Talk:Fermi energy

Fermi temperature

Fermi temperature is a stub right now, and doesn't contain any information that couldn't be easily shoehorned in here, especially since the concept is mentioned in this article. I'm pretty new here, does anyone have thoughts? -- 7segment 01:49, 21 March 2006 (UTC)

Good idea, I have just redirected it. Rex the first 12:35, 5 April 2006 (UTC)

Hmm, I think this was the wrong move. The discussion of the Fermi temperature in this article is pretty well useless; you could probably derive the only result presented (that T_f=E_f/k_B) simply by noticing that Boltzmann's constant was in units of joules per kelvin. Personally, I think the Fermi temperature should have its own article. I can contribute something to it...after I finish my physics assignment! Ckerr 14:34, 26 August 2006 (UTC)

Agreed. A separate article would be useful. Zebas (talk) 17:50, 14 December 2009 (UTC)

Dimensions

The article presupposes that people are interested solely in 3-D fermi gases, which is a grave error. LeBofSportif 17:15, 5 June 2006 (UTC)

I am a little confused about the vector term in the 3d model. Based on the since the n value has to be a positive integer, why is it necessary to have an absolute value of the vector? If all components of the vector are positive, how could the vector be negative? Methinks I am missing something. Good article btw. -Hellkyte

What?

What on earth is this article going on about? Can someone explain any of it so it makes sense to me? I have no idea about physics or anything like that.

Sorry; there's not a whole lot that can be done. The Fermi energy is not an easily graspable concept the way, say, momentum is, and I cannot off the top of my head see anything needlessly complicated in the way the article presents it. Your best bet is to probably get an idea about physics, then try to read the article. Ckerr 14:37, 26 August 2006 (UTC)

though there is nothing needlessly complicated about the introduction it assumes familiarity with many terms and concepts that are not 'required' to understand the concept of fermi energy. will try to remedy. --V. 18:23, 15 February 2007 (UTC)

Very nice and easy to understand article

In my opinion the starting part of this article (the introduction) is clear and very easy to understand. This is in contrast with many (most?) other physics related articles I have read so far. Most of them generally suppose that the reader actually already knows everything about the matters discussed in the particular article. This article doesnt suppose any more than very basic prior knowledge about it's subject. Congratulations to the author(s). It would be very nice if more of the physics related articles were like this one.

--- Hey, I totally agree. Compared to the German graduate book I've read before this really helped me a lot! Especially the derivation of the three-dimensional potential is very well introduced and structured!

About the Nucleus

Now since the fermi energy only applies to fermions of the same type, one must divide this density in two. This is because the presence of neutrons does not affect the fermi energy of the protons in the nucleus.

But in the following calculation, no difference can be seen. So what does "divide this density into two" actually mean? --Sandycx 08:09, 12 November 2006 (UTC)

Since it looks like the above was dealt with (area in the derived formula = .5 the norm), could someone please explain the justification for dividing by two, seeing as #protons != #neutrons in atoms? I feel like the derivation assumes that, which isn't true for many atoms. Doing a correction of the area for a specific nucleon based on the atomic makeup for an atom isn't terribly difficult, although it would add a couple extra lines. Just seems wrong to make this poor of an assumption to avoid a little extra derivation. Maybe I am missing something though, I am but a lowly chemist. -Hellkyte

Degeneracy isn't always 2, and more explanation of the derivation

In the derivation for 3 dimensions the degeneracy of the fermions has implicitly been assumed to be 2. Whilst this is the case for electrons, it is not true in general, so I think it should be mentioned. Also, I think the line $N={\frac {1}{8}}\times 2\times {\frac {4}{3}}\pi n_{f}^{3}$ is poorly explained. My thoughts are:

• the eighth is because we are in 3D and only want the eighth of the sphere in n_x-n_y-n_z space where they are all positive,
• the 2 is the degeneracy which we have assumed is 2, which is not always the case,
• the four thirds pi n_f is the volume of a spere of radius n_f,
• and most importantly n_f is never defined! It is ${\sqrt {\left(n_{x}^{2}+n_{y}^{2}+n_{z}^{2}\right)}}$.

