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Electron excitation and velocity 
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#1
Feb2813, 10:48 AM

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When an electron is excited from a valence band to a conduction band by photons, its velocity changes. Do photons change the velocity or it is another process that changes it?



#2
Feb2813, 11:06 AM

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PF Gold
P: 29,242

In a direct transition, the momentum vector doesn't change. So where is this change in "velocity"? Zz. 


#3
Feb2813, 12:23 PM

P: 272




#4
Feb2813, 04:05 PM

Emeritus
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PF Gold
P: 29,242

Electron excitation and velocity
Zz. 


#5
Feb2813, 04:37 PM

P: 272




#6
Feb2813, 06:37 PM

Sci Advisor
P: 1,659

Also try to doublecheck the results the formula gives for a direct transition (from a maximum of the valence band to a minimum of the conduction band) and rethink the difference in velocities you get. 


#7
Mar113, 01:23 AM

P: 272




#8
Mar113, 02:38 AM

Sci Advisor
P: 3,632

Interesting question. So you are creating an electronhole pair and they will generally separate with speed v_ev_h and some center of mass velocity depending also on the effective masses. But a transition will not only take place at +k but also at k, so you are also creating a pair with inversed velocities as long as the occupation of k and k states is symmetric (no electric field!). So no net current as long as there is no external field. On the other hand, with a field, you should be able to realise something similar to a photo diode.



#9
Mar113, 03:46 AM

P: 272

1) The electron which is excited to the conduction band at the point k with a special velocity v 2) The electron which is at the point k in the valence band with a velocity in the direction of v Therefore, there is a net current even without existence of external fields. 


#10
Mar113, 08:22 AM

P: 741

Initially, the photon has momentum in the vacuum. However, in the crystal the momentum becomes is "pseudomomentum". The pseudomomentum is conserved. The pseudomomentum of the photon has to equal the sum of the pseudomementum conductionelectron and valencehole. If you divide the pseudomomentum of each quasiparticle by the effective mass of that quasiparticle, you get the group velocity of each quasiparticle. 


#11
Mar113, 09:05 AM

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P: 3,632

Dear Darwin123,
Pseudo momentum, abbreviated as k, is conserved assuming a direct transition. However, group velocity ##\partial E/\partial k## is only equal to ##k/m^*## for bands with a quadratic dispersion relation ##E=1/2 k^2/m^*##. 


#12
Mar113, 09:10 AM

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#13
Mar113, 09:27 AM

Sci Advisor
P: 1,659

If you want to inject ballistical currents by optical injection, one way to do that lies in simultaneously having two beams present: one at your desired transition energy and one at half the energy. Then you can have single photon absorption and two photon absorption towards the conduction band. The interference of both allows you to create asymmetric electron distributions at +/k and therefore net current which depends on the phase difference between the beams. 


#14
Mar113, 09:28 AM

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#15
Mar113, 02:09 PM

P: 741

Although it is usually said in the text that k is the pseudomomentum, what author really mean is P is the pseudomomentum where ( 2pi) P=hk If we are talking about the center of the Brilloin zone (P very small), then what I said is true. P divided by the effective mass is the effective mass of the quasiparticle, whether it is electron or hole. Generally, we are talking about the center of the Brilloiun zone when the semiconductor has a direct band gap. That brings up an interesting question. What happens in semiconductors with an indirect bandgap? In those cases, P is usually at the edge of the Brillouin zone. However, surely the group velocity is not P divided by effective mass in that case. That would mean that the quasiparticle was always moving very fast! However, I stick to my original statement with two qualifications added. When the quasiparticle is near the center of the Brillouin zone, and the quasiparticle is either an electron or a hole, then the group velocity is the pseudomomentum divided by the effective mass of the quasiparticle. The two final quasiparticles move because both quasimomentum and energy from the photon were conserved. Anyway, I think that this is the answer to the OP's question. 


#16
Mar113, 03:30 PM

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P: 3,632

You are certainly right with your distinction between P and k although this is basically a question of units. Often ##\hbar## is set equal to 1 (natural units) and the difference appears.
To the second point you brought up, I don't see why the center of the Brillouin zone is special. When you shine light e.g. from a narrow banded laser on some material, excitations will take place at those k vectors where the energy difference between the valence and conduction band ##\Delta E(k)## equals ##\hbar \omega##, the energy of the laser photons. I don't think that the motion of the electrons is peculiar to the excitation. For all values of k with the exception of those where E has an extremum, the electrons have a nonvanishing speed. As Cthungha has shown, you need some sophisticated excitation mechanism to really generate a current. 


#17
Mar113, 05:40 PM

P: 741

In an indirect semiconductor, like silicon and germanium, one can adjust the frequency of the narrow band laser so as to excite freecarriers into the edge of the Brillouin zone without populating the center of the Brillouin zone. The photon has just enough energy to excite the electron to the band edge. The pseudomomentum is conserved by the emission or absorption of an acoustical phonon. One can excite the center of the Brillouin zone by using light with a high enough frequency to get the electron into the conduction band at k=0. This sort of thing can be very important when designing a photodetector or solar cell. It probably doesn't make a difference in solar cell applications or in a "slow" photodetector. It may make a big difference with an ultrafast photodetector. My work has included ultrafast spectroscopy and nonlinear optics. Therefore, I have met problems where the initial pseudomomentum of the electronhole pair is important. Now those electronic engineers who don't work in ultrafast electronics may quite rightly consider the initial impulse insignificant. However, there are applications where the initial condition of the electron and hole are important. Anyway, I think the OP was asking about the initial state of the electronhole pair. If so, then the answer is "yes". The electronhole pair for a few femtoseconds after being formed are moving due to the initial momentum of the photon. In at most 10 picoseconds after the electronhole pair are created, the initial pseudomomentum is gone. This is what I think you mean by "the motion of the electrons doesn't depend on excitation." To be more precise, you should have said, "After decoherence, the motion of the electrons doesn't depend on excitation." The momentum is very soon dispersed after the initial excitation. How long is an interesting problem that a lot of scientists are working on. There is cutting edge technology that is impacted by those few femtoseconds. Semiconductors are also used for a variety of optical applications. Some semiconductor crystals are used in optical retardationplates, optical waveguides, Pockel cells and optical switches. The terahertz technology being used at air ports involves ultrafast electrooptics where what happens before decoherence may be important. However, I got too excited. Let me agree with you, partly. Most electronic applications of semiconductors involve time spans after decoherence of the freecarriers. Therefore, the method of excitation does not affect the motion of the freecarriers significantly in terms of these common technologies. 


#18
Mar213, 05:20 AM

P: 38




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