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All right. I guess we can get started. Welcome back. It's the 15th lecture of the condensed matter, of course.

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When last we left off, we were talking about trying to approximately solve the shortage equation for electrons in solids.

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In the method we were talking about using was the nearly free electron model, nearly free electrons.

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The general idea being that we're going to write our Hamiltonian as some bear,

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just regular non interacting electron plane wave Hamiltonian plus some potential.

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So the eigen states of this in a plane wave k h r squared k squared over two m just your regular electron Hamiltonian.

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But then here we have some periodic potential which we're going to eventually consider to be weak, hence the name nearly free weak potential.

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And the idea that we took from scattering theory is that the matrix element between some K and K prime from this periodic potential has to be zero.

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If K minus K prime is not an element of the reciprocal lattice and we'll call it this of g, f, k, prime minus k is an element of reciprocal lattice.

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So then we treated the the potential V as a weak perturbation and in second order perturbation theory, second order third.

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We found that the energy of the plane way of K starts out as the bear energy

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of the plane with k plus v sub zero zero for m mode of the potential lattice.

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An overall shift of all of the plane wave energies that's uninteresting.

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People sometimes just drop it all together because it's just a constant. But the interesting pieces are some overall reciprocal lattice vectors.

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These are B squared over energy of K minus energy, not of K plus g.

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So this is our second order perturbation theory expression.

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And the thing that one has to worry about is that the denominator can be either zero or close to zero and that happens when.

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So this diverges when we have K absolute K being equal to absolute k plus g so that the energy of can and

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plus g are the same and that is equal to the same condition as K and K plus g being on on busy boundary,

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boron zone boundary. So in that case, we have a problem.

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Our second order perturbation theory doesn't work because of this degeneracy.

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And what we're supposed to do is we're supposed to use degenerate perturbation theory of degenerate theory,

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resisting all urges to make jokes about the word degenerate, but probably haven't made hundreds of times before.

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The general rules of degenerate perturbation theory are that you should take the states that are causing you trouble.

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The states that are giving you the divergence separate those out from all the other states in the

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system and just diagnose the Hamiltonian with this within the space of those problematic states.

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So for us, our the states that are causing us trouble are a states K and K plus G for some particular G.

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So we'll write our trial wave function, actually. Maybe we'll be a little bit more general and assume, assume, k and K plus g assume.

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These are near a zone boundary near a boundary.

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So it may not be an actual divergence, it can just be a near divergence.

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It's still bad enough to cause us to not trust our second order perturbation theory anymore and require us to use degenerate perturbation theory,

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even though if it's not a you know, it's not a zero in the denominator to something close to zero in the denominator.

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The general scheme is to write a trial wave function, which is going to be some some of the problematic cats.

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So K and plus G may take some linear, some of these too.

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And in the usual way, we will try to find the best linear combination of these two cats in order to get our lowest energy state within this space.

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So we've done this a couple of times before.

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The way one does that is you write down a Hamiltonian, an effective Hamiltonian equation sum over age and Phi Alpha equals E Phi.

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And this should look familiar from what we've did when we solved the tight binding model for the covalent for the covalent bond here.

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This H&M is a two by two matrix where an an m are chosen from K and K plus G.

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This is how you how you do the general perturbation theory. You just restrict.

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The space of states to the space of degenerates, states who are causing trouble in the second order perturbation theory.

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All right. So the only thing we need to do now in order to follow this prescription is we need to write out the matrix elements.

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So, for example, we have K, H, K, that's one of the four matrix elements we can write that is k,

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h, not K plus k, v, k, since our Hamiltonian is made up of the H, not part in the V part.

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This part is easy. This is just E not of K and this part is V, not just the overall constant energy shift.

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The second subscript is zero because k minus K prime is zero.

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Similarly plus g, h k plus g is equal to E, not of k plus g plus v not again.

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The more interesting term is the term the off diagonal terms of the two by two matrix k plus g, h, k.

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So let's write out the pieces of this. We have first K plus g, h, not K, and the second piece is K plus g, v k.

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Well, the first part, this is easy because this would be H, not K is in Oregon, State of H not so we can just pull out an E,

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not sub K and get K plus g k, but this is zero because these are two orthogonal plane waves.

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So at zero the first term is zero. The second term is we've defined that as being the sub g,

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so we're allowed to scatter from K to K plus G via the scattering potential G and up way up at the top.

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Oops I scrolled it off the top. Yeah.

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There it is on the top that the sub g of came on is k prime is an element of g, so we call this thing the G.

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So so what we have now is also so this thing is going to be overall this is going to be the sub g and we can do the other last matrix element h,

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k plus g, which using the same argument is the sub minus G or the sub g star.

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It's complex conjugate of its of its partner up here. So now we have our Hamiltonian R two by two matrix we can write down which is e not of k plus

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v not v sub g here star here v sub g down here and e not tables g plus v not so far so good.

