 So, let's start with the second lecture by Marcus. So, at the end of the first lecture, I had just started discussing how we will begin to repackage the kinematic information necessary to specify a scattering amplitude of massless particles, or gluon specifically. So, what we did was we took our four vector, p mu, and rewrote it as a 2 by 2 matrix, and then we observed that the condition p squared equals 0 was the same as the condition that the determinant of this matrix should be 0, which you can solve parameterizing it in terms of what we call spinor-helicity variables. So, the next step is I want to discuss how the information about the polarization of our particles can also be encoded in terms of these variables. So, gluons, like any massless spin 1 particle, come in two polarization states, or two varieties, plus or minus polarization, or left or right-handed, circularly polarized. So, what we can do is there's a relatively simple way to choose a canonical polarization vector for either helicity. So, let me just show you how that works for a negative helicity, or negative helicity gluon. We'll choose the polarization vector in the following way, epsilon a a dot. Now, again, I'll always be referring to four vectors, either in terms of four vector notation, or in terms of this a a dot notation, they contain exactly the same amount of information. So, there's a lot going on here, so let me remind you. Here, mu tilde is arbitrary, and we'll discuss it in just a moment, and let me remind you, this was, I mentioned this at the end of the first lecture, that the square bracket, is defined, contract the indices with the epsilon symbol. All right, so let's discuss whether this is a valid choice. So, first of all, so let's check that this is a reasonable, reasonable sensible choice. First of all, the polarization vector should be orthogonal to the momentum due to gauge invariance. That's relatively easy to see. This is clear, since epsilon dot p is, so recall, p equals lambda lambda tilde, and epsilon now is lambda mu tilde over lambda tilde mu tilde. Okay, and I'm using a condensed notation now, where I suppress the indices, because I know in my notation anything that has a tilde index carries a dotted, anything that has a tilde on it carries a dotted index, and anything without the tilde on it carries an undotted index. So there's no ambiguity here, I'm just saving a little bit of space. This factor in the denominator here, it won't affect my inner product. When you take the dot product of this, you get lambda lambda mu tilde lambda tilde. So the dot product factorizes in terms of an inner product of this part with this part, times the angle bracket of that with that, times the square bracket of that with that. Just, I remind you, this was also discussed at the end of last time. This again just follows from manipulating Pauli matrices and my definitions of these. So the Minkowski inner product between two vectors is given by the angle bracket of their, let's call this the holomorphic spinner, because it carries an undotted index, times the angle bracket of the anti-holomorphic spinner. Okay, but anyway, to finish my calculation here, this is obviously zero, because this is zero. These inner products are defined with the epsilon symbol, so obviously the inner product of lambda with itself is going to be zero. Okay, so I've explained my claim that it's clear that this choice of polarization vector is orthogonal to the momentum. Now let's discuss the ambiguity due to the choice of mu tilde. Because I want to claim that this choice is sort of canonical or essentially unique, and we'll see that it is if you appreciate the following. So first of all, mu tilde has two components, because the A dot index, just like the A index, runs over one and two. So we're trying to check that this thing is invariant under an arbitrary choice of a two component vector. So let me parameterize the possible space that this mu tilde can live in by checking what happens under the following deformation. So let's check what happens mu tilde A dot goes to mu tilde A dot plus alpha mu tilde A dot plus beta lambda tilde A dot. Okay, so the idea is the following, that generically, the reason why this all makes sense is generically mu tilde and lambda tilde will span the full two-dimensional space. So the point is, I said here you can choose any mu tilde. Obviously, you can't choose a mu tilde that's parallel to lambda tilde. So let me, I should have made that comment before, where mu tilde is arbitrary as long as lambda tilde mu tilde is not zero. Okay, that was implicit in my formula here, because that thing appears in the denominator. So I said mu tilde is arbitrary, but of course, you want to make sure that it's not parallel to lambda tilde. Because then that factor in the denominator would be zero, and my formula makes no sense. But as long as mu tilde and lambda tilde are not parallel to each other, then they span the full two-dimensional space. Here's