Alongside the formula I've seen for Fermi-Dirac statistics: $$\frac{1}{e^{\beta(E-\mu)}+1}=f(E)$$ I often see the addendum that this is only true for non-interacting electrons. By that, I sort of assume it means that two electrons will not interact and occupy the same state (otherwise, if it's simply a question of the forces exerted by the electrons, then this could be surely taken into account with the chemical potential term - or is the issue in the assumption of a constant difference between energy states?)

Thanks for your time!


Assuming non-interacting electrons means that we assume that the overall system obeys the Pauli exclusion principle, and that the electrons do not exert any force on or experience any force from anything else in the system. This is like imagining an ideal gas of electrons, and the system is called a Fermi gas.

If we allow the electrons to interact we have a Fermi liquid instead. As you can read in the link, you can consider this a system of free quasiparticles, which are collective excitations of the system (and do not correspond to any actual particles). These quasiparticles will obey Fermi-Dirac statistics, but the actual electrons will have their distribution modified by the "quasiparticle weight" $Z$, the exact origin of which requires a more involved discussion.

But if you want to learn more, look out for textbooks treating Fermi liquid theory or quantum liquid theory.

  • $\begingroup$ Thank you very much. I don't have time to tonight, but I shall check that out tomorrow! $\endgroup$ Mar 13 '19 at 21:39
  • $\begingroup$ Ok, well, it's not completely what I was looking for, but that's because I phrased my question badly rather than any fault on your part. I'll accept this, and I'll open a new, better-phrased, question! Thanks for your time! $\endgroup$ Mar 14 '19 at 17:14

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