# 17.3: We Postulate That the Average Ensemble Energy Is Equal to the Observed Energy of a System

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 $$E_i$$. In the collection of $$N$$ molecules, how many molecules ($$n_i$$) have the energy $$E_i$$?. This is the directly obtained from the Boltzmann distribution

$P_i = \dfrac{n_i}{N} = N \dfrac{e^{-E_i/k_BT}}{Q}$

This is because the fraction of molecules $$n_i /N$$ having the energy $$E_i$$ is

$\dfrac{e^{-E_i/ k_BT}}{Q} \label{BD1}$

which is the same as the probability of finding a molecule with energy $$E_i$$ in the collection. The average energy is obtaining 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

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

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 Varational method approximation.

Equation $$\ref{Mean1}$$ can be explicitly solved by introducting the Boltzmann distribution (Equation $$\ref{BD1}$$)

$\langle E \rangle = \sum_i \dfrac{1}{Q} e^{-E_i/ k_BT}$

The relations between Q and pressure and entropy are given by Eqns. (2.168) and (2.171) respectively.

The pressure $$p$$ can also be obtained as the ensemble average of the following partial derivative

$\left(\dfrac{-∂E}{∂V}\right)_T$

giving

$p =\sum_i - \dfrac{\partial E_i}{\partial V} P_i = \sum_i \dfrac{\partial E_i}{\partial V} \dfrac{e^{-E_i/ k_BT}}{Q} = \dfrac{k_BT}{Q} \dfrac{\partial Q}{\partial V}$

$= k_BT \dfrac{\partial \ln Q}{\partial V}$

The entropy is given by

$S = k_B \ln (Q + \dfrac{\langle E \rangle}{k_BT} \label{3.41}$

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