you can follow along with this presentation using printed slides from the nano hub visit www.nanohub.org and download the PDF file containing the slides for this presentation print them out and turn each page when you hear the following sound enjoy the show so on my first lecture a few people said there was an awful lot of material and some other people said I wish you'd talked about this and this and this too so I'll go relatively fast and you can stop me and ask me questions but it'll all be on the nano hub later and we have a nice set of references where you can see all of the details and mainly I I hope to to help you get an idea of what it's all about and how things work and we'll fill in the details later so as I said this is work that Tony Lowe has done and I'm up here trying to explain what Tony has done and I'd also like to acknowledge Supriyo and Yorg we spent a lot of time on these problems together so we'll start with the introduction it's a p-enjunction we we know how to draw energy band diagrams for p-enjunction so this might be an energy band diagram for silicon you can see this electron coming from the left over to the right there are states in the conduction band so it can flow freely then it hits a barrier there are no states underneath the conduction band there's a band gap so it can't flow so we apply a forward bias forward bias lowers the energy pulls the conduction band down and now they have more of a chance of getting over and current can flow so when we take this diode and apply a forward bias the current flows exponentially when we lower the barrier and this i0 in front of that involves things like ni squared which involves band gaps so ni squared goes as e to the minus band gap over kt that would lead you to believe that if you have a small band gap material you have a a very big ni and then if you apply a small forward bias you're going to get a sizable current you expect small band gap devices not to rectify very well graphene has zero band gap so you don't expect much rectification in a graphene p-enjunction but there are some really interesting effects that people are predicting will occur and people are doing measurements on graphene p-enjunctions that are also interesting so one of the basic ideas is and this is some of the some really nice work has been coming out of David Gold Harbor Gordon's group at Stanford these are measurements that show if you make an n plus I mean heavily doped n-type graphene junction with a moderately doped graphene junction you'll get a linear conductance doesn't rectify you don't expect it would rectify if you do a graphene p-enjunction it doesn't really rectify you get a linear conductance but you might expect that the conductance of a p-enjunction would be lower than the conductance of an n-enjunction even though there's no band gap and that's what's observed experimentally and we want to try to understand where's that coming from why why is that occur now there's also some other interesting things that can happen in graphene graphene has these long mean free paths so you can think of graphene electrons propagating over long distances ballistically and semi classical electron trajectories are really very analogous to the rays in ray optics and you can make some analogies between electron optics and and wave optics So there have been some interesting things happening, you know this we're all familiar with Snell's law so we have two different optical materials with two different indices of refraction light comes in at some angle to the normal theta one it refracts into the second material at an angle theta two and Snell's law says n1 sine theta one equals n2 sine theta two we can figure out the angle of refraction so some of you know that there's a lot of interest these days in artificially structured materials metamaterials that that can be structured to operate with a negative index of refraction when you do this with materials like that you find that the refraction goes on the other side of the normal and you can do interesting things like build lenses with perfect planes it's a lot of work in this field going on now why do I bring this up well first of all let me show you that there's a people in this field talk about a thing called a Veselago lens where the rays will diverge say from a point source then there'll be incident on a material that has an equal and opposite index of refraction and they'll be refracted back towards the normal and focus and you can do things like focusing with lenses like this and they're argued to to get around the diffraction limit of a conventional lens so there's a lot of excitement in optics about this and there's an interesting prediction that as far as I'm aware hasn't been has it been observed experimentally yet but I know with people that are trying to that if you build a p-n junction in graphene and you have a point source of electrons in say an n region diverging after incident on a p-n junction they'll be focused in much the way optical rays would be focused in materials with a negative index of refraction so we'd like to understand this you know where does this come about why does it happen but why is it predicted to happen because a lot of people are thinking that there could be some really interesting device possibilities that would exploit effects like this so you know one of the first things we have to think about is how do you make a p-n junction and of course this is not the not ultimately a manufacturable way but it's a way that you can study the effects and do the experiments and see what's there so the way these are made is we have a heavily dope silicon wafer we're going to use this as a back gate we just apply a potential to it there is a layer of SiO2 you can see it's 297 nanometers you know people like to use about 300 nanometers so that you can optically see one monolayer of graphene and we have contacts at the two ends while we have a four probe measurement here so here are some voltage contacts between and we have another gate on the top and a material PMMA another dielectric so that we have a gate on the top so we can get it from the bottom we can get it from the top if you apply a negative voltage at the bottom I'm applying a negative voltage no I'm thinking of a positive voltage I always use red for electrons so a positive voltage will attract electrons so you will electrostatically dope or induce electrons in the graphene so the back gate voltage has made the entire layer n-type we want to make it more n-type we just apply a larger back gate voltage but we have a top gate too and we could do something different with the top gate we could apply a negative voltage to the top gate and induce holes and that way we're left with an NPN structure we can apply a bias and we can we could look at the structures like this so it's actually easier and more typical to build NPN structures rather than single NP junctions but in most cases the feeling is that there is so much scattering going on inside this p region that you really think of it as two p-n junctions in series we don't have to worry about interference effects and things like that so that's how people make p-n junctions. Yes, the width of the interface how abrupt is the p-n junction so this is going to turn out to be really important and it'll depend on the electrostatic design the thickness of all of the insulators and everything so it's a it's a problem in 2d electrostatics and how many carriers there are in the graphene to screen it typical numbers are in the order of 50 nanometers. Now there are some devices and it'll turn out these negative index focusing devices that need junctions that are much more abrupt than that so one of the challenges to realize those devices is is going to be to make junctions that are more abrupt but that the width of that transition region turns out to be really important to understand the IV characteristics so we'll talk a lot about that. Okay so back to our p-n junction here it is remember we whenever the electrostatic potential goes up it moves the electron energy down whenever the potential is negative it moves the electron energy up so these bands all follow minus q times the electrostatic potential so I always have to keep that negative sign in mind so I could plot the conduction band the valence the valence band or the intrinsic level and it's just the upside down the inverse of the electrostatic potential versus position. Okay what would an energy band diagram for a graphene p-n junction look like? Well we're going to induce an electrostatic potential perturbation with these gates we're going to make the potential in one place lower so the potential whoops so the potential is more positive say on the left that lowers the electron energy and we'll make it maybe the top gate then we'll induce a negative potential pull the electron energy up so we've got a potential profile that looks like that. Now in a semi-classical picture then the potential profile is assumed to be slow enough on the scale of the lattice that it doesn't change the band structure all it does is to move these cones up and down they just follow the local electrostatic potential so over here in the n region the neutral point is right down here at this Dirac point or neutral point which is now spatially dependent and anywhere I go I can just imagine one of these graphene cones that's just being modulated with the local potential and somewhere there's a Fermi level so you can see here the Fermi level is above the neutral point so it's n-type in this region and you can see over here the Fermi level is below the neutral point on the right so it's p-type so we've made a p-n junction and we're interested in how current would flow all right so let's assume this one is abrupt we're raising we have a potential energy step up which means the electrostatic potential is negative yes how much voltage has to be applied to these gates in order to do this so you kind of saw when we just when I discussed earlier I guess it was Tuesday how high in the Fermi level we typically got how high the Fermi level got above the neutral point in typical experiments it was something like 3 tenths of an EV now but these these oxides can be quite thick 300 nanometers so you drop in order to move the potential 3 tenths of a volt in the graphene you may have a very large potential drop in the oxide so I don't know Tony is probably calibrated I think you know you see things 2050 even a hundred volts and on some of these back gates because it takes that much voltage on that thick oxide to move the potential a little bit in the graphene no it's simply a matter that that that electrode is so far away that in order to affect the potential in the graphene it's just we have we have a lot we have to apply a lot of voltage there now so ultimately if you want to make a high-performance device you're going to want to scale down the thicknesses of all of these oxides and operate with a volt instead of a hundred volts and things like that but if you read papers nowadays the gate voltages are typically very big just because the oxide thicknesses are very big but the the potentials we end up in the graphene are pretty modest. Yes, so as I understand you have to apply a positive voltage so one will have a positive voltage and one will have a negative voltage. Well I could do it either way. Well sure if I want to make a PN junction I might want to make an NN junction all right then I would apply different negative voltages or different positive voltages to both of them I might want to make a PP junction so people you know people do all different kinds of combinations. So my point is the channel length and the outside thickness so if there are comparable to one another like you said you have a huge voltage drop across the outside thickness. So how effective is the gate to length? If you don't have the outside thickness that it takes you