 So, I guess in the last lecture, I discussed things in terms of this density of states and this mode density that I mentioned without really getting into the issues of what determines the density of states. Also, if you know it, then from once you have the density of states, this is how you would calculate conductance. And just one point I think some of you asked me during the break that I thought I would mention is, you know, this discussion did not involve any quantum mechanics really. It was like density of states times velocity. And then where did the H suddenly come from? And the point I wanted to make is the way I was trying to do this in my lecture was that classically this dv times 2l could have been anything. I mean, it does, it could have any value. And what I tried to motivate this mode idea by saying that let's evaluate this for a one dimensional conductor. Using this standard way to you take ek relationships and impose bound, the periodic boundary conditions and all that, which I think brings in the quantum mechanics. But when you do that, then for a ordinary 1d conductor, this quantity comes out as 1 over h. And so you say, well, a 1d conductor, it makes sense to call it the number of modes. It's like 1d is like one mode, and that's 1 over h. And any complicated conductor could mentally think of as lots of conductors in parallel. So in general, then what you have would be 1 over h times the number of modes, number of how many conductors in parallel. But the way the h came was simply by evaluating this for a 1d conductor. Anyway, so what I want to talk about then for the next hour or so, next 45 minutes to an hour, is the ek relation for graphene and where that comes from. And specifically, I think what Mark had asked me to do was also introduce this Dirac equation. That is, when you look at the literature, a lot of the discussion of graphene usually starts from the Dirac equation. And in this context, where that comes from and how you relate it to this band structure, etc. So on the one hand, I want to talk about a few things that are like probably many of you may be familiar. It's in my book like chapters five and six is the graphene band structure. And those of you have taken my courses or it's on the nano hub. There are more detailed discussions. And the other hand, if I leave all that out and get straight into Dirac equation, I'm afraid I'll lose many of you because many of you may not, even if you have seen it, may not remember it. What we are talking about. So what I'll try to do is describe some of these principles of band structure and all, which are probably familiar to you and then connect up to the Dirac equation. But I will go through it relatively fast so that if you haven't seen it before, it won't necessarily be self-contained. I'll draw on this. Okay. Now the basic thing about band structure then in the context of doing here, I'd say is your starting point, your thing should be this Schrodinger equation. And that would usually look something like e psi equals some differential operator psi. Now usually, this differential equation is a very convenient thing to use when you are especially for analytical calculations. Almost all numerical calculations usually would turn it into an equivalent matrix equation, first thing. And then it would use the, solve the matrix version of it. That is, what you can do is you can use a set of basis functions. And you say that the function that I'm trying to find, the psi, is a superposition of many of these basis functions. So these are all known functions. And the expansion here has this coefficient psi m. So these are all the, these are all numbers. These are not functions. So this is the function I'm trying to find. And I write it as a linear combination of some known set of basis functions. And these coefficients then are found by solving a matrix equation, which looks something like this, then e psi m is equal to. So that's a differential equation which you can convert to a matrix equation like this. This is of course something you could visualize as a matrix. This e times some column vector psi is equal to the matrix h times psi. And how do you write the elements of this matrix? Well, there's the ab initio methods where you actually start straight from Schrodinger equation. Or there are semi empirical methods where you fit it to your favorite experiments and see if you can use those parameters to predict other things and compare with experiment, right? So, now in the context of graphene, for example, or any semiconductor, people have studied all this and they have a pretty good idea what all these h is to use. And the method I'll assume we'll be using here is this generally what's called the tight binding method. Where what these basis functions are, these basis functions are what you might call the atomic functions. That is graphene, as you know, is composed of carbon atoms arranged in a hexagonal lattice like this. And the basis functions you are using are atomic wave functions around each carbon atom. So typically, I guess for carbon, as you know, the atomic number is 6. So it's 1s2, 2s2. And then there's 2. So in the 2s and 2p levels, you have 4 electrons overall. And usually, you ignore the cord electrons and you use this 2s and 2p as your basis functions. So if you did that, then you'd have a total number of basis functions would be like 4 per carbon atom. So when you write this equation here, this matrix equation, the size of this matrix would have been like 4n by 4n, where n is the number of carbon atoms. That's what you'd end up with. And then you could try to solve this. Now there is a very important principle of band structure which helps solving equations like this. And that is that if you have a periodic solid, then actually you can write down the solutions of this almost by inspection. You don't actually have to go to a computer. And that's the principle that I'll just briefly review. As I said, there are lectures on the nano hub about all this that you could look up later if you haven't seen this. Now the basic idea is this, that the way this is written, these n's and m's would be running over every basis function, all 4n of them. But it's more