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STUDENT: [INAUDIBLE]
NIMA ARKANI-HAMED: Yes. Question time. Do I have to be looking at you as you're asking me?
STUDENT: No. You started off last time by saying there's no local observance of gravity and so that'd be easy to focus on how to identify the correct boundary observable of your program. But I guess--
NIMA ARKANI-HAMED: I mean, that's not part of the program that's trivial here. It's just the estimated principle. It's not trivial in others.
STUDENT: So the question I have is what happens if the universe is spatially compact?
NIMA ARKANI-HAMED: I think we're screwed. I mean, those are the really excellent questions.
STUDENT: So is this an argument that the universe is not spatially compact?
NIMA ARKANI-HAMED: This is nothing. Exactly. This is really-- as you know, people like to confuse themselves about quantum mechanics a lot. There's nothing to be confused about until you say the word cosmology. And in particular, closed universes are, I think, the most confusing thing of all, because in a closed universe, you can't separate aparatuses arbitrarily, you can't make them arbitrarily big. Both infinities are removed, and not in the subtle way like they are in internal inflation, just in an in-your-face way. So I think it's just trouble. It's just trouble.
And I think what quantum observables are in a closed universe-- they don't seem to exist. None to exist. Now, this is why I said it. I think the whole class of questions are the most important-- are the hardest conceptual problems that we have today. There's two attitudes. Depending on your temperament, you might think the attitudes as radical or reactionary. The reactionary attitude will say, good, we need quantum observables. Therefore, closed universes don't exist. We don't actually have a solution to the fundamental theory, blah, blah, blah. Which give us completely closed universes. That's why. That's, I think, the extreme quantum mechanical attitude.
By the way, that extreme quantum mechanical attitude has worked so far, all the time. However, in every situation where-- like in the black hole information problem, quantum mechanics help, no problem. But of course, it was a situation where quantum mechanics could hold. I mean, there was an observable there after all. There was S matrix. It wasn't predicting its own demise. OK, in this situation I'm giving a long answer because I have no idea. So it's possible, they just don't exist.
And that's the attitude you can really push. Like inside of ADS space they just don't exist. I mean, it's true. The observables live on the boundary. That's great. There's a boundary in theory. Perfect. The problem is we push this attitude further and further. We run into fewer and fewer things to talk about. There was imaginary situations like closed universes, there's our own inflating universe. And at some point, you seem to be left with nothing.
And then it's a real question. What do you do at that point? Perhaps quantum mechanics isn't quite right. Or if not quite right, it has to be reinterpreted, but in some very, very significant way. It is a strange feature about quantum mechanics that it needs this separation of the universe into two pieces-- infinite and not-- in order to make any prediction. It's just true. There's a reason why Lorentz and his class of friends were so bothered by all of this after all.
STUDENT: I'm a little confused. If I have a compact universe, and I'm just looking at a smaller piece of it, and presumably what happens locally is in some sense, decouples what happens very, very far away.
NIMA ARKANI-HAMED: No one is arguing that there aren't approximate observables. Of course there are approximate observables. And this is why, as I said, when you get your high theorist horse and tell the experimentals there's no local observables, the people who just won the Nobel Prize this afternoon will say, well, I got a Nobel Prize for my non-observable.
STUDENT: So if you can quantify--
NIMA ARKANI-HAMED: The problem is that because of gravity, there is no way, as I said yesterday-- it was a very standard argument. There is no way any experiment can be done to send the systematic error on that thing to zero. The reason that strug-- A, is just a fact. It's just an observational fact about the world, an empirical fact. Well, no one has done the experiment so many times-- in fact, we're extrapolating.
But we imagine if we do the experiment that many times, it wouldn't converge to a sharp answer. That's A. But B, the other reason it's troubling is that let's say you have a theory. Let's say I have a theory of quantum gravity. Great. So there's a bunch of equations. A bunch of numbers come out, right? What do those numbers mean?
Before you talk about the theory, you have to talk about what the theory is supposed to predict. It's a much more basic thing than having the theory. This is why the people who talk about very naive approaches to quantizing gravity, like [INAUDIBLE] gravity, are totally completely as misguided as they can possibly be. Because what do they talk about? They talk about discrete areas.
This is just nonsense. There's no such thing. There's no experiment that's going to measure anything to the requisite precision that could even tell you if an area was discrete or not. If someone comes up to you and says, I have a theory of quantum gravity and here is the local observables, you say, thank you very much, I won't pay any attention to your theory. It doesn't make any sense. There aren't any. There's no actual observables that correspond to it.
Now, the problem is that we can't be too high and mighty about this because we start running out of things to talk about in this attitude about the world. But like I said yesterday-- oh, thank you so much. We were talking trash about you the whole time.
STUDENT: It's all in there.
NIMA ARKANI-HAMED: Yeah, that's true. I'll have to have a conversation with you. I mean, if you want a really, really hard problem to work on, work on that one. It's way harder than any of the things that I'm talking about here. It's just that at least I have a sense that these problems are too hard, and that we have to first come to grips with some simpler things before attacking them.
You know, problems get solved when it's time for them to be solved. People like to make this analogy. It's a very good one. In the early 1900s, there were a whole bunch of things that you can be confused about-- black body radiation, radioactivity, superconductivity, OK? Very bad idea to work on superconductivity.
And in the end, it wasn't even a more fundamental problem than the other ones. It just that many other things needed to be understood before we attack that one. So you follow your nose. You work on the next hardest problem you can which you can hope to make some progress on. That's the attitude. OK, enough philosophy.
So we're interested in the Yang-Mills theory, computing what I called yesterday the color strip scattering amplitude. And so the actual scattering amplitude has a bunch of color indices on it-- A1 up to An-- but we can take every possible color structure that appears the trace of a product of tA1 up to tAn with various permutations of those indices. And the coefficient of each one of them is some object. Some permutation of M1, 2 up to n.
So we're just going to be interested in one of these things. It's called the color strip scattering, OK? And it also depends in a general theory on the felicities of the arc and the momentum, e1, e2, up to tn.
OK, our first task is to find the right set of variables to discuss these scattering amplitudes. And when I mean the right side, I mean a set where you finally see what they're actually functions of, rather than these redundant representations in terms of polarization vectors in momentum that we talked about. So in other words, we're after finding objects unlike polarization vectors, which don't exist, which, when you do a Lorentz transformation of them, transform as some Lorentz tensor on one side, but pick up the little group phase transformation on the other, OK?
This is a perfectly well-defined mathematical question. You think the answer would be in Weinberg volume 1, chapter 1, paragraph 1, and it isn't. It should be, it could be, but actually, no one has worked it out in general.
We fortunately have a very simple answer in four dimensions. It's been worked out last year in six dimensions. There's probably a very simple story in any number of dimensions, but it isn't in the literature yet. So if you want a very nice problem to work on, generalizing any number of dimensions would allow all of this progress in amplitude to start being ported to a higher dimension. So there's actually a nice paper yesterday about it, [INAUDIBLE] and friends, that started using the six-dimensional formalism to do the same kinds of games in six dimenions.
Anyway, we live in four dimensions, so let's introduce the variables. So the idea is I'm going to work with the ground spinor representation of everything. So in particular, a 4 vector I can represent as a 2-by-2 matrix, p alpha alpha dot, which is just sigma mu p mu alpha alpha dot. So if you remember, this is the matrix that's p0 plus p3 can go minus p3 1 plus i of p2. It can go minus 1 of 2. And the determinant of this matrix, p, is just equal to p mu, p mu. That was the point over p squared. And the Lorentz group is SL 2, c.
Now, let's say we have a massless particle so the determinant of p is 0. So p squared is 0, so the determinant of p is 0, OK? That means that p-- it's a 2-by-2 matrix There's a determinant of 0. Either both of its eigenvalues are 0, in which case it's 0, or one of its eigenvalues is 0. So that's the generic case where one of its eigenvalues are 0. That mean that it's a rank 1 matrix.
