 Okay, let's continue with the last lecture of today. Okay, so the last thing I want to talk about is trying to write down a dual description of a free Dirac fermion in two plus one dimensions. Okay, so this is another way in which we can extend the previous discussion. So we have something like a two-dimensional fermion that lives in two spatial dimensions. And this is just a two-component object just to make it completely clear. And we'd like to rewrite this in different ways. And, you know, we actually know enough now to do this. So I'll talk very briefly about a Bosonic description. And then I'll talk about a different fermionic description which will actually be very useful. This fermion fermion duality, if I have time, I'll try to show you one application of that to study quantum Hall physics. Okay, so that's the plan. Before I get into that, there were a couple of questions which I want to sort of maybe mention to everyone else. So one of the questions was the following. So we have, when we did the Boson-Vortex duality, this was a Boson duality, we said that there's a connection between the total flux of the emergent gauge field, little a. The flux we said was related to the Boson density. Okay, so if you calculate the magnetic flux divided by two pi, that is simply the Boson density, which is why Vortice is in the, if I were to insert some magnetic flux in the Vortex phase, that corresponds to the Bosons. Yes, so you can also go further, that's the thing that measures how many Bosons you have. You can try to write down the operator that actually inserts a Boson. If you want to create a Boson and destroy one, what is that on the Vortex side? And of course, what you need to do, if you want to insert a Boson, you need to change the magnetic flux, the flux of this little a. And the operator that does that, of course, is a monopole operator. Okay, so that's the dictionary. And today what we're gonna do is we're gonna, or at least in this class, what we're gonna do is we're gonna develop a strategy to introduce fermions into this model. And the strategy is very similar to what we did last time in one dimension. So if you recall in one plus one D, what we did was we took a charge and we bound it, so this was like sigma z, and we attached it to a domain wall, and that composite object was just our fermion. Okay, so the analog over here is we're gonna take a charge, we're gonna take this Psydaga, the Boson creation operator, we're gonna attach it to a Vortex, and we're gonna see that this combined object is just gonna be a fermion. So the thing that you can do over here is, since you have Vortices, you can continue this process. You can also attach Vortices to the fermions, make them Bosons, so you can do the reverse. So if I start with the fermion and attach a Vortex to this, this thing will be some Bosonic field, but you can do it two times. You can take the fermion and attach it to two Vortices, and this is gonna bring it back to some new fermion, which we're gonna call the composite fermion, and this will basically underlie the two things we're gonna do. This is gonna be the fermion duality, and this is the Bose fermion duality, basically. We have these new variables, and we insist on describing all the physics in terms of these new variables. Again, depending on the situation, it can be extremely convenient to use these new variables. Of course, if you're using them in the wrong context, life is gonna be really hard, so you need some judgment, knowing exactly when certain variables are the right ones. And we'll see at least one example of that, this half-in-lander level, where it'll be clear what the right variables are. Okay, so let's first begin with this. We'd like to develop a theory to describe just a free, direct fermion. Of course, the dual description will be extremely complicated. It'll be some strongly interacting theory, but there are objects in that theory, the monopole operators or something non-local like that, which are actually gonna be free. There's some sort of hidden free description, which is not apparent when you write up in the new variables. Okay, so it's an interesting way of making connections between things that are essentially free and things that look extremely complicated. Okay, so there are a couple of references. The one I like for this part is by Karch and Tong. It's for the Boos Fermi thing. There's also a nice paper by Seiberg and others. And then for the Fermi-Fermi stuff, the one I'll follow is myself. So this was 2016, I guess, or maybe I'm 15. Okay, so we'd like to come up with a theory for this model. And one way to think about a massless direct fermion, just like we did in one dimension, is to consider a phase transition. Okay, so think of fermions that are undergoing a phase transition. And the thing that's changing across this phase transition is the topology of the ground state. Okay, so in this particular case, let me again imagine changing the sign of the mass. So if you're at mass equal to zero, you have this gapless massless direct fermion. And on either side, you have a massive direct fermion. Okay, so I add to my direct Lagrangian. I have this external electromagnetic field just to keep track of the global U1 charge. These are direct fermions. They have some global charge. So now let's say I'm in one of these gap phases. All of the excitations are gapped. I can integrate them out and I can write down an effective action, just in terms of the external electromagnetic field. Which I obtain by integrating out these fermions. And so I don't know if this is familiar to all of you. That's what you're gonna get. You're gonna