As I'm new to wikipedia I thought I'd try to canvass opinion on whether adding this information would make the derivation too long before adding it. Uberdude85 01:30, 21 November 2006 (UTC)

Highest filled state

I'd like to take issue with the statement that Fermi energy is the energy of the highest filled state at T=0. If the highest filled state just happens to fill a band, then the fermi energy (the electrochemical potential) is mid-gap. Should this be changed? —Preceding unsigned comment added by 163.1.18.226 (talk) 17:56, 7 February 2010 (UTC)

The fermi energy (as the term is used in this article) is not the same as the electrochemical potential. I don't know how to make it any more clear, it's the second sentence of the article. Is it phrased in a confusing way? Did you just not read the second sentence of the article? Any other suggestions? --Steve (talk) 18:23, 7 February 2010 (UTC)

Chemical potential

This article claims that the Fermi energy is identically equivalent to the chemical potential. I believe this is incorrect. The Fermi energy refers to the energy at the Fermi surface, which is equivalent to the chemical potential, in a non-interacting theory. In the presence of interactions, the Fermi surface can become fuzzed out, so that the Fermi energy is not well-defined.

The chemical potential, on the other hand, is a thermodynamic concept which is not concerned with any microscopic model for a system. It is defined even in classical statistical mechanics, in which there is no such thing as a "Fermi sea." -- CYD

You may be right, at least about some of that. I went looking through my thermodynamics textbook (Sears & Salinger) for some ammunition, and was surprised to find a formula giving the chemical potential in terms of the Fermi energy:

$\mu =\epsilon _{F}\left[1-{\frac {\pi ^{2}}{12}}\left({\frac {kT}{\epsilon _{F}}}\right)^{2}+{\frac {\pi ^{4}}{80}}\left({\frac {kT}{\epsilon _{F}}}\right)^{4}+...\right]$

ε_F was introduced as an empirical constant and later defined in words as "the maximum energy of an electron at absolute zero". An F-D distribution function involving ε_F was given as a low-temperature approximation. However, the energy in the exact F-D distribution function is definitely the chemical potential from classical thermodynamics. Sears & Salinger gives a very detailed derivation of this. Although chemical potential was originally introduced in macroscopic thermodynamics, the relationship to microscopic theories (including Fermi seas!) is rigorously defined. You don't complain about temperature being invalid in microscopic theories, do you?

That said, in solid state physics, I have often seen the F-D distribution given involving E_F instead of μ -- hence my confusion here and on Fermi-Dirac statistics. I'll think about fixing the articles, but I haven't really got it straight in my head yet. -- Tim Starling

after several days calculation, i still can't obtain the coefficient in the second term of the series mentioned above, and i can't find the derivation in Sear and Salinger also. Would any body suggest me for further reaing?tobywhc 13:11, 10 December 2006 (UTC)

Ashcroft and Mermin has a fine derivation of the first two terms of the expansion on pp. 42-47. I also agree that the relevant quantity in the F-D distribution is the chemical potential rather than the Fermi energy. Reasoning naively, it makes sense that the finite T value of $\mu$ is lower than the Fermi energy because the band structure has spherical symmetry. This means that there are more states slightly above the Fermi energy than slightly below, and consequently that the point of half-filling drops as you smear out the Fermi edge. Csmallw (talk) 22:11, 19 May 2009 (UTC)

actually the chemical potential at zero temperature is (almost by definition) equal to the Fermi energy. this is true even for a system with a more complicated potential, there the fermi surfave can become complicated but the fermi energy, the energy of the highest occupied state (@ T=0) is still unique. --V. 18:23, 15 February 2007 (UTC)

Fermi energy in semiconductor

The very first line of this article states "The Fermi energy is (...) the energy of the highest occupied quantum state in a system of fermions at zero temperature". This definition does not work at all of you consider a semiconductor, in which case the Fermi energy is in the band gap, so it is in fact an unoccupied level. As for the difference between the chemical potential and the Fermi energy, the confusion comes from the definition of the Fermi energy itself, which some varies according to author. For example, in Eisberg & Resnick define it as the state with a probability of occupation of ½. According to this definition, the Fermi energy does indeed vary with temperature. However, the more accepted definition is that the Fermi energy is simple the limit of the chemical potential at T->0. See for example Ashcroft & Mermin; look in the back of the book for Fermi energy, and you get a direct reference to the difference with the chemical potential. Just don’t look in Kittel if you want to understand.

