Calculating Poincare recurrence time of a physical system I am interested in calculating the Poincare recurrence time of a physical system (i.e. a system with with continuous time evolution). I have seen physics papers giving estimations of the recurrence times but never any formula how they estimated it. I know that from Kac's lemma the number of time steps until the system returns to a region $A$ of phase space is $n_A = \frac{1}{\mu(A)}$ where $\mu$ is a measure and the whole measure space has measure 1. (Strictly speaking $n_A$ is the mean recurrence time over the region $A$.)
To construct a continuous time analogue, I divide time into increments of $\tau$, so that $n_A = t_A/\tau$. This leads to a recurrence time $t_A = \tau/\mu(A)$. But this encounters a problem which is that $t_A$ can be made arbitrarily small just by "fine-graining" time enough (by making $\tau$ arbitrarily small). Ideally I'd like to fine-grain time as much as possible but this would lead to $t_A = 0$ for any $A$. My interpretation of this is that if $\tau$ is too small, the system may not have the time to even leave the region $A$ to begin with. But by "recurrence" we normally mean leaving the region $A$ and coming back.
EDIT: Here is my attempt to deal with the above problem.
For a given $\tau$ partition $A$ into the points that leave $L$ and those that remain $R$, so that we can write
$$n_A = \frac{n_R \mu(R) + n_L \mu(L)}{\mu(A)} $$
where $n_R = 1$ is the mean recurrence time of the points that remain, $n_L$ is the mean recurrence time of the points that leave.
Let $\epsilon = \mu(L)/\mu(A)$ and $1 - \epsilon = \mu(R)/\mu(A)$. Using's Kac's formula $n_A = \frac{1}{\mu(A)}$ and rearranging we get
$$n_L = \frac{1}{\epsilon} \frac{1 - \mu(A)}{\mu(A)} + 1 .$$
Transforming this into real time and taking the limit of continuous time we get
$$t_L = \left(\lim_{\tau \to 0} \frac{\tau}{\epsilon(\tau)} \right) \frac{1 - \mu(A)}{\mu(A)}$$
which essentially tells us the recurrence time of points on the surface of the set $A$.
This can be applied to a quantum system in a $N$-dimensional Hilbert space. Let $\{ \psi_n \}_{n = 0}^{N-1}$ be the energy eigenstates of the system with eigenvalues $\{ E_n \}_{n = 0}^{N-1}$.
The state of the system can be expanded in terms of these states $\psi = \sum_{n = 0}^{N-1} \sqrt{p_n}e^{-i\phi_n} \psi_n$. The $p_n$ are constants of motion and $\dot{\phi_n} = E_n/\hbar$. Without loss of generality let $E_0 = 0$ and $\phi_0 = 0$ (consider only phases relative to the ground state phase). Also let $\omega_n = E_n/\hbar$. The state of the system is given by $(\phi_1, ... , \phi_{N-1})$, a point on a $(N-1)$-torus, $\mathcal{T}^{N-1}$. Suppose we consider a $L \times L \times ... \times L$ region on the torus. The measure of this region $A$ is $\mu(A) = (L/2\pi)^{N-1}$.
Let $V(L_1,...,L_{N-1}) =  \frac{L_1L_2...L_{N-1}}{(2\pi)^{N-1}}$ be the volume of an $L_1 \times L_2 \times ... \times L_{N-1}$ region on the torus. To first order in $\tau$, $$\epsilon(\tau) = \frac{1}{\mu(A)} \sum_{n=1}^{N-1} \frac{\partial V}{\partial L_n} \omega_n \tau = \frac{1}{\mu(A)} \sum_{n=1}^{N-1} \frac{V}{L_n} \omega_n \tau.$$ Letting $L_n = L$ for all $n$, then $\epsilon(\tau) = \frac{\tau \Omega}{L}$ to first order in $\tau$ where $\Omega = \sum_{n=1}^{N-1} \omega_n$. From this the quantum recurrence time is
$$t_L = \frac{L}{\Omega } \left( \frac{1 - (L/2\pi)^{N-1}}{(L/2\pi)^{N-1}} \right)$$
$$t_L = \frac{L}{\Omega} \left( \left( \frac{2\pi}{L} \right)^{N-1} - 1 \right)$$
This is a fairly simple result. There's a problem though. If you let $N$ tend to infinity, $t_L$ tends to infinity ($\Omega$ scales roughly with $N^2$). This would imply that infinite-dimensional quantum systems do not recur, which is something that is not true!
 A: You are completely right. Continuous time has usually this problem.
What people usually do is to bypass somewhat, as follows. Assume that $A$ is a ball (you'll see below how this makes things easier). Then the return time needs to be defined as the first entering time, after the orbit leaves the set $A$. You see that if $A$ is only measurable, these times (of leaving and entering) could be undefined (or zero).
An alternative is to consider a special flow/suspension flow and to associate to your continuous time system a natural discrete time system with induced quantities from the suspension. Then you can pass to the discrete time and use Kac's lemma as usual (in the direction that you describe).
Summing up, I suggest that you look at suspension but a complete answer depends on your system (or at least on the type of system that you assume to have or that you expect to have). I cannot be more detailed otherwise.
In case you know French: I warmly suggest http://www.math.univ-brest.fr/perso/benoit.saussol/art/hdr.pdf, no matter if it is a habilitation.
