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Welcome back. It's now the 14th lecture of the condensed matter course.
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More or less, we've finished talking about scattering.
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There's a couple of little bits and pieces of scattering that I'm going to try to pick up later, maybe at the end of some other lectures.
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Instead, I'd like to move on and use what we've learned when we study scattering in order to go back and study other types of waves in crystals,
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the type of waves we were interested in originally where electron waves and vibrational waves are phonons.
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So let me remind you a few things we had learned about electron waves in one dimension,
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and then we're going to try to generalise up to higher dimensions from what we've learned.
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So let's recall the type finding model type, binding model for electrons, binding model in one day, in one day.
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And we're going to consider the case of two orbitals with two orbitals per unit cell orbits per unit cell.
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And that can be two atoms, one orbital in each atom, or it could be two orbitals on the same atom, one atom per unit cell.
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And we can, we can draw out the dispersion in reduced zone scheme.
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Here's K, here's energy, here's pi over a, here's minus pi over a that's the barren zone, the big green zone.
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We can write, draw the dispersion kind of like this and like this two modes at each wave vector K in the bronze zone.
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Or we can unfold that picture in the extended zone scheme.
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Looking like this. Here's K, here's pi over a year minus pi over a year to pi over A and here's minus two pi over a.
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So we have the same picture for the first mode and then for the second mode, we're going to shift things over, shift the thing on the left,
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over to the right by two pi over eight to make it look like this and shift the
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piece on the right over by two pi over a to the left to make it look like this.
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So we have what we called the first boron zone here.
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First we're on zone busy here, and then these pieces here where the second bronze zone, second bronze zone.
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And so it's useful. There's chemicals there right in the middle.
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It's useful to try to generalise this concept to three dimensions or higher dimensions.
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So let's let's do that. So generally, the Britain zone for one zone, since we're thinking about electrons in three dimensions,
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is going to be always defined as a unit cell in reciprocal space.
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And we're going to have to figure out what we mean by first bronze owned, second bronze on and so forth.
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So it's general and bronze on general, general and bronze zone.
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We define as follows. So we'll start with the first bronze on points in space.
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That are closer to. That are. Closer to.
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That are. Closer to. K equals zero, k equals zero.
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Then do any other any other reciprocal lattice point receptor lat point are the first boron zone are the first boron zone first B's.
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So all these points in the first boron zone are closer to the point of k equals zero, then to the other reciprocal lattice points.
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Here's the reciprocal last point here. At two minus two pi over eight,
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here's another reciprocal lattice point here at two pi over a and everything between minus pi over and pi over a
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constitutes the first boron zone because it's closer to K equals zero than it is to any other reciprocal lattice point.
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This definition might look a little familiar to you because in fact it's the same definition as the big nearside cell.
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So this equals the big inner site cell sites.
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Cell of the cake was zero point of k equals zero.
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In reciprocal lattice. In reciprocal lattice.
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Does that sound familiar? I hope all points that are closer to a given reciprocal lattice point than to any all points that are
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closer to a given lattice point than to any other lattice point constitute its bigger site cell.
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So all points that are closer to K equals zero than to any other reciprocal lattice
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point constitute the vendor site cell of the cake or zero point in reciprocal space.
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Good. Yes. Okay, good. So how about the second foreign zone?
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Points where k equals zero is the second close.
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This is the second closest. Recep lat point.
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What point are the second brands on?
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Are the. Second.
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Second to B.C. So, for example, if we pick pick a point here.
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X call point X, the first closest reciprocal lattice point to this point X is here.
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The second closest reciprocal last point to the point x is equal zero.
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Therefore, this point x is in the second burn zone. Good.
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Okay, that's the definition. So a couple of notes about these definitions.
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First of all, one bronze zone boundary.
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Two opposite to opposite opposite busy boundary.
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Is a reciprocal lattice factor is a reciprocal factor.
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Sep lad back.
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Let's see. So here's a bronze. On boundary is the boundary between the first bronze on the second bronze zone and minus pi over a.
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Here's another one. It's the opposite one from minus pile raised plus pile ray.
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Distance between them is two pi over a, which is indeed a reciprocal lattice vector.
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Similarly, here's a boundary between the second Barone zone and out here actually would be the third bronze zone.
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And so it's a bronze zone boundary, the distance from it to the opposite Barone zone boundary over here at plus
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two pi over a is four pi over A and that is also a reciprocal lattice factor.
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That is generally a true statement. Okay. There's a similar so one, a sort of caveat, a corollary of this is that these boundaries.
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Or maybe, maybe it's not a corollary, but it's almost equivalent statement.
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These boundaries are points.
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Where? Absolute value of K equals absolute value of k plus g for some g for some reciprocal lattice vector g.
