# How does band structure lead to metals/insulators?

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1. Oct 22, 2015

### NCD135

I almost "get" it, but not quite. Can someone help?

Here is my current understanding.

Say you solve the Kronig-Penny model. You find that you get bands of N closely spaced energy levels, with large gaps between them.

Under normal conditions, electrons fill the band to some level (let's say we have a metal, so the electrons only fill part of the band). Half of the electrons have positive wavevector k, while half have negative wavevector -k. Thus, there is no current. Now, if we apply a field, at any given point in space, all of the energy levels shift up or down, according to the potential at that point. Here is where it gets hazy. Let's say we are in 1D, and the electrostatic potential is higher to the left than it is to the right, due to the applied field. Electrons moving to the left can lower their energy by moving to the right. Therefore, if we get a thermal fluctuation (technically the energy levels in the band are still discrete), an electron moving to the left, near the Fermi level, can reduce its energy by moving up into a higher kinetic energy state and moving to the right, thereby lowering its potential energy. Is this accurate? Is there a better way to explain this, because I still struggle with this explanation, which may be completely wrong.

In the case of an insulator, none of this can happen, because there will be no thermal fluctuations larger than the gap

2. Oct 22, 2015

### Simon Bridge

Welcome to PF,
Metals have overlapping valence and conduction bands... insulators have a large band gap.

3. Oct 22, 2015

### NCD135

Hi Simon,

I understand that much. My title might be misleading, but I hope my post was a little more clear about that! My question is much more subtle.

In insulators, the bands do not overlap and the gap is so large no thermal fluctuation can allow move to higher kinetic energy states, allowing them to conduct.

But in metals, we still have discrete energy levels (technically). At literally zero temperature, would we even have conduction? Aren't thermal fluctuations literally the only thing that can move an electron from one state to an excited state? Otherwise, wouldn't we be able to apply a huge field and get conduction in an insulator? And I struggle to picture any of this spatially. There is no "energy barrier" for electrons to overcome to move to the state with lower potential. There is just the sort of quantum energy level availability argument, which I guess I could sort of picture as a barrier.

4. Oct 23, 2015

### Simon Bridge

Yes. There are electrons in the conduction band.

No. And not relevant to the discussion.

you can do this. Its just that, usually, you reach the material breakdown potential first.
In between you get semi-conductors... conduction and valence band do not overlap but the gap is "small" enough that it is easy to promote electrons to the conduction band. There are lots of ways to do this, not just thermal.

This is not the situation with conductors. In conductors, some electrons which would otherwise be valence electrons occupy the conduction band in the ground state. Chemically they are shared to the whole crystal rather than stuck in individual atoms. There is no need to excite electrons to a conduction band state because they are already there.

The band gap is everywhere in the material, a spacially specific energy barrier is a different thing arrived at by having some locations with different materials. The band gap is a range of energies in which the probability of finding an electron is zero. It is not, and should not be pictured as, a barrier or a spacial regeon.

5. Oct 23, 2015

### ZapperZ

Staff Emeritus
I need to intervene here because I don't think the discussion is going in the right direction (DrDu, please interject if you think it is necessary).

First of all, there is no "overlap" of bands when an insulator becomes a conductor (there is a caveat here which I'll explain later). In other words, the conduction band doesn't overlap the valence band and then, it becomes a conductor. In band structure picture, a metal is when there is at least one band that crosses the Fermi energy. When you dope an insulator to become a metal, within the naive rigid-band model, all you're doing is moving the location of the Fermi energy with respect to the band, and eventually, it crosses either the top of the valence band or the bottom of the conduction band. Again, this is the naive picture, but it demonstrates the fact that you don't have overlapping bands, at least within the band conductors/insulators model. This is the description for most of the common materials/standard metals/standard insulators that we deal with.

