 OK, so thanks for coming back and hanging in there. So you might ask, OK, so what's so special about graphene? Why can't we do the same thing? Why don't we see another materials? And in fact, the rock cones are all over material. Many, many materials have the rock cones. High Tc, all high Tc materials have the rock cones in the band structure. The difference is that unlike all these other materials, in the case of graphene, there's nothing else at the Fermi energy except the Dirac points. So the density of states, as we'll see in a moment, at the Dirac point is 0. If I send in an electron at the Dirac point into graphene, the density of state is 0, it just goes through. So the electrons that you send in have no choice but visit the Dirac point. There's nothing else. So the Fermi surface is really two points, k and k prime points. And every electron in the material at charge neutrality has to visit that point. So all the physics is controlled by those points. Now, when you go to high Tc material, to any kind of material that has Dirac points, oh, there's plenty of choices. Why should they visit this very narrow channel where they can go anywhere? So the macroscopic physics completely misses the Dirac point because the weight is so small, because they can go around it. So what's special about graphene is that there's no other game in town. You have to go through the rock cones or you don't go at all. OK, so let's remind ourselves what are the ingredients that go into the special band structure of graphene. So there's only three ingredients, two-dimensional, honeycomb structure, and identical something per unit on every sub lattice. And this gives you the band structure. Now, notice that in this list of ingredients, it doesn't say carbon at all. It's purely geometric. It doesn't say carbon. So then you can go to the periodic table. Let's look at the fourth column here, where the geometry is going to be the same. The bonds are always, if you manage to have it as P2, you're going to have bonds that are at 120 degree. That's what goes into the band structure. And people have tried that, and they actually succeeded making graphene-like structure out of silicon, silicene, germanine, stunning. And then they said, OK, but it doesn't say that we have to sit on the fourth column. All we have to do is make it a honeycomb structure and put identical atoms on the two sub lattices. Let's try something else. So people were able to make graphene-like borosene here, borofene here, phosphorine, right? But all these structures are artificial, and they're unstable, takes a lot of work to get them. And they are very fragile. But one can make graphene-like materials out of other atoms. But then you can think, actually, in fact, nowhere did I actually say that these have to be atoms. They can be something else, as long as they're identical. So people have done graphene out of other things. For example, this is surface of copper. This was done by Harry Manoharan's group, and he had very patient students. So they picked up carbon monoxide and put it and arranged it in a triangular structure, triangular, not honeycomb structure. And the wave function is the complement of that. So the wave function of the electrons of those is honeycomb. So that's like graphene. And then they took an STM and measured the band structure. And lo and behold, it looks exactly like graphene. It has a density of states that goes to 0 at a Dirac point. Now, everything is different here. The hopping parameter is smaller. The Fermi velocity is small. The lattice spacing is larger. The hopping parameter is smaller. So the Fermi velocity is 1 third of the Fermi velocity in graphene. So this is a graphene-like material made by assembling atoms. But then it doesn't have to be atoms at all. People manage to make graphene out of what they call the atom structures, and they have exactly the same band structure. And it doesn't even have to be atoms at all. People have patterned gallium arsenide with holes and looked at the electronic structure. People have used microwaves, interference between microwaves. Again, they got the band structure of graphene. So as long as you have those three ingredients, because the only thing that goes in there is geometry, two-dimensional honeycomb, and you have to have some sort of interaction between the sides. OK. So one other thing here, so when we had this perfect structure, all the hopping parameters from one atom to its nearest neighbor were identical. Now, if we change these hopping parameters, if we don't make them identical, we can change the band structure. And there's many, many ways to do that if you relax the condition that the all distances are the same, so you can stretch it. You can impose an external potential as we're superposing to graphene layers or putting in a boronitride. You can put an electrostatic potential. You can introduce defects, all those things we have tried. And you can get all sorts of very interesting band structure by