Diamond/Silicon's band structure

[voila]

Homework Statement

I am asked to discuss the band structure of diamond. I saw the band structure of diamond has 4 filled valence bands and then 4 conduction bands. Silicon, the same.

Homework Equations

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The Attempt at a Solution

I'm feeling really silly because I don't understand why it is that way. Each carbon in diamond has a filled 1s and then 4 half filled sp3 orbitals. Therefore, I expect a filled valence band arising from the 1s orbital, and 4 other bands arising from the sp3 orbitals which in turn cross the fermi level, not 8 bands, 4 below and 4 above with a band gap.

EDIT: Don't answer yet, I'm updating the post.

 

[voila]

Just found around that each sigma sp3 bond splits into a bonding and an anti bonding state, the former being full and viceversa. The bonding states correspond to the therefore valence bands and the anti bonding to the conductions band. I've got no problem with that, but it seems I have a fundamental misunderstanding:

I usually thought that each atomic orbital gives rise to a single band. Comments I've seen for band structures in the past usually mention "this band arises from a p orbital" or whatever, but I never saw two bands being attributed to the same orbital, as in the sp3 above each giving two bands.

Take the simplest example of hydrogen atoms: they have half-filled 1s orbitals that also give rise to bonding and anti bonding states. Following the previous reasoning, the electron from each atom will go towards the bonding states, which therefore should completely be filled. Hydrogen should therefore be an insulator. And there would be another empty conduction band. However, I remember hydrogen was a very simple example we did when studying bands in solids, and we found there would be just one half-filled band, so hydrogen had to be a conductor.

Can you help me understand what's the right approach and why?

 

[ItsGameTime]

I think a Carbon atom alone has 4 unfilled electron states, but inside a diamond crystal, all states are filled. The adjacent Carbon atoms provide the additional electrons to fill the octet and to provide the balancing charge. With all states filled, and the delocalized conduction band being much higher in energy, is what gives diamond its high electrical resistance.

In the case of silicon, all states are filled, but the delocalized conduction band its much lower energy, which is why thermal fluctuations can excite electrons into the conduction band in a non trivial number, and give rise to its semiconductor properties.

 

[voila]

Just to update with the answer:

In the case of diamond, there's a two atom basis. In this atom basis, we can think of the orbitals as the bonding (filled) and anti bonding (empty) orbitals, which are the ones to give rise to their respective bands. In the case of hydrogen, there's just one atom in the basis with just one orbital, so we just get that one band. If we described it as a two atom basis (with a lattice point every other atom), then we'd indeed get two orbitals (bonding and anti bonding) and two bands. How is it that the same system now has two bands? because we're now changing the Brillouin zone. By doubling the size of the lattice vectors, the Brillouin zone gets folded, and what was one band in the case of the one-atom basis now folds into two bands.