Because one of its eigenvalues is 0, that means that I can write this 2-by-2 matrix as some vector times another vector. It's rank 1, OK? So we'll do that. We're write p alpha alpha dot as lambda alpha lambda tilde alpha dot. So, I mean, very, very concretely, if we had something moving in the z direction, then p would be like, 2, 0, 0, 0. That would be the matrix. So lambda could be group 2, 0. Lambda tilde could be group 2, 0. So I'm just writing p equals lambda lambda tilde.
Now, for real momentum, p alpha alpha dot has got to be a Hermitian matrix, right? That's Hermitian there. So that means that for real momenta, lambda tilde is actually lambda star. It's actually its complex conjugate.
A very big part of small story is that reality is not very important. In fact, we're going to complexify everything inside. Everything is going to be complex. We're going to be writing down integrals, everything is going to be a contour integral, everything is holomorphic, everything is complex all the time, OK?
So just remember that that's what reality is. That's what the reality condition is. But we're going to consider general lambda alpha and lambda tilde alpha dot general and unrelated. So what we're really doing is just talking about general complexified momentum and writing these light-like vectors. So these vectors are always light-like, but now if lambda tilde is not equal to lambda, the momentum will, in general, be complex.
Furthermore, on this complexified space, the Lorentz group isn't SL 2, c. It's SL 2, c cross SL 2, c. There's two SL 2's that act on the lambda and the lambda tilde, OK? So the Lorentz group-- the complexified Lorentz group is bigger. It's SL 2, c cross SL 2, c. And since everything is complex all over the place, I'll be dropping the c's all the time, so the Lorentz group is SL 2 cross SL 2. One SL 2 acting on the lambdas, the other SL 2 acting on the lambda tildes. The diagonal part of that is the real Lorentz group when you identify lambda tilde to be lambda star, which is still a complex group. It's still an SL 2, c. It's the diagonal SL 2, c of the product of H2, SL2.
Sometimes it will be useful to-- and this is largely for intuition purposes, and we're doing some explicit computation. It's useful to imagine that lambda lambda tilde are independent, but independent and real. Both of them real, OK? Lambda and lambda tilde independent and both real actually corresponds to doing physics in 2, 2 signature.
So imagine we were doing things in 2, 2 signature, not 3, 1 signature. Let's say we're doing things in 2, 2 signature. Then it would be useful for us to group of 4 vector into p0 plus p3, p0 minus p3, p1 plus p2-- no i's-- p1 minus p2. This is otherwise known as a completely general 2-by-2 real matrix, OK? A totally general 2-by-2 real matrix, but notice that its determinant is still p0 squared minus p3 squared minus p1 squared plus p2 squared, OK? So it looks like 2, 2 signature-- two pluses and two minuses, OK?
So if we think of lambda and lambda tilde as being independent and real, then what we're doing is describing a null momentum vector in 2, 2 signature. So that's just one particular case. We will sometimes do this, especially when we talk about going to a twister space, which involves doing various Fourier transforms relative to these guys. Just so we can think naively about doing Fourier transforms over real variables, we will imagine that we're doing things in 2, 2 signature. Very often it just doesn't matter what signature you're doing anything in. You just take care of going back to 3, 1 signature at the very, very, very end of the calculation. So the signature-- everything is going to be floating around constantly. Everything is going to be complex, OK?
STUDENT: So generally, lambdas are all complex in whatever.
NIMA ARKANI-HAMED: In general, the lambdas are complex, the lambda tilde's are complex with totally independant variables. I've just given you two interesting slices. One of them is lambda tilde equals lambda star, both complex, but one the complex conjugate of the other. That's 3, 1 signature-- real momenta in 3, 1 signature. The other one is lambda real-- lambda tilde real and independent-- and that corresponds to null momenta in 2, 2 signature. OK? Very good.
Now, notice something wonderful about this, which is that the lambdas and the lambda tildes are actually not uniquely associated to a given p. I can rescale lambda by something, and lambda tilde by its opposite, and it leads p invariant. Let's say for the moment we go back to 3, 1 signature.
So the rescaling that we're talking about, if it's going to keep lambda and lambda tilde complex conjugates of each other, would be to send lambda to e to the i theta lambda, and lambda tilde goes to e to the negative i theta lambda tilde. So the lambdas and the lambda tildes are not uniquely associated to get a given p. That's obvious just from counting. There are three independent components of a null momentum-- four in general and one that forces the energy to equal the magnitude of the momentum.
So there's three parameters. The lambda is a too-complex dimensional vector, so it has four parameters. We're forcing lambda tilde to be the complex conjugate of lambda. OK, so there's still four parameters. We haven't added any new ones. But there's four there, not three. And the fact that there's four is reflected in this rephasing variance. I can rephase lambda by any phase, do the opposite to lambda tilde, and I have the same p.
This should remind you of something. What else is there that keeps a null momentum invariant but rephases? That's what the little group does. That's exactly what the little group does. So in fact, that freedom on these guys is precisely the action of the little group-- exactly the action of the little group.
STUDENT: The v1?
NIMA ARKANI-HAMED: Yeah.
STUDENT: v1.
NIMA ARKANI-HAMED: Yes, because that's all that leaves p invariant. You can actually go through a whole argument if you want to do this more apparent. Let me just sketch it just to make it look identical to the little group. It actually literally is the little group. What you can do, just like you're used to in discussions of the little group, is you can say, let me pick some canonical k null vector, and for that canonical k, let me pick some canonical way of writing it as lambda times lambda tilde. This is my choice, OK? Let's put stars there. This is my choice. Of all the possible ways I can write it that way, I'm going to pick one way of doing it, OK?
And then I'm going to define-- if I define p to be some Lorentz transformation acting on k, I'm going to define the lambda-- so this is the lambda. This is the lambda for k, which is my choice. I choose whatever I want. But I'm going to define the lambda for p to just be the l alpha beta lambda beta acting on the lambda for k.
So if I take a reference null vector, I write a general momentum and some Lorentz transformation on the null vector, and I just define the lambda for these other vectors to be that Lorentz transformation in its SL 2 representation acting on lambda. So now I'm done. I've defined canonically a lambda and a lambda tilde with every null momentum in the world.
That's fine, but what's not true is that if I now do some Lorentz transformation on p, this is not equal to lambda alpha beta lambda beta on p. Instead, it's equal to that up to a phase. And that action of the phase is exactly the little group action for the spin-1/2 representation.
So these are the objects we're looking for. On the Lorentz transformations, they transform like the nice spinner, and they pick up a phase. So that means that if I were to build-- so what kind of Lorentz invariants could I build out of these objects? The sorts of Lorentz invariants I can build are just contracting the lambdas and the lambda tildes with the epsilon symbol. Just as we're used to all the time. So just as a notation, if we have a lambda 1 and a lambda 2, that angle bracket refers to contracting them with the epsilon symbol. And I will also sometimes write this as just 1, 2 when there's no confusion. Similarly, lambda tilde 1, lambda tilde 2 refer to contracting them which I will write as 1, 2.
But you see, these objects are Lorentz invariant, but they have little group transformation properties. This object on the Lorentz transformations picks up a phase as if we were talking about a negative helicity particle by convention. A negative helicity particle for 1 of spin-1/2, and negative helicity spin-1/2 for 2. This would be positive spin-1/2 for 1, positive spin-1/2 for 2. This makes it amazingly easy to write down objects that deserve, at least from the transformation properties, to be called scattering amplitudes.
Let's give a simple example. So let's say we have 1, 3 to the 4th divided by 1, 2, 2, 3, 3, 4, 4, 1. This happens to be the Park-Taylor amplitude for two gluons of negative helicity, and 1 and 3 and two gluons of positive helicity 2 and 4. but let's see how we would discover that. You just look at it and you see that under rescaling 1-- by the way, in 3, 1 signature it's rephasing, but in general, it's a completely general complex lambda goes to t lambda. Lambda tilde goes to t inverse lambda tilde.
