📃 semiconductor vs Mott insulator

Finally a nice picture for distinguishing the two:

“In a band insulator, as illustrated on the top-left figure, the valence band is filled. For N sites on the lattice, there are 2N states in the valence band, the factor of 2 accounting for spins, and 2N states in the conduction band. PES refers to “Photoemission Spectrum” and IPES to “Inverse Photoemission Spectrum”. The small horizontal lines represent energy levels and the dots stand for electrons. In a Mott insulator, illustrated on the top-right figure, there are N states in the lower energy band (Lower Hubbard band) and N in the higher energy band (Upper Hubbard band), for a total of 2N as we expect in a single band. The two bands are separated by an energy U because if we add an electron to the already occupied states, it costs energy U.

Perhaps the most striking difference between a band and a Mott insulator manifests itself when the Fermi energy EF is moved to dope the system with one hole. For the semiconductor, the Fermi energy moves, but the band does not rearrange itself. There is one unoccupied state right above the Fermi energy. This is seen on the bottom-left figure. On the bottom-right figure, we see that the situation is very different for a doped Mott insulator. With one electron missing, there are two states just above the Fermi energy, not one state only. Indeed one can add an electron with a spin up or down on the now unoccupied site. And only N−1 states are left that will cost an additional energy U if we add an electron. Similarly, N−1 states survive below the Fermi energy.”

arXiv:1310.1481 [cond-mat.supr-con] (2013)

17 July 2019