nanoHUB U Fundamentals of Nanoelectronics: Quantum Transport/L1.9: RiciprocolLattice / Valleys
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[Slide 1] Welcome back to Unit 1 of our course. This is the ninth lecture. 

[Slide 2] Now in the last lecture, we discussed these energy levels of graphene. And what we were trying to do is, this is a two-dimensional problem because this graphene is a two-dimensional material. And you have this kx and ky. And the energy is a function of kx and ky. And what you want is those energy levels which are close to the fermi energy. And what we discussed is how we could plot out a certain function and wherever that function is small, those are the regions where the eigenvalues are close to the fermi energy and those are these dark spots. And these are what we call the valleys. And just looking at this picture it looks-- look like you have six valleys but in practice, it's actually two not really six. And why it's two and not six? That needs this discussion with what this lecture is about. We'll talk about how you construct a reciprocal lattice and thereby see why it's really not six valleys but two. And this is of course a general principle that applies not just to graphene but to all materials in general. Now, just to recap, the basic idea for valley is something like this. But supposing you had a simple 1D problem, you had the E-k relation looking like this. There's a fermi energy somewhere here. And so, what determines conduction is the energy levels around the fermi energy. And so, you would say, OK, here we have got one valley. This is where all the conduction takes place. Well, if you have an E-k diagram that looks like this, you would say there's two valleys. Now, what might be a little more-- little confusing is, if you have a E-k relation that looks like this, in other words, kind of the inverse of this. Have it-- Fermi energy is here. And looking at it, you might think that there are two valleys, one here and one here but actually it's really one valley. And one way to say it is, yeah, there's a half a valley here and a half a valley there so overall it's one. Another way of saying it is that well here what we are drawing is the first Brillouin zone. But as you know in principle, we could take a window anywhere. We could use a different Brillouin zone. And this function is periodic so you-- actually, goes on like this. So instead of using this Brillouin zone I could have used a Brillouin zone like this. And in that case of course there would obviously be one valley right in the middle. OK. But if we want to use the conventional Brillouin zone, that's this one, the red one, then we should think of it as two half valleys. But what I'll try to show you in this lecture is in graphene we have something similar. What we need to do is construct this Brillouin zone. Now in simple 1D problems or simple rectangular lattices, the constructing-- the Brillouin zone was relatively straightforward whereas in graphene, you have to do it more formally. And that's what we'll be doing in this lecture. And when you do that, you'll find that the Brillouin zone looks something like this. See? And now you can kind of see why the six valleys became two. Well, it's because this valley you only get 1/3 of it within the Brillouin zone. Similarly, you'll get only 1/3 here and 1/3 there. So overall, it's like six 1/3 valleys. Or there may be ways of choosing your Brillouin zone or moving it around so that you get two full valleys but the conventional Brillouin zone at least you get six 1/3 valleys. OK. So let me explain then how that Brillouin zone is constructed. 