convenient to think of it something like this. E psi n. Now let me explain what I mean by this. So instead of thinking of it like this, where each of these n's runs over all the basis functions, what is more convenient is to think of it as if these indices run only over unit cells. One unit cell after another. Now what do I mean by a unit cell? The idea is that if you take this thing, what you mean by unit cell is the basic periodic unit, the one unit that you could replicate over and over and create the entire solid. Now the point is that if you, your inclination may be to take one carbon atomizer unit cell. I could say, well, I've got this carbon and I replicate it. That won't quite do. Why? Because this carbon looks different from that carbon. Here it's got two things on this side, one on that side. Here it's one on this side, two things on that side. So one carbon atom is not a good unit cell. Wouldn't work. You need to consider two of them together. But once you consider two of them together, then it makes a nice unit cell. In the sense you stand here, look around you, whatever it looks like. You go someplace else, look around you. It will look exactly the same. There will always be two things on the left, two things on the right. It will look exactly the same no matter where you go. And so the idea is then that you should think of, you are writing this entire matrix equation as something like this where this summation is over all the unit cells. But within a unit cell, when I write the wave function here, when I write that, it is not just one number like this one. In other words, it's not just the coefficient of one of them, but it is the coefficient of all the basis functions in that unit cell. So how many basis functions would I have in general? Well, in general I've got two carbon atoms, four for each one. So I'd have eight basis functions. So in general, the way you'd think about this one is that this is an 8 by 1 element. So for every m, what you have here is an 8 by 1 thing. Then for every n and m, this thing is an 8 by 8 matrix. This one is an 8 by 1. And this is kind of what you run into when you do band structure calculations of any semiconductor usually. In fact, people often use more basis functions. So instead of 8, it becomes 10. And then if you include spin, it becomes 20. And it gets quite complicated. It's not something I could do on a blackboard easily after this. Now what makes graphene very interesting and special is that because of this nice planar thing of these 8, it decouples nicely into 2 and 6. In other words, the basis functions I mentioned were this S, Px, Py, Pz. And you've got these two carbon atoms per unit cell. Each one has S, Px, Py, Pz. And so you would have eight components to it in general, the eight coefficients. And the thing is these six are separate from that one, the kind of decouples separate, as long as it is perfectly flat. Now if there was some curvature to it, mixing which people worry about, the thing is lots of actual research papers are written ignoring these completely. This is what you call the sigma bonds and so on. So when you look at the density of states, I guess I wiped that out. If your chemical potential is somewhere here, for current flow what you really are interested in is the density of states around the chemical potential. And these are all described by just these Pz orbitals. Whereas all these others, they contribute to states deep down here and states way up there. Which are very important if you're interested in mechanical properties of this, for example. Because that's where a lot of the binding energy, the strength of the binding and all that comes from. But if you're talking about current flow, then all that matters is the states right around the chemical potential or the Fermi energy. Because when you put a little voltage, these are the states that will carry the current. So there's lots of research quality papers that are regularly published, which basically use just the two by two part of it. Where the basis functions are really just one Pz orbital on each carbon atom. And so every unit cell basically has two of these things. One Pz orbital here, one Pz orbital there. And so the equation you'd solve would look like this. So I'll write it up here. Just write this. So this is the set of matrix equations that one has to solve. And where all these indices, they just run over all the unit cells. And within each unit cell then, like if I say take one two, and I ask what is the size of this matrix? Well it's two by two. Two by two because there's two things here and I have to tell you how those two things are connected to two things there. So H1, if I'm looking at the H between these two unit cells, it will be a two by two. Now as I said, the principle of band structure is the following. It goes like this. That as long as this is a periodic solid and these m's are all the unit cells, the solution can be written in this form. That's it. So what you can show, this is the part that of course is differential equations. It's hard to motivate exactly how one sees that something will work. But if someone tells you something will work, you can always kind of substitute it in there and check if it works or not. So here also I'd say, well if this is the solution, put it in here. You'll see what happens. And what you'd get is, if you take this and put it back in here, you'd get something like this. So if I just take this, substitute it in there, then what happens is, it's phi e to the power ik dot rm. So that's what you get here, phi e to the power ik dot rm. And then on this side, I had a e to the power ik dot rn, which I have taken over here and written it this way. So the bottom line is, if I substitute this solution in here, I would get a 2 by 2 matrix equation like this. So this is a column vector with two components. That's a column vector with two components. And these are all 2 by 2 matrices. And for any given k, you'd have to do this summation. But after you do this summation, what you'd get is a 2 by 2 matrix. Now, why is it important for this solution that the structure be periodic? The idea is that you see, this summation that we are doing here, if you look at