nanoHUB U Fundamentals of Nanoelectronics: Quantum Transport/L1.4: Dispersion Rellation
========================================
 

[Slide 1] Welcome back to unit 1 of our course This is the fourth lecture. 

[Slide 2] Now, in the last lecture, we talked about this hydrogen molecule and how you could use this idea of basis functions to turn the differential equation into a two by two matrix in this case because you had these two basis functions and the general form of the matrix equation usually looks something like this. E, then there's a matrix s called overlap matrix, and then there's this H matrix. Now, this, as we discussed in the last lecture, you could take -- multiply through by the inverse of this matrix so that it looks like E times Psi equal to a matrix times Psi. So in other words, instead of having two matrices, s and H, you put it into one matrix like this. And the reason I've put a prime here is so that one could now drop the primes and just call this matrix in terms of unprimed things. So it would look like E-Psi equals H-Psi. Now, the reason I mention this is, you see, if you are trying to do a first principles theory, then you might want to calculate the H and the s and start from here. On the other hand, what we say to us, we were just going to adjust the parameters so as to fit certain known experiments. So in that case, instead of starting from here, we might as well start from here and create Epsilon and t as our fitting parameters. See? And that's what we'll be doing from here on. I won't really bother to write the s anymore. We just work with this. E-Psi equals H-Psi and H then is described in terms of certain parameters, which you can adjust. 

[Slide 3] Okay. Now, what we'll be talking about next then is that, well, we have this hydrogen molecule, where we talked about how you get these two levels. What we want to talk about is something that's more like a solid, that is consider for example, like, a hydrogen solid. Like, one-dimensional solid with hydrogen atoms arranged in a row. Lots of them. Well, running from one to capital N. So now, you got n basis functions. So the corresponding matrix will now be N by N. So it will look something like this. E-Psi-1, Psi-2, up to Psi-n, and then, the matrix here with some element down the diagonal. And since it's uniform solid, you'll expect to see whatever here, it will be the same number there and same number and so on. And then, there will be a number that will be connecting the nearest neighbors and then I put a zero there. You might wonder how do I know that's zero? Well, if you look at how you're supposed to calculate these matrix elements, you'll notice it involves the product of these basis functions. Now, if you're trying to find this element, let's say one and three, then this one would be like u1 and that would be like u3. So this is u1 and that's u3. And you see, one and three don't have much overlap in the sense that wherever u3 is big, by that time u1 has more or less become zero. So when you multiply the two, it's a very small thing. So in general, it's true, then, that as you move away from the diagonal, these matrix elements get smaller and smaller. And in all these -- some empirical models, you see where we treat matrix elements as fitting parameters, you want to keep the parameters down to a minimum. And usually, there is one uses, what's called a nearest neighbor model. In other words, only if two basis functions are right next to each other, we assume there is a non-zero element there. On the other hand, if they're far from each other, like one and three, we put a zero. If it's one and four, we assume it's all zeros. So in that model, you would have something running down the diagonal. Diagonal, of course, involves the same basis function both places. So that's non-zero. And we assume something that's nearest neighbor, one, two, two, three, three, four, et cetera. And the rest you assume to be zero. Of course, the principal of band structure that we'll be talking about, that would apply even if you went beyond the nearest neighbor. All that is needed is that things should look periodic. That means, in this entire solid, solid,  if I am standing here or if I am standing there, it makes no difference. The world looks the same whether you stand here or there. As long as that is true, you can use this principle of band structure. But for the moment, let's just assume we have a nearest neighbor model and you could draw a picture like this to illustrate your model. That is, you have these basis functions, one, two, three, up to n, and for -- there's a diagonal element, Epsilon. That's what goes on the diagonal. And then, there's the t, which couples, nearest neighbors. There's the t. So in general, as you go to more complicated things, like two dimensional solids and so on, it gets more and more messy to actually write out this matrix. So it's much easier to visualize it like this. So one thing you should get used to is how to kind of connect this with this. So if I draw this, mentally you should kind of picture this. Because this is the matrix whose Eigenvalues we are trying to write down. On the other hand, writing out this huge matrix is always a pain. It's much easier to just draw that picture. Right? So we know what you're talking about. 

