 OK, so I continue today by, first I give you the reference, I forgot to give you yesterday. And so what did we do yesterday anyway? Yeah, so the main thing just is a brief recapitulation. So yesterday we deduced this, the composition of unshel momenta and four dimension. I'm going to deduce this phase ambiguity. And similar for the tilde under which a elicity state transform in this way. And we used this to constrain. We saw that this gave a very strong constraint on scattering amplitudes. And in particular, for four particles, we had the very simple formula. So useful references in this context. So there is a textbook by Wang and Elvang, which is on the archive. There's an older long paper by Arkan Ahmed, Kachazo and Kaplan, I think, which discussed a lot of what I've mentioned, and I will mention today. And for integrability in general, the connection with amplitudes, there's a very useful review by Beiser and many others, 10, 12.398. So today the plan is to discuss how it goes from trees to loops using unitarity. And tomorrow we'll also discuss unitarity again, but in the context of understanding renormalization. So what we understand about renormalization of quantum field theory from this unshel perspective. So one thing I will do tomorrow is derive, for example, the QCD better function. But I will also show you how, more generally, you can recognize this 2 to 22, there's the dilatation operator of the theory. And you may not have recognized this object, but this object is the Eisenberg XXX, Pinchaine and Miltonian. In disguise, so I will explain how to read that off tomorrow, and I will discuss integrability more generally, that's the plan. So before we get going to loops, to be useful to tool up about supersymmetry, because the loops will not be much simpler in supersymmetric theories. So we want to use, many of you are familiar with superspace, which you can use to simplify supersymmetric theories. For one supersymmetric, it's relatively simple, but still you have to introduce auxiliary fields and things like that. For one supersymmetric, it becomes more and more complicated and not feasible. The good news is we only need unshel Susie, which is much simpler. So when you have supersymmetric theories, you have supersymmetric generators, which are spinners. So you have additional symmetry in the theory labeled by spinners. Let's consider n equals 1 Susie, for example. So you have one of these spinners. So we need to understand how it acts on a non-shell state. By the Lorentz environment and the fact that things depend only on lambdas and lambdatildes, this must be proportional to lambda alpha times under the state with the same momentum. And if that states at illicity h, because of this little group rescaling, you know that this states as illicity plus a half. So that we know immediately. So if we look, for example, at a gauge multiplet in n equals 1 Susie, it has spin 1. It has a state with spin a half. It doesn't have any scalar. It has a complex conjugate of the gauge, you know. And it has a minus characteristic group. So these are the states in your theory. And q takes you up. So if q acts on this state, you would get on this fermion. So that would be the gluron. That would be a fermion. It would be a fermion. It would be a gluron. q acts on this fermion state. You get a positivistic gluron. But if q acts on a positivistic gluron, you just get 0, because there's nothing else to get. And similarly, there's a complex conjugate field, which gives lambda tilde alpha dot, which moves you down. These are consistent with the anticommutation relation of the q, because if you act on pH, one takes you up. The other takes you down. You get just lambda lambda tilde, which is p alpha alpha dot times pH. And you don't get a factor of 2, because in this product, there's always only one term which contributes. So you go up. So that's unshelved Susie. It's very simple. And in n quarts 4, so I could give you the Lagrangian of the n quarts 4 super symmetric Yang-Mills theory. But all I will need really is just the matter content in it. So I will not bother give you the Lagrangian. And in n quarts 4 symmetry, we have the same story, but now we have four of these q's. So we have q alpha a, and a runs from 1 to 4. And this a is called the SU4 r symmetry of the theory. The states now, there's a plasticity gluon. So if you make a table of the spins and the number of states, the plasticity, you have the gluon. Now we have four fermions, because you can act with four different q's. We have six scalars, four of the complex conjugate fermions, and one. So in n quarts 4, you only have, is it better? So already one thing you can see is simpler in n quarts 4, is that everything fits instead of one multiplet, which is very nice, because we can stop using all these discrete labels, like plus, minus, for the amplitude. Everything is one single super object. So how do we use this in practice? Well, we make a generating function, though, how to identify the states together into a