get a Chern-Siemann's term for the external electromagnetic field. And the sign that accompanies this is the sign of the mass term. It's either plus or minus. And the difference between these two, the plus and the minus mass term is going to be one over four pi. If I take the difference between them. That's what I would get. Now it turns out that you never actually have just a single direct fermion. They always come in pairs in two dimensions. If this is an intrinsically two-dimensional system, direct fermions always come in pairs. So there's another direct fermion sitting somewhere else. And we have presumably gapped that out with some mass term that looks like this. So that should also have its own response. So the usual convention is to say that there's a background which I've kind of implicitly neglected, but I'll just add it in here with a particular sign. It's a way to regularize this theory if you like. It's like a polyvillage regularization of this theory. And what this does for you is it tells you that this phase transition that you have is between a theory that has a trivial response and a theory that has a response which is given by minus one over four pi. I'll use this compact notation, a wedge, dA, to represent this product of these three terms. Again, this coefficient is completely key. That's the one place where you don't want to mess up your factors of two and pi. I can't guarantee anything about my signs. My signs will be all over the place. Okay, so this is the phase diagram we want to reproduce. Okay, so just before we get into that, let me discuss the physics of having such a term in the effective action. Okay, so you all know that, okay, so when I have a Chern-Samen's term in the external electromagnetic field, we later have Chern-Samen's term in the internal gauge fields as well. But let me think about this when I apply an external field. I get a response. I want to understand what it means to have an effective action that looks like this. Okay, so of course the physical things that you measure are electrical currents. So we can figure out what the electrical current is. So if I have JMU, that is, okay. So the derivative of this effective Lagrangian with respect to the gauge potentials are the currents. Okay, so of course in a Lagrangian, you'll have this sort of minimal coupling. So if you take the derivative with respect to the vector potential, you get the currents. Okay, so you can take the derivative over here for this effective action, or if you like, over here. And of course when the derivative hits, it hits these two locations. There are two contributions, and it's a good exercise to check that they both give you the same answer. Okay, there could be some factors of minus one that come in, because you're gonna shuffle the indices around, but there's also integration by parts that comes in, and those signs cancel. They give you the same contribution. So this gives you a factor of two. Okay, so you applied an external electromagnetic field, and you discover that you get a current. Okay, so we'd like to know exactly what the relation between the current and the field is. Let me pick some particular component. Let me say I take JX. Get the current in the X direction. Okay, so then this is, yeah, I picked it to be like that. So here of course there's no term in A. Actually there's, really speaking, a Maxwell term. I didn't, yeah, I didn't actually put that in. Strictly speaking, if you work out what is the effective theory over here, you get a Maxwell term for A external. That's actually just the effective theory of any insulator. You take an insulator, or a dielectric, and you look at the effective action for the external electromagnetic field. The only thing that changes is the dielectric constant. Okay, so the Lagrangian just looks like E squared minus B squared, with some coefficients that are changed. Okay, so that's what you get over here. It just looks like a regular dielectric. But over here you get an addition to that. You get some other response, which is more singular, which is this churned simons kind of response. And you want to understand physically what does that mean? What does it mean to have an effective action that looks like this? So if I apply an external electromagnetic field, I would discover I have a current, and the strength of that current is, well, there are two terms. So there's zero and y, E zero, A y, minus D y, A zero. So this is actually just the electric field in the y direction. So D by Dt of A y is just negative of the electric field. Okay, so this is just one over two pi. Yes, if you apply an external electric field along the y direction, you discover that you get a current along the x direction. Okay, so this is just a Hall conductance. Yes, so if I give you a current along some direction, and it depends on the electric field in the opposite direction, in the perpendicular direction, sigma xy is what we have over here. We have found that sigma xy is one over two pi. Okay, so that looks like slightly strange units for conductance. You measure resistance in ohms, right? So what is one over two pi in ohms? It's not very clear. So actually that turns out because we've used these theoretical physicist units where we set the charge of the particle to one, and we set h bar equal to one. Okay, so the right way to convert this into physical units is the sigma xy is actually e squared over h bar times this number. Okay, that'll convert it into inverse ohms, and you put this together, you get e squared over h, which is the unit of Hall conductance. Yes, so this is I think one over 25,000 ohms, something like that. Yes, so