Number of electrons in conduction band

  • #1
EmilyRuck
136
6
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 [itex]g_c(E)[/itex];
2) the Fermi probability distribution [itex]f(E)[/itex] for the material at its temperature [itex]T[/itex].
(as a reference, this link can be used). So, the number of electrons between the energy level [itex]E[/itex] and the energy level [itex]E + dE[/itex] is given by

[itex]n(E)dE = g_c(E) f(E) dE[/itex]

where [itex]g_c(E) dE[/itex] is the number of states between energy level [itex]E[/itex] and energy level [itex]E + dE[/itex] and [itex]f(E)[/itex] 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 [itex]E[/itex] 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 [itex]E[/itex] and [itex]E + dE[/itex]), how should we use the value of the Fermi probability for each of them?
 

Answers and Replies

  • #2
Vagn
518
64
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
EmilyRuck
136
6
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.

Yes, a factor of 2 is included in the density of states. My question is: if an energy level [itex]E_1[/itex] is such that (for example) [itex]f(E_1) = 0.1[/itex] and there are two possible states with opposite spins for the electrons at [itex]E = E_1[/itex]: will both the states have the same [itex]0.1[/itex] probability of being occupied? Or what else?
 
  • #4
hokhani
465
8
Yes, a factor of 2 is included in the density of states. My question is: if an energy level [itex]E_1[/itex] is such that (for example) [itex]f(E_1) = 0.1[/itex] and there are two possible states with opposite spins for the electrons at [itex]E = E_1[/itex]: will both the states have the same [itex]0.1[/itex] probability of being occupied? Or what else?
All degenerate states with energy [itex]E[/itex] have the same probability of being occupied, namely [itex]f(E)[/itex], no matter what the spin is.
 
  • #5
absalonsen
5
4
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
EmilyRuck
136
6
Ok, that's right now, so thank you both!
 

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