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I can understand that direct band gap semiconductor doesn't need phonon to absorb light. But in the following formula:

enter image description here The optical absorption coefficient is proportional to the square of $h\nu$ in the indirect band gap and to the square root of $hv$ in the direct band gap.

Doesn't it means that optical absorption coefficient gets a larger derivative in indirect band gap than in that of the direct band gap?

Why does the direct band gap semiconductor get higher optical absorption coefficient than an indirect band gap semiconductor?

enter image description here

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2 Answers 2

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The three-dimensional nature of a crystalline solid leads to a directional electron population. If a valence band maximum and conduction band minimum do not occur in the same electron-momentum direction, light absorption that would otherwise occur at the bandgap energy will be very unlikely. This is because the electron that hops up in energy (taking up the photon's energy to do so) must also change its momentum to complete that transition. Photons have enough energy, but do NOT contain enough momentum to make that quantum jump occur. So, the direct gap material has a higher photon absorption rate, because the 'easy' short-hop transition is not excluded by conservation of momentum.

There will be absorption due to more complex interactions, involving a third particle (usually a phonon being absorbed or generated), but those occur at a very low rate compared to the simpler direct electron promotion.

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  • $\begingroup$ I think you're trying to say that for a given gap of energy $E$ to be considered "indirect", the momentum offset $p$ is larger than the momentum of a photon of energy $E$. In other words, $p \gg E/c$. $\endgroup$
    – DanielSank
    Commented Sep 15, 2016 at 3:13
  • $\begingroup$ The Fermi surface is a representation in reciprocal space (kinda like a 3-D schmoo plot) of the momentum versus energy character in a crystal. The orientation of the bulges on this (symmetric) surface determines whether the bandgap is direct or indirect, irrespective of magnitude. $\endgroup$
    – Whit3rd
    Commented Sep 17, 2016 at 5:15
  • $\begingroup$ Ok now ask yourself what happens if the bulges are offset in momentum by less than $E/c$ where $E$ is the energy gap. $\endgroup$
    – DanielSank
    Commented Sep 17, 2016 at 13:12
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Adding to the answer of @Whit3rd: I think what OP is trying to ask is more how does it come, that absorption in the direct semicondutor is higher from qualitative reasons, but the formula tells us, that the derivative $d\alpha / d (h\nu)$ is lower for the case of direct band gap?

What one has to keep in mind is that these formulas are only valid for photon energies close to the bandgap energy. And close to the bandgap energy, indeed absorption coefficient of indirect semiconductor scales faster with energy, but only in a certain range. If we increase the photon energy even more, at some point direct transitions will also be allowed and the formula for the absorption coefficient will change.

Also being proportional does not automatically tells us whether it is higher or lower, so faster scaling does not contradicts qualitative arguments given by @Whit3rd.

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