this is a universal constant. It just depends electric charge. If this were electrons and the plant constant. So people measure this in the quantized Hall effect. They go and measure this at low temperatures. They find the Hall conductance that's just related to the fundamental constants. In fact, people define the fundamental constants now by this measurement. Okay, so what this is telling you is that this stuff over here is in a quantized Hall. This is an integer quantum Hall state. And this is just a trivial insulate. So if you take a material that's just a regular insulator, two-dimensional material made of let's say electrons, and then you convert it into an integer quantum Hall state. In the process, when you undergo this phase transition, at the transition point, your electrons will have the dispersion of a single Dirac cone. Yes, so this is the way we're gonna model a Dirac cone. We're going to find a different theory that will reproduce both of these phases. And then we'll conjecture that if I set the parameters in that theory right at the critical point, it's very likely it corresponds again to the single Dirac cone. And I'll have a dual description of the Dirac theorem. At least that's one strategy we can adopt. Yes, so again, I think you should note the analogy to the one-dimensional case. There we had Majorana fermions, gapless ones that occurred when I made a topological transition. The topology was different there. It had these Majorana zero modes. But nonetheless, there's a change in the topology of the fermions. At the critical point, you get some gapless fermion. And then you try to come up with a different theory that will reproduce those phases. Okay, so what is that theory? Well, previously we wrote down this vortex theory and we can ask what do we do with that third of the third theory of bosons? So if you remember, this vortex theory was a vortex field. Okay, and then we said that there was a Maxwell term as well. We write it as F squared. And of course this cannot be the right theory because of course the dual of this theory was just a bosonic theory. Okay, it didn't have fermions doing this interesting thing. But we can ask what is the minimal modification that we can do to make this into the theory that we want? Okay, sorry, there was one more term which was this BF coupling, which told me that told me that the flux of little a was carried the charge of the bosons. Okay, so the idea that we use is just like in one dimension, we're going to bind the charge to the vorticity. You're going to bind the boson charge to the vorticity. So I want to write down some term over here that's going to make sure that if I have some vortex, so if I have a vortex current, I want it to, at every point, I want it to be equal to the boson current. You want to fuse these two together so that everywhere where I have a whirl line of a vortex, I also attach to it a whirl line of the boson and that composite object will be a fermion. Okay, by those, by that logic. And then I can test whether that theory which I get from this kind of logic, I can test whether that theory is going to pre-produce these two phases for me. And if it does, then I can finally say that at the critical point, that's a dual description of the drag fermion. Okay, so that's the logic. Okay, so what's the term that will do this for me? Well, I know that in my effective action, I have a term which is the j vortex times little a. Okay, I already have that. It's sitting here in this theory. And I want to take this boson, the boson current we set, you can read it off from this. It's one over two pi. Okay, this is the boson current. Okay, so we want to do that attachment. And one way we can do the attachment is to introduce a Lagrange multiplier. When you integrate out the Lagrange multiplier, you'll get the attachment of these two. And the most convenient way to do that is to use this little a as your Lagrange multiplier. Okay, so you can try to write down something like a times Jv minus J boson. And when you do the integration over a, you just get these currents would like to be together. Okay, so this is just to motivate the theory. So you can put this in. So that would give you this term with an additional factor of a. There's a factor of four pi because I have two helpings of a. This thing is already there in the theory. So this is the additional term that I'll add. So just a churned Simon's term for little a. So it turns out that in order to get everything right, get the background contribution correct, I need to add a background term to this as well. So this is not very important, but just do that over here. It just involves the external electromagnetic field. It shifts everything by one unit. So these are the two terms I'm going to add. And I can combine them all with this term over here down the combined term. So that's the modification that I make to my previous theory. I add these two other terms. So before we actually compute the phase diagram of this model and check that it matches that one, maybe it's useful to take a step back and try to understand a little better why this churned Simon's term actually changes the statistics of your particle. So you can forget about the big a. We're just looking at some intrinsic property of this. So let's say I have this churned Simon's term, one over four pi, a wedge d a. Okay, so what you can do is you can imagine that you have a sphere. And let's say you just have one unit of flux that's piercing the sphere. It's one unit of gauge flux. Okay, so this is like a very small system. You have the magnetic flux of a, which is dx, ay, dy, ax. So we're