 Okay, for those of you who are looking for the notes, this lecture will follow rather closely lecture five of my GGI notes. So if you've downloaded those already, go to lecture five, which has the same title, the Goldstone-Boson Equivalence Theorem, and you'll be able to follow along. Or, afterward, it's fine. Okay, so in the previous two lectures, I've told you a little about the foundations of weak interactions, the experimental basis for the theory of weak interactions, and the precision electoral weak experiments of the 1990s, and actually even the more recent ones on the measurement of the W mass. And I told you that this picture of SU2 cross U1 for the weak interactions has a great deal of theoretical and observational coherence. That, what I wanted to do in this lecture is to talk about one of the subtleties of the SU2 cross U1 theory, which actually comes up all the time, so it's worth understanding it well. And that is the nature of the longitudinal polarization state of the W boson. So this is, let me maybe state the problem in the following way. In the previous lecture, we wrote down the polarization vectors for left-handed, right-handed, and longitudinally polarized Ws in the rest frame of the W. And these were, so this is epsilon equals zero here, one over the square root of two, zero one minus i zero, and zero zero zero one. So this is just simple quantum mechanics, the vectors that represent the various states of spin one, S3 equals plus one, zero, and minus one. Now what I'd like to do is to boost these vectors so that I have a W with a momentum vector E00p. So I'm going to boost very strongly in the end, in the three direction. And the polarization vectors for this W boson are given just by the boosts of these vectors. And for the top and bottom ones, you can immediately recognize what they should be because the boost has no effect on a transverse vector. However, in the middle case, there's a surprise. What you get is P over Mw, zero zero E over Mw. And what's interesting is that this vector, asymptotically, as the momentum gets large, becomes proportional to P mu over Mw. That is to say it's parallel to the momentum vector of the W boson. Now all of these vectors over here and over here satisfy epsilon dot P equals zero. They have to be orthogonal to the momentum of the W boson. But remember, this is Minkowski geometry. So in Minkowski geometry, as a vector gets light like, it can be very close to another almost light like vector. And yet at the same time have zero dot product with it. So the dot product of these two in Minkowski geometry is zero. Nevertheless, there's some weirdness about this because this can potentially be a very large number. If you have a one TEV W boson and Michelangelo and his colleagues at the LHC see those all the time, this number is 10 or 12. And so that can be an enhancement of the amplitude and then the rate that can get squared. And let me just give you a simple example of this. And it's one that we'll come back to a little later in the lecture. If I imagine that I have a virtual photon that decays to a W plus W minus pair, the structure of this amplitude will be something like E epsilon of the W plus dotted into the epsilon of the W minus times K minus minus K plus mu. That is to say there's a piece of the vertex. This is the whole Young Mills vertex, but there's a piece of the vertex, which is just like the coupling to a scalar particle times the product of the two polarization vectors. But if the polarization vectors are approximately P over MW, this thing is P plus dot P minus over MW squared, which is to say it's about one half S over MW squared. And then this is in the amplitude and it gets squared. So potentially you have a factor which is energy squared in the amplitude larger than what it can be. Now for a scalar field, if you drop this factor, that's exactly consistent with unitarity in the S wave. So this contribution is going to violate unitarity unless we do something about it. So this is kind of, this is definitely weird. This is definitely the correct polarization vector, but it potentially gives effects that are on physical grounds much too large to expect. And the question, the next question is, well maybe there's some way that this large number here, or which is present in the square here, maybe that always cancels. And the answer is that it doesn't always cancel. Sometimes the large factor appears, sometimes the large factor does not appear, and you have to know on physical grounds when that's gonna happen and when it's not gonna happen. And this enters many interesting processes at high energy, so it's something that we'd really like to understand very well. Now, what's the resolution of this? Well, it probably won't surprise you if I say that the resolution of this has to do with gauge invariance. The gauge invariance and the compatibility of gauge invariance and massive vector bosons is, if you like, very marginal. And it requires some special properties of the theory, in particular spontaneous symmetry breaking through the Higgs mechanism in order to make those two things compatible. And so I think if you understand the role of these longitudinal W bosons, it's really that somehow encodes the physics of the Higgs mechanism in a way which is very interesting and important. And so I'm gonna devote this entire lecture to the