Now, it doesn't mean that the "overlapping" of these bands do not occur. This overlapping of the empty conduction band and fulled valence band is what is described for a Mott-Hubbard insulator. This occurs when you change the "U" in the Hubbard Hamiltonian, which signifies the onsite repulsion/bandwidth. So in this case, the "gap" between the valence and conduction band shrinks and eventually, the two bands overlap.

Another scenario is the creation of Zhang-Rice singlet right at the Fermi level when you start changing the insulator. In this picture, as you change the insulator, you are creating a new state/band right at the Fermi level.

The study of metal-insulator transition isn't trivial. So we need to be careful how we explain this because different materials have different mechanism for such evolution. However, the typical, common description that we often start with is the band insulator/conductor that I described in the beginning.

Zz.

6. Oct 23, 2015

### NCD135

Okay, so I have some sort of fundamental misunderstanding here. I was taught that, at any given point in space, all an electric field can do is shift all of the energy levels by a constant. Even within a band, there are still technically discrete allowed energies. Therefore, absent small thermal fluctuations, there would be no way to promote an electron, even within a band, to a higher state.

Which part of this is wrong?

7. Oct 23, 2015

### DrDu

In fact, the two statements aren't contradictory as the overlapping of bands may lower the Fermi level.
This is what happens in earth alkali metals like Mg. The band originating from the s states in the isolated atom overlaps in the metal with the band originating in the p states. Hence electrons from the full s band can get to the lower states of the p band with the effect that both bands are only partially filled. Of course it is a matter of convention whether to speak of one band or of two overlapping bands here. Nevertheless this explanation is often found in text books. Clearly, it is not necessary as for example in the case of alkali metals like Na, the s band is only partially filled even at infinite separation of the atoms, so there is no overlapp of bands necessary to describe the conductivity of alkali metals.

8. Oct 23, 2015

### ZapperZ

Staff Emeritus
I'm not sure we're addressing the same thing.

I'm fully aware that, in band structure calculations, one does consider the different orbital overlaps, pretty much similar to the tight-binding approximation. However, this is not what I was addressing. I was tackling Simon Bridge's claim that "... Metals have overlapping valence and conduction bands..." This is not necessarily true in band metals and insulators. Once these conduction and valence bands are established, you do not get metals by making the conduction and valence bands to overlap. This happens only in Mott insulators under certain situations.

Also note that I specifically restricted myself to the rigid-band scenario, where doping, etc. does not change the band structure.

Zz.

9. Oct 23, 2015

### NCD135

Okay, so I have some sort of fundamental misunderstanding here. I was taught that, at any given point in space, all an electric field can do is shift all of the energy levels by a constant. Even within a band, there are still technically discrete allowed energies. Therefore, absent small thermal fluctuations, there would be no way to promote an electron, even within a band, to a higher state.

Which part of this is wrong? I just don't get it...

10. Oct 23, 2015

### ZapperZ

Staff Emeritus
Can you cite a source that actually says this?

A conduction band has delocalized electrons. It does not belong to any particular site. Look at the Bloch wavefunction. Does it have a well-defined position?

Zz.

11. Oct 23, 2015

### NCD135

12. Oct 23, 2015

### ZapperZ

Staff Emeritus
13. Oct 23, 2015

### NCD135

Hahaha. I'm not that crazy. I've taken classes/I read Kittel. Sometimes when I have a question like the above, my friends can't answer either, so I turn to the forums ;)

Can you help with this particular confusion? I can "derive" band theory (e.g. Nearly free electron model or Kronig-Penny), but I can still get held up by basic questions like the ones above! It's unbelievably frustrating! If all the states shift due to the applied field, how is it that electrons can move between states if not via thermal energy? And as you point out, the Bloch states don't have a well defined position--the electrons could equally well be at any site. So how would you, mathematically or even verbally, describe the situation if not how I did above (the potential is lower in one direction than the other, so electrons will transition into the states that will allow them to move there, if those transitions are possible--what defines this other than thermal energies?)

Thanks!

