Fluctuations in Fluctuation Dissipation theorem

In the derivation of the Fluctuation-Dissipation theorem. We encounter an identity

$$\langle\delta A(t) \delta B(0) \rangle = \langle A(t)B(0)\rangle-\langle A \rangle\langle B\rangle$$ where $$\delta A(t) = A(t) - \langle A\rangle$$ and $$\delta B(t) = B(t) - \langle B\rangle$$

$\langle\cdot\rangle$ denotes ensemble average.

How is this derived?

• I found the notation hard to use and wrote $A_{t}-\bar{A}$ for $A(t)-\langle A \rangle$ and so on. Then I expanded the left hand side into four terms each of which needs an ensemble average doing on it. But $\bar{A}$ and $\bar{B}$ are simply constants. I assumed that $\bar{A_{t}}=\bar{A}$ and $\bar{B_{0}}=\bar{B}$. I found this did the trick. Commented Mar 20, 2018 at 11:25
• @PhilipWood Yes we have to assume the relations you mentioned in last sentence, to get the result. That is the stationarity criterion. Thanks ! Commented Mar 20, 2018 at 11:29
• Use either \langle \rangle or \left< \right> for grouping/bra-kets/averages. PLain <> are typeset as operators and have (much!) too much space around them for such uses. The second form will automatically grow with the size of the contained material. Commented Mar 20, 2018 at 20:06

This is just the regular formula for covariance - $$Cov(x,y)=\langle xy \rangle-\langle x \rangle \langle y \rangle$$ Two assumptions were made here; (1) that the system has time translational symmetry, that is, the correlator $\langle A(t_1)B(t_2) \rangle$ depends only on the time difference $t_1-t_2$. Thus we can arbitrarily set one of them to time $t=0$ and hold all the difference on the other one. (2) Expectation value of a single operator is assumed to be independent of time, that is $\langle A(t) \rangle=\langle A(0) \rangle\equiv \langle A \rangle$. This assumption is due to treatment near equilibrium.