I am considering the theorem in a statistical mechanics context, but I suppose the question could be extended to other fields where it applies as well.

If we have a system with property $A$ and apply a small perturbation $f_0A$ to its Hamiltonian, the dissipation-fluctuation theorem relates the decay of the average $\langle A(t)\rangle$ to its equilibrium value $\langle A\rangle_0$ as such: $$\langle \Delta A(t)\rangle=\beta f_0 R_A(t)$$ Where :  $\Delta A(t)=A(t)-\langle A\rangle_0$ is the deviation of A from equilibrium, $\beta=1/(k_bT)$, and  $R_A(t)$ is the autocorrelation function of A defined as $R_A(t)=\langle A(\tau+t)A(\tau)\rangle$, where any $\tau$ can be chosen.  

The theorem takes its name from the fact that the autocorrelation is related to fluctuations. Now, while I can think of quite a few examples of links between some autocorrelations and some fluctuations, I am still wondering: is there a general relation between the autocorrelation of A and its fluctuations/variance? Is there a general relation in the form $R_A(t)=f(\mathrm{var}(A),t)$ ? Can I rewrite the dissipation-fluctuation theorem with actual fluctuations/variances, rather than autocorrelations?  

If we assume that the average of $A$ is 0 for simplicity, then $\mathrm{var}(A)=\langle A^2\rangle$, and: $$R_A(t=0)=\langle A^2(\tau)\rangle=\mathrm{var}(A)$$ But that does not tell me anything about later times t.

Alternatively, if I define $I(t)$ to be the time-primitive of A: $I(t)= \int_t A(t')dt'$, then I can write: $$\langle I^2\rangle=\left\langle \int_t A(t’)dt’ \int_t A(t’’)dt’’\right\rangle=\int_t \int_t \langle A(t’)A(t’’)\rangle dt’dt’’=\int_t \int_t R_A(t’-t’’)dt'dt'',$$

which is kind of connecting the autocorrelation of $A$ to the variance of $I$, but that is quite unsatisfying…

Can the dissipation-fluctuation theorem not be written literally in terms of fluctuations?

EDIT: an example of what I would ideally like to formulate if possible:

If I have an expression for the thermal fluctuations of, say, my volume about equilibrium: $$\langle(\Delta V)^2 \rangle=k_b TV \chi_T$$

Can I use that to predict the decay rate of small volume perturbations? It would require reformulating the above expression of the theorem to include that information which quantify fluctuations in my system


2 Answers 2


As you noted, $var(A)=\langle A^2(\tau) \rangle = R_A(t=0)$, which is time-independent for an equilibrium system. It does not allow access to the time variation of the fluctuations of $A(t)$. Still, such time variation exists. Therefore, the way to get the required information is precisely the use of the time autocorrelation function of A: $R_A(t)=\langle A(t) A(0) \rangle$.

Keeping for granted the mathematical derivation of the fluctuation-dissipation theorem $$<\Delta A(t)>=\beta f_0 R_A(t), \tag{1}$$ a conceptual description of its content may help in understanding the reason for the presence of the autocorrelation function.

Equation $(1)$ says that the time decaying variation of an externally induced perturbation of the observable $A$ (usually described as due to dissipative effects) is proportional to the correlation of fluctuations of $A$ at different times. This is basically due to the practical impossibility, at the linear order, to distinguish the decay of an externally induced change of $A$ after the end of the external perturbation, from the decay of a spontaneous fluctuation in the equilibrium system. This last decay may be measured by projecting the variable at time $t$ on the initial value $A(0)$. In other words, the time autocorrelation function is the natural (and unavoidable) way to extract the intrinsic time decaying of a time fluctuation of an observable $A$ at equilibrium.

  • $\begingroup$ Thank you for your answer. My question was indeed to ask if we could express the autocorrelation in terms of variance to reformulate the theorem with variances. So your conclusion is that there is no general way to relate autocorrelation to the most direct way to quantify fluctuations (variance), because the autocorrelation actually carries more information than the variance? Is it because the variance contains information about non-time resolved statistical fluctuations while the autocorrelation does carry a time resolution? $\endgroup$ Commented Jul 16, 2021 at 17:33
  • $\begingroup$ I have edited my question to give an example of what I have in mind, in case it helps illustrate it more precisely $\endgroup$ Commented Jul 16, 2021 at 17:45
  • $\begingroup$ @BarbaudJulien About your first comment, yes, the time autocorrelation function contains more information than its $t=0$ value (the variance of $A$. Now, if we are interested in the time behavior of the fluctuations, there is no alternative to $R_A(t)$. $\endgroup$ Commented Jul 16, 2021 at 19:35

You can think of the autocorrelation function as representing the variance-covariance matrix for the quantity $A$ at different times.

If we discretize time, $t_i=t_1, t_2, \cdots t_N$, then the autocorrelation function is \begin{equation} \langle A(t_i) A(t_j) \rangle = C_{ij} \end{equation} where $C_{ij}$ is the variance-covariance matrix for the set of variables $A(t_i)$.

For equilibrium systems, we can assume that $A$ is stationary, meaning, in the continuous case, \begin{equation} \langle A(t) A(t') \rangle = R(t-t') \end{equation} In other words: the correlation only depends on the time difference between $t$ and $t'$, and not on their absolute values. (For non-equilibrium systems this need not be true, for example there may be transients after I kick the system at some particular time).

In the discrete case, this imples that the variance-covariance matrix is a Toeplitz matrix, meaning that the matrix elements $C_{ij}$ only depend on how far away the matrix element is from the diagonal (all the $C_{ij}$ coefficients with the same value of $i-j$ are the same).

The point is that the auto-correlation function does encode the variance of $A$ (as well as the covariance between $A$ measured at two times separated by a time interval $\Delta t$).


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