multiplet. So the unshelved super space is defined as follow. A state with p and era is defined as, we start with a plus to the stick gluon. Then we add the states with fermion. So we introduce grassman variable era, four of them, and we make series expansion like that. And these A, B, are anti-symmetrical. They parameterize. They label the six scalars of the theory. Three-factor order fermion. And then, I guess, the next term, all the eras, the gluon at the bottom of the multiplet. So we make this. And that is the unshelved super space. We have momenta, and we have these eras. Now Susie is very simple. It's just a differential equation. If we take q on this, we move up. And that's equivalent to multiplying the whole multiplet by era. So the representation of the Susie now is simply q alpha A is equal to lambda A, when it acts on pn, peta. And q tilde moves you the other way, moves you the other way. And to move the other way, you need to take a derivative. The index is down. So that's q and q tilde in this representation. So that allows you to move up and down in a multiplet. That help us write the amplitude, because now, when we scatter four of these super states, we get a super amplitude, which is a function of the momenta and these. So it's only a function, if you wish, this super momenta, this super space. And now, we no longer have any of these discrete labels. But this amplitude is some polynomial in the era, because all the states are polynomials. And the fact that this theory is supersymmetric is just a statement that qA0 is equal to q tilde. That's what we get from Susie. A very efficient way to solve this constraint, remember that another similar constraint that we have is that the, and qA is the total momentum. So it's the sum over all particles of these generator acting on that particle. We already have seen one such equation, is that sum over i of Pi of A is equal to 0. And the way we solve this constraint is saying that the amplitude is proportional to a momentum conserving delta function. So the way we solve this constraint is similar. We just say that this 8 of these q's, there was 4 of these p's, now this 8 of these q's, 4 indices here, and spinor index here, delta 8, sum over i times something. And we call this curly 8 for the moment. So the amplitude has to be proportional to this factor. You could try to do this. You could try to add another delta function for q tilde. But because q tilde isn't really commute with q because it's a derivative operator, writing a delta function of derivative doesn't really work in practice. So we can't do that. So that's the best we can do. We can make the q supersymmetry manifest by pulling out this factor. And then the q tilde supersymmetry is just some differential equation that A happens to set psi. So that's Ancher-Sussi. Let's work an example. So for the four-particle amplitude, grassman delta function, remember, are just polynomial. So this delta function here, definition, is the product spinor index goes from 1 to 2, product r symmetry index goes from 1 to 4 of some verticals of qi alpha A. It's a product of things. That's what Susie delta functions are. Grassman delta functions are. So each of these q is dimension 1 r. So this whole thing is dimension 4. So this Susie amplitude would be related to this thing, but having this dimension 4 factor strip. And I'll just tell you the answer. The Susie amplitude is equal to the Grassman delta function divided by the denominator, which was there. So let's check this formula in some examples. So the idea of this super-emptitude is that it encodes simultaneously the amplitude for all the varial solicities in the theory. This is delta function of q. Yeah, the sum over all particles of their lambda times eta. And this lambda has a spinor index. This A has a r symmetry index. So there's eight different things you can write down. That's why this delta is 8. So let me suppress them. So let's study this formula in some example. Let's first reproduce this case, minus, minus, plus, plus. To get minus, we have to extract the component where eta 1 will occur to the fourth power. So the amplitude for minus, minus, plus, plus is equal to the super Susie amplitude, where you extract four powers of eta 1, four powers of eta 2, and no powers of the rest. Remember, this thing's a polynomial of degree 8 in the eta's? So we extract this particular coefficient of it, all the eta 1's and all the eta 2's. How do we do that? Let's just look at the, and yeah, this here really means eta 1 r symmetry 1, the 1 r symmetry 2, 3. I hope the labels are a bit confused when you start writing them out. But that's really what this means. Because each of the eta squared to 0, you cannot have one of them twice. So let's extract this. Let's just look at one r symmetry component. And let's just look at the eta 1 and eta 2 parts. So we have lambda 1 eta 1 plus lambda 2 eta 2, say r symmetry 1. And lambda is a spinor. So we get two cases. We have a product of this 2 and this r symmetry index. So