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Okay. So for example, you need, you need to have K and K plus G and see what's K and k k posted having the same magnitude for,
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for G being some reciprocal lattice vector.
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So if you take minus pi over a, you add a reciprocal lattice vector to pi over eight to it you get plus pi over a those have the same magnitude.
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So you're on the bronze on boundary. Similarly, you take minus two pi over a at a reciprocal lattice vector to it.
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For pi over eight you get two plus two pi over a that has the same magnitude.
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So you're on a bronze on boundary. The second statement, which is important about bronze zones,
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is each DC bronze zone has the same the same area or volume in three dimensions, same area and and represents.
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Represents each k point. Each crystal momentum.
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Each crystal wave vector, I guess. Wave vector k once.
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That was our general idea of constructing the bronze zone in the first place,
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that every physically different wave is represented exactly once in the bronze zone.
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The second bronze zone is equivalent to the first bronze zone.
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We've just removed the pieces around into different places, but each wave vector K is represented exactly once inside each bronze zone.
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Incidentally, is it obvious from the second bronze zone that you would define?
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The third bronze zone analogously points where the archaic zero is the third closest reciprocal at this point would be the third bronze zone.
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So you have the first, second, third. You can have as many bronze zones as you want.
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The final the final statement, which is useful, I think this is a homework assignment on the third homework set that the number of K states,
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number of K states, we even prove this in lecture in each busy B.C. equals the number of units cells.
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Unit cells. In system. In system.
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Does that sound familiar? We prove this in one dimension. I think for homework, you're supposed to prove it in three dimensions.
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It is also generally a true statement. So actually, let me do an example of this in two dimensions to make this more clear.
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So this we can start with a square lattice and real space.
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The reciprocal lattice of a square lattice is also a square lattice. So I've drawn a square last is supposed to be the reciprocal lattice.
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So it's a square lattice. Again, I've coloured some of the points differently just to make them easy to talk about.
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The point of k equals zero. I've coloured black in the centre and so first we're going to try to construct the
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first boron zone that is the inner site cell of this black point in the centre.
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So how do you construct the bigger site?
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So you put down perpendicular by sectors to the nearest points, perpendicular bisected to the red points here we'll make this point.
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That's a pretty good bisect here. Then this this plane, this perpendicular by sector,
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these two also giving all the perpendicular bisect is to the red points and then you colour in the area that is the first boron zone.
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Okay. Now the distance across the first boron zone is two pi over a which is a reciprocal lattice vector the way it's supposed to be.
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That distance there from perpendicular, bisected or perpendicular by sector is a reciprocal lattice vector.
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Now let's look for the second brian zone.
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How do we do that? It's a very similar construction. You just start putting down more perpendicular by sectors.
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So here the second closest sets of points are these blue points here.
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So let's put down perpendicular bisected to those like this, like this, like this, like this.
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So these are the perpendicular bisected to the blue points out here. We'll colour in the walled off area blue and this is now the second branch zone.
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Now it's this check that affects the definition.
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If you take a point like here in the second Barone zone, the first closest reciprocal lattice point is this red point here.
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The second closest reciprocal lattice point is k equals zero.
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Therefore, this point here somewhere in this blue region is in the second Brian zone.
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Good. Okay. Now it may not be obvious that the distance across the bronze zone from brown
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zone boundaries to bronze on boundary here is a reciprocal lattice vector,
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but I'll show you that it is this is a reciprocal lattice vector from zero to the blue point.
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You just shift it like that. It's indeed the distance across from beyond zone boundary to bronze on boundary so we can keep going on looking for the
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third baron zone and to construct the third bronze zone we have to take put perpendicular by sectors to these green points.
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So let's do that, put down those green lines which are perpendicular, bisected to the green points and you fill in the area now green.
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And let's check that it satisfies the definition.
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If you pick a point here in the green area, the first closest reciprocal lattice point is this red point here.
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The second closest is the blue point out here. The third closest is the cable zero point in the centre.
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So this is in the third Brian zone.
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Now also the distance from Brown Zone boundary to brown zone boundary across the system is four pi over a, which is also a reciprocal lattice vector.
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Now, another thing that you've been promised from these definitions is that each Barone zone has the same area,
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looks like the kind of might and that they should each represent each k point once.
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So the bronze zones represent every possible point. Exactly.
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Wants to see this. What we're going to do is we're going to take the pieces of this puzzle and with some very crude animation,
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we're going to move them by reciprocal lattice vectors and show that they're actually the same shape.
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So let's see how that works.
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So I'm going to take this red section, the first boron zone, and I'm going to get that exact same shape of the first bronze zone,
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same area, same shape, exactly by moving these pieces of the second bronze zone over.
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So this piece, I'm going to move by a reciprocal lattice vector like that, two pi over a over.
