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Wikipedia states the definition of Fermi energy as for "a system of non-interacting fermions". If we have to assume free electrons in a solid behave this way before we are able to calculate Fermi energy, how can Pauli exclusion be justified (because electrons are non-interacting)? Can Fermi energy be similarly defined for electrons confined to a single atom?

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Nuclear physicists have no trouble defining a Fermi energy for interacting nucleons, it's just more complicated than the non-interacting case. See the first chapter of Walecka's book among others. – dmckee Mar 20 '12 at 0:55
I assume you mean "because electrons are interacting" in your parenthetical. This is justified by Landau Fermi-liquid theory, and it applies to crystal quasi-particles. There is a similar qualitative idea in atoms, but it isn't used much (as far as I know). – Ron Maimon Mar 20 '12 at 4:00
The definition "a system of non-interaction fermions", i.e. a Fermi gas exists in a metallic solid after the Pauli principle has been taken into account. So roughly: 'free' electrons in a metal + Pauli = Fermi gas. For this Fermi gas you can define a Fermi energy. – Alexander Mar 20 '12 at 8:39
I am seeking for an answer to the less emphasized question too : Can Fermi energy be defined for electrons confined to a single atom? – Hiran Mar 20 '12 at 19:02
up vote 1 down vote accepted

You might say, of course, that in some sense the fermions do interact (and it is even called exchange interaction). However, it is physical forces, like Coulomb ones, that are understood to be absent. A relevant discussion has taken place here: Degeneracy Pressure, What is it?

Concerning the second question about single atoms, the answer is no. Firstly, Fermi statistics, as any statistics, can only be applied to macroscopic objects. Secondly, even if you were able to create a giant nucleus of a large charge and cover it with a macrosopic number of electrons, the electrons would be interacting with each other through Coulomb forces, hence will not represent a degenerate gas and hence will neither follow the statistical distribution of degenerate gases nor possess Fermi energy.

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