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In classical statistical mechanics, the Liouville's theorem tells that for a system in equilibrium the Poisson bracket $$\{\rho(q,p),H(q,p)\}=0.\tag{1}$$

1. Does it necessarily imply $\rho(q,p)=\rho(H(q,p))$? In other words, I'm not asking whether $\rho(H(q,p))$ satisfies (1). It of course does. But is it not possible to have other functional forms $\rho(q,p)$ which are nor necessarily of the form $\rho(H(q,p))$ but still satisfying (1)?

2. In the answer to question 1 is in negative, is it possible to mathematically derive or restrict the possible functional forms of $\rho(H)$? If not mathematically, can we rule out physically some functional forms of $\rho(H)$ (if not all)? After all, apparently any function $\rho(H(q,p))$ will satisfy (1).

A coherent, simplistic answer addressing all the questions above will be highly appreciated.

SRS
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    1 - no. Consider $H = \frac{\mathbf{p}^2}{2m} + V(|\mathbf{r}|)$. Note how ${\mathbf{L}^2,H} = 0$, thus any sufficiently smooth function $f(\mathbf{L}^2,H)$ will also have a zero Poisson bracket with $H$. This generalizes in a straightforward fashion to all conserved quantities. – Sean E. Lake Jun 26 '17 at 15:32

1 Answers1

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  1. (1) implies $\rho$ is a constant of motion, and nothing more or less. Since a system of phase space dimension $2N$ can have up to $2N$ functionally independent constants of motion (and in fact locally always has this maximal number of such constants, cf. e.g. this answer), $\rho$ need not be functionally dependent on $H$.

  2. Not only does every function $\rho(H(q,p))$ satisfy (1), every constant of motion satisfies (1), so, at least locally, the most general solution of (1) is a function of $H$ and $2N-1$ functionally indepedent constants of motion $I_1(q,p),\dots, I_{2N-1}(q,p)$. Note that this does not actually tell you anything new because it was a function of the $2N$ functionally independent $q,p$ to begin with! Since you have made no restrictions about what sort of physical system we are considering, it seems impossible to restrict the form of $\rho$ in any way.

ACuriousMind
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