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Then I'll take this piece, I'll move it to pi over a over this way and I'll take the top piece,
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move it to pi over eight down and the other one to pi over a up.
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And it fills exactly the first boron zone, so. Well, okay.
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My animation skills are not not perfect, which you see, the other one is even worse.
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But. But you get the picture that it's supposed to fill the first bronze on exactly one side we by shifting things
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by two pi over a we actually didn't change the waves at all because if you shift by two pi over eight,
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you're getting back exactly the same crystal momentum, the same physical wave by doing this.
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So the second bronze zone is effectively exactly equivalent to the first bronze zone.
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Now let's do the same thing for the third bronze zone. We'll shift this over by two pi over a this over by t by over a okay.
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Now it's really going to get bad, but but you get the idea that it's supposed to is supposed to fit perfectly.
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And had I been better with my animation skills, it would have fit perfectly.
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Okay. So that's the general general idea of O'Brien's answers.
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A couple facts about Brian Jones in 3D.
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So in 3D, the lattices that we're interested in right down a list of the lattices we're interested in are mainly in.
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And we have simple cubic, simple cubic.
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We have BTC and we have FCC. So the corresponding reciprocal lattice is very simple.
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That is well, a simple cubic lattice has has a reciprocal lattice which is also simple cubic.
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The VC lattice. This is something I mentioned and I think it's a revision homework problem.
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The reciprocal lattice of a BC lattice is FCC and the reciprocal lattice of an FCC lattice is C.
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So the shape of the first Barone zone, first being B.C. shape is well for a simple cubic lattice.
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The Fichtner site cell is a cube. So the first Barone zone for the simple cubic lattice is a cube.
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For the BC lattice, the reciprocal lattice is FCC.
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So we get something that's the shape of the bigger site cell for FCC, which was a funny shape.
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It was a truncated octahedron. I'll show your picture in a second. And for FCC lattice, the reciprocal lattice goes back.
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And so you get the bigger site cell for BBC here.
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Actually, I said this. This one is the truncated octahedron.
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Now this one is a truncated octahedron. And this is the let me just show it on the slide.
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Okay. So this is the first Bruns on an FCC lattice.
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This one's a chunk. This one is a truncated octahedron. This one is the truncated the dodecahedron.
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This one is 12 sides. This one has 14 sides. All right. Whatever. These are the shapes of the of the of the bronze zone.
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So in this case, the FCC lattice has a reciprocal lattice, which is boxy.
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And this is the Vicknair site.
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So the BC lattice, and this is the bigger site cell of the SCC lattice, which is also the first boron zone of the BC lattice.
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Now you'll notice that these pictures are drawn in K, space, K, X, Y and Z.
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And you'll also notice that there are some letters dropped on to these pictures.
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These letters are there because people typically, when they describe points on in the bronze zone, they usually say,
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we're talking about the Gamma Point or the X point or the L point, and that's just shorthand nomenclature that you don't need to know.
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But it's worth knowing that they have these when they say X point, they mean some particular point in the bronze zone.
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So the point in the middle of a square face is called the X Point.
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This point, this point at this point, the point in the middle of one of the hexagonal faces is called the L point.
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The point in the centre is always called gamma. You don't need to know these things.
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All right. So now that we we've we have these pictures of the bronze zone for, you know,
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for our various lattices, we can start thinking about the excitation spectra,
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either the electronic expectation spectra or the phone on excitation spectra for various different crystals.
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And so everywhere in the bronze zone there will be different like in the reduced zone scheme,
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you'll have different excitation modes at each possible value of K in the in the bronze zone.
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So let's, let's do an example being this Valentine's Day happy Valentine's Day everyone is we're going to do diamond
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for s all right so well we do diamond every year so just happens to be a Valentine's Day this year.
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So Diamond, you'll recall, is an FCC lattice with two atom basis, one carbon and 000 and one carbon at one quarter, one quarter, one quarter.
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This is the shape of the bronze zone as we had in the previous slide.
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Now the first question and opportunities, wind, chocolate, how many phonon modes should there be at each k vector in the first boron zone?
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How many phone unloads? No, no, no, no.
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Oh, my gosh. So there's there's two atoms in the.
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I'm going to eat this one myself. All right. I'm going to gain weight this year.
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Now, there's two atoms in the primitive unit, so each atom can move in three possible directions.
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So that means you have six modes. Yeah. So. Let's see if we have a picture of it.
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Yeah. So here's here's a picture of the diamond phonon spectrum.
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So the way it's drawn is that you take cuts through the barren zone along certain directions that you're interested in.
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So, for example, this line. So here's the frequency vertical and then the K vector is horizontal, but this given to you in a cut.
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So this is cutting from Gamma to X along the 100 direction gamma two X along 100 direction along this dotted line here.
