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QM - another particle in a box

  1. Sep 30, 2005 #1

    quasar987

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    Consider an infinite well of lenght L. We measure the position of the particle and obtain x=L/2. After the measurement, what is the probability of finding the particle in a given eigenstate?

    My solution: WhaaaaaAaaT ?! Neither Gasiorowicz and Griffiths talk about the probability of the particle being in a certain wave function! I tought the wave function was that one thing that was determinate in QM. Worst, the one clue I have is contradictory to the statement of the problem; it is when Griffiths says that if we measure the position of a particle, then its wave function must crumble into a dirac delta about this position so that another measurement performed right after will be guarenteed to return the same position. A Dirac delta function is no eigenstate of the particle in the box, hence a contradiction. (unless the answer to the problem is in fact 0% for that very reason, but I doubt it.)

    Eidt: I kinda recall someone on this forum saying how the wave function can be any linear combination of the eigenstates, but once a measurement is made, it must settle into one of those, but I can't find back the thread.
     
    Last edited: Sep 30, 2005
  2. jcsd
  3. Oct 1, 2005 #2
    If you do the measurement again really fast it'll be in the same state with a probability of 1. I don't know if this is what Griffiths meant but delta functions are actually eigenfunctions of the position operator.

    And regarding the edit, yeah that's correct.
     
  4. Oct 1, 2005 #3

    Galileo

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    Well, this is just some of the phrases that is often used by practising physicists that can throw you off, because it doesn't sound right.

    You do not measure in which eigenstate the particle is (you cannot), you measure observables. But, if you measure the energy of a particle in the well, you get, say, E_2 with a certain probability. There is only ONE (energy-)eigenstate corresponding to that eigenvalue. Because of this unique characterization of the state by an eigenvalue (or quantum number), instead of 'find the probability of measuring an energy of E_2' it is often sloppily said 'Find the probability the particle is in the state Psi_2' (<- eigenstate corresponding to E_2).
    It's wrong and confusing, but deal with it. Notice that you can only unambiguously talk about it this way if the quantum number uniquely characterizes an eigenstate. If you have degenerate eigenvalues, this is no longer possible. I won't go into that too much.

    There's another problem with the question that is of a more serious nature. Although the direc delta function is an eigenstate of the position operator, it is NOT a viable state for a particle to be in. For one thing, it isn't normalizable. If you measure the position, you always have some uncertainty (this is classical uncertainty. You simply cannot measure with infinite acuracy). This happens with all observables with a continuous spectrum. So the particle will not collapse into a delta function, but a function that is sharply peaked about the position, but which still has some spread because of the uncertainty in measurement.
    Read this post for an example of how wrong it can go if you use delta functions:
    https://www.physicsforums.com/showthread.php?t=86205
     
  5. Oct 1, 2005 #4

    quasar987

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    I seem to have found a pretty potent trail! pp.54 reads

    Ok, this applies to us, since we're making a measurement of the position (<-- the observable). The eigenstate of the observable are the solutions to

    [tex]X_{op}\mu = \frac{L}{2}\mu[/tex]
    [tex]\frac{\partial \mu}{\partial p} = \frac{-iL}{2\hbar}\mu[/tex]
    [tex]\Rightarrow \mu(p) = Ce^{iLp/2\hbar}[/tex]

    C can be determined by normalizing.

    This is the wave function of the state in p-space. To find if it corresponds to an eigenstate of the energy in x-space, I can either fourier transform the eigenfunctions u_n(x) into p-space and show that they are not of the form of mu. Or inversely, I can fourier transform mu(p) into x-space and show that it is not of the form of any u_n.

    Thumbs up or thumbs down?
     
  6. Oct 2, 2005 #5

    quasar987

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    One problem, how do I normalize mu ?!

    [tex]\int_{-\infty}^{\infty}\mu \mu^*dp = \int_{-\infty}^{\infty}CC^*dp[/tex]

    which diverge for all C. :grumpy:
     
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