might have a very smooth voltage drop. So your questions all relate to the electrostatic design and these are the kind of questions that anyone who designs a high performance transistor has to worry a lot about you know the scale that your potentials vary at two dimensionally dependent how thick the oxides are how close you get the gate to the channel. The structures that are being done experimentally these days are relatively thick so the transition regions the insulators are relatively thick. So the transitions between the N and P type are relatively broad. The channel lengths are still relatively long so ultimately if you want to scale these down and look for ballistic effects and high performance devices you're going to have to pay much more attention to all of those electrostatic design issues. Yes. And the amount of charge carriers which get increased are much more compared to that of the bag gates simply because the thickness of the bag gate is much more. So if you apply like two to three volts on a dog gate you get carriers which are in the order of 10 by 30 whereas in the bag gate you've been using 10 by 11 or 10 by 30. So that automatically takes care of whether you have N type carriers in the junction even if you apply like a different voltage. So it's much easier to have a thin top insulator so the top gate voltage can be much smaller than the bottom gate voltage. Yes. So if you didn't have the top gate then this would be an SOI in some way. And so in that case the offset thickness shouldn't matter that much. Is it that you need 100 volts because you have a top gate sitting there which is holding the potential. So you can get to the very thin offset. Yeah. But we're mainly talking here about operating above threshold. So the way you think of it you want to induce some charge say on the order of 10 to the 13th in the graphene. That's just C insulator times VG. If the bottom oxide thickness is very thick C insulator is very small and you need a high back gate voltage to induce some charge. Now your sub threshold swing might be ideal but we're inducing a lot of charge. Okay good. So let's do an abrupt one. So this one assume none of those two dimensional electrostatics will assume it's abrupt. And we've applied say a back gate voltage here on the left to lower the neutral point. The Fermi level is way above it. This is N plus. The Fermi level is way up in one of those cones. Over here on the other side we don't have the potential as high. So the Fermi level isn't as far above the cone. But it's N plus on both sides. N type on both sides. So I would call that an N plus N junction. Okay. And at low bias this thing doesn't rectify so it's not like a silicon diode where you can apply a volt across it. It doesn't rectify. So you apply a small voltage you get current immediately. So we're only going to look at the linear conductance near V equals 0. And all of the current flow will happen then near the Fermi level. So to understand this we're just going to ask what are the states near the Fermi level. Because those are going to control the current flow using all the ideas that Professor Dada talked about. So the picture is an electron here on the left comes in from the contact. It's coming in with a positive velocity if it's coming into the device. So the velocity is related to the slope of the EK. So this side of the cone here has a positive velocity. The magnitude of the velocity is always 1 times 10 to the 8th centimeters per second. But this side has a positive velocity. So it comes in and fills up that state, goes across the barrier, moves down relative to the bottom of the barrier. But it total energy stays the same and occupies that positive velocity state. Then it goes out to contact, comes around the other side and keeps going. Okay what if we have a PN junction? So the left side is similar. What we've done here I think is to keep the barrier the same, but we just lowered the Fermi level a lot. Maybe the back gate voltage was reduced. So now the Fermi level is much lower. Now when I go over this step you can see that on the left side I'm N type, on the right side I'm P type. So what will happen? An electron will come in from the contact. It has to occupy a positive velocity state just like before. There's no barrier because there's no band gap to stop anything. There's states everywhere, right? So it can flow right through that barrier and it has to occupy a state at the same energy because I'm assuming ballistic transport is no inelastic process. So it's got to be at the same energy, but I have two choices of where I can be in that cone. I have to pick the one that has a positive velocity physically. That's the one that's going to flow through. So if you look at what's happened here, the electron has gone from the conduction band to the valence band. It goes out to contact, back around and we just have continuous current flow. From the conduction band to the valence band. So it's something like interband tunneling. There's no band gap here yet to tunnel through. Very important for the future. Notice that we're always going to think about, you know, whenever we deal with one of these ellipsoids or valleys we think about K relative to the bottom of the valley. On the N side we have a positive velocity state that has a positive KX. On the P side we have a positive velocity state that has a negative KX. So we've just gone across, stayed at the same energy, stayed at the same velocity. But KX relative to the direct point has changed from positive to negative. And how did that happen? This force here, due to the barrier. DPDT is equal to the force. It just changed KX from positive to negative. Okay, so we have these various junctions that we can think of depending on how we bias it. N plus N, that would give us a good strong conductance. NP, we're going to see that that gives us a lower conductance. We're going to talk about why. We could conceive of changing this barrier height on the right side such that the direct point lines up with the Fermi level. So ideally if there are no fluctuations and everything is uniform, now there are no states to carry current. So in the ideal situation under that point, N type intrinsic junction, I wouldn't have any current. Okay, so that's sort of the introduction and now we want to understand in a little more detail. First I'm going to go back to the case of this electron optics to see why we would expect graphene PN junctions to behave like they have a negative index of refraction. And then we'll go and apply a bias and talk about the IV characteristics and try to understand why an NP junction has a lower conductance than a PN junction. And as Professor Appenzauer pointed out, we may have these junctions when we're trying to electrically characterize material, the conductance of graphene sheets. These junctions might be present and might be affecting the measurement. And we'll just take this standard simple picture of graphene. We're only going to work near a direct point. We have two equivalent valleys. The E is plus or minus h-bar Fermi velocity times magnitude of the wave vector and we're in two dimensions, so it's square root of kx squared plus ky squared. Density of states goes linearly with energy. Number of conducting channels goes linearly with energy. All of the concepts we had before. This is all that we'll need. Now there's one thing that I'll mention here that we're going to need a little bit later. When we think about electrons propagating in semiconductors, sometimes we use an effective mass equation for the electrons. If the potential is uniform, we just have a plane wave propagating in the xy plane and we might normalize that to some area. So when we do psi star psi, we get one. The area drops out of the final calculation. We've heard from Professor Dada's lecture that graphene has this two-component wave function associated with this a and b sub lattice. So this version of this effective mass level wave equation looks a little different. It's got two components. And I'll just show the answer here. It's got this plane wave part. But the two components, we'll write one as one component just as one and the other one is s times e to the i theta. So s is positive if you're in the conduction band, negative if you're in the valence band. And the phase is determined by the angle that k makes in the xy plane. So that's theta. So this leads to something very remarkable that was first noticed for carbon nanotubes, which are purely one-dimensional, but that plays a role in graphene too. So if I think about an electron, say that comes in incident along the kx axis. And so it comes in with kx non-zero, ky is equal to zero. It's just going along the kx axis. And I ask, what is its probability of back-scattering and going some kind of perturbing potential that scatters it and flips it around and sends it in the minus kx direction? Well, the angle is pi. I'm still in the conduction band, so s is positive in both cases. So the pre-factor here now is one minus one. So those two wave functions are orthogonal. When I try to do a psi star psi and send which is scattering potential in between to figure out the transition rate with Fermi's golden rule, if the perturbing potential is slow on the scale of the lattice between these two sitting on the A and B sites, I don't see the same potential and when I do psi star psi, I'll get zero. So there should be no back-scattering. You see that in carbon nanotubes. In graphene, you will also see that. There should be zero probability of back-scattering identically. Another valley would be located at a big k distance away. So again, it would take a very rapidly varying potential in order to do that kind of transition between valleys. Now there are mechanisms. So I understand that surface edge roughness scattering varies on an atomic scale. And that can give you a perturbing potential that's rapid enough that it can flip and cause you to do back-scattering. But when we think about phonon scattering or ionized impurity scattering, we expect the back-scattering to be suppressed. Now it's not quite suppressed to the extent that it is in a carbon nanotube because these are two-dimensional. So the perfect suppression only occurs for exact back-scattering. But when you average over all of the angles, it ends up suppressing the overall average scattering rate. So when you calculate scattering rates for graphene for acoustic phonon scattering or whatever, you'll get smaller answers than you would with the same perturbing potential that you would in silicon, say. So it's part of the reason that you get higher mobilities. Okay, so now we can talk about optics. All right, so let's understand this negative index of refraction business. As I said, I'm not aware of any experimental confirmation yet, but I know of people that are trying to. So it's a very interesting effect. So we have a graphene p-enjunction. And so one of the things to look at is, let's think, we have an electron came in from the contact. It's a positive velocity electron. It goes across the p-enjunction. We still want it to be a positive velocity electron because it's propagating from the left to the right. So the velocity on this side of the cone, the right side of the cone is positive. The velocity on the left side of the cone is positive. I mean, this is two-dimensional, so these are actually the cones. The cone is on the top. In the conduction band, the velocity when ky is equal to zero is just plus along the kx direction, but everywhere it just points normal to the constant energy surface and out. So whatever direction k is, the velocity points parallel to it out. Now if you look at the p-side, however, you can see that if I do a cone of constant energy there, the left side is the one that has the positive x velocity. So over here, the velocity vector is pointing positive. It's on the opposite side as in the conduction band. So