properties of these longitudinal W bosons. I'll give you first an example where this term doesn't cancel, and I'll explain the reason why. And then I'll give you an example where it does cancel, and I'll explain the reason why. And then there'll be another application that I hope is interesting to you. So maybe first I should talk about some properties of this term. So when you study the Higgs mechanism, you find out that roughly you start with a massless vector boson, and the massless vector boson only has two polarization states, left and right. So this is what you find in electrodynamics. You solve Maxwell's equations, and there are only two polarization states for a propagating wave. The ones with, they always have transverse polarization, that is the E field is always transverse to the direction of motion. And you can write the two possible polarization states as left-handed and right-handed in states of definite angular momentum. Now what you have to do in order to get a massive, so a massive boson, necessarily has three polarization states because you can go to the rest frame and take one of the transverse polarization states and just rotate it by 90 degrees, and now it has zero, three component of angular momentum, so it's the longitudinal state. So another state has to be introduced. And the way that that's introduced, as one would explain to you the first time you see the Higgs mechanism, is you add the Higgs field, you spontaneously break the symmetry. The Higgs field gives up a goldstone boson because the Higgs field, before you gauge it, has a continuous symmetry, and that symmetry is broken, and there must be a goldstone boson. And then somehow the vector field eats the goldstone boson, and this is transmuted into the longitudinal component of the field. Now, when the vector boson is at rest, it's very unclear how to divide the three polarization states into two that are, if you like, primordial, and a third one which is, came from the goldstone boson. But I hope you find it intuitive that when the vector boson is highly boosted, one can actually make this distinction. A very boosted particle resembles a massless particle. So I think it's natural to say that these two polarization states are mainly associated with the pieces that came from the original massless vector boson. But that means that this polarization state must have been the one that represents the goldstone boson which was eaten. Yes. This expression, if you just take that and you compute, let's say E plus E minus to W plus W minus, it would violate S wave unitarity. Yes. We're gonna see what happens. You have to wait a little while. Okay, so now this has to do also with the subtlety of what gauge you work in when you do calculations in a spontaneously broken theory. In the Feynman-Tolft gauge, the vector boson propagator is something like this, minus i g mu nu over q squared minus m squared. And this has four polarization states. So in addition to a l, a r, and the longitudinal state, you add a time-like state. So this thing has four non-zero eigenvalues. The time-like state has negative norm. So it actually has negative probability. And the goldstone boson is still around. And then word identities or BRST word identities tell you that these things always cancel their contributions, but it's a very subtle cancellation. On the other hand, in unitarity gauge, a massive vector boson propagator has the following structure, q squared minus m squared, g mu nu minus q mu q nu over m w squared. And this quantity here is the projector onto the three physical polarization states. So this epsilon mu epsilon nu with these three vectors adds up to this polarization tensor. And here again, you can see there's potentially a problem. This thing is enhanced by the momentum squared over m w squared. And if that somehow gets caught when you compute the Feynman diagram, you're gonna get an answer, which is two, three, 10 times larger than you expect. Now, Toft and Weltman proved that these two calculations always give the same result for gauge invariant quantities. But nevertheless, it's a little hard to see intuitively how that can possibly happen. So people thought about this very hard. And so this is assisted by a very interesting result called the Goldstone-Boson Equivalence Theorem. So the Goldstone-Boson Equivalence Theorem was first denunciated soon after the Toft and Weltman work by Cornwall in Tectopolis and by Viunakis. And what they said was there should be a theorem that makes manifest what I was saying about this correspondence over here. Namely, let me try and write this theorem. Let me write the theorem for a w plus very carefully. So let me imagine that I take my Higgs field and I write that as pi plus one over the square root of two v plus the Higgs plus i pi zero. So pi plus, pi minus and pi zero are the three Goldstone Bosons, which are in the Higgs field of the standard model. And they're eaten when the w bosons get mass. And what Cornwall, Tectopolis and Viunakis said is there's a theorem in spontaneously broken gauge theory that says that if I wanna calculate x goes to y with the emission or absorption of a very high energy w boson, which is longitudinally polarized, the amplitude for that is equal to the amplitude for x goes to y plus the emission of this now unphysical Goldstone Boson up to corrections, which are of order