Why is $\langle{\dot A}A^* \rangle=0$ for dynamical variable $A$?

In page 179 of Hansen and McDonald's book, Theory of Simple Liquids, 3rd edition, 2006, an identity of correlation functions was deduced.

Here $$C_{AB}(t)$$ is the time correlation function between dynamical variables. My question is how to get $$\langle{\dot A}A^* \rangle=0$$ after $$\langle{\dot A}B^*\rangle=0$$?

if $$B=A$$, $$\langle{\dot A}A^* \rangle=-\langle A{\dot A^*}\rangle=-{\langle{\dot A}A^* \rangle}^*$$ . So if $$\langle{\dot A}A^* \rangle$$ is a complex number, e.g., $$a+bi$$, then $$a+bi=-(a+bi)^*$$, what we get is only $$a=0$$, while $$b$$ can be any real number. This is different from $$\langle{\dot A}A^* \rangle=0$$.

I wonder if the statement that $$A$$ is a dynamical variable can give any restriction on the correlation function $$\langle{\dot A}A^* \rangle$$ that it should be real number.

• you build the correlators so as to be real (because you identify a measure with one real number). The correlator of $\dot {A}A*$ is the derivative of a real function (the correlator of AA*) so it has no imaginary part. Apr 25, 2019 at 11:34
• @france95 I see. Thanks a lot for your help!
• Please use '\langle' and '\rangle' ($\langle$ and $\rangle$) instead of '<' and '>' ($<$ and $>$). Apr 25, 2019 at 11:52
• @france95 I checked it again, and I found that as $(\dot A A^∗)′=(AA^∗)+A(\dot A^∗)$, And since $(\dot A A^∗)=(A(\dot A^∗))^∗$,$(\dot A A^∗)$ can still be a real number even if $(\dot A A^∗)$ is a complex number. Would you please give me more detail explanation? Thanks!
I believe that there is an extra subtletly. Almost always we choose dynamical variables with a well defined signature $$\varepsilon$$ under time reversal. This is mentioned at the start of that chapter (in the 3rd edition). This means that (their eqn (7.1.9)) $$\langle A(t) B^*(0) \rangle = \varepsilon_A \varepsilon_B \langle A(-t) B^*(0) \rangle = \varepsilon_A \varepsilon_B \langle A(0) B^*(t) \rangle .$$ It follows that the autocorrelation functions are both even and real functions of time (irrespective of whether $$\varepsilon_A$$ is $$+1$$ or $$-1$$). In other words, the above equation shows that $$\langle A(t) A^*(0) \rangle = \langle A(0) A^*(t) \rangle \quad\Rightarrow\quad \langle \dot{A} A^* \rangle = \langle A \dot{A}^* \rangle$$ as well as the relation you have seen already from stationarity $$\langle \dot{A} A^* \rangle = -\langle A \dot{A}^* \rangle .$$ Hence $$\langle \dot{A} A^* \rangle=0$$, as you wanted.