# I Number of electrons in conduction band

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1. Jun 20, 2016

### EmilyRuck

Hello!
In order to obtain the number of actual electrons in the conduction band or in a range of energies, two functions are needed:
1) the density of states for electrons in conduction band, that is the function $g_c(E)$;
2) the Fermi probability distribution $f(E)$ for the material at its temperature $T$.
(as a reference, this link can be used). So, the number of electrons between the energy level $E$ and the energy level $E + dE$ is given by

$n(E)dE = g_c(E) f(E) dE$

where $g_c(E) dE$ is the number of states between energy level $E$ and energy level $E + dE$ and $f(E)$ is the probability that they are occupied.
This would work if the number of states in that range is just 1 or 0. But what if there are multiple available states?

Fermi probability distribution gives the probability that a state at a certain energy $E$ is occupied: it is just one state. If we need to handle multiple states (like two electrons with opposite spin) at the same energy (or in the same infinitesimal interval of energies, between $E$ and $E + dE$), how should we use the value of the Fermi probability for each of them?

2. Jun 20, 2016

### Vagn

Shouldn't degenerate states be taken care of in the density of states? For spin up and spin down, a factor of 2 is included in the density of states, and doesn't affect the Fermi-Dirac distribution.

3. Jun 20, 2016

### EmilyRuck

Yes, a factor of 2 is included in the density of states. My question is: if an energy level $E_1$ is such that (for example) $f(E_1) = 0.1$ and there are two possible states with opposite spins for the electrons at $E = E_1$: will both the states have the same $0.1$ probability of being occupied? Or what else?

4. Jun 24, 2016

### hokhani

All degenerate states with energy $E$ have the same probability of being occupied, namely $f(E)$, no matter what the spin is.

5. Jul 23, 2016

### absalonsen

Fermi-Dirac distribution f(E) gives the probability that a quantum state with energy E is occupied. So, as long as there is no symmetry breaking that lifts the degeneracy of spin up and spin down electrons, they have the same energy, thereore f(E) for spin up and spin down electrons are the same. That's why we have a factor of two. Furthermore, if there are more available states with the same energy, f(E) for all those states would be the same.

6. Jul 26, 2016

### EmilyRuck

Ok, that's right now, so thank you both!