Mw over the energy of the w. So I think you can see that this is quite non-trivial. This thing can have the P over M enhancement. This term here doesn't because it's a scalar emission. But nevertheless, the gauge invariance, the structure of the theory given by gauge invariance should guarantee that this relation is true. So I'm not gonna prove this theorem because the proof is rather technical. But if you'd like to see the proof, there's a very beautiful paper by Chanowitz and Guy Ard. The references in the notes, maybe I should just give you the date if you wanna look it up, 1985, in which they kind of pull out the full structure of BRST and they prove this theorem for any number of w boson emissions in the full renormalized SC2 cross U1 theory, actually in any renormalizable gauge theory. So this is a true theorem. And I wanna try and explain it to you through some examples. So now what I'd like to do is to show you in some calculations how this correspondence works, how we might or might not have an effect of this term and how we interpret that in terms of the emission of a Goldstein boson. So let me start with the simplest case or a relatively simple case, which is the decay of the top quark. Now, in his slides, I think Michelangelo didn't quite get to this today, but in his slides he gave you this as an exercise, but I didn't see the slide, so now I'm gonna do the homework for you. So at least you can thank me for that. So let's think about the decay of the top quark. The top quark decays to a w boson and a bottom quark. This vertex here is the v minus a interaction, u dagger of b left, sigma mu u of t, epsilon star of the w boson. So that's the v minus a interaction that we've been talking about all week. To get an accurate result, you should keep the bottom quark massive, but I'm going to treat the bottom quark as massless, which is fine for what I wanna tell you about. And that's gonna make the calculation a lot easier. So the spinners, let's set up the top quark. So it's spinning in the plus three direction. So the ut left will be the upper component of this. Let me try and get the rest of this oriented correctly. Ah, please excuse me. I'm gonna do something a little different. I'm gonna take c and c here for an arbitrary spin direction. So it's just easiest if we let the w boson go along the three axis. The bottom quark go in the opposite direction and point the top quark spin in some relative orientation for which this angle is theta. And then u of t left is equal to the square root of mt. U left of t is the square root of mt times cosine theta over two, sine theta over two. So that's the effect of this rotation angle. And the u left of the b will be what you get for a left handed particle going in the wrong direction. So it'll be the square root of two eb times one zero. Okay, now we're all set. And now we just have to compute this matrix element, dot it with the w polarization vector and we can calculate all of the helicity amplitudes. So our three choices for the w polarization vector are unfortunately the things that I erased. So let me not write them again. The right handed, left handed or longitudinal polarization. Okay, so let's now calculate the helicity amplitudes for these decays, there are three of them because the top quark can potentially go to a right handed w boson, a left handed w boson or a longitudinal w boson. And so let's calculate those in order. So first of all, for the w right, we get ig over the square root of two, the square root of two mt times eb from the product of these factors, a one zero, a one minus sigma and cosine theta over two, sine theta over two. And then this thing dotted into the right handed polarization vector. So sigma dot that vector, it's in the final state so I have to complex conjugate it. So that is going to give zero, zero, two and zero and one over the square root of two. And so unfortunately this doesn't have any overlap with that and so for this process we're gonna get zero. For the w left, it's the other way around. The polarization vector, the epsilon star is one i zero. So we're gonna get this subject and now there's an overlap and we're going to get then ig over the square root of two, there's a square root of two as you see here. There's a factor of sine theta over two. There's this factor of the square root of two mt. Okay and I think I actually got that right. Okay, now finally the longitudinal polarization vector. So sigma one sigma dotted into the longitudinal polarization vector which you remember is p over m zero zero e over m is going to be this matrix. P plus e over m, please excuse me, this is mu, mu. So there's an extra minus sign from the Lorentz metric when I dot this with this. P minus e over m zero zero. And now I have to take the matrix element of this thing between the spinners which I wrote here. It's going to pick out this term. So I'm going to get ig over the square root of two. There's no extra square root of two, a two mt eb. P plus e over mw. And the cosine of theta over two. And so now we can directly simplify this if you just remember again the kinematics of a decay of a massive particle to a massive and a massless particle. The e of the w is mt squared plus mw squared over two mt. The p which is also the energy of the bottom quark is mt squared minus mw squared over two mt. So this factor here is actually