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I am trying to understand Liouville's theorem physically.

It says that $\frac{\partial \rho}{\partial t} + \{\rho,H\} = 0$.

Thus, we have $\frac{d \rho(q(t),p(t),t)}{dt}=0$.

I would like to understand this by looking at harmonic oscillators, cause they are easy: Now, this somehow means that if we started with a bunch of harmonic oscillators that were rotating with a particular distribution of amplitudes, this distribution will somehow be conserved, is this correct? Unfortunately to me, this just sounds pretty similar to energy conservation, so could anybody explain to me where the difference between these two concepts is?

Also, we said that our system is in equilibrium if and only if we have that $\frac{\partial \rho}{\partial t}=0$. This sounds alright, as it means that we are fine with differences in the $q,p$ coordinates, but the density should stay the same. Talking about oscillators, I would understand this as: We are fine if our oscillators rotate, but there should not be a increase or decrease in their total rotation energy. Is this understanding correct?

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    $\begingroup$ If you read the Wiki article carefully, you will find that the generator of the symmetry whose conserved current $\rho$ is is indeed the Hamiltonian, so this is indeed nothing else but conservation of energy. $\endgroup$
    – ACuriousMind ♦
    Commented Nov 3, 2014 at 19:24
  • $\begingroup$ but I guess there must be something wrong with this notion. If your Hamiltonian is time-dependent, then the Liouville equation still holds. $\endgroup$
    – Xin Wang
    Commented Nov 3, 2014 at 19:27
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    $\begingroup$ Related: physics.stackexchange.com/q/64496/50583 $\endgroup$
    – ACuriousMind ♦
    Commented Nov 3, 2014 at 19:33

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