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And then from here to here is Gamma two l along the one, one, one direction from here to here, along that dotted line.
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This one is a little more complicated. It's X to gamma along the 110 direction.
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So that actually goes from this X through this K into the gamma in the centre of the next brown zone over.
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So that was a little complicated, but. All right. Now you'll notice that if you pick a point in the centre here, you will count exactly six modes.
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One, two, three, four, five, six. If you count over here, you'll discover there's only four modes.
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Why is that? Well, the reason is because two of them are degenerate. And you'll see these two modes coming in to this point here.
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And then they have the same energy all the way down from here to here.
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And then two modes here are degenerate up here, and they have the same energy or same frequency all the way over here.
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And the reason for the degeneracy is because you're looking along a particularly symmetric direction where
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vibrations transverse in this direction and vibrations are transverse in that direction look identical.
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Whereas along this direction, it's a very symmetric direction.
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So all directions of vibration actually have different energies.
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You can see a couple of interesting things in this. In this figure, the gamma point is K equals zero.
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You can see the acoustic modes coming down to 22k equals zero linearly.
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And these are the optical modes up here at at high frequency.
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And one of these modes is longitudinal. Two of them are transverse.
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Now we can ask the same questions about an electronic excitation spectrum.
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And you when you calculate an electronic excitation spectrum in this type binding approximation,
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you have to decide how many orbitals you're going to consider on each on each atom.
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So for example, in that picture up there, we considered two orbitals per unit cell.
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So given that a typical type of calculation for carbon, you don't want to keep all the orbitals out to infinity.
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So generally you just keep the orbitals you're interested in.
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What is usually interested in is the orbitals where there's some sort of interesting action going on that the, the one, the one orbitals.
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There was a core orbitals that just completely filled. Nothing interesting going on there.
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The orbitals that are three three has three p and so forth.
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Those are just completely empty, totally uninteresting. The ones that are interesting are the ones they're sort of partially filled, partially empty.
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Those are two S and the 32p orbitals. So given that we're considering the three S and and the 32p orbitals,
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how many excitation modes eigenvalues should we find at each k point in the bronze zone?
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One more chance. We're considering four orbitals per atom.
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What was the answer? Eight. Did you say? Yeah. Okay, so almost.
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I got that. All right. Eight. Yeah. So there's two orbitals in your account.
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What you need to do is you need to count the total number of orbitals per primitive unit
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cell and that will give you the total number of excitation modes per per k vector.
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Now this diagram, this is an excitation spectrum, an electronic excitation spectrum of diamond.
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These things are frequently known as spaghetti diagrams for obvious reasons,
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because it looks like some sort of complicated ball of spaghetti that you can't make any sense of.
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But but it's actually it's fairly simple. Well, not that not that complicated.
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What's again, you're looking at cuts through the barren zone. So this is a line between the L and Gamma Point along this lot dotted line.
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This is a line from Gamma to X along this dotted line, then X to K, K to gamma.
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So it's basically just taking cuts to the branch zone.
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And at each K point, there are eight modes one, two, three, four, five, six, seven, eight and some in some directions.
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It looks like there's only six. Let's see, for example, this direction.
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One, two, three, four, five, six. Because two of them have become degenerate here.
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See, these two came together. Oh, no, hang on. These two came together here to make only one.
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That's because are actually two modes with the same energy. So you have to be a little bit careful about that.
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Four Now carbon has a valence of four, which means each carbon donates four electrons and that means with two carbons per primitive unit cell,
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we have donated eight electrons total. So you fill the bottom four bands with both spins.
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So the bottom four bands are filled and there's a gap in the upper four bands.
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And in fact, as we'll discuss in a couple of lectures later on,
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that means what we have is we have a really good insulator filled bands, then a big gap and then a bunch of empty bands.
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It's also what makes Diamond nice and transparent, so it makes Diamond beautiful, appropriately enough, for Valentine's Day.
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All right. So let's move on from here.
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I think that's enough of diamond for today.
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So one of the things that's fairly important about this, this story is that with our tight binding spectra,
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we always have the principle that if you take K to some K plus G,
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you get back exactly the same wave and you have the same principle for vibrations as well.
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Now, you might ask whether this are the principle that if when you shift K by a reciprocal lattice vector,
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whether that is actually special to the tight binding approximation we use or whether this is actually a general statement,
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let's remind ourselves why it is that we found this for an exit, for r for a tight binding model, basically the tight binding wave function.
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This is a slightly different language than I wrote it before. But you can you can write it as either the I KKR times electrons sitting on
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lattice point R is this we used to call this thing phi and we gave this a no.
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I think it's more or less I guess the difference is now we're doing it in three dimensions rather than one dimension.
