Why can not the density of states of a metal at Fermi level be zero? Thanks!
Any reason you would expect it to be zero?
I have no reason, but what does it happen when the gradient of the energy diverges logarithmically at Fermi level, which implies that the density of states at Fermi level will be zero?
Thanks a lot for answering nasu!
When does this happen in a realistic energy band structure?
I want to know because when you consider a model system of interacting electron gas in the so-called jellium model, the thermodynamic potential to first order of interaction, it turns out, that stopping at first order leads to the density of states at Fermi level vanishes. So, it argues that no such behavior is observed in metals, due to that, it needs to go beyond first order. But, which are the physical implications of having the density of states at the Fermi level vanishes?
Thanks for your answer.
Looking at this in terms of the single-particle picture and without considering exceptions (roughly):
Having a non-vanishing density of states (DOS) at the Fermi level means that at a finite temperature there will be a possibility for an electron to change its state from the one just below the Fermi level to the one just above the Fermi level at virtually no energy expense - in other words, there will be free states available to the electron at an arbitrarily small energy difference. Hence the electrons will be able to conduct at any (even very small) external electric fields and the system is a metal.
If, on the other hand, the density of states is zero at the Fermi level the electrons (at low enough temperatures) will not be able to reach any new states and will therefore not conduct. Hence the system is an insulator, not a metal. Roughly. (Of course there could be exceptions if the DOS is zero only at a single point, where the Fermi level just happens to lie, but non-zero [large enough] in the neighbourhood of that point ...)
More precisely: using Fermi's golden rule you find that the probability for a transition to occur is proportional to the DOS (of the non-occupied states, in the single particle picture, or just the DOS of the multi-particle states) at the final state of the transition "DOS(E_final)". Small perturbations (at least approximately and at large enough times) preserve energy, so that "E_final ~ E_Fermi" and therefore the probability for a process to occur (to first order) is proportional to "DOS(E_Fermi)". If that density of states at the Fermi level is zero the process is suppressed (~forbidden to first order). If you think of the external electric field as the perturbation in question this means that a small enough electric field will not change the state of the system (approximately and to first order), hence cause no electrical conduction, hence the system is an insulator (at low enough temperatures) and not a metal.
Something like that.
(I would have to think a bit more about transitions where the DOS is zero only at a single energy, which also happens to be the Fermi level. An exception could be graphene, when the Fermi level lies exactly in the middle of the Dirac cone and the DOS there is exactly zero (but not in the neighbourhood, where it varies linearly with energy), but the system can still conduct due to the presence of nearby-enough states on the Dirac cone ... Wikipedia actually calls graphene a "zero-gap semiconductor" [not a traditional metal]. :) I would have to think a bit more about jellium, though.)
Graphenes is special. It has an incredibly high mobility at room temperature because of the Dirac dispersion. The quasiparticles are pseudorelativistic massless Dirac fermions. However, the carrier density is very low.
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