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Everyone happy with this? More or less this go by too quickly.

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Yeah. Happy. Okay, good. All right.

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So for homework problems at four, I think you're going to actually solve this in general for a K near the Bronson on boundary,

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which makes the algebra a little bit more complicated.

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But what we're going to do here is we're going to simplify our algebra a little bit and assume K is right on the brown zone boundary.

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Assume K and K plus G are on busy boundary, on busy boundary just to make the algebra a little bit easier.

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So that means that E not of K equals E not of k plus g at the same energy.

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So right on the degeneracy point, in which case this is a very simple matrix,

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has two elements which are the same along the diagonal and then complex conjugate matrices,

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elements on the off diagonal and the energies are then e not of k plus v not plus or minus absolute v.

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And that's her answer. So what is this? What does this mean?

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So maybe draw it here. So we have here's K, here's energy E and before we add the perturbation, we have a nice parabola.

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We have the usual. Drop this. So this is the usual parabola which is e not of k equals h where squared k squared over two m.

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So that's the free electron parabola. Then we can put down the periodic potential.

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It has some r on boundary for this.

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So it has a lattice constant,

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a spiralling zone boundaries at pi array and minus pi over six at more browns on boundaries at minus two pi over a and two pi over a here.

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So here at this point there's bronze on boundary minus pi over a and pi over.

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You can scatter between them by this reciprocal lattice vector g equals to pi over a can get you from minus primary to pi over a.

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So in second order of perturbation theory, you would discover that if you if you analyse this point here,

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at this point here, you would get a divergence in second order perturbation theory.

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And we would have to use degenerate perturbation theory in order to figure out what happens to these points.

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And what we just derived is, in fact,

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they split in energy in the way it's going to look is that this is going to shift down and then this will shift up a little bit so that,

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in fact, you open up a little gap between the things coming in from the left and the things coming in from right.

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And its size is too absolute big here.

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And the same thing happens over on this side like this.

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Opens up a gap of size to Fiji.

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And the same thing would happen up here where you can scatter from bronze on bounded to bronze on boundary by g equals four pi

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over a and here a gap would open up at the bronze on boundary here and the gap would open up with the bronze on boundary here.

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Okay. So since I've drawn that fairly badly, I was prepared for this and I have a slide.

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So here's the same picture and extended zones scheme. This is the it starts with the bear parabola, the the nice free electron parabola.

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And then when you add the periodic potential, the major effect of the periodic potential is near the bronze on boundary where it opens up this gap.

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You notice that the states inside the bronze zone boundary get pushed down in energy by

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a little bit and they get pushed down more if they're closer to the bronze on boundary.

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Because as you're close to the bronze and boundary, if you think in second order perturbation theory,

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that's where the divergence is getting worse and worse and worse. It doesn't actually diverge.

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But still, the perturbation is becoming more and more important as you get closer to the bronze on boundary.

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So the gap on things get pushed down more as you get close to the bronze and boundary.

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And then if you're just outside the bronze zone boundary, things get pushed up and you open up this little gap at the bronze on boundary.

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Same thing over here. Now we're perfectly entitled to view the same thing in reduced zone scheme where we take this piece of the

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spectrum and we shifted over by two pi over A So it's over here and we take this piece of this fixed spectrum.

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We shifted over by two pi over eight in the other direction until it's over here and we view everything within the first bronze zone.

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But now we have two bands within the first bronze zone. So we have a lowest band and we have a higher band.

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And this looks a little bit like what we found when we studied the tight binding model a couple weeks back.

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We had a four when we had two orbitals per unit cell.

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There were two bands at two values of of two iron states at each possible value of K, which is what we have here.

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And the structure looks more or less similar. We have a low band, a high band, a gap opening at the bronze zone boundary.

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However, we're really coming at this from a very complimentary direction.

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When we did a tight binding model, we started by taking a bunch of orbitals and then we weakly coupled them

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together and allowed the electrons to hop from one orbital to the next to next.

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Here we did the opposite. We started with a plane wave and we weakly perturbed it with a periodic potential.

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But either way you look at it, you get the same physics of separate bands gaps at the Baron's own boundary.

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If you look at it from the nearly free electron model, though, really this picture you can see in this picture,

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it's it's actually just a free electron parabola with small perturbations at at the bronze on boundary.

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So this parabola is sort of continued in the second band. So this this piece of the parabolas are reflected back and continue in the second band.

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They don't quite touch because of the the gap opening up, but the curvature of this band near the bottom is actually given by the free electron mass.

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So the thing that causes the curvature at the bottom of your parabola that's dependent on the free electron mass, not on anything else.

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So we really should view this this bottom of the band, which looks like a parabola as just being essentially free electrons.

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Okay, so a question that's fairly important at this point is maybe I'll put it over here is why the gap?

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Why the gap? Mind the gap.

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Why the gap? Okay. So in order to understand why it is that this gap is opening up at the bronze on boundary in a different language,

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it's useful to, to, to give an example.

