Fermi energy for a fermion gas

Question

I came across an old exam question for my statistical mechanics course:

Let the energy levels of a gas of $N$ fermions be given as $$\varepsilon_n = (n+1)^{\alpha} \ \varepsilon_0 \ , \ \ n = 0,1,2... \ \text{and } \alpha > 1 $$, with each level having degeneracy :

$$ g_n = (n+1)^{\beta} ,\ \ \beta > 1 $$

Calculate the Fermi energy , and the total average energy in the limit $N >> 1$.

What I've tried

Since $N \rightarrow \infty$ we can use the grand canonical ensemble. The average number of particles that occupies the $n$th orbital is (?):

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

Hence the total number of particles is:

$$ N = \sum^{\infty}_n \langle N_n \rangle $$

If $g(n) = g$ (i.e. same degeneracy for every level) , the sum can be written as an integral over the momenta $\vec{p}$:

$$ \sum_n = g\sum_{\vec{p}} = g\dfrac{V}{h^3} \int d^3p $$

In the limit $T \rightarrow 0$ , we find:

$$ \dfrac{1}{e^{\beta{(\varepsilon_n + \mu) }} +1 } \rightarrow \Theta(\varepsilon_F - \varepsilon) $$ ($\Theta$ being the Heaviside stepfunction, and $\varepsilon_F$ the Fermi energy), hence in the same limit, for $g_n = g$ we have:

$$ N = \sum^{\infty}_n \langle N_n \rangle = g\dfrac{V}{h^3} \int \Theta(\varepsilon_F - \varepsilon) d^3p $$

The integral is just volume of the ball with radius $p_F = \sqrt{2m \varepsilon_F}$...

What to do if $g_n$ is not constant ?

We get the integral (?):

$$ \dfrac{V}{h^3} \int d^3 p \ \Theta(\varepsilon_F - \varepsilon) g(n) $$

I'm not quite sure what to do here...

Answer 1

In the case of zero temperature, fermions occupy states with lower energy. Thus, the equation for the number $n_{\text{max}}$ of the occupied state with the highest energy has the form $$ N \approx \sum_{n=0}^{n_{\text{max}}} g_n $$ Condition $N \gg 1$ leads to $n_{\text{max}} \gg 1$. The relative change of the power function is slow in the domain of large arguments: $(n+1)^\beta / n^\beta \approx 1$ for $n \gg 1$. Therefore, the equation for $n_{\text{max}}$ can be written as $$ N \approx \int_0^{n_{\text{max}}} (n+1)^\beta\ dn \approx \frac1{\beta+1} (n_{\text{max}}+1)^\beta. $$ So we have $$ n_{\text{max}}+1 \approx (N(\beta+1))^{\frac1{\beta+1}} $$ and $$ \varepsilon_{\text{F}} = \varepsilon_0 (n_{\text{max}} + 1)^\alpha \approx \varepsilon_0 (N(\beta+1))^{\frac{\alpha}{\beta+1}} $$