Talk:Fermi energy

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:

<math>\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] </math>

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