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Wave packets spreading out

  1. Aug 4, 2004 #1
    Why does a wave packet spread out in space as time passes?
    What difference would it make to the universe if it did not?

    And have a look at this link which shows a movie of "the time evolution of a quantum wave packet."
    http://webphysics.davidson.edu/mjb/acs_transformations_qm/packet.html

    Is the movie lacking any important information that I should know about a wave packet?
     
    Last edited: Aug 4, 2004
  2. jcsd
  3. Aug 4, 2004 #2
    That is a wave packet loses linear momentum as time progresses by spreading out and lowering the peak hence increasing its wavelength. The decrease in momentum is related to a decrease in kinetic energy of the wave. If it is a light wave, in contrast to matter wave traveling in vacuum, then the speed remains constant.
     
  4. Aug 4, 2004 #3
    I think of wave propagation in relation to its source. As the wave moves farther away from the source, it has to fill the volume vacated by the preceeding wave. Of course, this process changes if or when it incounters another system and is consummed.
     
  5. Aug 5, 2004 #4
    Antonio Lao:
    That is a wave packet loses linear momentum as time progresses by spreading out and lowering the peak hence increasing its wavelength. The decrease in momentum is related to a decrease in kinetic energy of the wave. If it is a light wave, in contrast to matter wave traveling in vacuum, then the speed remains constant.

    Kurious:
    Does kinetic energy become potential energy in the case of matter wave?
     
  6. Aug 5, 2004 #5
    This is a very good question to which I don't have good answer at the moment but I can make a few educated guesses.

    We know that all matter slow down when set in motion. This will include the planetary motion around the sun. Matter can slow down when it acquires and gains mass. Matter can slow down when it is challenge by opposing forces (frictions, field resistances, etc.).

    But in QM, when the exact location of the electron is found, the wave packet collapses, i.e., the wave nature of the particle disappears and only the particle nature is left behind. But according to the Heisenberg's uncertainty principle, we can never find the exact location of the electron and the solidity of atom is just a balance of forces.

    One can make the assumption that even energy has to have its relativeness. So for matter losing relative kinetic energy gaining back relative potential energy.

    For all practical purposes, the photon is purely kinetic. Its rest mass is zero but its momentum is [itex] \frac{E}{c}[/itex].
     
  7. Aug 5, 2004 #6
  8. Aug 5, 2004 #7
    The conservation of energy is true only for an isolated thermodynamic system (mass and energy cannot enter or leave the system boundary). And it makes the following statement:

    Total energy of the system is a constant and is equal to the sum of potential energy and kinetic energy.

    For a closed thermodynmic system there is no mass crosses the system boundary. But its volume can vary. And also energy can enter or leave the system boundary.

    The statement above is the simplest case of time independent Hamiltonian function. And the Hamiltonian approach was used in the beginning of quantum mechanics before Feynman replaced it with the Lagrangian formalism.
     
    Last edited: Aug 5, 2004
  9. Aug 5, 2004 #8
    Last edited: Aug 5, 2004
  10. Aug 6, 2004 #9
    It has to do with how each of the two formalisms deals with the concept of "path" in physics. Hamiltonian works good in classical mechanics but Lagrangian can work good both in classical and quantum mechanics.
     
  11. Aug 6, 2004 #10
    By the principle of continuity with its equation given by

    [tex] \frac{\partial \rho}{\partial t} + \nabla \cdot \left(\rho \vec{v}\right) =0 [/tex]

    the concept of density is implied to exist but not equal to zero. So the path of flow can be analyzed.

    But in quantum mechnanics (discrete systems), the path of a quantum particle such as an electron has no meaning since each point in the path is associated with an uncertainty of position and velocity. There is no way to know whether the electron has actually move from here to there or there to here. No way of knowing whether the motion is forward or backward in time. The Hamiltonian is soluble if it is time independent but if time need to be analyzed then a direction of its flow must be accounted by the analysis. But in Lagrangian, since it is the difference between potential and kinetic energy, the description of the direction is not necessary, but the result is a principle of least action which is a more elegant form of Fermat's principle of least time.
     
  12. Aug 6, 2004 #11
    Since the Lagrangian is really a continuum formalism, Feynman came up with path integral method for sum over histories of all possible paths and each of these paths is associated with a probability function. The most probable path is the classical path.
     
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