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Reading about bands dispersion, I came across the following (Computational Chemsitry of Solid State Materials):
The amount of covalent interaction looks small, which may be deduced from the,
likewise small, dispersion of the oxygen 2p bands which is less than 3 eV.
How is the dispersion obtained (in this case 3 eV) just by inspecting the band structure diagram?
The energies of electrons in tightly bound states (some of which might be referred to as "core states", depending on circumstances) are very nearly the same as the energies of the free molecule. These electrons do not interact very much with their neighbors, so their properties are nearly as if they are isolated. In such cases, with little or no overlap between states of adjacent molecules (little contribution to bonding), the corresponding energy states on each molecule are the same, that is they are degenerate, and there is no interaction to break the degeneracy.
In the extreme case of no interaction whatsoever, the energy eigenstates can be taken to be either the localized molecular states themselves, or extended Bloch states, but in either case the result is the same: the states all have the same energy. In the Bloch picture, that means a perfectly flat band with no dispersion.
If there is some interaction, the Bloch states are the energy eigenstates, not the localized states. The interaction will break the energy degeneracy; the states no longer all have the same energy. To the extent that there is an interaction, there will be some dispersion in the band: the more interaction, the more dispersion. The greater the covalent interaction, the greater the dispersion.
I can't say more than that without knowing more about your crystal, for example is 3 eV large or small? Do we expect the 2p oxygen levels to participate in bonding? And so on.