a two-dimensional space, so mu tilde points that way and lambda tilde points that way. They're not parallel to each other. So then you can parametrize your full ignorance about what this object might be by considering this two-parameter deformation. So when we do that, the beta parameter obviously drops out of the denominator. Let me phrase this better. The beta deformation, epsilon to epsilon plus beta p. You see, if we plug that in here, nothing in the denominator, if we make the beta deformation, nothing in the denominator is going to change because the additional term that we would pick up in the denominator is beta times the angle bracket of lambda tilde with lambda tilde. That's zero. Now, the extra contribution we get in the numerator from the beta deformation is beta times lambda times lambda tilde. But that's exactly p. And this is just a gauge transformation. We reviewed last time that gauge invariance requires that the choice of your polarization vector be, you know, that you should get the same physics under a transformation where you shift your polarization vector by a multiple of the momentum. I haven't discussed that. That is a crucial point. I certainly will complexify everything. What motivates you to ask that question at this point? Yeah, it's fine for everything. I think there's no problem with everything here being complex, including alpha, beta. Yeah, I'll discuss the real versus complex in just a moment. The alpha deformation amounts to an over, well, is trivial. This one is really trivial. Because what happens is it amounts in the numerator to rescaling by a factor of one plus alpha. And in the denominator, it also amounts to rescaling by a factor of one plus alpha so they cancel out. The alpha deformation is trivial. It has no effect on epsilon. So the conclusion is that this choice of polarization vector for a negative felicity gluon is almost canonical, but it has an ambiguity that's precisely the ambiguity that we expect and require a polarization vector to have, namely the ambiguity of being able to add any multiple of the momentum. Now, I haven't cluttered the blackboard with subscripts i, but remember that this entire discussion holds individually for each of the n particles in the scattering amplitude, because each one is specified by a null momentum and each one has its own polarization vector. Yeah, that's right. If you want to, for real gauge transformation, you would want beta to be real, but very soon I'm going to complexify momenta. Let me just comment here that the analogous formula for a positive felicity gluon, polarization positive felicity gluon, what would be epsilon aA dot, it's exactly the conjugate of that. And again, where mu is arbitrary as long as that inner product is nonzero. Okay, now let's discuss dimensional analysis for just a moment, because we can get a lot of mileage out of dimensional analysis. Okay, so the idea is in any n particle amplitude, we know there is an overall factor, epsilon 1 mu 1 epsilon 2 mu 2 dot dot dot up to epsilon n mu n times other stuff. My comment here just means that you know just by thinking about Feynman diagrams that any scattering amplitude can only depend on these polarization vectors by an overall factor. Okay, you'll never have more complicated dependence on the polarization vector. And this is a statement that obviously holds true at any loop order, just thinking about the structure Feynman diagrams. You'll never have a term that has a 1 over some inner product of polarization tensors. You'll never have crazy complicated functional dependence on these epsilons. They only ever can possibly enter by an overall factor. So that's very highly constraining. But we also know, oh and more important, I'm sorry, let me finish this here, other stuff that only depends on the Pi not epsilon i and hence only on products lambda, lambda tilde. Let me explain why that comment is important because this is something I didn't mention last time but this is an appropriate point. The relation P equals lambda lambda tilde is invariant under a rescaling where you scale lambda by a factor of t and lambda tilde by a factor of inverse t. Obviously if you do any rescaling like that, you won't change P. So the point I'm making here is all of this stuff here will be invariant under such a rescaling because it can only depend on the momenta of the scattering particles. But this factor can depend on such a rescaling. If we go back to the formulas here, I've deleted, let me put it back on the board, I've deleted it. Let's look at this positive helicity glue on. What would happen under such a rescaling? This thing is not invariant. The lambda tilde gets scaled by a factor of 1 over t. The lambda gets scaled by a factor of t but it's in the denominator. So overall this gets scaled by a factor of 1 over t squared. So each epsilon for positive helicity scales as 1 over t squared. And similarly each epsilon for negative helicity scales t squared. So this also places