nanoHUB U Fundamentals of Nanoelectronics: Quantum Transport/L1.5: Counting States
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[Slide 1] Welcome back to unit one of our course. This is the fifth lecture. 

[Slide 2] Now, last lecture we talked about how one could make use of this periodicity of a lattice, the fact that you're considering a very long, one-dimensional solid, but it's periodic in the sense things look exactly the same anywhere along the chain. You could make use of that to write down the eigenvalues of this N by N matrix where N can be a huge number, but you can write it down analytically. And you've got his E, K relation where for a given K you could -- there is a corresponding E. Now one point of course is not clear, and that is we started with the N by N matrix which naturally has N discrete eigenvalues. But what we have here is a continuous function which looks like I could give you any K and you'd find a corresponding E. So looks like we got lots of energy eigenvalues instead of just the N that we are looking for. Now actually the situation is worse. 

[Slide 3] That's because you see the equation we got was E equals epsilon plus 2t cosine ka. And what I plotted earlier was just the range from minus pi to plus pi. But of course ka could keep increasing. You see, it doesn't have to stop at pi. You could go to 2 pi, 3 pi, and so on. Right, so -- and if you plotted out the cosine function of course it would all be periodic. So it is not just that we have an infinite number of eigenvalues of energy we could find here. There's also energy eigenvalues we could find out there. Right, but then at the end of the day we are looking for N eigenvalues, right? So what we are talking about in this lecture is this rule for counting states, how we extract this discrete number out of this continuous result. So the first point is, of course, why you don't really need to come to consider anything outside this range of minus pi to plus pi. This range which is usually called the first Brillouin zone. And the reason you do not need to consider anything outside goes something like this. You see, when we derive this dispersion relation we say that because it's a uniform solid you could write the wave function that -- at the Nth point -- Nth lattice point in this form. For a given k we have a function looking like this. Now supposing someone says, "Well, I don't quite like your k. I've got a different k, and it's bigger than yours by 2 pi." So you had a ka, and he has a k prime a. And it's ka plus 2 pi. Question is does that represent a new solution? Does that represent a new wave function? And the answer is that well, corresponding to his k prime if you calculate the psi N prime what you'd get is what we had before and then E to the power in times 2 pi. That comes from this extra 2 pi. Right, so nk prim A is actually nka plus n times 2 pi. And the point is since small n is an integer that quantity is always one at all those discrete points of interest. And so psi n prime for every n is equal to psi n. So in other words, this other person, he or she, who came up with a different k -- k prime is not really describing a new solution. He or she just has a different way of expressing the same thing. It's the same size, you see. What that means is that on this picture if you have a solution with some k and somebody adds 2 pi to it that's not a new solution. It's the same one. So you could restrict your attention to a range of 2 pi anywhere on this axis. And the convention is to do it symmetrically around zero. You need not have done that. As I said, a window of 2 pi anywhere would be fine. But what we'll do is make it between minus pi and plus pi. And note that this argument works only because the lattice is discrete. Because we are only interested in the wave functions at these discrete lattice points, at point 1, point 2, point 10, 11. Not at 10.7, for example. There's nothing in our model about psi 10.7, because if there was then of course with n equals 10.7 that wouldn't be 1. But because the discrete is lattice you need to consider Ks only in the first Brillouin zone. 

[Slide 4] Okay, now the next point then is that okay, we understand that we are restricted to this Brillouin zone, but then it still looks like a continuous set of values. Why do -- how do we extract these N discrete numbers? Now this is where the periodic boundary condition comes along. And that is because you're using periodic boundary condition that is as if this whole thing is in the form of a ring, so point 1 normally would come to point zero, and point zero must be the same as point N. So it's a ring. So as you go around the ring what would have been psi zero should have been the same as psi n. But psi n, according to our model -- according to our solution is psi zero E to the power INka. And this condition would be satisfied only for specific values of k such that -- you see, this is psi zero; that's psi zero which means E to the power INka needs to be equal to 1. And that will be true only if Nka is an integer multiple of 2 pi. So that nu stands for some integer. So in other words, only those value of ka should be considered which are integer multiples of 2 pi over N. And now you can see how you get the N discrete eigenvalues, roughly speaking. You see, this range here is 2 pi, and only values of ka that are separated by 2 pi over N are allowed. So it's like there's a point here, another one, 2 pi over N, and then 2 times 2 pi over N, and 3 times 2 pi over N. And so the spacing is 2 pi over N, overall thing is 2 pi. So that's how you pick up the N values. So we'll see this with a concrete example. So overall then the point is the allowed values of ka are integer multiples of 2 pi over N. Or you could say the allowed values of K are integer multiples of 2 pi divided by the length of your solid. 

