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Consider the simplest case - a gapped system where the electron in the valence band is excited to the conduction band. In this process, is the spin conserved? Or to word it differently, can the excited electron have a different orientation of spin compared to the hole?

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Spin-orbit coupling
Solids are composed of atoms, where orbital and spin momentum are coupled via spin-orbit coupling. While in some cases this interaction can be neglected or considered as a higher-order correction, strictly speaking it should be taken into account when calculating band structure, and may have significant impact. Thus, the electrons in conduction band are not in pure spin states.

Still, as an approximation one often uses the band structure without spin-orbit interaction, including the latter via empirical terms, whose coupling constants either follow from more precise band calculations or estimated experimentally. These coupling constants typically turn out to be much bigger than what one could expect by simply including the spin_orbit term isnpired by Dirac equation into the effective mass Hamiltonian. Specifically, one often speaks of Elliot-Yafet interaction (bulk SO coupling), Dresselhaus or Dyakonov-Perel interaction (in lattice-asymmetric materials), and Rashba interaction (in non-symmetric semiconductor nanostructures).

Spin of holes
If we think of a hole as a vacancy left by an electron, excited to the conduction band, this description has to compensate both for electron spin and the orbital momentum that the electron ahd before being excited. The orbital momentum traces back to the atomic orbitals forming the corresponding valence band. In typical semiconductors one has three valence bands, and the corresponding holes are assigned different total spin and its projection on the quantization axis; $(1/2, \pm 1/2)$, $(3/2, \pm 1/2)$, $(3/2,\pm 3/2)$ (see Kittel's book, but also the comments to this answer). This nomenclature is extremely important when discussing the selection rules for optical absorption.

Many-particle effects
It also has to be kept in mind that neither electrons in the conduction band, nor the holes in the valence band are single-particle excitations, but excitations of a many-body system with strong Coulomb coupling. While the Coulomb coupling commutes with spin operator and conserves the oevrall spin, one has to add spin-spin interaction, particularly between electrons and holes (simplest inclusion of it is done via Bir-Aronov-Pikus Hamiltonian). The situation becomes even more complicated when considering composite particles, such as excitons, whose spin is badly defined (although this fact is often missed when one adopts approximate Hydrogen-like description, see Knox's Theory of excitons for a deep discussion.).

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  • $\begingroup$ Vadim, thanks for the great answer. So if I understand correctly, in theory, the spin can be different between these quasiparticle excitations (one being the hole and the other being the electron that was excited)? $\endgroup$
    – livars98
    Jan 9 at 20:14
  • $\begingroup$ Firstly, it depends on how they are generated - if it is via excitation by electric field across the band, or via photon absorption, or via some kind of other process. Secondly, after excitation their spins may change independently via spin-orbit coupling or scattering by other electrons/holes. $\endgroup$ Jan 9 at 20:24
  • $\begingroup$ That makes sense. Thank you. $\endgroup$
    – livars98
    Jan 9 at 20:28
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Yes, the Spin of the electron will be the same (as it is essentially the same electron of the same shell).

[ Moreover, the spin of the electron will not matter at all, as the other electron of different atom 'lives' in splited subshell (different). So, Auf Bau'S principle will not be violated
]

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