nanoHUB U Fundamentals of Nanoelectronics: Quantum Transport/L1.8: Graphene
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[Slide 1] Welcome back to Unit 1 of our course. We are now on the eighth lecture. So in the last two lectures we talked about this how to go beyond one day and how to take into account the lattice with a basis. Now what we have is an example, a real word example graphene, 

[Slide 2] which requires us to think of a 2D lattice with a basis. So what's graphene, well that's something I already explained in the introductory lecture, that graphite has these two-dimensional layers which are weakly coupled and this coupling is weak enough that people are able to peel off a mono layer of carbon atoms from the surface of graphite and that's what's called graphene. And what you have here then is this two-dimensional array of carbon atoms. So here I've shown this red and blue but they are actually all carbon atoms and I'll explain in a minute why I've drawn some as red and some as blue, we'll get to that. But what I've shown here is a-- like a unit cell and its surroundings, of course the solid itself goes on forever. It's like this infinite lattice here, I've just drawn a small section of it for our purpose, OK. Now, as I'd explained in the introductory lecture for graphene if you're trying to understand its electrical properties you can use just one basis function per atom. That's this pz orbitals and that's a non-obvious statement but we had discussed it a little bit in the introductory lecture. For the moment we'll just assume we got one basis function per atom which means if one atom where a unit cell, then the unit cell would just have one basis function, right? And then the theory would be a little simpler because then you wouldn't have to consider the lattice with the basis it would be just a 2D lattice. But the thing is that unit cell though actually needs to contain two atoms and the reason is you see the definition of a unit cell is that if we consider the unit cell anywhere, they are all identical that is when you stand on one and you look around you can't tell where you are on that lattice. Now the point is that this atom is a little different from that atom in terms of its environment. If I stand here, I see two people on my left and one on my right. If I stand here I see one on my left and two on my right. So in that sense the environment that this carbon atom sees is different from the environment that this carbon atom sees. And so you can't call one of these atoms a unit cell because the next one then would be different but unit cells no matter where they are it should all look the same. So what you need to do is take two of them and lump it in one and think of that as a unit cell, then you see it's all identical, that is if I stand here in this unit cell I see two on my left, two on my right and you go anywhere on this lattice and any other unit cell and you look around you will see exactly the same environment. And so what we do is when you want to calculate the band structure as you know we have to evaluate that quantity so you stand at some unit cell n and do the summation over the all other unit cells m. So what we do is we stand here, call that n and do our summation around there. And because it's periodic what it means is I could have stood anywhere on graphene, on any unit cell and done-- perform that summation and got the same answer. So we stand here then and we need to do this sum, how many terms in this sum? Well, there'll be five terms because you could put m equal to n or you could put m over here, over there, over here, over here. So those are four unit cells next to it, so those four and m equals n. So there'll be five terms. So what we need to do is write out those five terms and add them up. So the first one is m equal to n that's basically just the matrix describing this unit cell isolated from everything else and there your diagonal elements will call it epsilon and the t is the element between the red and the blue on that-- inside a unit cell. So it would be-- look like a matrix like this and since n is equal to m just like before there's no phase factor involved. This is just e to the power i0. Well, now let's put the m on this one. So you're now talking of that one now what you get is again a matrix that describes the coupling of this two component thing to that two component thing. What does the coupling matrix look like? Well, here you have a red and a blue, here you have a red and the blue and the coupling then is between the blue of this and the red of that. So that's why you have a t there and the rest are all zero. What about the phase factor? Well remember that rm minus rn basically means a vector running from this unit cell to a corresponding point on that unit cell. So you could draw a vector between this red and the red or you could draw a vector between blue and blue, or from the midpoint to the midpoint. So anyone of them they're all the same vector. So how much is this vector, well it has a little bit of an x component and a little bit of a y component. The x component is like that much which is half of what I have denoted here as 2a. So 2a is that distance what do you want is that much, like half of it. So that's a. So it is a in the x direction and in the y direction it's actually that much which is half of what I've labeled as 2b. So that's b in the y direction. So it's a e to the power ik dot that vector, that's this vector it is written