huge constraints on amplitudes. In other words, if someone hands you a formula and claims, hey, I've computed the 17 gluon scattering amplitude, here's my formula. Do you think I'm right? The first thing you should do is check that it is consistent with this relation. Specifically, it follows that A scales as t to the power minus 2 times the number of positive helicity gluons plus 2 times the number of negative helicity. Well, you have the choice. I actually changed from a stronger statement to a weaker statement here. You can do this scaling individually for each of the particles. Let me put that comment here. And over there I wrote it as a true but weaker statement about what the scaling would be if it simultaneously transformed all. It's all right here under a simultaneous transformation. But anyway, perhaps a better way to express this is the following. We can write a differential equation. This is using Euler's theorem on homogeneous functions. What we've concluded is that under such a rescaling, we've concluded that the scattering amplitude has to be a homogeneous function of the lambdas and lambdatildas with scaling that depends on the helicities. Let me finish writing it and then you'll see what I mean. H i equals plus or minus 1 is the helicity of gluon i. So this is an application of Euler's theorem for homogeneous functions. This differential equation expresses the fact that if you simultaneously do a rescaling that rescales all of the lambdas by a factor of t and that minus sign because you're scaling all the factors of lambdatilda by a factor of 1 over t, then the entire amplitude has to scale by a factor of t to the minus 2 times h i. And this is true for each i independently. So we now have enough information that we can actually determine some of the simplest scattering amplitudes completely with nothing more than this scaling property and Lorentz invariance at our disposal. So let's discuss three-particle scattering. And here's where we will see an important distinction between real and complex kinematics. So three-particle scattering is a very special kinematic configuration. When you have three massless particles, on the one hand you have energy momentum conservation which says that these three-four vectors have to sum to zero. And that implies, for example, that p1 plus p2 squared has to equal p3 squared. But that's equal to zero for an on-shell massless particle. On the other hand, this is equal to 2p1 dot p2. The other terms vanish. These are also zero for an on-shell massless particle. So the conclusion is, and this is again something very special for three-dimensional, sorry, three-particle kinematics, p i dot pj equals zero for all i and j. This is very special three-particle kinematics. Now, what does this imply in terms of our spinor-helicity variables? Now recall again that 2pi dot pj is ij ij. So we need ij ij equals zero for all i and j. There are two branches of solutions. Either lambda 1a, lambda 2a, lambda 3a are all proportional to each other. It implies that ij equals zero for all i and j. Or the lambda tildes are all proportional. Or all three lambda tildes are proportional. That implies that ij equals zero for all i and j. Here I'll mention the issue of real versus complex. We have p equals lambda lambda tilda. So for real momenta, lambda and lambda tilda should be complex conjugates of each other. This is for real momenta in Lorentzian signature. But if you choose lambda and lambda tilda to be complex conjugates of each other, obviously the only solution to this equation is for them all to be zero. You don't have these interesting branches where some of them can be nonzero and others are zero. So for complex p, lambda and lambda tilda can be independent complex variables. So implicitly whenever we talk about three particle scattering amplitudes, we have in mind that we're working with complex spinor-helicity variables, spinor-helicity variables that are independent complex numbers and correspondingly complex values of momenta. So let's continue and let's give these two branches two different names. The names will become clear in a moment. The names won't make any sense for right now. Let's call this branch MHV bar and let's call this branch MHV. Now we're going to use the scaling property to try to write down a possible MHV three-point amplitude. So V MHV three-point amplitude. So here the only things we have to work with, so by Lorentz invariance, we only have three things to work with. One two, one three and two three. All the angle brackets are zero. So these are the only three things we have to work with. So we have to construct an amplitude out of them. Moreover, let's start with the tree-level amplitude. In fact, we'll get the right answer that's true at arbitrary loop order. But at tree level, well actually it's not necessary to make this assumption, but it will make my presentation easier. We need to get a rational function of these three. So it must be one two to some power