Optical process in semiconductors

AI Thread Summary
The discussion focuses on the optical absorption in semiconductors, particularly the difference between indirect and direct bandgap materials. A key point raised is the relationship between the momentum change of an electron after absorbing a photon, specifically why \hbar\vec{q} is much smaller than \hbar\vec{k}. This is attributed to the significant difference in wavelengths, with typical optical photons around 500nm and electrons having much smaller DeBroglie wavelengths in the picometer range. The mathematical relationship k = 2π/λ illustrates this disparity, emphasizing the relative values of k and q. Understanding this relationship is crucial for grasping the optical processes in semiconductors.
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Homework Statement


I'm trying to understand this part of my notes where the optical absorption in both indirect and direct bandgap semiconductors. This part specifically mathematically describes the change in energy and momentum of an electron in a semiconductor after it has absorbed energy from incident photon.

Qn: Why is \hbar\vec{q} << \hbar\vec{k}?


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http://img222.imageshack.us/img222/9665/qnmj5.th.jpg

This is the part I am referring to.

The Attempt at a Solution

 
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Hmm, I am currently taking Solid State myself and am currently learning this material, but I'll try to offer some insight.

I would think that \hbar\vec{q} << \hbar\vec{k} would have to be true when you consider the wavelengths of typical electrons and photons.

Remember that the magnitude of the wave vector for either electron or photon is:

k=\frac{2\pi}{\lambda}

Where, for electrons, lambda is the DeBroglie Wavelength.

An average optical photon has a wavelength of ~500nm, while the DeBroglie wavelength of an electron is in the picometer range, much, much smaller. So, how does this effect the relative values of k and q and thus, the values of \hbar k and \hbar q?
 
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Oh yeah, you're right. That should be the explanation. Thanks.
 
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