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Wave function of a particle

  1. Feb 20, 2007 #1
    Hello

    I know that every particle has attached a wavefunction in quantum mechanics.

    How does a free particle move in quantum mechanics?

    The wavefunction has periodic zeroes, and the |wavefunction|^2 gives the probability of finding that particle...so does this mean that in space the particle has certain places in which it cannot exist?
     
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  3. Feb 20, 2007 #2

    jtbell

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    A free particle with a definite momentum has a complex wave function:

    [tex]\Psi(x,t) = Ae^{i(kx - \omega t)}[/tex]

    Its probability density function is

    [tex]\Psi^*\Psi = A^*e^{-i(kx - \omega t)}Ae^{i(kx - \omega t)} = A^*A[/tex]

    which is uniform and does not have any zeroes.

    This is an idealization because there is no such thing as a free particle with a perfectly definite momentum. A realistic free particle wave function is a wave packet whose p.d.f. is maximum at the position where the particle is most likely to be found, and falls off to zero as you go further away from that position. It's rather like taking the functions above and making A a variable instead of a constant, rising from zero to a maximum and then falling back to zero as you move along the x-axis.
     
    Last edited: Feb 20, 2007
  4. Feb 20, 2007 #3
    I see. The wavefunction is complex! The real and imaginary parts have any physical significance?

    When do I use the current density in quantum mechanics?
     
  5. Feb 21, 2007 #4

    dextercioby

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    No.

    In problems involving the tunnel effect, for example.
     
  6. Feb 21, 2007 #5

    George Jones

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    The probability (for 1-dimensional problems) of finding a particle in the region between [itex]x =a[/itex] and [itex]x =b[/itex] is

    [tex]\int_{a}^{b} \Psi^* \Psi dx,[/tex]

    so the probability of finding a particle at [itex]x = a[/itex] is

    [tex]\int_{a}^{a} \Psi^* \Psi dx = 0[/tex]

    even when the wavefunction at [itex]x = a[/itex] is non-zero.

    (Delta functions aren't really allowed as states.)
     
    Last edited: Feb 21, 2007
  7. Feb 21, 2007 #6

    vanesch

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    Careful! A complete system has a wavefunction attached to it! This is frequent error in learning quantum theory, it was a historical error too, and the way things are introduced in most textbooks could give the impression that with a particle, goes a wave.
    But it is only in the case that our "complete system" is a single particle, that there is a wavefunction attached to a particle.
    In the case we have several particles, there is a SINGLE wavefunction which is describing the SET of particles ; most of the time, this single wavefunction cannot be split into N "single-particle" wavefunctions.
     
  8. Feb 21, 2007 #7

    dextercioby

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    That's true only if one buys the Copenhagen Interpretation. The Statistical Interpretation denies it.
     
  9. Feb 21, 2007 #8

    Demystifier

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  10. Feb 21, 2007 #9

    dextercioby

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    At that time i wasn't even aware of the existence of the statistical interpretation. I was taught Copenhagian in school and didn't bother into MWI and BMech.

    I did some more reading inbetween, i guess...

    EDIT: And i'm 24 yrs old. So far...
     
    Last edited: Feb 21, 2007
  11. Feb 21, 2007 #10

    Demystifier

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    You learn and change your opinion very fast. You must be rather young. Let me guess: 26?
     
  12. Feb 21, 2007 #11

    George Jones

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    Entanglement accounts for much "spookiness" in quantum theory.

    I keep meaning to get a copy of Ballentine and look at this; on my "to do someday" list.
     
  13. Feb 22, 2007 #12
    Sorry for the probably silly and/or already treated question: if an electron is point like, should this mean the probability to find it in a very small volume of space is not (almost) zero?
     
  14. Feb 22, 2007 #13

    dextercioby

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    It all depends on the wavefunction, so it can take any value between 0 and 1, including 0 and 1.
     
  15. Feb 23, 2007 #14

    George Jones

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    I'm not sure what you're asking. As dextercioby has said, the probablity could be any number between 0 and 1.

    Note that the position operator has no eigenstates, so it is impossible to prepare an electron in a state that is localized at a single point.

    Does this thread help?
     
  16. Feb 23, 2007 #15
    Define "Statistical Interpretation"

    (I am familiar with MWI and I assume BMech is Bohmian Mech.)
     
  17. Feb 25, 2007 #16

    dextercioby

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