the velocity is anti-parallel to k below the conduction band and it's parallel to k above the conduction band. This relates to this change in sign we were talking about earlier. So that's very important for us to remember. In the conduction band, the group velocity of electrons is parallel to their wave vector. In the valence band, the group velocity is anti-parallel to the wave vector. So is this unusual? Well, let's draw the ek for any semiconductor. It might be parabolic like this. The group velocity is always the gradient of the energy band. So if I have an electron in the conduction band, the velocity is going to be parallel to k. Plus kx will give me a plus velocity. If I have an electron in the valence band, the derivative is negative. So the k vector is still positive. So the velocity of the electron is going to be in the opposite direction of the k wave vector. We never worry about that because we usually think about holes in the valence band and the hole is going in the opposite direction. So the hole is going in a positive x direction when it has a positive kx. But we're going to think about everything in terms of electrons. So we're going to follow these electron trajectories and watch them focus. All right, so any small band gap semiconductor would behave similarly. All right, so now we have, let's try to do Snell's laws for graphene. So we have these pictures. We have an incident wave vector. We have it partially reflected at some angle theta one prime, partially transmitted at some angle theta two, and we want to figure out what the angles are. We could draw, these vectors could be velocity vectors. If I'm doing optics, right, the velocity, normal optics, the velocity is in the same direction as k. So the arrows would look the same if I was thinking of them as the velocity of the photons or the wave vector of the photons. All right, now we'll do graphene p and junctions. Same kind of situation. We have an electron with an incident wave vector coming in at an angle theta one. It can reflect at an angle theta one prime, or it can transmit at an angle theta two. Well, these trajectories are semi-classical trajectories. So they follow this semi-classical law. Force is the PDT, and the force is minus q times the electric field. So if I have a uniform junction and I have no variation in the y direction, there's no electric field in the y direction. So when this electron goes across the interface, k y cannot change. There's no electric field in the y direction to change it. Now there's a junction, there's a step in potential there, so k x can change, but k y can't. So one of my boundary conditions is going to be that k y is conserved, the incident y component of the wave vector, the reflected wave component of the wave vector, and the transmitted wave component of the wave vector all have to be the same. And energy is going to be conserved too, because I'm not having any scattering processes here. The electron isn't going to gain or lose any total energy when it goes across. And that's enough to give us the Snow's Law. So on the n side, I have an incident wave vector and a reflected wave vector. We're going to assume that we're near t equals zero, so we're only going to worry about energies that are near the Fermi level. So it means k's that are at the energies near the Fermi energy, or wave vectors at the Fermi level. The magnitude of the wave vector is always going to be k f, that's never going to change. So if I just do this on the n side, I have the y component of the incident wave, and the y component of the reflected wave have to be equal. So we immediately see that the angle of incidence has to equal the angle of reflection. No surprises there. Now let's look on the other side, see what happens in the p side. So remember I'm going to... the electrons come in with some wave vector k f, that depends on how far the Fermi level is above the neutral point on the n side. They transmit across, and they have some wave vector, that depends on how far the Fermi level is below. And I'm going to assume a symmetrical junction, just to make things simple, it doesn't have to be. So that on the n side, this energy difference between the electron and the neutral point, epsilon 1, is equal and opposite to what it is on the p side, just a symmetrical p-n junction. And all I have to do is to adjust the potential step correctly so that I make it symmetrical. So the incident Fermi wave vector is going to be equal to the transmitted Fermi wave vector. And then I have to worry about these two choices. I can conserve energy in two different ways, but only one of those has a positive velocity and will give me a transmitted wave. So I have to choose the negative one. So we'll use the same boundary conditions as before. The picture that we have now is that the electron comes in at some angle in the kx, ky plane, coming in with a positive bx velocity, transmits across the y component of the wave vector cannot change, because there's no force in the y direction. However, the x component has to go, has to be minus whatever it came in with in order to give me a positive velocity. So the wave vector has to flip over here such that this velocity arrow is pointing in the plus bx direction. So that's what it's going to look like in the k space. So the sign of the tangential component of the wave vector stays the same across the interface, but the sign of the normal component of the wave vector flips and inverts. So if I'm going to draw the picture, it sort of looks non-causal. It looks like the wave vector on this side, the ky component is the same as it was on the incident, but the kx component is negative, so it comes in from the other direction. But if I want to do it in terms of velocities, on the p side the velocity is anti-parallel to k, so everything looks like it makes sense if I think about it in terms of velocities. So we end up, we argue that the angle of incidence equals the angle of reflection. You can also see that in order to make kx transmitted minus kx incident, I have to have theta 2 equal to minus theta 1. The minus sign means it's refracted on the opposite side of the normal, so it has to bend down this way. And if I were thinking about that in terms of Snell's law, that's what would happen if n2 is equal to the opposite of n1. And that's why people say there's analogy with a negative index material. So it behaves as though it has a negative index of refraction. If you want to make sure you understand all of this, you could relax the assumption that this is a symmetrical junction with the same energy on each side, same kf on each side, and you could do it more generally. And the relation you would get is that angle of incidence is always equal to angle of reflection. But we would get epsilon 1. This energy difference between the Fermi level and the neutral point on the n side is equal to the epsilon 2 on the p side, sin theta 2. So this energy difference is playing the role of the index of refraction. And since it's positive on the n side and negative on the p side, we get a negative index on the p side. And you could ask yourself questions about what's the critical angle for total internal reflection such that you could use these materials to say guide electron waves. And you can figure out what that would be. You get the smallest critical angle, which means everything is internally reflected. When epsilon 2 is zero, that means you put the direct point at the p side right on the Fermi level, so there are no states to transmit into. Okay, so people are looking at ideas like this. It causes you to think about all kinds of device ideas that you might think about, and a lot of people are thinking about those right now. Who knows? It could be a lot of fun. Alright, so we've done angles. The thing we haven't talked about is what the magnitude of the reflection and transmission coefficient. We just know what angles things are going at. So if you remember, when you do this in electromagnetics, you match boundary conditions. So you match the tangential component of the electric field and the H field at the interface, and then you can compute the reflection and the transmission coefficients. It'll depend on the polarization of the incoming wave and the indices of refraction on both sides. So we would do something different here. And, you know, we have a set of references at the end of this lecture where you can see how some of these are worked out. And it's not too difficult to work this out. You just have to do it with this two-component wave function. So we've got an incident wave coming in. Remember this angle theta is the angle that k makes in the xy plane. That gives me the two components of the incident wave. I have a reflected wave, and it's going to have a minus kx here and go out with some amplitude r. And then I have a transmitted wave. And here I just have to be a little bit careful about my sine on kx, or I could get into trouble. Normally when I say I have a minus kx, that's propagating in the minus x direction. But on the transmitted sine, a minus kx is going to propagate in the plus x direction. So I have to be careful about that. So you match boundary conditions at the interface. You can find the magnitude of the transmission and reflection coefficients. You can square them to get the intensity or the reflection coefficients. And you can compute the transmission coefficient. So as I said, I think we have a reference at the end where a paper where people have worked this out. And you can see how it's done. And the answer is this. The transmission coefficient goes as cosine squared theta incident. So let's see how we think about this. So the way you interpret this plot is you start here, and if your transmission is one, then you're all the way at the arc. So you can see if you go in at an angle zero, this red curve is all the way up there. It has a magnitude of one. If you come in normally incident, you transmit perfectly. The transmission coefficient is one. If you come in at an angle, the transmission coefficient is smaller than one.  So you get this cosine squared theta. Why does that happen? You can think about it in terms of this picture. You have an electron coming in in the conduction band with some wave vector. The velocities are parallel to the wave vector there. Being transmitted into a p-region where the velocities are anti-parallel to the wave vector. And you can see that these velocities, they're sort of a representation of the wave function. It has different shapes. People also represent the pseudo-spin with vectors that look identical. So you can see that things aren't matched very well between the conduction and the valence band. The wave function mismatch is giving you this reflection. Except for one place. If you come in right along the KX direction, and then you transmit right along the KX direction, everything is matched nicely. And you get perfect transmission. Now that's also another manifestation of this absence of back-scattering. Perfect back-scattering is prohibited. You don't get that. Now remember, this is an abrupt junction. Or something. And this perfect transmission for theta equals zero is something that people think is very interesting. Now, if we have a p-n junction and we're measuring its conductance, we're going to have carriers coming in with all different angles, and we'll have to average the transmission over all of these angles. And because the transmission is reduced by this wave function mismatch, we'll have a lower conduction for a p-n junction than for an n-n junction. So this effect will be there, but it's not nearly enough to explain what you see experimentally. I think Tony tells me it explains about 10% of it or so. So there's something else going on in the real experiments, and it turns out to be related to the non-abruptness of the junction. So these are the kinds of junctions that people are going to have