it, you're standing at some n. So let me draw that thing again. So you're standing at some n, and then you have to do this summation over all the neighboring m's, all the neighboring unit cells. So you have to stand here, do the summation over this, this, this, and this. And if you wrote it out, there will be five terms in the summation. Five because you put m equal to n, then you put m equal to that, and then you put m equals that. So that equals that. You'd have five terms, and you write it all out. And the thing is that this solution will work if, after doing that, what you come out with is independent of the n that you used, independent of where I stand. In other words, if someone decides that instead of standing here and doing his summation, he's going to stand here and do his summation around that, he should get exactly the same matrix. Because if not, then of course, this solution isn't really working because this thing gives you a number that is dependent on n, and then I can't solve it. I can't get any further. But in a periodic structure, of course, the important thing is that because no matter where you stand, the surroundings look the same. So if you stand here and do the sum, or if you stand here and do the sum, you'll always get the same matrix back. And that's what this matrix is. So that's this principle of band structure that I'm kind of, as I said, not really doing justice to it because normally if I were teaching this, this would be kind of the whole lecture. I want to do more important things. So I'm just sort of saying enough so you can see the rest. So the bottom line then is that when you try to find the band structure, the way it works is that you evaluate this H of k. So the basic equation that you then solve looks something like this. E phi. Phi has two components. And those two components denote the wave function at this carbon atom and at that carbon atom, like the A atom and the B atom. The two of them. It gives you those two components. And this is this. Here you have this, what I'd called H of k. And that is obtained by taking H and M and doing that sum over all that. That's what you get here. Then you put phi. The way it works is the way you're supposed to find the band structure is that you go through k. So this let's say a particular direction in k. Choose any k. Evaluate this quantity. It's a 2 by 2 matrix which has two eigenvalues. And you find the two of them. Go to another k. Solve it again. Find the two eigenvalues. Connect. Draw all these eigenvalues. Different k's. And all these when you join up, it looks something like this. When you join up, it looks something like this. And that's how you get your two bands. That's how the thing is supposed to work. And in general semiconductors, what happens is you have 8 or 10 components which is why band structures look very complicated. For every k, there's about 8 or 10 branches in the whole thing. That's why it's normally band structures look like. It works very well. It involves just two of them. You just get two branches. And the fermi energy or the chemical potential is right in the middle. The number of electrons is such that if there were no extra electrons from anywhere, it would exactly fill up half the states. So half the states means every... And what I'll show is the states come in pairs so that any time you get one here, you get one there. So half the states basically means everything below this middle is filled. Everything above is empty. So this h of k looks something like this. You have 0 here, 0 here. You have some complex number here and the conjugate of that complex number over there. That's this h of k. And what does this h0 look like? That's this minus t times 1 plus 2. So this is the result that I'm kind of stating without proof. This is what, as I said, normally would take one or two lectures to go through. But this is in my book and it is lectures out there that you could look at. But the bottom line then is that for any k, if you want to find the two allowed eigenvalues, this ek relationship, what you do is substitute the k you want, write down this matrix, find its eigenvalues, plot it. Go to the next k, plot it. Go to the next k, plot it. And if you connect it all up, you'll have the ek relation. That's the basic idea. And that's what this expression looks like and you'll notice the nice thing about it is there's only one parameter involved in this whole thing. That's this t here. What's this t? That's like in this carbon lattice, when you write this Hamiltonian matrix, it tells you the degree of the ease with which electrons can go from one PZ orbital to the next. The overlap, they say overlap integral between these two PZ orbitals next to each other. And people believe it is somewhere between 2.5 and 3 electron volts. Based on all the experiments, all the work people have done in this area, I think it's about 2.5 to 3 volts. And that's the only parameter involved. You see everything else that A and B, those are just the parameters of the lattice. B is this distance well-known exactly what it is. A is that distance well-known exactly what it is, et cetera. This is the whole thing. Now, the first point I want to make then is that if you look at the bands here, if you plot it on this scale, this is kx, ky, and at every point here, you could calculate two eigenvalues. And those two eigenvalues will come in pairs because what you can show is when you have a matrix like this, its two eigenvalues are given by E is equal to plus and minus magnitude of H0. That you can show easily. This 2 by 2 matrix, those are its two eigenvalues. So you'll have these two eigenvalues. So your chemical potential is somewhere around here 0 and then there's a plus H0. That's one eigenvalue minus H0. That's the other eigenvalue. So which ones are the most important ones? You'd say those regions in k space where the eigenvalues are close to 0 because that's where the chemical potential is. So for some value of k, if this happens to be 3 volts, it really doesn't matter. Unless you put a huge voltage across a ballistic section, it won't matter. So what matters is those regions where this is really small. So the first thing you do is you say, well, what