doing that. OK, so here's the summary of the first part. So there's the three ingredient. The Hamiltonian is the so-called Dirac-Weil Hamiltonian. It's simply the Pauli matrix projection on the momentum, or this is the helicity. The energy is linear in momentum, so you have ultralativistic quasi-particles. The wave function is a plane wave multiplied by a spinner. This pseudo-spin is parallel to the momentum in the electron band. It's anti-parallel in the whole band, and everything is reversed at the k prime because they are time reversal symmetric brothers of each other. The electrons, they have a very phase of pi, so you actually have to go twice in the reciprocal space to come back to the same place. This is one way of looking at it. And the density of state, which I didn't tell you about, I don't know why I didn't tell you about, the density of state is linear in energy. So the density of state is linear in energy and vanishes at the Dirac point. And we are going to use that tomorrow when I'm going to talk about scanning tunneling microscopy, spectroscopy rather. Why choose? Why two dimensions? This is a two-dimensional Hamiltonian. We don't have a three-dimensional Hamiltonian. If you have three-dimensional, you're going to get the bi-semi-metals, or the three-dimensional version of this. It's two-dimensional because I'm talking about graphene. If you make it three-dimensional, you get a wild semi-metal. Much more complicated. But yes, sure. You have one that actually goes from one surface to the other, right? But this is more advanced stuff. OK, so we're done here. I'm going to move to another presentation, which is the second part. You see that I'm really falling behind here, but that's OK. You know, I had no idea how long this will take, and I had no idea who the audience is. I taught this in Chengdu in China this summer. And I was told that these were students who knew about solid state, and they had quantum mechanics. And when I came to a class, they had no idea what quantum mechanics. They didn't even know, haven't heard, Schrodinger equation. But it turns out that there is a Chinese word for it, and they didn't know any solid state. So I ended up teaching a solid state course. But I don't think here I have to do that. But anyway, I'm not going to get to where I wanted to get. And I think I realize that the best thing to do is not to ask. OK, so now we're getting to the second part. And here we're going to put a magnetic field. What happens when you apply a magnetic field? So the sort of things that I'm going to tell you about, I'm going to talk about Onsager relation. This is not a traditional way of getting to Landau levels, but it's a very elegant way where you really, I think there's not too much algebra involved. And then I'm going to compare this way of getting to Landau levels to getting them out of the Hamiltonian. And quantum Hall effect, I'm going to talk about integer quantum Hall effect and fractional quantum Hall effect, the difference between non-relativistic electrons and relativistic electrons. And I probably won't get there today. OK, so let's take a classical electron in a magnetic field. So in a magnetic field, we have a shooting electron with a velocity v. We know that there is a Lorentz force that goes like this. And when the electric field is 0, then the equation of motion looks like this. So the acceleration is perpendicular to both the velocity and the magnetic field. So the motion is in a circle. So you have a cyclotron orbit. The frequency is proportional to the magnetic field, inversely proportional to the mass. I'm putting in a star here, because this is a preparation to put it on a lattice. And the cyclotron radius, the radius here, it's inversely proportional to the magnetic field. And you see, when you do it classical, you can have any orbit you want, any radius you want. Now what happens when we add to this a lattice and quantum mechanics? So we want to add a lattice and quantum mechanic. So what we're going to do, first we start by looking at the effect of the lattice on the electronic motion. And then we're going to use a semi-classical approximation based on the Onsager relation to show that the orbits are actually quantized. You cannot have any orbit that is physically possible. And that's because the wave function has to be single valued, so when you go around it has to change. The phase has to change by a multiple of 2 pi. And this will allow us to calculate a lambda level sequence in a very nice and elegant way. So here is a little bit reminder from your solid state book. So let's say that we have a wave packet moving in a solid. So the average location of the packet is at r. You have a wave vector k. So the velocity dr by dt for this wave vector k is proportional to the derivative of the band with respect to k. So this is basic stuff. The velocity is proportional to the slope of your band in k space. And the equation of motion, the time rate of change of the momentum is basically