The expected energy in the canonical ensemble is given by

\begin{equation} \begin{split} \langle E \rangle &= \frac{\displaystyle\sum_{i=1} E_i e^{-\beta E_i}}{\displaystyle\sum_{i=1} e^{-\beta E_i}} \\ &\hat{=} \frac{1}{\mathcal{Z}} \displaystyle\sum_{j=1} E_j g(E_j)e^{-\beta E_j} \end{split} \end{equation} Here, $i$ runs over all microstates, $j$ runs over all distinct energy levels, and $g(E_j)$ is the multiplicity of states with energy $E_j$.

In a thermodynamic system, the expected energy $\langle E\rangle$ converges toward a single macroscopic value in the thermodynamic limit, i.e., $\langle E \rangle \rightarrow E$ as $N\rightarrow \infty$. However, thermodynamic systems satisfy specific Euler homogeneity relations. Particularly, $E$ is homogeneous of the first order with respect to the system size (number of particles):

\begin{equation} E \propto N \end{equation} Therefore, $\langle E \rangle$ should be homogeneous of the first order in the macroscopic limit.

Q: What restrictions must $g(E)$ fulfill for $\langle E \rangle$ to homogeneous of the first order when $N\rightarrow \infty$, thereby qualifying as a thermodynamic system?

Alternative formulation Let $\mathbf{g}$ be a vector of all multiplicities, $\mathbf{e}$ be a vector of all energies, $\mathbf{I}$ be an identity matrix, and $t$ be a vector of all exponential terms. Then, equation (1) simplifies to \begin{equation} \langle E \rangle = \frac{\mathbf{g}^\mathsf{T} \text{diag}(\mathbf{e}) \mathbf{t}}{\mathbf{g}^\mathsf{T} \mathbf{I} \mathbf{t}} \,. \end{equation}


1 Answer 1


Instead of thinking in terms of multiplicity of states, it's best to think in terms of density of state. After all, you'd expect to have a continuous spectrum of energy in the thermodynamic limit. Essentially, you want to write for any function $f$: $$ \sum g(E)f(E) = \int f(E)D(E)dE $$ Your partition function: $$ Z = \int e^{-\beta E}D(E)dE $$ is therefore the Laplace transform of $D$ (setting the ground state to $0$).

To get an extensive mean energy $E=N\epsilon$, you therefore expect the scaling: $$ D_N(N\epsilon) \asymp e^{Ns(\epsilon)} $$ with $s$ having a well defined limit in the thermodynamic limit when $N\to\infty$. In other words you expect to have the well defined limit when $N\to\infty$: $$ \frac1N\ln(D_N(N\epsilon))\to s(\epsilon) $$ You interpret this limiting function as the intensive entropy. You get from it the intensive energy by the Laplace method: $$ \beta = \frac{ds}{d\epsilon}(\epsilon) $$ You can check this explicitly for independent two state systems, harmonic oscillators, or free particles and other similar simple systems. However, the point of the canonical ensemble is precisely that you avoid directly estimating $D$. Rather, you directly compute $Z$ and then deduce from it $D$.

Hope this helps.


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