As I understand the Hubbard model in quantum condensed matter theory, we only consider electron-electron Coulomb interactions for two electrons (with opposite spins) in the same single-particle orbital since these are "short-range". However, the Hubbard model is then used to model a Mott insulator, which includes a conductive phase. When we model other conductors (ie metals) we end up with delocalized electrons, where there can be no "short-range" interaction. How does the concept of delocalized electrons fit with the Hubbard model?
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1$\begingroup$ The Hubbard model exists in many phases, depending on parameter's choice. Leaving out intricacies for very small on-site repulsion, I expect there to be a phase where the system is a band insulator (there is certainly such a phase for $U=0$). Several aspects depends on dimensions. Which dimensions are you talking about? For example in $D=1$ , $U=0$ is a critical point (of a KT transition if memory serves me well). So the behavior I just told you about only exist at $U=0$ there. $\endgroup$– lcvCommented Feb 27 at 12:04
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$\begingroup$ Mott insulator is an insulator, so there is nothing conductive.... $\endgroup$– Meng ChengCommented Feb 27 at 12:15
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'The Hubbard model is an approximate model used to describe the transition between conducting and insulating systems.' (Wikipedia)
It does so by considering two parameters, the hopping term t and the onsite repulsion U. Consider half filling, that is, one electron per site. If t is small / U is large then U prevents the electrons from moving. Electrons will reside in localised orbitals and the system is an insulator. t then does not contribute to the energy. If t is large / U is small then electrons delocalise and the system is a conductor. Now U does not contribute to the energy.
