Fermi energy for a Fermion gas with a multiplicity function ##g_n##

[phos19]

TL;DR

Fermi energy for arbitrary multiplicity

I ran across the following problem :

Statement:

Consider a gas of $N$ fermions and suppose that each energy level $\varepsilon_n$ has a multiplicity of $g_n = (n+1)^2$. What is the Fermi energy and the average energy of this gas when $N \rightarrow \infty$ ?

My attempt:

The average occupation number for a state of the $n$th level is:

$$\langle N_n \rangle = \dfrac{1}{ e^{\beta(\varepsilon_n + \mu)} + 1 }$$

Usually if the system has a fixed degeneracy, say only the spin degeneracy $g = 2s +1$ , one can write the total number of particles $N$ as an integral over $\vec{p}$:

$$
N = \sum_n \langle N_n \rangle = \dfrac{gV}{h^3} \int d^3 p \ \dfrac{1}{ e^{\beta(\varepsilon_p + \mu)} + 1 }
$$

One can than find the Fermi energy in the limit $T \rightarrow 0$.

But this is not the case when $g = g(n)$... Any hints on how to do this ?

 

[DrClaude]

The generic equation for the total number of fermions is
$$
N = \int_0^\infty f(\varepsilon) D(\varepsilon) d\varepsilon
$$
where
$$
f(\varepsilon) = \frac{1}{e^{\beta(\varepsilon + \mu)} + 1}
$$
is the Fermi-Dirac distribution and $D(\varepsilon)$ is the density of states. The degeneracy factor $g$ is part of the density of states, so it will stay inside the integral if is dependent on $n$ (so dependent on $\varepsilon$).

You should however be looking at the equation for the average energy. In the limit $N \rightarrow \infty$, the energy levels can be considered continuous and an integral similar to the one above is obtained.