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. – Philip Wood Mar 20 '18 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 ! – Chetan Waghela Mar 20 '18 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. – dmckee Mar 20 '18 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.