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I am having trouble understanding the physics of band gap insulators.

Usually in undergrad solid state physics one looks at non-interacting electrons in a periodic potential, with no disorder. Then, if the chemical potential lies in the gap between two bands, the material is insulating. At least in this derivation, the individual electronic wavefunctions composing the bands are not localized.

However, when talking about insulators, people often think about localized electrons.

Do the electronic wavefunctions become localized in band gap insulators?

If they are, is it because of interactions? I was thinking that perhaps, since screening is not effective in insulators, the role of interactions is increased, and therefore perhaps the entire non-interacting, single-particle picture used to construct the band structure breaks down. Similarly, an impurity potential will not be screened and could localize the states. So which is it?

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The deep insight of Anderson is that the difference between insulators and conductors is not the energy spectrum. In fact the entire picture we are taught in introductory courses is highly misleading. [Note: Everything I am going to talk about will be about single particle effects, so no interaction.]

First lets just remember the introductory picture. We have a perfect crystal, so we get energy bands. We fill those bands up with electrons. In the case when a band is partially filled we get a conductor. In the case when all of our bands are completely occupied, so that the Fermi level lies in the gap, we get an insulator.

Now that problems: finite conductivity is entirely dependent on impurities. In the absence of impurities momentum is completely conserved. If I give the carriers any momentum, they will never lose it. Therefore a finite current can never dissipate, which is the same as saying the resistance is zero. Since there will all always be some carriers at any non-zero temperature, in the absence of impurities all materials will be "perfect conductors".

So it is clear that to make any sense we need to add impurities. However if we add impurities the nice energy band picture disappears. Since we just added random stuff to our Hamiltonian there is no reason we shouldn't be able to to find a state of any energy if we look hard enough. Obviously there will be more states in what used to be the bands, but there will also be states in the gap. In short the bands will blur together.

But if the bands blur together then there is no longer any notion of a gap - so what could possibly separate insulators and conductors? It is not the electronic energy spectrum, it is the electronic wavefunction themselves. Since there is no longer translational symmetry these are not restricted to the Bloch form. There are two main possibilities:

1) The wavefunctions near the Fermi level are extended, i.e. their magnitude is roughly constant over the entire system, like a plane wave. This is a conductor.

2) The wavefunctions near the Fermi level are localized, i.e. their magnitude decays roughly exponentially as you go out from some point. This is an insulator.

This is what actually distinguishes insulators and conductors. Going back to the band gap classification of materials - why does it basically work? The reason is if one adds disorder to a perfect crystal, the states that are added in the gap and near the band edges are usually localized states, so thinking about the gaps leads to the correct answer. But this is not the direct physical mechanism.

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Interesting. I didn't downvote but I am curious if there are physics problems with this answer. Seriously downvoters: leave a comment explaining. @Bebop, could you cite any good literature on this? Solid state is not my forte so I only followed you up to this: "there is no reason we shouldn't be able to to find a state of any energy if we look hard enough." Of course impurities distort the bands, but it's not obvious to me that they blur them together in any serious way. Can't density of states/spectral functions be directly measured anyway? – Michael Brown Aug 18 '13 at 4:32
@MichaelBrown: It is clearest coming from the insulating side - think of disorder as providing all these potential wells which have localized states. If your system is infinite and your disorder is not tuned in some weird way, then you will be able to find a potential somewhere hosting any energy level you want. When I couple the wells together the energy levels hybridize, but this won't change the fact in an infinite system I can find a state that with any energy I want. – BebopButUnsteady Aug 20 '13 at 16:04
From a more technical view, a random Gaussian potential just introduces a finite lifetime for all states. A state with a finite lifetime can be measured at any energy. Philosophically, it is another manifestation of "Everything which is not forbidden is mandatory" - Once there is disorder there is no conserved momentum so your system is just a big box of states, and there is no reason to expect a gap. Experimentally, states in the gap are measured all the time, sometimes they are called band tails, or impurity bands. Here is an RMP – BebopButUnsteady Aug 20 '13 at 16:23
As to the downvote, I've also always been grumpy that this, of all my answers, has net downvotes. This is just Anderson localization,, so it is hard to imagine what physics problems there are. I'm not sure what other references I can add. – BebopButUnsteady Aug 20 '13 at 16:42
Thanks. That clears it up. I've done what I can about the downvote. :) – Michael Brown Aug 20 '13 at 23:48

No, the electrons are not localized. Insulation is an effect coming from energy/bandstructure properties. Kohn-Sham orbitals (the orbitals in a bandstructure) are in general delocalized.

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Thanks for your reply, but I'm a little surprised at how confident you are. What do you say about this ? – jjj Apr 8 '13 at 11:29
I can't read through all of it, but the first sentence in the conclusion references a "sum of disconnected functions", not localized. Also, Wannier orbitals are mentioned, which are localized, but not Kohn-Sham orbitals. Fig. 5 has localized electrons, but they are on a ring, not an infinite crystal. Thus, I can't see a reference to localized Kohn-Sham orbitals. Maybe I'm missing something :) – Rafael Reiter Apr 8 '13 at 16:58

I think you don't 100% understand the "simple case" of a perfect crystal without electron-electron or electron-phonon interactions.

Let's say this crystal has full bands, a full valence band and empty conduction band. Say there are N electrons in the valence band (N is some huge number), one for each of the N valence-band states. In linear algebra terms, the electronic states in the valence band form an N-dimensional space of kets. This space, like any space in linear algebra, has infinitely many different bases. It has a basis of the N bloch states, which are delocalized, and it also has, say, the basis of N Wannier orbitals which are all localized.

You can say "There's an electron in each delocalized bloch state of the valence band". Yes, you're right. I can say "There's an electron in each localized wannier state of the valence band." I'm right too. Electrons are indistinguishable, it's meaningless to assign individual electrons to individual states in this situation, and to say whether or not they're localized.

Therefore, the material itself, in its perfectly insulating state, does not reveal to us whether it makes sense to think of electrons as localized or not. On the other hand, if there is (say) an electron in the conduction band, you can look at whether or not it's localized.

Insulators like sapphire are usually described as having localized electrons because when current is moving through them, it is usually via electrons which happen to be occupying localized wavefunctions during the process of moving. It's not because Bloch's theorem doesn't apply. (Although it might not apply.) They may have some current due to electrons occupying delocalized states too, but it's usually a much smaller contributor to the current than electrons occupying localized states during the course of their motion (hopping / polaron / anderson-localized, whatever).

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