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So examine a particular potential.

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I'll choose V of X and we're going to do this in one dimension. Actually, this is one dimensional.

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So I should have mentioned that this isn't one dimension. This is more this is a more general equation which hold in any dimension.

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We're going to do this example in one dimension also. So VAX will choose it to b to v twiddle cosine to pi over a times x.

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That's a nice periodic potential. Maybe I'll even draw that periodic potential.

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Everyone probably knows what it looks like, but I'll draw it anyway. So here's our axis x.

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Maybe we'll put a here minus a out here and a cosine potential.

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Looks like this. And looks like. This.

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So this is V of X and the lattice constant is a, the periodicity is a,

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and the reason I've chosen this particular simple cosine potential is because it has nice Fourier modes.

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V sub two pi over a equals v sub minus two pi over a equals v twiddle and all other all other v sub g are equal to zero.

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So it only has two four modes which are non zero and they're both given by V twiddle.

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All right. So now let's think about what this is going to do.

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What this can do is it can scatter by v sub g only for g is two pi over a.

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So in this picture it means it can scatter from this point to this point four by g.

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These are the two points in the general perturbation theory. They can mix with each other.

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This point of k equals minus pi over a and k close plus pi over a.

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So the scattering we have to worry about is from k equals minus pi over a and to k equals plus pi over a,

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being scattered back and forth by this potential.

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So it can scatter back and forth between those two way vectors, which are right on the Bruns on boundary.

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So in real space, if I think about x k representations of these of these cats,

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these things are either the I pi over a x and you know, the minus I pi over a x.

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I'll use that notation if we then diagonals this Hamiltonian in that in that's in this space.

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What we'll get is we'll get our two eigenvalues or our two eigen functions are the CY Plus

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will be the I pi over a x plus even minus I pi over a x and its energy e plus is is pi

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over a not plus v twiddle and the other I can state sine minus is either the i pi over x

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minus the minus I pi over a x and its energy e minus is e not pi over a minus b total.

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So one of these linear combinations of these of these plane waves is getting pushed up in energy.

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One of them is getting pushed down in energy by V-twin.

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Also, again, we have these two plane ways, one going left, one going right.

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They're getting mixed together by the by this by this scattering potential, by this periodic potential.

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And one of the two I can say is has been pushed up in energy.

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One of them has been pushed down in energy. Why is that? Well, let's see if we can figure that out.

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Let's take five plus square it so we get the probability density.

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So this is the density, the problem probability density prob density cy plus squared which is proposed passing all to cosine squared of pi over a

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x and sine minus squared will be proportional to science squared of pi over a x in this plot those on this same figure.

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So let's see cy plus looks like this cosine squared of the cy plus squared and then cy minus looks like this.

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Sy minus squared looks like that so they have their amplify their probability

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density is in different places along the in the positional space they're different

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places along the axis and because of this the CI plus squared CS I'm putting

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that in quotes c is mostly positive v v of x so you see where CI plus is great,

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where it's where it's very large. That's where V of x is positive.

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And when CI plus is small is where V of x is negative.

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So that means CI plus e plus is pushed up and energy is pushed up up because it's experiencing mostly the positive potential.

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Whereas in comparison CI minus squared sees mostly negative v v of x because CI

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minus is large where v of x happens to be negative and so it's pushed down.

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E minus is pushed down down in energy.

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So let's sort of think about this a little bit more carefully.

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So what's going on is we have this space which has a left going wave and all right, going wave.

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That's the space of the degenerate perturbation theory.

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And with the variational principle, what we're supposed to do is we're supposed to put them together in a way that will,

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you know, to get the ground state.

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We're supposed to put them together in a way that will minimise the total energy and the way you minimise the total energy to get the ground state

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as you put them together such that you get a sign minus cosine minus has its maximum where v of x is negative and that pushes down its energy.

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Therefore it's in it pushes down as much as possible.

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Therefore it's the ground state, eigen state and conversely, C plus has pushed up its energy as much as possible,

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as high as possible by putting its density where the potential is positive.

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So is that clear why that's going on? All right, good.

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So do you ever read the Examiner reports people read these reports even though notice they're really nasty?

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I think it's an unwritten requirement. The examining reports have to be really nasty.

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I've never been on the examination committee yet.

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One day my time will come and I'll probably read the rule, the unwritten rule, somewhere on some tablet that says you must be nasty.

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Okay, so there was a question that came up about this nearly free electron model came up about four years ago and it really threw a lot of people.

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So I'm going to go through the physics of what was involved in there, which isn't too much more complicated than what we did.

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But but it's something that, you know, a lot of people found rather unexpected.

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So I'm going to explain more or less what what was going on in that problem.

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So when you run into it, someday we'll trouble you. So we're going to consider a square two dimensional lattice.

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And I'm drawing reciprocal space. This is a reciprocal lattice. So a square 2D lattice, square 2D lattice.

