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

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?

• "My question is why can I assume $g_i=1$ in this case since wouldn't that affect the answer?" You haven't given us enough context to understand why you, in this situation, should or should not use $g_i=1$. If the context is in the linked pdf, it would be nice if you could include that context in the body of your question, since the link is subject to rot.
– hft
Commented Dec 19, 2022 at 21:30

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.