[Slide 3] So for that you first start from the real lattice and construct the reciprocal lattice from it. But before we talk about graphene, let's see how you do it if you had a simple rectangular lattice. Because in this case, you don't really need the formal method but we'll do it the formal way, that way you'll have it for using it for graphene later. OK. Now, the first thing is this is a rectangular lattice and every point on this lattice can be written as a linear combination of two what are called basis vectors. Now, this is a little different from the basis functions that we talked about earlier in this course. Right here, we're talking of basis vectors. And this is a two-dimensional vector which can be written in terms of the two basis functions, by two basis vectors. By contrast, the basis functions are kind of like in a-- is a much higher dimensional space which is why you need many, many basis functions to represent your function in that whereas here, we're just talking about any two dimensional vector on this plane that we are trying to write down. And you could write it as a linear combination of this one and this one on this lattice. OK. So this a1 and a2. What's A1? It's this vector in the x direction. a2, that vector in the y direction. Now, the way you construct your reciprocal lattice is you first write down the basis vectors for the reciprocal lattice that is the reciprocal lattice is also rectangular in this case and it's-- all the points here can be written as linear combinations of two basis vectors. One of them in the x direction 2 pi over a, another the Y direction 2 pi over b. Now, the defining relation though between the basis vectors and reciprocal space and the basis vectors in real space, the defining relation is this one. That small a1 should be orthogonal or perpendicular to big A2 and small a2 big-- perpendicular to this big A1. Because when you know, two things are perpendicular their dot product is zero. So, first thing is that this guy, this one here, is perpendicular to that. And this one here is perpendicular to that. Next point is that if I take the dot product of this one and this one, the ones that are not perpendicular, then I'll get a nonzero number and that should be 2 pi. Now, we didn't quite need this general prescription to write things down in the rectangular case because in a rectangular case, it's basically simple. This was a, that's 2 pi over a. This is b, that's 2 pi over b. But in nonrectangular cases, it isn't as straightforward. And there, we need to follow this formal prescription of how given small a1 and a2, you can find big A1 and big A2. And once you have big A1 and big A2, you can construct this lattice by taking linear combinations of that. OK. Now, why is this important? Well, as long as these relations are satisfied, what you can show is that if I take any reciprocal lattice vector, capital K, and dot it with a vector in this real space, the answer will always be a multiple of 2 pi. Why is that? Well take the dot product of this with that. Now, small a1 dot capital A1 gives you 2 pi. So that's this 2 pi small M capital M. That's this. Now when you take m A1 dot it with the last one, you get zero. Why? Because A1 dot A2 is zero. You take the A2 dot it with the A1, again you get zero. And then, you take the A2 and dot it with the A2, that's how you get this second term, the 2 pi. So the point is you could have any integers capital M and capital N any integers small m and small n. But if I take a dot product of a vector in this space with the vector in that one, the answer will be a multiple of 2 pi. Why is that important? Well, what it means is when I write down a wave function, as you know, when we're describing the dispersion relations, when we're obtaining the dispersion relations, we say that in a periodic solid you can write wave functions in the form of plane waves E to the power ik dot r. And the point is that if I add the reciprocal lattice vector capital K to that K, makes no difference, it's the same function. So when you're classifying your states in terms of the small K, the point is any K is exactly equivalent to a different K, which differs from it by a reciprocal lattice vector. And so, if someone has a solution with some K and another person has a solution with small K plus capital K, the point is they are really the same solution, they shouldn't count as two separate solutions. So what that means is that in this K-space, when we're trying to classify our solutions, we're looking at how many solutions we have, you should only look within a Brillouin zone, the region which is like around that point obtained by taking half these distances. So it's like you bisect all of these and get this region here. And the idea is that if you have another point K out there, there's always an equivalent point inside which is equivalent, which means you don't need to look for a solution corresponding to the red X, it's really covered by the blue one because they differ by a reciprocal lattice vector that much. So this is this idea of the Brillouin zone, the region, the domain in K-space where you need to look at your eigenvalues. Anything outside basically is equivalent to something inside it. OK. So we'll now use this general principle to construct the reciprocal lattice for graphene. 

[Slide 4] Graphene as you know has this hexagonal thing. And what we have shown here is that one unit cell connected to, you know, two unit cells on this side, two unit cells on that side. And as you know, we are supposed to think of the red and the blue as one unit. OK. Now, first question is what are the basis vectors? Well, the basis vectors are obtained by taking this unit cell, let's say, connecting it to that one so you can draw it between the red and the red, or the blue and the blue, same thing, same vector. So that's one basis vector, this is another basis vector. And any location on this entire lattice can be written as number times a1 plus another-- I mean, an integer times a1 and an integer times a2. OK. So those are the two basis vectors, here. Well, next question is how do you write them down in terms of your x and y coordinates? Well the a1 has a x component which is that much which is what we had labeled a, that's like half of that distance, and-- to the y component of that much which is half of this. So you have ax plus by. A2 is again the same x component but the negative y component so it's ax minus by. And you'll see in a minute why I'm writing this. For the z direction, of course, these are two-dimensional solids so the z direction is kind of irrelevant. But let's write down a basis vector in the z direction as well which I'll just write it as z. Now question is, how do I write capital A1 and capital A2 so that I can construct the reciprocal lattice? And there's a very simple general solution to this here and that is capital A1 is given by that expression. What you can show is, if A1 is given by that expression then automatically, it will ensure that A2 dot A1 is zero. And that A1 dot A1 is 2 pi. So those two basic relations will be satisfied if I write my capital A1 this way. How do I see that? Well, this basically says that capital A1 should be perpendicular to A2. And this of course ensures it. Why? Because capital A1 is the cross product of a2 and a3. And as you know, when you take the cross product of two vectors the resulting vector is perpendicular to both of them. So, I have automatically satisfied this. Well what about small A1 dot A1 equals 2 pi? Well that's easy to see. Dot this with small A1. Then you see the numerator becomes exactly the denominator, so cancels out and you're left with 2 pi. So next step then is, take that expression, use this a1, a2, a3 and evaluate it. First thing is the numerator a2 cross a3. So you have to take the cross product of those two things. Now x cross z is minus y. So you get a minus ay. y cross z, that's plus x. So you get a minus bx. Now we need the denominator, so you have to take that quantity and dot it with a1. So when you take the dot product of two vectors what you do is multiply their x components multiply their y components, add it up. So it's plus a times minus b so that's minus ab and plus b times minus A that's again minus ab. So you add it up, you get minus 2ab. So now, if you divide it, you'll get the expression for capital A1. This pi over a in the x direction plus pi over b in the y direction. Similarly, we can write down the other reciprocal lattice vector a2. It is a3 cross a1 divided by that times 2 pi. Similar algebra, if you work through it. You'll get something looking just like this but with the second one with its sign reversed. So, we now have the reciprocal lattice vectors, now we can try constructing the reciprocal lattice itself. 

