# Don't electrons have discrete energy levels in semiconductors?

I am having a hard time understanding the relation between the fermi distribution of electrons in a semiconductor, and the fact the electron energy states are discrete.

The fermi distribution is supposed to give us the probability of an energy level being occupied by electrons. It looks like this: yet electrons are supposed to have discrete energy states, and no electron is supposed to be in the band gap: I must be missing something here, how is it that electrons have a fairly high probability of being in the band gap ?

## 3 Answers

There are two important contributions: First the density of states, which tells you the states, which can potentially be occupied. Then there is the Fermi-Dirac distribution, which tells you, which energies are occupied. The Fermi-Dirac distribution does not include allowed and forbidden states. You must fold it with the density of states, which is then called combined density of states. This function then tells you the distribution of occupied or unoccupied states.

• I think this did not answer the original question completely. Electrons do have a discrete energy, and the corresponding Fermi-Dirac distribution gives you the probability (= average occupancy, in this case) that a fermionic quantum state occupies a given (discrete) energy. But this is unhandy for calculations. Therefore one performs the so-called continuum limit where one treats discrete energies as quasi-continuous. Then the FD-distribution tells you the average occupation of a given quantum state with (continuous) energy. The rest is explained by engineer. – RogueDodecahedron Feb 18 '15 at 21:53

In a (perfect) semiconductor, there are no electronic states available in the band gap. Therefore, despite the fact that the Fermi-Dirac distribution "predicts" (or better: accounts for) electrons with energies lying in the band gap, they cannot exist there because no states are available.

If the Fermi level lies in the band gap, the undoped semiconductor behaves like an insulator because all electronic states in the valence band are occupied and the electrons can't move. If $E_\mathrm{F}$ was smaller than the highest energy level in the valence band, the material would become conductive because free states in the valence band become available. If $E_\mathrm{F}$ was larger than the lowest energy level in the conduction band, the material would also become conductive for the same reason.

The Fermi function or Fermi-Dirac distribution was developed as a quantum mechanical expansion of the Boltzmann distribution. With it, you can describe the free electron gas appropriately, whereas the Boltzmann distribution fails by predicting zero energy for all particles at absolute zero (see Pauli exclusion principle). The fermi-dirac distributiom arises from statistical mechanics considerations only and is not influenced by the density of states. Even if there is a probability that a state be occupied by a fermion within the bandgap, if there are no states to be occupied, what do you have?