values of k is this H0 close to 0? Where does it have its minimum values? And I think in the handout, I had a plot that's like a grayscale plot. What it is is this kx and ky, and what I plot there is if it's dark, it means the eigenvalues are small and if it is white, that means it's big. So the important regions are the dark places. So that I think you have in your handout and what you'll find is the dark regions are somewhere here. And you can see why. Where do these dark regions occur? Well, this is a point where kx is 0. So if you look at this expression, supposing kx is 0, what that means is this factor is 1. So what value of ky would make this quantity 0? Well, it's 1 plus 2 cosine. So if the cosine happens to be a minus a half, then of course it would just cancel and it gets 0. So when is cosine minus a half? The angle is 120 degrees or 2 pi over 3. So that's these two points. The cosine 2 pi over 3 and the cosine minus 2 pi over 3. Similarly, if you look at the points where kx is equal to pi, if you look in there, kx equals pi means e to the power i pi. That's actually a minus 1. So now what I have is 1 minus 2 cosine ky. So when will that be 0? So whenever the cosine happens to be plus a half, not minus a half, because I already picked up the minus from there. So now I need a plus a half. So that's like when this is pi over 3. So that's where these are. It's like halfway to that point. So those are the six regions which are most important. And as you know in band structure, ek relation may have a very complicated relationship with the entire kx, ky, kz in general plane. But what matters is these values. And what you try to do is try to get a simple expression for e versus kx, ky around those important points. Because you say, well, that's what will matter. So the usual thing we do when we write e equals ec plus h bar square k square over 2m, you could view that as kind of a Taylor series expansion around one of those minima, one of those values where it happens to be. And in a similar spirit you could say, well, what I'd like to do is figure out what this ek relationship looks like, how I could write it in a compact way around those points, for example, around those values, those places. Because in general, of course, it's given by this. That's it. It's nothing more. What we're saying is could we get an approximate relation around those points? What does it look like? Now the one point I'd like to make, and this again invokes some concepts that if you have not seen it before wouldn't be clear, is this idea of a reciprocal lattice or the Brevan zone. What I mean by that is this has to do with this number of values again. So in silicon it has six values also. You might say that, well, it looks like graphene also has six values, just like silicon. There are six regions in K space where you have those things. But actually graphene really has only two values. Overall. And the reason is this, that if you draw the Brevan zone, it looks something like this. And you are only supposed to consider states within a Brevan zone. Now the 1D version of this is kind of relatively easy to understand. That is, if you had an EK relationship for a 1D solid, for example, it would look say something like this. And you usually run from minus pi over A to plus pi over A. And if you actually just take that EK relation and plot it out, it would go on forever like this. But of course when you count states, you only count everything within a Brevan zone. Why? Because the argument is that a point here is the same as a point there. If you wrote down a solution, if somebody wrote down one of these solutions with a particular K, and someone else wrote down another solution for which instead of this K, he used that one, the wave functions at each of the points on your lattice would still be the same. That's what you can show. That it won't make any difference. And it's a little bit to do with, I mean it's somewhat analogous to discrete Fourier transforms, that idea also. That anytime you have a discrete lattice, it gives you this periodicity in K space. And just as the periodicity in real space gives you discreteness in K space, et cetera. So those are things probably seen before. But in this case, the important point to notice that when you're counting states, you should only consider one Brevan zone. Not those out here, because this is really equivalent to that one. But you could always choose your Brevan zone appropriately depending on your convenience. You could have, for example, counted it from here to here. That would be okay. There's no reason why you should do it symmetrically like this. You could have done it this way. As long as you cover, go over one Brevan zone so that no two points inside are connected by a reciprocal lattice vector. So in this context, what it means is the reciprocal lattice vector is kind of like this. And what you can show is that this is the Brevan zone and when you look at these valleys, every one of these only has one-third of a valley there because the other two-thirds is outside. So what you really have is like six one-third valleys, which is the equivalent of two valleys. And the way you can count things a little more conveniently is by taking this guide, for example, and moving it up here because this and this are connected by a reciprocal lattice vector. So what is here? You could just as well put him there. Similarly, what is here? You could just as well be here. So the advantage then is instead of having three one-third valleys like this, you'd have one nice full valley sitting there. So you're bookkeeping and all that is a whole lot easier. And lots of times when you're counting states, that's a whole lot easier to do. So you'd have to move that. Similarly, instead of having these three guys sitting in three places, you could have them all kind of sit right there if you want. But the bottom line is finally six one-third valleys make two valleys and those two valleys, you can move around as you see fit, you know, depending on the problem at hand. So I think next day when you look at the sub bands in CNT and all, you'll see that sometimes it's in terms of understanding the CNT sub bands in carbon nanotubes. So