So you see, this object picks up four powers upstairs and two downstairs, so it goes like t squared under rescaling 1. So that's like twice helicity negative 1/2. So it's helicity negative 1. So particle 1-- this amplitude, it has some weight under rescaling lambda 1, and its weight just gives you the helicity of the particle. That's all. So the scattering amplitudes are homogeneous functions of these invariant brackets and their homogeneity tells you what the felicity is.
STUDENT: So they're not invariant? I'm confused. They're not invariant under this scaling?
NIMA ARKANI-HAMED: Definitely not. Amplitudes are not invariant under the Lorentz transformations. They're not invariant. They pick up a little group action on the external particles that tells you what velocities they have. So they better not be invariant. OK? And they're not invariant. That's how you tell. That's the difference between this being gluons, that's another amplitude that would just be for scaling. So particles one and three are felicity negative one. Particles two and four are felicity plus one.
STUDENT: So I'm confused. Probably a very simple question, but confused what you say that the amplitude is not invariant under-- I would think the little group is like the u1 of magnitude. That's what I had--
NIMA ARKANI-HAMED: No, no. That's not what we thought. It's a u1, but it's like saying for rotations. The spin of half particle's not invariant under rotations. It rotates. It transforms into some representation.
STUDENT: What did you mean when you said that these bracket 1, 2s are Lorentz invariants in some ways?
NIMA ARKANI-HAMED: No, no. I've completely contracted all the Lorentz indices. So under the Lorentz transformations, they're invariant. But they pick up to the phase which is the little group. Right? And that's because of this fact. Once you define them, once and for all in some way, then Lorentz transforming their corresponding momenta doesn't just take them back to the ordinary, to a Lorentz transforming method, they're in the phase. This is exactly what happens with the little group argument. If there's [INAUDIBLE] here who haven't gone through this, this you really should go through. This isn't one book, volume one, chapter one, paragraph two. OK? But this is very important. This is what particles are. The particles are just defined by how they transform under the Lorentz group and the fact that they pick up these auxiliary transformation properties of the little group.
But the point is that using these variables, it's incredibly simple to keep track of that. OK? In a way that was not possible with polarization factors. And this is why, as I said, you've never seen this kind of formula in a standard class. And that's why you've never seen an actual scattering amplitude. Here, we're actually staring at the object itself. That's what it really is. It's really a function of these variables in a non-redundant way. Yes?
STUDENT: Can you go through why one and three helicity and two and four of the other from this one?
NIMA ARKANI-HAMED: Yes. Great. So let me say here. General statement first. If you have an amplitude that depends on a bunch of lambdas, I'll always use a convention where a runs from one to n is the number of particles. This has half the property, but if you rescale like so, it's equal to t to the power of negative 2ha m of lambda, lambda 4. In other words, it's separately homogeneous under rescaling every one of the particles. It's not some overall thing. And you think any one of the particles, you rescale lambda, lambda tilde, it has to have a particular way.
And also, if someone goes and gives you this formula and adds another term-- let's say they added a term with some square bracket one, three to the fourth upstairs. That just doesn't make any sense. It doesn't have a well-defined homogeneity. All the terms, everything has to have a well-defined homogeneity.
So this is what scattering amplitudes are. Just grouped theoretically. It's very simple. You build them out of lambdas and lambda tildes, out of the Lorentz invariant in a product that you can build out of [INAUDIBLE] after particular functioning.
Any questions about this so far? All right. So now I'm going to-- let me write that back up again here. Like it's a product over all a.
So let's begin our discussion of scattering amplitudes by talking about the most basic scattering amplitude which is a three particle amplitude. This one of the things that when you first learn a new subject, you tend to become obsessed with some things that are just awesome and are really simple, but totally awesome. This is one of the really simple, but totally awesome, things. Because it actually goes to the heart of what's so powerful about this way of thinking about things.
You see, we normally don't ever write down the three particle amplitude because it normally doesn't exist. Quantum field theory books don't tend to begin by talking about, let's consider the three particle amplitude. For gluon one goes to two gluons. Because there's no face base for it to happen. The amplitude vanishes. OK? A signature for real momenta, the three particle amplitude vanishes. However, that just turns out to be a complete accident of real momenta in three, one signature. It doesn't even vanish in two [INAUDIBLE]. It's just a real accident of three, one signature. An accident that, if you like, delays progress for a long time until you realize it.
STUDENT: The amplitude is vanished.
NIMA ARKANI-HAMED: The amplitude itself vanishes.
STUDENT: The face phase.
NIMA ARKANI-HAMED: The face phase also vanishes. The amplitude itself vanishes. The amplitude itself straight up-- in fact, it would be very disturbing if the amplitude did not vanish. Precisely because--
STUDENT: The face phases don't let you able to 2a.
NIMA ARKANI-HAMED: Right. But the actually, physical amplitude vanishes. Right? Physical amplitude just vanishes. If you just plug in, you plug in the vertex and the Yang-Mills or something. Of course I'm talking about it vanishes in the Yang-Mills. It doesn't vanish-- in phi-cubed theory, it doesn't vanish. But it actually vanishes in Yang-Mills.
STUDENT: Including felicity?
NIMA ARKANI-HAMED: Yep. But anyway, it doesn't matter because it doesn't actually vanish. So let's actually write it down. Just to show them-- we'll do it more systematically in a second-- but just so you see very, very explicitly--
Here are three nice generic momenta that sum to zero, they conserve momentum, and they're all on shell. They're all on shell, but this last one is on shell because there is an i there. So one squared from five squared is equal to zero. So there are whole sets of three momenta that are complex which allow this process to be on shell. OK.
Well, let's actually look at the statement of conservation of momentum is for three particles more carefully. So lambda one, lambda one tilde plus lambda two, lambda two tilde plus lambda three, lambda three tilde is equal to zero.
Now there's a very important general fact, but surprisingly useful in this business over and over again, which is that it's possible to take a two dimensional vector and expand it in the basis of any other two dimensional vectors. OK? That very basic fact, we're going to use again and again. In fact, there is a-- one constant to it is, just in case you heard these funny words, they're something known as the Schouten identity which is a statement of this form plus cyclic, so ca negative b alpha plus bc lambda a alpha physical to zero. Now this statement simply reflects the fact that I can always state lambda c and expand it terms of lambda b and lambda a. OK? That's all. If you just write the formula lambda c equals some linear combination of lambda a, lambda b, that's exactly that identity. All right?
So the fact that these are two dimensional vectors is going to be very, very important, constantly. But anyway, in this case, we don't need you to do something so fancy as that. Let's just take this equation and dot it into let's say, lambda one. If I dot it into lambda one, then from here I see that one, two lambda tilde two plus one, three lambda tilde three on the dot has got to equal zero, similarly, for the other ones. And similarly, for the equation if I replace-- I just do it with the conjugate variables.
Now let's look at this equation for a second. Let's say that one, two is, actually, not equal to zero. If one, two is not equal to zero, then one, three it is also not equal to zero. Because otherwise, we'd be setting lambda tilde two to zero. That's just zero momentum. We're trying to find generic momentum. So if one, two is not equal to zero, then one, two and one, three are both not equal to zero. That's fine. But it means that lambda tilde two is proportional to lambda tilde three.
So you see that-- and similarly, if you now apply this to the other equations, we find that if one, two and one, three and two, three, either they're not equal to zero-- their angle brackets are all not equal to zero-- but then all of the lambda tildes have to be proportional to each other or the other way around. So there's only two ways of conserving momentum. Either all lambda tildes are proportional which means that all the lambda tilde i and lambda tilde j vanish. And the lambda i's are generic. So i, j are not equal to zero or the other way around. OK? That's extremely powerful.
Let me go back for a sec, just coming back to the question because this highlights the point. I'm going back to the question, why does the amplitude have to be zero in Yang-Mills. Five cubed is special. The amplitude can just be a constant. In Yang-Mills, you know it can't be a constant. It's got to depend on momentum in some way. Right? But what could it depend on? Are there analogs of s, v, and u for three particles? There are no analogs at s, v, and u because p1 plus p2 is equal to p3, negative p3. So p1 plus p2 squared is zero. So p1 dot p2 is zero. All the pi dot pj are zero. There's simply no Lorentz invariants. There are no Lorentz invariants you can build out of the momentum.