to learn to produce if you want to do these electron optics. But most of the junctions these days are much more graded. So let's look at a graded symmetric p-n junction. So the only difference we have here now is that everything looks equal and opposite on the left and right-hand sides, the n and the p-type side, but the transition does not occur abruptly. So you can see if I'm following electrons through at the Fermi level, the conduction is occurring at the Fermi level. We're going to talk about the conductance now of these. So we're interested in the states at the Fermi level. Those are going to control the conductance. Now you can see here I've got a point. This happened infinitely in the abrupt one, but now it happens gradually. There's a point there where the potential goes to the Dirac point. There are no states there. So how is the current going to get conducted? Something is going to happen. If this Fermi energy was way up and both sides were n-type, then I would never have any problem. I would always have lots of states to flow through. But when it makes a transition from n-type to p-type, your Fermi level has to go through the center of that cone. And that's going to reduce the current. And we'll see how because it happens in a kind of interesting way. So let's think about each of these waves that come in at an angle. Let's think about them each separately. So we have some particular incident wave. It comes in at an angle theta to the normal of this junction. It has a magnitude kf. So in the y-direction it comes in with a y-component that's kf sin theta. And we want to find out how does that flow across this junction. How does it transmit across this junction? In that particular way. Well remember when we draw these band diagrams, it's really a two-dimensional band diagram. I'm plotting it versus kx or ky or magnitude of k, but there are two dimensions there. So let me say that kx equals zero and plot this versus ky. I just get these two crossing bands like this. Now ky is fixed because I'm only going to talk about an incident electron that comes in at this angle. So ky has a specific value now. It's never going to change because momentum is conserved. The ky is always conserved across the junction. So it comes in with this particular value of ky. That means if it's in the conduction band it has that red energy, h bar kf vf sin theta. If it's in the valence band it has this energy minus h bar kf vf sin theta. Maybe I should have drawn it on the other side if I want them both to have positive velocities. But the point is that there is some energy associated with the fact that this finite k puts them somewhere on that cone. Now our total energy then is h bar vf times the square root of kx squared plus ky squared, which is fixed now for this particular rate that's coming in. Ky is going to change across the junction because h bar dx, h bar dx dt is equal to minus q times the electric field. So there's a force on kx it's going to change. But the energy has to stay the same. So if I were to plot this ek plot now versus kx because kx is going to change when I go across the junction. So I'm interested in e of kx. ky is fixed. So as kx gets smaller and smaller and goes to zero, the energy can never get smaller than this value h bar kf vf sin theta. And then when kf gets negative, kf squared will get positive and the energy will go up. So I've got a band gap. For this particular incident angle when I'm plotting e versus kx there's a band gap there. Or people will call it an apparent band gap because there's only a band gap at this angle. And if I made a different angle I'll have a different band gap. So each ray that comes in will see a band gap unless it comes in with theta equals zero. And then there's no problem. It goes all the way down. So now if I draw the energy band diagram for this particular ray coming in at an angle the energy band diagram sort of looks like a typical semiconductor with a conduction and a valence band and a band gap. So when I think about an electron coming in the conduction band to the inside going across the junction I can see here now it's in a forbidden region. So it's got to be an effervescent wave. And when it finally comes out it's over here in the minus kx part of the valence band. So this looks like band to band tunneling in any semiconductor. Let's see what did I want to say here. Yeah that's about it. And the band gap is a different band gap for each ray that comes in with a different angle. Each one will have a different band gap. Okay now if you know this is one of the nice plots that Tony has in his paper I'm going to refer you to this paper if you want to really see how this works out quantitatively and how nicely you can explain measured data. Tony has a very nice paper where all of these calculations are done. Professor Dada talked about how you do the NEGF simulation and how you extract the spectral function and that's the density of states. You just do the NEGF simulation and plot out the density of states. Dark blue means low density of states. So here he's plotting out one where you're coming in at some energy in the y direction equal 0.15 ev and you see a band gap. See a low density of states that's the band gap. You know you can see electron waves coming in and reflecting off of that potential barrier and interference patterns being set up also. Okay. Alright so how would we handle this? Let's see. We have an electron that comes in. Epsilon 1 is its energy with respect to the middle of the band gap now or what's the direct point. So epsilon is a function of position now and it's just h bar vf squared of kx squared plus ky squared given by the graphene band structure. Okay. But ky squared is fixed. We're just talking about one ray coming in. So epsilon of x is equal to h bar vf kx squared plus kf squared sine theta. So everything works out nicely. So over here on the end side I have a big positive kx and it adds to my ky squared and I get the total energy. Over here on the p side I have a big negative kx. kx squared adds to ky squared and I get the same energy. But if I look in the middle and ask myself what is kx? I could solve this equation for kx squared. And you can see that as you go into the depletion region, epsilon of x is going to get smaller and smaller and smaller and right in the middle it's zero. Okay. If it gets too small, smaller than kf squared sine theta, then kx squared is negative. Kx squared negative means it's imaginary. Meaning I've got an evanescent wave, not a propagating wave. So inside this transition region, there will be some region there where the wave vector becomes negative. I have an exponential wave and you're tunneling through that region. And you can treat that with standard WKB theory. People do band to band tunneling with WKB theory all the time. We know what kx is as a function of position. The wave function goes as e to the ikx. When kx is positive it's a propagating wave, but when kx is negative it's an evanescent wave. There's this familiar technique that you may have used before to compute transmission coefficients. When the potential varies slowly enough that there aren't quantum reflections, then you can do this WKB approach to compute the tunneling current. And it basically ends up integrating kx across the region where it's imaginary, from one side to the other. Alright, and you do the algebra and it's in one of the references that we have. And the results you get is that the transmission probability is e to the minus pi kf v, the width of this transition region, times sine squared theta. Now if you want to, you can recast it into a different form so it might look like something you may have seen before. Remember this band gap that this incident ray at angle theta sees is related to the angle and to kf. And the potential step that we've induced in order to make it go from n-type to p-type, the potential step was 2 epsilon 1 divided by q. The transition region was d, so that's the electric field in the transition region. So I could just rewrite this expression as e to the minus pi eg squared over some constants times the electric field. And if you've done band to band tunneling and semiconductors, you've probably seen this. You see expressions like this for silicon, for anything. When you do it for silicon, for the silicon dispersion in three dimensions, I think it's eg to the three halves power or something. But it goes exponentially, it's e to the minus band gap to some power. The electric field is always on the bottom because the stronger the electric field, the thinner the distance over which you have to tunnel. So it's a very familiar expression for band to band tunneling. Okay, so what if you plot this transmission coefficient up? For the abrupt junction, we got the red line and that we argued was due to this mismatch in these two component wave functions. We had perfect transmission for normal incidents, but we had reduced transmission for others. If you do it when you have a finite transition region, it's much more pronounced. When you go at a bigger angle, you have a bigger band gap and it's harder for the carriers to tunnel through. However, when you go at zero angle, there is no band gap, so there is no problem. So you still get, no matter how wide that transition region is, you still get perfect transmission, normal incidents, but it just becomes very sharply focused. So now when you want the total conductance and they're coming in with a spread of angles to the p-n junction and you have to integrate over all of those angles, only a very small cone of them is going to be able to get across to the p-n junction. So the current is going to be reduced substantially. And now, as Professor Dada mentioned, you could just do an NEGF simulation and all this would be there. And what happens now, what we've done is when we do WKB, we ignore reflections. When you do the step part, you're ignoring the fact that if there's a transition, you've got a tunnel through a region. In a real junction, you're going to have both things occurring at the same time. If you run an NEGF simulation, you'll automatically get it. But I think Tony tells me if you just multiply the abrupt one times the graded one, it works pretty well. Is that right? So if you want to know what happens in a realistic junction, you could do it this way. All right, so now we can talk about it. The discussion is only one slide. So we have the main point coming. Yes? The incident angle is like zero. So in that part, there is no band gap along that direction. So in that case, the relationship is like linear. So when you're close to that point, the density gets smaller. Yeah, so I guess the way, so your question is how are you going to get to that little bottleneck there? And also the dilemma is actually if you have some ever normal parabolic band, in that case, the density is actually independent of that. So in this way, I mean, the condition is supposed to be updated to what you're helping. So I think the first part of your question is what happens in, at normal incidents, you're still going through this direct point. There's still a point where the density of states is zero. So I guess the way I would think about that is you're tunneling through something that's infinitely short. So it's not hurting you. You're tunneling from one side to the other side. It's like a band gap where there are no states. But you have states on one side, you have states on the other side, and D is essentially zero, so you can tunnel right through it. Yeah. Because of the parabolic states. So maybe we can come back to this because we're going to talk about how these densities are states in different parts of the junction next. Maybe we can come back to this question. Okay. So now we want to do the conductance. Now, so what I'm going to do, maybe this won't get exactly at your question. I'm going to go back to an abrupt junction