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

phos19
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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 ?
 
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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.
 
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