going to integrate this over the entire sphere. This is just two pi. Let's say I have that configuration. It's like having just one monopole in the center of this sphere. So one monopole. So really this is like a charge, original Boson charge. I just integrated one Boson density. So if you have one monopole, this term is going to induce gauge charge on the surface. So remember that the gauge charge, rho is just the 0th component of, is just j zero. So this is minus dL over d little a. And if you have a term like that, this is going to be minus one over two pi. So you have exactly two pi worth of flux. So this will give you a gauge charge of minus one. So this act of putting in one monopole at the center induces gauge charge on the surface because of this chance I have instead. Now it's not a good idea to have gauge charge, unbalanced gauge charge in your system. You're always looking to have gauge neutrality. If you have a very large system, you cannot have an unscreen gauge charge. So the low energy configuration is for one particle with positive gauge charge to come and sit on the sphere. Yeah, so you basically ask for one of these bosons, these cyvortex particles to come and sit on the sphere. Cyvortex dagger, for example, neutralizes. Yeah, this is really going to be the low energy configuration of the system. So you have a particle with gauge charge that's sitting on the surface of the sphere and that's seeing the magnetic field of a monopole. No, it's, so it's all over the surface. It's spread out on the surface. Yeah, so there's no center, actually. I mean, just drawing it like that. There's just a surface and there's magnetic flux that's piercing that surface. And the chance I'm in some is such that every time there's a magnetic flux going through, it sucks in gauge charge. And the net amount of gauge charge that it sucks in is exactly the coefficient over here, which gives you minus one. Yeah, so one way to think about it is if you put in a magnetic, as you change a magnetic field, if you're trying to put in flux, you get electric fields that circulate around it just by, I guess it's Faraday's law. So you have a circulating EMF and because of the Hall current, if your electric field is tangential, you get current that's perpendicular so that's radially directed. So in the process of setting up a magnetic field, you suck in, you get current that flows inwards and that accumulates charge. And you can integrate that over time and see what charge you get and that's exactly this amount. We'll also be localized, that's right. That's right, yeah. But here the magnetic field is considered to be uniformly spread on the surface of the sphere. But then you have to neutralize everything so you add this positive charge to it, positive gauge charge to it. So now you have the problem of a particle on a sphere in the field of a monopole. And I'm sure you've probably solved problems like this. So a charge plus monopole, well, for example, this combination is a fermion and you can see it, for example, by the special functions you use to describe the orbits of this particle are these monopole harmonics. They have half integer charge, they have half integer spins and those correspond to fermions as well. So this bound state of a gauge charge and a monopole will have half integer spin, which again tells you that it's a fermion. You can also just look at the particle moving around the equator. You can figure out that it has this minus one sign. There's got to be at least a pair of these states. You can't have a unique ground state. There'll be at least a pair of them which corresponds to spin a half. So anyway, that's sort of a long-winded way of saying why. This term is actually giving you, you shouldn't be surprised that you get fermions from this construction. That's maybe the point. Okay, so the last thing that we need to do is to verify that we recover the phase diagram that we had over here by taking two different limits for this theory. Okay, so for example, let's take, let me take the analyze first when this r prime is greater than zero where everything is gapped. Okay, so the vortex field is gapped. So I can simply integrate out the vortex field. And then I see that I'm left with these terms over here. I have a Maxwell term and I have this Chansamans term. Chansamans term has one fewer derivative than the Maxwell term. It's more relevant. So let me just focus on this Chansamans term. I see in this theory, I can actually absorb this external electromagnetic field into little a. I can just redefine this little a with this thing absorbed. I can do the integration of a little a and there is basically no effect of this big a on the theory at long distances. Okay, so this r prime greater than zero theory is you can integrate. And essentially you don't get any Chansamans term for big a. This actually corresponds to the trivial. It's just an insulator. You probably get some Maxwell term for big a, but this is what you expect just for a regular insulator. Everything else has disappeared. And this is all you're left with. Okay, so that actually corresponds to this trivial phase. So we have solved half the problem. And now we'd like to see whether we get this whole response in the non-trivial phase. Of course, that's the more interesting thing. Can a Bosonic theory with the appropriate Chansamans term can it actually give me this whole response? Okay, so let's look at this second case, which is this r prime less than zero. So now I've got to condense my vortex field again within quotes. This is a non-local operator. And if I do that, I end up getting, and I've got to analyze this