14. Oct 24, 2015

### ZapperZ

Staff Emeritus
But you should never use another forum post as a "reference source" to counter a point. I personally find it frustrating that I have to "correct" something that someone read from a dubious source, or even Wikipedia! You used Kittel. So try to find the same thing there!

I actually don't quite understand what exactly your problem is. Are you trying to figure out why a metal can still conduct at 0 K, when there's no thermal fluctuation?

The problem here, I think, is that you are still thinking that there are some "discrete" states here, and that an electron in the conduction band needs to jump up above the Fermi energy at 0K for it to "move". If this is exactly the issue you are having, let's tackle this one step at a time.

First of all, what made you think that there are discrete states in the conduction band? If we remove that silly post that you read in another part of PF, do you have ANY valid sources and references to indicate such a thing? When you solved for the Kronig-Penny model, did you look at the dispersion curve? Are there any discrete states in each band? If there is any, show me.

Zz.

15. Oct 24, 2015

### NCD135

I would answer yes to both of those questions. Griffith's Quantum Mechanics writes it as: "Evidently there are N states in each band, so closely spaced that for most purposes we can regard them as a continuum."

You have the boundary condition that ψ(x+Na) = ψ(x), which quantizes the crystal momentum, right? In other words, it makes the RHS of this equation: (P/Ka)Sin Ka + cos Ka = cos ka, not continuous, but a set of discrete values, since k can't be just anything. Or do we ignore the quantization because that's a made up constraint?

16. Oct 24, 2015

### NCD135

PS I'm not trying to counter any of your points! I'm just trying to explain my confusion, because I know that you are correct, and I want to understand why!

17. Oct 24, 2015

### ZapperZ

Staff Emeritus
But here's your dilemma. N is large, but you don't want to consider them to be so big that it is continuous. Yet, in solving your equation, you have explicitly assumed that the crystal is infinitely long, meaning that the bulk boundary condition is so far way, it may as well not exist. So at what point to do you select which one to choose and which one to discard?

The reason why we do what we do is that (i) because it is simple; (ii) because it matches very well with our measurement.

Also note that even at T = 0 K, while you may not have, in principle, any thermal fluctuations, you still have quantum fluctuations, if we want to head down that not-so-easy little road.

Zz.

18. Oct 24, 2015

### NCD135

Thanks! This makes sense to me. Now I see the contradiction and can believe that it is more accurate to view the band as "literally continuous" than "technically discrete but effectively continuous."

19. Oct 25, 2015

### DrDu

That'a all ok. With Born - von Karman boundary conditions you consider a conductor with the topology of a torus. Or, in one dimension, of a ring. This is quite natural for the consideration of conductivity as you need a closed circuit. You can produce a driving electric field in a ring with a changing magnetic flux in the middle. The discrete states with given k already carry current, but as long as the band is filled, the sum of the current invariably sum up to 0.

20. Oct 25, 2015

### Daz

It seems to me that this discussion hasn’t really addressed the OP’s fundamental confusion. For an electron in a solid to be moving it must have kinetic energy and therefore posses more energy than it would have in equilibrium. i.e. It is in a higher energy state and the OP’s difficulty could be summed up as “How did it get there?” Not unreasonably, the OP proposes thermal excitation as the mechanism, but that’s not right.

Imagine what a free electron would do in an electric field — it would simply accelerate. Unless something stops it, it will simply continue getting faster and faster ad infinitum. The free electron acquires energy from the field.

In the semi-classical treatment of conduction in solids we assume that the electrons in a solid behave in a similar way to free electrons but with a modified E-vs-k diagram. It’s that E(k) relationship that encompasses all of the crystal lattice effects and as such it’s periodic in k.

It doesn’t really matter whether you view the states as discrete but finely spaced, which will be the case for any real, finite solid or a continuum. The latter only arises from the consideration of Bloch’s theorem in the idealised case of an infinite, perfect crystal. It’s the external applied field that the electrons are reacting to, not thermal excitations.