Heavy/Light Holes, what of Electrons?

In summary, the valence band in some materials has sub-bands known as heavy and light holes with different effective masses. The reasons for this lifting of degeneracy are either anisotropy in the crystal or a coupling effect. However, in the conduction band, there are no sub-bands of heavy and light electrons. This is because the majority of electrons in various bands are quasiparticles with different effective masses than the bare electrons. Some materials, such as ruthenates superconductors, have effective masses hundreds of times the bare mass. This is known as the "heavy-fermion" effect. The separation between the heavy and light hole bands is due to the different angular momentum states of the outmost
  • #1
jeffbarrington
24
1
So some materials have sub-bands of the valence band, known as heavy and light holes (they have different curvature so different effective masses, this I understand). Sources seem to give different reasons for this, either because of anistropy in the crystal or some sort of coupling effect, but suffice to say the degeneracy is lifted, somehow.

What about the conduction bands? I have never seen anybody talk about 'heavy' or 'light' electrons (save for people talking about muons or hypothesising about some less-massive version of an electron - but that's particle physics and completely different).

Why aren't there sub-bands of the conduction band? Or can there be such a thing, and I just haven't looked hard enough?
 
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  • #2
Most of the electrons in various bands are quasiparticles, and they certainly have different effective masses than the bare electrons. In fact, some of the ruthenates superconductors have effective masses hundreds of times the bare mass. They are not called “heavy-fermion superconductors” for nothing.Zz.
 
  • #3
ZapperZ said:
Most of the electrons in various bands are quasiparticles, and they certainly have different effective masses than the bare electrons. In fact, some of the ruthenates superconductors have effective masses hundreds of times the bare mass. They are not called “heavy-fermion superconductors” for nothing.Zz.

Does that mean to say there is a lifting of degeneracy though? I know the effective mass of an electron is different to an actual electron mass, but is there an analogy in the conduction band to the degeneracy lifting in the valence band? I suppose there is an infinity of bands at ever higher energies, but from my limited understanding the separation between the 'heavy hole' and 'light hole' bands is something different.

Thanks
 
  • #4
jeffbarrington said:
Does that mean to say there is a lifting of degeneracy though? I know the effective mass of an electron is different to an actual electron mass, but is there an analogy in the conduction band to the degeneracy lifting in the valence band? I suppose there is an infinity of bands at ever higher energies, but from my limited understanding the separation between the 'heavy hole' and 'light hole' bands is something different.

Thanks

Is this due to "degeneracy" or due to those holes being in different bands?

In those "spaghetti" band structures, one can see that, even at the same binding energy, different bands may have different curvatures, resulting in different effective mass. I don't consider this to be an issue of degeneracy.

Zz.
 
  • #5
It would help to know your background. It seems to me that you are coming from a semiconductor physics background (or learning from a book with a focus on semiconductor physics) because this question often arises in the context of semiconductors, especially direct gap III-V semiconductors such as the "standard" material GaAs.

If you have a look at the states, which make up the valence band, you will find that for most III-V semiconductors the outmost valence band electrons can be traced back to p-type orbitals, which already sets the angular momentum you need to consider (L=1). Considering total angular momentum (J=L+S), this means that you have two possibilities for total angular momentum: |J|=3/2 and |J|=1/2. The latter carriers will form the spin-off band. It is shifted away from the other bands due to spin-orbit effects. The ones with |J|=3/2 have 4 possible J_z components: +/- 1/2 and +/- 3/2. The states with z-projection 3/2 form the heavy hole band. The ones with z-projection 1/2 form the light holes. If you have a look at the conduction band, you will find that the underlying electron states are primarily s-type for most semiconductors (L=0). This means that you have only one possible value for |J| and therefore no heavy/light/split-off conduction band states. This would be different if you had a material, where the conduction band is mostly p-like (or some other higher angular momentum), but for standard semiconductors, you will only find few exceptions to the p-like valence band/s-like conduction band scheme.

Considering all materials out there, this configuration is of course by far not the only one possible, but many semiconductor textbooks do not even include other cases - this is why I am asking about your background.
 
  • #6
Cthugha said:
It would help to know your background. It seems to me that you are coming from a semiconductor physics background (or learning from a book with a focus on semiconductor physics) because this question often arises in the context of semiconductors, especially direct gap III-V semiconductors such as the "standard" material GaAs.

If you have a look at the states, which make up the valence band, you will find that for most III-V semiconductors the outmost valence band electrons can be traced back to p-type orbitals, which already sets the angular momentum you need to consider (L=1). Considering total angular momentum (J=L+S), this means that you have two possibilities for total angular momentum: |J|=3/2 and |J|=1/2. The latter carriers will form the spin-off band. It is shifted away from the other bands due to spin-orbit effects. The ones with |J|=3/2 have 4 possible J_z components: +/- 1/2 and +/- 3/2. The states with z-projection 3/2 form the heavy hole band. The ones with z-projection 1/2 form the light holes. If you have a look at the conduction band, you will find that the underlying electron states are primarily s-type for most semiconductors (L=0). This means that you have only one possible value for |J| and therefore no heavy/light/split-off conduction band states. This would be different if you had a material, where the conduction band is mostly p-like (or some other higher angular momentum), but for standard semiconductors, you will only find few exceptions to the p-like valence band/s-like conduction band scheme.

Considering all materials out there, this configuration is of course by far not the only one possible, but many semiconductor textbooks do not even include other cases - this is why I am asking about your background.

Ah thanks, I began to suspect I might have been learning about GaAs specifically. I've done a year of undergraduate condensed matter physics plus almost another year, and this heavy/light hole stuff came up when we started photonics.
 

1. What are heavy and light holes in a semiconductor?

Heavy and light holes refer to the two types of valence band states in a semiconductor. The valence band is the highest energy band that electrons can occupy in a solid material. Heavy holes have a larger effective mass and lower mobility, while light holes have a smaller effective mass and higher mobility.

2. How do heavy and light holes affect the electronic properties of a material?

The presence of heavy and light holes in a semiconductor affects its electrical conductivity and optical properties. This is because the effective mass of the holes determines how easily they can move and contribute to electrical and optical processes. In general, light holes are more mobile and contribute more to electrical conductivity and optical absorption than heavy holes.

3. What is the significance of heavy and light holes in optoelectronics?

In optoelectronics, the presence of heavy and light holes in a material is crucial for its performance. Light holes are responsible for the absorption of light and generation of electron-hole pairs, which are essential for photovoltaic devices. On the other hand, heavy holes play a crucial role in the transport of charges and contribute to the electric current in optoelectronic devices.

4. Can heavy and light holes be manipulated in a material?

Yes, heavy and light holes can be manipulated in a material through various techniques such as strain engineering, quantum confinement, and doping. These techniques can alter the energy levels and effective masses of the holes, thereby changing their electronic properties. This manipulation can be used to improve the performance of optoelectronic devices.

5. How are heavy and light holes related to the bandgap of a material?

The bandgap of a material is determined by the energy difference between the valence band and the conduction band. Heavy and light holes are part of the valence band and can contribute to the bandgap. In some cases, the presence of heavy and light holes can lead to a split in the valence band, resulting in a direct bandgap material, which is desirable for optoelectronic applications.

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