# Electron excitation and velocity

1. Feb 28, 2013

### hokhani

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. Feb 28, 2013

### ZapperZ

Staff Emeritus
Really?

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

Zz.

3. Feb 28, 2013

### hokhani

Yes, the vector $k$ dosen't change but the velocity is $v(k)=1/\hbar dE/dk$ 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.

Last edited: Feb 28, 2013
4. Feb 28, 2013

### ZapperZ

Staff Emeritus
Carry out the derivative one step further, and look at the "sign" on the effective mass.

Zz.

5. Feb 28, 2013

### hokhani

Ok, but according to the formula, the velocity doesn't depend on the effective mass.

6. Feb 28, 2013

### Cthugha

First of all the formula clearly says that the velocity does NOT depend on the curvature of the conduction band. The effective mass, however, does.

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.

7. Mar 1, 2013

### hokhani

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.

8. Mar 1, 2013

### DrDu

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 realise something similar to a photo diode.

9. Mar 1, 2013

### hokhani

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.

10. Mar 1, 2013

### Darwin123

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.

Initially, the photon has momentum in the vacuum. However, in the crystal the momentum becomes is "pseudo-momentum".

The pseudo-momentum is conserved. The pseudomomentum of the photon has to equal the sum of the pseudo-mementum conduction-electron and valence-hole. If you divide the pseudomomentum of each quasiparticle by the effective mass of that quasiparticle, you get the group velocity of each quasiparticle.

11. Mar 1, 2013

### DrDu

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 $E=1/2 k^2/m^*$.

Last edited: Mar 1, 2013
12. Mar 1, 2013

### DrDu

Yes, but what I meant is that the electromagnetic field will excite equally likely an electron at k and one at -k.

13. Mar 1, 2013

### Cthugha

Nevertheless the situation is fully symmetric and you will simultaneously excite the transitions of electrons at +k and -k in equal proportion leaing to no net current at all.

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. Mar 1, 2013

### DrDu

Cool!

15. Mar 1, 2013

### Darwin123

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.

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. Mar 1, 2013

### DrDu

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 non-vanishing speed.
As Cthungha has shown, you need some sophisticated excitation mechanism to really generate a current.

17. Mar 1, 2013

### Darwin123

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.

In an indirect semiconductor, like silicon and germanium, one can adjust the frequency of the narrow band laser so as to excite free-carriers 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.

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.

My work has included ultrafast spectroscopy and nonlinear optics. Therefore, I have met problems where the initial pseudomomentum of the electron-hole 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 electron-hole pair. If so, then the answer is "yes". The electron-hole 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 electron-hole 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.

Not all interesting questions have to relate immediately to technology. Not every technological application directly relates to electric current.

Semiconductors are also used for a variety of optical applications. Some semiconductor crystals are used in optical retardation-plates, 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 free-carriers. Therefore, the method of excitation does not affect the motion of the free-carriers significantly in terms of these common technologies.

18. Mar 2, 2013

### zhanghe

which kind of velocity are you talking about, phase velocity (k) or group velocity (dE/hdk)? different velocity, different answer.

19. Mar 2, 2013