equal to the Lorentz force here. So e is the external field. It does not include the lattice field. The lattice field is all included here in this epsilon here. Everything that has to do with the bands is buried in here. And the range of validity of this, and this is important, is when you're sitting inside one band. This will no longer be true when you have transitions between bands. You have zener transitions or things like that. So in graphene, we don't have to worry about this. There's only one band, basically. OK, so if you put a block electron in a uniform magnetic field. Now I took the electric field to be 0. Then we get that the equation of motion here e vanishes. So the time rate of change of the momentum is proportional, it goes like v cross b. So immediately you see that the change of k is perpendicular to the magnetic field. So if magnetic field is in the z direction, the change of k is going to be in the plane perpendicular to it. And k parallel does not change, because b does not act on the parallel motion. And the velocity, which is the slope here, so you can rewrite this. You can rewrite this dk by dt dot vk. You just take the derivative respect to time, multiply by vk here, you get this. And this has to be 0. And therefore, the energy epsilon k here is a constant of motion. The energy is a constant of motion. So if you have a magnetic field pointing in the z direction and you don't have an electric field, the energy is going to be constant. It doesn't change when the electron moves in a cyclotron orbit. This is just like in the classical case. And these two facts determine uniquely the electron orbit on the Fermi surface. OK, so let's see what's the cyclotron orbit in real space. So we have this equation here. You use a little bit of this identity here to find that the velocity perpendicular to the field is this expression here. It's proportional to the time rate of change of k cross product with a magnetic field. And then vk is just dr by dt. So you integrate over time from 0 to t. So you see that the position at time t is proportional to the position in k space multiplied by this factor here, which is the square of the magnetic length. Let me put that down here. So magnetic length is the characteristic length scale when you have electrons in a magnetic field is square root of h bar over eb. And I'm sure you've seen that. So the orbit in real space is equal to the orbit in a reciprocal space multiplied by the square of the magnetic length. That's all there is to it. So it's very easy to go from one to the other. One other caveat is the cross here tells you that it's turned by 90 degrees. So if you know the orbit in the k orbit, you immediately know the r orbit. You turn it by 90 degrees, and you multiply by the square of the magnetic length. So this is the connection between the two orbits. We're going to need that for the Onsager relation. OK. Now here comes the derivation of the Onsager relation. So since the electron wave function has to be single valued, if you take the momentum along a closed cyclotone orbit, this is simple Bohr-Sommerfeld quantization. So p dot dr, you take it on a closed orbit, has to be equal to an integer of times 2 pi. So this is simply the old Bohr-Sommerfeld quantization that we know from elementary physics. Now if you have an electron in a magnetic field, so the magnetic field is a curl of a vector potential. And p here is not the simple momentum, but you have to put in here the mechanical momentum or the invariant momentum. So p goes to the momentum that we usually use, minus ea, where e is the charge of the particle for an electron. Of course, it's going to have a negative sign. So we have to use this mechanical momentum because in order for the physics to be gauge invariant. So you can always add a gauge to your vector potential, but for the physics to be gauge invariant, you have to use this momentum here. This is always true. So whenever you have an electromagnetic field, this is the form of the momentum. OK. So when you take this integral, it looks like it has two terms. The first term is the kinetic term, and the second term is called the field term. So the first term is the kinetic or the Brouille phase. So you take the integral of p. Now I'm just doing the first term here, and I'm using this relation which I had before, hk dot equals to er dot cross b. It's just a little bit of algebra, homework, if you want to prove it. But this is how it goes. p dr, you write it p over h bar, k dr. You can write it like this, r cross b. And you find you take this integral, it's equal to e over h bar times 2 phi, which is the total flux. Basically, this is related to the Stokes theorem, not quite. But basically, OK, do this as a homework because I'm not going to go through all the steps. So the kinetic term here is equal to the flux that threads the orbit. So it's the magnetic field times the area multiplied by this term e over h bar and 2. And now we can do this similarly an integral