that's alpha equal 2, alpha equal 1. Sorry, this is a spinor index. So this product contains this. And we want to extract the component proportional to eta 1, 1, eta 2, 1. So we can get this component from two places. We can get this guy times this guy. That give us lambda 1 alpha equal 1, lambda 2 alpha equal 2. Or we can get the other product. And because they anticommute, we get a minus sign. And this is the determinant of this product. This is what we call 1, 2 yesterday. So remember that lambda is the two spinor and this product is just the determinant of this 2 by 2 matrix lambda 1, lambda 2. So at the end of the day, extracting eta 1, eta 2 just mean to take a factor of 1, 2. So in practice, it's very simple. So when we take delta 8 e lambda eta, and we extract eta 1 to the 4, eta 2 to the 4, we've done it from one r symmetry index. If we do it for the 4 guys, we just get it to the 4th power. And that's how you see that this formula reduced to that one as a special case. But this formula contains much more information. For example, now we can scatter scalars and so on. If we want to scatter a scalar, for example, let's consider the amplitude for negative elastic gluon, a plus elastic gluon, scalar 1, 2, and it's complex conjugate scalar 3, 4. That will be just equal to the super amplitude with eta 1 all the 4th power, then eta 3, 1, eta 3, 2, eta 4, 3, eta 4. This is how we extract that particular scalar component. And let's look at the r symmetry 1, we have 1 and 3. So we get the factor of 1, 3. r symmetry 2, we get 1 and 3 again. r symmetry 3, 1 and 4. So it's completely trivial to convert this expression to spinner brackets. And downstairs, we have, again, 1, 2, 2, 3, 4, 4, 1. This guy cancels up to a sine. So that's the amplitude for scalars. And that, what is this amplitude? So we could have computed it in Feynman diagrams. This includes this diagram, so scalar gluon, gluon, plus this diagram, gluon exchange. And also the diagram at the four point vertex. So without actually doing them, we've computed the sum of these three Feynman diagrams. And we can also similarly extract amplitude with fermions and so on. And the amplitude will be the same as in QCD. Even though we've done n equals 4, the same diagrams will occur in QCD. So that's punctual Susie. Let's discuss loops. The basic method to relate loops to trees is unitarity. It's called like that because it's related to the statement that the S matrix is a unitary matrix. So S is dagger equal to 1, means you can derive that imaginary part of t is equal to t dagger t. And if you draw this dagger, if you expand this equation at weak coupling, it gives you something like the imaginary part of some one loop amplitude is equal to a product of three amplitudes. And on this cut, all the propagators are on shell. The Kutkowski formula allows us to be more explicit. This imaginary part is, if we imagine that only this channel is time-like, Kutkowski give us that this imaginary part is equal to, for example, this integral when it's cut. And the cut means to replace each propagator by, well, I have a p square plus epsilon minus it goes to 2 pi delta of p square with all propagators. Yes, sorry, that's too small. Well, what I want to say is that propagators are replaced by 2 pi delta of p square times, normally, you retain on the positive energy. So the way we use that, so this is, unitarity is normally, in this equation, it's a statement about the integrated expression. If you take the imaginary part of this loop, it's equal to replacing this propagator by delta function and putting the three amplitudes on the left and on the right. The unitarity method extends this to a statement you can check at the level of the integrand. So that was done by, it was pioneered by Bern, Dixon, Dunbar, and Kossover in the mid-90s. And the idea is simple is that if an integrand, integrand, then it's the correct integrand. So what does that mean? All the correct cuts mean that on all cuts you replace 1 over p square by delta function. And you do that for all the possible channels you can imagine. So you take your integrand, so you put these two propagators on shell, and you compare what you get with a product of two, three-level amplitudes. And you do the same thing for the other channel and the cross channels. One, two. You do it for all channels, and it matches in all cases. Then you got the full integrand. And that's enough. That's a Bern, Dixon, Dunbar, and Kossover. Let me quickly illustrate this in a simple example, and then I will mention what are the sum of caveats and solidities in extending this approach to wire loops and in general. Let me just quickly give an example to make it less abstract. And the example, I will do the example in n-course four. As you will see, because the integrand will be much simpler there, it will allow us to do it on the board. Let's consider the amplitude for one loop amplitude for minus, again, this one minus, two minus. Three plus, four plus. And now let's look at this cut