a, one three to some power b, and two three to some power c for some powers a, b, c to be determined. Now the only possibility, the only possibility consistent with scaling, I haven't yet specified what the helicities are, but that will emerge from this argument, the scaling property that was just on the blackboard here a few moments ago, and honest dimensional analysis. And I regret now that I called that earlier analysis dimensional analysis because I was really only using the scaling property of the lambda and lambda tilde variables. But here I also need to input the fact that scattering amplitudes should be dimensionless. That gives us a constraint on a plus b plus c. The only possible solution is, oh well, okay, there are three possible solutions depending on permutations. So I'll write the answer. I'll change this to be completely honest. The only possibilities are its permutations. Let's identify, let's identify what that solution is. Evidently, so we can read off our scaling formula, this object corresponds to a scattering amplitude where particle one and particle two have negative helicity and particle three has positive helicity. And there might be a concern about an overall sign. In the standard way of writing this, you would write three one, which has an opposite sign of what I've written. And the other two possibilities correspond to the permutations of this. So if you had particle two being the one with positive helicity, this would be three one cubed over one, two, two, three. The last possibility where particle one is positive and two and three are negative would be two, three cubed over one, two. Any other possible choice? So the point is, let's check that this has the correct scaling. Let's check that this has the correct scaling under particle one, say. If we scale particle one, if we scale the spin or helicity variables of particle one by a factor of t, then the numerator here will grow by a factor of t cubed and the denominator will grow by a factor of t. So the whole thing will scale like t squared, as it should for a negative helicity particle. On the other hand, for example, if we scale the spin or helicity variables according to particle three by a factor of t, then we get a factor of t squared in the denominator, which is correct for a particle of positive helicity. So the claim is that these are the only possibilities that are consistent with that scaling behavior. So the scaling has to be either zero or two for every single particle. You can't have some particle scale with a factor of t to the one and another particle scales as a factor of t cubed. No, that makes no sense. Every single particle individually has to scale either with a factor of t or with a factor of t squared or t to the minus two. So in particular, oh, well, I haven't gotten there yet. Hold on just a second. I was about to make another comment, but it's premature. The M-H-V bar branch gives you another set of amplitudes. Remember that on the M-H-V bar branch, the only non-zero spinner interproducts that you have to work with are the ones with square brackets. So here you have these amplitudes. And I'll just write bicyclic permutation 1, 2, 3. OK, now we're ready for the comment that I was going to make. This exhausts the possible three-point amplitudes. In particular, 1 minus 2 minus 3 minus, and A, 1 plus 2 plus 3 plus, are zero. Because there's no, from the point of view of this analysis, there's zero because there does not exist any possible Lorentz invariant function built out of the allowed interproducts that has the right scaling properties to possibly be consistent with what you would get from any Feynman diagram. Yeah, I know I've been stressing about that. That was clearly wrong. It should have dimension of 1 over energy. Because the overall, yeah, I was a little bit stressed. Each factor here, each bracket or bracket has dimensions energy, if you will. Because remember this formula, p i dot p j is i j i j. So take the square root of both sides dimensionally. Three-point scattering amplitudes have dimension. Yeah, go ahead. I think so. Oh, sorry. Thank you, thank you, thank you. See, you're keeping me honest. These students, I hear all the students in the back and they're murmuring. And that means they're too shy to ask questions or point out my mistakes. So are there any other mistakes anyone would like to point out or questions one would like to ask? Yes. Okay, excellent. So these again, I'm using capital A's not script A's. So these are the color ordered partial amplitudes. So as an aside, the full three-level three-point amplitude would be, let's see if we can reconstruct it. Trace TA1, TA2, TA3, A123. Plus trace TA1, TA3, TA2, A132. And then on top of this, you could specify your helicities, whether who has plus and who has minus. So for three particles, these are the only two independent permutations, mod-cyclic transformations. Every other permutation of one, two, and three is cyclically equivalent to either