sometimes convenient to think of the Brevon zone a little differently. Instead of using this one, actually moving this up here and moving that there. So on. Okay. Anyway, so for this discussion then, let's consider these two valleys. So these are what people usually write as the K valley and the K prime valley two valleys. And what you're saying is we are only interested in the EK relationships around those two valleys. Okay. So the way to find, to simplify this, you have an H is equal to, this quantity then is a zero if you are right at that value. And it's also a zero if you're right at that value and right at that point. And what you'd like is an expansion of this right around those points. And the way you can do it is you can do a what you might call a Taylor series expansion. That is you could write it as something times Kx plus something times Ky. And the question is how do I find these some things? What is this and what's that? And the idea is that this one should be the derivative of H with respect to Kx evaluated at that value. That's this Taylor series expansion. Simple. So if we do that, what I'll get is take del H del Kx that's equal to minus taking derivative with respect to Kx, so I pick up this Ia minus Iat to e to the power i KxA cosine Kyd. So all I've done is taken partial derivative with respect to Kx assuming Ky is constant. That's it. Now, we want it around that value. So I can now put KxA equal to 0 and Kyd equal to 2 pi over 3. And what you get then is 0 means this is 1, cosine 2 pi over 3, that's the minus a half. And so the minus a half and the 2 minus 2 that cancels so you get Iat. So I can put this Iat here. This is the first one. Now the second one, that's this one, I'd have to find del H del Ky and that would be minus T 2T e to the power i KxA and now I have to take cosine Ky, I have to take the derivative so I get sine KyB and I'm taking the, I'll put a B. So cosine when I take the derivative sine right? So this minus would be gone. No, there's minus there. Yeah. And I pick up the B. And once again I put in KxA equal to 0 KyB equals 2 pi over 3 and so what you get is 2 pi sine 120 degrees the square root of 3 over 2 so that's square root of 3 T B. Now what you can show is square root of 3 times B is actually A. See if you, that's just a little bit of geometry. If you take that A and this B for a hexagonal lattice you can show square root of 3B is really A. So basically you get an 18 here. Now what if I were doing this at the other valley? I'd get much the same except that you see at the other valley the KyB would be minus 2 pi over 3. Makes no difference to this one because it's a cosine. This would still be exactly what it is but this one then would have picked up a minus sign if I went to the other valley. Okay. So the bottom line is then that H can be written as IatKx plus AtKy which means I could write this as At times Ikx plus Ky. This is in the one valley I'm sorry I guess I called it K it's the K valley and if you do the K prime valley it is At times Iks minus Ky. So I should put approximately equal to because this is a Taylor series expansion. So that then could be written this way if you want it. So this is often the basis for like you see the usual expansion that usually the way you use the this effective mass equations is that you can take this off now here. The way you normally use effective mass equations is to say that E is equal to Ec plus H bar square K square over 2m and then we turn this into a differential equation by saying E psi is equal to Ec minus H bar square over 2m del squared because the idea is that if you had started from a differential equation like this and assumed a solution like a plane wave you would have got this dispersion relation and the effective mass equation kind of works in backwards you say that since I know Ek looks like this I'm going to replace K with minus IH bar del and get an effective mass equation like this and then I'll use this equation for treating all kinds of slowly varying potentials so while the first uses of the effective mass equation in semiconductors was to impurity levels in semiconductors where the idea is that you take this equation and then you add an impurity potential to it and try to find the impurity levels and so on that was one of the first uses back in the sixties of the effective mass equation and in general whenever we try to calculate particle in a box level in the conduction band what we're using is a equation like that what have you gained instead of starting from the original Schrodinger equation well the original Schrodinger equation would have this enormous atomic potential in it this one has no atomic potential in it that's all part of the effective mass story that's all part of that this now almost looks like an electron and vacuum with a different effective mass that's what you're normally doing now in the graphene context of course we don't have something like that but what we have is something looking like this where we could approximate H with this so we could write this E as so I could write let me write it up here so let's say I'm working with the K value then the equation would look something like this as I said this band structure equation that was this one what I did was expanded this around the K value and now it looks like that and now instead of K I can put in derivative you see so K would be like minus I d dx sorry thank you thank you yeah this is actually a complex conjugate of this so I should write minus I kx plus K y whereas if you went to the other value then what would have happened is instead of plus K y I would have minus K y but at the K value this is the equation and the thing is I can now turn that into a differential equation and in the same spirit that people turn this into a differential equation what do they do well they just take K and replace it with like if you have a Kx you replace it with minus I d dx if you have a K y replace it with minus I d dy etc that and it goes both ways if I had a differential equation you could write a dispersion relation if you give me a dispersion relation I can turn it into a differential equation and go on and use it for other things so same spirit you