here. So we need to calculate the product of three amplitudes on the left and on the right. On the left, we have, so let's call the momentum the cut L and L tilde. And here, we have this four-particle amplitude, which is two minus, three plus. Then on the right, we have some other three level four-point amplitude, four plus, one minus. And we have to sum over the illicities in the cut. Let's first do the case where we have a plus air and a minus air. So it's a minus plus on the other side. So this three amplitude, we know. It has this Park-Taylor denominator, which is two, three, three L tilde, L tilde, L, L2. Then the similar thing on the other side. This gets squared. Then we have L tilde, four, four, one, one, L. But the important thing would be the numerator, air. The numerator takes the form. So the two minuses are this. So we get L2 to the four of this guy on the left. And we get L tilde, L tilde one on the right. So that comes from gluon loop. So I'm having gluon in the intermediate state. And there's another case for gluons where it's minus plus. And that case gives, when I flip this, I get simply one into exchange, but N and L tilde exchange. So now it's L1, four. That's the gluon contribution to this loop. Now let's include the fermion contribution. So in N equals four, there's four fermions. So we have a minus four. And remember, each time as we saw here, to take a fermion amplitude, we essentially move one of these eras from side one to side two. And that will simply change one of these products. So let's call this product eight to the four. Let's call this B to the four. We get simply AQB plus A. That's what we get from fermions. And that's because this fermion amplitude here, if this is middle in more detail, if I put this, this amplitude is simply the minus half times the two gives two L tilde three. And this gives L to L to L to the one. So somehow changing, moving your list is around. Just change whether you have L tilde or L. So that's what we get from fermions. And for scalars, now we get plus six A2. And you see that something nice happens when you have precisely this combination. And now we get A minus B to the fourth. And A minus B, that we write it in full, it's L2 L tilde one minus one two. And remember there's indeed density when you are antisymmetrize things in the two-by-two in a two-dimensional vector space. So that gives simply L, L tilde, one, two, four. So in N course four, you get an extremely simple expression for the cut. Whereas in QCD, you have this miss. And so now let's continue with the statement. So if the integrand allows the correct cuts, then it's correct integrand. So this expression is valued when N and L tilde are on shell. We want to uplift this to an expression which is valued everywhere, even when they are off shell. So let's rewrite this a little bit. And it's useful to this whole expression to pull out the three amplitudes. So this will be one, two, the fourth, one, two, two, three, four, four, one, time something. And then we've got two L tilde's on top. I'm scared of this. Two, three, and four, one are there. One, two, the four is there. But we have extra one, two, and three, four. And here we have L1, L2, L tilde three, L tilde four. One thing I will, that's a technical statement that you can explain how there's some identities of us to simplify that further. The physical expectation is that what can be in the denominator should be written as normal propagators that you expect on physical grounds. And indeed, you can show that this expression, the denominator, is equal to 1 over L minus P1 squared L over P2 squared. So the way to rewrite this, the point is that this L tilde is not an element from L on this unitarity code because of momentum conservation. So you can rewrite this expression in terms of normal propagators. But you always expect to be able to do that on physical grounds. And the numerator turns out to be just S. So what have we learned? We've learned that on this cut, the integral is the same as what you get from cutting integral d4L of S t L squared L minus P1 squared L minus P1 minus P2 squared L plus P4 and P2 squared, which is just scalar. So these are the five propagators you will have in box for L is here, 1. How did I label these things? 1, 2, these are the four propagators in the scalar box. So the integral on this cut is exactly what you get from cutting this box. Now we have to check the other cuts. So we have to check, in principle, that determines the integral up to things which don't have both of these propagators at the same time. So it determines the integral up to just product of, essentially, these other propagators. But the other cut, trust me, because the symmetry of this problem gives exactly the same expression. And that's the full answer. So if you were to compute the other cut, you would find exactly that it's consistent with that. And that proves that's the one loop integral in this theory. In other words, a1 loop is equal to the three amplitude times this box integral. So