particular term. So I write my psi v as some amplitude and a phase. So the low energy physics is really just the fluctuation of this phase. So I get some coefficient times delta mu phi v minus the gauge field. That's the part of Lv that involves this vector potential. And then of course I have this Chansamans term, a wedge a minus big A wedge D. And there's also a Maxwell term. It's not going to play any role. Okay, so this is a rather different situation because over here it's actually very expensive to have a vector potential, to have any non-trivial vector potential. Okay, so if this vector potential was just the gradient of a phase, you could cancel it off by an appropriate choice of this phi v. But if I have a transverse part of this vector potential, so let me write it like that. This is essentially the part that is cannot be canceled off by that. Something that's not a pure gauge. Then this is expensive. Okay, so essentially what this system is trying to do is to expel any vector potential that's transverse. It's like a superconductor. This is like the Meissner effect. It's expelling any vector potential. So what you would like to do if you can, a quick way of analyzing this theory is to just set this transverse part of the vector potential to zero. That's what's gonna minimize this part of the action. Okay, but then you look over here. The only thing that's gonna contribute here is the transverse part of the vector potential. Of course, over here you take the curl and then you take the dot product with this. In both cases, you only get the transverse part. So essentially I'll set little a to zero over here. Okay, so the effective response is just a wedge dA. So, but that's exactly the response we wanted over here. Even the sign came out right. It's this response that we wanted. Okay, so by tuning the sign of r from r greater than zero and to r less than zero r prime, I could get both sides of the phase diagram. Okay, so then the proposed duality is that this thing without the r term, we write it as the critical theory, is equivalent to a single Dirac fermion. So this is in the low energy limit, of course. All of this is assuming some renormalization group to the infrared massless Dirac. Okay, so one of the things we kind of live off in these kind of conjectures is there are not very many fixed points. There are not many RG fixed points. So if you have two different theories that describe the same transition, they're very likely to be the same. It's hard enough to have one fixed point, one continuous transition. And so if you have two different ways of describing the transition, there should be parameter values where they're the same. When you say three dimensions, three space time. Two plus one, yeah. Yeah, you can add that if you want. Yeah, I mean, I'm thinking of this as the fixed point theory. I'm not being strictly speaking at the fixed point. If you just take the Wilson Fisher fixed point, all of those terms are actually there, right? Take it in three dimensions. So I'm thinking of a model starting with this model and flowing to the fixed point. Yeah, it's some sort of shorthand notation. It's understood that the actual fixed point is much more complicated than that. Yeah, so that was this Masonic description. So I don't actually know of an application of this yet. So in the remaining few minutes, actually how many minutes do I have? 25 minutes, wow, okay. I think you may be making some calculation mistake, but that's fine. I'm not gonna say anything. What I wanna talk about is this fermion duality to another fermionic description. And I'll describe that very briefly, because that's somewhat involved, but the idea is very simple. Another way to get fermions is to attach, to start with the theory of fermions, and attach a pair of vortices. Okay, not just one, but two vortices. So let me just tell you what the final result is. And this duality is extremely useful. It's somehow, you know, thing about fermions is that they are less strongly interacting than bosons in some sense, even at the fixed points. So you can actually make some progress with that. So there's a fermion fermion duality. Yes, you have electrons, let's say that's your fermion field. And then I will, let me just use psi. And I'm gonna go to these new variables, which are gonna be a product of two vortices. Times the fermion field. And I'm gonna call this the composite fermion. Okay, so this composite fermion is a non-local object, because it's attached itself to these non-local fermion fields. They are vortex fields. Yes, so you can kind of guess what this duality might look like. And there's some subtleties over here that I won't really have time to go into. So that's the left-hand side. This Dirac theory with this normalization where one side is trivial, let's say. And in terms of these new variables, of course, because they're non-local, they involve these vortex fields. I'm gonna have the gauge field appearing. But because there are two of these vortex fields, I'm going to have twice a mu. Okay, it couples with twice the strength. But the rest of it is the same. I have this in a coupling of one over two pi. Just add a minus sign. A wedge, the external. And that's gonna be it. Okay, so it's a free Dirac fermion is dual to, again, a Dirac fermion coupled to a gauge field with strength two, gauge charge. And the flux of that gauge field is the original U1 global charge. Okay, so there's an important caveat. I think many of you seeing this are worried about the parity anomaly. So you're taught never to write down a fermion theory like this without a minus one over eight pi times this gauge