In section 4.3 of "Statistical Mechanics of Nonequilibrium Liquids" by Evans and Morriss, the following identity is noted:

$$ \langle \dot{A} A^* \rangle = 0,$$


$$ \langle A B^* \rangle = \int \mathrm{d} \Gamma f_0 A(\Gamma) B^*(\Gamma) = (A,B),$$


$$ \dot{A} = iL(A). $$

Here $*$ denotes complex conjugation $L$ is Hermitian on the inner product: $(A, L(B)) = (L(A), B),$ so $iL$ is skew-Hermitian.

For completeness, the "true adjoint" on the phase space integral is $L^\dagger = - \mathcal{L}$, and the identity $\mathcal{L}(f_0 B) = f_0 L(B)$ holds for the equilibrium distribution $f_0$.

Attempts at a Solution

It's fairly straightforward to show that the object $ \langle \dot{A} A^* \rangle$ must be pure imaginary:

$$ (A, iLA) = -(iLA,A) = -(A, iLA)^*$$

Alternatively, from an integral point of view:

$$ \int \mathrm{d} \Gamma f_0 \dot{A}(\Gamma) A^*(\Gamma) = \int \mathrm{d} \Gamma f_0 iL({A}(\Gamma)) A^*(\Gamma) \\ =-\int \mathrm{d} \Gamma {A}(\Gamma) i \mathcal{L}(f_0 A^*(\Gamma)) \\ =-\int \mathrm{d} \Gamma {A}(\Gamma) f_0 iL(A^*(\Gamma))\\ =-\int \mathrm{d} \Gamma {A}(\Gamma) f_0 (\dot{A(\Gamma)}^*),$$ (noting that $iL$ is real.)

For real functions this solves it, however the identity is given for arbitrary phase functions.

I've searched for some other way of showing, for example, that the inner product is also equal to its conjugate, therefore rendering the expression zero. But I can't find any way of doing so - approaching from the physical properties of Liouvilleans only leads to agreement with the operator algebra. I don't think there's an answer here purely from the operator properties, but if someone with experience in this formulation of classical statistical mechanics can see what's going on, I'd be grateful.


1 Answer 1


I'll start with what it says in my go-to reference for this kind of stuff, Theory of Simple Liquids by J-P Hansen and IR McDonald, and then fill in some details. I only have the second edition, but hopefully the same material appears in later editions. In their chapter on "Time-dependent Correlation Functions and Response Functions", they say:

Equation (7.1.10) shows that autocorrelation functions are necessarily even functions of time. If $A$ is a complex quantity, the autocorrelation function is conventionally defined as
$$ C_{AA}(t) = \langle A(t) A^* \rangle \tag{7.1.11} $$ which ensures that $C_{AA}(t)$ is a real function of $t$ for all times.

Their eqn (7.1.10) deals with the signatures of the dynamical variables under time reversal, and applying it to an autocorrelation function gives \begin{align*} C_{AA}(t) &= \varepsilon_A \varepsilon_A C_{AA}(-t) = C_{AA}(-t) && \text{even} \\ &= \langle A(-t)A^*(0) \rangle = \langle A(0) A^*(t) \rangle && \text{stationary} \\ &= \langle A^*(t) A(0) \rangle = C_{AA}^*(t) && \text{real} \end{align*} since $\varepsilon_A=\pm 1$ according to whether $A$ is an even or odd function of the momenta.

The two properties (being even in time, and being real) seem to be linked, as they should be, because quite generally the autocorrelation function of a complex variable is a Hermitian function $$ C_{AA}(-t) = C_{AA}^*(t) $$ The "extra" feature here is that we are discussing dynamical variables which are functions of position and momentum, which have a characteristic signature under time reversal.

So it follows that your average $\langle \dot{A} A^*\rangle$, which is the linear coefficient in a Taylor expansion of $C_{AA}(t)$ about $t=0$, being both imaginary (as you have shown) and being an odd-order term in the expansion, must vanish.


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