[Slide 4] Okay. Now, the point is then we want to write down the Eigenvalues of this huge matrix, and it's a huge matrix because this solid probably has, let's say, a million points down. So it is a million by a million matrix. But what you can do is make use of the fact that every row looks the same. There's a t, Epsilon, t. t, Epsilon, t. No matter where you go. And because of that, you can actually write down their Eigenvalues analytically. So what you could do is, as you know, a matrix equation like this actually a set of n algebraic equations. That is, we have e-Psi-2 is equal to t-Psi-1. You know, the rule for matrix multiplication. T-Psi-1, Epsilon-Psi-2, and then t-Psi-3. Similarly, the next one, e-Psi-3, would be t-Psi-2, Epsilon-Psi-3, t-Psi-4. Okay? So in general, you could write one of those equations as e-Psi-n, n being the -- the small n being -- telling you which row you are on, and that's equal to Epsilon-Psi-n, because Epsilon is on the diagonal, and then t-Psi-n minus one, because t connects the one just before it, and then t-Psi-n plus one because that t connects the one right after it. So here, if you put n equals two, you'll get the second row. N equals three gets you the third row, fourth row, and so on. So this equation kind of describes this entire matrix -- set of matrix equations. Right? So instead of algebraic equations, written like a matrix. So one thing you should be able to do is see these connections. That is, that one line, how it is equivalent to many, many algebraic equations, which in turn is equivalent to this matrix equation, which is E-Psi equals H-Psi. Okay? Now, this equation, we could try to solve in this way. Divide through by Psi-n. So it looks like E is equal to t, Psi-n minus one over Psi-n, plus Epsilon, because that's Psi-n divided by that Psi-n goes away. And then t, Psi-n plus one over Psi-n. Now, the way you can -- the reason you can solve this analytically is that just as we talked about in lecture two, about differential equations, that as long as the coefficient does not vary, you can write the solutions in the form of plane waves, exponential IK.R. Here also, as long as these coefficients are the same in each row, you can write down the solution in the form of plane waves. That is, Psi-n is equal to Psi-zero times e to the power i-n-k-a. And so, it's kind of like the e to the power i-k-z, because the z is now n times the latis spacing a. So n-a is like the location of the z, the z location. Right? It's now a discrete latis, so that's why it's an integer times a. Now, if you use this -- so how do I now this works? Well, as with differential equations, you put it in and see if it works. Basically, we check if it's working. So if you are -- accept this, then you can see what Psi and minus one over Psi-n. It's e to the power i, n minus one, k-a over e to the power i-n-k-a. And the n cancels out, leaving e to the power minus i-k-a. Similarly, if you are to do Psi-n plus one over Psi-n, you would have got n plus one there. So you would have got e to the power plus i-k-a. So if you put that in, this becomes t to the power minus i-k-a, plus epsilon, plus t to the power, plus i-k-a. So e is equal to that. And as long as your k and e are related by this relation, the point is you have found a solution to this entire large set of equations. You see, analytically. Now, one little thing that might bother you is that -- you see, I say that every row looks the same. But what about the last one here? You know, here is Epsilon and then the t here and the t there. But here is just Epsilon and t, nothing to the left. See? And that's why, of course, usually, if you thought of a solid that ends here, then what I just said won't actually work. This won't really satisfy the full set of equations. You'll get in trouble at the ends. But what people do is they assume this periodic boundary condition. That is, they assume that the solid is in the form of a ring, Okay, so that this end connects up to this end. And here, I made this point in part A also, that the reason you normally use periodic boundary conditions is not because real solids are periodic. Real solids are usually in, like, boxes. They end here and start here. But the feeling is that as long as you're talking about the bulk properties of a large solid, the actual boundary conditions don't matter. Physically, you know that. That the actual properties of the solid don't depend on the conditions at the surface. And so, in that case, you might as well use whatever matter -- whatever boundary conditions makes your life mathematically the simplest. So that's the usual justification for using periodic boundary conditions. That is mathematically simplest and the actual things don't matter. On the other hand, when you get to small things, of course, the boundary conditions do matter. And interestingly, I mentioned graphene in the last lecture. Graphene is one of those materials, I mean, one of those rare materials, which is actually available -- or has been studied both in the form of a flat sheet like this and in the form of a rolled up sheet, which is called a nanotube. You see, where it's like a straw. In that case, the periodic boundary condition is actually the correct physical boundary condition. You know, not just a convenience, but it's the actual boundary condition. And what people find is, indeed, when you compare graphene to nanotubes, if the circumference is very big, it doesn't matter. The properties don't really matter whether it's graphene or a nanotube. But once it gets small, the circumference gets to be, like, 10 nanometers or less, then it does matter. Okay? Anyway, but for this principle of band structure that we are talking about here, the point is we'll always assume periodic boundary condition. In other words, as far as this matrix is concerned, what you are kind of doing is taking point n and connecting it to point one. So it's as if you are adding a t at this end and a t at this end. So as long as you do that, whatever we have done is exactly true and you could have taken this big matrix and gone to Matlab or -- I mean, to a computer and found its Eigenvalues and it will match exactly what we are getting analytically. But the -- if you are doing it numerically, the important thing is you will have to remember to add those because the theory -- the analytical theory, corresponds to periodic boundary conditions, not otherwise. Okay. So under those conditions, then, you can write the Eigenvalues in terms of the values of k. Now, 

