Computing heat capacities in Bohmian Mechanics secur Member Posts: 165 Threads: 0 Joined: May 2016 Reputation: 0 05-27-2016, 11:42 PM That's perfectly clear, thanks! Quiet reader Junior Member Posts: 3 Threads: 0 Joined: May 2016 Reputation: 0 05-28-2016, 12:25 PM 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. secur Member Posts: 165 Threads: 0 Joined: May 2016 Reputation: 0 05-28-2016, 01:56 PM (This post was last modified: 05-28-2016, 02:00 PM by secur.) Hello Quiet Reader, Yours is a good suggestion but at this time such an experiment seems out of reach. In 1993 David Bohm speculated that such could be done by 2093 - at best - and that still seems reasonable. So far both theoretically and experimentally dBB is indistinguishable from standard QM, AFAIK. If you google the topic you will find a lot of discussion. Valentini has done the most work in this area, here's a paper from 2011 http://rspa.royalsocietypublishing.org/c....2011.0598. As you know they're doing some amazingly precise investigations into behavior of both fermions and bosons, using lasers as a key tool. Are you familiar with Nonlinear Quantum Optics? They're learning about photon behavior at a detailed level undreamed of last century. And many other areas of materials science and condensed solids are making great advances. Considering this, I have one possible suggestion along the same lines as yours. In addition to attempting to design experiments specifically to test for disequilibrium, perhaps it would be useful to review the many very small-time-scale experiments that are already being done, looking for any unexpected phenomena which might have something to do with disequilibrium. Maybe someone has already done such an experiment, is struggling to explain the results with standard QM, and doesn't know about dBB! If nothing else one would find out about many fascinating new results. Schmelzer, I'm sure, knows more details. But I'm rather confident that my main point is correct: experimentally observing disequilibrium is probably still beyond our technology. Schmelzer Administrator Posts: 215 Threads: 31 Joined: Dec 2015 Reputation: 0 05-28-2016, 04:08 PM The problem is, does it go ever outside the equilibrium? Once whatever we actually do is nicely described by some QT state, it follows that it is nicely described by a dBB state in quantum equilibrium, which is there already from the start and never leaves it. For me, the question is not about the real technical possibilities - there is not even a theory which gives us a hint what we would need to reach some non-equilibrium. Quiet reader Junior Member Posts: 3 Threads: 0 Joined: May 2016 Reputation: 0 05-29-2016, 03:24 AM If there would exist no way to knock a quantum system out of equilibrium, then it would make little sense to even talk about non-equilibrium quantum systems. In that case, the overhead of Bohm's formalism would in fact be obsolete. If there is anything behind all that, then non-equilibrium quantum systems have to exist. @secur: Thanks for that link, I shall look into Valentini's work! Experimentalists are smart guys, they will find a way if the laws of physics allow them to proceed. If we have to wait until 2093, then so be it :-) Schmelzer Administrator Posts: 215 Threads: 31 Joined: Dec 2015 Reputation: 0 05-29-2016, 06:17 AM (This post was last modified: 05-29-2016, 07:26 AM by Schmelzer.) The situation is the reverse one:  The hypothesis that there may be non-equilibrium states simplifies dBB theory, and gets rid of an additional assumption - that the state has to be in quantum equilibrium.  That it is, can be derived.  A. Valentini, H. Westman, Dynamical Origin of Quantum Probabilities, Proc.Roy.Soc.Lond. A 461, 253-272 (2005), arxiv:quant-ph/0403034 is the relevant paper about the relaxation to equilibrium. user7348 Junior Member Posts: 12 Threads: 1 Joined: May 2016 Reputation: 0 05-29-2016, 05:39 PM (This post was last modified: 05-29-2016, 07:06 PM by Schmelzer.) I don't know if I'm convinced on the heat capacities, but there is one result I'm certain Lubos is full of it, casting doubt that anything the man says is reasonable. He claims that in the Hardy experiment Bohmian Mechanics can't reproduce P = 1/16. Could somebody post the solution to that on physics.stackexchange? I know it's a simple calculation. The hilarious thing is that Lubos recognizes that it's the local realist theories that can't produce that result, but dBB is nonlocal and Lubos also recognizes that as well. But, he thinks dBB is ruled out because only a nonlocal theory could reproduce