nanoHUB U Fundamentals of Nanoelectronics: Quantum Transport/L3.5. Graphene
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[Slide 1] Welcome back to Unit 3 of our course on quantum transport and this Lecture 5. Now, in the last two lectures we developed a general method for finding the self-energy and what we want to do in this lecture and the next is look at a couple of examples of how you apply this method to different problems. 

[Slide 2] Now, earlier if you remember, that was in Lecture 2 of this unit, we talked about how you could find the self-energy using this method of basis transformations, so the idea is if you have a contact like this and you think of it as 1D chain with these alphas connected by the betas where alpha represents the entire column, beta is what connects one column to the next and then for a square lattice like this beta looks like an identity matrix in the sense it's not only diagonal but all diagonal elements are equal. Now, alpha has a non-zero upper diagonal and lower diagonal but what you could do then is do a basis transformation here and make this diagonal and beta, of course, still stays diagonal because it is essentially an identity matrix and so once you have done a basis transformation both alpha and beta became, can become, diagonal but this is not in general possible because there are problems like say for example graphene where we don't really have a square lattice, we have this hexagonal lattice and in this case, if you look at alpha, I guess you could try to use this unit for example you could think of that as one column. So, this is one unit, that's another identical unit and they're connected by these three betas, so beta would be these three things here and this would be your alpha and that would be another alpha. Now, in this case beta is not an identity matrix. Here beta was an identity matrix because every element in this column was connected to a corresponding element in that column, so 1 got connected to 1, 2 got connected to 2, 3 got connected to 3 and equally. Here it isn't quite like that. When you look at this column, you got many elements here in the 1 there, 1 here, 1 here, 1 here, etcetera and this one is connected to the that one but this one's not connected to anything directly; I mean, that point on this column is not connected to any point on that column at all. Again, this one's connected to that so what that means is when you try to write the beta you won't get something like this, you won't get an identity matrix, you'd get connections for certain points and 0s for many others. So, and now if you try to diagonalize this alpha, the alpha you get for one of these units, you know we this all this hexagons that you could write down, if you try to diagonlize alpha, beta would become something else. I mean, to start with it as diagonal but because it wasn't an identity matrix when you do a basis transformation it won't stay diagonal anymore and, of course, if alpha and beta are both diagonal then you can visualize the problem as lots of 1 D problems in parallel because they're both diagonal but if one of them is not diagonal then you don't have that luxury anymore; then all this different chains get connected up, you see. So, for graphing that's why you couldn't really use the method 

[Slide 3] we had in lecture 2 but what you could use is the method we just discussed in lectures 3 and 4 because in lectures 3 and 4 what we said is, any structure that's described by alphas and betas you could first find the surface Green's function by solving this matrix equation iteratively and then once you have the surface Green's function, you could write the self-energy by taking this tau g tau dagger. So, numerically that is how you could handle this problem. So, if you think of a sheet of graphene where, you know, carbon atoms are at these points as you know and you could think of current flow in this direction so that way you'd a contact here and a contact there so you'd have a Sigma 1 here and a Sigma 2 over there and those Sigmas you could write down using this principle, okay. And if you're interested in the details you should look at the matlab codes that are in the notes which describe how this problem is done. Now, so you could apply it to this problem or you could apply it to, I guess, a different problem where the contacts are put in a different direction for example. Now here, we called this usually the armchair edge, now why do you call it the armchair edge, well if the current is flowing in this direction and if you look at this edge, I guess with little imagination, you could see why someone might think of that as an armchair, see it's somewhere, I guess you could imagine laying down on that so that's why they call this an armchair edge. Now, when you go at 90 degrees to it which means supposing instead of the current flowing this way, you put your contacts in this direction so the current flowing this way, then you'd call it a zigzag edge because now it doesn't require much imagination. If you we look at this edge it sure looks zigzag, right. So, often these are the 2 limits that people talk about, of course, you have to be careful because, of course, if the, if this direction, you know, if in the transverse direction you have an armchair then the longitudinal direction you'd have a zigzag and so you have to be careful when you're referring to something as armchair which side you are talking about, this one or that one, okay. And of course, you could always have all kinds of other orientations because your contacts could be at an angle like this, it doesn't have to be like this, you know, either this way or that way like the 2 limits that are shown; these are the most common ones but there can be all kinds of different angles in each case though, you could write down these alphas and betas that is what you'd have to do is set it up, figure out you know 1D chain, what each unit should be, so figure out the alphas and betas, then you'd have to solve this iteratively to find the surface Green's function and then you could find the self-energy. 