The entire amplitude is pure polarization vector, if you like. All it depends on is felicity. Once you know it, there is no way of modifying it by multiplying by some random function of Lorentz invariants. OK? And we're seeing that. We're seeing that quite explicitly here. So we can now talk about the-- this has a very powerful consequence which is that the amplitude is either a function of angle brackets or a function of square brackets.
STUDENT: So how do these to get four?
NIMA ARKANI-HAMED: We'll do four in a second. This is very special to three. We'll see the structure of the three particle is completely nailed nonperturbatively period. The one thing you can be and we're going to write it down. Starting a four particle--
STUDENT: So you say zero?
NIMA ARKANI-HAMED: No, it's not zero. No, we're going to write it down. It will turn out to be zero for real momenta, but it's not, in general, zero. So let's actually-- let me just write down the answer. Let me write down the answer. And you can trivially generalize it for any situation you want. But let me write down the answer in the particular case for all the particles of the same spin, s. And only they have different felicities. OK? So there's only two configurations. Either two are plus and one are minus or all are plus and the other way around.
So let's start by talking about the case where say two of them are plus and one of them is minus. And they all have spin, s. Then, this amplitude is the following.
STUDENT: Excuse me, I'm a bit confused. You say spin and felicity. We are talking about--
NIMA ARKANI-HAMED: So particles of spin one, but they have a felicity--
STUDENT: So we can--
NIMA ARKANI-HAMED: No, no, no.
STUDENT: Only one.
NIMA ARKANI-HAMED: No. For whatever the spin is, if it's a spin one particle-- everything is mass-less here. Right?
STUDENT: Yes.
NIMA ARKANI-HAMED: If it's a spin one particle, it can still have felicity plus one or minus one. I'm just giving you the general formula for--
STUDENT: What other spin they could have?
NIMA ARKANI-HAMED: Any spin. Any spin you want. Spin two, spin 17. We're going to write down the formula for all of them.
STUDENT: All integers?
NIMA ARKANI-HAMED: Sorry?
STUDENT: Integers.
NIMA ARKANI-HAMED: Oh. Integer or-- sorry, so this is--
STUDENT: You know some [INAUDIBLE] also for spin half?
NIMA ARKANI-HAMED: Also for spin half. Yes. For anything. Of course, there is a three-point function with all spin half. And this is a nice way that you will discover it. I don't want to spend a long time on this, but playing with a three-point function and a four-point function, you can reconstruct every standard fact about quantum field theory-- spin statistics, anti-particles, the whole she-bang. But the fact that Yang-Mills theory at spin one has got to be describe the Yang-Mill, spin two has got to be described by gravity. There is no consistent higher spin theories. All that stuff goes follows very powerfully and beautifully just like playing with both my functions.
STUDENT: Even if you-- it can be everything.
NIMA ARKANI-HAMED: That's right. But here, I'm just doing kinematics. I'm just writing down the possible objects that have the correct Lorentz transformation properties to be deserved to be called the three-point function to be particles. And I'm telling you, all the spins are the same, it's 1,2 cubed over 1,3 2,3 to the power of s. Let's see why that's the case.
The important point here is that because of this fact, it can either be a function of the lambda tildes or a function of lambdas, but not both. In particular, that reflects the following fact I should have mentioned earlier. When we talk about something like a standard Lorentz invariant, on P1 dot P2. What is P1 dot P2? P1 dot P2 is nothing other than 1, 2 times angle bracket 1, 2-- square angle bracket times 1, 2. See, that makes sense in this vanishing little group way.
STUDENT: Is there a factor of 2 somewhere?
NIMA ARKANI-HAMED: There is a factor of 2, yes. 2 equals 1 in these lectures, OK?
[LAUGHTER]
I'm working mod mistakes, OK? Yes, there's a 2, isn't there? But damned if I know where it is. Is it here?
STUDENT: Yes.
NIMA ARKANI-HAMED: Probably there. Well, there you go. I'm good. All right. Thank you. OK, so these are general Lorentz invariants. The Lorentz invariants we're used to asking you things like that are products of spinor brackets like this with vanishing little group weight. And you see exactly what we saw before in manifesting itself, because one or the other of these are 0. All of these sorts of objects are just 0 here, or be part of, in the spinor case, of three parts.
So that means that it can only build the invariant out of one kind of bracket or the other kind of bracket. But that's it. Now, the actual powers that can appear are completely fixed by getting the correct homogeneity under all the particles. See, this funny 2 you've got over 1, 32, 3-- this guarantees the net weight under 1 is there's three 1's upstairs and one downstairs, so it's 2x upstairs, which is correct.
So that's positive helicity. That tells us that particle 1 has positive helicity. Particle 2 has particle helicity. Particle 3 has negative helicity. The general formula looks identical to something anyone who has looked at 3 point bundles in conformal field theory is familiar with. The formula, if you have s1, s2, s3, is 1, 2 to the s2 plus s3 minus s1 2, 3 to the other way around and so on. So you can work it out for yourself. This is what it is in the general [INAUDIBLE].
Now you might have said, wait a minute, I could've written down something else. I could've written down 1, 2 2 over 1, 3; 2, 3 to the negative s. These have exactly the same weights. So those are the two allowed objects that you could have. Those are the two allowed objects that you could have.
But we have one more physical requirement, which is that the amplitude should be smooth in the limit where the momenta become real. In the limit where the momenta become real, this goes to 0, and that blows up. So we don't use this one.
So that's the plus, plus, minus amplitude. What is the minus, minus, plus amplitude? Well, just exactly-- times some constant. Minus, minus, plus would be angle bracket 1, 2, over 1, 3; 2, 3 to the power of s times some constant. And those are all the possible plus, plus, minus and minus, minus-- all the amplitudes of two particles of one helicity and one of the other.
Let's volunteer already and make a number of additional comments. First, this is a completely nonperturbative statement. There's no-- this is just, kinematically, can't be anything else. So three particle amplitudes are completely fixed by the Poincare invariants.
That's wonderful, because we know in the Lagrangian, somehow the cubic vertices are the most important things. But in Lagrangians, they come along in quartic vertices and all sorts of other stuff. But when you're talking about the amplitudes, that invariant statement is that the 3-point amplitude is completely fixed by Poincare symmetry, its structure. It's totally fixed by Poincare symmetry. We can read off-- yes?
STUDENT: When you talk about having these amplitudes for complex momenta, should I think of those as being defined by analytic continuation?
NIMA ARKANI-HAMED: Absolutely. Absolutely.
STUDENT: But then, if you start off in the [INAUDIBLE] why did you get 0?
NIMA ARKANI-HAMED: Well, this is-- you see that all the functions we're talking about are actually meromorphic. They're not going to be pure analytic. So they have interesting pole structures, which is why you don't necessarily see them. So they're all defined-- they're all analytic up the poles and branch cuts. And the poles and branch cuts are very important, but yeah.
OK, so that's-- so you see the pole right here, right? Now of course, when you uniformly go to 0 momenta, there is more 0 upstairs than downstairs. So it vanishes. But if you can't analytically continue at any one of the momenta, you would encounter some poles in that [INAUDIBLE].
OK, so secondly, just the units of this object, you can just read straight off. Right? Just to prevent one, it's all spin-1 back up, which is dimensionless. If it's spin-2, the coupling is units of length squared, and so on. Third, notice something totally remarkable about this. For example, the famous statement that Henry made famous, that gravity is gauge squared, is completely obvious here. This is the one place where it's dead obvious, that it hits you over the head. The 3-point amplitude for gravity is literally the square of 3-point amplitude for spin-1. Period.
Now notice-- let's compare this to the Lagrangian. You stare at the Yang-Mills Lagrangian and the gravity Lagrangian. The cubic vertex of gravity has 80 terms in it. The same as-- if you took the Lagrangian, shoved in all the external polarization vectors, there'd be gigantic cancellations in those 80 terms, and that's the answer you found. But Lagrangian is horribly obscure in this.
