta-p_x.

    At least, the above is my guess, I never worked it out myself.

    cheers,
    Patrick.
     
  8. Oct 18, 2004 #7
    Integral calculation

    Might I suggest you redo your calculations? The uncertainty is p is desinqrt as
    sqrt ( <p^2> - <p>^2)

    where <p^2> = (nh/(4L))^2 and ,p. = 0 so the uncertainty in the momentum is not zero in the eigenstates of the particle-in-a-box.

    achemist
     
  9. Oct 18, 2004 #8
    Integration typos

    Hi, sorry for the typos. If you do the calculation for n=1 you will get
    delta P = Pi hbar / L
    delta x = .181 L
    so indeed we get delta x * delta p = .181 Pi * hbar
     
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