Valley Degeneracy of Semiconductor Band Systems

[Quasi Particle]

Hiya,

trying to learn something about the band system of semiconductors, I found that the conduction band is degenerated at the minimum by valley degeneracy.

Do you know where this comes from? In how far is it dependent of the form of the band gap (i.e. direct, indirect)?

I would be very glad for any explanations or references!

 

[Dr Transport]

The conduction band is doubly degenerate because there are 2 electrons. Never heard of valley degeneracy before.

 

[Pieter Kuiper]

If the minimum of the conduction band has a wave vector that is not at the center of the Brillouin zone (at Gamma, k=0), there will be several valleys of minimal energy in different but equivalent k-directions.

Maybe this is called valley degeneracy.

 

[QMrocks]

just consider the case of bulk onductor like Si. The valley minima along delta direction which in the momentum space is denoted by the direction vector [0,0,1] [0,1,0] etc. There are six such possible direction, resulting in a valley degeneracy of six.

 

[Quasi Particle]

Hm, I was thinking too one-dimensionally. Cheers.

But wait. Why are there only different valleys (or directions) if the band gap is indirect?
There is no valley degeneracy in GaAs, and it has the same crystal structure as Si except that it has a basis of two different atoms and Si a basis of two equal atoms.
There seems to be a dependency from the width of the band gap. If it is much greater than the thermal energy, the semiconductor is called "not degenerated". I don't see how that fits in.

@ Dr. Transport: Do you mean spin degeneracy? I think that occurs at the valence band (?)

 

[QMrocks]

Hm, I was thinking too one-dimensionally. Cheers.
But wait. Why are there only different valleys (or directions) if the band gap is indirect?
There is no valley degeneracy in GaAs, and it has the same crystal structure as Si except that it has a basis of two different atoms and Si a basis of two equal atoms.
There seems to be a dependency from the width of the band gap. If it is much greater than the thermal energy, the semiconductor is called "not degenerated". I don't see how that fits in.
@ Dr. Transport: Do you mean spin degeneracy? I think that occurs at the valence band (?)

There are only degeneracy for indirect semicon because the conduction valley minima can be allowed to have minima in k-space which are symmetrically the same. Direct bandgap means the minima is at [0,0,0] and there is no other accompanied valley minima.

Except for the case of valence band where it is triply degenrate due to symmetry. But when one consider spin orbit coupling, it reduces the degeneracy to 2+1. This is different from the 2-fold spin degenracy in absence of magnetic field as pointed out by Dr transport.

You may want to elaborate more on the bandgap vs degeneracy thing.. i.e. the context.

 

[Pieter Kuiper]

There seems to be a dependency from the width of the band gap. If it is much greater than the thermal energy, the semiconductor is called "not degenerated". I don't see how that fits in.

It does not fit in, because it is something very different.

A gas of fermions is called degenerate when the Fermi energy is smaller than the thermal energy. This is the case in metals.

In semiconductors, the gap is usually much larger than the thermal energy. The electron gas in the conduction band can then be treated with Boltzmann statistics.

 

[Dr Transport]

Spin degenereracy happens when taking the spin-orbit band into account, each state will have 2 states, a spin up +1/2 and a spin down -1/2 in the conduction band. The valence bands are a whole different game.
Si and GaAs have different symmetry properties, Si has the symmetry group $$O_{h}$$ which is has 48-fold symmnetry at the center of the Brilloun zone. GaAs transforms as $$T_{d}$$ which is 24-fold symmentric. They may be cubic, but after that the similarities end. The $$T_{d}$$ is a sub-set of the $$O_{h}$$ group.
The conduction band minima for silicon is not at the center of the zone as it is for GaAs, but out along the $$<100>$$ direction, thus reducing the symmetry of the conduction band even further to 12-fold symmetry, this is why there are 6 ellipsoids in the band structure. The conduction band minima is at the center of the Brillouin zone in GaAs, it has only a doubly degenerate s-state and graphically the band will be a single ellipsoid (each spin has its own, but they have the same energy and they overlap completely). QMrocks and Pieter Kuiper said the same thing a little more succinctly.

In the valence bands, if we ignore spin and the spin-orbit interactino for the moment, we have p-states, which are triply-degenerate. Include spin and the spin-orbit interation, we get two sets of states $$p_{3/2}$$ (4-states) and $$p_{1/2}$$ (2-states). At the center of the zone all the $$p_{3/2}$$ are degenerate, and the $$p_{1/2}$$ is separated from the otehr states by the spin-orbit splitting.

I did a lot of work in Si, Ge, GaAs amongst a bunch of other more exotic semi-conductor band structures for my dissertation so this is bringing back some memories of grad school.

 

[Quasi Particle]

So I seem to have mixed up the 2fold degeneracy of the conduction band with the 3fold (2fold) degeneracy of the valence band.

Thank you for your answers, it all makes more sense now.