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The questions is mostly in the title, but might expose another of my misunderstanding of the band structure of solids and how that leads to metals and insulators.

If we have a solid, and the fermi energy lies at the top of one of the bands, it will be an insulator, because there are no thermal fluctuations large enough to allow the electrons to move up to the next available state.

On the other hand, if the fermi level is in the middle of one of the bands, the solid will be a metal. If we apply a field, the electron can easily move up to a higher state within the band, and we will have conduction.

Here's my question: doesn't it take thermal fluctuations to "smear" the energy levels within a band? A band isn't really a continuous $\epsilon(k)$ spectrum; it's a series of discrete $\epsilon_k$ values. Does that not mean that at truly zero temperature, we would have the same (insulating) situation as above?

Thanks

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    $\begingroup$ (1) you never get to T=0K, (2) while perhaps 'discrete', those energy levels in the bands are ~1E-22 eV apart, so you have to be many orders of magnitude closer to T=0 than we have reached. $\endgroup$ – Jon Custer Oct 21 '15 at 14:40
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    $\begingroup$ And tunnelling effects would still allow relevant conduction for voltages that are not ridiculously small. $\endgroup$ – Sebastian Riese Oct 21 '15 at 14:47
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You are right. Perfect metal without interaction and impurity will not conduct direct electric current at zero temperature. It will do Bloch oscillation. However impurity scattering or thermal relaxation will destroy the Bloch oscillation, and lead to finite conductivity. With interaction, the metal may go superconducting at low temperature which is beyond the simple band theory consideration.

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    $\begingroup$ This is true, but I don't think it is quite what the OP is asking about. $\endgroup$ – Rococo Oct 24 '15 at 0:59
  • $\begingroup$ I do not think this answer is correct. Take an infinite perfect crystal at 0 K. As you say, in this case if one applies an electric field, there will be Bloch oscillations (so one would have an AC). However, if one now removes the electric field, the Bloch (quasi)electrons will be left with a non zero velocity that will persist forever (so infinite conductivity), unless the electric field is removed exactly when the Bloch electrons were having a vanishing velocity in their harmonic motion. A more realistic case with boundaries and impurities would prevent Bloch oscillations to occur even $\endgroup$ – thermomagnetic condensed boson Jan 15 at 10:08
  • $\begingroup$ at 0 K. Please comment on this. @Rococo I also invite you to provide your opinion. $\endgroup$ – thermomagnetic condensed boson Jan 15 at 10:08
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    $\begingroup$ @tttt Interesting observation. From a quick survey of the 4 condensed matter textbooks I have on hand, I found one (Ashcroft) that characterizes the perfect lattice as a perfect conductor, one (Marder, 16.2.1) that characterizes it as an insulator, and two that don't give a definitive statement. Clearly, one must be careful about how one defines the conductivity in this situation. I can't get into this in a comment but if you wanted to ask a separate question about this issue I would probably give an answer. $\endgroup$ – Rococo Jan 19 at 21:11
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    $\begingroup$ @tttt I think that, generally speaking, saying that something is an 'insulator' refers to the DC conductivity. It does not mean that the conductivity at every frequency vanishes. $\endgroup$ – Rococo Jan 19 at 23:46
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A metal can only conduct if it is part of a circuit. There need to be contacts with the medium that provides the carriers. This will add electrons or holes. These can propagate unhindered through the metal. So the metal is a conductor with zero resistance.

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I'm not sure if this is how experts think about the issue, but in general phases of matter are only rigorously defined in the so-called "thermodynamic limit"- that is, taking infinite volume while keeping the density fixed. When applied to a band structure, this results in the spectrum within a band being truly continuous, and there is no energy gap.

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