Abstract
Results of the electronic structure calculations for the compound Bi2Sr2CaCu2O8 are discussed and compared to those obtained for YBa2Cu3O7. An analysis of the contribution of the densities of states at different atomic sites shows that the states at the Fermi energy, EF, have a strong two-dimensional character due to the presence of the CuO2 planes. Moreover, for the bismuth compound, the contribution of the Bi-O planes at EF is substantial. The elements Y and Ba in YBa2Cu3O7 and Ca and Sr in Bi2Sr2CaCu2O8 act essentially as electron donors, and their corresponding densities of states at EF are negligible. An analysis of the electronic charges at different atomic sites shows that the CuO3 chain units in YBa2Cu3O7 and the BiO2 plane units in Bi2Sr2CaCu2O8 play the role of electron reservoirs for the creation of holes in the CuO2 planes.