Physics, asked by grraj3196, 1 year ago

Why does a degenerate band create a non-analytic band?

Answers

Answered by brainlyboytopper
0
I'm reading a paper (Rev. Mod. Phys. 84, 1419 (2012) - Maximally localized Wannier functions: Theory and applications p 1423) and I'm trying to understand a certain part. They show this figure:


Answered by Anonymous
0
When used in reference to a semiconductor, the term "degenerate" means that it is doped so much that the material has metallic properties such as a Fermi surface and high electrical conductivity.

To be strictly correct, it is more appropiate to say that the semiconductor "has degenerate electron statistics" when said statistics are only appropiately described by a Fermi-Dirac distribution at thermal equilibrium, which is the case when the semiconductor is highly doped. However, the phrase "degenerate semiconductor" is considered acceptable shorthand for the term "semiconductor with degenerate electron statistics".

Evidently, this concept of degeneracy is different from the degeneracy of energy states, in which the term is also often used in solid-state physics. However, the degeneracy of the energy of the eigenstates of a Hamiltonian is just an instance of a broader mathematical concept of degeneracy, which applies when multiple members of a set share a common characteristic, such as the value of their energy. Off the top of my head, a classical system of coupled harmonic oscillators may have degenerate oscillation modes, sharing a common frequency. Propagating electromagnetic modes in a waveguide can be said to be degenerate if they have the same propagation constant.

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