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That's why it looks a little different. But this is basically the type binding wave function that we wrote down before and
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this thing is obviously invariant undertaking K to K plus g because either the I k,
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either the Igot R when you shift k by g g that R equals one because g is in the reciprocal lattice and R is in the direct lattice.
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So if cake is shifted by g, it doesn't change this factor here and you get back the same wave.
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So but you might ask whether this is just a property of the tight binding model or if this is actually something that's general.
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And this is actually a really important question and it's a question that came up very shortly after the discovery of the shortening equation scoring.
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I discovered his equation in 1926, in 1928, that people were already worried about this issue.
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And it was resolved very famously by Bloch, Felix Bloch, and what's known as Bloch's Theorem.
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Block's theorem. All right. Okay. Which is frequently.
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Referred to as the underpinning of all of material science.
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It's an underpinning of all of semiconductor physics.
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It's an extremely important principle which may not seem so important until we talk about some of its implications.
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But let me state the theorem first, then we'll prove the theorem. Then we'll talk about some of the implications.
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So the statement is an electron in a periodic potential.
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In periodic potential potential.
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Periodic vrr, I guess.
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Periodic means V of our equals vrr plus v of capital r with capital being a lot of spectre has eigen states of the form as eigen states.
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Of the form. Okay.
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There's a lot of symbols here for a second size sub K of R equals e to the i k
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dot are times you superscript alpha some k of our where where you is periodic.
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So it's the same if you shift it from you, you've r equals your R plus r and ks.
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And KS. Can be chosen in the first bronze zone, can be chosen in first bronze zone.
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In other words, K is crystal momentum. K is crystal wave.
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Vector. Crystal wave vector. This form is known as this form of SCI.
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OSSI is known as a modified plain wave is known as as a modified plain wave.
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A modified plain wave. Modified plain wave.
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That that should be fairly obvious. It's a plane wave, the iCar, and it's modified by being multiplied by a periodic function.
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You. That's the modification of the plane wave.
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The index. Alpha. Alpha is the band index.
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And the band index is meant to describe.
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You know, you get a different wave function if you're talking about this point here or this point here at the same k vector,
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at the same K vector, you can have more than one eigen state.
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So the energy's eigen energies. R e superscript alpha of k so e one and e two depending on which which band we're talking about,
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which which which energy level, the lowest or the next lowest or the next next lowest and so forth.
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Okay. So this is a pretty big bunch of statements we've made here,
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but it's fairly important because it tells you about the general structure of iron states in a periodic potential.
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So let's first prove this theorem and then we'll talk about why this theorem is so important
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and why it underlies so much of material science and semiconductor physics and so forth.
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So first, okay, so this is sort of a quasi proof. It's actually fairly rigorous, but we'll we'll not fill in the details.
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But the quasi proof goes like like this.
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First we'll consider a Hamiltonian, which is h, not just the free electron Hamiltonian plus or potential V, which is periodic.
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So V is periodic and H not is the usual p squared over two m.
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Now, if we throw away the periodic potential, we would know the iron states.
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Those are just plain wave eigen states.
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Plane waves are e not k k where enought you know of k is just the squared k squared over two m we usually have.
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So so far so good. We know we have the plane waves, then we have to add the periodic potential.
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Now, what can the periodic potential do? The periodic potential can scatter you from some K to some K prime via a matrix element of this form.
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Now we know a lot about matrix elements of this form. From what we learned in scattering theory, this thing here,
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this matrix element from a plane wave K to a plane wave k prime via a periodic potential
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v is the for you transform for a transform a vrr evaluated at k minus k prime.
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And that is the following it is zero.
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If K minus K prime came out as k prime is not a reciprocal lattice vector is not equal to g and we'll call it v sub g, f k minus k prime equals g.
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I mean it's called k prime minus cable's g that.
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And this is just Allawi's condition.
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You can scatter from a wave vector K to a wave vector k prime only if the difference between and K prime is a reciprocal lattice vector.
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We learned this when we studied scattering of X-rays and neutrons and so forth.
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So this gets us almost to the proof. So we can almost conclude the result here.
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Why? Well, the point here is that the Hamiltonian with the scattering potential is blocked diagonal.
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No, no pun intended. BLOCK, diagonal.
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BLOCK of the K diagonal. What I mean by this is that you can scatter from K to a k prime if k k prime are
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separated by G and then from k prime you can scatter to another k double prime.
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If those are separated by a g and you can scatter to another k prime and so forth and so on and so forth.
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But you can never change the crystal momentum.
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All of the ks that you can get to by scattering by the via the periodic potential, all of them differ from each other by reciprocal lattice vectors.
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So the Hamiltonian breaks up and if you go into these sectors, if you like.