this one or this one. So thanks for asking that question. So again, everything henceforth will always be with these color-ordered partial amplitudes. Any other questions? Yes. Right, so my motivation here, okay, this turns out to be true in complete generality, but in order to simplify things, I said at tree level. And that's just because if you think about Feynman diagrams, tree-level Feynman diagrams are always rational functions. You've got their momentum factors in the numerator coming from the three-point Yang-Mills vertex, there are propagators in the denominator, but you never have some complicated transcendental function of the momenta. So yes. That's right. So that's something that's not obvious, and that's this very special feature of three-particle kinematics. We'll discuss the n-particle generalization of this in just a minute, but remember, three particles is very, very special. So if you would go to loop level, at loop level, and I will discuss loop amplitudes in my last two lectures, but at loop level, of course, you can have complicated transcendental dependence on your momenta. So you still need these overall factors to give you the right scaling. You could conceivably imagine having these dressed by complicated functions of the kinematics. However, that can't happen for more than three particles. It can't happen for three particles, because remember that everything else, other than this factor that's determined by the polarization vectors, can only depend on the full Lorentz invariance, p i dot p j, but those are all zero for three particles. So these formulas are actually exact to all loop order, but that, again, is a special property of three particles. I think I have enough time in this lecture to discuss... Oh, was there another question? Excellent question. It doesn't, it can't, but I haven't proven that, yeah. Yeah, what's the argument why it can't remind me stuff? Yeah, it doesn't. Okay. So, oh, so by the way, let's celebrate a little bit. I've computed all possible three particle scattering amplitudes of gluons with no work. Okay. We didn't even draw any Feynman diagrams. We just discussed what are the possible properties that Feynman diagrams have that leave an imprint on the answer. Okay. Now we're going to generalize this and compute various endpoint amplitudes, again, with minimal work. And the tool that will allow us to carry this out is something called the BCFW recursion relation after Brito-Cachazzo-Fang and Witten from 2004 and 2005. So the idea here is the following. Let's consider a one complex parameter family of scattering amplitudes parametrized by a single complex parameter z with the following, let me explain how it's defined, specified by lambda one goes to lambda one hat equals lambda one plus z lambda two. And lambda two tilde goes to lambda two tilde equals lambda two tilde minus z lambda one tilde. Actually, maybe a better way to write this is the following. So what we're going to do is we're going to take our scattering amplitude, an n particle scattering amplitude, and we're going to deform the kinematics by one complex variable z. So obviously at z equals zero this gives us the amplitude we started with, the undeformed amplitude. The second important observation is that this deformation preserves energy momentum conservation. If you write it out, you need the sum from i equals one to n of p i mu to be zero, but that's just the sum from i equals one to n of lambda i a lambda i tilde a dot. So this is, I'll suppress the indices, save space. When we deform this, we get lambda one plus z lambda two lambda one tilde plus lambda two lambda two tilde minus z lambda one tilde plus dot dot dot. So these two terms cancel out. Here we get plus z one lambda two lambda one tilde minus z lambda two lambda one tilde. So if we start with a collection of kinematics that satisfy momentum conservation, energy and momentum conservation, then this one parameter family of deformations will preserve that condition. That's crucial because it means that we can still talk about honest on-shell scattering amplitudes that in particular are gauge invariant. So you know that when you calculate individual Feynman diagrams, they are not gauge invariant. They will in general depend on what gauge choice you've made. You can also make field redefinitions in your Lagrangian which can completely change individual Feynman diagrams. But when we talk about scattering amplitudes, we talk about quantities that are independent of gauge, gauge invariant, and independent of any possible field redefinition in the Lagrangian. So it's crucial that this one parameter family of deformations keeps us within the world of, you know, the much tamer world of honest on-shell scattering amplitudes rather than the wild, crazy world of individual Feynman diagrams. Okay. So now that we have this one parameter family deformation, let's consider a contour integral. Around some contour, around a circle, let me specify small circle