If you are advocating changing the first line to "lowest unoccupied state" Im willing to go with that. That seems to be consistent with defining the fermi energy as the limit of mu for T->0. I think it does not make sense to to start this article with a definition based on the chemical potential as this would probably not help anyone who does not already know what the fermi energy is. Also, the Fermi energy is independent of temperature in any definition (the fermi-dirac distribution function is 1/2 for E=Ef at any temperature). --V. 02:05, 21 July 2007 (UTC)

Actually, there is the same problem with "lowest unoccupied state". In an intrinsic semi-conductor, the lowest unoccupied state at 0K is the first level at the bottom of the conduction band, whereas the Fermi level is in the middle of the band gap. That’s why in solid state physics the Fermi energy is defined as the chemical potential at 0K. I don’t mind keeping the part about the highest occupied state, but it should be seen as a visualisation of what the Fermi energy is, not as a definition. It should also be noted that this is only valid for metals or a fermions gas. Maybe keeping the introduction the same, and add a note for a clarification later in the article? Also, it is actually not true that that the Fermi distribution is always equal to ½ at E=Ef. That is because it is the chemical potential which should be used in the Fermi-Dirac distribution, not the Fermi energy. Since they are almost the same (see equation in article for the equation relating the two), a lot of text books just use the Fermi-energy, but this is wrong. The reason why it is wrong is that by definition, as you mentioned, the Fermi energy is independent of temperature. However, in many cases in a semi-conductor, the chemical energy, thus the energy where the fermi-dirac is equal to ½, changes with temperature. This is why it can’t be said that the Fermi distribution is always equal to ½ at E=Ef. --22 July 2007

I would probably start with the statement that it's where the f-d is half, and then clarify the problem for more complicated density of states later, as I don't know that the full explanation could fit clearly and concisely in the lead. — Laura Scudder ☎ 03:00, 26 July 2007 (UTC)

I see what you mean. So lets keep the intro the way it is, it seems to get the concept across. Feel free to add a section on "Fermi-Energy in Semiconductors" to explain some of the subtleties.

The problem is not unique to semiconductors. It occurs whenever you do not have a continuum of states, but rather a gap between energy levels. I recommend specifying that Fermi energy is chemical potential at T=0, and explaining that this is equivalent to highest filled state in cases where there are a continuum of energy levels.

What about this definition of the FE at finite Temperatures

$N=\int _{-\infty }^{E_{F}(T)}g(E)dE=\int _{-\infty }^{+\infty }g(E)f(E,\mu ,T)dE$

i.e. the upper cutoff of the integral over the density of states. I saw this in several books on SSP... In the case of free electrons, this also exlains the expasion of the chemical potential in terms of EF at the end of the article (look for example at the Sommerfeld expansion in Ashcroft) ...

Incluse general spin states

I think article should be expanded to include any half integer fermion case, instead of concentrating only on spin 1/2 particles —Preceding unsigned comment added by 203.197.196.1 (talk) 23:27, 23 April 2008 (UTC)

Quasi Fermi levels

I have started an article on quasi Fermi level, it needs work and is not ready for a "prime time", but when it is I think a link would be appropriate. --Thorseth (talk) 21:40, 20 May 2008 (UTC)

Chemical Potential = Fermi Energy at 0K?

I think it's a serious error to state that chemical energy (but not electrochemical energy) is equal to Fermi energy at absolute temperature. Does the author mean ELECTROCHEMICAL energy?

Because that makes sense.

!! I see the error now. Throughout the article CHEMICAL potential is used for ELECTROCHEMICAL potential. But they are very different concepts.

Chemical potential has to do with doping, whereas electrochemical potential gives rise to the electromotive force.

I suggest we fix this error.

