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

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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.

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    $\begingroup$ 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. $\endgroup$
    – AoZora
    Apr 25, 2019 at 11:34
  • $\begingroup$ @france95 I see. Thanks a lot for your help! $\endgroup$
    – FaDA
    Apr 25, 2019 at 11:51
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    $\begingroup$ Please use '\langle' and '\rangle' ($\langle$ and $\rangle$) instead of '<' and '>' ($<$ and $>$). $\endgroup$ Apr 25, 2019 at 11:52
  • $\begingroup$ @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! $\endgroup$
    – FaDA
    Apr 25, 2019 at 12:55

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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.

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  • $\begingroup$ I got it this time. Thanks a lot for your help! $\endgroup$
    – FaDA
    Apr 25, 2019 at 15:54

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