# The Onsager Fluctuation Regression Theorem


Suppose that $$F_e (t)$$ is of the form

$F_e(t) = F_0e^{\epsilon t}\theta(-t)$

which adiabatically induces a fluctuation in the system for $$t < 0$$ and the lets the system evolve in time according to the unperturbed Hamiltonian for $$t > 0$$. How will the induced fluctuation evolve in time? Combining the Kubo transform relation with the linear response result for $$\langle B(t)\rangle$$, we find that

\begin{align*} \langle B(t)\rangle &= \int_{-\infty}^0ds\, e^{\epsilon s}\int_0^{\beta} d\lambda\langle \dot{B}(-i\hbar\lambda)B(t-s)\rangle _0 \\[4pt] &= -e^{\epsilon t}\int_0^{\beta}d\lambda \int_t^{\infty}du \,e^{-\epsilon u}{d \over du}\langle B(-i\hbar\lambda)B(u)\rangle _0 \end{align*}

where the change of variables $$\underline {u=t-s }$$ has been made. Taking the limit $$\epsilon\rightarrow 0$$, and performing the integral over $$u$$, we find

$\langle B(t)\rangle = -\int_0^{\beta}d\lambda\,\left[\langle B (- i\hbar \lambda) B (\infty )\rangle _0 -\langle B(-i\hbar\lambda)B(t)\rangle _0\right]$

Since we assumed that $$\langle B\rangle _0 = 0$$, we have $$\langle B(-i\hbar\lambda)B(\infty)\rangle _0 =\langle B(-i\hbar\lambda)\rangle _0\langle B(\infty)\rangle _0 = 0$$. Thus, dividing by $$\langle B(0)\rangle$$ , we find

${\langle B(t)\rangle \over \langle B(0)\rangle } = {\int_0^{\beta} d \lambda B (-i\hbar \lambda) B (t) \rangle _0 \over \int_0^{\beta} d \lambda B (-i\hbar \lambda) B (0) \rangle _0} \rightarrow _{\hbar \rightarrow 0 } {\langle B(0)B(t)\rangle _0 \over \langle B(0)^2 \rangle _0}$

Thus at long times in the classical limit, the fluctuations decay to 0, indicting a complete regression or suppression of the induced fluctuation:

${\langle B(t)\rangle \over \langle B(0) \rangle }\rightarrow 0$