could say I can take this replace all the K's with I d dx and turn it into a differential equation that I use here after it's a little more complicated than that one because it's got two components now only thing the differential equation will now be a couple differential equation with two components two components that you should interpret as the A component of your wave function and the B component of your wave function so the wave functions over the unit cells are all plane waves e to the power I Kx type of thing but within a unit cell is two components the one that tells you the A what's on the A side what's on the B side that's it so that's how you could get a differential equation of this sort and what I'll talk about for the next few minutes then as we finish up is the connection to the Dirac equation that is a lot of people who are doing calculations which start from something like this you see and this looks a little bit like the Dirac equation that is used in relativistic quantum mechanics and of course if you are familiar with the solutions to that then the one advantage is you can take certain things and apply it here if you're not familiar then it probably doesn't help a whole lot you could as well start from here or you could have even started from here I suppose you could have just gone from the without doing any expansions you could do just whatever we had in the beginning okay okay now in order to talk about this Dirac equation though I need to explain a little bit about something called the spin matrices the poly spin matrices are in your notes and it's kind of late in the day for this but let me try to explain this anyway the spin the basic idea if you have not seen much I'm not familiar with spin I always say that the closest analogy is the polarization of light that almost as if just as light if it's propagating in a certain direction it has these two orthogonal polarizations it can be x or y and you think of these two polarizations of light similarly with spin you can have two components this very simple analogous in that sense now there's one important difference though it works like this supposing with light you had a polarizer you see that let's light in in this direction and here you have an analyzer that's in some other direction or let's say you let in light at some angle theta and then you have an analyzer which only lets out light in this direction now what would get through and the answer is it would be given by cosine squared theta and everyone understands why that is so because this is vector it has a component of cosine theta along that and sine theta along the other direction so the components you write as cosine theta and sine theta and when you square that that's what gives you the intensity that's how you get cosine squared theta now spin is very similar except that instead of being cosine theta and sine theta it's actually cosine half theta and sine half theta and that's I guess has a very direct experimental consequence that you can see these days that as you know people have done experiments these days with devices that have magnetic contacts you know two magnets so one magnet injects a certain direction that depends on what direction the magnet is in and the other one acts as an analyzer which looks at what which allows some component to come out right that's it now with light if one polarizer was in this direction and you wanted to block it you make the other polar analyzer in this direction 90 degrees with spin if one magnet is in this direction and you want to block it you make the other magnet in the anti-parallel direction so if one is this way the other is what 180 degrees you see with light if you had put a polarizer and an analyzer that was anti-parallel it wouldn't have blocked the light it would have let just as much to go through but with so the point is that with light what is like 90 degrees with spin is like 180 degrees you have to put here 180 degrees in order to get the same effect as 90 degrees that's the point and so in thinking about spin the important point here is that just as we learn about vectors which way usually the components of a vector the way you write it is cosine theta along z if you had a like a unit vector pointing in some direction theta and phi you write the z component as cosine theta and the x and y components as sine theta cosine phi and sine theta sine phi so this is how you would write the components of a vector that's something we learned somewhere in first year college I guess now if you are trying to write the components of a spinor that is the spin it has got two components an up component and a down component and the up component is given by cosine half theta e to the power minus i phi over 2 and sine half theta e to the power plus i phi over 2 so when writing the components of a vector you have these three real components when writing the components of a spinor of course mentally we are still thinking of something that points in some direction but when you write its components instead of writing three real things we write two complex things and that's what kind of goes into your equation that's how you represent it and so on now this of course again if you haven't seen this before it takes some time to sink in and again on the nano hub I think I have a couple of lectures on this just this concept of spins and all that you can look up but here I just wanted to say enough so that you can see the connection to the Dirac equation and how it goes how the Dirac equation works now the next point I wanted to make and that has to do with this poly spin matrices and that is that the poly spin matrices there are three of them sigma x which is 0 1 1 0 sigma y which is 0 minus i plus i and sigma z which is 1 0 0 minus 1 now you might ask that well how do I write these components here that I wrote down you know I said that so where does this come from and that of course by proper discussion takes a lot more time but the point I wanted to make is that this in a way is like the eigenvectors of these matrices so what I mean by that is let's look at say the components of this for a spin that points along x so let's say this is the z direction that points along