# Fermi level as a function of temperature

1. Aug 20, 2009

### TheDestroyer

Hello guys,

very short question :), why does Fermi level go down with temperature? is there some physical explanation or discussion for this? apart from mathematics (well, some mathematics would be cool if necessary

Thanks guys :)

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2. Aug 21, 2009

### TheDestroyer

Oh! come on guys! is it that mysterious? at least tell me why the question is that difficult?!?!?

3. Aug 21, 2009

### Mapes

The slope of the curve $(\partial \mu/\partial T)_P$ at any point is $-S$, the negative value of the entropy, and entropy is always positive.

4. Aug 21, 2009

### sokrates

What is your y-axis in that plot?

As far as I know, Fermi energy (Ef) does not move with respect to temperature. It stays at a constant value but for increasing temperature fermi function is broadened because fermions are thermally excited and therefore the probability of occupying higher energy states is increased (thus the broadening)

See this:
http://hyperphysics.phy-astr.gsu.edu/HBASE/solids/fermi.html

5. Aug 21, 2009

### TheDestroyer

This is not the answer pal, I understand the graph very well. I'm asking physically, I've just put the graph to illustrate the situation.

I got the answer, for who's interested, I thought it over and got something, if you think the answer is wrong tell me.

We're going to use the definition of chemical potential, that it's the energy required to add a particle to the system.

The Fermi distribution at T=0 is a step function, meaning if we want to add any other electron to the system we need to put it after those levels, meaning we need to insert it after the step function, and so the energy required to add this electron to the system is precisely EF(T=0) (Fermi energy at T=0).

http://www.doitpoms.ac.uk/tlplib/semiconductors/images/fermiDirac.jpg

At higher temperature, the distribution loses the step function profile (as we can see in the link), and therefore adding another electron is possible before EF(T=0) since there are vacancies.

At very high temperatures, kT becomes very large that it dominates over EF(T=0), and therefore without consuming any energy, we can find vacancies for new electrons as in the following figure

and so, continuing to increase temperature makes the system extract heat upon adding electron, in other words we get negative Fermi level or chemical potential. The heat extracted is because the system tries to sustain the Fermi distribution function.

I hope this is right, tell me what you think guys.

6. Aug 22, 2009

### TheDestroyer

7. Aug 22, 2009

### TheDestroyer

Hey guys! come on! is this physics forum or wheat farm!??!?!

8. Aug 22, 2009

### Mapes

I gave you the quantitative value and interpretation of the slope in my post #3 above. If your physical picture makes more sense to you, use it!

Note, though, that your description of the chemical potential is a little less precise than the standard definition, which is $(\partial E/\partial N)_{P,T}$, the infinitesimal increase in system energy when matter is added (usually at constant pressure and temperature). The differential definition is a bit more precise because only energy differences are meaningful (energy can only be measured relative to a reference value).

9. Aug 22, 2009

### TheDestroyer

But in the thermodynamic limit we can consider adding 1 particle a continuous event, right?

10. Aug 22, 2009

### Mapes

For individual particles, we necessarily have to replace $(\partial E/\partial N)_{P,T}$ by $(\Delta E/\Delta N)_{P,T}$, it being exceedingly difficult to add a fraction of an electron. But my point isn't with $\partial N$ vs. $\Delta N$, it's with making sure your description of the chemical potential specifies an energy change of a system. "Energy required" is vague because it's not clear who supplies this energy: the particle? the system? an external entity? Know what I mean?

11. Aug 22, 2009

### TheDestroyer

Well, is it that big problem where the energy will come from? we just give it, let's say as heat! isn't that good enough?

12. Aug 22, 2009

### olgranpappy

Hi TheDestroyer,

Maybe you would have got a better/quicker answer if you posted this in the Condensed Matter forum... then again, maybe not.

The standard explanation of the *decrease* of the chemical potential with temperature is that because the Fermi function becomes broader, and because there is typically a higher density of states at higher energy (think of the DOS for a free particle), the chemical potential must decrease in order to keep the total number of particles constant.

In general the chemical potential will not *always* decrease, but rather it will generally "be repelled" by the region of higher density of states--typically, as stated above, the DOS is higher at higher energies, but not always.

See, Callen's books for an explanation of this effect. Cheers,

13. Aug 22, 2009

### TheDestroyer

14. Aug 22, 2009

### olgranpappy

I don't think so... but your explanation is a little hard to follow.

Check out the reference I gave. Another good one which works out how the chemical potential depends on temperature in great detail is Ashcroft and Mermin's book "Solid State Physics"

15. Aug 23, 2009

Thanks :)