field, chance I'm in stone. Okay, so this by itself is anomalous. There are ways to fix that. I think Cyberg will talk more about that. They have a way to fix it by introducing another gauge field. We have a different way in this paper to fix it where we put this on the surface of a three-dimensional system. Okay, three plus one-dimensional system. Okay, so solve by making this a surface theory and then you forbid certain monopoles from the bulk of that system. There's a natural monopole over here which is gonna give you problems. Okay, that is really the content of this parity anomaly. If you put in the natural monopole, it gives you some half-gauge charge. So we're gonna solve this by putting this on the surface and forbidding certain monopoles. There's a formal way of doing this. You know, sync odd monopoles, one one, strength one, single monopoles. Yeah, so it turns out all these fixes, they don't really affect the basic analysis of this theory. So if you're careful and you know what you're doing, you can kind of get away with this, this kind of sick-looking theory. Yeah, so this was the duality that is extremely useful because it has these fermions on either side and you can sort of use this for various things. Yeah, so let me give you one example of what you can use it for, which is actually related to the quantum Hall effect. So it turns out that these fermions are actually closely related to what people use to describe the fractional quantum Hall effect. These are called composite fermions. They're bound states of electrons and double vortices. And pretty much the entire physics of fractional quantum Hall effect, you can understand very simply in terms of these variables. Yeah, so all of the people, everything that people see in experiments, you can basically understand from using these variables and making some simple, you know, putting them in some simple states. But because the variables are complex and, you know, non-local, those simple states in reality turn out to be rather complicated fractional quantum Hall states. Yeah, so I want to have time to describe all of that. But in some sense, the fact that this is a very stable kind of composite is what underlies a lot of the fractional quantum Hall effect. Okay, but what is new that was discovered in this duality is that it's not just a bunch of composite fermions. They also have this kind of Dirac structure. Yeah, so one thing you can do is you can ask the following question. Let me apply external magnetic fields. Okay, so you set up this A external in such a way that there's a uniform magnetic field, which is just the curl of this A external. Yeah, so that's just a situation you'd like to be in to understand, look at quantum Hall physics. So quantum Hall physics, we have this two-dimensional layer of electrons, and then you apply a magnetic field. A strong magnetic field, you go to low temperatures, and then you kind of study that system. Yeah, so we can introduce a magnetic field by simply having the right configuration of A external. Yeah, so we pick a gauge for this A external so that the, you know, pick a form so that this curl of A external is just filled. Okay, so what does that do to my left-hand side? I just have a Dirac fermion in a magnetic field, okay? So this is a very famous problem. If I have a Dirac fermion, you can think of all the negative energy states as being filled, all the positive energy states as being empty, you put on a field. Yeah, so this is like your classic lander problem, but now for Dirac dispersion. And what you find is that, you know, you get these Landau levels, flat bands. There's one exactly at E equal to zero. And then there are, you have all of these filled. These are all empty, but you have a particle-hole symmetry over here. Half the bands are filled, half of them are empty. So you said everything in negative energy is filled, everything at positive energy is empty, but there's one exactly at zero energy. Okay, so the only way this can be particle-hole symmetric is if this is half-filled. Okay, so there's a very natural problem you can study, which is if I take this lowest Landau level, I fill it half of it with electrons. You can model that simply as taking a Dirac fermion and applying a field. And the question is what happens? So this is a highly degenerate situation. You have twice as many orbitals as electrons. Okay, so you can choose any half of these orbitals. They give you distinct states. You can make linear combinations of them. It really depends on the interactions between the electrons. We have not really specified. There should be some short-range interactions as well that have not written out explicitly. You get some state. Okay, so experimentally it's known that the state you get is actually a metallic state. Although you have a very high field, you get a metallic state. Okay, and how do you understand that? Well, you just look at the right side of this duality. This is a very easy way to understand where you get a metallic state. You now have an external magnetic field. So DA external is the B field. Okay, so what that means in the dual theory, applying a magnetic field. So the indices work out such that the magnetic field multiplies the zeroth component of A zero. So basically you have, so there's basically gate charge. You have that much background gate charge when you apply a magnetic field. Okay, so you're gonna screen this gate charge. You're gonna get the number of composite fermions. Each of them has gate charge two. So you get one half times B external over two pi. This many composite fermions that are gonna be generated to get gauge neutrality. Okay, so