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And so the wave function must be of the form, cy k must be of the form,
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some over g of some coefficients of column AMG plus k times either the i k plus g dotted with are.
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Now why should it have to be of this form? Well, you start with some of the iCar and then you mix into that.
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The Hamiltonian mixes into that different plane waves whose wave vector K differs by a reciprocal reliance vector g.
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And you can mix in as many of those as you want and they all get some coefficients a.
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But at the end of the day, the wave function must be a sum only of things that have the same crystal momentum k plus some reciprocal lattice vector g.
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There's no way that I potentially can take you can change your crystal momentum.
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That's the important realisation. Good. Yeah.
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Happy? Good. All right. So from here we're pretty much done because you can just factor out the, you know,
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the I k dot are and then you have some over g of a of g plus k into the I,
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g r and this piece here is the is the periodic function u u of R which and you can tell this thing must be periodic because if you take e to the I,
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g g dotted with R plus r, move it by a direct space lattice vector.
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This thing is the same as the i k that are.
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So this thing is periodic. As as claimed.
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Okay. Furthermore, we can also check what happens if you take and you shift it by a reciprocal lattice vector g.
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So let's let's actually take this form of K up here and let's shift K by G.
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So I have K plus g, I actually I've already used the letter G, so let's call this a G prime shifted by K to Kate plus g prime.
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So we'll write it as some over g a sub g plus k plus g prime e to the i r k plus g, prime plus g.
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But the sum is over a dummy index so we can define a dummy index.
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It's called g twiddle equals g g twiddle of vector equals g plus g prime vector.
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And then if you substitute that in, that becomes sum over g twiddle vector a of g twiddle of vector plus k into the i k r k plus g
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twiddle vector dot r which is exactly the same expression that we started with up up there.
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So if you shift K by reciprocal lattice vector G, you get back exactly the same wave function,
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which was what we were trying to answer in the first place.
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Is it special that when you shift K by reciprocal last vector you get back the exactly the same wave function?
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Is that special to the tight binding model? It's not. It's a property of the shortening equation in the periodic potential.
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Okay. All right. So the implications of this theorem, which is viewed as so important, is one implications.
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One are an excitation are all excitations can be described all excitations.
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Can be described be described in one bronze zone in one B.C., because whenever you go outside of the bronze zone,
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you can just zip by a reciprocal lattice vector to get back into the first back into that bronze zone.
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So that's an important statement. The second important statement, maybe more important is that a periodic potential,
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a periodic potential does not change, does not scatter crystal momentum.
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And this is something that I mentioned before when we discussed the type finding model that the crystal momentum is conserved.
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The crystal momentum is always conserved in the fact that we have this very strong periodic potential doesn't ruin that.
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The reason this is so important, remember, we had this puzzle when we studied the Sommerfeld model that you have this mean free path of electrons,
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which is enormously long, hundreds of atoms or thousands of atoms long.
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And we couldn't understand it because there's a big nucleus that the atom can hit every few angstroms.
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It could scatter off of tons and tons of nuclei. The point here is that the you don't have pure plane waves.
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What you have is modified plane waves.
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You have these plane waves times a periodic function, but once you modify the plane wave by multiplying it by this periodic function,
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that is perfectly good plane wave to go all the way across the system without scattering at all.
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So if you take a, you know, an electron, you put in a little wave packet of this modified plane wave,
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it would travel clear across the system as long as it doesn't hit any impurities or or nasty things like that.
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But it could go all the way across the system without scattering at all.
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So this sort of resolves our our initial issue of why it is that electrons can travel so far without without scattering.
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Okay. And where did this this result come from?
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It's important to realise that this important result really came from Louise condition.
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It came from the fact that a periodic potential only scatters you by reciprocal wave vectors.
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All right, so the one thing that this doesn't do for us.
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Is it does not manage to. It does not manage to help us actually solve the shorthand equations.
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So solving solving SchrÃ¶dinger equation is still hard and we've already discussed one way to do it.
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One way to do it is tight binding. And by type binding, what we're really doing is we start with.
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Start with atomic orbitals. With atomic orbitals. And allow week hopping, allow week hopping.
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But that might sound a little non generic because maybe the hopping is strong from one place to another.
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So with useful too,
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it's actually very useful to think about it from a very complementary viewpoint where instead of starting with a type binding model,
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we start with plane waves, start with with plane waves, plane waves and add week periodic potential, the air.
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So it's entirely the opposite way of looking at things.
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This approach, approach to two is known as the nearly free electron model, nearly free electron model,
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and is a very useful model for understanding a lot of the physics of semiconductors and band structure in solids.
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So let's let's go about that. Let's start it. We probably won't finish it today, but the so how do we do this?
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We start with our H0 on k are plane waves without the periodic potential.
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So is zero k on k where is zero is the usual h bar squared k squared over two m?
