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Computing heat capacities in Bohmian Mechanics
Quote:Schmelzer: In quantum equilibrium, the probability distribution of the particles is completely defined by the wave function.  In this sense, these additional degrees of freedom are irrelevant for anything what matters in physics.  Thermodynamics is only about rules of approximation if one is no longer able to distinguish microscopic variables but has to handle only averages.   If an equivalence is proven for all physical variables, one does not have to proof a separate theorem for averages.  

We do not need any additional assumptions here.  Once we have quantum equilibrium, the wave function defines the complete state completely.  So all the functions which matter physically can be computed from the wave function.  

And forget about counting the degrees of freedom.  This is meaningful and correct if we have the usual classical formulas for energy of type \(H = \frac{1}{2m}p^2 + V(q)\) for these classical degrees of freedom.  The formula for energy in dBB theory in quantum equilibrium is, of course, the same as that of quantum theory.  Which is what matters.

Then it would be interesting to generate situations in which the system might be out of quantum equilibrium. In atomic physics, experiments exist with fs LASER pulses. It should be possible, somehow, to create situations in which the dbb theory offers different predictions, because it has that additional possibility of treating systems that are  out of quantum equilibrium. How much time does it take for a system to equilibrate? Perhaps, with a clever experimental setup, systems could be designed which equilibrate particularly slowly. I would suggest to do research in this direction since it offers the opportunity to distinguish between these different approaches to quantum mechanics.

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RE: Computing heat capacities in Bohmian Mechanics - by Quiet reader - 05-28-2016, 12:25 PM

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