Abstract
The electronic structures of Sb and Pb-covered Bi(1 1 1) are studied by means of relativistic DFT calculations. The Bi bilayer built of Bi(1 1 1) atomic planes is found to be semiconducting, with the indirect band gap of 0.51 eV. It is suggested therefore that the closing of the gap, leading to semimetallic properties of slabs built from Bi bilayers as well as of a bulk Bi crystal, stems from the interaction between Bi bilayers and not just from the surface states. One monolayer of adsorbed Sb or Pb leads to dramatic redistribution of electronic densities in the surface region and related metallization of the Bi(1 1 1) surface, with a high density of states at Fermi level. Adsorbed Sb and Pb bilayers, in contrast, do not destroy the electronic distribution pertinent to the layered Bi crystal. However, while under Sb bilayer the Bi(1 1 1) film again becomes semimetallic, the surface with adsorbed Pb bilayer remains metallic. This difference in behavior can be explained by a significant charge transfer, that is, electronic donation from Pb bilayer to the substrate, which is absent for Sb bilayer. The adsorption of Sb or Pb on one face of the Bi(1 1 1) film destroys the inversion symmetry of the system and leads to the appearance of Rashba splitting of the bands. For a clean Bi(1 1 1) film, the spin-splitting is absent, which means that the existence of the surface itself cannot cause the required break of the inversion symmetry.
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