nanoHUB U Fundamentals of Nanoelectronics: Quantum Transport/L1.6: Beyound 1D
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[Slide 1] Welcome back to Unit 1 of our course, this is the sixth lecture. 

[Slide 2] So what we talked about in the last couple of lectures is how to write down the energy eigenvalues  for this one D solid. And what we now want to do is bring in a couple of additional complications, generalize the results so that we can do real solids. Okay? And the complication we'll be addressing in this lecture is consider a two-dimensional solid, you see? And here also you can visualize the matrix in terms of a diagonal value you call epsilon and a nearest neighbor coupling which I'll call t1. So it's t1 in the x direction and t2 in the y direction. Those could be equal, but they don't have to be. Now here again if I have to actually write out the matrix, that would get messy, because you see in 1D, you know, this -- this -- these three are kind of next to each other when you write them out. In 2D, you know, those are next to each other, but this one isn't, so writing out the matrix itself can be messy. But the point is we shouldn't have to write out the matrix, we can just kind of visualize the structure of the matrix by writing down what's on the diagonal and what couples to the nearest neighbors. And go on from there, the question is, in this problem, then, can we write down the Energy Eigen values analytically? 

[Slide 3] Now, the point is, then, if we take the Schrodinger equation, it's this E psi equals H psi, and it will look something like this, E psi n. So if you stand at one point n and E psi N will be equal to there will be this Hnm connecting to various psi m's. So if you stand here it will be connected to this one, that one, that one and the one up here that I haven't drawn, all right? So you could write it as summation over m, Hnm psi m. So this is what the matrix equation would look like. In 1D also we could have written it this way. In 1D that summation actually had three terms in it, the diagonal term, and then epsilon and then the t to the left and the t to the right. Here this summation has kind of five terms to it, a diagonal term and then a t1 to the left, t1 to the right, t2 below and t2 above, right? Now again, because of the periodicity of the solid, the fact that you could go to any point n and the matrix structure would look exactly the same. Whether you stand here, whether you stand there, whether you stand there, doesn't matter, wherever, any form of solid. You can stand anywhere, you wouldn't know the difference. They all look the same. As long as that property is true, the solutions to that equation can be written again in this form of plane waves that we talked about. Psi n equals psi 0, E to the power ik dot rn where rn is the location of the nth atom, so wherever that is. Now if you take this and substitute it in here, you'll get, instead of psi n we put in the psi 0 E to the power ik dot rn, and here also you put in the psi 0, and like before the psi 0's cancel out and you could take this exponential from here and put it on this side, with a minus sign. And therefore you then have the expression for the energy in terms of k. So just like before, Ek is now equal to Hnm e to the power ik.rn minus rn. So what we'll do next is apply this to this problem. This is a general relation. Actually would work 2D, 3D, doesn't matter. In general, we just stand at one point and do the summation of all the neighbors that it is connected to and it should work. And what we did for 1D was a special case where you just had three terms. So here what we'll do now is evaluate this for that 2 dimensional solid. 

[Slide 4] So as I said, there's a lot of terms here, five of them, because you're standing at a particular point, n, that index, and then doing the sum over all the m's. And of course the reason it works is because it's this periodic solid, so it doesn't matter where you choose your n. You could choose it there, you could choose it there, choose over there, doesn't matter, you get the same answer. The neighborhood looks the same. Okay, so let's try to write out all those five terms. So the first term would be when m -- you have to do the summation over m, when m is right on top of n. Well in that case, that phase factor is zero, so it's just HNM and the diagonal element is epsilon, so that's it, that's the first term. Now let's try to write down the next term corresponding to that point. So supposing your m is here. So of course the connection that's t1, that's the t1, what takes a little time to figure out is this phase factor, you know, the rm minus rn. So, you see, you have this m and n here, so you want this rm, so if your origin is here, rm is that vector. rn is that vector. And you use this triangle law of adding vectors you can show that RM minus RN is essentially the vector joining this point to this point, so it is like if you draw a vector from the point n to the point m, that's really rm minus rn. So what is that vector? Well, it's a vector whose length is a and it points in the direction x, so that's what I've written here, xa. So that's this phase factor when m is located at this point. So now when you have k.x, that's the x component of k. Because, you know, if k is a vector, so if you want it's x component, what would you do? Take it's dot product with x, so that's the kx. So this then becomes t1 e to the power ikxa. Okay, so we have done two terms, one with the m right on top of n, and the other with the m sitting there. Next term is with the m sitting here. Okay? And in this case, of course it's exactly like before, same t1, but now the vector is this one, which is still of length a, but in the negative x direction. And so instead of ikxa, you'd get minus ikxa. So that's three. Now we have got two more, one below and one above, and these would have the phase factor corresponding to this vector or that vector. And these would be vectors whose length is b pointing in the plus y direction or the -- sorry the plus y direction or the minus y direction. And so it's plus minus b in the y direction and the corresponding factor would be t2 e to the power plus minus i but now it would be y component of k times b. So those are the five terms, one, two, three and then two of them, five. So if you add them all up, of course this and this would add up to give you twice the cosine kxa and this, if you add up the two terms, you'll get twice the cosine kyb. So that's the dispersion relation for this two dimensional lattice. 

[Slide 5] One other question you could ask is, what is the Brillouin zone. Remember when you're counting states there's a certain region that you have to look at and anything outside that would be double counting, so you only look for states within those case. In the 1D case we saw that the range of k goes from minus pi over a to plus pi over a. For this rectangular problem you would have a similar situation with respect to ky. It will run from minus pi over b to plus pi over b. And the first Brillouin zone would essentially be this rectangle, whose bounded by those limits, see? So in this rectangular lattice it's fairly straightforward to extend the concept of Brillouin zone from 1D to 2D and on to 3D. But later in these lectures we'll be talking about graphene where it's not quite so straightforward and you have to do this more formally, which you'll be doing this in a later lecture, I believe the ninth lecture. And that's because it's not a simple square lattice or a rectangular lattice like what you have drawn, it's got this hexagonal structure. So that we'll come to later. In this problem, this is it, really. This is the Ek relation and if you want to actually count the states, you look at the first Brillouin zone  and then you have to look at discreet points in k. And what determines those discreet points depends on the size of the solid. If the solid is big, the points are much closer together, which is usually the case, if the size is small, you'd look at the points would be further apart, so just like before. 

[Slide 6] So what you want to do next is, see we started with the 1D problem, talked about the complications, one additional complication, that's the 2D, and what you want to now talk about is another problem, another complication, which is what you see here. It's a perfectly periodic solid with one element within a unit cell. But more generally you could have multiple elements within a unit cell. So what we'll consider next is a 1D solid but this unit cell actually has like two types of atoms, or something that's -- I mean, so the basic unit cell doesn't -- is not just one thing like this, but like two things are denoted with a x and a zero, so that's what we'll do next. Thank you.