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The wave packet description

  1. Nov 5, 2006 #1
    "The particle and the wave picture are both simplified forms of the wave packet description, a localized wave consisting of a combination of plane waves with different wavelength."

    it's confusing. Can somebody explain it? particle is an object, it seems wave is also an object.How to combine the two thing together?
  2. jcsd
  3. Nov 5, 2006 #2
    flowerew: How to combine the two things together?

    The quantum state vector combines wave and particle properties; one representation (in particular, the position representation) of the state vector is the wavefunction.
  4. Nov 6, 2006 #3
    i dont think thats what hes asking. your (masudr) are refering to the wavefunction ie shrodingers equation or wavefunction for position and probility whereas i think flowerew is asking how perpendicular EM and electrical fields can be construted as a particle, which is a good question.
  5. Nov 7, 2006 #4


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    I think of it as a localized wave.

    Imagine the waves in a pool combining in such a way that there is a big bump in the middle of the pool that smooths out to the boundaries of the pool. A bump, that is a localized standing wave.

    Pool waves never do this, because the wavenumbers are pretty much random. But for particles, there is an average wavelength, a Gaussian distribution (for a free particle) that the wavenumbers mostly hang around. Because of this, when you add together an infinite set of waves then, you get a localized standing wave "envelope" that is the particle.

    Its comprised of waves, so it has wave properties, but the waves form an overall localized wave that gives it its particle properties then.

    As far as EM fields and quantization, you'll need to read some Quantum Field Theory and search some "second quantization" stuff. In short, one can quantitize any normalizable field, which then becomes a QFT. The EM field is normalizable, which is why QED is so successful. But the Einstein Field Equations are not normalizable and hence the trouble.
  6. Nov 7, 2006 #5
    Many people on this forum agrees on the fact that the wavefunction is just a mathematical representation and not a "real" wave.
  7. Nov 7, 2006 #6
    How dare you even speak such words! You obviously are not smart enough to comprehend quantum mechanics. You should be placed under house arrest and go back to studying inclined planes.
    Last edited: Nov 7, 2006
  8. Nov 7, 2006 #7
    there are no particles, there are only waves
  9. Nov 7, 2006 #8
    How would you prove it?
  10. Nov 7, 2006 #9


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    The wavefunction of 2 particles already doesn't "live" in ordinary space, but in a 6-dimensional configuration space...
  11. Nov 7, 2006 #10
    The question is asked the wrong way round. It's not that photons (or any excitations of a fundamental quantum field) are constructed out of EM fields. Instead, it's that EM fields are constructed from many zillions of photons.
  12. Nov 7, 2006 #11
    where in the schrodinger equation is there anything but a wavefunction?

    what we call a "particle" is just the localized wavepacket in the classical limit
  13. Nov 7, 2006 #12


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    I prefer to say that there are no (pure) particles, nor (pure) waves. Quantum-mechanical objects have both particle-like and wave-like aspects.
  14. Nov 7, 2006 #13
    AFAIK you need to assume point particles in order to write down the Schrodinger's equation. So, it seems strange to deny the existence of those point particles, isn't it?
  15. Nov 7, 2006 #14
    No you don't.

    You need to assume a classical system that possesses a Hamiltonian (or equivalently a Lagrangian).
  16. Nov 8, 2006 #15
    For a hydrogen atom we assume the electron and nucleus to be point charges (we know this is not true for the nucleus and there are deviations for the calculated energy because of that) so the force between them is given by the Coulomb law.

    The squared amplitude of the wave function gives the probability of detecting those point charges in a certain place. So, we have particles and their motion is statistically described by a wave.

    The motion of water molecules can be described by a wave but this doesn't mean that a water molecule is a wave.
  17. Nov 8, 2006 #16
    Well, technically speaking, the Hydrogen atom is treated as a system with Hamiltonian (I'll write it in Cartesians and their conjugate momenta since they are the usual canonical variables in the electrostatic case, although it'd be more concise in spherical polar co-ordinates):

    \hat{p_z}_2) =[/tex]

    (\hat{x}_1 - \hat{x}_2)^2+
    (\hat{y}_1 - \hat{y}_2)^2+
    (\hat{z}_1 - \hat{z}_2)^2}}

    You haven't assumed anything of the system apart from the fact that it has the above Hamiltonian. To get to that expression, you might use some classical physics of point particles and the Coulombic interaction between charged particles. However, note that the quantum mechanics of the system starts here, at the Hamiltonian.

    You haven't assumed anything more, specifically, you haven't talked about particles - you've talked about systems. I'm not being pedantic: in QFT, you don't deal with particles directly, you deal with oscillating fields, where particles happen to emerge.

    P.S. You usually eliminate 6 of the variables by transforming to so-called "centre-of-mass" and "relative" co-ordinates, and while this has physical content in this case, it didn't have to.
    Last edited: Nov 8, 2006
  18. Nov 8, 2006 #17
    particle physics would disagree
  19. Nov 9, 2006 #18
    really? i dont think so

    tell me what happens to your particles when their average seperation is on the order of the de broglie thermal wavelength? are they still "particles"?
  20. Nov 9, 2006 #19
    Then please tell me what is the meaning of x1, px1 and m1? What is their physical significance?

    Is there any other way?


    I have no training in QFT so I cannot contradict you here, but my understanding was that QFT treats fields as particles and not the other way around.
  21. Nov 9, 2006 #20
    The SE can't be solved until you specify the potential energy function, U. For the H-atom, U(r) = -e^2/r in which you assume point charges. Is this not a contradiction to current quantum interpretations?
    Last edited: Nov 9, 2006
  22. Nov 9, 2006 #21
    Or perhaps you misinterpret wave functions? Does a wave function correspond
    to a single quantum object or to an ensamble of single quantum objects?

    If the second is true then QM has nothing to say about particle or wave character
    of a single quantum object, a single electron for example. QM describes ensambles.

    If the first is true then how come that electrons appear as points on the screen
    in Young-like experiment although the their wave functions spread across the
    whole screen? In other words if QM applies to single quantum objects then one
    should be able to predict WITH CERTAINTY (not just probability) an outcome of a
    single run of an experiment with one electron!

  23. Nov 9, 2006 #22
    Is this a polish local realist speaking ? :wink:
  24. Nov 9, 2006 #23


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    What is sure, is that the wave function generates correct probabilities for an ensemble. Now, as whether or not one can give a meaning to the wavefunction of a single quantum object depends on what philosophical preferences one has about ontology.
    Personally, I think it is impossible not to consider that a wavefunction describes individual systems, given that there aren't always ensembles.

    This also has foundational problems of course. Where does the ensemble come from if you only do a thing once ? It gives problems in cosmological considerations. Now of course, without "ensemble", probabilities don't make sense either and hence the "single-event" description is only a description, without predictive value, but at least it is a description.

    The question is: is the ensemble, the ensemble of objects, or the ensemble of observers ? If you do a thing only once, but this generates an ensemble of observers, then you have your probabilities there...

    This is because one has taken as a hidden ontological assumption that determinism of observation should be true, in other words, that it is not possible to generate an ensemble of observers.
    Last edited: Nov 10, 2006
  25. Nov 10, 2006 #24
    Hehe, the true vanesch style!
  26. Nov 10, 2006 #25
    Yes. Not sure if realist, though.

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