# Calculating Poincare recurrence times

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, inspired by octonion's answer. 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!

Yes I think your interpretation of the situation is correct. For a very small time step there is probability near one that all the points in $A$ remain in $A$. So there is a probability near one that the recurrence time is just one time step.

However for the tiny measure of points that do leave $A$, those points stay outside $A$ for a very large number of time steps, and they increase the expected recurrence time in such a way that the total expected recurrence time is just $1/\mu(A)$.

To see how this works let's take a simple model where your set $A$ has half of the measure of the space, and given you start the system in $A$ there is a probability $\epsilon$ you leave next timestep $\Delta t_{small}$. Let's approximate the ergodic system as a Markov chain (this is the approximation there are no correlations between timesteps).

If we consider a large time interval $\Delta t_{big}=N\Delta t_{small}$, by ergodicity there will be a probability of near $1/2$ (exact in the limit $N\rightarrow\infty$) that you leave $A$ the next step. So for $\Delta t_{big}$ the expected recurrence time is $$\frac{1}{2}+2\left(\frac{1}{2}\right)^2+3\left(\frac{1}{2}\right)^3+\dots=\frac{1}{2}\frac{1}{(1-1/2)^2}=2,$$ which makes sense since the measure $\mu(A)=1/2$.

For $\Delta t_{small}$, there is a probability near one of having recurrence time one, $$(1-\epsilon)+\epsilon(2\epsilon+3(1-\epsilon)\epsilon+4(1-\epsilon)^2\epsilon+\dots)$$ $$=(1-\epsilon)+\epsilon^2(1+(1-\epsilon)+(1-\epsilon)^2+\dots+1+2(1-\epsilon)+3(1-\epsilon)^2+\dots)$$ $$=(1-\epsilon)+\epsilon^2(\frac{1}{\epsilon}+\frac{1}{\epsilon^2})=2.$$

So both big and small time step have the same recurrence time.

This is just a gloss on octonion's answer.

There is a nice proof of Kac's Lemma here and an example of Ehrenfest that I think is instructive. If (a return to) $\mu(A)$ is the state of interest, the largest possible transition time consistent with (distinguishing between) meaningful states seems easiest to use.

The probability of a given state is independent of time, and the time is arrived at by taking the inverse of the region of interest, $n_A=1/\mu(A)$, and assigning a plausible time interval to that number. Then on average it will take $n_A$ time steps to reach that state. If you decide to use a time associated with some larger region with $\mu(B)=(2) \mu(A)$ then the return time is $n_B=1/\mu(B)=1/2\mu(A)=n_A/2.$ Then we need $2n_B$ steps on average to return to region A. The time-steps are evidently smaller, but it doesn't add anything.

If there is a continuous-time example it would be nice to see. I didn't find one.