I am trying to read Kittel for a project, and he mentions the properties on silicon and germanium so briefly, that I don't understand it at all. He talks about p states, and I don't really know what that means. I've taken quantum mechanics, but I don't understand what is actually being said. I hope someone with experience in solid state could re-write this paragraph into something more digestible, here goes.
The conduction and valence bands of germanium are shown in Fig. 14, based on a combination of theoretical and experimental results. The valence band edge in both Si and Ge is at k = 0 and is derived from $p_{1/2}$ and $p_{3/2}$ states of the free atoms, as is clear from the tight-hinding approximation (Chapter 9 ) to the wavefunctions.
The $p_{3/2}$ level is fourfold degenerate as in the atom; the four states correspond to $m_J$ values $\pm 1/2$ and $\pm 3/2$. The $p_{1/2}$ level is doubly degenerate, with $m_J$ values $\pm 1/2$.The $p_{3/2}$ states are higher energy than the $p_{1/2}$ states; the energy difference $\Delta$ is a measure of the spin-orbit interaction.
The valence band edges are not simple. Holes near the band edge are characterized by two effective masses, light and heavy These arise from the two bands formed from the $p_{3/2}$ level of the atom. There is also a band formed from the $p_{1/2}$ level, split off from the $p_{3/2}$ level by the spin-orbit interaction.
I just don't get it. What are these states? I looked at germanium, it looks like the outer shell has four electrons. Apparently, two are in the S orbital, and another two in the P orbital. If I put these two in three bins, there are six choices right? (1,1), (2,2), (3,3), (1,2), (1,3), (2,3). Is that even relevant? I still don't understand what the p states mean. Can anyone explain this, and how spin-orbit coupling comes into this?
Best Answer
This answer is drawn from Harrison's "Electronic Structure and the Properties of Solids".
Imagine starting with a sample of semiconductor in which the atoms are spaced much farther apart than in reality. As you note, there are 4 valence electrons in each case (Si and Ge), 2 of which fill the $s$-states and the other 2 partially filling the 6 available $p$-states, so 4 of the 8 possible valence states are occupied. For these high-Z atoms, the $s$-states' energy is well below the $p$-states'. Each atom in the sample has the same states, and there is no interaction between atoms, so the energy levels in the sample are discrete and highly degenerate.
Now reduce the lattice spacing, so interaction occurs between atoms. The $s$ and $p$ states are now no longer energy eigenstates, but the designations are still used as labels for the energy levels. The initially discrete (and highly degenerate) $s$ and $p$ energy levels broaden into bands, with the entire $s$ band and the bottom third of the $p$ band initially occupied. (This partial filling is characteristic of a metal.) The energy difference between the top of the $s$ band and the bottom of the $p$ band falls with the lattice spacing.
At some spacing, the top of the $s$-band crosses the top of the occupied portion of the $p$-band, and an energy gap opens between 4 occupied bonding (valence) bands and 4 unoccupied anti-bonding (conduction) bands. (The high symmetry of the lattice is key to this transition.) (For a semi-conductor, the gap is small, on the order of 1 eV.)
Finally, the spin-orbit interaction separates the top of two $p_{3/2}$ and $p_{1/2}$ valence bands at $k=0$, which would otherwise be equal. That separation amounts to 0.29 eV in Ge (as noted in the figure).