Photon absorption by electrons at opposite k-points

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SUMMARY

The discussion centers on the photon absorption process by electrons transitioning from the valence band (VB) to the conduction band (CB) at specific k-points in the Brillouin zone. It establishes that the momentum of a photon is insufficient to alter the crystal momentum of an electron, yet it affects the real momentum due to the relationship defined by the group velocity formula, v_g = (1/ħ) ∇_k E. The conversation also highlights the significance of effective mass, noting that while holes exhibit negative effective mass in the valence band, electrons retain positive effective mass during transitions. The Moessbauer effect is cited as an analogous phenomenon illustrating these principles.

PREREQUISITES
  • Understanding of band theory in solid-state physics
  • Familiarity with the Brillouin zone concept
  • Knowledge of group velocity and effective mass in semiconductor physics
  • Basic principles of photon-electron interactions
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  • Study the implications of the Moessbauer effect in solid-state physics
  • Explore the mathematical derivation of group velocity in relation to energy bands
  • Investigate the role of effective mass in semiconductor behavior
  • Learn about the k-space representation and its applications in electronic transitions
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Physicists, materials scientists, and electrical engineers interested in semiconductor physics, photon interactions, and electronic band structure analysis.

hokhani
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Transition of an electron from the valence to conduction bands (direct transition at a k-point near the band edge) would change the momentum of electron because the sign of the group velocity in valence and conduction bands are opposite. Could one infer that the direction of radiation is a determining factor in the transition? In other words, by radiating light from a specific direction, if we have transition at k we can not have the same transition at -k because if the radiation could change the group velocity at k from v to -v, it would change the group velocity at -k from -v to -2v while according to band structure the transition at -k requires changing from -v to v!
 
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How large is the momentum of a photon compared to the size of the Brillouin zone?
 
M Quack said:
How large is the momentum of a photon compared to the size of the Brillouin zone?
The momentum of photon is too small to change the crystal momentum of electron but it changes the real momentum of electron because ##v_g={1/\hbar} \nabla_k E## and assuming parabolic valence and conduction bands (with the same effective mass) this formula says that the velocity changes from, say, v to -v in the optical transition.
 
The assumption is always that the real momentum is taken up by the whole crystal, which has a huge mass, and therefore the resulting velocity is nothing.
A very good example for this is the Moessbauer effect (OK, photons are absorbed/emitted by the nuclei, but the principle is the same).
 
M Quack said:
The assumption is always that the real momentum is taken up by the whole crystal, which has a huge mass, and therefore the resulting velocity is nothing.
A very good example for this is the Moessbauer effect (OK, photons are absorbed/emitted by the nuclei, but the principle is the same).
No, this problem occurs even if we don't suppose the phonon effect.
 
When a transition happens that an electron jumps from VB to CB by being excited only by a photon, the electron always jumps almost vertically in the E-k diagram. The momentum conversation is still held though the election velocity direction becomes reverse. But do not forget that its effective mass also becomes from negative into positive. As usual people prefer to use h with bar *k to describe the momentum instead of mv for simplicity. In the way. You just need to check k for the direction of momentum of electron.
 
zhanghe said:
But do not forget that its effective mass also becomes from negative into positive.
Thanks. But I don't agree. The hole effective mass is negative (valence band) but it is not true for an electron which is still in the valence band. To describe the behavior of the whole valence band which has lost an electron, we use the hole language and consider a negative effective mass for the hole.
 
Last edited:
hokhani said:
Thanks. But I don't agree. The hole effective mass is negative (valence band) but it is not true for an electron which is still in the valence band.
Well, the m* of hole is positive.
 

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