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And the reciprocal space of a square lattice is also square.

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So I'm going to draw the bronze zone. So this is K equals zero here in the centre and the first bronze zone is this square here.

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And this point here in the corner is pie. Over a pie.

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Over a cake.

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Now, the first thing this question asked you to do is find out what happens when you add a period of potential.

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What happens to a point here on the bronze on boundary?

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Okay. So that's basically exactly what we just did. That point there, due to the periodic potential, it can mix in with this point over here.

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These two points are separated by a reciprocal lattice factor. This is a reciprocal lattice vector to pi over two, pi over eight.

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So these two can mix together via scattering. They have the same energy because they're the same distance from k equals zero.

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They have the same amplitude of their K. So.

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So this condition holds x equals e, k plus g.

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So they mixed together. And it's exactly the same calculation we just did.

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It's just, you know, it's really a two dimensional model, but we're just considering this cut through two dimensions.

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There will be a gap opening up at the bronze on boundary at this point.

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At this point. So that much is simple or should be simple. I hope it's exactly like what we did in two dimensions, in one dimension.

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But now in two dimensions. The thing that was complicated is what happens if you consider this point up here in the corner.

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So let's call this .0.1. So now what we have to do is we have to ask where did the where did the divergences come in second order perturbation theory?

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Well, this point, one can scatter by a reciprocal lattice vector to this point two,

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but it can also scatter by reciprocal lattice vector to this point three.

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And this is also a reciprocal lattice vector to this point four.

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And in fact, four can kind of scatter the three. Working together to two can scatter to three.

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They can all scatter to each other by reciprocal aspect is those vectors in case space are all reciprocal as vectors

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and they're all allowed to mix to each other and they all have the same energy before you add the perturbation.

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So when you do your degenerate perturbation theory, you'll find a bunch of these things, the bunch of these denominators that are all zero.

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If you're considering this point scattering here, there'll be a denominator zero.

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Scattering here, scattering here, all zero. So the way you handle this is you have to write your trial wave function.

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That includes trial. That includes all of these wave vectors, cy one, k one plus five, two, k two plus five, three, k three plus five for k four.

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All of these things have to be included in your trial wave function and then your Hamiltonian is then a four by four matrix,

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which includes all possibilities of how you scatter back and forth between all of those four points.

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Okay. So that was what was what was asked on the in the in the exam.

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And a lot of people got thrown by it, even though it's not really that much more complicated.

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Okay, good. Happy with that? All right.

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Let's do it. Let's do a real example here. So in three D, in real example, real life, we have things like FCC lattices.

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This is as we discussed a couple days ago, this is the branch zone of the FCC lattice.

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The gamma point is appointed k equals zero in the middle.

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The X point is that the barren zone boundary in the middle of this square face here in the direction the l point is in the bronze on boundary,

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in the middle of this hexagonal phase, for example. And the material we're going to think about is the material.

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Silicon Carbide. Silicon carbide.

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So actually, it's another important industrial material. It has this structure here is FCC with a basis.

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So the yellows are silicones and the and the blues are carbon.

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The silicon can be taken at 000 and the carbon at one quarter, one quarter on quarter.

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We've seen this structure before. This is a zinc blend structure. Zinc sulphide, gallium arsenide takes the same structure.

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Silicon carbide takes this structure as a very common structure. Okay, now.

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All right, so here let's see if we can we can understand how this picture,

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this complicated picture of all the eigen states for electrons can come out of nearly free electron models.

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So let's look down at the lowest energy, the lowest energy. Here we have the gamma point.

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That's a cake with zero. Gamma is another name for cable zero.

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And as you go out in the bronze zone towards the point X, you see this thing that looks kind of like a parabola.

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And if you back up that oops, back up back here.

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So it looks kind of like this parabola here. And that's exactly what you're seeing here.

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This is a free electron parabola coming out from K equals zero.

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And when you get to the bronze zone boundary, there's a gap opening up.

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And this is the second half of the parabola going up as back, back, back there.

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Here it is. So here's the second half of the parabola up here, just like the second half of the parabola is here.

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And he could go off in another direction.

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So going from Gamma out to the L point, again, you have a parabola and there's a slightly bigger gap in the L direction.

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And here's the second half of the parabola here. So it's basically free electron model.

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And if you measure the curvature down at the bottom of the band and your gamma equals zero, you get basically the mass of the free electron.

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It's not exactly the mass of the free electron, but it's it's pretty close to the mass of the free electron within 30% or 20% or something.

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So. So it really tells you that what's going on near the bottom there is basically free plane way physics just perturbed

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slightly by the scattering potential that at the X point scatters you across the branch zone and opens up this gap.

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Is that clear so far? All right.

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So when we're on the subject of really nasty exam questions, about six years ago, there was an exam question about this.

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Then not a single student got right and the examiner went berserk on that and ranted about the awful students and so forth.

