Spin-orbit Hamiltonian in tight-binding

In summary, the spin-orbit interaction is a correction to each atomic Hamiltonian and is neglected in standard textbooks.
  • #1
rmanoj
3
0
Hi,

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?

Thanks,
Manoj
 
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  • #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.

http://deepblue.lib.umich.edu/handle/2027.42/26719

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.

http://prb.aps.org/abstract/PRB/v57/i3/p1620_1
 
  • #3
rmanoj said:
Hi,

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?

Thanks,
Manoj

Hi, Which textbook has a very comprehensive introduction to tightbinding theory (including tight binding model with spin-orbital interaction under external magnetic field)?
Thanks
 

Related to Spin-orbit Hamiltonian in tight-binding

1. What is the Spin-orbit Hamiltonian in tight-binding?

The Spin-orbit Hamiltonian in tight-binding is a mathematical model used to describe the interaction between the spin and orbital angular momentum of an electron in a solid material. It takes into account the crystal structure and atomic orbitals of the material to calculate the energy levels of the electrons.

2. How does the Spin-orbit Hamiltonian affect the electronic properties of a material?

The Spin-orbit Hamiltonian can cause a splitting of energy levels in the electronic band structure of a material, known as the spin-orbit coupling. This can have various effects on the material's properties, such as altering its magnetic properties or influencing its optical properties.

3. What factors influence the strength of the Spin-orbit Hamiltonian?

The strength of the Spin-orbit Hamiltonian is influenced by several factors, including the atomic number of the element, the crystal structure of the material, and the distance between atoms. Generally, heavier elements with a larger atomic number have a stronger Spin-orbit Hamiltonian.

4. Can the Spin-orbit Hamiltonian be used to study spintronics?

Yes, the Spin-orbit Hamiltonian is an essential tool for studying spintronics, which is a field that focuses on the manipulation of electron spin for use in electronic devices. The Spin-orbit Hamiltonian helps to understand and control the spin behavior in materials, which is crucial for the development of spintronic devices.

5. How is the Spin-orbit Hamiltonian calculated in tight-binding models?

The Spin-orbit Hamiltonian is calculated by taking into account the overlap between atomic orbitals and the hopping energy between neighboring atoms in the crystal lattice. These parameters are then used to construct a matrix representation of the Hamiltonian, which can be solved to obtain the energy levels and associated wavefunctions of the electrons in the material.

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