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It's a good place to start. And then let's assume this is weak, assume weak.
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V All right.
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So what do you do when you have a weak perturbation to a Hamiltonian? Well, you use perturbation theory, which you presumably learned last year.
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So you remember what happens to the energies in perturbation theory.
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So let's start with first order of perturbation theory. First order, order, purge the energy this probably look familiar the energy of the wave.
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The Eigen State K is the bare energy without the perturbation plus the first order of perturbation theory term which is k,
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v, k and this matrix element here when I scrolled it off the top of the board.
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No, maybe I didn't, maybe I did. I think I did. We defined it as the sub zero because k minus k is zero and this v sub zero is the same.
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Same for all K for all k.
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So this just gives an overall constant energy shift of the the eigen state.
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So the the I can say t to get shifted up in energy or they get shifted down in energy.
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This is sort of the the zero thorium mode of this potentials where a constant potential is added on top of any fluctuating potentials,
390
00:42:35,950 --> 00:42:42,999
things that change in space. So the zero three mode of the of the potential just shifts up and down the overall energy of K.
391
00:42:43,000 --> 00:42:46,030
Does this expression look familiar first or probation theory. You just take the exponent.
392
00:42:46,030 --> 00:42:52,209
Yeah. Okay, good. So this is not interesting. And as a matter of fact, people frequently drop V zero because it's so uninteresting.
393
00:42:52,210 --> 00:42:57,550
They just get rid of it. I'll try to keep it, but maybe I'll forget it by mistake at some point.
394
00:42:57,850 --> 00:43:01,240
So second order is more interesting.
395
00:43:01,570 --> 00:43:08,469
So it's second order you have ic equals e k not that's the they are ps plus v zero.
396
00:43:08,470 --> 00:43:14,500
That's the first order piece. And then you have the second order piece which is sum of k prime not equal to k.
397
00:43:14,920 --> 00:43:24,810
Then we have k prime v k squared over e sub k not minus e k prime not.
398
00:43:24,820 --> 00:43:28,959
Is that the right order? Yeah. Does that look familiar from second order of probation theory?
399
00:43:28,960 --> 00:43:33,010
Good. All right, now, what do we know about this second order term?
400
00:43:33,010 --> 00:43:41,320
Well, again, we have lousy condition for the upstairs matrix element that in order for that matrix element to be non-zero,
401
00:43:41,590 --> 00:43:47,020
you must have K prime minus K must equal a reciprocal lattice vector g.
402
00:43:48,130 --> 00:43:54,550
So this term here can be rewritten as some over g of reciprocal.
403
00:43:54,550 --> 00:44:10,120
Last factor is g of v sub g squared where g is k prime minus k divided by e sub k not minus e sub k plus g not.
404
00:44:11,630 --> 00:44:14,960
Good luck. Happy with that? Somewhat.
405
00:44:15,830 --> 00:44:22,510
Okay. So this is our expression for the second order shift in energy of the plane waves that we started with.
406
00:44:22,940 --> 00:44:25,940
But we have to be a little bit careful about expressions that look like this.
407
00:44:25,940 --> 00:44:34,040
And the reason we have to be careful is because you might have a situation where the two terms in the denominator are either equal to each other,
408
00:44:34,040 --> 00:44:39,140
in which case you get zero downstairs or very close to each other, in which case you get something very small downstairs.
409
00:44:39,290 --> 00:44:43,640
And in either case, the secondary term would blow up, in which case you have a divergence.
410
00:44:44,030 --> 00:44:48,080
And that means the second order perturbation theory is not correct.
411
00:44:49,220 --> 00:44:52,670
So you must be careful.
412
00:44:53,510 --> 00:44:59,360
Be careful when he sub k.
413
00:45:00,050 --> 00:45:05,900
That's just right. Is SMK not approximately equal to E sub k plus g?
414
00:45:06,530 --> 00:45:12,439
Not now. When does that happen? Well, e sub k is actually just proportional to k squared.
415
00:45:12,440 --> 00:45:17,150
So that will happen when absolute k equals absolute k plus g.
416
00:45:17,660 --> 00:45:23,090
Well where's that. You'll remember somewhere here.
417
00:45:25,130 --> 00:45:29,360
There it is. It's the condition for being on a bronze on boundary.
418
00:45:29,600 --> 00:45:36,290
When you're on bronze on boundary e sub K and it's a K plus g are actually the same.
419
00:45:36,380 --> 00:45:50,390
Let's actually draw a picture of that. So what we have is we have a this is our bare spectrum K and E, the bare spectrum is just a parabola.
420
00:45:52,120 --> 00:46:04,450
So this is e sub k is h bar squared not k squared over two m and here is pi over a here is minus pi over a.
