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I am studying statistical mechanics and have a question on these separate definitions. The number of possible states of a system at an energy E is given by

$$\Omega(E)$$

and the probability of the system being found in a particular state is therefore given by

$$\frac{1}{\Omega(E)}.$$

The notes then claim that $\Omega(E)$ is the weight of each state and $\Omega$ is the statistical weight of the ensemble. In this context is $\Omega$ the total number of states of any possible energy? What exactly do we mean by "statistical weight of the ensemble", I am confused by the use of the word ensemble here not statistical weight.

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    $\begingroup$ Perhaps $\Omega$ refers to the distribution function of $E$. $\endgroup$
    – Superfast Jellyfish
    Commented Mar 1, 2020 at 21:50
  • $\begingroup$ Are you talking about the distribution that will include the Boltzmann factor? $\endgroup$
    – Charlie
    Commented Mar 1, 2020 at 21:59
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    $\begingroup$ It's impossible to give a complete and correct answer without links to the notes for context. However, with the fundamental assumption of statistical mechanics that all allowed microstates are equally probable, it's certainly true that $\Omega(E)$ is (proportional to) the probability that the system has energy $E$. In other words, $\Omega(E)$ counts the degeneracy of states at energy $E$ (sometimes the degeneracy of states is also called $g(E)$). In that sense, you can think of $\Omega(E)$ as a statistical weight for the (sub-)ensemble of states with energy $E$. $\endgroup$
    – Andrew
    Commented Mar 1, 2020 at 23:27
  • $\begingroup$ I can see the ambiguity of my question given lack of context. Your comment is helpful though, thank you. $\endgroup$
    – Charlie
    Commented Mar 1, 2020 at 23:32

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In the microcanonical ensemble one considers the energy of the system to be fixed or at least in an infinitesimally thin shell $[E,E+\delta E]$. The phase space volume of this shell is a measure for the amount of states with that energy, therefore you can just integrate $$\Omega(E) = \int_{E\leq H\leq E+\delta E} d\Gamma$$ where $H$ is the Hamiltonian (For example $H=\frac{p^2}{2m}$ for a non-relativistic, free theory) and $d\Gamma = d^{3N}q\ d^{3N}p$ integrates over the $3N$ dimensional phase space ($N =$ #particles). In this sense, $\Omega(E)$ is the "number" of states with the energy $E$ and my guess it that by statistical weight, we mean something like "how large is this volume compared the all of phase space/the rest of the microcanonical ensembles". I'm not sure if that answers your question but it may help a little :)

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