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But but, you know, that's just, you know, what that's required of them to do.

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But actually, the question was actually quite difficult.

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And I think it was I'm going to explain to you what the question was about.

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It's something that if you hadn't seen it before, it would be very, very hard to figure it out.

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But I guess the expectation was you were supposed to be taught it, and I don't think anyone taught it.

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So now I'm going to teach it. So if it comes up, you'll all know it.

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So the question was about comparing something like this to something like silicon.

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So this is silicon here. It looks almost exactly like silicon carbide.

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The only difference is instead of carbon at one quarter, one quarter, one quarter, you replace those carbons with another silicon.

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So it's all silicon silicon in 000 in silicon and one quarter, one quarter on quarter.

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We've seen this structure before, too.

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This is the same structure as diamond silicon and diamond structure are are equivalent, same, same crystal structure.

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And actually, if you look at the two two energy eigen energy spectra here of K across the bottom and energy up top,

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they they're actually silicon and silicon carbide they look almost well, I wouldn't say identical, but extremely similar.

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A lot of the features in one are there in the other as well.

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So for example, here you have the same parabola,

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but you'll notice there's one one sort of shocking difference here that there's no gap at at the bronze on boundary here at this point.

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The X points here, the parabola is come together and they they don't they don't split at all.

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And so what one is supposed to deduce is, in fact, that there's no no back scattering at the X Point at all.

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For some reason in silicon, you do not get scattering at the X Point.

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So it's now we're going to try to figure out why that is.

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So, first of all, for for the chocolate bar,

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the scattering at the X Point goes from an X Point on one side of the barren zone to the opposite side of the barren zone.

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What are the Miller Indices associated with that reciprocal lattice vector?

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What? Not. No, this is really good.

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This 1001 is a very good guess, but it's not right.

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Does anyone want another guess? No, no, no, no, no, no.

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So 001 is the right direction. Or I guess it's in in the X direction or 100, if you prefer.

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If you're doing the X direction,

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it would be that would be pointing in the right direction because the x is in the x direction and the minus x would be in the opposite direction.

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So be in the 100 direction that you're scattering or the 001 if you're going the other way.

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So that's the right direction, but it's not 100. And the reason it's not 100 is because 100 is not a reciprocal lattice vector for FCC lattice.

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And you'll remember back to this picture here, they remember that 100 is a reciprocal lattice for the simple cubic lattice.

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If you define the planes by the 100 mil index, you get these plains here, but the SCC lattice has some additional lattice points in the middle.

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So 100 is does not get all the last points. Therefore it is not a reciprocal lattice vector for the FCC.

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Lattice 200 is the smallest reciprocal lattice vector in that direction.

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So here is scattering from x to minus x. I'll even write that down.

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So x to minus x scattering is to zero zero for silicon.

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Well, actually for any two FCC lattice x two minus Xs to zero zero.

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Okay, so we've gotten that far. Now, the question is, why is it that we don't get any scattering at 2009 index is anyone's guess of that.

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I have one more of these left. What is it?

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I can't hear who the master. The the mass.

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Yeah, well, they're both silicones, so. Yeah, so. So that doesn't.

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That doesn't do it for you any other. It's not an excuse.

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The rest of it is. It is FCC latest FCC list with a basis.

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So I'm not going to give you give it to you for that, but it's close to that.

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So maybe I'd give you a half of this, but I don't know how to do that. So. So it's not PR, it's not just an FCC loudness.

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And that's the point that you have to keep track of the lattice and the basis to find out if they're scattering.

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So let's think back to I mean, is the FCC lattice which allows the silicon the silicon is a lattice plus of eight times a basis.

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So if we think about what we learned about scattering from from x ray scattering, remember that s is s lattice times s basis, right?

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And S is what gives you the amplitude of scattering, right?

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If we square as you get the amplitude of scattering this work for x rays and it will work for electrons as well.

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So s lattice that enforces selection rules and force selection rules and to zero zero actually satisfies the selection rule for an FCC lattice.

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All even are all odd for FCC to zero zero is all even.

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So we're okay with that. So that's not going to that's not going to vanish.

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But let's look at s basis as basis you'll recall is some over atoms alpha in units l of e f sub alpha the form factor it is a i g that are alpha.

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And if we do this for silicon, we'll get f silicon, and then it will be one for the point at 000.

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And then we'll get either the two pi,

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I mean the Miller indices of the scattering to zero zero times the position of the lattice point one quarter of 1.2 one quarter of that.

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Does this look familiar from our x ray scattering adventures? Yeah.

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Okay, so you'll notice that this thing here is actually minus one, so as a basis is zero.

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So there's no back scattering in silicon in the 200 direction because you know, I'll give you I'll give this to you anyway because oops,

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because I don't want to keep myself because I'll eat it and I've been gaining weight.

