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Consider the outer-most s orbitals for Mg or Ca solid. They will form one band. The 2 electrons in the outer-most shell fill this band. Shouldn't the next p orbital band be above the fermi level and make Mg and Ca insulators?

One can also think about free electron model where the electron-ion interaction is treated as perturbation to the free electrons Hamiltonian. In that case, the first order perturbation theory will lift the degeneracy at the boundary of the first brillouin zone. The number of states in the first band (depending on the size of supercell) should exactly allow twice the number electron to occupy the band full, and make the crystal an insulator.

The above arguments are obvious wrong/incomplete due to the fact that Mg and Ca are metals in reality. However, I want to know what went wrong with my argument. What makes the two metals?

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    $\begingroup$ Atomic orbitals do not translate directly to Bloch wave functions. $\endgroup$
    – Jon Custer
    Commented Nov 14, 2022 at 2:01
  • $\begingroup$ I was talking about the tight binding model in the first paragraph, Isn't it the way to relate atomic orbitals to bloch wave function? The number of states in the s orbital bands equals the number of atoms, which allow the 2*(number of atoms) valance electrons to fully occupied the s orbital bands. $\endgroup$
    – Bohan Xu
    Commented Nov 14, 2022 at 2:28
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    $\begingroup$ Why do you think the 3P band lies entirely above the Fermi level? homework.study.com/explanation/… $\endgroup$
    – Buzz ♦
    Commented Nov 14, 2022 at 2:51
  • $\begingroup$ @Buzz I see. That's my wrong assumption. For some reason, I thought s orbital bands are much lower in energy than p orbitals bands. $\endgroup$
    – Bohan Xu
    Commented Nov 14, 2022 at 2:54

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I think I know the answer to my own question now. s orbital and p orbital have the same energy if we ignore shielding. For element like C, N, O, F, the 2s orbital are closer to the nuclei, so it is less shielded and have lower energy than 2p orbitals.

However, in alkali and alkaline earth metals, the s and p orbitals are both very diffused so the difference in shielding is small. This means the energy difference between them are small, compared to the band width of their corresponding bands. This means the band from different orbitals type will crossover and form metals.

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