How Does the Density of States Apply to Electrons in Bands?

Vanush
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"the density of states (DOS) of a system describes the number of states at each energy level that are available to be occupied. "

But I thought there can't be more than 1 electron in a state? How does DoS have any meaning when dealing with eleectrons?
 
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My understanding is as follows:

The density of states, g(E), tells you the number of possible states at each energy. Since these states are degenerate, you can have one electron in different states at the same energy.

The expected number of electrons in a given energy state, f(E), is calculated using Fermi-Dirac statistics.
http://en.wikipedia.org/wiki/Fermi-Dirac_statistics
This can be no more than 1 because of the Pauli exclusion principle.

So then the total number of electrons at a given energy would be f(E)g(E).
 
nicksauce said:
My understanding is as follows:

The density of states, g(E), tells you the number of possible states at each energy. Since these states are degenerate, you can have one electron in different states at the same energy.

More precisely, g(E) = (# of states between E and E+dE) / (dE)

In a finite system, it is always a series of delta functions.

As the system size gets bigger so that we can assume that it is in the thermodynamic limit, we smooth out the delta functions to get a continuous version of g(E).
 
Why do electron states split into bands in solids if states exist for electrons that have the same energy level
 

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