mt squared minus mw squared, the square root of that. And this factor here which is the sum of the energies of the w and the bottom quark is the mass of the top quark. So this is now developing in a somewhat interesting way. The three amplitudes have the following structure. They have an ig over the square root of two. They have this factor of mt squared minus mw squared to the one half. And then for right, zero and left, this is zero. This one is cosine theta over two and it has an extra factor of mt over mw. This one is sine theta over two with a square root of two in front. Can I explain this pattern in terms of angular momentum? Well, most of the factors can actually be explained with angular momentum. The top quark, let's say looks like this and let's consider the forward production of a w boson. The bottom quark is always left-handed. So it's spin points that way. Here, let me leave this ambiguous for the moment. The w, the spin depends on what polarization state I pick. So, oh, please excuse me, that way. So if I pick a right-handed w boson, this is j three equals three halves, that plus that. So I can't produce that state from a spin of half object. So I get zero. That's the result of angular momentum. If I have a left-handed w, that's compatible with the left-handed spin of the top quark, the backward spin of the top quark, and you get the sine theta over two distribution. That's perfect. And if I have a longitudinal w, then I get a forward w, so I get a cosine theta over two. So far so good, everything's working out. The only problem is this factor here, which I think is, at least naively, it's totally unexpected. So let's not just put this together. The total width of the top quark is one over two mt, one over eight pi, the integral d cosine theta over two of the amplitude squared. And when you work it through, you get the following result. Alpha over 16, alpha w over 16 pi, the mass of the top quark, because there's the square of this factor, a one minus mt over mw squared. Oh, please excuse me. There's a two p over mt from phase space. So you get one factor of this from the square of that. You get another factor of this from phase space. And then there's the following factor. mt over mw squared from this term and two from the left-handed polarization state. And I think I now got this completely right. Good. Okay, now, what we expected was that the width of the top quark should be something like alpha w over something times the top quark mass. But what appeared is an extra factor of mt over mw squared. And that's a little of a surprise. Maybe if you'd done this calculation in another way, you would still have been surprised, but there would have been a different origin. Namely, if you'd used the polarization sum, epsilon, epsilon star, equals g mu nu minus q mu q nu over mw squared, you would have found out that to your surprise, this thing didn't cancel, but instead it actually gave the largest term. And this thing would turn into this expression in the final rate. So the first question is, is this real? And the second question is, can we understand it? So first of all, I'd just like to say that this enhancement here, the enhancement of the longitudinal polarization state, is exactly what's predicted by the Goldstone-Boson equivalence theorem. Because the Goldstone-Boson equivalence theorem tells you that the production of a longitudinal w boson, the amplitude is equal to the amplitude for producing the Goldstone-Boson, which was eaten when the w boson became massive. But interestingly, this vertex here is g over the square root of two. But this vertex is yt, the top quark-ukala coupling. And this diagram is naturally enhanced over what you would naively expect for that one by yt over g over two in the rate it would be squared. That would be m top squared divided by g squared over two, sorry, yt squared. Now m top is yt over the square root of two, mw times v, gv over two. So if I just put v squared everywhere and then divide by two, this is actually m top squared over mw squared. So the Goldstone-Boson equivalence theorem explains this enhancement of the longitudinal polarization state because the very fast w, the highly boosted w, which is emitted in top decay, when it's longitudinally polarized, it isn't really a w boson, it's really a Higgs boson. And a Higgs boson couples more strongly to the top quark than a w boson because it couples with the top quark-ukala coupling instead of the su2 coupling. And so in that way, the Goldstone-Boson equivalence theorem, well, first of all, tells you that this extra term, q mu q nu over mw squared, doesn't cancel out. It's actually the most important contribution and it's a contribution that actually is essential to getting the right answer for this expression. Okay, well now the most interesting part. Is this actually correct observationally? Well, interestingly it is because, and we can test it in the following way. You remember that in the previous lecture, I told you that you can measure the polarization of a w by looking at the direction of the lepton from the w in the w frame when it decays. For a right-handed w, so this is now the cosine of this angle, you go to the w frame, the w is moving this way, you boost it to rest, the lepton in that frame goes off this way and you measure that angle. The distribution is one plus cosine squared theta