because it's kind of easier to see what happens in an abrupt junction. And when we go to a graded junction and we want to do it quantitatively, I'm just going to show Tony's simulation results so we can see what happens there. So how do we think about the ID characteristic here? So I have these junctions and I can do all kinds of things with the front gate and the back gate voltages. So my back gate voltage can control the location of the Fermi level in the inside region. If I make the back gate voltage positive, I pull this potential down and I make it more n-type. So one of the degrees of freedom I have is on this Fermi level in the n-side. If I make it less than zero, the left region is p-type. If I make it greater than zero, the left side is n-type. And my back gate voltage will let me do either. Now, the other degree of freedom that I have is on this potential step. The height of that potential step is minus q times the junction potential. And I can control that with a top gate voltage. So I can move that around. So if I make the height of that step, which is minus q times the potential bigger than EF, then I've made the right side p-type. And if I make the height of that potential step less than EF, then my direct point is down below the Fermi level and I've made the right side n-type. So I've got these four combinations. And when people do the measurements, they make these plots where they vary the Fermi level and they vary the top gate voltage. And they'll plot the two results out on a two-dimensional plot because we can ask the question, what happens for any possible combination of those two variables? So you can draw a picture like this. We can move the Fermi level on the left side. We can move the potential for the barrier height with the top gate voltage. And we have a point. The dashed line here tells me that the Fermi level is above the, is above zero, above the drag point, then the left side is n-type. If the Fermi level is below, then the left side is n-type, p-type. So that's what the dashed line tells me. Fermi level above it n-type, Fermi level below it p-type. Now the height of the barrier that we put is minus q times the junction voltage. If the height of the barrier is equal to the Fermi level, then I've lined the drag point on the right side up with the Fermi level. There are no states there, so there should be no conductance. So along that line, along this diagonal red line, I've got the Fermi level on the right side lined up with the drag point. So now I can just go in and look in each region here. If I'm above the dashed line, I'm n-type. If minus q times the junction potential is less than the Fermi level, or I've written it the opposite way here, then I'm n-type. So I would have n-end when I'm above. If I have my Fermi level below, then I change from n on the left side. If I have my Fermi level on the left side below the dashed line, then I change from n-type to p-type on that side. And if I go on the other side, if I'm the yellow side here, if the Fermi level is above the dashed line, I'm n-type. If it's below the dashed line, I'm p-type. But if the height of the barrier, which is minus q times vj, is bigger than the Fermi energy, then the right side is p-type. So you can go through in this picture like this and even determine which region is n-type, which region is p-type. And when people do experiments, they map this all out and they try to understand when I measure the conductance here, what happens, when I measure it in the p-p region, what happens. So we want to compute the conductance. So this is where it gets a little difficult to do, because we have this expression that Professor Dada introduced to us. This shows us this Landauer expression that the conductance is just 2q squared over h times the number of channels at the Fermi level. The trouble is the number of channels at the Fermi level is now position dependent. So which one do we put in this? So really this formula doesn't apply when the density of states varies with position, the density of channels varies with position. We should do an NGF simulation, or there are other techniques where we could essentially solve a Boltzmann equation and spatially resolve it, and we could do it. But we're looking to try to get the answer in a much simpler way. And it would seem reasonable to think that wherever the number of conducting channels is the smallest, that's going to be a bottleneck, and the current is never going to be able to be more than what's available at that point. So what I would do is I would look on the N side, and I would look on the P side, well I'm sorry, I would look on the N side and then I would look on the P side. In each case I would figure out how far above the Dirac point I am, I would figure out how many conducting channels I have. Whichever one is smaller, that's the one I would put in my expression. I would say that's what the current is going to be for these bias conditions. Now what happens here in the abrupt step, where it's going abruptly from one to another, I'll say that's so thin, anything that happens there, the carriers will just tunnel right through that zero distance region, and I won't have to worry about that. For a graded region, I have to worry about it. And Tony's NEGF simulations will give us the answer. Okay, so in this case that I've sketched, the number of channels inside two is smaller than the number inside one, so that's the number that I would use. So you can write a little MATLAB script, and that's how Tony made this plot. And you can vary the Fermi level, and you can vary the height of this step. So we have to worry about signs. Everybody likes to use different signs. VPN here is the energy barrier, not the voltage. So we're plotting it negative on the left side, because you would get this barrier with a negative gate voltage. But the height of the barrier is positive. So it's the same plot that we've done before, with the same axis. And you can see the various regions that we identified before. You can see the diagonal line down the middle. So if I were to take, for example, if I were to set my back gate so that the Fermi level was 0.3 EV on the left side, and then say, okay, I'm going to vary the height of the junction step, and just go from a large positive one to a negative one. And I'll look each point at where the minimum number of channels is, and I'll compute the conductance that way. And I guess what you're seeing here, I guess I didn't say, is a color-coded plot. When it's red, it means the conductance is high. When it's dark blue, it means the conductance is low. So if I were just to plot the conductance along that point, it's pretty high over here on the left. It goes through this dark blue band, which is a diagonal one, and then it drops down. Then it comes out of that region. It goes back up, and it goes back in very high again. And what's happening there? Well, I fixed the Fermi level at 3 tenths of an EV, and I'm just varying the height of this step. So when I make this step 3 tenths of an EV, then the drag point on the right side is exactly at the Fermi level, and that's why it goes down to 0. And when I get close to that, I just get a smaller and smaller number of modes. The density of states goes linearly with energy. And then when I come out the other side, if I increase the barrier even more, then I have a density of states that gets bigger and bigger. It's p-type, but it gets bigger and bigger. And eventually, if I increase this barrier way up, I have a lot of states here because I'm very far away from the drag point. It increases energy. But now the minimum number is over here, and it doesn't change anymore. So you can map these out and see how it should look. And if you want practice on understanding this, fix the barrier height at 6 tenths of an EV, and then sweep the Fermi energy and see if you can explain why the plot looks like this. And just ask yourself, where's the minimum number of channels in each case? Alright, so people do these kinds of experiments. So this is the PN junction structure we showed before, and they plot the data this way. So here you can see it. We have to look a little bit closely. Now they like to plot resistance instead of conductance. So invert everything. You can see a diagonal white band. That means high resistance. That was the dark blue band, low conductance. And you can see the dark bands in the corners. Those were the bright spots with high conduction. Now they're the dark spots with low resistance. So the point I want to make is that the measured data look qualitatively like that simple argument. Now if we do one of these plots, all of these horizontal lines here, these are like the plots that I showed you. So if you pick one of them, let's pick this, what is this, teal-colored one here. This is like the one I showed you at three-tenths of an EV for the Fermi level. And we just plotted the conductance as we varied the barrier height. So here we're plotting the conductance as you vary the top gate voltage. This is what varies the barrier height. So we had the conductance going high, down to a minimum, coming up again. We're plotting resistance here so everything is flipped upside down. What you can see is that the resistance is low. It goes up very high at that point where our conductance went very low and then it turns around and comes down. Again, qualitatively it looks similar. But there's one very important thing is that if you look on the side where you have a P-P junction, or if you look on the side where you have an N-P junction and on the side where you have an N-N junction, you'll see that they're not symmetrical. So you get a lower resistance. Whoops, I guess this is the N-N side. You get a lower resistance when both sides are N-type than you do when there's a P-N junction. That's the experimentally observed asymmetry between P-N and N-P junctions. And when Tony does these NEGF calculations, so this is the plot that came out of the NEGF simulation. Looks very much like that MATLAB plot where we just said what are the minimum number of modes. There's structure and everything. But the numbers are, things are smooth and rounded out a little bit different. And what you see here is that over on the P-side, so when the built-in potential is big, say 0.8, means we have a big built-in potential. The Dirac point is way above the Fermi level. So this side is P-type. You can see a conductance. The NEGF treatment gives us a conductance that is significantly smaller than that simple minimum modes argument. However, if you make the potential barrier small, so you bring it down here, then the conductance goes up to a much higher value. And if you keep bringing it down, the conductance actually starts to drop. So what's happening there is you're having a bigger and bigger potential step, so you're getting reflection of the electron waves off that step. So you're getting quantum mechanical reflection and exploring things. Okay. So the simulations also show us this asymmetry that you see experimentally. All right. Now, I guess we should be careful about how much we go through here when I'm looking at the time. On the N side, we could just compute the conductance from the minimum number of modes which would be given on the N side. On the N plus side, there would be more modes. So who would use the number of modes at the Fermi level on the left side? If we want to do the PN junction problem, then what we have to do is to compute the transmission coefficient for each of the KY's and we have to integrate over all of those KY's. Remember, we had a transmission coefficient that went as cosine squared theta. So if you do that calculation, you'll find there's a factor of two thirds, which is just what the simulation shows. If you look at the place of the NN and NP, the NP junction has a conductance that's two thirds of the NN junction. This is an abrupt junction, so that's entirely due