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Gibbs paradox in the Bohmian interpretation

  1. Nov 19, 2013 #1

    atyy

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    The Gibbs paradox in statistical mechanics is usually resolved by saying that particles do not have distinct trajectories, and so are truly identical.

    For example, http://ocw.mit.edu/courses/physics/8-333-statistical-mechanics-i-statistical-mechanics-of-particles-fall-2007/lecture-notes/ [Broken] says "The above treatment of identical particles is somewhat artificial. This is because the concept of identical particles does not easily fit within the framework of classical mechanics. To implement the Hamiltonian equations of motion on a computer, one has to keep track of the coordinates of the N particles. The computer will have no difficulty in distinguishing exchanged particles. The indistinguishability of their phase spaces is in a sense an additional postulate of classical statistical mechanics. This problem is elegantly resolved within the framework of quantum statistical mechanics. Description of identical particles in quantum mechanics requires proper symmetrization of the wave function."

    In the Bohmian interpretation identical particles have distinct trajectories, and so are not truly identical. How is the Gibbs paradox resolved in the Bohmian interpretation?
     
    Last edited by a moderator: May 6, 2017
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  3. Nov 19, 2013 #2

    Demystifier

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    Easily! Even though the particles are not identical, their wave functions are. The entropy to which the Gibbs paradox refers is the entropy of the density matrix, which corresponds to an uncertainty of the WAVE FUNCTION, not an uncertainty of the particle position.

    More specifically, assume that the system is in the thermal equilibrium. This means that the density matrix is thermal. But particles are not in the thermal equilibrium; particles are in the quantum equilibrium.
     
  4. Nov 19, 2013 #3

    atyy

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    Is it permissible to think of quantum equilibrium as a sort of thermal equilibrium? If it is, can we say that there are multiple time scales of equilibration, with the wave function coming to equilibrium more slowly than the particles?
     
  5. Nov 19, 2013 #4

    Jano L.

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    It should be also said that the term "Gibbs paradox in statistical mechanics" is a misnomer. Gibbs did not refer to the result of entropy calculation as to a paradox. The factor ##1/N!## is no paradox in classical statistical mechanics. Jaynes explained this very well in his paper

    Jaynes, E. T., 1992, `The Gibbs Paradox, ' in Maximum-Entropy and Bayesian Methods, G. Erickson, P. Neudorfer, and C. R. Smith (eds.), Kluwer, Dordrecht

    available on the website with Jaynes' works

    http://bayes.wustl.edu/etj/articles/gibbs.paradox.pdf



    From the paper:


    As I understand him, whether we calculate entropy with assumption that permutations count, or that they do not, depends on how much information we admit to the calculation. If we do not have any means of distinguishing the particles (molecules of ##\text{N}_2##) experimentally, we do not have reason to count their permutations. If in future some bright physicist finds out that all nitrogen molecules are little bit different, we will count their permutations into the entropy. This can happen in principle, the present state of knowledge does not preclude it. The entropy will be different than what we use today, but that is alright, because there is no such thing as ultimate value of entropy. It depends on the available information and particular situation we choose to describe.
     
  6. Nov 20, 2013 #5

    Demystifier

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    Good point!
     
  7. Nov 20, 2013 #6

    Demystifier

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    Something like that. Both quantum and thermal equilibrium are examples of the general concept of equilibrium.
     
  8. Nov 20, 2013 #7

    atyy

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    So in attempts to derive statistical mechanics from quantum mechanics in which the entire system is pure, but the reduced density matrix is mixed and thermal, the quantum equilibrium hypothesis effectively enters when using the Born rule to get the reduced density matrix?
     
  9. Nov 21, 2013 #8

    Demystifier

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    Yes.
     
  10. Dec 20, 2013 #9

    HPt

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    In Eur. J. Phys. 35 (2014) 015023 http://dx.doi.org/10.1088/0143-0807/35/1/015023 (also available as arXiv preprint) you will find the resolution of the Gibbs paradox for distinguishable particles (classical or not).
     
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