Third, you can write down a 3-point function for spin-100. There's no obstruction at the level of the 3-point function to having higher-spin theories. In fact, there's a cottage industry of people who've spent years developing a sort of BRSP-like formalism for writing down complicated nonlinear Lagrangians of the cubic order for massless higher-spin theories.
That's completely pointless. We know it's possible. This is the beauty of talking about the amplitude. The amplitude doesn't have any off-shell gauge redundancies, any excess baggage. It's just the answer. You can write down the answer. It's consistent, it's Lorentzian, and everything is fine. So the fact that they succeeded after all these years to write it down in cubic Lagrangians that were consistent with their BRSP stuff or what is not very impressive. We just derived that fact in one line.
What's hard is the next stage in their grand program, is to extend this to the quartic level. Find this to the quartic Lagrangian, and that's where all the difficulty begins, because you see, in this way of thinking about things, what we're making manifest from the get-go is the fact that amplitudes of the correct transformation properties-- we're looking at them as they are and so on. But then, if you want to then build some theory, you'd start having to have to imagine what a 4-point function would look like. But-- let's actually do that.
Actually, before I move on to the 4-point function and what's wrong with higher-spin theories, let me just make another quick comment here, just to finish the story. What would be plus, plus, plus amplitude look like?
STUDENT: It can't be 0.
NIMA ARKANI-HAMED: No. That's 1, 2-- that's even easier-- 2, 3; 3, 1 to the power of s. Everybody appears upstairs twice. So that's the all-plus amplitude, and the all-minus amplitude is the same thing with angle brackets. OK?
STUDENT: So can you repeat again-- where do I see the divergence if you were to use the other brackets with minus s?
NIMA ARKANI-HAMED: Sorry?
STUDENT: If you were to write this one as the other bracket with the minus s?
NIMA ARKANI-HAMED: Ah. That also has a divergence.
STUDENT: Right, can you repeat how you see this--
NIMA ARKANI-HAMED: Yeah, the reason is going to real momenta. You see-- right. I should have mentioned this. Here, what we saw, even for complex momenta, the lambdas that satisfy momentum conservation are a little bit singular. Not singular, but they're a little bit constrained. All the way up until this has got to be proportional, but the lambdas can be generic, or the other way around.
Now going to real momenta, real momenta means that lambda tilde must be lambda star. So if all the lambdas are proportional, all the lambda tildes are also proportional. So real momentum in this language means that you're extending the angle bracket to 0 also the square brackets to 0.
So that's why, if I send all the square brackets here to 0, this thing vanishes. Well, there's 2 in the denominator, 3 in the numerator, so it vanishes. If I flip the other way around, if I lead with the angle brackets, 1, 2 cubed over 1, 3; 2, 3, to the power negative s, then it'll diverge.
STUDENT: OK. Here is the one we look at.
NIMA ARKANI-HAMED: That's right. Exactly. So if for this amplitude, I attempted this one, then I would also have a divergence. So I don't do that one. But of course, the all-minus amplitude is 1, 2; 2, 3; 3, 1 to the power of s.
Now, you just look at these things, and you see that they have different units than those do. In fact, what does correspond to? That corresponds to writing down some trace f cubed term. It's a higher-dimension operator. A higher-dimension operator can give you this thing, can give you that amplitude.
You also learn that no matter what higher-dimension operators you write down, whatever the heck is going on, all the possible cubic terms you could write down, with more and more derivatives on them, wherever you want-- in the end, these are the two possible forms of the 3-point function.
You also learn-- remember, in the case of gravity, you see that the first kind of correction you can have, the other kind of 3-point function you have, this would come from an R cubed term. In fact, the two-loop counterterm in gravity is exactly that. It's exactly giving you that structure for the scattering.
STUDENT: But can you think about them in different spin, or in math--
NIMA ARKANI-HAMED: So you have s1, s2, s3?
STUDENT: Yeah.
NIMA ARKANI-HAMED: Yes, yes. As I said, the formula, then-- which I encourage you to work out-- is that it would be 1, 2 to the s1 plus s2 minus s3; 2, 3 to the dah, dah, dah, and the other way around. So it has a very simple generalization.
OK, this is just incredibly beautiful, right? You see that the 3-point function is just a slave to Poincare invariance. It can't be anything else.
All right. Now let's briefly talk about the 4-point function. See? So let's do a simple example. Let's say, imagine we want to do a gravity scattering amplitude. So we have particles 1, 2, 3, and 4. Let me just tell you ahead of time what the answer must look like.
That's the most general possible thing it could look like. I've pulled out this factor 1 and 2; 3, 4 to the 4th. This now makes these have the correct-- makes sure that 1, 2, 3, and 4 occur with the correct weights to interpret it as gravitons, massless gravitons of the appropriate helicities. Whatever is left is going to have vanishing little group weight, so it can only be a function of the actual Lorentz-invariant kinematics-- the standard ones, s, t, and u.
So once again, in complete generality, you could write any 4-point function by stripping off these trivial angle, square bracket factors that take the helicities into account multiplied by some function s, t, and u. In this way, it's extremely easy to make ansatzes for possible scattering amplitudes. 3-point, you're nailed; 4-point, here's an ansatz, higher-point, you can make ansatzes. And you could--
STUDENT: You mean gravity here, no?
NIMA ARKANI-HAMED: How do you know it's gravity? The fact that it's gravity, is that you read off from here. What is the homogeneity of this thing when you're rescaling particle 1? You discover, if you rescale 1, it goes like t to the 4th. Therefore, does helicity 4 times 1/2, which is 2. So the formula's telling you what's involved. It's telling you it's two particles of helicity plus 2, and two particles of helicity minus 2.
STUDENT: If you raise that--
NIMA ARKANI-HAMED: Yeah.
STUDENT: If you raise that to the power of 2 instead, would that be a spin-1?
NIMA ARKANI-HAMED: That would be spin-1. That's right, absolutely. You can raise it to the power of 97-- uh-oh.
[LAUGHTER]
[SIDE CONVERSATION]
Yes, you can raise it to the power of 98, as well. Yeah?
STUDENT: Quick question. So for 4-point, is it besides the invariance s, the t, u-- do you need to worry about the antisymmetric tensor contracting with all the momenta?
NIMA ARKANI-HAMED: Do you need to worry about antisymmetric tensor contraction with all the--
STUDENT: For all momenta. It's also Lorentz-invariant.
NIMA ARKANI-HAMED: So here, for the moment, I'm assuming parity, just to make my life a little bit easier. You're right. There is one other object that I could do But let's just-- thank you.
Now you could do the following exercise. But now there are physical requirements. A physical requirement is that the poles of this 4-particle amplitude have to have a particular physical interpretation. If there are poles, if there's any pole in s, t, or u, then-- let's say there's a pole in s-- it has to be of the form 1 over s.
And the residue of that pole has to be the product of two lower 3-point amplitudes. That's a statement that the pole is reflecting the physical process of space-time. Poles should have the physical interpretation of something taking place over a long distance in space-time.
So this is the famous factorization properties, even at the-- let's just say we're talking about tree-level now. For the 3-point function, even the word tree-level didn't make an appearance. Let's imagine that some lowest order of approximation-- we're talking about the scattering of some lowest order of approximation. This is the sort of factorization property that they have.
So that's now a constraint on what those functions of s, t, and u can look like. And now you can do the following exercise. I think the exercise-- well it now occurs in Tony Zee's quantum field theory book in the second edition. So I'll just quote what the exercise is.
Put in arbitrary spins there. Let's say for simplicity, we're doing all the same spin. But put in the requirement-- put in this requirement, that if it has a pole, then it has to factorize in this way. And we know what the lower particle amplitudes are, right? We know what the 3-point amplitudes are.
Here we have to sum over all the helicities that could run in the middle. Put this plus. There's a minus there. There's a minus. There's a plus there. But summing over all those helicities, you should get-- it should factorize appropriately.