to wave function mismatch. But as I mentioned earlier, that's not nearly enough. That explains about the experimentally observed asymmetry is 10 times more. So it's much more pronounced than this. And the experimentally observed asymmetry is due to the fact that there is some transition region that we have to tunnel through. So if we use as our transmission coefficient the expression we got from this WKB treatment and if we do the integration over all of the KY's that come in, we'll find that the conductance is reduced by this factor square root of Fermi wave vector times the depletion region thickness. So the thicker it becomes, the smaller this conductance becomes. And typically this number is quite a bit bigger than one sort of on the order, rather than the order of 10 or more. So you get quite a pronounced conductance drop. So I'll refer you, Tony has a rather a much more extensive comparison of his simulations with the measured data. And you can see that if you account for the depletion region width and all of these effects, you can do a very nice job of explaining the data including some things that we did in the experiments. So it can be done. All right, so I have one slide left, I believe. And here there's a lot that could be said about this but I'm not, but I don't have much to say about it. You know, why is this relevant? If you're not making PN junctions, Professor Appenzeller talked about the fact that the standard characterization techniques that we do, we have a big sheet of graphene, we modulate the Fermi level with a back gate and we measure the conductance versus back gate and we try to do something about the scattering properties and what's going on in this graphene film. And we haven't really talked about the junctions. So always when you make a device you hope that the contact is nice and you're basically measuring the resistance of the film of interest and not of the contact. But the simplest way you can think about these contacts is there's some metal with some work function. Graphene has some work function. You put the metal on the graphene, there's some charge balance, some charge transfer, either from the graphene to the metal or from the metal to the graphene. The net result will be that underneath that contact the Fermi level is either above the direct point or below the direct point. So it's either N-type or P-type. So depending on what metal you've used, when you do this measured conductance and sweep the piece in between from N-type or P-type you'll either be the same material as underneath your contact or you'll be opposite. So on one side you'll see that reduced conductance of the P-enjunction and on the other side you won't. So people have a couple of explanations for what this asymmetry might be caused by. It's still not something that's completely resolved but it is very likely that the contacts are playing an important role in these.  Whether you see it on one side or another side might depend on what metal you use for the contacts. This is a very nice paper that I'll recommend to you. They also talk about how if you carefully do four probe measurements you can eliminate these effects and always get symmetrical characteristics. Okay, so I'll summarize. Just by saying if you have an abrupt P-enjunction which we don't experimentally but if we want to realize some of these electron optics devices we're going to try to make then there's this mismatch in the wave function because you're tunneling from the conduction band to the valence band and that reduces the transmission. For a graded junction we assume it's graded so slowly that there are no quantum mechanical reflections and we have this problem with this apparent band gap that we have to tunnel across and the distance you have to tunnel across can be sizable. It increases when you increase the angle or it increases when you increase the transition width. In both cases if you have a normally incident ray it goes right through and that's one of the special things that people always remark about about graphene. So the net result is that these P-enjunctions are considerably less than that of an NN junction and finally what does this all mean to people? Well if you're doing experiments this might be a nuisance and something that's clouding the interpretation of your experiment but you also might be using these to realize some type of electron focusing or electron guiding device and there's a lot of interest in that these days. Okay so I'll end there thanks for being a good audience if there's a question or two and we have one more added bonus that we're going to do here in a minute. Professor Dada is going to make a few more remarks about any GF which sort of came out of some discussion I think that that occurred after his last lecture. So let me see if there's a question or two and I think we're going to rearrange the cameras for that. Okay any questions before we make this switch? I didn't want to discourage questions so go ahead and please stop to see the rectify. So you're worried about what might happen here if the interaction between the metal and the graphene actually opens a band gap in the contact? Yeah yeah and I don't know enough to speculate about that and there are all kinds you know so this is a problem that I think we don't clearly understand I don't know if anyone does another thing that we haven't talked much about and that Professor Dada and his student Roxana did some recent work on you know you may be aware that when you put a metal on a semiconductor you get states called metal induced gap states that tunnel a little bit and people worry a lot about these in shot key barriers and things they tunnel a very small way into a semiconductor because the band gap is very big. In graphene when you put a metal down on graphene so these metal states can tunnel in and you can get these metal induced gap states which affect the electronic structure and which it seems under some conditions can be quite important. So there's a lot of things that need to be thought through carefully here about exactly what do these metal to graphene contacts do and I don't think it's fully resolved yet. Not even partially resolved. Yes. Having a thick oxide so it will give us a less charge control. So in that case I assume that to turn off and turn on the transistor will be really high. So ion and all pressure will be so small. So how do you solve that problem? So the question is these thick oxides are going to lead to poor transistor characteristics and that's right these are not transistors that we're making. So these are test structures to explore some physics and it's convenient by these scotch tape methods of producing graphene to have a 300 nanometer thick SiO2 back oxide so that you can see that you have one monolayer of graphene but this is not ultimately a high performance device. We'll have to build the devices differently. The top gate it's much easier to make a thin oxide. It's much easier to induce the doping in the source in the drain but in any kind of realistic transistor structure that would give you all kinds of parasitic capacitance that you don't like. So these are all the long ways from real devices.


Righty! Wonderland has dropped to their press release for the 2023 season, so you know what that means I'm gonna be talking about everything new for 2023 This might exclude a few minor details that we'll have to wait till the park opens to find and see and talk about We also have a planned video with Canada's Wonderland that should come out around mid-march That'll go into more detail But let's just dive right into it So for those of you that don't know there was a lot of tea dropped in Wonderland's press release today A lot of it centers around the two new rides coming to Canada's Wonderland 2023 And then a new food festival and halloween haunt receiving one of its biggest upgrades in a really long time I'm super ecstatic to talk about the halloween haunt things in this video. So let's get right into it So we obviously can't talk about What's new in 2023 at Canada's Wonderland without talking about snoopy's racing railway snoopy's racing railway it Snoopy's racing railway is a multi launch tire launch coaster for the kids in planet snoopy It'll feature a launch that takes you over through a course. It's very twisty I almost have started calling it the maverick the kids junior maverick coaster at Canada's Wonderland And then it'll go through a second tire launch that'll then bring you to the final part of the ride Rumor does have it that it might only complete one cycle at Canada's Wonderland and not do the two cycle Run like you see on thunder run. This is for capacity reasons the same reason why the train has one extra car It's a really fun and unique coaster and the minimum height requirement for this ride I believe is 36 inches I will fact check that and put it on the screen if I am incorrect But nonetheless, it's a really beautiful ride and honestly, it really suits Wonderland's lineup extremely well I Absolutely adore the decision to add this coaster in planet snoopy and you will require an adult Or sorry, you will require a child to ride this ride if you are an adult. It is located in the kids section I know that's a little bit unfortunate But it's a really smart decision to keep the line slightly shorter for the kids The other attraction for the adults or the thrill seekers at Canada's Wonderland is tundra twister Tundra twister is a fascinating ride first of its kind in Canada Well, actually it's first of its kind in the world similar to sledgehammer. Canada's Wonderland likes building these attractions that can't be found elsewhere Further more so Canada's Wonderland is going to focus on building attractions that don't already exist in Canada So look forward to future attractions Similar to this or coaster wise aren't in Canada Nonetheless the big stuff Canada's Wonderland is upgrading their halloween haunt event Halloween haunt is going to feature eight haunted houses this year instead of the four or sorry the five that we saw last year Three brand new ones one is from a pre exist Sorry is a pre existing halloween haunt attraction from knotsbury farm So i'm really excited about that because knotsbury farm has a really good halloween haunt event But nonetheless, uh, so there'll be three brand new ones and one Upgrade or change from what I understand I'm going to show you guys the one brand new one called dark ride that is from knotsbury farm So that's going to pop up right on the screen right about now Face to right part 46 face to right part 46 come in Yeah, hey, what's up? There's a disturbance reported at the abandoned dark ride. Please report to the location Yeah, yeah, I'm on it Another new thing coming in 2023 is the new food festival taste of brazil running from august 19th to the 20th um, it is replacing brew and barbecue and um, honestly, it really I love these Multicultural food festivals it really plays on world expo I'm going to make a whole video about why I think world expo is such an important thing coming to canada's wonderland Um, and how I really think wonderland could chase after the whole epcot vision with world expo and I'd love to see Um, them do something like that. So look forward to that video in the next, um Couple of days working on that but that is so far everything that's new There are obviously winter fest is returning other food festivals and other shows as well um There's going to be a old show returning where uh tundra was in the castle in medfare Um, and all the loved old ones are coming back as well another new feature will be at lazy bear lodge They'll be having a vip fireworks and dining experience limited availability obviously So this will be an upcharge attraction and I'm assuming it'll be sitting right on the patio Um, eating and enjoying the fireworks with the best view in the house. So You know, I'm definitely going to try and check that out. That sounds absolutely amazing. Um, and yeah I'm surprised it didn't take advantage of it of the first year But then again, I remember lazy bear kind of opened after all the fireworks. So that makes sense So super excited to check that out. I believe that's everything new for 2023 so far from the press release I probably missed a few things but very minor things and then obviously when the park opens We'll be covering everything new for 2023 in greater detail. Anyways, thanks so much for checking out this video Hope you're having an amazing week and uh guys, we're getting really close to wonderland openings So get those season passes ready. All right. Have a good one guys. Bye