I see that my concern has been brought up before by CYD. He's right and he has put the problem better than me. If nobody is willing to correct this serious error, I am going to do it

128.46.213.219 (talk) 20:45, 22 August 2008 (UTC)

Well, the quantity that actually enters calculations is:

$e^{(E-\mu )/k_{B}T}$

where E is the (say) electron energy level is and μ is the (electro??)chemical potential. It's the energy difference that counts. So if E is measured relative to the energy of an electron in vacuum at that point, then μ also must be measured relative to the vacuum energy at that point, i.e. μ is the chemical potential. If E is measured relative to the energy of an electron in vacuum at a fixed point in space (or at infinity), then μ needs to be likewise, so μ would have to be the electrochemical potential.

Do you agree so far?

In my field (semiconductor materials science), E is virtually always measured relative to the vacuum energy at that point (or in a given material, relative to the conduction band edge at that point, which amounts to the same thing). So chemical potential, not electrochemical potential, would be the right thing to use. Perhaps other fields are different though. At the very least, this could be explained better. What do other people think? --Steve (talk) 02:58, 23 August 2008 (UTC)

Using the definition of Fermi energy in the first line of this article, the limit as T$\rightarrow$0K of the chemical potential $\mu$ is not always the Fermi energy. For example, in an intrinsic semiconductor, $\mu$ is in the gap between the valence band and the conduction band. In this case, the limit as T$\rightarrow$0K of $\mu$ is in the gap, which is not at the Fermi energy which is at the top of the valence band, according to the definition in this article. For the article to be self-consistent, this should be fixed.

However, I have come across at least one place where the definition of Fermi energy is the chemical potential at T=0K. It was explained that all states at or below the Fermi energy are occupied, which is true even if the T=0K definition of Fermi energy is in the gap and the highest energy state that is occupied is at the top of the valence band. But that is not the definition stated in the first line of this wiki. --Bob K31416 (talk) 00:16, 6 May 2009 (UTC)

Bob, this is incorrect. See my comment from April, 8 + refs here.--Evgeny (talk) 16:59, 7 May 2009 (UTC)

I think your point regarding an intrinsic semiconductor's chemical potential µ, is described by this excerpt from the Shigelski 2004 abstract that you mentioned,

"...the Fermi–Dirac distribution function ceases to be accurate for the thermal occupation probabilities when the temperature is so low that the number of electrons in the conduction band is of order unity. The use of the correct occupation numbers results in µ going to the bottom of the conduction band as T-->0."

Please note that this supports the basic point that I mentioned for the Fermi energy of an intrinsic semiconductor (although according to Shigelski 2004, I got the zero-temperature limit of µ wrong). According to the definition of Fermi energy in the first line of this wiki, the Fermi energy for an intrinsic semiconductor is at the top of the valence band. Thus, if we use the zero-temperature limit of µ at the bottom of the conduction band, from Shigelski 2004, the Fermi energy is still not the limit as T$\rightarrow$0K of the chemical potential $\mu$, for this case of an intrinsic semiconductor. --Bob K31416 (talk) 19:39, 7 May 2009 (UTC)

The approach I've seen most often is that the books initially define EF as the "highest occupied state / lowest unoccupied state", and then they get to semiconductors/insulators and say "oh wait, that definition is ambiguous when there's a band gap", and then they say "OK the real definition is mu at 0K". :-) --Steve (talk) 06:36, 6 May 2009 (UTC)

Thanks for contributing to the discussion. Regarding the Fermi energy EF, I presume you meant the energy of the "highest occupied state..." as T$\rightarrow$0, or equivalently, for lowest energy configuration of the system. I don't see how that "is ambiguous when there's a band gap". Are you sure that was in the books that you were recalling? --Bob K31416 (talk) 14:49, 6 May 2009 (UTC)

Relationship to Maxwell distribution

If I understand correctly, doing an analogous derivation at finite temperature for sufficiently low density would yield a Maxwell distribution. If this is correct, it seems worth mentioning. It connects it to something more familiar to many. It can be explained where that approximation fails, and why it matters that the particles are fermions. 72.75.67.226 (talk) 03:55, 8 October 2009 (UTC)