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Bohmian mechanics description of a large number of interacting atoms would require a large phase space due to the large number of classical degrees of freedom. The entropy per atom is given as the logarithm of the volume of the phase space of states that are accessible at thermal equilibrium. An atom's heat capacity is close to $k_B$, but Bohm's theory seems to be in conflict with this. If there is a way to compute an atom's heat capacity in Bohm's theory in a natural way that doesn't include some ad-hoc solution which will conflict with other kinds of physics experiments, I'm curious to see the calculation.

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    $\begingroup$ Entropy of an atom or entropy per atom? $\endgroup$ – John Rennie May 17 '16 at 5:52
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    $\begingroup$ What's the difference, @JohnRennie ? Well, there could be a difference. The naive application of Bohmian theory doesn't even agree with the fact that the entropy of M+N atoms is additive - because the phase space isn't the Cartesian product of the two spaces (one needs the space of the pilot waves which is a tensor product, not the Cartesian product needed in a classical theory). $\endgroup$ – Luboš Motl May 17 '16 at 7:52
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    $\begingroup$ Dear John, well, if you're picky in some way, you can't. You may still define the probability distribution on phase space (or, in QM, density matrix) for a single atom at temperature $T$, right? If you do so, in the canonical ensemble, every atom will individually contribute the same (in correct theories) - the amount "per atom". In a microcanonical ensemble, the low number (one) of atoms will hurt but in the canonical ones, all the extensive quantities are uniformly distributed among atoms. $\endgroup$ – Luboš Motl May 17 '16 at 8:20
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    $\begingroup$ Otherwise I won't answer this question officially because I think that there are way too many users on the server who are completely irrational. But the answer obviously is that the entropy of 1 atom or many atoms in any Bohmian theory is clearly infinite as the system has infinitely many classical degrees of freedom. Even with any truncation of them or discretization, the heat capacity will parameterically exceed $O(R)$. It's also the case that there are no papers that make such calculations - the result is inconvenient so the Bohmians, who are dishonest activists, find it better to be silent $\endgroup$ – Luboš Motl May 17 '16 at 8:22
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    $\begingroup$ QM is the unique theory that may predict low heat capacities because at low enough temperatures, the number of states (energy eigenstates) whose $E-E_0$ is low enough relatively to $kT$ is small - often one (the ground state is unique). That's why the system carries (almost) no information/entropy, and why the heat capacity is so low. This is only possible because "infinitesimally changed" ground state wave functions don't count as new states - only orthogonal, and therefore very different (and much higher-energy) states are independent, and therefore different. $\endgroup$ – Luboš Motl May 17 '16 at 8:25
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One should distinguish here de Broglie-Bohm theory for the general situation outside the equilibrium, and that for quantum equilibrium. Entropy is defined as usual by $H=-\int \rho \ln \rho dq$. Outside the quantum equilibrium it is useful to split it into the entropy relative to the quantum equilibrium $H=-\int \rho \ln (\rho/|\psi|^2) dq$. This relative entropy has been used by Valentini to prove a "subquantum H-theorem" that a general initial distribution will tend toward quantum equilibrium, see for example, http://arxiv.org/abs/1103.1589 for details.

In quantum equilibrium, we have $\rho=|\psi|^2$, so that the formula becomes $H=-\int |\psi|^2 \ln (|\psi|^2) dq$, thus, the standard quantum-mechanical one. After this, you can apply standard quantum theory.

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    $\begingroup$ This is not the quantum mechanical formula for the entropy at temperature $T$. The result is clearly wrong. Just note that it's nonzero even for $T\to 0$. The quantum mechanical for the entropy is the von Neumann entropy $S=-{\rm Tr}(\rho \log \rho)$ and it has the interesting property that $S\to 0$ for $T\to 0$ - something that was known as the third law of thermodynamics even decades before quantum mechanics was born. That's the thing that no classical theory may ever reproduce. Instead of $S\to 0$, you get $S\to \infty$ at all times, a maximally wrong result. $\endgroup$ – Luboš Motl May 17 '16 at 10:19
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    $\begingroup$ Are you saying that you can go through the ordinary thermodynamic argument from this point and derive the correct heat capacities despite not doing it by reducing to the quantum mechanical calculation? If that's the case then you should probably produce the argument here (or link to something) because "after this you can apply standard quantum theory" is wrong as stated. You're not at standard quantum theory yet; as LM noted your entropy is completely different. $\endgroup$ – Robert Mastragostino May 17 '16 at 18:36
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    $\begingroup$ @Schmelzer - As lubos points out, $\rho$ and $\psi$ are dimensionful, so your logarithms are ill defined. Flagging your answer for very low quality is something I considered, but it is really reserved for formatting errors, so I gave you a pass. Sorry, but your answer is utter nonsense, and 10th grade students are familiar with enough math to know that your logarithms are ill defined. $\endgroup$ – user7348 May 25 '16 at 13:12
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    $\begingroup$ @Schmelzer No, they are not equivalent. They have different mathematical formulations. They agree for a large, but relatively small subset of quantum mechanical phenomena, e.g. double slit for example. BM doesn't work as a QFT, and as Motl pointed out it doesn't work for heat capacities. $\endgroup$ – user7348 May 25 '16 at 19:21
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    $\begingroup$ @Schmelzer Actually, it is the place to show that Motl's argument is nonsense. That's precisely what my question asks. I think what you meant to say was that you can't show how to compute heat capacities in BM, hence the excuse. $\endgroup$ – user7348 May 25 '16 at 19:23

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