- #1
hokhani
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How a bound electron-hole pair (exciton) can move together while the velocity of the free electron in the conduction band is opposite to that of the corresponding hole in the valence band?
Excitons: Opposite Movements of Bound Electron-Hole Pairs
- Thread starter hokhani
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In summary, the valence-electron and the valence-hole are traveling together, but the valence-electrons corresponding to the valence-hole are moving in the opposite direction.
- #2
Cthugha
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This is kind of a common confusion as it is hard to keep track of all the signs when discussing electrons and holes.
It turns out that the velocity of the hole is not opposite to the one of the electron. It also obviously cannot be as considering an electron and a hole picture are two sides of telling the same story.
The group velocity is defined as proportional to the derivative of the energy with respect to the crystal momentum:
[tex]v_g=\frac{1}{\hbar}\frac{d E}{d k}[/tex].
When you excite an electron from a full valence band to the conduction band, it will have momentum k. The valence band (and therefore the hole) will be left with a momentum of -k, so electron and hole have opposing crystal momentum. However, also the dispersions of the bands are different. Putting zero for simplicity between the two bands and assuming the parabolic approximation, you will find that both dispersions will be almost the same, but differ by the sign, so [tex]E_h=-E_e[/tex].
Plugging both into the equation for the group velocity, you find that both E and k change sign and therefore the velocity is the same for both and does not change. Other quantities, like effective mass indeed change their sign.
As a second comment, in semiconductors excitons are quite delocalized anyway and the crystal momentum and the velocity of electrons and holes change quite often.
It turns out that the velocity of the hole is not opposite to the one of the electron. It also obviously cannot be as considering an electron and a hole picture are two sides of telling the same story.
The group velocity is defined as proportional to the derivative of the energy with respect to the crystal momentum:
[tex]v_g=\frac{1}{\hbar}\frac{d E}{d k}[/tex].
When you excite an electron from a full valence band to the conduction band, it will have momentum k. The valence band (and therefore the hole) will be left with a momentum of -k, so electron and hole have opposing crystal momentum. However, also the dispersions of the bands are different. Putting zero for simplicity between the two bands and assuming the parabolic approximation, you will find that both dispersions will be almost the same, but differ by the sign, so [tex]E_h=-E_e[/tex].
Plugging both into the equation for the group velocity, you find that both E and k change sign and therefore the velocity is the same for both and does not change. Other quantities, like effective mass indeed change their sign.
As a second comment, in semiconductors excitons are quite delocalized anyway and the crystal momentum and the velocity of electrons and holes change quite often.
- #3
hokhani
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Many thanks.
Please look at this picture.
http://postimg.org/image/kgr5cbpev/
Suppose the electron at k is excited from the valence band to the conduction band. According to the picture if we attribute hole to the electron in the valence band at -k, free electron and hole are moving in the same direction but we consider the empty place as hole which is moving in opposite direction compared to free electron.
could you please help me if I am wrong?
Please look at this picture.
http://postimg.org/image/kgr5cbpev/
Suppose the electron at k is excited from the valence band to the conduction band. According to the picture if we attribute hole to the electron in the valence band at -k, free electron and hole are moving in the same direction but we consider the empty place as hole which is moving in opposite direction compared to free electron.
could you please help me if I am wrong?
Last edited:
- #4
Cthugha
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The picture does not show up.
But according to your comment, you want to apply the simple one-particle picture to excitons. This is doomed to failure. If you excite an exciton, you do not create a free electron with momentum k and a free hole with opposite momentum, but you create a bound pair of these with total momentum k given by your excitation conditions. It is like having one process creating hydrogen and a second process creating free protons and electrons. These are very different things. Expressing a genuine two-particle state in a single-particle basis will most often go wrong.
On the other hand, you can excite free electrons and holes which may afterwards lose energy and form excitons. But in this case the electron/hole created by one photon are not necessarily the ones which will form the exciton.
It is unfortunate that some books tend to draw the exciton dispersion and the free electron hole dispersion into one graph as that may lead to misunderstandings. The book of Pelant and Valenta (luminescence spectroscopy of semiconductors) gives a good summary of why that is so.
But according to your comment, you want to apply the simple one-particle picture to excitons. This is doomed to failure. If you excite an exciton, you do not create a free electron with momentum k and a free hole with opposite momentum, but you create a bound pair of these with total momentum k given by your excitation conditions. It is like having one process creating hydrogen and a second process creating free protons and electrons. These are very different things. Expressing a genuine two-particle state in a single-particle basis will most often go wrong.
On the other hand, you can excite free electrons and holes which may afterwards lose energy and form excitons. But in this case the electron/hole created by one photon are not necessarily the ones which will form the exciton.
It is unfortunate that some books tend to draw the exciton dispersion and the free electron hole dispersion into one graph as that may lead to misunderstandings. The book of Pelant and Valenta (luminescence spectroscopy of semiconductors) gives a good summary of why that is so.
- #5
hokhani
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Excuse me. I edit my last post sending the image. However I attached it here.
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- #6
hokhani
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By this, do you mean that my earlier comment (according to the picture) is correct?Cthugha said:
- #7
Darwin123
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hokhani said:
I think that you are conflating the valence-electrons with the valence-hole.
The valence-electron is not the same as the valence-hole. The valence-electron moves in the opposite direction the corresponding valence-hole. The kinetic-energy of the valence-hole is the negative of the energy of the valence-electron relative to the top of the valence-band.
In an energy-level schematic, the valence-hole smiles :-) while the valence-electron frowns :-( If you see the valence-band turned down, then this is a valence-electron band. If you see the valence-band turned up, it is a valence-hole band.
The picture of the exciton is of a conduction electron bound to a valence-hole. The conduction-electron and valence-hole have to travel together. However, the valence-electrons corresponding to the valence-hole are moving in the opposite direction.
The valence-electrons are tunneling from one atom to the other. The hole is really the gap in the valence-orbital, not the electron. So the valence-hole is traveling in the opposite direction.
It is like watch a air bubble under water. If the air bubble goes right, the water near the bubble is going left. If the air bubble goes up, the water near the air bubble is going down. The valence-hole is like the air of the bubble, and the valence-electron is like the water near the air bubble.
- #8
1. What are excitons?
Excitons are a type of quasiparticle that consists of an electron and a hole bound together by the Coulomb force. They behave like a single particle and are created when an electron is excited to a higher energy state, leaving behind a positively charged hole in the lower energy state.
2. How do excitons move?
Excitons move through a material by the opposite movements of the electron and hole. The electron moves in the direction of the applied electric field, while the hole moves in the opposite direction.
3. What is the significance of excitons in materials?
Excitons play a crucial role in the optical and electronic properties of materials. They can transfer energy and carry electrical charge, making them important for applications in optoelectronics, photovoltaics, and light-emitting devices.
4. How are excitons created and destroyed?
Excitons can be created through various processes, such as absorption of light or injection of electrons and holes. They can be destroyed through recombination, where the electron and hole recombine and release energy in the form of light or heat.
5. Can excitons be manipulated?
Yes, excitons can be manipulated by controlling the properties of the material they are in, such as the strength of the electric field or the presence of impurities. This allows for the tuning of excitons' properties, which is important for designing new materials with desired functionalities.
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