for the field portion. So now this integral is e over h bar, a dr. And this a is a curl of b. And now you directly use Stokes theorem. And you end up that the second integral is e over h bar times the total flux. So basically, the Bohr-Sommerfeld integral here has two parts. Both of them are multiples of the flux that threads the cyclotron orbit. So let me summarize. So on one hand, Sommerfeld Bohr tells us that this integral has to be 2 pi times an integer. On the other hand, we just calculate it. You take the sum of the two. That it has to be 2 pi e over h times the total flux that goes through your orbit. And now you make the two of them equal to each other. And then you find that the total flux that goes through this orbit, it's an integer n multiplied by this quantity h over e, which we call a fundamental unit of flux. So what this means is that in order to have the wave function be single valued, the total flux for the cycle of the orbit has to enclose an integer number of quantum flux lines. You see that here. You see that in superconductivity, when you have vertices that penetrate in a magnetic field, you see that in superfluids, when you have vertices, when you rotate a superfluid, it's all required by the single valuedness of the wave function. And this number here, h over e, it's a universal constant here. It's a very tiny number. Notice something curious. It's 4 times 1, 4 times n minus 7, Gauss times centimeter squared. These are CGS units. What is that going to be in MKS units? What is it going to be in MKS units? Sorry, sorry, Tesla meters, I mean. So there's something curious that I just noted a lot so long ago. A Tesla is 10 to the fourth Gauss. And I was always wondering why this 10 to the fourth. It's because of this. Yeah, yeah, yeah, exactly. So a Tesla is 10 to the fourth Gauss. The centimeter squared is 10 to the minus 4 meter squared. So this is the same units Gauss times centimeter squared or Tesla meters squared. So I think that's where the definition comes from. I don't know, but this makes sense. OK, so this is a quantum flux, very important quantity. So the magnetic flux enclosed by cyclone orbit is quantized in units of phi naught. Very important. Just by knowing that, there's a lot of physics that we can get out without a lot of work. OK, so this is our answer to quantization condition. So the total flux is equal to an integer number of flux line that goes through the orbit. Now I want to go from the real space orbit to the k space orbit. And the reason is when I know the k space orbit, I can immediately relate k to the energy and I can calculate basically the quantization of energy. So I want to know, I want to go from this to the k space orbit. So the real space orbit now, it's just the area multiplied by b. So I'm dividing by b here. And the real space orbit is equal to an integer number times the magnetic length squared times 2 pi. So this is the area of the orbit in real space. This is the n-cyclotron orbit. So now we want to go to the k orbit. Now remember, k goes like r over lb squared. So all I have to do is the area in real space is the area in k space multiplied by the magnetic length to the fourth. So the bottom line is that if I want to go from one to the other, the same quantization condition gives me that the area of the n-cyclotron orbit in k space is sks times the square of the magnetic length is 2 pi n. This is very important. This is the Anzager relation. So the Anzager relation tells us that the area of the n-cyclotron orbit in k space multiplied by the square of the magnetic length is 2 pi times an integer n, which counts what orbit this is. Now this is a semi-classical relation, and it misses a lot of the quantum mechanics. So if you do it right, what you get is a correction for n. And the correction for n, so instead of n here, we have n plus lambda. And this correction has two terms. One is the one-half, which is also called the Maslow phase. If you might recognize, this is exactly the zero point motion. It's the one-half that you have in a harmonic oscillator. And the second is the Berry phase. So the Anzager relation has an extra term here, which is a half, which always comes in here, plus the Berry phase, gamma divided by the units of 2 pi. So this is the generalized Anzager relation. So let's now take an example. So we know the Anzager relation, and now I'm going to show you how we can use it to just get out the quantization of energy level in a non-relativistic two-dimensional electron system. So again, I've written down the generalized Anzager relation. My pointer is dead again. OK. So now for electrons that don't have any special topology like electrons in hetrostructure, gallium aluminum arsenide, or if you spray electrons on top of helium, or if you have electrons in a MOSFET, they are trivial, topologically trivial, which means that the Berry phase is zero. So if I put zero here for the Berry phase, I get. And another thing that I'm assuming here, let's assume that