[Slide 5] N times the lattice spacing. Okay, let's look at a specific example which is actually also of practical interest, and that is if you had a benzene molecule, benzene molecule has carbon atoms that are actually arranged in a ring. So here of course the periodic boundary condition comes naturally; it's not a matter of convenience here. Right, this is benzene ring. So if you wrote out its matrix for something like this you would have these epsilons down the diagonal and then this t, t. So you know, exactly like our 1-D solid with periodic boundary conditions. So energy eigenvalues would be given by epsilon plus 2t cosine ka. But this is a solid of length 6 with periodic boundary conditions. So what that means is the allowed values of ka will be integer times 2 pi over 6. So that will give you, if you look at the values -- so you would put nu equals zero. That would give you that point. Minus 1 and plus 1. You know, when it's 1 this is like pi over 3, so it's somewhere there. And then twice that, thrice that. And then you're at the edge of the Brillouin zone. Now if you look at that you might say, "Whoa! Don't you have seven values there?" How did I get the minus three? Well, it's 2 pi over 6, so in order to get to minus pi you need minus 3 there. And plus pi is plus 3. And so you might look at that and say, "Well, you found me seven eigenvalues. There's supposed to be only six." Well, the point is minus 3 and plus 3 are really the same. Because remember, the principle is any two values of ka that are separated by 2 pi, exactly 2 pi, are equivalent. So this is minus pi, this is plus pi. So they're really the same. So you could choose either one, but not both. 

[Slide 6] Now if you had a solid of length N then your allowed values would be 2 pi over N times an integer. So you run from minus N over 2 to plus N over 2. Now if N was an even number, let's say 20 then this would be like minus 10 to plus 10, and that would tend to give you 21 values. But the point is you keep either this one or that one. But then if your N was an odd number like say 21 then you actually go from minus 10 to plus 10. And minus 10 then is not quite at the edge. It's a little bit inside because you see N is 21, so when you put 10 there it doesn't quite give you pi. So it's a little less than the pi. So minus 10 is here, plus 10 is over there. So -- and then overall you got 21 eigenvalues. So that's how you'd get these discrete results, you see, out of this continuous theory that we had done. Right, and those should match exactly what you'd have got if I'd just taken the original matrix and numerically solved it. So it's a very powerful method that allows you to write down these eigenvalue of a huge matrix just on a piece of paper. Now those of you who are familiar with discrete Fourier transforms they would notice this similarity. You see, in Fourier transforms usually you have a signal in time and in frequency. Here it is space z and k, so z and k are kind of like this time and f, t and f. And you see the fact that in z you have a definite range, that's what gives rise to the discreteness in k. So because it's of length -- that's the length of your solid, that's what determines the spacing of these points. And the spacing of the lattice here is what determines the range in K. So this Fourier transform relationship, which you may be familiar with if you are familiar with discrete Fourier transforms -- see it's kind of in a similar. On the other hand, if you have never heard of discrete Fourier transforms, you know, don't worry about it. That's just an incidental remark. 

[Slide 7] Okay, so we started with the -- this 1-D problem, and we talked about how you could write down the energy eigenvalues of this infinite matrix analytically. What we want to do is now introduce two additional complications so that we can do real solids. And the first one is instead of a 1-D solid let's think of a 2-D or a 3-D. So in how to go to -- go beyond 1-D that's what we'll do next. Thank you.