nanoHUB U Fundamentals of Nanoelectronics: Quantum Transport/L1.7: Lattice with a Basis
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[Slide 1] Welcome back to Unit 1 of this course. This is the seventh lecture. 

[Slide 2] Now in the last lecture we saw how we could take the simple 1D concept and extend it to 2 dimensions or 3 dimensions. Now, what we'll do in this lecture, we'll see how we can extend it to what you might call a lattice with a basis. Now, what do I mean by that? You see what we considered here, and we're still talking about a 1 dimensional example, here what we had is these X's represent the basis functions and it's periodic, so every unit cell has only one basis function, where as what we are going to consider now is something more general where if we looked at a unit cell, it has two basis functions. What do I mean by that? Well, for example you could have a solid where there's an A atom and B atom, and then again, A atom and a B atom, so that the unit cell has two atoms. And so you could have one basis function on each one of those atoms. That would be one case. Another possibility is it's the same atom but you're including a S orbital and a P orbital as your basis functions. So those are all possibilities that could give you more than one basis function for a unit cell, okay. And because it's -- this is different from that, the point is if we look at -- if you think of one basis function only, it doesn't quite look periodic. You know, it's a X and a 0 and an X and a 0. On the other hand, if we take the X and the 0 and put them together, think of it as one unit, then you see it's periodic. And that's the conceptual jump you need to make to handle real solids, because real solids have more than one basis function per unit cell. Indeed, often it's far more than two. So if you're doing a silicon band structure, you might well have 20 basis functions per unit cell, okay. But what we'll try to do in this lecture is to get it clear, like, having two basis functions, what difference it makes because there's an important qualitative difference with what you have when you do only -- take only one basis function. Now, of course when it's 20 the bookkeeping gets more complicated, but no new concepts are added. So what we'll talk about in this lecture then is how to conceptually, kind of take two of them, the X and the 0 and think of it as one unit, so that you now have a periodic lattice spaced by B, but each unit of that periodic lattice is two things. So that's this conceptual jump that I mentioned. Now, here, like every diagonal element was this number epsilon. In this picture, the diagonal element has these two things in it. So the diagonal element itself is now a two by two matrix. How do you write that down, this two by two matrix? Well, think of this block, this little square cut off from everything else. What that means is just an X and a 0. So what would be the matrix representing that? Well, epsilon on the diagonal and t between them. So, this then, itself becomes the diagonal element of this composite lattice, this lattice with a basis. What about the connection? Well, here the connection was a single number. But now the connection itself is a two by two matrix. Why, because it's connecting an object with two components to another object with two components. So you have an X and a 0 here and an X and 0 here. And the connection is actually between the 0 of this and the X of that, so between the 0 of the left one with the X of this right one. And that's what we have written as t prime, which is different from the t, so some other number, which denotes the coupling between 0 and X when the 0 is on the left, here, okay. So that would be the two by two matrix then, describing that connection. What about that connection? Well, similar, another two by two matrix, but now you'll -- it is actually transposed. The key is actually up here in this corner rather in the other corner. In general, whatever you have here will be the conjugate transpose of this one. And since these are real numbers, there is no conjugate to worry about, but you still have the transpose. So what was here would go there. But you don't need to invoke that idea if we just think about it where this should go, whether it's connecting the 0 of this with the X of that then you could figure out where the non-zero element should go, okay. So now that you have these three things, these are what describe this lattice with a basis, what you could write the wave function at the nth unit cell in this form. And just like before, because it's a uniform periodic solid, you can write it in the form of plane waves. So it's exactly the same, but now the wave function in the N unit cell is not a number, but has two components, an X component and a 0 component. So that's what we have indicated with the column vector, so two by one, same here, a two by one. 

[Slide 3] So how do you get the dispersion relation, proceed similarly, that is we first write the overall equation, the matrix equation, which looks like E psi equals summation, H nm psi, just like before again, except the sides as you know, this side represents the wave function in the mth unit cell and it has two components, an X and a 0. This psi is the wavefunction in the nth unit cell, again it has two components. And Hnm, which is the nm element of the H matrix, that's now got a two by -- that's like a two by two matrix. It's not a number. So there's a two by two matrix, multiplying a two by one given you a two by one. And if you substitute this into that equation, you can put that in there and you'll get again, looking just like before. And you could take this exponential ik dot rn and pull it over to that side with a negative sign and define a matrix H, which is Hnm E to the power ik dot rn minus rm. And E psi 0 equals H of K times psi 0. Now, if you look through, every one of these equations looks exactly what we had before. With this subtle difference or very important subtle difference, and that is the psi's are two by one column vectors, these things are two by two matrixes and because of that, you see, if these had been numbers, if psi 0 had been a number, then you see I'd have just cancelled the psi 0. If that's a number just take it out from both sides and you have the energy eigenvalue. We'd be done. This would be a one by one matrix, which means basically a number an E would be equal to that. We'd have the dispersion relation, move on. But now, I can't cancel out this column vector with that one. What I have to -- what I have is a two by two eigenvalue equation. So what I have to do is take this two by two matrix, find its two eigenvalues. So for every k, I'll get two values of energy. Unlike before, where for every k there was a corresponding energy. Now, for K there are two values of energy. And that's what I'd have to do next, okay. But otherwise, again, the steps are very similar to what we had before. 

