Free electron theory of metals works unreasonably well in spite of the fact that we neglect the Coulomb repulsion between the electrons. Is there deeper reason why this should work? Somewhere I heard that this has to do something with Fermi liquid theory but I'm not familiar with it. Can someone explain in simple terms why free electron theory works? The underlying explanation should also be able to guess situations where it should expected not work.

  • $\begingroup$ have a look hyperphysics.phy-astr.gsu.edu/hbase/Solids/band.html $\endgroup$
    – anna v
    Commented Feb 8, 2017 at 10:23
  • 2
    $\begingroup$ Free electrons theory not always works for metals. It fails for example for ferromagnetic systems. As far as I know, the way one can avoid Coulomb repulsion in using the Landau-Fermi liquid theory. I don't really know much about this model but as I understand this theory maps the interacting electrons into non-interacting quasi particles. In this sense, the Coulomb repulsion is hidden by some sort of "change of variables". $\endgroup$
    – Diracology
    Commented Feb 8, 2017 at 13:07

2 Answers 2


In simple terms, the reason that free electron theory works is that the (non-interacting) electrons in a metal form a fermi-surface. If in the absence of interactions the fermi surface is the ground state of the system.

If you add coulomb interaction this will enable electrons to interact. However, due to the Pauli exclusion principle (and conservation of momentum) there are very few electrons that actually are capable of scattering against each other. Basically since there is a Fermi surface only electrons close to the at the edge of the surface can actually scatter.

The tendency to scatter also goes to zero, as one approaches the Fermi surface from the outside. This vanishing tendency to scatter means that excitations of the non-interacting fermions are almost a correct description of the excitation also in the interacting case.. as long as you are close to the Fermi surface (and we typically are).

As to when this picture fails, is for instance the case of superconductivity. If there are small attractive (instead of repulsive) interactions between the electrons, this destabilizes and destroys the fermi-surface, invalidating the whole picture.

  • $\begingroup$ @MikealFremling If one is looking for more details about the stability of the Fermi surface, one can refer to the references in the following post physics.stackexchange.com/q/69358/16689 $\endgroup$
    – FraSchelle
    Commented Feb 10, 2017 at 15:04

The free electron model was introduced before quantum mechanics theory of solids was developed. The following shows what happens to the electrons bound in a solid state lattice


The whole lattice is involved in the formation of the bands, there is no individual identity to the electrons, they just occupy the available energy levels in each band, while as fermions they also obey the Pauli exclusion principle: one electron at each energy level.

An important parameter in the band theory is the Fermi level, the top of the available electron energy levels at low temperatures. The position of the Fermi level with the relation to the conduction band is a crucial factor in determining electrical properties.

Most solid substances are insulators, and in terms of the band theory of solids this implies that there is a large forbidden gap between the energies of the valence electrons and the energy at which the electrons can move freely through the material (the conduction band).

The electrons bound in the valence band cannot move freely unless energy excites them into the conduction band, where their mobility is through the whole lattice, no longer bound in around a specific nucleus . The difference between insulators and conductors lies in the energy necessary for the transition from the valence band to the conduction band. In insulators it is large, in metals it is very small, and the transition energy is available from the black body radiation, the vibrations of the lattice. Thus in the conduction band they are as free.

It should be noted that coulomb interactions are already included in the quantum mechanical boundary value problem of "electrons in lattice", for the qualitative solution shown in the image above.

  • $\begingroup$ I'm not sure I understood your last sentence. Band theory does not consider electron-electron interactions. Both in the nearly free electrons and tight binding models, the only interaction is between electrons and the ions. That is why band theory fails in some cases (ferromagnetic systems, Mott insulators...) $\endgroup$
    – Diracology
    Commented Feb 8, 2017 at 12:56
  • $\begingroup$ @Diracology I am just answering the statement in the question "in spite of the fact that we neglect the Coulomb repulsion between the electrons" $\endgroup$
    – anna v
    Commented Feb 8, 2017 at 13:51
  • $\begingroup$ @Diracology I think Anna means not that band theory includes the interactions — instead that solid state physics operates in mean field approximation, and thus band theory, working in this framework, automatically has the interactions included. $\endgroup$
    – Ruslan
    Commented Feb 8, 2017 at 14:58
  • $\begingroup$ @Diracology what ruslan said $\endgroup$
    – anna v
    Commented Feb 8, 2017 at 16:24
  • $\begingroup$ I understand that there are interactions included but those interactions are only with the lattice. There still are Coulomb repulsion being neglected. I am not saying that Anna's post is wrong. It is correct but I am not sure it actually fully answers the OP question. $\endgroup$
    – Diracology
    Commented Feb 8, 2017 at 16:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.