Statistical Mechanics - Calculating Equilibrium Averages

According to statistical mechanics, the probability that a given state with energy E is occupied in equilibrium at constant particle number N, volume $\cal{V}$, and temperature T (constant $N\cal{V}\it {T}$, the `canonical' ensemble) is proportional to $e^{-E/k_B T}$, the `Boltzmann factor.'

$$probability \propto e^{-E/k_B T}.$$ (1)

The equilibrium value of any observable O is therefore obtained by averaging over all states accessible to the system, weighting each state by this factor.

Quantum mechanically, this averaging is performed simply by summing over the discrete set of microstates (Figure 1):

$$\langle O \rangle = \frac{ \sum_n O_n e^{-E_n/k_B T} }{Z},$$ (2)

where Z is the partition function:

$$Z = \sum_n e^{-E_n/k_B T},$$ (3)

and $O_n$ is the expectation value of the quantity O in the $n^{th}$ energy eigenstate:

$$O_n = \int \Psi_n^* O \Psi_n \vec{dr}.$$ (4)

Classically, a microstate is specified by the positions and velocities (momenta) of all particles, each of which can take on any value. The averaging over states in the classical limit is done by integrating over these continuous variables:

$$\langle O \rangle = \frac{ \int O e^{-E/k_B T} \vec{dp} \vec{dr} }{ \int e^{-E/k_B T} \vec{dp} \vec{dr} },$$

where the integrals are over all phase space (positions $\vec{r}$ and momenta $\vec{p}$) for the N particles in 3 dimensions.

When all forces (the potential energy V) and the observable O are velocity-independent, the momentum integrals can be factored and canceled:

$$\langle O \rangle = \frac{ \int e^{-K/k_B T} \vec{dp} \int O... ...\int O e^{-V/k_B T} \vec{dr} } { \int e^{-V/k_B T} \vec{dr} },$$ (5)

where $K = \sum_{i=1}^{N} p_i^2 / 2 m_i$ is the total kinetic energy, and $E = K + V$. As a result, Monte Carlo simulations compare V's, not E's.