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Spin-orbit Hamiltonian in tight-binding

  1. Jan 24, 2010 #1

    In the usual tight-binding Hamiltonian for semiconductor materials, say GaAs, the basis in which the Hamiltonian matrix elements are specified are the atomic wavefunctions for each atom in the basis. So for GaAs, including just the valence wavefunctions 2s,2px,2py,2pz, we have 8 basis functions (4 from Ga and 4 from As) in the case of spin degeneracy and 16 basis functions when we bring in spin.

    However, in standard textbooks, I find that the spin-orbit interaction Hamiltonian is presented as 8x8 matrix. Shouldn't this be 16x16? Is it the case that the spin-orbit interaction is a correction to each atomic Hamiltonian and that interatomic spin-orbit interaction is ignored? In that case the true 16x16 Hamiltonian would become block-diagonal with two 8x8 blocks each of which block might be what the textbooks say. But in that case, how is it that both Ga and As atoms have the same coupling parameters?

    Is there any literature that does a good job in explaining bandstructure methods in reasonable detail and in anticipating student questions like these?

  2. jcsd
  3. Jan 24, 2010 #2
    I figured out the answer. As I had guessed, it is a correction to each atom in the basis and the inter-atomic spin-orbit coupling is neglected in standard, pedagogical treatments. This is enough to explain the breaking of the 6-fold valence band degeneracy in GaAs to the heavy-hole, light-hole and split-off bands.


    However, it is possible to be more accurate. For example, in III-V semiconductors, there is no true inversion symmetry. Therefore, the band-edges shouldn't really lie at the [tex]\Gamma[/tex] point in the Brillouin zone but slightly off it.

  4. Jan 15, 2011 #3
    Hi, Which textbook has a very comprehensive introduction to tightbinding theory (including tight binding model with spin-orbital interaction under external magnetic field)?
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