How is the number of microstates $\Omega (E)$ related to the partition function $Z$? I know the mathematical relationship. I want to know the physical interpretation.
My understanding is that $\Omega (E)$ represents the number of microstates of a system with a given fixed energy $E$ and the partition function $Z$ represents the number of microstates for all possible values of $E$ of the system. Am I correct or missing something here?
Best Answer
Indeed, the notation $\Omega(E)$ signifies the number of microstates with energy $E$. On the other hand, the partition function $Z$ is mostly conveniently described as a sum over microstates such that the probability $p_i$ of being in a state of energy $E_i$ is,
$$p_i = \frac{e^{-\beta E_i}}{Z}.$$
We can think of $\Omega(E_i)$ as being a degeneracy factor when we consider the partition function,
$$Z = \sum_{\{E_i\}} \Omega(E_i)e^{-\beta E_i}$$
so as to properly count all states, even those with possibly the same $E_i$. The partition function itself though does not represent a number. You wrote,
which I interpret as,
$$\sum_{\{E_i\}} \Omega(E_i) \neq Z$$
which is clearly not the definition of the partition function. However, we could possibly recover this from the partition function. Writing $\Omega(E_i) = \Omega_i$, if the limit,
$$\lim_{\beta \to 0} \Omega_i e^{-\beta E_i}$$
exists and $\sum_i \Omega_i e^{-\beta E_i}$ converges uniformly, then the limit and summation may be exchanged by the dominated convergence theorem, and we have that the total number of microstates,
$$\sum_{i} \Omega_i = \lim_{\beta \to 0} Z(\beta).$$