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According to this Wikipedia article, the $H$-theorem was Boltzmann's attempt to demonstrate the irreversible increase in entropy in a closed system starting from reversible microscopic mechanics. However, a list of objections have apparently been made to his original approach (see the section "Criticism of the $H$-theorem"), so there's probably still a controverse today. The article also describes Gibbs' try to achieve the same goal (see the section "Gibbs' $H$-theorem"). At this point, the discussion turns to a modern version of the theorem, including a quite odd "blurring" argument, which I'm not sure to understand correctly (and which doesn't seem to be totally conclusive, at least, from what one can read on this article). I'm citing (and I assume this depiction of the argument is not very rigorous):

The critical point of the theorem is thus: If the fine structure in the stirred-up ensemble is very slightly blurred, for any reason, then the Gibbs entropy increases, and the ensemble becomes an equilibrium ensemble. As to why this blurring should occur in reality, there are a variety of suggested mechanisms. For example, one suggested mechanism is that the phase space is coarse-grained for some reason (analogous to the pixelization in the simulation of phase space shown in the figure). [...] Edwin Thompson Jaynes argued that the blurring is subjective in nature, simply corresponding to a loss of knowledge about the state of the system. In any case, however it occurs, the Gibbs entropy increase is irreversible provided the blurring cannot be reversed.

My main concern is: does this blurring/loss of knowledge come from any well-known physical law/principle? For instance, should we link the quantization of the phase space to $\Delta x \Delta p \geq \frac{\hbar}{2}$, or to some kind of observer effect? I don't think this is the reason, as this principle should only hold in the context of quantum mechanics, but I don't know of any similar (semi-)classical mechanism to be considered.

More generally, is quantum mechanics needed (as suggested by this answer to a related question) to explain completely how the irreversible dynamics observed in nature at the macroscopic scale emerge from reversible laws?

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My main concern is: does this blurring/loss of knowledge come from any well-known physical law/principle? For instance, should we link the quantization of the phase space to ΔxΔp≥ℏ2ΔxΔp≥ℏ2, or to some kind of observer effect?

The description in the Wikipedia article is misinformed and misguiding. The blurring, or coarse-graining, is merely one possible formal mechanism to get $H$ decrease in time in the (Boltzmannian) $H$-theorem.

This procedure introduces artifical scale below which details are discarded and thus it has little to do with mechanics where nothing gets discarded or the 2nd law of thermodynamics, where we are not concerned with such details at all (we only use macroscopic state variables).

Also $H$ (a function of probabilities and therefore time) is not related to thermodynamic entropy (function of macroscopic state variables) in any simple way.

In fact there is no blurring of the probability function needed to explain the 2nd law. Jaynes has shown how to derive the entropy formulation of the 2nd law for thermally isolated system without any blurring, only using Hamiltonian mechanics, principle of maximum information entropy and the assumption that results of macroscopic experiments are repeatable.

Let the system evolve from thermodynamically equilibrium state $A$ to thermodynamically equilibrium state $B$. We will describe the whole process also by means of probability density $\rho(t)$, starting at time $t_A$ with initial function $\rho(t_A)$ and ending at time $t_B$. The initial density $\rho(t_A)$ can be anything as long as it respects constraints of the macrostate $A$. The subsequent densities $\rho(t)$ for $t \geq t_A$, however, are entirely determined by the Hamiltonian and the initial condition; we are not free to choose them.

Due to the Liouville theorem, the information entropy (often called Gibbs' entropy)

$$ I[\rho] = \int -\rho\ln\rho \,dqdp $$

remains constant in time, there is no blurring of $\rho$ of any kind . (1)

It will be shown that thermodynamic entropy in the final equilibrium state $S_B$ is greater or equal to the initial thermodynamic entropy $S_A$.

This result is possible because thermodynamic entropy of a macrostate, in general, is not simply proportional to information entropy of the time-dependent probability distribution $\rho(t)$. Its relation to the concept of information entropy is this:

Value of thermodynamic entropy of macrostate $X$ is given by the value of information entropy for that probability distribution $\rho_X$, which is both consistent with macrostate $X$ and gives maximum possible value to the information entropy. (2)

Now, obviously $\rho(t_B)$ is consistent with macrostate $B$ but $I[\rho(t_B)]$ is not necessarily the maximum possible value of $I$ for all $\rho$'s compatible with the macrostate $B$.

The probability density that is not only consistent with macrostate $B$ but also maximizes information entropy is, in general, different from $\rho(t_B)$. Let us denote this maximizing density as $\rho_B$; then the relation of the two information entropies is

$$ I[\rho_B] \geq I[\rho(t_B)]. $$

Now, based on (2) we can write this as $$ S_B \geq I[\rho(t_B)], $$ that is, thermodynamic entropy in the final equilibrium state is higher or equal to information entropy of the evolved probability distribution.

Based on (1) we can write this also as $$ S_B \geq I[\rho(t_A)]. $$

This means that whatever the density $\rho(t_A)$ is chosen to describe the macrostate $A$, thermodynamic entropy of the final state $B$ is equal or higher than information entropy of $\rho(t_A)$. And so if the density $\rho_A$ is chosen such that it maximizes $I$ under constraints of the initial macrostate $A$, $I$ attains value of thermodynamic entropy $S_A$ and we obtain the inequality

$$ S_B \geq S_A. $$

This means we derived entropy formulation of 2nd law of thermodynamics from the principle of maximum information entropy, using constancy of Gibbs' entropy as one of the ingredients.

