5
$\begingroup$

Recently I came across this idea of Gibbs that, it is the coarse-grained entropy that always increases, whereas the fine-grained remains a constant. So classically, coarse graining refuses us some microscopic information which leads to increase in entropy (due to loss of that information).

In quantum mechanics however the phase space has some inherent quantisation, so can this be addressed as coarse graining of phase-space in that scale ? If yes, then the increase in entropy is inevitable fact of nature, since quantisation of phase space inherent in nature irrespective of the measurement we perform ?

PS : I am sure coarse graining and quantisation have entirely different origins, my question here to is to only address their effects rather than their causes.

$\endgroup$
8
$\begingroup$

The von Neumann entropy, written in terms of the quantum mechanical density operator, is a constant of the motion if you keep track of everything (including entanglement with the environment) and don't have any collapse events (which, depending on your favorite interpretation of quantum mechanics, might not exist anyway).

The thing is that this fact already accounts for quantum mechanics. There's no additional fact of inherent quantization of phase space that needs to be, or even can be, added to this. The density operator is already a fully quantum mechanical operator in Hilbert space, and the structure of the Hilbert space and the behavior of other operators (such as the Hamiltonian) on that same Hilbert space is implicitly there. So energy quantization, angular momentum quantization, and so forth are already implicitly taken into account in this theory. But the von Neumann entropy is still constant with respect to time.

However, the von Neumann entropy does give us a way to understand the increase in entropy of a non-isolated system that can gradually become entangled with a thermal or quasi-thermal environment, in a process called decoherence. If you try to look at the density operator of just the not-quite-isolated system you were trying to study, you have to "trace out" (perform a sum over) all of the states of the environment that it's entangled with that you're not explicitly paying attention to. When you do this, increase of entropy comes out pretty naturally.

This answer is only a very brief summary of a huge topic that's also an active area of research. But the key words that I think will head you in the right direction are:

von Neumann entropy

entanglement

decoherence

$\endgroup$
2
  • $\begingroup$ an excellent, precise, clear answer $\endgroup$ Nov 27 '15 at 18:28
  • 2
    $\begingroup$ I will only add that for Gibbs, and all classical treatments, "coarse-graining" is subjective. It is not something that happens in Nature, it is something we adopt in changing how we will describe Nature: it is our decision to ignore some of the information---e.g., because it would be impractical to collect that information $\endgroup$ Nov 27 '15 at 18:30
1
$\begingroup$

However, there is at least one theory (not complete) where the "coarse graining" is not subjective but rather natural result of thermodynamics re-formulated in terms generalized covariant theory of many weakly interacting subsystems (as opposed to non-interacting particles of the Boltzmann's ideal gas). In that theory proposed by Carlo Rovelli at all the coarse graining corresponds to the natural gauge symmetries. So there is a way to derive the second law without introducing the "lost information", but based on the transfer of information to the environment resulting from the entanglement. It may be equivalent to decoherence but it is based on Gibbs formulation of entropy.

$\endgroup$
2
  • $\begingroup$ That is an interesting insight, can you please elaborate it a little more (few details of Carlo's theory). $\endgroup$
    – user35952
    Jul 6 '18 at 5:38
  • $\begingroup$ @user35952 the details can be found in a series of papers going back to 1999. For example arxiv.org/abs/1309.0777. In a nut shell, Rovelli reformulated laws of thermodynamics in a context of quantum gravity. $\endgroup$ Jul 7 '18 at 9:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.