the result, but he knows dBB is nonlocal. [too personal] Schmelzer Administrator Posts: 215 Threads: 31 Joined: Dec 2015 Reputation: 0 05-29-2016, 07:05 PM (This post was last modified: 05-29-2016, 07:09 PM by Schmelzer.) (05-29-2016, 05:39 PM)user7348 Wrote: ... but there is one result I'm certain Lubos is full of it, casting doubt that anything the man says is reasonable. He claims that in the Hardy experiment Bohmian Mechanics can't reproduce P = 1/16.Why should I care about claims of people who do not believe in well-known, published theorems? He has a blog which is sufficiently popular, so I think there is a necessity to answer claims he makes about dBB theory. But only as far as an answer is necessary. Once a reference to the equivalence theorem, as given by Bohm 1952, is sufficient to answer the claim, I do not see a reason to do more. If he accepts this or not is his problem, not my. In the past, he has raised more reasonable and more interesting arguments - nothing new, but answering the well-known classical objections is also something worth to be done. But if he does not believe published theorems, so be it, nothing to care about. If he succeeds to publish something about this in some good journal, this would be another thing. Then this would be a chance to get a publication in a good journal by refuting him. So I have, with a cheap refutation of a refutation of Bell's theorem, got a publication in Annalen der Physik, a quite famous name even if the most important papers have been in the past. BTW, the point he has made which I remember was not not completely off - it was that even if the proof of impossibility could not be applied because dBB is nonlocal, it does not follow that dBB predicts the correct result. This is ok. That it gives the correct result follows not from such a counterargument, but from the equivalence result from Bohm 1952. Quiet reader Junior Member Posts: 3 Threads: 0 Joined: May 2016 Reputation: 0 05-30-2016, 01:11 AM (05-28-2016, 04:08 PM)Schmelzer Wrote: The problem is, does it go ever outside the equilibrium?   Once whatever we actually do is nicely described by some QT state, it follows that it is nicely described by a dBB state in quantum equilibrium, which is there already from the start and never leaves it. For me, the question is not about the real technical possibilities - there is not even a theory which gives us a hint what we would need to reach some non-equilibrium. (05-29-2016, 06:17 AM)Schmelzer Wrote: The situation is the reverse one:  The hypothesis that there may be non-equilibrium states simplifies dBB theory, and gets rid of an additional assumption - that the state has to be in quantum equilibrium.  That it is, can be derived.  A. Valentini, H. Westman, Dynamical Origin of Quantum Probabilities, Proc.Roy.Soc.Lond. A 461, 253-272 (2005), arxiv:quant-ph/0403034 is the relevant paper about the relaxation to equilibrium. I would agree here - it is nice for a theory to have this additional option of non-equilibrium. Eventually, the experiment has to show whether that part is just an overhead or the find of the century. Of course, I have no clue. My first guess would be that a quantum system runs out of equilibrium whenever a sudden and global change of the wave function takes place. A kind of switch in the experimental setup which changes a global symmetry. Then, one has to custom design a system which equilibrates as slowly as possible. Perhaps some sort of topological restriction which prohibits a very rapid equilibration. Finding such a hint for non-equilibrium would be dramatic, it would simply open a completely new path toward the understanding of quantum systems, and it might have dramatic implications for quantum field theory, too. user7348 Junior Member Posts: 12 Threads: 1 Joined: May 2016 Reputation: 0 05-30-2016, 06:13 PM But, you know the mathematical structure is certainly different in the two theories. In quantum mechanics, let H be a complex Hilbert space of countable infinite dimension. The state of a quantum mechanical system is a unit vector in of H up to scalar multiples. The observables are given by self-adjoint operators A on H. The expectation value of an observable A for a system in a state phi is given by the inner product (phi, Aphi). This is how quantum mechanics works, but BM is "not" this formulation. One can claim it's observationally equivalent, fine. But, it's not mathematically equivalent, and I'd really enjoy watching someone quiet Lubos by performing the P = 1/16 calculation from the framework of BM. « Next Oldest | Next Newest »