Is Liouville's theorem usually applied to a "macroscopic" single-particle or to a many-particle phase space?

Question

Consider an isolated system of $N$ identical nonrelativistic classical particles that move within a $d$-dimensional spatial manifold according to a given Hamiltonian function.

Liouville's theorem states that "the phase-space distribution function is constant along the trajectories of the system." But I'm not entirely clear on how "the phase-space distribution function" should be interpreted, and whether "the phase space" under consideration is the full "microscopic" $2Nd$-dimensional phase space or the "macroscopic" $2d$-dimensional single-particle phase space.

There seem to be two conceptually quite different formulations of the theorem:

In version 1, the phase space under consideration is the $2d$-dimensional single-particle phase space. In this case, the phase space distribution function $\rho({\bf q}, {\bf p})$ has a very simple interpretation. It's simply a number density: $\rho({\bf q}, {\bf p})\ d^d{\bf q}\ d^d{\bf p}$ equals the number of particles in the infinitesimal real-space volume $d^d {\bf q}$ at position ${\bf q}$ whose canonical momenta lie within the infinitesimal volume $d^d {\bf p}$ in momentum space at momentum ${\bf p}$. So $\int d^d{\bf q}\ d^d{\bf p}\ \rho({\bf q},{\bf p}) = N$. We can also trivially divide $\rho({\bf q}, {\bf p})$ by $N$ to get a normalized probability density function that gives the probability (density) that a particle selected uniformly at random has position ${\bf q}$ and momentum ${\bf p}$.

In version 2, the phase space under consideration is the full $2Nd$-dimensional phase space. In this case, the interpretation of the phase space distribution function $\rho({\bf q}, {\bf p})$ is more abstract: it corresponds to an abstract statistical uncertainty over microstates that does not correspond to any kind of number density or count of microstates (since the individual points in the full $2Nd$-dimensional phase space themselves are the microstates).

To pick a few random examples: this article seems to be taking the first perspective (since it talks about the "density of system points" - although the article is inconsistent in its use of the variable $n$, so it isn't entirely clear). This article, this one, and this one explicitly say that the distribution function is a number density (and one of them clearly implies that it is defined on the single-particle phase space). This article identifies the distribution function with a count of microstates, so it seems to be taking the first perspective as well.

On the other hand, this article seems to be taking the second perspective, because it says that the phase space is "$6N$-dimensional" (although it doesn't define $N$).

Each formulation seems to have its pros and cons. The first formulation is concrete and conceptually straightforward to me, since we're just talking about the density of particles evolving over time. But if "the Hamiltonian" is just defined on the single-particle phase space, then it isn't clear to me how it can incorporate interactions between the particles. So it seems to only apply to noninteracting ideal gases, which is a pretty serious limitation.

The second formulation can handle arbitrary interacting many-body Hamiltonians, but I have a hard time conceptualizing exactly what it means. It deals with the time-evolution of an abstract "phase-space distribution function" that (I think) cannot represent any kind of number density over microstates (since we're already operating at the "microstate" level). So what exactly is the theorem saying?

My problem is that some explanations seem to conflate these two pictures, by first describing the phase-space distribution function as representing some kind of number density function over macrostates, but then when actually doing the math, suddenly switching to a $2Nd$-dimensional "microscopic" phase space that seems to already be working at the microstate level. At this microstate level, shouldn't the state of the system just be single deterministic point in the "microscopic" phase space?

At some level, I guess the answer is just "The same mathematical derivation works in both cases, so either interpretation is valid." But this answer isn't fully satisfying to me. Is either version the "better" or "more useful" interpretation of Liouville's theorem? The first version seems extremely limited, since it can't handle interparticle interactions, but the second version seems conceptually confusing and less useful in practice.

Answer 1

Is Liouville's theorem usually applied to "macroscopic" single-particle or to a many-particle phase space?

Classical statistical mechanics is about many-particle systems, with $N$ (number of particles) that is a very high integer number. The Liouville theorem is usually meant for probability density $\rho$ in phase space for such high $N$, but it applies to phase space for any value of $N$, even $N=1$.

In case the system is just one particle described by a Hamiltonian (which can describe external force acting on the particle), the Liouville theorem is still valid even if statistical physics is inapplicable; then the theorem is about probability density in 6-dimensional phase space.

The second formulation can handle arbitrary interacting many-body Hamiltonians, but I have a hard time conceptualizing exactly what it means. It deals with the time-evolution of an abstract "phase-space distribution function" that (I think) cannot represent any kind of number density over microstates (since we're already operating at the "microstate" level).

$\rho$ in the Liouville theorem is probability density in phase space, so integral over the whole phase space is one.

So what exactly is the theorem saying?

The Liouville theorem says that Hamiltonian evolution of systems implies that for any valid function $\rho(q,p,t)$ (obeying the Liouville equation, inferred from Hamilton's equations), value $\rho(q^*(t),p^*(t),t)$ for any phase trajectory $q^*(t),p^*(t)$, is constant in time. One can interpret this as follows: imagine that for a microstate $q_0,p_0$, there are many identical systems (ensemble), each with different microstate that is somewhere in a hyperball centered around this microstate, with uniform density in the phase space. The theorem states that as the systems evolve in time, their representative points move to different points of the phase space, and the hyperball may deform wildly, but the points maintain this density. Thus they do not clump somewhere and get removed from somewhere else; the ball deforms and mixes into a larger region of the phase space, so it may look as if the points expand into larger hypervolume, but when enough points are used, the cloud is so dense so that filaments separated by regions with zero representative would be "seen". If we then calculate volume of the occupied filaments, we would get the same volume as volume of the initial ball.

A different interpretation could be like this. If we represent density value $\rho$ in 2D phase space by brightness, zero represented by perfect black, and positive value represented by a proportionally bright white, then the theorem implies that a ball that has initially brightness 1, in the course of time, deforms into a twisted whirl of filaments in black space, with thin filaments separated by black filaments, but brightness of the filament, howsoever thin, is the same as brightness of the ball at the start; the filaments becomes thinner and more mixed with black filaments, but they do not lose their brightness.

At this microstate level, shouldn't the state of the system just be single deterministic point in the "microscopic" phase space?

Microstate of one $N$-particle system is represented by one point in $6N$-dimensional phase space. Probability density $\rho$ on this phase space, induced by some assumptions (constraints) about the macroscopic system (such as internal energy, volume, number of particles) gives, for any simple continuous region of phase space, probability that a system obeying those assumptions is in a microstate that is somewhere in that region.