[Slide 4] So, that would be the overall procedure. Now, I'll just show you some result that you could get and as I said the matlab codes are in the notes, so if you did the armchair edge you get this black curve and if you do the zigzag edge you get the red curve and what are we plotting here? Well, this is the function of energy, the transmission and this is a ballistic conductor, that is you're assuming here you have a ballistic graphene, you have 1 contact here another contact there. So, and so plotting this transmission and we choose a specific width in this case, I think it's about like 15 units or so in this direction and the 2 armchair and zigzag where chosen such that the width is approximately the same, so in this case the width corresponds to this direction, in this case the width corresponds to that one and, again, the dimensions that are chosen they're about the width and what you find then is that when you look at the transmission as a function of energy it's minimum right when it's in the middle and then as you go in this direction the number of modes increases, the density of states increases and if you go below, again, same story density of states increases, number of modes increases and one point to note is that there's hardly any gap here. This is the important problem with graphene in general that people talk about that unlike other semiconductors there's no gap in the sense there isn't an energy range where you have like 0 modes ordinarily because it's a to start with graphite is semimetal, so it's like there's density of states on this side, density of states on this side, it's 0 only right there and that, I guess, detracts from certain applications like if you wanted to make a transistor and you wanted to turn things off, you wanted to put your fermi energy in the gap and so not having a gap from that point of view is disadvantage. But, all this of course would come out automatically from this NEGF method that we have been discussing, as I stated before, that the power of this NEGF method is that if you know what you're doing, if you understand the procedure clearly you can get numerical results without necessarily understanding all the physics of what is happening and so you can use the numeric to improve your intuition and understanding, so what you should do is set it up properly, make sure you can calculate things numerically and then see if you can understand what you calculate. So, here for example one thing one could do is that is look at the semi-classical answer. So, what do I mean by semi-classical answer? 

[Slide 5] Well, the semi-classical one looks something like this. What we have discussed in unit 1 is that when you look at the band structure of graphene there are these multiple valleys, so this was discussed in unit 1 and if you go to one of these valleys right around there you have a linear dispersion relation where E is proportional to k, k being measured from that point, okay and so if you want to look at one of these valleys, you could write E as a constant times ka. Now, how do you calculate the number of modes? Well, roughly speaking again the idea is number of modes is the number of wavelengths that fit into the cross-section, so if you have a graphene sheet whose width is W and in the semi-classical description it doesn't matter whether it's armchair or graphene or zigzag because we are treating it as kind of a continuum thing. So, if it's a width W then the number of modes is kW/pi and that comes from this idea that how many half wavelengths fit into the width. These are the semi-classical arguments that we went through in part A of the course, okay. So, if you do that this kW/pi, now why the 2? Well, 2 is because in graphene there are these 2 valleys and as we discussed, again, in unit 1 there are kind of 6 valleys within a first Brillouin zone but each valley is a 1/3 valley, so overall it's like 2 valleys, 1 around here and 1 around here, so the number of modes is 2 times kW/pi and now if you use this expression you could replace k in terms of E and then you get this expression, so you get the number of modes as a function of energy and it's a straight line so that's what is plotted here, that's this straight line. So, this is what you'd have got from a simple semi-classical argument, you'd get that straight line and you can kind of see that the actual result is, follows that straight line roughly, at least at low-energies and we don't expect it to fit at high-energies anyway because this dispersion relation only holds at for small energies around that valley, so you don't really expect it to holdout here, but the interesting point is, until here it holds quite well, see there's the straight line and, of course, what the straight line isn't showing is all these steps because a straight line is assuming, I mean treating this just as a number not taking its integer value or anything so you don't expect the steps you just get the straight lines. Now, again the point to note is that when you do the numerical method you don't have to make all these arguments. You don't have to understand how many valleys graphene has for example. So, here in order to get this semi-classical answer are to put this 2 in; where did the 2 come from? Well, 2 because there's 2 valleys. How do I know there is 2 valleys? Well, because when you look at the first Brillouin zone there's actually this six minima and each one is a 1/3 valley and so on. Remember, we had a long discussion, so it takes a lot of understanding, a lot of discussion to come up with this expression, alright. That's kind of what we did in unit 1, but numerically you see, this plot came without understanding any of that. The numerical how did we get it? Well, we just setup graphene, wrote down the alphas and betas, found the surface Green's function, found the Sigma and numerically that's how we got it. So, when you have a new material where you don't understand all this yet, you could still use the numerical method but then what you should do is go through all the arguments properly and see if you can understand what you're calculating. 

[Slide 6] Now, numerically the other thing you can do is, easily is, from instead of graphene you could be looking at carbon nanotubes, you see the difference between graphene and carbon nanotubes is graphene is just a sheet like this, whereas in carbon nanotubes it's rolled up like a straw, right. So, you have this periodic kind of boundary condition along the width. So, here what I've shown is graphene, in carbon nanotubes what would happen is this end would get connected to that end again. Now, numerically that's easy enough to do. That is when you write the alpha you just connect this end to that end that's all. Similarly, for zigzag edge, you'd connect this end to that end, you'd think of it as if it's rolled around, so numerically you could just change the alphas and betas and if you look at the matlab  code, just takes one line to do that, you know, where you make that change and if you do that you'll get a different transmission, you see so we are plotting the same thing here, transmission but now you'll notice the steps are not quite as fine as this one; roughly speaking, with periodic boundary conditions you tend to get less number of steps but the steps itself are bigger, so again, it follows roughly the same semi-classical line but the details are now different and when people look at small structures indeed they see these differences 

[Slide 7] experimentally as well. So, that was I guess intended to illustrate how you apply this new method that we talked about in Lectures 3 and 4, this general method 4 finding the self-energy and that's this graphene example. What we'll do in the next lecture is talk about a different example involving magnetic fields, so we'll go back to the square lattice that we had and for a square lattice I had mentioned earlier you can usually get the simple answers by this method of basis transformation but not when there's a magnetic field involved, so what we'll look in the next example, in the next lecture we'll look at is how you can do problems that involve magnetic fields, small magnetic fields or high fields but in a square lattice.