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About Wave Functions

  1. Sep 1, 2009 #1
    We know, a particle may be represented by a wave PACKET which has a group velocity.
    But most of the time, we are concerned with the wave FUNCTION of a moving particle.
    In the de broglie hypothesis, it is said that a wave equation y=e^i(kx-wt), CANNOT be associated with a moving particle as the phase velocity comes out greater than that of light.
    But , in *deriving* / or arriving at the schroedinger's eqn, we assume a wave function of a free particle as the gen. wave travelling wave eqn.(of course, we assume it to be complex)
    I don't know, I am getting confused between the relationship between the wave packet of a particle and it's wave function.
     
  2. jcsd
  3. Sep 1, 2009 #2

    olgranpappy

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    A wave packet is a wave function. we just say "packet" so that you remember it is made up of a linear combination of different free particle solutions.

    E.g., If the free particle hamiltonian is given by
    [tex]
    H=(-1/(2m))\frac{d^2}{dx^2}
    [/tex]
    Then, we have stationary solutions of the form [itex]e^{ikx-iwt}[/itex] for any k, as long as w=k^2/(2m). But these are unphysical (unnormalizable) solutions and can not represent the wave function of a real particle.

    On the other hand a linear combination of these plane wave can be normalized and has the form:
    [tex]
    \Psi(x,t)=\sum_{k}a_k e^{ikx-iw_kt}\;.
    [/tex]

    Which you can call a "wave packet" or a "wave function".
     
  4. Sep 1, 2009 #3

    dx

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    The approximation of quantum mechanics in which particles wave-functions can be treated as wave packets (also caleed quasi-particles) is called the semi-classical approximation. The group velocity corresponds to the particle concept of velocity, and the phase velocity corresponds to the particle concept of momentum. Just like the energy and momentum of classical particles are constrained by the equation E2 - p2 = m2, the angular frequency and wave numbers of these particle-like wave packets satisfy the relation

    [tex] \omega^2 - k^2 = \frac{m^2}{\hbar^2} [/tex]

    (called a dispersion relation). This prevents the wave packets from having a group velocity greater than light, just like E2 - p2 = m2 prevents ordinary particles from having a velocity greater than light. There are hypothetical particles called tachyons, whose dispersion relation is

    [tex] \omega^2 - k^2 = -\frac{m^2}{\hbar^2} [/tex]

    and which have a group velocity which exceeds c.
     
    Last edited: Sep 1, 2009
  5. Sep 1, 2009 #4
    I'd like to ask a question here.
    1)I have understood that a "packet" localises the particle. It is a superposition of waves of different wave lengths. But De broglie hypothesis says that a moving particle has with it associated a single wave length h/mv. How is this possible?
    And then how do you explain the successful interpretation of this "single wavelength=h/mv"concept in the Davisson Germer experiment
    2)And then, why do we use as the wave function[tex]\Psi=e^{-i(kx-wt)} [/tex] of a "FREE particle" while arriving at the schrodinger's time dependent formula. The assumption itself is wrong since this is not physically realizable(unnormalizable) wave function of a particle ?
     
    Last edited: Sep 2, 2009
  6. Sep 2, 2009 #5

    dx

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    In general, quantum particles do not have a definite energy and momentum, like classical particles do. They will be in a superposition of such states (with definite E and P). These are the states that the de Broglie's formula talks about. These states are plane-wave wave functions, characterized by their a definite wave number and an angular frequency.
     
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