Liouville's theorem

The first important result that we want to prove is Liouville's theorem; this is a very important theorem which will be useful in the following, and ultimately justifies why we microscopically describe systems in phase space and not in other possible spaces[1]. It essentially states that the phase space volume is locally conserved, or in other words time evolution doesn't change it. Let's show it explicitly.

Theorem

Let${\displaystyle {\mathcal {H}}}$ be the Hamiltonian of a physical system of ${\displaystyle N}$ particles, and ${\displaystyle \rho (\mathbb {Q} ,\mathbb {P} ,t)}$ a generic probability density in phase space which can in general also depend explicitly on time. Let ${\displaystyle (\mathbb {Q} _{0},\mathbb {P} _{0})}$ be the initial conditions of the system and call ${\displaystyle (\mathbb {Q} (t),\mathbb {P} (t))}$ their time evolutions, namely the solutions of Hamilton's equations:

${\displaystyle {\dot {\mathbb {Q} }}(t)={\frac {\partial }{\partial \mathbb {P} }}{\mathcal {H}}(\mathbb {Q} (t),\mathbb {P} (t))\quad \qquad {\dot {\mathbb {P} }}(t)=-{\frac {\partial }{\partial \mathbb {Q} }}{\mathcal {H}}(\mathbb {Q} (t),\mathbb {P} (t))}$

Then ${\displaystyle \rho }$ is constant along${\displaystyle (\mathbb {Q} (t),\mathbb {P} (t))}$, namely:

${\displaystyle {\frac {d}{dt}}\rho (\mathbb {Q} (t),\mathbb {P} (t),t)=0}$

Proof

Since ${\displaystyle \rho }$ is a probability density it must satisfy the continuity equation[2]:

${\displaystyle {\frac {\partial }{\partial t}}\rho (\mathbb {Q} (t),\mathbb {P} (t),t)=-{\vec {\nabla }}\cdot {\vec {J}}(\mathbb {Q} (t),\mathbb {P} (t),t)}$
where the probability flow is given by[3]:
${\displaystyle {\vec {J}}(\mathbb {Q} (t),\mathbb {P} (t),t)=\rho (\mathbb {Q} (t),\mathbb {P} (t),t){\begin{pmatrix}{\dot {\mathbb {Q} }}(t)\\{\dot {\mathbb {P} }}(t)\end{pmatrix}}}$
The continuity equation can be rewritten as:
{\displaystyle {\begin{aligned}0={\frac {\partial }{\partial t}}\rho (\mathbb {Q} (t),\mathbb {P} (t),t)+{\vec {\nabla }}\cdot {\vec {J}}(\mathbb {Q} (t),\mathbb {P} (t),t)={\frac {\partial \rho }{\partial t}}+\sum _{i=1}^{3N}\left[{\frac {\partial }{\partial q_{i}}}\left({\dot {q}}_{i}\rho \right)+{\frac {\partial }{\partial p_{i}}}\left({\dot {p}}_{i}\rho \right)\right]=\\={\frac {\partial \rho }{\partial t}}+\sum _{i=1}^{3N}\left[{\frac {\partial }{\partial q_{i}}}\left({\frac {\partial {\mathcal {H}}}{\partial p_{i}}}\rho \right)+{\frac {\partial }{\partial p_{i}}}\left(-{\frac {\partial {\mathcal {H}}}{\partial q_{i}}}\rho \right)\right]\end{aligned}}}

If we now compute all the derivatives, since ${\displaystyle {\mathcal {H}}}$ is a sufficiently regular function we can use Schwartz's theorem to cancel all the terms with the second derivatives in ${\displaystyle {\mathcal {H}}}$. We therefore get:

${\displaystyle 0={\frac {\partial \rho }{\partial t}}+\sum _{i=1}^{3N}\left({\dot {q}}_{i}{\frac {\partial \rho }{\partial q_{i}}}+{\dot {p}}_{i}{\frac {\partial \rho }{\partial p_{i}}}\right)}$
However, by definition of total derivative we have precisely:
${\displaystyle {\frac {d\rho }{dt}}={\frac {\partial \rho }{\partial t}}+\sum _{i=1}^{3N}\left({\frac {\partial \rho }{\partial q_{i}}}{\dot {q}}_{i}+{\frac {\partial \rho }{\partial p_{i}}}{\dot {p}}_{i}\right)}$
So in the end:
${\displaystyle {\frac {d}{dt}}\rho (\mathbb {Q} (t),\mathbb {P} (t),t)=0}$

As a consequence of Liouville's theorem we have that the probability density in phase space of an isolated system in equilibrium is such that:

${\displaystyle {\frac {d}{dt}}\rho _{\text{eq}}(\mathbb {Q} (t),\mathbb {P} (t))=0}$
(in fact since the system is in equilibrium its probability density doesn't depend explicitly on time). Therefore ${\displaystyle \rho _{\text{eq}}}$ is a conserved quantity; we have however seen that the only conserved quantity for an isolated system at equilibrium is its energy[4] so ${\displaystyle \rho _{\text{eq}}}$ must be some function ${\displaystyle F}$ of the Hamiltonian ${\displaystyle {\mathcal {H}}}$:
${\displaystyle \rho _{\text{eq}}(\mathbb {Q} ,\mathbb {P} )=F({\mathcal {H}}(\mathbb {Q} ,\mathbb {P} ))}$

From what we have seen when we studied the microcanonical ensemble we have that the function ${\displaystyle F({\mathcal {H}})}$ is ${\displaystyle F({\mathcal {H}})=\delta ({\mathcal {H}}-E)/\Omega (E,V,N)}$, but this is just a particular case (there is nothing that urges us to suppose so). Therefore, the a priori equal probability postulate is compatible with Liouville's theorem but does not necessarily derive from it.

1. For example, if we consider a system of particles of the same mass we could have equally well described it in the space of positions and velocities rather than with positions and momenta. In this case, in fact, a collision between particles can be seen as an exchange of velocity between the two and since all the particles have the same mass this is also equivalent to an exchange in momentum. However, if we consider particles of different mass this is not true any more and the only quantity exchanged between colliding particles is momentum. Therefore, position-velocity space does not have the same properties of phase space (in particular, Liouville's theorem does not apply).
2. This is a general property of probability distributions: in fact probability can't just "disappear" and "reappear" in different parts of the phase space.
3. This can be better understood through an analogy with fluids: if ${\displaystyle \rho ({\vec {r}},t)}$ is the fluid density, then it must satisfy the continuity equation ${\displaystyle {\dot {\rho }}=-{\vec {\nabla }}\cdot {\vec {J}}}$ where ${\displaystyle {\vec {J}}=\rho {\dot {\vec {r}}}}$.
4. We are considering the system fixed and still, so that its momentum and angular momentum are zero.