that's the idea of the uniterity method. You compute the integral of all the cuts, and you match that integral with some, and you try to uplift these cuts to a full integral, which is valid away from the cuts. And that's completely general. We can apply this idea in QCD also. There's nothing to do with supersymmetry. But some important things happen in supersymmetric theories. That's the principle. And one thing is that the integral needs regularization. So here I've done all the calculation in d equal 4. But in general, for example, if you use dimensional regularization, you want the full expression. And dimensional regularization is what people normally use in this field. And you're forced to introduce a regulator because, as we'll see eventually, but this integral has infrared divergences. If it doesn't have UV divergences, then QCD will also have UV divergences. You need to regulate. And conventional choice is dimensional regularization. And a useful way to think about it in this context is that your momenta, your d-dimensional momenta, consists of a four-dimensional part where mu goes from 1 to 4 plus an extradimensional part. It's called a mu-perp, which has index mu equal 0 minus 2 epsilon. So there's a normal part and this extradimensional part, such that the full d-dimensional p-square, let me remove this d, is equal to p-square plus mu-perp square. And that allows you to write d4 minus 2 epsilon as d4p d minus 2 epsilon mu. And now we have an integral, which depends on this four-dimensional part and this extra part, which is essentially a mass. You can do that. Yeah, you can always do that. And that is a very useful way to think about it. And yeah, it's discussed in the Wang and Elvin reference. It's a useful notation. But the point is that what we've computed here, spin or elicities. So there's language. So we're using the master's amplitude and this lambda and lambda tilde notation. We only get the amplitude when this mu-perp is equal to 0. This whole formalism is intrinsically four-dimensional. And we don't get these extra dependence on this. And that can be dangerous because sometimes you can have naively the dependence on, there's only epsilon of these. Naively, it should be order epsilon. But sometimes you can have integrals, for example, of this type, l-square plus mu-perp cube. And because of this, my symmetrical integration becomes epsilon l-square. So we think that's order epsilon because it's only epsilon of these guys. But this integral diverge so you get epsilon over epsilon, which gives something order 1. So because of such epsilon over epsilon effects, you have to be really careful. And here you could worry that this is not the complete answer. You could add extra terms portion to mu-square. What this gentleman proved in the mid-90s, Bern Dixon, Dunbar and Kossauer, they proved that in Susie theories, the UV behavior is better. Remember that's one of the point of Susie's, to make the UV behavior better to solve the hierarchy problem. So the UV behavior is better in Susie theory in such a way that Susie theory, these mu-square terms never produce epsilon over epsilon. So you don't need them to compute. They're not needed to get A to order epsilon 0, which is what we really care about, the four-dimensional limit of the amplitude. Just one. One's enough. Yeah. Yeah. So that's an important statement. And this is why this calculation was correct and sufficient. One thing I would like is that these epsilon over epsilon effects are called rational terms. So it's kind of a interesting, they're really painful to calculate in general. And this is where the calculation of one loop amplitude by now is sufficiently understood that people are automating computer programs for LHG calculation. But these computer programs typically spend all of their time getting these epsilon over epsilon effects. Because all the other effects you can get very efficiently using this four-dimensional language. So it's a very important, in practice, it's all very important terms. Very difficult questions associated with them. One way to think about them, for example, they're not there in supersymmetric theory. And you can also imagine a softly broken, if you start from, so one way to get per Yang-Mills is to take n equals 1 super Yang-Mills. We add a fermion. But then softly break the Susie so that you take the mass of the fermion to infinity. And because you have supersymmetry in the UV, it's good enough. You can still work in four-dimension. But now you need to have this extra massive particle. So this is why I introduced this notation here, because somehow this mu is very similar to the mass of this Gay-Geno. And what you need in practice, you want to do a per Yang-Mills calculation. To get the rational terms, you need to include a Gay-Geno loop where the Gay-Geno goes to infinity. So people have discussed these loops. You can find this in the literature. There are techniques to get that. But the basic reason these things don't, the couple, is that you have