[Slide 5] you could combine that minus i-k and plus i-k and write it as Epsilon plus two-t, cosine-k-a. All right? Now, here, I assume that the t was the same, both one-two and two-one were the same. In general that need not be the case. You see, the matrix has to be Hermitian. So -- and what that means is H-m-n will be equal to H-n-m complex conjugate. So if you have t here, the number here must its complex conjugate, but they don't have to be equal. So in general, you could have say t to the power of plus I-Phi and t to the power of minus I-Phi. Well, what would be the dispersion relation then? This e-k relation? That's straightforward because from here, you see the basic rule is simple. You just take t to the power of minus i-k plus Epsilon plus t to the power plus i-k-a. So now, you would have t to the power of minus I-Phi, minus i-k, t to the power of plus I-Phi and you do plus i-k-a. And now, if you combine the two, you would get Epsilon plus two-t, cosine k-a plus Phi. So what you want to do is get used to visualizing this matrix in terms of a picture like this. Because once you have this picture, it's very straightforward to write down the E-k relationship. It's just E is equal to the diagonal element Epsilon plus t times appropriate phase factor, t times a appropriate phase factor. 

[Slide 6] Similarly, if you wanted to include the next nearest neighbor, as I mentioned, our models were -- are usually nearest neighbor. But supposing you wanted to use a model in which the point is also connected to the next one over, and we call that t2, what would it -- how would the dispersion relation be affected? Well, it will just add two more terms to it. Because now, Psi-n would not only be connected to n minus one and n plus one, but also to n minus two and n plus two. And what that would do is just add a term here, which is t2 to the power of minus i2ka and the t2, e to the power plus i2ka. And those two, you could combine to get 2t2, cosine 2ka. So once you visualize it this way, it's fairly straightforward to write down the corresponding dispersion relation. 

[Slide 7] Now, how do I -- what does it look like this dispersion relation? Well, if you plot it, so this axis is this E minus Epsilon, so E is equal to this constant plus a cosine. Now, the way I have drawn this, you'll notice at k equal to zero, it's the minimum, and that k equals to Pi; it's a maximum. Now, that's the correct picture if t is a negative quantity. Because you see, when k equals to zero, the energy is Epsilon plus 2t. So if t were a positive, it would actually have been the -- have been a bigger number like that one. But because I am assuming t is negative, which is what usually corresponds to the cases we are interested in, that's why at k equals zero, it's a minimum. Whereas at k equals Pi, that's where cosine Pi is minus one. So it's Epsilon minus 2t. But since t is negative, that gives you a bigger number. Okay? Now, how would this -- how would we make this model correspond to the parabolic dispersion relation that we discussed for the wave equation? Remember, back in lecture two, we had a differential equation and there we said that as long as it was -- all had constant coefficients, you could write the E-k relation in this form, this parabolic relation. So that is not a parabola. This is a cosine function. And of course, there's no way you can make this lie on top of that. But what you could do is, if you chose your parameters carefully, you could make it fit reasonably well within a small range of energies. And often, you might be interested in problems where the electrons of interest, that are conducting electricity, are confined to this range of energies. In which case, this model would fit the parabola quite well. See? And of course, in solids, often, the actual E-k relation is not necessarily a parabola. I mean, it's in vacuum, it's always a parabola. In a solid, it's usually not exactly a parabola anyway. So the main point then is choose your parameter such that what your model gives fits the known E-k relations for that solid. So how would I make it fit a parabola? Well, you could use this Taylor series expansion. The idea is cosine x for small x is one minus x squared over two. So you could write it as one minus x squared over two. And then, there's x4 and so on, but those we are not worrying about, leaving it out. So if you want to make those two things correspond, then you can see t must be equal to minus h bar squared over 2m a square. I mean, t a square is that. So you can see. You have to choose your t like that. And as far as the other parameter, Epsilon, you should choose it equal to Ec minus 2t. So in other words, if you know that in your solid the electrons of interest have energy dispersion relation like this and you're trying to get this discrete model going with -- and you have to parameters to fool around with, Epsilon and t, then this is how you should choose them. And in that case, in your model, electrons will behave as if their energy levels follow that curve, the black one, whereas what you really want to do is the red one. But then, as long as they match reasonably well in your energy range of interest, you would be okay. So that's the general philosophy here. 

[Slide 8] Now, in the next lecture, what we'll go on to is another important concept here that we need to discuss, and that is -- you see, I started from this n by n matrix and said that because it's periodic, you can write down the Eigenvalues analytically. But you might be wondering that, well, you know, this is n by n matrix. That has only n Eigenvalues. Here, what we got was this continuous function. So how do I get n Eigenvalues out of this? Because after all, they should correspond to what that matrix would give. Well, that's what we'll talk about in the next lecture on counting states. Thank you.