From the Fermi-Dirac statistics, it can be derived that the probability of an electron filling a level with energy $\epsilon = \mu + \delta$ and the probability of an electron not filling a level with energy $\epsilon = \mu - \delta $ is the same, with the expression $1/(e^{\delta \beta} + 1)$.
We are given a density of states which divides energies in 3 bands: negative energies, energies between 0 and $\epsilon_g$ and energies larger than this $\epsilon_g$. This way, a simple calculation lets us figure out the number of occupied states, since we have a probability of it being occupied and the number of levels.
The g($\epsilon$) function used in this problem is:
\begin{equation} g(\epsilon) = \left\lbrace \begin{array}{lll} (\epsilon - \epsilon_g)^{1/2} & if & \epsilon > \epsilon_g\\ 0 & if & 0<\epsilon < \epsilon_g\\ (-\epsilon)^{1/2} & if & \epsilon <0\\ \end{array} \right. \end{equation}
This generates $n_e$ states occupied by fermions with positive energy and $n_h$ unoccupied states with negative energy, these numbers being:
\begin{equation} n_e = \int^\infty_{\epsilon_g} \frac{(\epsilon-\epsilon_g)^{1/2}}{e^{\beta (\epsilon - \mu)} + 1} ; n_h = \int^0_{-\infty} \frac{(-\epsilon)^{1/2}}{e^{\beta (\mu -\epsilon)} + 1} \end{equation}
Here comes my question: knowing the expression for the density of occupied states with positive energy and the density of unoccupied states with negative energy, how can the chemical potential be calculated? It should be noted that the density of states g($\epsilon$) is symmetric with respect to $\epsilon_g/2$, leaving us with the form of a semi-conductor.
I am inclined to believe that the answer to this lies on understanding the concept of chemical potential, but its relation to the micro-states still remains a mystery.
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