Problem in understanding the band structure

Question

I am reading the Scientific Background on the Nobel Prize in Physics in 2007, The Discovery of Giant Magnetoresistance. In Chapter 2, paragraph A it is written:

In the free atoms, the 3d and 4s atomic energy levels of the 3d transition elements are hosts for the valence electrons. In the metallic state these 3d and 4s levels are broadened into energy bands. Since the 4s orbitals are rather extended in space there will be a considerable overlap between 4s orbitals belong- ing to neighbouring atoms, and therefore the corresponding 4s band is spread out over a wide energy range (15–20 eV). In contrast to this, the 3d orbitals are much less extended in space. Therefore the energy width of the associated 3d energy band is comparatively narrow (4–7 eV). In practice one cannot make a clear distinction between the 3d and 4s orbitals since they will hybridize strongly with each other in the solid. Nevertheless for simplicity this two band picture will be used here and the 3d electrons will be considered as metallic – i.e. they are itinerant electrons and can carry current through the system, although they are still much less mobile than the 4s electrons.

Can somebody explain this paragraph to me in a simpler way? I would be really grateful!

I understand that valence electrons are on 3d and 4s shells in 3d transition elements. But what does it mean that these levels are broadened into energy bands in a metallic state? Don't all the electrons create the energy band? And how do you imagine these orbitals in your mind? I know that orbitals are the solution of Schrodinger's equation, but when I read that 4s orbitals are rather extended in space, I totally don't know how to imagine it and how to think about it.

Answer 1

But what does it mean that these levels are broadened into energy bands in a metallic state?

The left panel shows some of the energy levels (eigenvalues of the Hamiltonian operator) for a single isolated atom. The levels are labeled by the standard notation of atomic orbitals (3d and 4s in this case). If you have $N$ independent atoms (atoms very far apart), then these two levels are $N$-fold degenerate. The right panel instead shows what happens on a lattice: i.e. when the atoms are so close to one another that their atomic orbitals overlap. Now the atomic states are no longer degenerate, and the degeneracy is lifted by the presence of orbital hybridization between neighbor atoms. As you can see, the energy levels are scattered around the original atomic level, and form a "band". This is what they mean with "levels broadened into bands".

They say that the typical width of the 4s-band is $15-20$eV, while the typical size of the 3d band is $4-7$eV. This is why 4s orbitals are wider in space (simply the average electronic position is large for a large principal quantum number as 4), and they overlap much more with 4s orbitals of adjacent atoms, thus lifting the degeneracy more. On the other hand 3d orbitals are more localized around the atom (average electronic position smaller than in 4s), they overlap less with the 3d orbital of the neighbor atoms, and the bandwidth is smaller.

The resulting system is metallic because, due to the spatial orbital overlap, electrons can jump from an atom to a neighbor atom very easily. On the other hand, if the atoms are very far apart, there is a region of space between two atoms with very low probability of finding an electron, so electrons don't tunnel from an atom to another.

As a final remark notice that the right panel is not very precise, because typically the energy difference between 4s and 3d is smaller than $20$eV. This means that the two bands are most likely overlapped, but I have introduced a gap between the two for graphical purposes.