How do you determine the value of the degeneracy factor in the partition function?

In the partition function, expressed as $$Z = \sum_j g_je^{-\beta E_j}$$ I'm wondering what determines the $g_j$ factor. I've been trying to look around the internet for an explanation of it but I can't find one. I guess it is the number of degenerate states in a given energy level? How do you determine how many degenerate states there are? A simple example involving how to determine the $g_j$ factor would be great. Thanks for the help!

Answer

Consider a quantum system with state (Hilbert) space $\mathcal H$. For simplicity, let the Hamiltonian $H$ of the system have discrete spectrum so that there exists a basis $|n\rangle$ with $n=0,1,2,\dots$ for the state space consisting of eigenvectors of the Hamiltonian. Let $\epsilon_n$ denote the energy corresponding to each eigenvector $|n\rangle$, namely \begin{align} H|n\rangle = \epsilon_n|n\rangle \end{align} Now, it may happen that one or more of the energy eigenvectors $|n\rangle$ have the same energy. In this case, we say that their corresponding shared energy eigenvalue is degenerate. It is therefore often convenient to have a the concept of the energy levels $E_j$ of the system which are simply defined as the sequence of distinct energy eigenvalues in the spectrum of the Hamiltonian. So, whereas one can have $\epsilon_n = \epsilon_m$ if $n\neq m$, one cannot have $E_n = E_m$ if $n\neq m$. Moreover, it is often convenient to label the energy levels in increasing index order so that $E_m < E_n$ whenever $m

The degeneracy $g_n$ of the energy level $E_n$ is defined as the number of distinct energy eigenvalues $\epsilon_m$ for which $\epsilon_m=E_n$. For simplicity, we assume that none of the levels is infinitely degenerate so that $g_n\geq 1$ is integer for all $n$.

The partition function of a system in the canonical ensemble is given by \begin{align} Z = \sum_n e^{-\beta\epsilon_n} \end{align} In other words, the sum is over the state labels, not over the energy levels. However, noting that whenever there is degeneracy, sum of the terms in the sum will be the same, we can rewrite the partition function as a sum over levels \begin{align} Z = \sum_{n} g_n e^{-\beta E_n} \end{align} The degeneracy factor is precisely what counts the number of terms in the sum that have the same energy.

As for a simple example, consider a system consisting of two, noninteracting one-dimensional quantum harmonic oscillators. The eigenstates of this system are $|n_1, n_2\rangle$ where $n_1,n_2 = 0, 1, 2, \dots$ and the corresponding energies are \begin{align} \epsilon_{n_1,n_2} = (n_1+n_2+1)\hbar\omega. \end{align} The canonical partition function is given by \begin{align} Z = \sum_{n_1,n_2=0}^\infty e^{-\beta \epsilon_{n_1, n_2}} = \underbrace{e^{-\beta\hbar\omega}}_{n_1=0,n_2=0} + \underbrace{e^{-\beta(2\hbar\omega)}}_{n_1=1,n_2=0} + \underbrace{e^{-\beta(2\hbar\omega)}}_{n_1=0,n_2=1} + \cdots \end{align} If you think about it for a moment, you'll notice that, in fact, the energy levels of this composite system are \begin{align} E_n = n\hbar\omega, \qquad n_1+n_2 = n \end{align} and that the degeneracy of the $n^\mathrm{th}$ energy level is \begin{align} g_n = n \end{align} so that the partition function can also be written in the form that uses energy levels and degeneracies as follows: \begin{align} Z = \sum_{n=1}^\infty g_n e^{-\beta E_n} = \underbrace{e^{-\beta\hbar\omega}}_{n=1} + \underbrace{2e^{-\beta(2\hbar\omega)}}_{n=2} + \underbrace{3e^{-\beta(3\hbar\omega)}}_{n=3} + \cdots \end{align}