# Out-of-equilibrium correlation functions

Let us consider a classical thermal ensemble $$$$\rho_\beta = \frac {1}{Z_\beta} e^{-\beta H}.$$$$ The Hamiltonian generates a mixing dynamics if $$$$C_\beta(t) = \langle A B(t) \rangle_\beta \to \langle A\rangle_\beta\langle B \rangle_\beta \hspace{10mm}t\to+\infty$$$$ If we prepare now a system according to an out-of-equilibrium distribution $$\rho_0$$, a generalization of the above condition would be $$$$C_0(t) = \langle A B(t) \rangle_0 \to \langle A\rangle_0\langle B \rangle_\beta \hspace{10mm}t\to+\infty$$$$ which is convergent to the above mixing condition in the limit $$\rho_0=\rho_\beta$$. Is it possible to show that the second long-time limit or a similar factorization of the averages hold out-of-equilibrium for a class of initial the distributions $$\rho_0$$? We assume that our Hamiltonian $$H$$ generates a mixing time evolution, that is, that the first limit in the equilibrium regime $$\rho_0=\rho_\beta$$ holds.

In general, are there results on this class of out-of-equilibrium long-time limits to which I can relate my question?

I will be dropping the temperature subscript and use $$\rho_0(x)$$ for the initial distribution over phase space $$x$$ and $$\rho_\infty(x)$$ for the equilibrium distribution limit at $$t\to\infty$$. Also, I am assuming that the functional form of $$A$$ and $$B$$ is independent of time. In this case, you can express the time correlation function as:

$$$$C(t) = \iint A(x_0)B(x_t)\rho_0(x_0)\rho_t(x_t|x_0)dx_0dx_t$$$$

Note that the only place you have time dependence is $$\rho_t$$, since $$x_t$$ is simply an integration variable independent of time. Then your assumption is that the conditional distribution decorrelates from the initial one and decays into an equilibrium one:

$$\lim_{t\to\infty}\rho_t(x_t|x_0) = \rho_\infty(x_t)$$

In this case the integral separates into:

$$$$\lim_{t\to\infty}C(t) = \int A(x_0)\rho_0(x_0)dx_0 \int B(x_t)\rho_\infty(x_t)dx_t \equiv \langle A\rangle_{0}\langle B\rangle_{\infty}$$$$

for any initial distribution $$\rho_0(x)$$. In the case of your initial distribution being the equilibrium distribution $$\rho_\infty(x)$$, this simplifies to your initial equation.

• Thanks, I didn't write the answer with any particular reference in mind, but some topics which you might be interested in are the Langevin equation and its path integral formulation and Markov processes. These show you that any time-dependent distribution can be decomposed into an infinite product of "transitions" and their expectation value is an expectation value over all possible paths during this time. In my answer, the conditional distribution $\rho_t(x_t|x_0)$ can actually be expressed as a very complicated infinite-order path integral, but this was not very relevant to the question. Jun 1, 2020 at 7:09