Ensemble averages

If the system is ergodic, ensemble averages of any physical quantity $A$ can be computed as time averages over a molecular dynamics simulation, and can be approximated as:

$$\langle A \rangle \simeq \frac{1}{L} \sum_{n=1}^L A (n \delta t),$$ (7.96)

where $A (n \delta t)$ is the value of $A$ evaluated with the particles at positions ${\bf r}_1 (n \delta t), \dots, {\bf r}_N (n \delta t)$. The root mean square fluctuation of $A$ is:

$$\sigma(A) = [\langle A^2 \rangle - \langle A \rangle^2 ]^{1/2},$$ (7.97)

with

$$\langle A^2 \rangle = \frac{1}{L} \sum_{n=1}^L [A (n \delta t)]^2.$$ (7.98)

If all $L$ samples of $A$ were statistically independent from each other, then the standard deviation of $\langle A \rangle$ would be obtained as

$$\sigma(\langle A \rangle) = \frac{\sigma(A)}{\sqrt{L}},$$ (7.99)

however, on the trajectory generated by the molecular dynamics procedure most of the evaluations of $A (n \delta t)$ will be similar to each other, because the necessity of making small time step increments to generate an accurate trajectory means that the configurations will be close to each other for some time. So, in computing $\langle A^2 \rangle$ over $L$ time steps we do not have $L$ independent samples, but only a fraction $L / L_c$, where $L_c$ is related to the correlation time $\tau_c$, which measures the number of steps that we need to wait to obtain a statistically independent sample of $A$. It follows that the statistical error on the average value of $A$ falls off as

$$\sigma(\langle A \rangle) = \frac{\sigma(A)}{\sqrt{L/L_c}}.$$ (7.100)

$L_c$ is not known in advance, but it is fixed, and depends only on the system and on our choice of $\delta t$, and so the statistical error on the average value of $A$ can be reduced as much as wanted by increasing the number of samples $L$.