Do more energetic orbitals have wider bands in crystals?

[go quantum!]

Let's consider we have a crystal and we search the energy relation of the bands as function of the reciprocal lattice vectors. We would do that theoretical by applying the Tight-binding method. Let's consider that each band is associated with some kind of orbital ( for instance, 3d orbitals). Is it true that we should expect that the bands which are wider in energy are associated with the most energetic orbitals since those are more delocalized in space and the overlapping with the same orbitals of the neighbour atoms are more significant?

Thanks for your attention and for spending your time helping me.

 

[kanato]

Yes, that is correct.

 

[go quantum!]

Thank you for your reply!
Continuing the discussion of the OP...
I, then, heard that when searching for superconductors with higher critical temperatures, people would like to find materials with high density of states at Fermi level, and for that the best candidates would be materials composed of transition metals for their narrow bands. My question is the following. Since the bands of transition metal come from d-orbitals I would say according to OP that they will be wide instead of narrow which would imply low density of states. What's wrong with my reasoning?

 

[kanato]

When considering valence orbitals, usually the order of the size of the orbitals are, from largest to smallest, s p d f. So for something like copper, the valence orbitals are 4s4p3d and the 3d will be the smallest orbitals and have the narrowest bands.

 

[go quantum!]

Thank you!
Can you also give me some justification or idea that explains this fact?

 

[kanato]

The radial part of the wavefunction for 4s has 3 nodes, so that it is orthogonal to 1s,2s,3s wavefunctions. These nodes are always near the origin, which will mean that the majority of its density is outside these nodes and located farther away. Mathematically, you see this is because there is a polynomial with more terms in the 4s wavefunction than the 3s, 2s, 1s. For 4p, it only has to be orthogonal to 2p and 3p, so it only has two nodes. So it is not as extended. 3d does not have any nodes so it is more localized. You can see the differences if you plot the radial wavefunctions.