integrals of the type. Integral d4l m squared. And it matters only at l goes to infinity. So let me write all the propagators the same. If you have this integral, for example, if you take the limit m to infinity, you might think you have m8, m 4 over m8. My thing is just 0. But that integral, I put a pi square here, is just 1 over 6. It's not 0. And so you could say that the limit m goes to infinity. But if you take the limit under the integration sign, you will miss this term, because it looks like you get 0. So you have to carefully keep the mass and compute these things. Of course, the more matemons correct statement is that this limit is a distribution, which is delta function at the bit solid. It's l to infinity here. But this is not developed. So one would like to have a framework in which one can make such statements. And that should be the computing. Instead of having to calculate this whole massive amplitude and take the limit, effectively all we want is the coefficient of this delta function. But this is not developed yet. So I wish I would be able to talk about it, but it doesn't exist. But hopefully next few years will be developed. And this leads to all sort of questions at 1 and 2 loops. How do we deal with this? How do we use these d equal 4 amplitudes to compute? You would like to be able to mass less d equal 4. We're doing calculation in, for example, in periangulars, which has these mass less gauge fields. They are three-level amplitudes that are especially simple when we have mass less particles in d equal 4. And we would like to be able to use this simplicity to completely build loops. And right now it's failing because of these contact terms. But hopefully these contact terms, or if you wish, these epsilon or epsilon effects. So I think it's an important open question. And as you go to wire loops, keeping track of all these effects become increasingly painful. So when did I start? How much time do I have? 10 minutes? OK. So what I will do in these 10 minutes is I will discuss the extreme simplicity of this result, that the 1, 2 is proportional to just the scalar box. And the simplicity turns out to not be an accident. It's a consequence of a, it has a symmetry explanation. It's a symmetry which is special to planar in course 4, superhand middle. It's called dual conformal invariance in planar in course 4. What is this invariance? Let's see what's the special symmetry of this object. If you just stare at this, you probably don't see it. The symmetry is revealed by introducing this notation to make it look more symmetrical. So 1, 2, 3, 4. Instead of having a loop momenta, let's introduce region variables, y1, y2, y3, y4. And write this integral as integral for y0 over y0 minus y1 squared, y0 minus y2 squared. So I haven't done anything. I'm just saying that each of these probators is L or y0 minus something. And I haven't told you what these somethings are, so I haven't done anything yet. But you get the right answer, provided that all you need is that yi minus yi minus 1 is equal to pi. So if every time you go from one probator to the next, you add a pi, then that's the same expression. But now it looks much more symmetrical. And actually, the numerator is important. sT, that's s, for example, is t is p2 dot p3. That's y3 minus y1 squared. So the integral is this. Now it looks more symmetrical. And now we can start to ask what the symmetries are. These are four vectors. The claim is that if you take yi mu goes to then, if you take an inversion, this thing goes to itself. It has an inversion symmetry in momentum space. Let's just check the claim. We take yi minus yj squared. It's a simple exercise to show that this goes to yi minus yj squared over yi squared. I just need to check that. There's three terms. You can check it. This. So let's check that this is invariant. So from this factor, we apply the inversion to this factor. We get an extra y1 squared, y3 squared. But from the denominator, we also get one factor of y1 squared here, one factor of y3 squared here. This cancels. From the inversion, we also get y0 squared to the fourth. But this exactly cancels the Jacobian of y0. So when you include the Jacobian, you do the inversion of y1, y2, y3, y4, y0. This goes exactly to itself. And it's a standard result about conformal symmetry that when you have translation symmetry and inversion symmetry, you have a full conformal symmetry, which is that case is SO2 conformal. So I, this whole symmetry. So one can, yeah. So the, yes. So that's the first question you could ask. Is this a one loop accident? Just look at one loop four point. It's not a lot of data. No, it's not an accident. It's there are total loop orders. And for all number of external legs, it's there. Yes, and even the three amplitude have them, which I will actually discuss tomorrow. So what I want to say, yeah. So they don't have time to properly discuss the physical interpretation of the symmetry today. I can mention a bit of history. So in this very specific context, this