In elementary quantum mechanics, we know that when the system possess continuous rotation symmetry, the angular momentum is conserved, and the Hamiltonian of the system commutes with angular momentum operators ( $ [H, J^2 ] = 0 $ and $[H, J_z]$ ), and vice versa (I believe). In this situation we can use their eigenvalues ($j$ and $m_j$) to refer to the states since they are preserved. However, in some papers I read (for example, this paper), these notations are used in semiconductors to refer to different bands (LH and HH), while, to me, continuous rotation symmetry obviously does not exist in crystals. Therefore I wonder what am I missing here?
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$\begingroup$ Light hole and heavy hole have nothing to do with angular momentum. $\endgroup$– Jon CusterCommented Dec 8, 2020 at 2:24
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$\begingroup$ @JonCuster I haven't really read theories about LH and HH, so I apologize for not knowing a lot about it. However if they have nothing to do with angular momentum, then why is it used to denote those bands (in the paper I mentioned in the question, and also in the comment in this question (I have tried reading the paper linked in this question but I found it difficult to understand :( ) $\endgroup$– Frank WangCommented Dec 8, 2020 at 2:35
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Thinking within the framework of tight-binding approximation, one can tie different bands to the states of isolated atoms, which are labeled by their angular momentum (i.e., s-states, p-states, d-states, etc.) This terminology thus penetrates into labeling semiconductor bands, even though the angular momentum is not conserved. If I am not mistaken, this language is used even by Kittel.
This analogy is sometimes taken even further, e.g., when discussing the total momentum of holes (i.e., their spin plus their "angular momentum"), and when discussing the selection rules for light absorption and exciton formation. As I have already said, this makes perfect sense within the tight-binding picture, but needs not be taken too literally.
Unfortunately, I am not in a position to comment about the relation of this notation and the crystal symmetries (some of which are rotational symmetries).
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$\begingroup$ Thank you! Can you list some elementary reading about the fact that you mentioned: “one can tie different bands to the states of isolated atoms, which are labeled by their angular momentum”, I would like to know how are the bands related to the atomic orbitals. $\endgroup$ Commented Dec 8, 2020 at 10:52
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1$\begingroup$ Simple tight-binding model is discussed in many places - you will probably find a simple introduction by googling it. On the other hand, tying the bands to actual atomic states is in the domain of actual band structure calculations, which is not simple at all. I think I learned about this from the Kittel's book, but it is not easy to understand. $\endgroup$– Roger V.Commented Dec 8, 2020 at 10:56