More generally, is quantum mechanics needed (as suggested by this answer to a related question) to explain completely how the irreversible dynamics observed in nature at the macroscopic scale emerge from reversible laws?

If we wanted to explain the process completely down to elementary particles, a theory of these particles such as quantum theory would be needed. But if the goal is merely to show that irreversible evolution of macroscopic variables is consistent with reversible evolution of microscopic variables, then the answer is no.

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    $\begingroup$ As I understand it, the principle of maximum information entropy is a useful starting point for making predictions about a system that you know very little about (that can later be compared with experiments to iteratively refine the model). Relying too much on maximal uncertainty perspectives makes it difficult to appreciate the importance of dynamics in statistical mechanics, and the failure of some systems with large numbers of conserved quantities (e.g. solitons) to equilibrate to the distribution associated with the canonical ensemble. I would be interested in hearing your thoughts. $\endgroup$
    – TLDR
    May 18, 2016 at 0:43
  • $\begingroup$ Of course there is no guarantee maxent principle is always well working. But in the view I explained above, the probability distribution does not evolve (equilibrate) to the canonical one, it is governed by the equations of motion. Canonical distribution is just the best to use when all we know are the macroscopic constraints. $\endgroup$ May 18, 2016 at 7:19
  • $\begingroup$ An interesting answer which solved some of my misunderstandings, thank you! Just to make myself clear, is Boltzmann's $H$ then the same thing as the information entropy? $\endgroup$
    – dahemar
    May 18, 2016 at 8:50
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    $\begingroup$ @DavidHerreroMartí, you're welcome. Boltzmann $H$ function is originally defined as function of probability density defined on 6-dimensional space of states of single molecule (the independent dimensions are due to 3 coordinates and 3 components of velocity of the molecule). Information entropy in general is much more general than that, it refers to any probability distribution on any space. In the above explanation, information entropy is so-called "Gibbs entropy" and refers to probability distribution $\rho$ defined on the phase space of the whole system. So they are very different things. $\endgroup$ May 18, 2016 at 18:29
  • $\begingroup$ I agree that Jaynes' derivation of the second law is quite elegant and does not require any special assumptions about the microscopic dynamics. However it seems to lead to Loschmidt's paradox on similarly like many other derivations: Jan, although you stated explicitly that t_B is after t_A, we can change that and say t_B is earlier than t_A and everything in your argumentation still holds. Dynamically, we start with a final condition which develops uniquely from t_A to t_B. With the same arguments as above we derive that thermodynamic entropy in the past is equal or higher than in the present $\endgroup$
    – Martin_S
    Dec 18, 2021 at 22:17
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The physical principle being invoked is the finite resolution of any experiment, independent of the value of $\hbar$, together with coupling between observable and microscopic degrees of freedom, i.e. it applies to both classical and quantum systems. Technically energy conservation and Liouville's theorem, or unitarity in QM, are also needed to prevent distinct trajectories from collapsing to some sort of attractor. Classically, the result follows from sensitivity to initial conditions, and the fact that most systems have a relatively small number of globally conserved quantities (for an important exception, see the Fermi Pasta Ulam problem). The cutoff of $\Delta x\Delta p\geq \hbar/2$ is only natural because of our additional empirical knowledge that the resolution of any classical description breaks down at this scale. If we used an ordinary ruler to measure position, a pendulum to measure time, and the equilibrium displacement of a fixed spring to measure forces, then our resolution and predictive power would be limited by the accuracy of the ruler and pendulum. In quantum mechanics, the finite resolution of any experiment implies that any initial measurement ignores degrees of freedom associated with small scales. These degrees of freedom usually do not have a significant effect on trajectories over short time scales, but cumulatively lead to an apparent loss of unitarity over long times. In nuclear magnetic resonance, you can predict to reasonable accuracy the state of a single spin shortly after a pulse is applied, but over time, many-body interactions with ambient particles lead to a gradual propagation of coherence to increasingly larger length scales, and eventually to complete dissipation.

In both classical and quantum dynamics, the increase in entropy can be thought of as coming from incomplete initial data: as the missing information becomes experimentally accessible, it leads to an error term, or non-zero Shannon/von Neumann entropy associated with the observable part of the system.

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  • $\begingroup$ Although, there's something that still bothers me a bit: the difference between thermodynamic and information entropy. How can a subjective, observer-dependent effect give rise to something as "objective" as thermodynamic entropy? Am I missing something? $\endgroup$
    – dahemar
    May 17, 2016 at 7:26
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    $\begingroup$ Thermodynamic entropy is observer dependent to some extent. The Gibbs paradox is usually explained with respect to identical or indistinguishable particles, but it also applies to distinguishable particles that are left undistinguished (a very appropriate term that I first saw used by James Sethna in his stat mech textbook) either because the experiment cannot distinguish the particles, or because even if visible differences between particles existed, it would be near impossible to use the differences in osmotic pressure to extract work. $\endgroup$
    – TLDR
    May 17, 2016 at 13:04

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