What you find, if you're talking about a single spin s particle, is that there's two solutions-- s equals 0, and s equals 2. And that's it. Everything else is just impossible. It's impossible to find a function of s, t, and u that factorizes properly. It's a very simple exercise. I don't have time to go through it now, but it's a very simple exercise.
This is how you discover that the only consistent theories of a single self-interacting massless spin-s particle is either phi-cubed theory or gravity.
What about spin one? Well, remember that was for a single spin s party. So if we allow ourselves to have cubic interactions with three-point couplings with many different species of particles, then I can put in an a b and a c. So I have 1 2-- for example, if this is minus minus plus, I have 12 cubed over 1323. 123. Let's say I'm doing spin one times some coupling that I can call fabc.
STUDENT: What is SDU? I missed that.
NIMA ARKANI-HAMED: SDU are the Lorentzian variants of p1 dot p2, p1 dot p3, p1 dot p4.
STUDENT: All right, you said that s can be either 0 or 2, but it's not the--
NIMA ARKANI-HAMED: It's not that, I'm sorry. It's the spin of the particle. Its not that. Let's call it just subconstant K. You'll find if you go through exactly the same exercise, now this F would have to have four indices on it, and so on. You find that that F is got to be made out of the square of something, the square of the k's. And it's only consistent with the k satisfying Jacobi identity. So you discovered that the only existing interactions are Yang-Mills. This is all the same as the content of Lambert's theorem, that the only consistent massless spin one particles are Yang-Mills, massless spin two particles are gravity. But the logical structure of the argument is very different. But actually the result is much more direct. And you get the answer much, much more quickly.
The actual arguments are different because the starting point are totally different. You see, Weinberg, more or less starts with some underlying picture of a Lagrangian. Even though it's set to be some s matrix argument. He's really imagining drawn Feynman diagrams, and talking about soft limits. And demanding precisely that when you replace the polarization vector by p mu and dot it into the amplitude and the soft limit, it vanishes. That's a very strong constraint that he uses to establish that gr is the only theory-- massless spin two, Yang-Mills for massless spin one.
Our starting point in these arguments is totally different. We are guaranteed to be dealing with objects with the correct Lorentz transformation properties. We don't have to impose it at the end. We start from that, but because we don't even have an underlying picture of Feynman diagrams in mind, the thing that we have to check is things like unitarity, factorization, and so on which is the check that we do. Of course you'll arrive at the same result.
STUDENT: I don't understand, where do you pose factorization? How do you know that they fulfill that property?
NIMA ARKANI-HAMED: Purely from the point of view of formula zone in the sense of writing down objects, or the correct Lorentz transformation properties, they don't. But there's more to physics than group theory. And these factorization problems are what telling you, are what encoding that there was locality in space time. It's telling you that there's really some underlying local physics here. Even though we're not talking about Lagrangians or anything like that. The underlying local physics is reflected in very particular sorts of factorization properties of the amplitudes.
STUDENT: Yes, but don't you have to go through some Lagrangian or [INAUDIBLE] diagrams to get that?
NIMA ARKANI-HAMED: No, you can rather say it the other way around. You can say that, for theories that have these nice factorization properties. If someone handed you an amplitude like that, you would look at it and you'd know it's something very interesting happens. It has poles, but the poles don't have random interpretations. The poles look always like 1/p square. You can imagine Fourier transforming that in your mind, and realizing that it corresponds with singularity along the light cone.
And further more that the residues of those poles are the product of two lower point things. So that's allowing you to reconstruct the picture in space time. Something happening over there, problem getting for a long time on the light cone and making something go on over there. You can go slightly off shell and getting a singularities that goes exactly on shell.
STUDENT: I'm rather confused because it seems like we are imposing locality here.
NIMA ARKANI-HAMED: At the moment we're just learning to talk about amplitudes in this other language.
STUDENT: So this part doesn't emerge?
NIMA ARKANI-HAMED: No, absolutely not. I'm just telling you how to look for it, though. You see, this is very nice. If someone handed you a bunch of functions like this, they all have the correct Lorentz transformation properties. How do you know whether they're just cropping something, or it's corresponding to a local theory. There are things that you can check. You hunt around for its poles, and you see if its poles have a good interpretation.
At higher orders there's not just poles, there's branch-cuts, there's more things that you have to take into account. But the poles and discontinuities have always a physical interpretation to relate to lower point processes. That's how locality and unitary are imprinted on to these objects. When we play with Feynman diagrams, and polarization factors, and redundancy, those things are built in. The unitarity and the locality are built in. We suffer, on the other hand we never see the object. We're always giving them these redundant representation so we never see the object. So it's just this sort of orthogonal starting point for thinking about all of these things.
STUDENT: What happened to the fermion?
NIMA ARKANI-HAMED: You see, in this entire argument, I was assuming that we were only talking about interactions between all three spin s particles. But you can't do-- if its three [INAUDIBLE]. That's actually a way of discovering that you can't do because we got these fractional powers of the angle, square, square [INAUDIBLE]. But it'll take me a little too long to explain.
STUDENT: But here it's saying it's really the 1960s, the old asymmetric approach. No Lagrangian, no field theory.
NIMA ARKANI-HAMED: That's right, some of the velocity sounds the same, but there is a one very critical difference. I'm just repeating something I said yesterday, but there's two critical differences. One sort of practical, and one more probable than the other. The one that's sort of practical is that the '60's people from our point of view now, were working on the hardest possible theory to give an asymmetric description for. A theory of scalars, which already suck, and are massive which suck even more. What they should have been doing, had they been playing with the s matrix of Yang-Mills theory of gravity they might of gotten a lot further, A. But B, their basic attitude was that the room to impose causality and unitary, and locality and get the answer uniquely out of it.
An attitude that seems absurd today, because we know we can write any one of 10 to the 500 Langrangians or more. And those run unitarity wonderful, Lorentz invariants possible theories. So it's ridiculous, you'd think that you're going to get a unique answer out of that. The attitude instead is that we're going to try. We're still trying to eviscerate the local space time description. But we're going to look for other principals. It's not going to be just unique or out of the blue. We're going to have to find some theory, find some principles that tell us what to do. And have the causality and the locality, unitarity emerge out of that. They know too soon that you're just going to impose these things and we want to see that they're coming out. Not only do not impose them. They are so important they were just ancillary consequences of some secondary starting point. That and of course the fact that we actually got somewhere. And there is a variety of other differences by the way, and important tactical differences. They were very obsessed with two to two scattering.
And the two to two scatering-- it pays to quickly complexify, not the momenta, but the [INAUDIBLE]. So what they did was talk about extending the functions and the complex STU plan, and so on and so forth. Here, we're actually complexifying the momentum directly. It makes a very big difference. Especially as you go to a higher particle. So it's also very technical. OK, so you've seen the three-point function. [INAUDIBLE] As I mentioned, there's an infinite class of two point functions, n point functions in the Yang-Mills that are from just pure Yang-Mills that we can talk about where particles i and j have negative helicity, and everyone else has positive helicity. It turns out it'll understand in a moment why that the all plus scattering amplitude and the one minus scattering amplitude are equal to 0. And if you have two of them, this amplitude in the Yang-Mills is this beautiful formula to be written down already a number of times.
STUDENT: For [INAUDIBLE].
NIMA ARKANI-HAMED: For everything.
STUDENT: But for the three it was not 0.
NIMA ARKANI-HAMED: No, this is in the Yang-Mills now, for all plus. Remember it was possible to write the point couplings with all [INAUDIBLE]. But they have difference units. You don't get them from pure Yang-Mills. You get then if you have some [INAUDIBLE] coupling, but if you don't have [INAUDIBLE] you don't have them, OK? So the all plus amplitude really just vanishes [INAUDIBLE].
STUDENT: What are the extra ingredients you are putting in?
NIMA ARKANI-HAMED: Now I'm talking about scattering amplitudes in Yang-Mills, standard Yang-Mills, two derivatives, no higher dimensional operators, nothing but standard Yang-Mills. Not into super Yang-Mills yet. No super Yang-Mills.
