The Onsager fluctuation regression theorem (Optional)

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

$$\langle B(t)\rangle = \int_{-\infty}^0ds e^{\epsilon s}\int_0^{\beta} d\lambda \langle \dot{B}(-i\hbar\lambda)B(t-s)\rangle _0$$

$$= -e^{\epsilon t}\int_0^{\beta}d\lambda \int_t^{\infty}du e^{-\epsilon u} {d \over du}\langle B(-i\hbar\lambda)B(u)\rangle _0$$

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

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

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

$${\langle B(t)\rangle \over \langle B(0)\rangle } = { \int_0^... ...} {\langle B(0)B(t)\rangle _0 \over \langle B(0)^2 \rangle _0}$$

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

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