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So okay so so you so you don't get that back scattering that in the 2005200 because there is what this is telling you is that

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there is exact destructive interference between the scattering of the two silicones in the unit cell with the both scatter,

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but then you get exactly destructive interference by them. And so in fact, you get no net scattering at the end of the day.

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And this this physics is actually it occurs for x rays in silicon.

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There's no scattering of x rays for silicon in the to zero is the direction it occurs for electrons.

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There's no back scattering of the x point for electrons in silicon and also occurs for phonons.

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Let's see, here's the phone on spectrum. For silicon.

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This looks a whole lot like the phone on spectrum for Diamond, not coincidentally, because it's the same crystal structure.

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So here's the silicon photon spectrum it has.

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So we went we had this a previously there's six phonons at every K vector because

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there are two atoms per primitive unit cell and each one moves in three directions.

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So there's six photons one, two, three, four, five, six, and three of them are acoustic coming down here.

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Now, you'll notice if you look at this point here at the X Point in this picture,

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silicon carbide, a gap has opened up in this picture, a gap has not opened up.

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And it's coming from essentially the same physics, even with phonon waves that in the 200 direction, there's not back scattering.

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Therefore you don't have an opening of a gap,

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whereas in Silicon Carbide you don't have perfect cancellation because this term would have f silicon, this term would have f carbon.

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Those don't equal each other. So you would still have back scattering and you have a gap opening up at the branch on boundary.

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Okay. So I've realised this is, this is a little bit complicated but it was on an exam and hopefully,

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you know if it comes up on an exam again people will be able to get it right.

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But I do admit that if you've never seen that before, it's something that's pretty hard to figure out.

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So. All right. So having basically discussed everything that we need to know about the nearly free electron model,

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we can now move on to talking about generally band theory, band structure and theory of electrons.

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So these are complicated pictures here.

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There's silicon carbide again with it's complicated picture of all this I can say that every every K so let's I'm.

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We started talking about about band theory when we talked about the tight binding model.

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Maybe let's draw a picture of some bands I like over here.

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So here is energy. Here's K in this direction. So I'll draw the bronze zone.

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There's a bronze zone from minus pi over a two pi over a and then you might have some bands that look like this,

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and then a higher band that looks like this and maybe some higher bands up here as well.

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Now, some things that we mentioned when we talked about the tight binding model is that if you have a filled band,

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filled band and a gap plus gap, you have an insulator equals insulator.

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Repeating this because it's fairly important. So if we fill a band, we fill, say, this band.

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And a gap. So the next band, we have an insulator. And the reason this is an insulator is because there's basically no freedom of where

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to put your electrons unless you overcome the gap to the next to the next band,

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which could be a lot of a lot of energy required to get up to the next band.

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You can't rearrange the electrons at all. They can't absorb any heat.

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They can't change. They can't change their momentum because all the momentum states are already filled.

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So it's basically just completely inert. So I'll write that down. It's inert.

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And if you have two bands that are completely filled, again, you'll have an a gap to the next band.

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It will also be an insulator. A little bit of nomenclature, which is useful is the highest fill band.

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Highest filled band is usually called the Valence Band.

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Valence Band. So here this would be Valence Band and the lowest empty band.

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The lowest empty band.

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Is known as the Conduction Band. Which would also be a really good name for a rock group.

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So if anyone makes a rock group, I'm suggesting a name induction band.

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Okay. So another important statement this insulator was small gap with a small gap.

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Gap is known as a semiconductor is small here is sort of less than about four electron volts equals.

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This is known as a semiconductor.

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And the reason for this nomenclature is that if the gap is sufficiently small, then at zero temperature this would still be a really good insulator.

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But at room temperature 300 Kelvin, if the gap is less than about four electron volts,

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then a few electrons will still be able to get thermally excited from the filled band into the empty band.

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So you'll have a couple electrons running around free,

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just thermally occupying the conduction band and they will carry some amount of current that you'll,

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you'll get a couple of electrons running around up here and that will allow them to carry some small amount of,

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of current, some small amount of conductivity. So it's a semiconductor conducts sort of semi conducts poorly, but it conducts similarly.

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It can absorb small amounts of heat, but not a lot of heat because only a very few electrons can be rearranged.

375
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Okay, that makes sense. Yeah. Okay. So actually, so this picture of silicon carbide.

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Silicon Carbide, these bands here are all filled. The Fermi energy or the chemical potential lies in the middle of this gap.

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So you would say that it's an insulator or semiconductor.

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And if you sort of look at the difference in energy between this point, the lowest point of the conduction band and this point here,

379
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which is the highest band point of the Valence band, it's a little less than 40.

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So we say silicon carbide is a semiconductor. Okay.

381
00:40:49,800 --> 00:40:56,600
There's something else that we mentioned when we talked about tight binding model is that counting electrons

382
00:40:56,990 --> 00:41:03,500
is counting electrons and states is rather important to figuring out if you have a field band or not.

383
00:41:04,760 --> 00:41:10,880
If you have any unit cells in your system, then you have empty states.