x and that angle let us say is would be 90 degrees so if you put that in phi is 0 so it will be cosine pi over 4 sine pi over 4 so this would be like 1 1 with a 1 over root 2 for normalization so if you have a spin that points along x then its components are given by 1 1 now that also is the eigenvector of this matrix so another way of finding the spin along x is to find the eigenvectors of this similarly if you want to spin along y you find the eigenvectors of this if you want to spin along this you find the eigenvectors of this and one of the things you will see in the literature in this context a lot is this quantity sigma dot n now what does that mean this you know which has these components so what you are supposed to do is take sigma x and take the x component of n which is sine theta cosine phi then take sigma y sine theta sine phi and then take sigma z and cosine theta so sigma dot n is like sigma x times the x component of n sigma y times the y component of n sigma z times the z component of n that's what I did took those things wrote it down and each one of these is a 2 by 2 matrix so if you add all that up you will get a nice 2 by 2 matrix and whose eigenvector will actually be that that's the point so if you want the eigenvector if you want to know how to represent the spin in a particular direction it's like you take sigma dot n find its eigenvectors and one eigenvector is the up along that direction the other eigenvector would be the down along that direction that's how it works generally and so this is a quantity that you see often now this is very useful once you get used to this idea because what it does is it allows you to represent these quantities that have 2 complex components as a vector visualize any quantity with 2 complex components as some kind of a vector pointing in some direction then it works both ways vectors can be represented as 2 complex components or the other there's a mapping clear mapping and this often allows you to visualize these objects this is this pair of complex quantities ordinarily you couldn't visualize it but using this mapping once you get used to it and it takes some time of course you can visualize them and that's this concept generally you see you often hear the word pseudo spin what do you mean by pseudo spin it often has nothing to do with spin it means that in a given physical problem for some reason your wave function has 2 complex components and as long as that 2 complex components I could mentally use this mapping to think of it as a vector pointing in some direction so that would be the pseudo spin direction sort of that mapping you see and that's again a visualization aid let's say if you don't want to use it you can keep using this there's no reason why not now the 2 important properties I wanted to mention here one is the sigma dot n that it's eigenvectors give me the give me the components pointing in that direction along n and the other property is that if you take sigma dot n and square it you get the identity matrix you take that matrix multiplied by itself you'll get and these are again in the notes I handed out so you can easily check this for yourself I think I wrote down that matrix and I also mentioned that the square is i now this is something I'll be using for discussing this Dirac equation ok now the Dirac equation the way it works is the following you see the Schrodinger equation worked something like this that people said that we know that e is equal to p square over 2m that was the classical energy momentum relationship and how do you get a Schrodinger equation out of it well instead of e you put ih bar d dt and instead of p you put minus ih bar del and so that's how you get this so that's how usually you say heuristically anytime you have a ek relation or ep relation energy momentum relation you can write the corresponding wave equation and once this was understood I guess back around like 1930 or so Dirac was then trying to figure out what to write for the relativistic energy momentum relation and the relativistic energy momentum relation works something like this looks something like this so relativistically energy and momentum are related this way so if you have 0 momentum e is equal to mc square the most famous equation in science yes and when you plot this e versus momentum momentum it will look something like this e versus p and this is mc square and if you do a binomial expansion around that then you can show that this will be something like mc square plus p square over 2m so people say that the non relativistic version is sort of like a binomial expansion around that anyway but the point here is then given this how do I write down a differential equation to go with it given that I know e becomes ih bar d dt and p becomes minus ih bar del the whole problem of course is I've got a square root here you see because if I write e that's like a square root now how exactly do I write the square root of a differential operator and this is where Dirac came up with this brilliant idea he saw that he said that you see I can't write what I can write is a matrix like this so this is actually a 4 by 4 matrix where this i is a 2 by 2 identity matrix and sigma dot p I just explained these are all 2 by 2 matrices that's this this and this minus mc square i so overall these are 4 by 4 matrix and the point is the property of this matrix is if you square it that is if I multiply it by itself what I'll get is exactly this times the 4 by 4 identity matrix in other words if I multiply this by itself then the final result will be if I multiply those 2 things final result will be an identity matrix with that thing in front so in a sense he had managed to take the square root of this you see because e is equal to square root of this and so the final Dirac equation is ih bar d psi dt is equal to this times psi and this is almost like the square root of that operator because as I said you know if you multiply these 2 things and that's easy to show actually we can check you know you can see that all the diagonal stuff will be 0 because you know when I multiply these 2 things supposing I multiply this by this and this way by this you can see mc square sigma dot p sigma dot