Talk:Fermi energy

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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:

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)

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)

Contents 1 Fermi temperature 2 Dimensions 3 What? 4 About the Nucleus 5 Degeneracy isn't always 2, and more explanation of the derivation

[edit] 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)

[edit] 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)

[edit] 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)

[edit] 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)

[edit] 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 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 .

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)