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

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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 $$\langle E \rangle$$.

17.3: We Postulate That the Average Ensemble Energy Is Equal to the Observed Energy of a System is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.