around the origin. And the word small has a specific meaning. Okay. The integrand here obviously has a pole at z equals zero. So here's what I'm going to draw the poles in the complex z plane. Pole at z equals zero. Then there may be other poles. This thing. Okay. Oh, and here it's important that I specify that we're talking about tree level amplitudes. Or at tree level, A of z can only have poles in z. Of course, once you include loop corrections, you could have branch cuts. But at tree level, Feynman diagrams are always rational functions of momenta, so you can't possibly get any dependence that's more complicated than poles in z. So this thing here might have some other poles at some other locations in the complex plane. And the word small means that I'm taking the contour sufficiently small that it only includes the pole at z equals zero. We know this thing has no problem at z equals zero. At z equals zero, it is exactly just the n particle scattering amplitudes. So certainly it's a fine, well-behaved object. It might have some poles elsewhere in the complex plane. Those are the ones that we're going to discuss in a moment. Okay. So then by Kossi's theorem, you can imagine closing the contour in two ways. On the one hand, this is equal to A of zero. Let me put a one over two pi i here. By Kossi's theorem, the value of this contour integral picks up the residue of this at z equals zero. That's A of zero. Or you could imagine closing the contour the other way. You pull this contour off to infinity, picking up all the additional poles with a minus sign. Or you can think about it another way, change variables from z to one over w, and then this thing becomes outside the contour and all the other ones become inside the contour. So I'll write here minus the sum over all other residues, all other poles, the residue of A of z over z. Okay. So this is our main formula. This tells us that we can calculate our n particle scattering amplitude by identifying all of the residues in this formula. So that's going to be my next topic. How many more minutes do I have? Five, ten? I don't remember when I started. Three. Okay. Excellent. So then all I will do is identify where those poles are, and we won't compute the residues yet. We will postpone that till the next lecture. A of z is a rational function which can have poles, or does have poles, when an internal propagator in some Feynman diagram goes on shell. So for example, let's, okay, so there are many individual Feynman diagrams that contribute to this. Let's look at some particular Feynman diagram, and I'm just going to make something up here because it doesn't really matter. I'm being intentionally vague here. Well, it's important that it's tree level. Okay, so let me be a little more careful here. Okay, here I've just invented some Feynman diagram. Okay, there's one of the many Feynman diagrams that contributes to this scattering amplitude. And this has a property that it has a pole. Okay, it has that propagator. Here, there are many other Feynman diagrams that have that propagator. So before when I was just drawing random dots in here, what I had in mind was that you should consider all possible Feynman diagrams that have this propagator and anything else. Now when we do the shift, this gets shifted to P2 minus Z lambda 1 lambda 2 tilde plus P3 plus dot dot dot plus Pk squared. Right, because I'll write it over here. P2 is lambda 2 lambda 2 tilde. So it gets shifted to lambda 2 lambda 2 tilde minus Z lambda 2 lambda 1 tilde. That's what I wrote over there. Now let's multiply this out. Okay, so this is a quantity squared. So we get this thing, P2 plus P3 plus up to Pk squared. Then we get the cross term. The cross term is written like this. I'll explain this notation in just a second, plus zero. There's no quadratic term because lambda 1 lambda 2 tilde is a null momentum. It squares to zero. It's a null momentum because thinking again, thinking of it as a 2 by 2 matrix, it's a matrix with determinant zero. Okay, so this thing has a pole. This is my final sentence. This or any such diagram of this form out of space, I'll write it here, has a pole Z being one half P2 up to Pk squared over this thing. And I haven't defined that notation yet, but it's implicit in what's written here. So in my last negative minute of honesty, I'll say that one P2, well, in general, I pjk means Ij jk. And then you extend by linearity. So I pj plus pkl means Ij jl plus ik kl, et cetera. Or if you like the 2 by 2 matrix notation, think about this as a 2 by 2 matrix. It's a sum of 2 by 2 matrices, so it's a 2 by 2 matrix. And it's sandwiched on the right by this thing, lambda 2 tilde, thinking about it as a column vector, sandwiched on the left by lambda 1, thinking about it as a row vector. Okay, so sorry for going a little over time. We've identified an example of where this kind of thing can have poles, and next time we'll compute the residues and finish our calculation of the BCFW recursion. Thank you.