[Slide 4] So how would we evaluate this matrix? Well, do the summation over m. This is a 1D lattice, so the summation has three terms. That is you can either put m, you see you stand at one point, n and then vary the m. So if m is equal to n there is no phase factor, that's this diagonal element, that's this one. If m is the one on the right, then you get this matrix. That's that one. And the phase factor is E to the power ikb. And if we take m and put it on the left, then you have this element with the phase factor of E to the power, minus ikb. So this H of k now is a two by two matrix and you obtain it by adding those three terms. Again, very similar to what we had with the -- for the lattice without the basis, the simple one. The main difference is again, instead of being numbers, we have matrixes. And the size of the matrix is equal to the number of basis functions per unit cell. Okay, if we add this up, what do you get? Well, you'll notice the diagonal element is just epsilon and epsilon because it's 0 here and 0 there. So when you add it up, you got epsilon and epsilon on the diagonal. If we look at the upper right corner, what you have is that term t and this one, t prime E to the power, minus kb. That one is 0, so that's what goes there. If we look here it's that term T and the T prime E to the power, plus ikb, which goes here. Note that this matrix is hermitian, meaning it's equal to its conjugate transpose. And that is if you take the conjugate of this and transpose it, you get back the same thing. And what that requires is whatever is here has to be equal to the complex conjugate of that of what is here. And you can see that works. So you could write it in this form h0, h0, star, h0 is that quantity. Now when you have a two by two matrix like this, you can check it out, as long as the diagonal elements are equal actually the eigenvalues are very easy to write down. It's the diagonal element, plus minus the magnitude of h0. So how do you find the magnitude of h0? Well, you multiply it by its complex conjugate. That gives you the magnitude squared. So if you want the magnitude, you take the square root of it. So what's inside if you multiply it out, you'll get that number. So this is the ek relation then. So previous remember, 1D lattice, we got a ek relation, for every K there was energy, E. Now, for every K there are two energies, epsilon plus h0, epsilon minus h0. 

[Slide 5] So what does it look like? We can try to plot it out. So your kb, k times b is the lattice size, that's the distance between each unit cells, runs from minus pi to plus pi, that's the first Brillouin zone, just like before. And for each K then there are two eigenvalues, one here and one there. And this is epsilon plus h0, epsilon minus h0, so it's kind of symmetric around epsilon. So for every point up there, there's one down there. So what's this gap here in the middle? Well, this is the point where k is 0, so you can put k equal to 0, cosine of 0 is 1. So that, you see, looks like t plus t primed squared. And so when you take the square root, you just get t plus t prime. So that's what this one is. And if you go to the edge of the Brillouin zone, that's where kb is pi, so cosine of pi is minus 1. So then it looks like t squared, plus t prime squared, minus 2tt prime, which is t minus t prime, all squared. So when you take the square root, you get t minus t prime. So here they are separated by the sum. Here they are separated by the difference. Now, one interesting thing to check out is what 

[Slide 6] if t prime were equal to t? Say, you know, we have here this -- this is different from that and that is why this lattice, you know, looks like one where the unit cell has size 2, and because, as you know, we assume the epsilon on the diagonal to be the same. In principle of course, those could have been different, but in what we did, we just put them equal to keep it simple. So the only thing that was making us treat the unit cell as having two basis functions is that t is different from t prime. Now, what if t prime equal to t? Well, if you just do what we did, you'd get a plot looking like this. The gap would disappear. There would be no this t minus t prime will become 0. This gap would disappear and you'd get two branches still looking like this. But what you might say is well, if t prime is equal to t, it's really just a 1D lattice without a basis, so we could do what we did two lectures ago in the fifth lecture. And we don't need all these additional ideas. And if we just did that actually, what you'd get is something that would look like this solid curve. You see? Remember, with just 1D without a basis, what you get is this epsilon plus 2t, cosine ka. So that's this one. And this runs from minus 2 pi to plus 2 pi because remember this axis is kb and b is the distance between the axis. But if the -- if it's a lattice without a basis, then what we usually use as a, would be that distance. So a is like half of b, which means when ka is pi, kb is actually 2 pi. Anyways, so the bottom line is that if you had done that problem, the special case of a lattice with t equals t prime. If we done that problem the old fashion way, namely what we did in lecture five, you would have got this solid curve. If you do it as a lattice with a basis with two atoms per unit cell, you know, following what we did in this lecture, you'll get two branches to the curve, but the branches will stop here. At the end of the day you'll get exactly the same number of eigenvalues. In one case it is these eigenvalues, now it is these and that. It's as if whatever was here kind of got folded back in there. And whatever was here got folded back in here. But of course, if we had taken the matrix and gone to a computer and asked for it to give you the eigenvalues numerically, you'd have got exactly the same eigenvalues. These are just, almost like, two different ways of classifying the same eigenvalues. But the point I wanted to make here is whatever we did here, when you put t prime equals t, you get back the old results. But of course the reason we are doing this, bringing in all these new ideas of having a lattice with a basis is because in general, when you try to find the energy levels in a solid, the unit cell has more than one basis function. And here we just illustrated it with two basis functions so that there were two branches to the band structure. In general there could be 20 branches to a band structure, 30 band branches depending on how many basis functions you have per unit cell, okay. Well, we are now ready to move onto the next topic. You see in the last two lectures we talked about how you can take the simple 1D model and extend it to 2D and extend it to 1D lattice with a basis. Next what we'll do is this example of graphene, 

[Slide 7] the topic of great current interest, and which kind of exemplifies what you could call a 2D lattice with a basis. So it kind of puts together what we did in this lecture and the previous lecture. So see you then in the eighth lecture. Thank you.