# Why do we have heavy and light hole bands in semiconductors?

On the topic of the valence band of a semiconductor (in this example GaAS), it is the case that the valence band has some structure to it. As shown in the illustration below, we see that at the $\Gamma$ point there are three visible bands: the spin orbit split off band (due to the effect of the spin orbit interaction) and above that the heavy and light hole bands.

Now, my question is rather simple: why do we observe these different light and heavy hole bands? I'm aware of the fact that heavy and light refers to their effective mass (inversely related to the curvature of the bands), and these are thus bands of holes with a different effective mass, but what is the cause of their masses being different?

• They arise from different atomic levels, which isn't a great answer. On the other hand, assuming they would all be the exact same thing doesn't work... – Jon Custer Jan 13 '16 at 22:17
• @JonCuster Hm, yeah. I just stumbled upon this powerpoint from MIT web.mit.edu/6.730/www/ST04/Lectures/Lecture24.pdf in which the summary (last slide) states that the spin orbit effect lies at the base of the difference between heavy/light hole bands, but without additional text I don't understand it from the slides that cover this. – user129412 Jan 13 '16 at 22:26

To delve more deeply into the origin of the various bands, you should go look at the literature where these bands are calculated. The classic reference for silicon and germanium is Energy-Band Structure of Germanium and Silicon: the k.p Method. Since this is still fairly early in band structure calculations, they do walk you through how the Hamiltonian is built up from the various terms.

It should also be noted that the parabolic approximation is only valid near the symmetry points - in general the bands are not simple parabolas.

• I suppose it should be somewhere in there, indeed. However, and I might be mistaken, I don't really see any literal mentioning of the mechanism that causes these bands to behave differently. It is the result of the diagonalisation and such that they perform, but it isn't really interpreted, which is what I'm trying to find. – user129412 Jan 13 '16 at 23:45
• ActualIy suppose the above derivation suffices. It seems that the introduction of the spin orbit effect is what leads to these two bands being different, which is related to how only total angular momentum $J = L + S$ is conserved due to spin orbit containing an LS product type of term. This means we have $j = 3/2$ and $j = 1/2$ states (the latter being the split off band), and the former being of two types; $(l,l_z) = (3/2,\pm3/2)$ and $(3/2,\pm1/2)$. Why exactly the $\pm$ is irrelevant for the energy is not clear to me entirely, but $l_z = \pm3/2$ is the heavy band and the other the light. – user129412 Jan 14 '16 at 14:47