we don't have any scalar potential. And then the orbits in reciprocal space are circles. So the area is simply pi k n squared. So pi k n squared times Lv squared is equal to this. This is just Anzager's relation. And I'm going to put zero for the Berry phase. OK, gamma equals to zero. And now I want to write this in terms of energy rather than in terms of k. So now I know that the energy is quadratic and no momentum for these trivial electrons, h bar squared k squared over 2m. So I'm plugging in for k squared, 2m e over h bar squared here. So I get the energy of the nth orbit. I'm just plugging this in here. And I'm doing a little bit of algebra. It's h bar eb over m times n plus 1 half, where eb over m is the cyclotron frequency. I immediately got the quantization of energy levels for non-relativistic electrons. So it is really exactly like a harmonic oscillator. These are called Landau levels. They're equally spaced in energy. They are linear. The energy is linear in field. And the energy spacing is constant. One important thing is that there is no level at zero energy. You have a 1 half there. And that is because we have zero point motion. This is a harmonic oscillator. The electron cannot come to a full stop. So you need the 1 half there, unless you have weird things happening where the zero point motion is swallowed by the very phase. We'll get to that. So very nice. We just got it. This is just the present. I mean, no work at all. All you need to know is the dispersion. So this is the energy Landau level sequence for non-relativistic electrons. How does it compare to a full quantum mechanical solution? Now there, I mean, full quantum mechanical solutions, you have to do some work. Now it turns out that we're going to get the same energy sequence for the full quantum mechanical solution. But notice, if you use on-sucker relations, you can get the energy sequence, but not the wave function. In order to get the wave function, you have to put in the work to solve Schrodinger's equation or Dirac equation as it may be. OK, so let's do a little bit of an exercise here. So let's put a two-dimensional electron magnetic field. So we can choose any gauge we want for the vector potential. So as long as you can add any term that looks like this, the divergence of some vector lambda. And so, for example, I'm going two common gauges that we use for two-dimensional electrons. One is called the lambda gauge, where so you take the magnetic field, which is parallel to the z-axis, and the lambda gauge is like this. So it's 0 in the y direction, a you have in the x direction, you have b, and 0 in the z direction. Another popular gauge is the so-called symmetric gauge. And that, so a is 1 half b cross r. So a is b times a unit vector in the minus y and x here. Or you can write b times r e vector in the polar direction. So these are two examples of different gauges. But all the gauges, the physics has to be invariant. If you've chosen a correct gauge, the physics should not change, certainly not the energy levels. So the energy levels are going to be the same regardless of the gauge that you choose. And therefore, we can simply use the Onsaker relation to derive the energies. But if we want to know the wave function, the wave function does depend on the gauge. So it's different for different gauges because the wave function reflects the symmetry that you chose for your gauge. So this has a symmetry where you have something in one direction and nothing in the other, where here you have a radial symmetry. The wave functions are going to reflect that. So here is our Hamiltonian for the free particle with the various names to it. It's either called covariant momentum or mechanical momentum. This one is called canonical momentum p that we always use when there's no vector potential. So this is our Hamiltonian. We use now the Landau gauge written like this. Now you see immediately, if you put the Landau gauge, the Hamiltonian turns, you have a p squared, px squared over 2m. And you have a quadratic term here. And what we're going to be looking out for is we're going to bring it to the form of a harmonic oscillator. Because a harmonic oscillator is one of the only three problems that we know how to solve analytically in quantum mechanics. So we try to bring everything to harmonic oscillator. We can do this here because we have a square. So this is how we transform it. So px squared. So I'm just plugging it a here so that I only have a term in here in the y direction. So this is my Hamiltonian. Now immediately, you see that p y commutes with the Hamiltonian. So p y is a good quantum number. Therefore, I can replace the operator p y with the momentum in the y direction. So that's what I'm doing here. I'm replacing p y by h bar k y over m omega c squared. I've multiplied and divided by omega c squared. So I'm bringing it to a form which we all love, know and love, because this is simple. Here are some, again, notations. Cyclotron frequency, magnetic