p minus mc square just cancels out when you look at the diagonal elements you got mc square times mc square that's m square c4 and then you got sigma dot p whole square and that of course because of this property would give you just p square etc so you can do this algebra but the important point is you can show that this is almost like the square root of that operator and so the Dirac equation looks something like this ih bar d dt of psi and now his psi actually has 4 components is equal to this 4 by 4 differential operator times psi so that's the standard 4 component Dirac equation actually you had to thank you, thank you very good yeah thank you you should be a c here very good so this is a 2 by 2 matrix that's a 2 by 2 that's a 2 by 2 that's a 2 by 2 now if you apply this to something that's moving in 2 dimensions so the momentum only has x and y components and no z component then you see that equation becomes something like this ih bar d dt of this 4 component guy 1 2 3 4 is equal to mc square mc square 0 0 and then this c sigma dot p we are saying p only has x component and y component so this is c sigma x px plus c sigma y p y and I think if you write that out it would look like 0 c px minus i p y c px plus i p y 0 and then here we will have minus mc square minus mc square c px minus i p y and this is in your notes actually I have written this out so you can look at that now if you actually had the particle also moving in the z direction that would have complicated it because then you would have picked up pz components here but because we are in this plane in x y plane you leave out the pz components so those are out those are out and then you see this 4 by 4 kind of neatly decouples into a 2 by 1 2 by 2 that's this and the other one that's this because all the other stuff is 0 so you could I can just rearrange rows and columns and kind of write it as 2 separate 2 by 2s and any one of them I guess you'll notice looks a lot like that that's the point I'm making that you see this at that's of course h bar times the velocity in graphene because d e dk that's really that guy so this is h bar v and that looks a lot like this with the masses equal to 0 because these 2 are 0 so with the mass terms equal to 0 so you would say it's a massless Dirac fermion will be the massless Dirac equation so that's the equation we are talking about now one thing you'll notice is that it's not quite the same the way we have done it namely you see I had here kx plus ky whereas what appears here depending on which one you look at looks like px plus ipy looks a little different and this is where the choice of coordinate system determines exactly what you will get see so you see in a lot of papers where they initially start with a coordinate system like I have drawn this way and the y going that way and but then when they actually connect up to the Dirac equation they rotate the coordinates because otherwise it doesn't quite come out right you see you rotate the coordinates accordingly and so usually the rotated coordinates the ones they will use would be something like this so if instead of using my x and y like this I put my I just turn it 90 degrees so this is x prime and this is y prime I have just turned it 90 degrees then I think what will happen is that equation will become like 0 h bar vf times you see what was x is kind of now like minus y so I would write here iky prime remind myself I have rotated coordinates and what was y is now x similarly here of course I just get the complex conjugate so now you see it starts looking like kx plus iky and kx minus iky looks like that one for example now if you wanted to look like this where you have the kx minus iky up here and the kx plus iky down here then what you have to do is interchange the a and b whatever you call the a component you should start calling it the b interchange them so but the thing is that once you have adjusted all that then you would have a direct like equation which you could write as as if this can be written as c sigma dot p c sigma dot p or h bar c sigma dot k so you could write the interaction and write the Hamiltonian kind of compactly like that see and that is what a lot of the literature does a lot of literature because it was the work was done by physicists who are already familiar with the Dirac equation so that's where they start almost that's what we use in this context and so when you decide on things like when you try to do problems like particle in a box levels and things like that just as an ordinary semiconductor you take the effective mass equation and do things you would usually start from the Dirac equation and do things but it's not really necessary I mean even if you have not heard of Dirac equation don't know any of this you could just have done it with the tight binding model that we started from it's really not necessary on the other hand if you want to follow the literature and connect with them then it's really important to be able to translate and connect up accordingly and the one important concept though that is useful kind of way beyond graphene and anything else that whole idea of pseudo spin that which I didn't really do justice to it I kind of skimmed over it that whole idea that a two-component complex thing can be visualized as a vector and reverse and I've seen that work you know people who do like optical polarization in fibers I've seen them represent even optical polarization as a two-component complex thing so use this thing in reverse I've seen them do that in reverse that whenever you have a two-component wave function with complex components you can visualize it as a vector and that way you have a pictorial way of thinking so in graphene you have an electron going from one state to another that goes from say conduction band to valence band and the wave function changes somewhat so it goes from one complex two-component thing to another complex two-component thing and mentally it's useful if you can think of it as if there was some pseudo spin like this this is that they have a nice picture pictorial way of thinking about it so that's a concept that's useful in graphene and in many other contexts it's a much more general thing thank you