# Fermi Dirac distribution and degenerate energy states

In Quantum Mechanics and in semiconductor materials, the number of electrons $N$ in conduction band is usually computed as follows:

$$N = \int_{E_c}^{+\infty} g_c(E)f(E)dE$$

where $g_c(E)$ is the density of states of electrons with respect to energy and $f(E)$ is the Fermi-Dirac distribution.

When the density of states is computed, it is taken into account that each energy level can have two electrons with opposite spins: even if the level is single, the electrons may be two.

$f(E)$ is always said to be «the probability that an electron actually occupies a state with energy level $E$»: but what if the state is a "double" state? Is this probability halved of doubled? In other words: which is the approach followed to obtain the above integral?

This is mentioned in Wikipedia, but without a proof. The cited source is too wide to be used.

$f(E)$ is the probability that a quantum state of energy $E$ is occupied. There are two quantum states (for two spin states) at each energy. The probability cannot be doubled, since that could then exceed 1. All that happens for a spin $1/2$ particle is that the number of available quantum states is doubled.
• Ok, and so each of the two quantum states at energy $E$ has the same probability $f(E)$ of being occupied? – BowPark Jun 24 '16 at 9:43