# Does Liouville’s theorem hold for any ensemble?

I’m having some difficulties with my Statistical Mechanics course.

In particular, I have some doubts about Liouville’s theorem and the various ensembles. Consider, for instance, the Canonical ensemble. Does this theorem hold here? Or it doesn’t because the Hamiltonian is not conserved.

• It seems to me that the statement of Liouville's theorem is true for any distribution (possibly with some smoothness conditions though), thus for any ensemble. – Raskolnikov Jan 24 at 10:01

An ensemble is nothing else that a probability density function (PDF) $$\rho(p,q)$$, where $$q$$ are the canonical coordinates and $$p$$ the momenta. There are several forms of this PDF, depending on the macroscopic conditions to which the system is subject:

• Constant number of particles ($$N$$), volume ($$V$$) and energy ($$E$$): mircocanonical ensemble (*)

$$\rho(p,q)= \begin{cases} 1 & E<\mathcal H (p,q) < E+\Delta E\\ 0 & \text{otherwise} \end{cases}$$

• Constant number of particles ($$N$$), volume ($$V$$) and temperature ($$T$$): canonical ensemble

$$\rho(p,q) = \frac{e^{-\mathcal H(p,q)}}{\int dp dq \ e^{-\mathcal H(p,q)}}$$

... and so on ($$\mathcal H$$ is the Hamiltonian). This PDF (or "ensemble") tells you how likely is it that, given the macroscopic conditions (the macrostate) you find your system in a certain microstate $$(p,q)$$.

Lioville's theorem is valid for any of these PDF, as long as they describe a system governed by an Hamiltonian $$\mathcal H (p,q)$$ which does not depend on any time derivative of $$p$$ and $$q$$.

This is because in such system a state point $$(p,q)$$ moves following Hamilton's equations of motion:

$$\dot p = -\frac{\partial \mathcal H}{\partial q}; \ \ \ \ \dot q = \frac{\partial \mathcal H}{\partial p}$$

These equations are invariant under time reversal and uniquely determine the motion of a state point at all times. As a consequence, the trajectory of the system in phase space is either a closed curve or a curve that never intersects itself (otherwise, where would the system evolve if $$t$$ is reversed?). Furthermore, the trajectories of two different state points can never intersect (for the same reason as above).

As a consequence of this, the probability distribution of these state points, that is, the ensemble $$\rho(p,q)$$ must move like an incompressible fluid. Exactly like in fluid dynamics, we can express this fact mathematically as

$$\frac{\partial \rho} {\partial t} + \nabla(\mathbf v \rho) = \frac{d \rho}{dt} = 0$$

or, using Poisson's brackets:

$$\frac{\partial \rho} {\partial t} + \{\rho,\mathcal H\} = 0$$

Main source: K. Huang, Statistical Mechanics. For an online proof, see for example here.

(*) Some authors, like Landau, prefer $$\rho(p,q)=\delta(\mathcal H - E)$$ for the microcanonical ensemble.