as ax plus by. Next, we can put the m here, so exactly the same coupling matrix, same story blue of this couple to a red of that same story, but now the phase factors are a little bit different. Why, because now the vector we are looking at is this one you see previously we have this one now we have this one. Same x component but the y component is now negative so it's ax minus by. Well, move on to this site. Here again now you'll see the matrix has actually reversed because it's now the red of that one coupled to the blue of this one. So if you think through it the t is now in the upper-right hand corner and the phase factor again now we want this vector and its x component as well as its y component are now negative. And finally for this one is the same matrix because it's red after this couple coupled to the blue of that same story, but now the phase factor again a little different, the x is negative but the y is positive. So these are the five things, the five 2 by 2 matrixes that we need to add up to get the h of k that will be whose eigenvalues we are suppose find. So to add these up, these are the five terms if you add it up, you can see as far the diagonal element is concerned is clearly just epsilon. Why, because none of these four have any non-zero elements on the diagonal. The diagonal elements are just zero. So when you add it up the diagonal is still epsilon nothing's changed if we look at the lower left corner that's where the contributions come from these two that's why you have something on the lower left. If we look here they contribute to the upper right corner and if we look at it carefully you'll see this is the complex conjugate of that which it must be usually, you know, this matrix has to be Hermitian and so whatever is here whatever number appears here is the conjugate of that one. So let's try to write h0 basically it is t, oh sorry, we're looking at the lower left so its this t plus that t times this quantity which is k dot xsk, x component of k 

[Slide 3] so that's kxa, k dot y that's y component of k, so kyb. Here also it's kxa but now it's minus kyb and all of them have a t in front so this is t that's t, that's t so we put it all together like so. And here you'll notice both of these have e to the power ikxa so I could pull that out an exponential ikyb plus exponential minus ikyb. When you add that you get twice cosine kyb. So with just a little trig you can get from here to that. So this is the expression then for this lower left and if you work out the upper right it will just be their complex conjugate of that. What that means is instead of plus ikxa you'll get a minus ikxa you'll notice on both of these the x component has that minus sign to it. OK, so now we are ready to move on to find looking at its eigenvalues, I'm thinking about that. 

[Slide 4] So we have this h which is a 2 by 2 matrix, as epsilon's on the diagonal h0 and as I've mentioned before in the last lecture, when you have a 2 by 2 matrix and the diagonal elements are equal, the eigenvalues are just given by the diagonal plus minus the magnitude of h0. And what is the magnitude of h0? Well, that's this quantity, so if you want its magnitude you'd have to multiply by its complex conjugate get and take the square root. Now, so these eigenvalues then will all look symmetric around the epsilon. So the epsilon is here. There'll be for some k to h0 depends on k so for some k you might get a level here, but if you get a level there, they'll be another one down here. So, it will always be like a mirror. Whatever you get up here the mirror image will be down here. They all occur in pairs. Now where is the fermi energy? Well, as I've mentioned before in graphene, I guess this was in the introductory lecture, the number of electrons is such that it fills up exactly half the levels. So what that means is to accommodate all the electrons, they can all go into this, this half. You don't need to use the upper half which means the fermi energy or the electrochemical potential at equilibrium would be right around here at epsilon. Now for current flow purposes as I've mentioned before, what matters is the energy levels right around the fermi energy. So what that means is what is really most important is the levels around here. And although I've shown a little gap here actually in graphene there is no gap in the sense, these kind of touch each other right there and what you're really interested in is the energy levels right around that region. What that means is, we are interested in those values of k. For which the magnitude of h0 happens to be zero. Now where is that? You might think well possibly around the origin, where kx is zero and ky is also zero, but not really if you put it in you can see there. If kx is zero, exponential zero that's one, ky0, well cosine zero that's also one. So you get 1 plus 2, so h0 is three times t and that's about as big as h0 gets. So, what that means is, at k equals zero actually h0 is huge, you're getting the levels up here, down here. So those are not the important points as far as current flow. So as current flow is concerned what really matters is those points where h0 is? Close to zero, small. So how do you locate h0? The zeroes of h0, well you could do it numerically which is what we have done here is-- this is the kxky plane. And what we have done here is what you might call a grayscale plot, that is we have plotted the magnitude of h0 at different points in k. So you choose a particular