12.6: The Onsager Fluctuation Regression Theorem

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

\[ F_e(t) = F_0e^{\epsilon t}\theta(-t) \nonumber \]

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 \( {u=t-s }\) has been made. Taking the limit \(\epsilon\rightarrow 0 \), and performing the integral over \(u \), we find

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

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

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

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

\[ {\langle B(t)\rangle \over \langle B(0) \rangle }\rightarrow 0 \nonumber \]