421
00:46:05,080 --> 00:46:13,990
And the energy of these two, these are separated by a reciprocal lattice vector to pi over a and their energy is identical.
422
00:46:14,440 --> 00:46:17,980
So this would give us a divergence in second order perturbation theory.
423
00:46:18,280 --> 00:46:25,180
Similarly, if we go to minus two pi over a and two pi over a,
424
00:46:27,910 --> 00:46:33,969
these are separated by a reciprocal lattice vector for pi over a and their energies are identical.
425
00:46:33,970 --> 00:46:37,480
So they will give you a divergence in second order perturbation theory.
426
00:46:37,690 --> 00:46:44,920
So the place where second order perturbation theory fails is exactly when you're on a bronze zone boundary.
427
00:46:46,150 --> 00:46:54,340
So what we're going to need to do is we are going to need to are probably going to have to do this next time.
428
00:46:54,880 --> 00:47:04,450
But what we're going to do to fix this problem, fix the problem by using using degenerate perturbation theory,
429
00:47:05,530 --> 00:47:15,880
degenerate perturbation theory, effort theory, which hopefully you've learned something about already.
430
00:47:17,320 --> 00:47:20,889
So we're not going to have time to launch into that today.
431
00:47:20,890 --> 00:47:25,510
So I'm going to actually take the opportunity to insert one piece of information or maybe two
432
00:47:25,510 --> 00:47:29,710
pieces of information about scattering that I promised you we would talk about some time,
433
00:47:30,070 --> 00:47:34,060
and that is scattering off of. So maybe I'll put that over here.
434
00:47:34,360 --> 00:47:41,310
Scattering off of x rays and neutron scattering x ray and neutron scattering.
435
00:47:41,320 --> 00:47:44,690
This is different subject just inserted in this last 5 minutes.
436
00:47:44,710 --> 00:47:49,450
Neutron scattering from.
437
00:47:50,380 --> 00:47:54,910
From liquids. Liquids and amorphous solids.
438
00:47:55,780 --> 00:47:58,790
Amorphous. Solids.
439
00:47:59,810 --> 00:48:04,610
So in liquids, in amorphous solids, we do not have periodic structures.
440
00:48:04,820 --> 00:48:12,110
We do not have periodic arrangement of atoms. But nonetheless, you get a lot of information out of the neutron X-ray scattering.
441
00:48:12,110 --> 00:48:20,929
If you think way back to when we did Fermi's Golden Rule, that the scattering amplitude is still the 48 transform of the periodic of the potential.
442
00:48:20,930 --> 00:48:24,230
But it's not a periodic potential, it's Fourier transform of the non periodic potential.
443
00:48:24,560 --> 00:48:27,920
But does that mean there's no information in it? No, there's still information in it.
444
00:48:28,130 --> 00:48:31,690
It's just that you don't get sharp peaks anymore. What you get is here.
445
00:48:31,820 --> 00:48:36,110
There it is. So this is X-ray scattering on liquid aluminium.
446
00:48:36,140 --> 00:48:39,290
It's heated up to some high temperature. So it's so it's a liquid.
447
00:48:40,220 --> 00:48:44,210
There's a function of a scattering angle or the reciprocal wave vector.
448
00:48:44,540 --> 00:48:48,350
That's the way vector along this axis.
449
00:48:48,590 --> 00:48:56,540
The amplitude of scattering it has peaks. And the peaks represent roughly the distance between between atoms.
450
00:48:56,550 --> 00:49:03,710
It's one over the distance between atoms. The same way the peaks represent the the periodicity of the unit cell when it becomes crystalline.
451
00:49:03,860 --> 00:49:10,430
Now what happens is as you cooled down a liquid and it gets more and more crystalline, that becomes more and more local order.
452
00:49:10,550 --> 00:49:12,770
And these peaks get sharper and sharper and sharper.
453
00:49:12,980 --> 00:49:21,090
And finally, when the thing actually crystallises, they become a delta function peaks the way we expected them to for for a crystal.
454
00:49:21,110 --> 00:49:25,490
But if it's a liquid, it's you can sort of think of a liquid as locally being like a solid.
455
00:49:25,880 --> 00:49:30,060
Each atom has a couple of neighbours, but if you sort of look farther away, it's not ordered any more.
456
00:49:30,080 --> 00:49:36,790
It sort of loses its order. But if you sort of look just in some small region, it probably has, you know, it's organised over a small region.
457
00:49:36,800 --> 00:49:41,360
So you get these sort of broad peaks in in the scattering spectrum.
458
00:49:42,510 --> 00:49:50,670
Okay. I think maybe I won't try to cover any further issues today and we'll start again on Wednesday, I think.
459
00:49:50,690 --> 00:49:51,380
Have a good weekend.