kx, particular value of ky. Look at the magnitude of h0 and ask the computer to plot it in this grayscale plot which means if it is small you make it black, dark. If it is big then you make it white. So you can see its whitest right there because it just as we discussed when kx is zero and ky is zero, h0 is kind of at its peak, it's 3ts. So that part's real white. So where are we looking? We want the black parts, so there are these parts, you know, where it actually goes to zero. So what are these locations in k, where actually h0 is zero. Well you can easily check this through that point course is kx equal to zero. But ky is not zero but it's 2 pi over 3 because if kx is zero, you see that's out, but cosine of 2 pi over 3 that's minus a half. Its cosine 120 degrees, that's minus a half. And so two times minus a half is minus one that cancels the one. So that's where the h0 is zero and the same works for of course minus 2 pi over 3, so those are two points there. Another place where this goes to zero is this one which is where kx is pi. So because kx is pi this is e to the power i pi which is like a minus 1. So it is 1 plus 2 times minus 1 and then pi over 3. The cosine of pi over 3 is half cosine 60 degrees, that's half. So one plus two minus one and half, that again cancels out, that's how you get the zero and you'll get a zero if it is pi and minus pi over 3. Or if it is minus pi and pi over 3 and so you have these 4 points. So those are the zeroes, those are the values of k where h0 goes to zero which means around there, if you look at the energy eigenvalues they would contribute right around here and they are the important ones as far as current flow goes. So what the people often do is rather than use this general expression that works for all k, what you do is, you go to one of these zeroes and do a little, what you might call a Taylor expansion of that function around that point. So try to write a simple function, some polynomial. Like a linear function of k-- ks, but which works right around there. So let's see how that is done. 

[Slide 5] So supposing you want to find an approximation to this, an approximate function that will work, that will represent this but right around there because exactly at that point, it's zero. But question is as you move away from that point, how does it change? Well the general principle is, when you want to do the this taylor series expansion is you look at the derivative of h0 with respect to kx at that point, multiply it by kx minus zero, that means kx minus the value where you're expanding. Similarly for ky, you look at the derivative with respect to ky and again evaluated at that one and multiplied by ky minus 2 pi over 3. And of course if kx is zero and ky is 2 pi over 3, you get zero, but as it moves away from that point, this is a linear function that will describe what you want fairly well. So, next step then is to evaluate those derivatives. So take the first one. I've to take the partial with respect to kx so there's a t in front that's fine and look inside. Derivative of 1.0, now when I take the derivative of this, again it's partial with respects to kx, so ky for this discussion is-- are like a constant. So when you have an exponential and you take the derivative, it's like multiplying by ia. So that's what you have here, 2ia. So ia's because I took the derivative with respect to kx. And then we have the old function, now we evaluate it at that point. So put in kx equals zero which means this drop's out, put in kyb equals 2 pi over 3. So you get to minus a half. So what you then get is minus iat times kx. So that's this one, next we need to do the second one. Again, similar story but now we need to take the derivative with respect to ky. So kx parts stays the same but cosine kyb when you take the derivative you get minus sine kyb and then multiply it b. So that's the b and then you have the minus sine kyb. Again, it needs to be evaluated at that point, so when you evaluate it you get this minus b square root of 3. The square root of 3 is because you're taking the sine of 2 pi over 3, sine 120 degrees which is like square root of 3 over 2. So when you put that and then you get this quantity and I've written this as Beta y rather than ky because it's actually ky minus 2 pi over 3. So what you want to know is not the absolute value of ky but how much it has deviated from that zero. So that's what I'm calling zero, so that's what I'm calling bty. So overall then, that function can be approximated with this what you might call a linear function, it's linear in kx and ky. Now, it looks like there's a different coefficient for kx than for ky, but actually those happen to be equal. That's because if we look back at the definition of a and b, this is what we call a and this is what we call b. And since this angle is 30 degrees, what you can show is that square root of 3 times b is actually the same as a. So over all then, you could write it as minus iat times kx plus i Beta y. And this is the result that you often use. So this is the expression for h0 around that zero, that location, location in k space where the h0 is zero Now you could do a similar extension around each one of those, but don't all look about the same wherever you go. The details would be a little different, but at the end of it you'd get an expansion like this. So I might say well there are these different pockets of electrons that conduct and anyone of them could be described with that expression. 

