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Does expansion a(t) affect QM probability density?

  1. Jun 16, 2015 #1
    I'm picturing a comoving particle, meaning at rest with respect to CMB? Does expansion affect the probability density, density current? I can't see how it wouldn't? But then it seems like there would have to be a uniform negative probability density current everywhere.
     
    Last edited: Jun 16, 2015
  2. jcsd
  3. Jun 16, 2015 #2

    mfb

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    I guess the definition of probability density current should be modified then. I'm not aware of setups where this would actually be relevant.
     
  4. Jun 17, 2015 #3
    This is the equation I am trying to understand the meaning of.

    [itex]j\quad =\quad \frac { \hbar }{ 2mi } \left( { \Psi }^{ * }\frac { \partial \Psi }{ \partial x } -\Psi \frac { \partial { \Psi }^{ * } }{ \partial x } \right) [/itex]

    The part inside the parenthesis is what i would like to get a feel for. I know it describes the difference between the product of of the wave function and the partial derivative of it's conjugate, and the product of the conjugate and the partial of the wavefunction (the "reversed product").

    The difference would represent.... the spatial non-commutativity of the wavefunction?




    Not that it's important but this is the paper I am trying to read understand some of.

    http://arxiv.org/pdf/quant-ph/0105112v1.pdf
    On Koopman-von Neumann Waves
    D. Mauro
    (Submitted on 23 May 2001)
    In this paper we study the classical Hilbert space introduced by Koopman and von Neumann in their operatorial formulation of classical mechanics. In particular we show that the states of this Hilbert space do not spread, differently than what happens in quantum mechanics. The role of the phases associated to these classical "wave functions" is analyzed in details. In this framework we also perform the analog of the two-slit experiment and compare it with the quantum case.
     
  5. Jun 17, 2015 #4

    atyy

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    The expansion causes particle production. I think the rate is low, bu it is similar to why the black hole produces Hawking radiation, in the sense that neither the expanding universe not the interior of the black hole is stationary.

    http://arxiv.org/abs/1205.5616
    Particle creation and particle number in an expanding universe
    Leonard Parker
     
  6. Jun 17, 2015 #5
    Intuitively, this formula is basically the particle velocity times the probability density. Recall that the momentum operator is ##\frac{\hbar}{i} \frac{\partial}{\partial x}##. We divide by ##m## to get the velocity.

    The difference is just needed to make the overall expression real-valued.
     
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