384
00:41:11,600 --> 00:41:15,980
K States. But then you have times to spin states.

385
00:41:16,790 --> 00:41:28,640
Spin states per k parquet means that two mn electrons will fill a band will fill band.

386
00:41:30,710 --> 00:41:41,270
And this to get to an electrons you need two electrons per unit cell so that you expect that if you have any,

387
00:41:41,690 --> 00:41:49,700
any, even number, even number of electrons per unit cell number of electrons per unit, cell per unit.

388
00:41:49,700 --> 00:42:03,140
So you would expect that you can fill can fill integer number of bands, integer number of bands.

389
00:42:05,870 --> 00:42:10,160
So let's take the case of silicon carbide, for example. So silicon carbide.

390
00:42:11,540 --> 00:42:18,860
Silicon Carbide. So silicon carbide both have valence four.

391
00:42:18,920 --> 00:42:23,570
So silicon and carbon both. Valence for both valence for.

392
00:42:27,990 --> 00:42:32,760
Valence equals four, which means we should count four electrons for each of them.

393
00:42:33,870 --> 00:42:38,700
If you wanted to count all of the so why am I only counting four instead of all of their electrons?

394
00:42:39,060 --> 00:42:43,890
The reason is because it's only we really generally only count the electrons in the furthest out shell,

395
00:42:44,130 --> 00:42:49,620
because only the farther in cells are completely inert. They just weigh buried at much, much lower energies.

396
00:42:50,520 --> 00:42:56,460
But if you wanted to count their filled shells as well, you would discover that the filled shells are all even as well.

397
00:42:56,760 --> 00:43:02,010
So. So we would still have an even number of electrons both for silicon and carbon.

398
00:43:02,400 --> 00:43:09,030
So the primitive unit cell here, primitive cell in this picture, it has four electrons.

399
00:43:09,690 --> 00:43:18,150
Four, sorry, eight electrons. Total, total, one silicon atom, one carbon atom, each carrying four electrons.

400
00:43:18,420 --> 00:43:22,110
So we should fill four bands. Four bands.

401
00:43:24,960 --> 00:43:29,370
I think I asked the same question about Diamond about a week ago or more.

402
00:43:29,700 --> 00:43:33,180
And in fact, if you count the number of bands here which are below the Fermi Energy,

403
00:43:33,420 --> 00:43:36,540
there are four of them one, two, three, four, and they're all filled.

404
00:43:36,810 --> 00:43:41,970
We do have this little bit of a confusion here that if you count here, it would look like there's only three.

405
00:43:42,270 --> 00:43:47,069
But if you look more carefully, you'll notice this fact. There's two bands here which come together and have the same energy.

406
00:43:47,070 --> 00:43:50,459
So this one is actually two bands which just happen to have the same energy here.

407
00:43:50,460 --> 00:43:52,920
They have different energy, so you can count that there is actually four of them.

408
00:43:53,220 --> 00:43:57,960
So in Silicon Carbide, all these four bands are filled and all these bands up here are empty.

409
00:43:58,860 --> 00:44:02,220
Good, then? Yes. Yes. Okay, good. All right.

410
00:44:02,550 --> 00:44:10,800
Now, in in contrast, we have metals have part filled bands.

411
00:44:11,850 --> 00:44:23,129
Part filled bands, which is almost always true, are mostly mostly true either today or next time.

412
00:44:23,130 --> 00:44:28,920
We'll discuss about why it's only mostly true. Mostly true for odd for odd number of electrons.

413
00:44:28,920 --> 00:44:37,620
Odd number of electrons per unit cell, for example, if we had a picture like this.

414
00:44:38,370 --> 00:44:42,269
So here's the bronze zone, here's a band, here's a band.

415
00:44:42,270 --> 00:44:47,730
Suppose we have one electron per unit cell that would be enough to half fill the lowest band.

416
00:44:48,360 --> 00:44:55,080
And so we have a part full band. This is one electron per unit cell part fell band here.

417
00:44:55,350 --> 00:44:59,070
And we would call this thing a metal because since the band is only part filled,

418
00:44:59,280 --> 00:45:05,340
we can make low energy excitations by taking an electron from the fill point to the unfilled point, arbitrarily low energy.

419
00:45:05,640 --> 00:45:12,209
We can also change the momentum of these filled states by just taking some of the electrons from over here

420
00:45:12,210 --> 00:45:18,000
and moving them over here at very low energy cost of just slightly shifting the entire the entire picture,

421
00:45:18,000 --> 00:45:23,490
slightly shifting it to the right. Fill a couple more states over here and a few less fewer states over here.

422
00:45:23,760 --> 00:45:26,999
And so as you're changing the momentum, you're actually adding current in that case.

423
00:45:27,000 --> 00:45:32,459
So by changing momentum of all these electrons, you actually change the total electrical current in that system.

424
00:45:32,460 --> 00:45:35,640
And so since it can carry current, we call it a metal.