on the right-hand side of this duality, this half-filled lander level simply corresponds to filling up the draxy of these composite fermions up to a level where you get the end composite fermions being some fraction of the magnetic field. Okay, there's a very simple rewriting. Magnetic field on this side is chemical potential on this side. And you just fill up more of these composite fermions. Okay, so it's quite clear how you get a metal over here. You just have a finite density of these composite fermions. Yeah, so this was sort of known before. You didn't need the duality to see this. People actually explained this half-filled lander level as composite fermion, Fermi C. But what they didn't realize is that these composite fermions, they actually have a direct character. That's actually the most natural thing for these composite fermions to be. So for example, if I just look at the wave function of these composite fermions, yes, if you just look at a drake equation, you know, the Hamiltonian of a drake theory is, if I just write down the single particle Hamiltonian, it looks something like sigma z px plus sigma x py, something like that. So if I go up to this Fermi C where I filled up my composite fermions, I look at the state of the fermions. For example, you're at px equal to plus some number and px equal to minus that number. You have different values of the sigma z. Actually, let me switch this around. You have different values of this sigma x state. So in one case, the spin is that way. The other case, the spin is the opposite way. You have this kind of structure on going around the Fermi C of these direct composite fermions. So then you cannot really scatter. It's hard for this particle to scatter over there. So this is kind of related to the Klein paradox. If your spin is pointing in this direction, you encounter some impurity. The impurity doesn't flip the spin. So you don't have this process. You should have a suppression of this kind of scattering. So this was completely new. There's no reason to expect this for regular fermions. Okay, regular fermions, you'll get all kind of scatterings. And you can go and look if this happens in reality. So actually, I had a couple of slides on that. So it's not, we're not yet at the stage where we can do this in an experiment, but we can do this in a numerical experiment. Okay, where you literally simulate the system. Works, okay, amazing. Let me just back up a little bit. So this was these large-scale numerical studies. You can actually do a very good job of taking just the slowest land level, making it half full and putting in interactions and looking at the ground state of this model. It's a highly non-trivial quantum-manybody problem. It requires these large-scale numerics, density matrix renormalization group. But at the end of the day, you end up seeing something that looks like a Fermi surface. This is supposed to show you that there's some sign of this Fermi surface of these fermions forming a Fermi surface. And that itself was quite spectacular. You take electrons in a very strong magnetic field. You have no reason to expect the electrons to form a Fermi surface. Okay, they don't have a dispersion. The field is much larger than any Fermi energy they may have. Okay, but there's some other objects that don't see the magnetic field at all that kind of move in straight lines and form this Fermi surface. Okay, those are these composite fermions. And then you can ask, do you see this back scattering? Okay, this is a thing that needs to be, should be suppressed if it's really Dirac-like. And it'll be there if it's according to the old theory, which did not notice this Dirac character. Okay, so you have this Fermi surface. You have all of these points. And they look at some correlation function as a function of the transferred wave vector. And you get these kind of kinks that correspond to certain events, certain scattering events across the Fermi surface. Okay, so for example, the first kink over here, C to D on the left, that's scattering fermions from the point C to the point D. Okay, it's that scattering. And there are various scattering like that. You can also see there are two particle scatterings, A and D combined, and jump across the Fermi surface to B prime, D prime. Okay, there are others, even three fermions being scattered. But the thing that's really striking in this graph is that the most obvious scattering is not there. What's the difference between green and red? So I think there are different operators that were used to measure the scattering P2 and P5. So they have the detailed structures different, but in both cases, you have this missing peak, which is just going from, for example, C to B prime. That's just scattering across the Fermi surface, so D to A prime. Okay, those are the guys that are missing. Yeah, and similarly in this one, it's a, yes, it's a somewhat larger system. You've got all kinds of scattering, including four particle scattering. You can detect that in this. These numerics are really, really clean, but again, the most obvious one, E to A prime is missing. Okay, so this gives you, it would be mystifying to see this if you had, if you didn't understand that there was like a drag structure. Yeah. Yeah. A to B prime is allowed. So if you look at what the spin looks like, it looks, there's A to B prime. Okay, so this is scattering from here to here. So the spins have some projection on one another. It's only here that it's zero. Yeah, it's only the diametrically opposite points that are zero. Okay, anyway, so let me stop over here. That was the application of this fermion duality.