Undergrad Fermi-Dirac distribution, when does it break?

[Telemachus]

Well, the question is if the well known occupation distribution of the energy levels for fermions does break, which means when it is not valid anymore. The Fermi-Dirac distribution reads:

$\displaystyle f_{FD}(E)=\frac{1}{exp\left({\frac{E-\mu}{k_B T}}\right)+1}$ And gives the occupation probability for a state of energy E, for a system of fermions, which obeys the Pauli exclusion principle.

So, the question is if this distribution is always valid or not (wikipedia says it's only valid for non interacting fermions). When does it break, and why. For example, an interacting gas of electrons does obey Fermi-Dirac distribution? if not why.

And how do you derive this distribution function? that probably would answer the question and shade light over it's range of applicability.

https://en.wikipedia.org/wiki/Fermi–Dirac_statistics

For example, in superconductivity something clearly happens, because electrons start to behave like bosons, and do not obey Fermi-Dirac anymore.

This is not homework, is just a question that I was disusing with a friend, who says that is only valid for a free electron gas. I think it is much more general than that, and that electrons (at least, non interacting electrons) should obey Fermi-Dirac in a more general situation, like for example in the occupation of the energy states of an atom (but there, the electrons actually interact, so, according to wikipedia, my friend is actually right).

Thanks!

 

[mfb]

If the fermions interact notably then the energy levels depend on the occupancies of the energy levels - a feedback not considered in the distribution. Otherwise the distribution works.
Good QM textbooks should explain why it looks like this.

 

[Telemachus]

Yes, now I saw it in the wikipedia, it explains the derivation using the grand canonical ensemble I think. Is it a good approximation to use Fermi Dirac distribution in atomic physics? I mean, can one neglect the electrons interactions for this distribution in the atomic orbitals or the interaction is too strong there? and the electron-nucleus interaction doesn't matter, only electron-electron interactions? so, I can place electrons on any arbitrary potential, while they are not interacting with each other, right?

 

[Orodruin]

Just to mention another (quite obvious) case where the FD distribution does not apply is if your system is out of thermal equilibrium.

 

[mfb]

The electron-nucleon interaction generates the potentials in the first place. The electron-electron interaction is still relevant (the inner electrons are shielding the charge of the nucleus for the outer electrons), but often you can find some approximation that works.

 

[Telemachus]

Ok, so the only interaction that matters is the electron-electron interaction, the electron nucleus interaction can be arbitrary, right? because that is the only way that the grand canonical assumption might be violated. So, for example, it might be valid for the outer shells of some atoms, where the electron-electron interaction might not be so important and one can model the (electron+nucleus)-electron like a screened potential of (a modified nucleus charge)-electron, but probably not for the inner shells. Is that right?

 

[mfb]

Well, the electron-nucleus interaction determines the energy levels and orbitals, and these determine how strong the electron-electron interactions are... it is a complex system.

With some effective approximation for the shielding effect, and at low temperatures (only a few energy levels play a role and not many electrons are excited) it should give a good approximation.