425
00:45:36,710 --> 00:45:49,640
However, metals can also occur. Metals also occur also occur with even number with even number of electrons per unit cell.

426
00:45:52,220 --> 00:45:56,240
Also something I mentioned a couple of weeks ago when we discussed trying to model.

427
00:45:56,690 --> 00:46:01,220
The way that happens is here's your boron zone. If you have a lowest band, it looks kind of like this.

428
00:46:01,430 --> 00:46:06,920
And then you have a higher band that dips down like this. You and you had, say, two electrons per unit cell.

429
00:46:09,110 --> 00:46:17,600
That would be just enough electrons to fill one band. But I mean, you could choose to fill this entire band if you wanted to, this entire lowest band,

430
00:46:17,840 --> 00:46:22,520
but that would be energetically unfavourable compared to partially filling both of the bands.

431
00:46:22,520 --> 00:46:31,550
So instead it's better to partially fill this and partially fill this instead of filling these higher energy states in in this lower band.

432
00:46:31,610 --> 00:46:37,040
So that makes sense. Yes. Yes. Okay, good. So so in this case, it bands overlap.

433
00:46:37,770 --> 00:46:48,140
All right. This if bands overlap overlap and it looks it looks a little unnatural when I draw it this way.

434
00:46:48,410 --> 00:46:52,280
But if you look back to the silicon carbide band structure here,

435
00:46:52,460 --> 00:47:00,500
you could certainly imagine your head that this minimum just dip down a little bit further and this maximum came up just a little bit further.

436
00:47:00,740 --> 00:47:06,379
And if that were true, so that this mat minimum went below this Fermi Energy and this went above the Fermi energy,

437
00:47:06,380 --> 00:47:13,970
then some of the electrons that was previously in this band that was filled would then go into this band which was empty to lower the energy.

438
00:47:14,540 --> 00:47:21,920
Okay. So one thing that we that's really important to keep track of is how big are the gaps between bands?

439
00:47:22,280 --> 00:47:26,990
So if the gaps between the bands are really huge, then you're not going to have this situation ever occur.

440
00:47:28,220 --> 00:47:31,760
You know, if the bands are sufficiently far apart, then they're not going to overlap.

441
00:47:32,000 --> 00:47:35,390
So how big are gaps? How big are gaps?

442
00:47:36,080 --> 00:47:40,700
Big gaps between bands, gaps between bands.

443
00:47:47,390 --> 00:47:53,550
Question Well, we had different ways of describing how we understand the band structure.

444
00:47:53,570 --> 00:47:58,010
One way of describing it was from tight binding. In the tight binding picture.

445
00:47:58,520 --> 00:48:09,950
Recall that a band, each band, each band comes from an atomic orbital is from an atomic, orbital, atomic, orbital.

446
00:48:14,600 --> 00:48:18,620
With energy. With energy. With energy.

447
00:48:22,250 --> 00:48:37,910
It's called Ace of I. And the bandwidth is from hopping, from hopping, hopping T.

448
00:48:38,510 --> 00:48:43,910
So if the original atomic orbitals are very well spaced apart from each other and the hopping is small,

449
00:48:44,120 --> 00:48:48,559
then the bands don't have a chance to overlap because initially start very far apart from each other.

450
00:48:48,560 --> 00:48:52,580
And the hopping is what spreads the bands out from a single energy into a band.

451
00:48:52,760 --> 00:48:56,630
So if the hopping is small and the energy spacing is large, then the bands won't overlap.

452
00:48:56,660 --> 00:49:04,220
So that's one way of understanding it. That's sort of a natural way to understand why it is that there something like noble gases.

453
00:49:04,370 --> 00:49:09,920
When they form crystals, they are insulators because the hopping between noble gases is essentially zero.

454
00:49:10,160 --> 00:49:13,910
So these are these original topic orbitals, they're spaced a little bit,

455
00:49:14,180 --> 00:49:18,800
but the hopping between that spreads out these orbitals into bands is negligible.

456
00:49:18,980 --> 00:49:22,010
So they don't form very wide bands and so the bands never overlap.

457
00:49:23,660 --> 00:49:35,180
So, so that's one way of understanding it. But in a nearly free model, nearly free the gaps at the Britain zone boundary at these boundary.

458
00:49:39,500 --> 00:49:47,610
Our from from the periodic potential from potential vivax.

459
00:49:48,470 --> 00:49:54,710
So if you know that the periodic potential is strong, you should expect gaps that are pretty big opening up at the bronze zone boundary.

460
00:49:54,890 --> 00:50:00,680
Whereas if the potential is fairly weak, then you expect not to have big gaps in the bronze on boundary.

461
00:50:01,100 --> 00:50:06,230
All right. I think maybe we have 2 minutes left.

462
00:50:06,260 --> 00:50:10,640
I think maybe we should stop there and we'll pick up more on Bang Theory tomorrow.