5.3: The Average Ensemble Energy

We will be restricting ourselves to the canonical ensemble (constant temperature and constant pressure). Consider a collection of $N$ molecules. The probability of finding a molecule with energy $E_i$ is equal to the fraction of the molecules with energy $E_i$. That is, in a collection of $N$ molecules, the probability of the molecules having energy $E_i$:

$$P_i = \dfrac{n_i}{N} \nonumber$$

This is the directly obtained from the Boltzmann distribution, where the fraction of molecules $n_i /N$ having energy $E_i$ is:

$$P_i = \dfrac{n_i}{N} = \dfrac{e^{-E_i/kT}}{Q} \label{BD1}$$

The average energy is obtained by multiplying $E_i$ with its probability and summing over all $i$:

$$\langle E \rangle = \sum_i E_i P_i \label{Mean1}$$

Equation $\ref{Mean1}$ is the standard average over a distribution commonly found in quantum mechanics as expectation values. The quantum mechanical version of this Equation is

$$\langle \psi | \hat{H} | \psi \rangle \nonumber$$

where $\Psi^2$ is the distribution function that the Hamiltonian operator (e.g., energy) is averaged over; this equation is also the starting point in the Variational method approximation.

Equation $\ref{Mean1}$ can be solved by plugging in the Boltzmann distribution (Equation $\ref{BD1}$):

$$\langle E \rangle = \sum_i{ \dfrac{E_ie^{-E_i/ kT}}{Q}} \label{Eq1}$$

Where $Q$ is the partition function:

$$Q = \sum_i{e^{-\dfrac{E_i}{kT}}} \nonumber$$

We can take the derivative of $\ln{Q}$ with respect to temperature, $T$:

$$\left(\dfrac{\partial \ln{Q}}{\partial T}\right) = \dfrac{1}{kT^2}\sum_i{\dfrac{E_i e^{-E_i/kT}}{Q}} \label{Eq2}$$

Comparing Equation $\ref{Eq1}$ with $\ref{Eq2}$, we obtain:

$$\langle E \rangle = kT^2 \left(\dfrac{\partial \ln{Q}}{\partial T}\right) \nonumber$$

It is common to write these equations in terms of $\beta$, where:

$$\beta = \dfrac{1}{kT} \nonumber$$

The partition function becomes:

$$Q = \sum_i{e^{-\beta E_i}} \nonumber$$

We can take the derivative of $\ln{Q}$ with respect to $\beta$:

$$\left(\dfrac{\partial \ln{Q}}{\partial\beta}\right) = -\sum_i{\dfrac{E_i e^{-\beta E_i}}{Q}} \nonumber$$

And obtain:

$$\langle E \rangle = -\left(\dfrac{\partial \ln{Q}}{\partial\beta}\right) \nonumber$$

Replacing $1/kT$ with $\beta$ often simplifies the math and is easier to use.

It is not uncommon to find the notation changes: $Z$ instead of $Q$ and $\bar{E}$ instead of $\langle E \rangle$.