[Slide 6] So what will the energy eigenvalues look like, well if you use this approximation then you'll see the magnitude of h0 is very simple, it's just square root of kx square plus beta y squared. So it's like epsilon plus minus at times this, so squared-- and this is like the magnitude of the ka vector which means out here as you move away, the further you go, the energy changes linearly. So this is kind of different because we are used to this parabolic dispersion that is a changes as kx squared, you know, usually you have this h-bar squared k squared over 2m, but graphene is famous for x linear dispersion where e is actually proportional to the magnitude of k. And because of that, the velocity of electrons in graphene is actually a constant, you see usually we think the velocity will depend on k in the sense you think of mv as being h-bar k. And so, the bigger the h-bar k, more the velocity, but in this case because energy is linear with k, you see the correct definition for v and this is something we talked about in the first-- in Part A of this course and the correct definition of v is derivative of e with respect to the momentum that is h-bar k. And because it is proportional to k, what you get is just at, dE/dk is just at. And so you get at over h-bar. And its independent of k, so it could have electrons in there with different values of k and the energy will be changing linearly, but the most important point there is that alluded-- the energy is changing linearly, but the velocity is the same at each k. It's the fixed number for graphene. And how much is that number, well you could estimate it, you could put in the values that a is of course that length very well known number in the sense this is the carbon, carbon bond length in graphene which is well known, I think its about 1.4 angstroms which is like 0.14 nanometers and this is like 1.5 times that. So that comes to about 0.2 nanometers. And then there is this one parameter t which of course if we're doing first principles theory you'd have been calculating by doing various integrals, but for our purpose it's a fitting parameter, right? It's something we adjust according to experiment. I mean, general for graphene people believe it's about 3 electron volts, you know, somewhere 2.5 to 3 electron volts. So if you put in those numbers you'll get a velocity of about 10 to the 6th meters per second, probably a little less than that, but the point here is that it kind of illustrates this whole semi-empirical approach very nicely. That in the semi-empirical approach we treat that t is a fitting parameter, what do we fit it to, well some well known experimentally measured quantity, it's like say, velocity. If you know that it's about say h times 10 to the 5 meters per second, well use an appropriate t so that at least comes out right. And then you can see whether you can use this simple model to understand other things. So as long as you have used one experiment to fit your numbers, and then you are able to explain 10 experiments then you are obviously doing something right, OK. Now, so that's the nice thing about graphene, that it illustrates this whole thing in a very almost textbook way. You can easily go through it in a simple lecture like this and see all the steps nicely. Now there's one detail about graphene though, we still haven't talked about and that's what the next lecture will be about. And that is, I mentioned these valleys and I say that you could-- that actually the electrons that conduct in graphene could view it as if they are in this pockets. And you might say that well, we got six little pockets like this so it is almost like six things in parallel. And this is something then that would have a very direct experimental evidence for that, that is if you measure the ballistic conductors as you know normally a ballistic conductance is 12.5 kilohms. It's like about 25 kilohms per level and 12.5 because usually there's this two spins. OK, this is what we talked about again in Part A of the course. If you have six pockets like this then you'd think the resistance would be 1/6 of that, OK. So instead of 12 kilohms it would be like 2 kilohms divided by six, but not really because actually although this looks like six, in graphene they are actually only two of this pockets that you called them valleys so in graphene they have actually two valleys. This six it's not their right number, it's actually effectively only two. And so the ballistic conductance in graphene and in carbon nanotubes is actually 12 divided by 2 which is about 6 kilohms and that's experimentally directly measured. So what we'll address in them next lecture then is how many valleys are there, why is it that apparently it looks like 6 but it's effectively only 2 and that's because of course you have to consider this Brillouin zone as I've mentioned earlier, you only look at the stage within one Brillouin-- within the first Brillouin zone. And what we'll find is when you do that this 6 kind of effectively become 2. That's what we'll talk about in the next lecture. Thank you.