[Slide 5] What it will look like is you stand at the origin, draw A1 for example. So, what's A1? It's pi over a in the x direction, pi over b in the y direction. And what I've plotted here though, you'll notice I'm not plotting kx and ky, I'm plotting kx times a and ky times b. So for a1, kx times a is pi and ky times b is also pi and that's why that point looks like pi comma pi. So that's A1. What's A2? That's this vector. So A1 is that vector. A2 is that vector. And this vector is just like the other one but it's pi comma minus pi. What other vectors-- What other reciprocal lattice vectors can I construct from standing from here? Well, various linear combinations of A1 and A2. So if I add the two in pi comma pi and pi comma minus pi when I add them, I get two pi comma zero, so that's this vector. This is A1 plus A2. This is just the negative of that. Similarly, this is the negative of that. And this again, negative of this is that. So that's how you can write all the nearest neighbor vectors. And you can have longer vectors, one going out there for example. And you can construct the entire reciprocal lattice. What we need is just this immediate vicinity because, if you remember, the Brillouin zone is obtained by drawing or kind of taking half of each one. So, drawing perpendicular bisectors. So when you do that, you see, you take this bisect it, take that bisect it, take that bisect it. And that's how you get the first Brillouin zone. And this is exactly what we're-- what I've drawn earlier. You can see, looks like, you know, this point is right in the middle of that region where you have the valley and this is also exactly in the middle of the valley, each one, which is why none of the valleys are really complete. So when you look at this region, you only have this 1/3 of the valley there, that's the A. See? Similarly here, you have just 1/3 of a valley, B. But of course, what you could do mentally is this point is equivalent to that point. Why? Because they are separated by a reciprocal lattice vector. If I add a reciprocal lattice vector to this, from here I can get that, which means all the states here are equivalent to states over there. So, instead of having a third of a valley here and a third of a valley there, I could kind of put them around the same place. Similarly, I could take the C, this valley from here, and move it up there, again by translating with this vector. So then you see, instead of having three 1/3 valleys, you'd have one full valley around here. Similarly, these three the ones here, here, and here they could also be combined into one full valley around here. So you could say OK, I got one full valley there and one full valley here. But bottom line of course is that at the end of the day, graphene has two valleys. And you could see it was a relatively mathematical argument, you know, we have to go through this reciprocal lattice construct things in Fourier space and so on. But the number of valleys has a very direct consequence in terms of measurable quantities, which people have done this is when people look at ballistic conductors, ballistic graphene or ballistic-- this carbon nanotube which is like graphene rolled up into like a straw. When people look at the ballistic conductance, as you know, usually the ballistic conductance is supposed to be 12 and a half kilohms roughly. Why 12 and a half? Well, the quantum of conductance is-- this quantum of resistance is 25 kilohms that are two spins in parallel. So, 25 kilohms in parallel with 25 gives you 12 and a half kilohms. That's the ballistic resistance usually. But when you measure it in graphene or in carbon nanotubes, what you get is more like 6 kilohms, half of that. Why? Well, because graphene not only has the two spins in parallel but also has the two valleys in parallel. And the point to note is it's two valleys in parallel not six valleys, it's two. And the argument of course why it's two is what we just went through. And the point that I was trying to make is although the argument is fairly mathematical it has a very direct experimentally verifiable consequence in the ballistic conductance. 

[Slide 6] Well that brings us to the end of things we wanted to talk about in this unit and now it's time to sum up. Thank you.