STUDENT: Pure Yang-Mills is a tree level statement?
NIMA ARKANI-HAMED: Pure Yang-Mills is a tree level statement. These are all tree level statements. This is a tree level statement. Obviously, I'm running out of amplitude here.
STUDENT: So the statement is that you use only [INAUDIBLE]. That's something you kind of use Lagrangian theory.
NIMA ARKANI-HAMED: Well, no. We could spend all day talking about this philosophy back and forth with or without Lagrangians. I could have spent a huge amount of time on it. But after a little while it gets boring, OK? But let me do one more. You could say, yes I take it from a Lagrangian, or you could have another attitude that you say, well, the cubic couplings are nailed by [INAUDIBLE]. Is it possible for me to build a theory that's completely nailed by [INAUDIBLE], entirely nailed by it. So all I do is put in the information about the three point coupling, go. I want to find the theory where all the 400 or 500 point functions, all of its properties are completely determined by that of a three point function, is that possible? It's very constraining because the four point function has to have the correct factorization properties. And we're led to the three point function. One you got the four-point function, we move to the five-point function, can we do that? Can you proceed in that way? In this way, in place of a Lagrangian, the seed that you're putting in is the three-point perhaps. That's a much better seed because that seed is completely nailed by the [INAUDIBLE]. Now we could decide, are we putting in both kinds of three point vertex? Only one kind of three point vertex? OK? And here we're putting the lowest dimension three-point vertex.
STUDENT: OK. But you can start from the--
NIMA ARKANI-HAMED: That's right you can just start-- and of course you can translate all of this to Lagrangian language, but these are just equivalent. Everything at this level is equivalent. It's just a question of looking at it from a very complementary point, but of course you can translate everything back and forth.
STUDENT: [INAUDIBLE]
NIMA ARKANI-HAMED: I'm sorry?
STUDENT: Is that 0?
NIMA ARKANI-HAMED: That's 0 for all. So let me make this statement more precise. So let's say if you look at this amplitude of all plus, or all plus and one minus. In Yang-Mills theory, ordinary Yang-Mills theory, this is 0 at tree level. And super Yang-Mills, it's actually 0 period. And it's true for any momenta. Yes?
STUDENT: For the three-point one that you wrote down on the 1 minus [INAUDIBLE].
NIMA ARKANI-HAMED: That's right. This is where everything hire than three points. It's for everything higher than three points. In fact the statement that it's zero, it is 0 for generic momenta, and anything higher than three part. The three-point amplitude is the special case. The three point amplitude is special because it's not quite, not 0 for completely generic lambdas and lambda tildes. There are no completely generic lambdas and lambda tildes. They have to satisfy momentum conservation. But yes this is for everything higher than the three-point function.
STUDENT: So if I have Yang-Mills, this is not true, right?
NIMA ARKANI-HAMED: At loop level it's not true, but it's true in tree level.
STUDENT: So right now everything we're doing is tree level?
NIMA ARKANI-HAMED: At the moment now we have been doing, is tree level.
STUDENT: To prove this you need something in addition to scaling. How many powers are--
NIMA ARKANI-HAMED: To prove the statement, to prove the fact that it's 0-- it's a direct consequence for super symmetry which we'll see in-- but if not a detailed statement.
STUDENT: But it's not a consequence of just the scaling.
NIMA ARKANI-HAMED: No, it's definitely not a consequence of just the scale, absolutely not. I'm just writing this down as-- so here's another example. Here's an infinite class of amplitudes that we know, OK? And you can easily see that we have the correct scale. Now let's talk about super amplitudes. Now we're talking about supersymmetry, and we're going to be talking about maximum supersymmetry. And this is yet another example of how much more invariant the amplitudes are as a way of talking about physics than the Lagrangian, OK? But actually want to motivate, why-- even if you've never heard of supersymmetry, human you'd be led to thinking about inventing it. And actually the maximal versions, n equals 4 and n equals 8, thinking along these lines. So now we have these beautiful amplitudes, they are labeled by velocities oddly depend on momenta, or lambda and lambda delta So their functions are velocity, lambda and lambda delta normally. But it's actually very annoying because they're labeled by this discreet series of plus ones and minus ones, depending on the velocities of the external particles. It's very choppy, it's very discreet. For n particles there's 2 to the n different amplitudes that you could be keeping track of here.
And there's-- OK. So if you want to do anything, if you want to write down the formula here, like in this factorization formula, you're always summing over these helicities. So there's these discrete sums all over the place.
And in general, we don't like discrete things so much. It would be great to be able to do all of this in a much more smooth and continuous way. It would be nice if the functions weren't functions of two smooth variables and one discontinuous discrete variable.
But that's actually exactly what maximal supersymmetry allows you to do. And it's actually a special thing of maximal supersymmetry. It's not true for n equals to-- or for gravity, it's not true for n equals 7 or n equals 6. n equals 4 and n equals 8 are special.
And the reason is that n equals 4, maximal supersymmetry puts all the space in one multiplet. That's not true for any lower supersymmetry. But if you start with a negative-helicity gluon, you can act on it with four supercharges. So I have the Q's and the Q tildes. And there's four of them, so I runs from 1 to 4, because it's n equals 4 or 8. So in general, we say that it runs from 1 to script n for the number of supersymmetries.
But if it's maximal, if n equals 4 or 8, then that can start from the negative-helicity gluon and that can go-- there's one of those and there's four states-- there's four gauginos, there's six scalars, four of the other gaugino of the opposite helicity, and one positive helicity gluon, so negative 1, negative 1/2, 0, 1/2, 1. But importantly, everyone is the same multiplet together.
So that means that there's actually a better way of labeling the amplitudes, not as the function of h, lambda, lambda tilde. In other words, instead of using helicity states as the external states, it's better to use something that makes the action of the supersymmetry a lot more manifest.
So we do that by using Grassmann coherent states. These are just things that are, if you like, you have the supersymmetry algebra, why don't we label things by momenta to begin with? We want to diagonalize as much of the Poincare group as we can. So we use momenta. They all commute. Just do as much as you can.
So we can also diagonalize sum of Q's. We can't diagonalize both of the Q's-- the Q's and the Q tildes-- because they don't commute. But choose one of them. Choose the Q's, say. I can choose to diagonalize the Q, that makes the action of Q manifest.
So I can choose [AUDIO OUT], which I'll call eta. But this is going to be lambda alpha acting on eta-- oh, sorry-- lambda alpha, eta I acting on eta lambda and lambda tilde. And I have to get the alpha index from somewhere, since it's coming from the lambda.
So that's what the state is. And I can write it explicitly as eta, lambda, lambda tilde is e to the Q bar, lambda tilde acting on, let's say, the negative-helicity gluons.
STUDENT: So, Nima?
NIMA ARKANI-HAMED: Yeah?
STUDENT: The point is to make eta continuous, or--
NIMA ARKANI-HAMED: Yeah, so eta is now a continuous but Grassmann parameter.
STUDENT: Continuous Grassmann?
NIMA ARKANI-HAMED: I mean, it is a Grassmann parameter. But we're going to integrate over-- just as usual on Tuesday, we treated it totally symmetrically with the other variable, but it's just Grassmann.
So this is equal to-- so let me be a little more precise. So there is a Q tilde, alpha dot I-- oops-- eta I. And actually the out vector is what I call W alpha dot, where W alpha dot satisfies the W lambda tilde, W tilde. And W tilde, lambda tilde is equal to 1 now.
But anyway, if you expand this out, what you get is-- so this state eta-- the state eta, lambda, lambda tilde is literally like a particular linear combination of all of these helicity states. So it's literally the negative gluon state, lambda, lambda tilde plus eta I times the negative 1/2 lambda, lambda tilde, because there's four of these, so there's an I index on there, plus 1/2 factorial eta I, eta J, times a 0 scalar, lambda, lambda, tilde. There's six of those, which are index symmetric. Hence, there's I and J plus the other one.
So it's literally some linear combination of the external helicity states, on which 2D acts nicely.
STUDENT: Nima?