was on this, this symmetry was obtained by Aldi and Medicina as a T duality, a strong coupling, of the ADS5 cross S5 string theory, which is due out to this n equals 4 theory, that's strong coupling. It was Aldi and Medicina in 2007. It was noted. So this one loop undergoing had been known for a while. Two loop and three loop had been known also for a while. It was noted, it's been known since the 90s, but it was noted for three loop, well, one, two, and three were available at the time. By for the three loop integral, people who noticed that are Aldi and Medicina. There was a Drummond and Kachemsky, Sarkachev. They noticed this explicitly before, but there was no real concept of understanding. It was a symmetry of the theory. But yeah, the T duality explained that. So yeah, it's noted that was noted for three loop integral. And for example, the three loop integral was expressed only in terms of the generalization of this box integral to two loops that you just get these two ladders. And these can also check that exactly the ladders which are invariant on their inversion. And at three loops, you get the three loop integral is a sum of permutations of these things. It's very simple. But the simplest stays explained because there's very few integrals you can write down which are invariant on their inversion. And basically, all of them which you can write are this and this and their permutation. And they all occur in the three loop integral. So it was noted there. This symmetry for the one loop integral and for the ladders was noted, say often in the 70s, noted that the box had this property. Many people in the 90s noticed it too for the ladders. The first, I did some research, the first people who used this symmetry are Kotkowski and Wick in 54. They studied the ladders for electron proton scattering where you exchange a massless scalar. They noticed it had this SO4,2 symmetry. They noticed it's a bit more subtle because there are masses. It's a bit more subtle, but it's there anyway. There's this masses. They noticed that there was two points there, Y1, Y3. If you have two points and you have SO4 symmetry, think of this as SO6, sloppily speaking. If I give you two points in a six-dimensional vector space, you break SO6 down to SO4. This SO4 symmetry, when you have an electron and a proton, they like to make bound states. This SO4 symmetry is the SO4 of the hydrogen atom. So you know that the states of the hydrogen then equal to 1, we have 1s. Then there's 2p, 2s, 3p, 3s, 3d. And the fact that they all degenerate is explained by a SO4 symmetry. And it's exactly this symmetry. That was noted by Wick and Kotkowski. What else can I say? Yeah, we can track it, of course, if we insist we can track it back in history, because this SO4 in the context of hydrogen atom is generated by angular momentum. And what it's called is Laplace-Rungen-Lenz vector, which I will not write down explicitly, but you can Google it and Wikipedia will tell you what it is. Rungen-Lenz vector. And the fact that the Laplace-Rungen-Lenz vector is conserved is the reason why the two-body problem in classical gravity, if you have two planets go around each other, they make elliptical orbits and the orbits don't precess. So if you don't have one of our potentials, the orbits will look like that over time. They will precess. The fact they don't precess in classical mechanics is explained by the conservation of this vector. And these two guys generate this SO4. That's a statement about classical mechanics. So this mysterious symmetry, this dual-conformer symmetry, is exactly the same symmetry which makes planets go around in ellipses. So yeah, it's still mysterious, but OK. So it's a generalization of it through this. Yeah, so in the real world, this is broken by, it's broken at order alpha 4 by spin orbit effects and so on. So the spectrum as the energy level order m alpha square over n square. And then there's a correction over the alpha 4. And this alpha 4 correction breaks the symmetry in the real world. In the n equal 4 model, the symmetry is present to all loop orders. So if you make a hydrogen atom in the n equals 4 model and you can have massive particles using the X mechanism. So yeah, I actually wrote a paper about that last summer. So that's why I say these things. If you actually make a hydrogen atom, hydrogen like atom in n equals 4, it will have the symmetry to all loop orders. This model also has it. Yeah, so it differs from, because it's not the real deal. It's not an actual photon. It's a scalar. And because of this effect, somehow it's a scalar. It's not a Dirac fermion. So this spin orbit interaction is gone. So this effect which break it in the real world are not there in this model. And they are not there in the n equals 4 hydrogen like atom. So it has this symmetry at all before. Yeah, so last summer we used this to compute the spectrum in terms of the cusp and the risk dimension. But I will not have, unfortunately, time to discuss that. But yeah, that's all for today.