# Fermi Level and density of states

1. Mar 30, 2015

### Wminus

Hi. Look at the picture on 1:28 and 1:37 in this video:

How is it possible that the fermi-level is between two energy bands? The fermi level is defined as the highest energy level that contains an electron 50% of the time, so how is it possible for the fermi level being in an area that is forbidden for electrons?!

Thanks for all replies

2. Mar 30, 2015

### kith

3. Mar 31, 2015

4. Apr 7, 2015

### Wminus

Hi again. I have a new question concerning this: What's up with the chemical potential being equal to the fermi level? Shouldn't the chemical potential be equal to the energy of the lowest available state?

BTW, isn't it a funny coincidence that the valence bands are exactly full in insulators? That there is an exact match between amount of states in valence band and number of free electrons? Or is real life nothing like the drawings I've seen in my book (Kittel)?

And one last question while I'm at it. How come the integral $$n = \int D_E(E) f_E(E) dE$$ is always constant, over all temperatures? I mean, the fermi energy and the density of states are derived separately, so how does it turn out the integral of their product is invariant? Is it because we assume that the electrons are bosons when finding their density of states expression, $D_E(E)$?

Last edited: Apr 7, 2015
5. Apr 7, 2015

### ZapperZ

Staff Emeritus
They are only equal at T=0 K in metals. They are not equal all the time in all types of material, even though many people (and I'm included in that) are often sloppy with their notation and often use those two interchangeably.

For band insulators, you calculate the band structure, and then you ended up with all these bands for different states of the material. If NONE of them crosses the Fermi energy, then you have an insulator. So it is the result of the band structure that produces an insulator.

No, it is because the number of particles is a constant. In such calculation, you assume that the number of particles isfixed, i.e. nothing is leaving, and nothing is coming in.

Zz.

6. Apr 7, 2015

### Wminus

But it's not just true for metals. It is also true for an intrinsic semiconductor at 0K. How is this stuff possible? Last time I checked the chemical potential is the energy needed to add one particle (of a certain type) to a system. This energy must be the one for the lowest available state, which cannot be the fermi level as it is located in the band gap..

Let me put my misunderstanding this way:

If the fermi level is in a band gap, all the valence bands are filled with electrons, right? OK, but say you add a few free electrons to the material. Then the fermi energy will increase as the fermi sphere becomes bigger, but none of the conduction bands will fill up with electrons until the fermi energy reaches one of those bands. Right? OK, then where do the electrons go?

I realize that $n$ is usually kept constant. What I meant was that it is a bit mystical to me how the DOS function and the fermi distribution are such that the integral of their product is independent of temperature, despite them seemingly not being linked in any explicit way.

7. Apr 7, 2015

### ZapperZ

Staff Emeritus
This is just statistics. For a semiconductor, there are many people who would rather use the term "chemical potential" rather than the Fermi energy, that resides inside the gap. It is a balance between the filled DOS and the empty DOS.

First of all, you can't just ADD electrons when there's nowhere for it to go! The probability of an electron being able to "go" into something depends on the probability of that transition happening AND the probability that that state is EMPTY. Review the tunneling phenomenon. It is similar to that! For a band insulator, all the states in the valence band are filled.

Change the temperature of the DOS and the Fermi function independently, and integrate them over all limits. Do you think the value of each one of them will change with changing temperature, assuming sum rule conservation? The Fermi function just broadens at the leading edge. Do you think its area under the Fermi distribution will change?

Zz.