Why is the degeneracy factor for Bose Einstein distribution set to 1 automatically?

Question

In https://scholar.harvard.edu/files/schwartz/files/12-bec.pdf, the article says "With Bose-Einstein statistics, we determined that using the grand canonical ensemble the expected number of particles in a state i is"

$$\langle{n}_{i}\rangle=\frac {1}{e^{\beta(\varepsilon _{i}-\mu )}-1}$$

And it goes on and derive the relationship between the total number of particles and the number of particles in the ground state.

$$N=\sum_{n_x,n_y,n_z=0}\frac{1}{e^{\beta\varepsilon_{1}(n_x^2+n_y^2+n_z^2)}\left(\frac{1}{\langle{n_0}\rangle}+1\right)-1}$$

However, as I understand, the equation for the expected number of particles in a state i is

$$\langle{n}_{i}\rangle=\frac {g_i}{e^{\beta(\varepsilon _{i}-\mu )}-1}$$ where $g_i$ is the degeneracy of energy level $i$. My question is why can I assume $g_i=1$ in this case since wouldn't that affect the answer?

Answer 1

The expected number of particles in a given state $i$ is given by $$\langle n_i\rangle = \frac{1}{e^{\beta(\epsilon_i-\mu)}-1}$$

If there are $g(\epsilon)$ states which all have energy $\epsilon$, then the expected number of particles with energy $\epsilon$ is given by $$\langle n(\epsilon)\rangle = \frac{g(\epsilon)}{e^{\beta(\epsilon-\mu)}-1}$$

In other words, in your second expression $i$ does not label a state but rather an energy level; $g_i$ is then the number of states with that energy.