05-27-2016, 07:47 PM

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.

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.