What's up, everyone? It's your girl Kenzie here from Lane sports cards, and I'm ripping a personal for Mike D We got the brand-new release 2022 TriStar hidden treasures autographed basketball jersey. All right, Mike. Let's see what these are all about Thanks for your order that cake honey.    Let's see what we can get no way. This one is sick How about a Tyrese Maxie? Heck yeah, I signed that nice and big there on the top of the zero Tyrese Maxie You see your fanatic COA right there. Very nice. He had a nice year last year I'm looking forward to seeing him play this year. Is this This is a Nike Authentic jersey too. Wow, man.   Thank you. Yeah second team all SEC all freshman team McDonald's all American and Mr. Texas basketball Whoo That's sick All right, congrats on that one Mike. Thanks again for hopping in and we will get this right out to you


Ortonal Polinomiel, Gina Brin ensemble with finite point charge. OK, hello. Yeah, thank you. First, I want to apologize. My voice today is very bad, so I'm going to use this and talk a very silent, small voice. First, I want to thank the organizer for giving me the chance to speak here. First time visiting Triste. And I want to speak about Plainer Othagonal Polinomiel. And this is the ongoing project with Menyang. So let me first describe the model. So this is Coulomb gas. The difference between what Professor Guionne taught us is that it's on the planer. It's on the full complex plane. So they repel with each other with beta equal 1 or 2, depending on the convention, and with some confining potential q. This is the probability distribution for n point particles at the complex plane. And I'm going to use Zn for the cap normalization constant called partition function. And so partition function is one interest. Another thing of interest is the characteristic polynomial, which is just degree n polynomial whose zeros are at the Coulomb gas. So it's a random polynomial, because Coulomb gas is randomly distributed according to this distribution. And if you take this random polynomial average, then I get this deterministic polynomial. And these are objects of interest. There's two objects, partition function and this polynomial, Pn. In the large n limit, that's the interest of the talk. OK, so I will use the specific potential, not general potential. Potential is Gaussian. If I have only this quadratic term, then it will be a Gini-Brown symbol, where the particle is on the unit disk, uniformly distributed. But I will have L point fixed charge at the location a1, a2 to al, with charge strength c1 to cl. So for example, in this picture, I fix 3. I can even make it move. So these are the blue dots, our Coulomb gas. And I fix 3 point charge with different strength. So this one I choose charge 4, this one charge 2, and this one charge 0.2 or something. And all the other are charge 1. So you can see there's a little more distance from all the other charge because it has more repelling force. And actually charge can be slightly attractive also, like charge can be any number bigger than minus 1. And for this model, we want to calculate the partition function, which is now explicitly written in this form. So for all the fixed charge, you also have the interaction term. And first, we will look at the polynomial, the average characteristic polynomial. What is known is this average characteristic polynomial is also orthogonal polynomial, which satisfies this planar orthogonality with respect to the measure given by the same queue that defines the probability distribution. So the same queue, this one. And the orthogonal polynomial more explicitly can be written in this form. So that's, yes? How did I make this animation? I just add Brownian noise. So this is the cheating because I don't know how much more noise I have to add to make the beta equal 2. So this is for some unknown beta. So each particle is moved by the force, deterministic force plus Brownian noise. Deterministic force is from the Gaussian and the force from all the other particles. But this is a cheating, not beta equal 2. C is not an integer. C can be any number larger than minus 1. So here is what is known so far is the asymptotics of the orthogonal polynomial. The message here is pretty less is known. For very simple potential, they could obtain the location of the zeros in the large and limited. Location of the zeros are usually concentrated on some one-dimensional structure even though the particles are supported on two-dimensional support. So there are few examples where this shaded part is where the Coulomb gas will stay, but the zeros of this average characteristic polynomial will stay on a one-dimensional, called skeleton of this domain. And most of the result is used using Riemann-Hilbert technique developed by DK MVC. Only M is here in this crowd, like Laughlin. But the goal is whether we can use this tool which was so successful for other complex orthogonal polynomial, whether we can apply the same technique to planar orthogonal polynomial, and that's so far almost what we have so far. Okay, and also there's a general result given by Heidenbaum and his student. This is the general asymptotics of the planar orthogonal polynomial for broad class of Q. So here Q can be any smooth potential, but there's also some subtle assumption about its shape of the equilibrium measure. But with subtle assumption, they obtain the asymptotics of this orthogonal polynomial. But here the asymptotics is only in the domain which is outside the support of the Coulomb gas. So outside, in our case, it's outside the unit disk. And all the zeros are some subset of some one-dimensional structure of the unit disk. So this doesn't tell you where the zeros are. The location of the zeros is still an unknown problem. Okay, so let me describe the answer for my system. So first I will define what is the Segel curve. The Segel curve is the curve defined by this equation, a very simple equation, which is this closed curve between yellow and dark region. So this closed curve is the Segel curve and that's actually the limiting zero locus of this truncated and little scaled exponential function. So exponential function doesn't have zeros, but if you truncate the exponential function, Taylor series truncation, then it will have zeros and zero converges to the Segel curve in the scaling limit. Okay, and I will describe the result when L equal one. So if you have just single fixed point, if you just have single fixed point, then the zeros of the characteristic polynomial goes to this Segel curve. That's already proven. But it's not just the Segel curve. It can be some generalized version of Segel curve, which I will describe soon. So depending on whether this fixed charge is inside the disk or at the boundary of the disk or outside the disk, you will have different shape of the Segel curve. Okay, so to describe the Segel curve for general case, I will describe the Segel curve in a different language. So again, the same equation for the Segel curve and this Segel curve can be understood as the following. So first I draw the graph of log of the absolute value as the yellow cone. That's the shape of the graph of log of C. And I draw the graph of the other parts, real part of A bar Z, and that gives, of course, the plane. And where this two graph meets is where this equation is satisfied. So I draw this yellow cone plus this blue plane and where it meets and those solution sets is where the Segel curve.  And the right-hand side is, if I take only, so at every point, if I take the maximal value of the two graph, then I obtain this right-hand side. So that's the maximal between these two yellow part and the blue plane. And actually, that's exactly the log of the polynomial in the limit. That's the result from when there's only one fixed point. Okay, now I describe when there's a finite number of fixed points. So here I will describe multiple Segel curves. So multiple Segel curve is the same. We consider log of C, the previous cone-shaped yellow curve. And then we consider many planes and for each plane coming from all new addition of the fixed point. So fixed point A1 gives the plane, which is directed in the A1 direction, another plane A2. So there's L planes if you add L fixed points. And you draw all the planes and all the cones and you look at the graph of the maximum. Then you see this kind of graph, which is for three fixed points. So here the red dots are the location of the fixed points. I call the maximal function, which is represented by this graph, capital phi. And I just define each region where the jth plane is dominating, is omega j, for example, here. And I also note that each plane can move up or down by the level which I denote by L. Lj, if you increase Lj, then plane goes up, Lj down, place goes down. And you have to match this height such that the plane is exactly intersecting this log of the other plane, other graph, at the location of this aj, the point. So once you match the height, such that the point, the fixed location of the fixed point is at the boundary, exactly at the boundary, then that determines the height Lj. And it's not very trivial, but you can show that even aj such multiple sego curve is unique. So all the Lj's, the height, is chosen uniquely. Here I try to show how this lemma can be proven. First, you choose one plane until a1 is at the boundary, and then move the second plane until a2 is at the boundary. Actually, this point can be at the boundary between the two planes also. So whenever two planes meet, the boundary is the straight line. So you see many straight lines, and aj's can be in the straight line. So such curve is unique, and the theorem goes as the same as in the one fixed point case. So this maximal function that I showed in this graph here gives the asymptotics of the polynomial. And actually Riemann-Hilbert problem is always overkill. So this is the demonstration of the result. So here you cannot actually see this line, but there's a white line between all the intersection of the planar graphs. And the location of the zeros is exactly matching on the intersecting lines. So it shows that the result is actually very good. These red dots are the location of the fixed points. Actually, one can put the red dots also outside. Then, yes, question. Computing the zeros, I have to go through the Riemann-Hilbert problem. And I need a lot of mathematics. No, away from the zero locus. But as Riemann-Hilbert problem always does, it gives stronger result everywhere. Also, as common in Riemann-Hilbert problem, you need special asymptotics. So you have different form of asymptotics outside, inside, or on the locus of the zeros. And you need special asymptotics called local parametrics near each fixed point, each fixed charge. And if you zoom in those fixed charge, then it's given by also truncated exponential function, but it's the other way. It's subtracting the finite truncation. And the other part gives the local behavior of the zeros. Actually, this is only for integer C, but if it is not integer, then it's replaced by some integral. So I will explain how I obtain this in a very crude way. I don't want to go to the detail. So I will define the multiple orthogonal polynomial