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hokhani
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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?
hokhani said: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?
Yes, the vector [itex]k[/itex] dosen't change but the velocity is [itex]v(k)=1/\hbar dE/dk[/itex] and the curvature of conduction band and valence band are in the opposite directions so the velocity in valence band is in the opposite direction of that in the conduction band.In a direct transition, the momentum vector doesn't change. So where is this change in "velocity"?
hokhani said:Yes, the vector [itex]k[/itex] dosen't change but the velocity is [itex]v(k)=1/\hbar dE/dk[/itex] and the curvature of conduction band and valence band are in the opposite directions so the velocity in valence band is in the opposite direction of that in the conduction band.
Ok, but according to the formula, the velocity doesn't depend on the effective mass.ZapperZ said:Carry out the derivative one step further, and look at the "sign" on the effective mass.
hokhani said:Yes, the vector [itex]k[/itex] dosen't change but the velocity is [itex]v(k)=1/\hbar dE/dk[/itex] and the curvature of conduction band and valence band are in the opposite directions so the velocity in valence band is in the opposite direction of that in the conduction band.
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Ok, but according to the formula, the velocity doesn't depend on the effective mass.
Observing the valence band and conduction band it is clear that the slope(velocity) is in the opposite direction for them. If the direct transition be from the maximum of the valence band to the minimum of the conduction band, the velocity is zero for both of them but direct transitions do not merely take place at the band edge and we can have direct transitions at other points.Cthugha said:Also try to double-check 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.
I don't agree. When an electron is excited from a point k, the hole is nothing but the electron at the point -k so that we have two charge carriers which are not canceled by the other carriers:DrDu said:Interesting question. So you are creating an electron-hole pair and they will generally separate with speed v_e-v_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 realize something similar to a photo diode.
It's basically a case of conservation of momentum. However, the periodicity of the lattice complicates the conservation law a bit. So to be precise, one has to talk in terms of pseudomomenta in stead of momenta, quasiparticles instead of particles. To understand the difference, you have to read up on Bloch's theorem. In simple cases, like you just described, there isn't much a significant difference. However, I want to state in precisely so as to avoid confusion later.hokhani said: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?
hokhani said:I don't agree. When an electron is excited from a point k, the hole is nothing but the electron at the point -k so that we have two charge carriers which are not canceled by the other carriers:
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.
hokhani said:I don't agree. When an electron is excited from a point k, the hole is nothing but the electron at the point -k so that we have two charge carriers which are not canceled by the other carriers:
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.
Cool!Cthugha said: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.
Actually, the letter "k" stands for wave vector not momentum. By the De Broglie relation, the momentum is Planck's constant time k divided by 2 pi.DrDu said: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^*##.
There is more than one extremum in energy. There is the extremum of energy in the center of the Brillouin zone (k=0). There are extrema of energy at the edge of the Brillouin zone(k=k_Reciprocal). The absorption spectrum of the semiconductor has threshold energies corresponding with both types of extrema. Thus, the frequency of that narrow band laser has a big influence with the photoconductivity spectrum.DrDu said: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.
The initial motion of the electron or hole immediately after excitation is affected by the method of excitation. The pseudomomentum of the quasiparticles have to be conserved. However, the initial pseudomomentum will in a very short time disappear due to collisions between quasiparticles. Sometimes this loss of memory is called decoherence.DrDu said: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 non-vanishing speed.
Not all interesting questions have to relate immediately to technology. Not every technological application directly relates to electric current.DrDu said:As Cthungha has shown, you need some sophisticated excitation mechanism to really generate a current.
hokhani said: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?
zhanghe said:which kind of velocity are you talking about, phase velocity (k) or group velocity (dE/hdk)? different velocity, different answer.
Electron excitation refers to the process in which an electron is given enough energy to move to a higher energy level or shell within an atom. This can happen through various methods, such as absorbing light or colliding with other particles.
When an electron is excited, it moves to a higher energy level, causing changes in the atom's properties. This can include changes in the atom's reactivity, color, and ability to emit light.
The velocity of an excited electron can be affected by factors such as the energy of the excitation source, the distance between the electron and the nucleus, and the presence of other particles or fields that may influence its movement.
The velocity of an electron is directly related to its energy. As the electron gains energy through excitation, its velocity increases. Conversely, as the electron releases energy and returns to a lower energy level, its velocity decreases.
Yes, electron velocity can be measured using various techniques such as spectroscopy or particle accelerators. The velocity of an electron can also be calculated using the known energy levels of the atom and the principles of quantum mechanics.