NIMA ARKANI-HAMED: Yes.
STUDENT: [INAUDIBLE]
NIMA ARKANI-HAMED: Sure, yes. So now, we have a choice, for any state, we have a choice of whether to label things by eigenstates of Q or eigenstates of Q tilde. So in general, we could talk about amplitudes a function of now eta a, lambda a, lambda tilde a, or something different, which is to label it by eta tilde a, lambda a, lambda tilde a. I do whichever I like.
I can go from this to that representation just by Fourier transforming the eta variables. It's just like diagonalizing Q versus diagonalizing Q, Q tilde. So just Fourier transform eta or eta tilde relates me from one representation to the other representation.
Now, here is just a very beautiful thing. The action of n equals 4 supersymmetry is completely well-defined on these amplitudes-- on those states. I know exactly what n equals 4 does. Well, it's just trivial. I know exactly what Q's do to it, any supersymmetry transformation. Everything is totally manifest.
This is an on-shell n equals 4 supersymmetry. And it's totally well-defined. Contrast it with off-show n equals 4 supersymmetry that no one has found a way of writing n equals 4 that makes that manifest.
In fact, this is-- as I said, it's one of the interesting things about supersymmetry, something that certainly bugs all grad students-- certainly bugged me-- when they first learn it, is that whether or not a theory has supersymmetry seems to depend on the cleverness of the theorist. It isn't the property of the theory. It's a property of the theorist to see if the theorist was smart enough to find the correct off-shell degrees of freedom, auxiliary fields to add in, to be able to see the supersymmetry, even if you forget the auxiliary fields, if we just talked about just the-- integrate out all the auxiliary fields.
The point is that what the supersymmetry transformations are depends on the Lagrangian. That's not true for other normal global symmetries. Like a U1 symmetry is delta phi is I phi for every theory. Doesn't matter what the Lagrangian is. What the symmetry is, is it is what it is.
Not so for supersymmetry. Supersymmetry depends on the theory. So it depends on the cleverness of the theorist, whether or not it has the symmetry. I think this is another hint that there is something wrong with the off-shell way of thinking about this, especially so because this beautiful and symmetric theory that we have for Yang-Mills theory, there is no Lagrangian for it. There is no Lagrangian that makes all of its symmetries totally manifest.
But the actual symmetries, the actual for n equals 4 symmetry acting on the physical object, the scattering amplitude is trivial. There is it. Using these variables, we make it totally trivial.
STUDENT: So--
NIMA ARKANI-HAMED: For n equals 8, it's exactly the same thing. For n equals 8, this is exactly the same thing. I mean, I've never seen the n equals 8 Lagrangian in my life. I mean, I can't bear flipping through so many pages. But it's possible to work on n equals 8 scattering amplitudes. No problem.
I mean, I've written papers on n equals 8, because it's really easy. The supersymmetry is really easy. And not only is it really easy, it's easier than other theories, because now, we don't have to drag these plus and minus indices around. The amplitudes are just beautiful, smooth functions of the Grassmann variable-- of Grassmann and the Poissonic variables.
For example, if I want to go back, instead of doing a helicity sum, what do think I do? Well, there's an eta there. I put an eta there. And I just do integral e to 4 eta. So I'm doing the Yang-Mills. That's it. That's the helicity sum.
STUDENT: Can you write down how to do it where you're transforming Grassmann space?
NIMA ARKANI-HAMED: Oh, it's just the integral e to the-- so integrals are way easier for Grassmann intervals than for ordinary integrals, right? Because they're the same as derivatives. But you just follow the rule that integral d eta d eta is 1. And you just do it. Everything truncates. So I write these as these exponentials just to remind me what they are-- they truncate.
So this is the object of interest. And so let me just end by writing the supersymmetry version of that. Oh, so perhaps I can just say one thing very, very briefly. So what does supersymmetry do? Well, let me say it in words.
A Q supersymmetry transformation, let's say on these guys, because the etas are eigenstates of Q, a Q supersymmetry transformation just multiples this by an overall factor. What overall factor does it multiply it by? It multiplies it by e to the-- well, by e to the lambda eta.
So a Q SUSY acts in this-- if I act on this with e to the zeta Q, I end up getting an overall factor which is e to the power of zeta times the sum over a of lambda a eta a, because Q is an eigenstate. So you see that, in order for this to be invariant under SUSY, in fact the amplitude has a double function of that factor in it.
It's exactly the argument for translations. When you do a translation, you get e to the I-- the translation a or d times the sum over all the momenta. That's what the amplitude is multiplied by. The only way it can be invariant is if the amplitude is a double function of that guy sitting in front. So similarly, we should expect to find the delta of the sum over a lambda a, eta a in terms of everybody.
The other SUSYs, what do the other SUSYs do? The Q tilde SUSYs shift eta. So if I multiply either the Q tilde zeta tilde here, it's clearly just going to commute through here-- the Q tildes commute. And it just shifts to the eta.
So that's what SUSYs do. They re-phase and they shift. And using this fact, it's incredibly simple to prove these statements just as a simple application. How would I, for instance, get the amplitude with all plus?
To get the amplitude with all plus, I have to take M, and I just have to integrate 4 eta 1 up to d 4 eta n of M of eta 1 up to eta n. All of these integrals pick out the eta to the fourth components, which are the plus helicity amplitude. Because they start with minus helicity, so the eta to the fourth components force me to pick up plus helicities.
Now, it's very easy to see that this has an edge. The simple argument is that using the Q tilde SUSYs, I have four supercharges, but I also have two directions-- zeta tilde. I'm doing e to the zeta tilde, alpha dot, Q tilde, alpha dot I. So I have two of these zeta tildes.
So I can use these SUSYs to actually send two etas to anywhere I like. I can translate them to anywhere I like. In particular, I can use them to translate them to the origin. Actually, for this argument, for the all-- well, I can send two of them to the origin.
So let's say I do that. I use my two SUSYs to translate two of these etas to the origin. But if I translate them to the origin, this integral manifestly vanishes, because it doesn't depend on the two guys that I translated. So it's just 0 of the Grassmann equation.
Similarly, for the case with one of them being minus, I just get rid of one of these, say. Once again, though, I can translate two of them to 0, I can still remove one of them, send them to the origin. One of the integrals gives me 0, and it's still 0. So we have to have at least two negative-helicity particles and positive helicity for the other argument.
STUDENT: We really have to go on.
NIMA ARKANI-HAMED: Yes. So finally, then this is what this Parke-Taylor amplitude looks like. It has a delta 8 of the sum over a lambda a, eta tilde a, delta 4 of the sum over a lambda a, lambda tilde a. These are the momenta-- the supermomentum conserving delta functions. And the rest of it is just 1, 2, 3, up to n1. So that's the super Parke-Taylor amplitude.
I should say, the supersymmetric statements indicate lambda, eta, and eta tilde-- sorry-- lambda, lambda tilde, and eta tilde. But if you take M of t lambda, t inverse lambda tilde and t inverse eta tilde, it's just equal to t to the minus 2 M of lambda, lambda, and eta tilde. So again, there's no helicities anymore, plus and minus or anything. But you've got to rescale the eta variables. So that's the supersymmetric version of Parke-Taylor.
So I think we got all the kinematics taken care of today. And tomorrow, I will talk about the-- this is the kinematics that takes care of Poincare symmetry and super-Poincare symmetry. And tomorrow, I'll extend it by talking about the nicest kinematic way of talking about the full conformal transformations and the superconformal transformations.
That'll motivate the introduction of twistor variables. And the analog of this conformal symmetry on the dual side will motivate the introduction of the momentum twistor variables. So we'll talk about twistors and momentum twistors. And then with all the groundwork set, we'll be able to get going with the problems sets. All right, thanks a lot.
[APPLAUSE]
The second in a 5-part series of technical lectures on scattering amplitudes given by Prof. Arkani-Hamed in conjunction with his Messenger lectures on fundamental physics at Cornell University. The focus is application to N=4 supersymmetric Yang-Mills Theory.