as exactly defined by Silva. So multiple orthogonal polynomial is, instead of just one measure, you have multiple measures to define orthogonal polynomials. So P, multiple orthogonal polynomials defined by index N1, N2, and L, not just one degree, so that their sum becomes the degree of the polynomial. And it requires L measure. So first measure, mu1, gives N1 orthogonality. The second measure gives N2 orthogonality. So in total, they give exactly N orthogonality that can define the N degree N polynomial uniquely. And this is the measure of this multiple orthogonality, which is given by some crazy integral. So I said L different measures, so difference is coming from this blue index. So if, oh no, ignore the blue index. Blue index ignore, just red index. So if k different, then one of the power in the integral changes. So sometimes you subtract one of the power, then that gives L different measures. So they are very, quite close to each other. And all in, so L different measure is in my problem, it's all living in the same contour. The contour is actually here. Contour is from going around each fixed point, each fixed charge once, while rotating the origin once. You can always arrange the contour in such way. So that's my definition of the orthogonal polynomial. And what we proved is that our planar orthogonal polynomial is actually same as the multiple orthogonal polynomial for such choice of index. So if I give a degree 100 and 10 fixed charge, then I divide the 100 into most uniform way into 10 decomposition, so 10 plus 10 plus 10 plus 10. So for such index, you define the multiple orthogonal polynomial with the previous measure, then it gives exactly the planar orthogonal polynomial. So having such multiple orthogonal polynomial, then we have immediately the Riemann-Hilbert problem, which is the matrix Riemann-Hilbert problem of size L plus 1 by L plus 1. So if you have additional charge, the size of the Riemann-Hilbert problem increases. And this is the corresponding jump condition and boundary condition. And the rest is the same steepest descent analysis developed by DKMBC. So I will just describe very quickly. So we want approximate solutions to the exact solution, which is unknown. The exact solution is unknown, but we want the approximate solution. To find the approximate solution, instead of looking at trying to find the solution, we look at the Riemann-Hilbert problem, exact Riemann-Hilbert problem, and compare it with the approximate Riemann-Hilbert problem that can be easily obtained from the approximate solution. And the theorem is the exact solution, the distance from the exact solution to the approximate solution is given by the distance of the problem, distance between the Riemann-Hilbert problem, exact Riemann-Hilbert problem to approximate Riemann-Hilbert problem. So if we can show that these two Riemann-Hilbert problems is close, then we can prove that these approximate solutions are also close. That's the main strategy of using Riemann-Hilbert problem. But there's other benefit. First, Riemann-Hilbert problem can give you a lax pair and it also can give a recurrence relation. And that's how I obtain actually the numerical calculation, the zeros of the polynomial. Because finding zeros by Gram-Schmidt is, I believe it's impossible, because there are many moments with exponentially small contribution, which actually the zeros are very sensitive to exponentially small quantity. And without this recurrence relation, it's impossible to plot the zeros. And the second thing is the partition function, which I will explain now. So partition function is related to the moment determinant of this multiple orthogonal polynomial system. Okay, so this is more thing about Riemann-Hilbert problem side. So I said the multiple orthogonal polynomial is supported on the contour that goes through all the A1 to A2, A3, A, all the fixed point. But some of the fixed point is actually inside the multiple sega curve. So first you fix the contour to go around the most outer boundary of the multiple sega curve. But whenever the fixed point is inside, you have to come back and go out following this branch cut, because all this charge strength Cj is not equal to integer. Then there's a nasty branch cut and that actually gave me so much headache. By the way, this talk has been similar talk has been given a few months ago and there's some people in the audience apologizing. So at the time, it's a few months ago, but non-integer C was still not completely proven, but I thought it's a small hurdle, but it took much more than that. So now non-integer C can also be handled. So that's how you draw the contour. So you have the contour that goes around the branch cut coming from the non-integer C. So some proof that I really worked and you have to write many notation with the general size matrix because it's for arbitrary size L plus 1 by L plus 1.  Okay, now partition function. We are interested in the partition function because we are actually interested in the moments of the characteristic polynomial, which has been appeared in other talks. So I remind you the characteristic polynomial is chi is the polynomial that has zeros at the Coulomb gas. And if I take the absolute value and take the power by C1, then it becomes the moment and I can add multiple moments of the characteristic polynomial. So we want to calculate this expectation value of the moments of the characteristic polynomial. So why do we look at this quantity? So physically that's the following. So if we have L fixed charge with certain charge, then of course each charge feels some force, electric force coming from the external source and also between each other. But the question is what if you throw in this background Coulomb gas? What will be the force? How the force of this fixed charge change? And the answer to that is after this background charge is added, they don't feel the force from each other. So the background electrons will completely screen the electric field between each other so they don't feel any electric field. That's exactly the statement of this conjecture by Weben Wang. And there's more exciting recent result. So Weben Wang actually when they make this conjecture, they prove this formula for single charge, electrical one. And further, so this is the case when the fixed charge is inside the bulk but when the fixed charge is near the boundary of this unit disk, then there's a special function, Pagnet V4 appears. I think basically given by this paper but for some slightly different model which is actually equivalent. And then less than a month ago, there's another few weeks ago, there's another paper that proved this conjecture for L equal 2. They used, okay, so their method is different so they can only do this strength to be integer. So that's good because I can still contribute. So they can do strength integer actually after they can also make one of them non-integer. So C1 integer then C2 non-integer, then they can prove this fact for 2 fixed point. So this is the plot of actually the correlation, two fixed points. So when we have two fixed point, the moment can be plotted for different number of background gas and it converges to this, the top line is the theoretical line and this color line is converging to the top line. And of course the top line diverges when the two charges meet but what they did is when they are very close they can see more fine structure of the Coulomb gas and it's regularized by Pangneve 5. So this regularization is described in their paper. So I will describe how we can also study these moments of the characteristic polynomial and a little bit of result because this is ongoing work. So first, the solvability of the Riemann-Hilbert problem because Riemann-Hilbert problem is related to the multiple orthogonal polynomial, it's related to the unique existence of the multiple orthogonal polynomial. Whenever there's an orthogonal polynomial, the existence is related to the non-vanishing determinant of its moment. So moment determinant, non-vanishing then some orthogonal polynomial exists. So in this case also there's a moment determinant corresponding to the multiple orthogonal polynomial. So since there's many, many measure, for each measure you write down the moment matrix, rectangular matrix of some size and if you combine it for each measure then you will get the moment matrix of N by N, the degree of the multiple orthogonal polynomial. And if that determinant of that matrix is non-vanishing, then it's equivalent to having unique multiple orthogonal polynomial and it's also equivalent to the solvability of the Riemann-Hilbert problem. Also in another context, solvability of Riemann-Hilbert problem is known to be related to so-called tau function and tau function is given any Riemann-Hilbert problem or given matrix ODE system like here. Then there's an associated tau function. Tau function is defined by one of the tau called Jinbo Miwa Ueno tau function is defined this way. Also there's other definition. So the property of the tau function is the non-vanishing tau function means the solvability of the corresponding Riemann-Hilbert problem. So in our case we have the Riemann-Hilbert problem for YN and we can make the lax pair system or linear ODE by multiplying some deterministic matrix B which I show for example here. And once you do that you can find all the quantities to define the tau function. Actually this is variation of the tau function. Tau function is defined up to constant with respect to the variation of the parameters in the problem. So you can find the tau variance variation if you know the solutions to the Riemann-Hilbert problem. So if you have the asymptotic solution then you have the asymptotic behavior of the tau function. And what is known is this tau function is equivalent to the moment determinant because both, so the right hand side tau function is telling you the solvability of Riemann-Hilbert problem. The left hand side is showing you the unique existence of the orthogonal polynomial so they must be related but they are actually the same. And that's shown for example by Bertola and here's the main idea how to show them. So whenever, okay, so instead of showing that so it's shown by induction by looking at the tau quotient. So tau n divided by tau one length is equivalent to this ratio between the moment determinant and you can easily show the tau at n equals zero is equivalent to the moment determinant at n equals zero and that shows the equivalence. And therefore now we have the tool to calculate the moment determinant. And moment determinant is actually equivalent, not equivalent but related to the partition function by some crazy relation which come out of nowhere. I don't have good explanation. And from that we can calculate the variation of the partition function which can be obtained with a quick calculation using the asymptotics we have to be this and once we integrate then we have the antiderivative up to constant the partition function is behaving as the conjecture says for arbitrary number of L and for arbitrary set of Cj's. Okay. Yeah, so my work is too little to fill the whole 45 minutes so it's finished but since I have too many too many times I will say one more possible future work. That worked on but just so if we have two fixed point and the fixed point was have the same strength similar order strengths as all the other Coulomb gas but if we make this strength much bigger than the other Coulomb particle then you can actually change the droplet, the support of the Coulomb gas can drastically change so it's no longer you need disk and in such case such domain becomes some special domain sometimes called quadrature domain and the location of the zero is then conjecture to be the mother body of such domains. So those can also be studied using the formulation in terms of the remandable problem in the future and that's all. Thank you.