lengths. And this is a harmonic oscillator basically with a shifted origin. So x naught here is lb squared times k y. I'm just translating it. So you see, immediately, I brought it to harmonic oscillator. The solutions are going to be harmonic oscillator, which is just the center is shifted. And now the energies have to be this, energy of harmonic oscillator. And lo and behold, it's exactly the same spectrum that I got from the Ansager relation. But now we can also calculate the wave functions. So we're going to do the wave functions in the Landau gauge. You have, so immediately, you see the x and y motions are the couple. In the y direction, we have a plane wave. In the x direction, we have a harmonic oscillator. So the wave function are a product of a plane wave and a harmonic oscillator. So this is a Hermit polynomial. And this is what they look like. So in the y direction, we have a plane wave. And in the x direction, we have a Gaussian. We have Gaussians with, OK, what I'm looking here. So in the first Landau level, this is what they look like. The wave function in the ground state. In the second Landau level, they look like this. So you have two nodes. So first Landau level, no node. This is a, in the second Landau level, you have two nodes and so on and so forth. Now if you do the same thing for the symmetric gauge, and I'm not going to go through all the work. But again, you're going to end up with a harmonic oscillator. The wave function is more complicated. So you get these Gaussian here. There's a Gaussian here, r over lb. And the n equals 0 level looks like this. The n equal 1 level looks like this. And you have nodes inside for an equal one. Here you'd have two nodes. And basically, you see immediately the wave function reflects the symmetry of your vector potential. So this is a circular vector potential. The wave function have the same symmetry. Now which one, and then you may ask, what wave function should I use? It will depend on your system. It depends on your boundary condition. You always want to choose the gauge that has the same symmetry of your boundary condition. And this way, you're going to find the simplest solution. If your boundary condition and the symmetry of your gauge are not consistent, you're going to eventually solve it, but it's going to be a total mess. And you won't understand what you've done. OK, Landau level degeneracy. How many electrons can I put in a Landau level? So for that, we really need to look at the wave function. We're going to go back to the Landau gauge here. So the wave function is a plane wave multiplied by harmonic oscillator. And you see that ky is the quantization in the y direction. How many? So ky is if I have a finite system of width Ly and length Lx, there's only so many ky's that I can have. So ky runs from 2 pi over Ly times n when n runs from 0 to our integers. So these are all the ky's that I can have in the y direction. And x0, remember, is Lb squared times 2 pi over Ly. This was how we defined this shift. So x0 is the position of each one of these. So how many of these can I have? So clearly, x0 cannot exceed the width of your sample here. So x0 is larger than 0, but it has to be less than the size of your sample. And you plug all that in, and you find that n has to be Lx Ly times eb over h. Basically, this is the area of your sample times b divided by the unit of flux. So the degeneracy per unit area, very simple to find, is equal to the number of flux lines that you can fit in your sample. So every flux line can accommodate one state. So you thread your samples with this flux line. You throw one electron per flux line. And this is as many as you can put in. This is called the orbital degeneracy. So the orbital degeneracy is just the number of flux line that you can thread through. However, we can have other degeneracies, internal degeneracy, like spin. That is 2. You can have valid degeneracy for graphene. That's another 2. So it's a total of 4. So the total degeneracy of the number of electrons that you can put in for flux line is g0, orbital degeneracy, multiplied by the internal degree of freedom, which is for graphene, it would be 4. For non-latavistic electrons in semiconductors, it would be 2 just for spin. OK, I think are we done? Or are you guys exhausted by now? Shall we just quit? OK, so why don't we quit and leave the quantum Hall effects for tomorrow? In fact, I'm not going to get to all this stuff anyway. So my plan for tomorrow was to talk about research topics like atomic collapse, condo effect, these twisted layers. So we can completely skip quantum Hall effect and go to that. Or if you guys would like to learn about quantum, if you don't know about it, fractional integer quantum Hall effect, we could continue. Can I have a show of hands? Who wants to go to research topics tomorrow? OK, I think I don't even need to count the rest. Who wants to go to quantum Hall effect? Oh, wow, OK. OK, I'm going to think of a compromise. OK, see you tomorrow.