for a right-handed w, one minus cosine theta squared for a left-handed w and sine squared theta for a longitudinally polarized w. And so from this example, what you would predict is the following. The branching ratio for the top to go to a longitudinal w is mt squared over mw squared divided by mt squared over mw squared plus two. And when you put in the numbers, that's about 70%. So if I made this distribution for the leptons from w bosons from top decay, then I should see a picture that looks like this. Mostly the central production and then just a bit over here from the left-handed production. So that's what that function should look like. The zero here is the absence of a right-handed coupling. So pure v minus a, there's no right-handed part of the coupling in the original Lagrangian. If there were a right-handed coupling so I could produce b-rights, then I would get something over here. So the zero is equal to the absence of b-right in this decay. Okay, now let's look at the data. So in the Tevatron and the LHC experiments, you can collect top quark events. Here's a very beautiful top quark event collected by Atlas. Top quark pair production events are especially straightforward to analyze. So in this case, there's one top quark, which is going up, which is decayed to three jets. And there's an anti-top quark going down which is decayed to a b-jet, a w which is decayed to an electron, and a neutrino. And so what you see is this jet coming down actually has a b-tag of the kind that I talked about in the last two lectures. Up top it's resolved into three jets and so this is the kind of signature that you're interested in for top quark production. Actually, if you look at it in the theta versus rapidity plane, that's the picture given here. Unfortunately, it's upside down. I don't know why they did that. I know who to blame, but I won't tell you. You see the dots are the calorimetric energy in the phi versus rapidity plane. And you see there are three jets which corresponds to the one up there. And then one jet and the lepton candidate which is shown at the top. Is there a better, yeah, this is a better view of it. Okay, so now what's interesting about this event is that we basically measure everything but the neutrino. If we have an idea that this is a top quark pair production event, then the neutrino, there's a lot of information that we have to reconstruct the neutrino because the lepton and the neutrino have to sum to the W mass. We measure the missing energy so we know already the transverse momentum of the neutrino. The electron and the neutrino have to sum to the W mass and the whole thing has to sum to the top quark mass. So I think you can see that from the information given, you can reconstruct all the four vectors, do the transformations, go to this frame and measure this distribution. And so let me show you a couple of pictures of distributions measured in this way. Here's the one from the tevatron and you see it roughly has this structure, although maybe the resolution leaves something to be desired, here's the one from the LHC and oh, please excuse me. I didn't, maybe I didn't draw it right. This is totally symmetrical. There's a contribution like this so it should look like that. And with some smearing, you can see that that's exactly the structure that you find. I'm sorry, there is a non-zero probability here to have cosine theta equals plus one, but that's entirely due to detector resolution and if you take the curve there and smear it with the detector resolution of Atlas, you get the histogram to which the data is fitted. So really one can directly measure these W polarizations and show that this number is indeed realized in the experiments to a couple percent accuracy at this point. Okay, well this is incredibly cool because now we've seen an example where the extra term, this enhancement that comes from the propagator of the polarization sum of W bosons including the longitudinal polarization state actually gives an enhancement in the amplitude. However, the next thing I have to tell you is this doesn't always happen and there are times when you have to actually cancel this enhancement. And probably the one which is easiest to understand is the one that I started with down here on the blackboard which was the case of E plus E minus to W plus W minus. So let's go back and think about that case in a little more detail. So E plus E minus to W plus W minus is generated by this Feynman diagram but of course it's not the only one. One can also have a Feynman diagram which involves virtual Z exchange and there's also a diagram where you have virtual neutrino exchange like this. So actually the full amplitude is given by the sum of these three diagrams and that's the thing that we should now try to understand. So let me consider first the case which is a little easier which is if I have a right-handed electron beam then the right-handed electron doesn't couple to the W in the neutrino and this diagram is absent and I just have to consider the sum of these two diagrams. Now each of these diagrams individually gives a result which is in the amplitude S over M W squared larger than you would expect. I already proved that to you. So now what I'd like to ask is what happens when you take these two diagrams and add them together is there some mysterious conspiracy which is going to cause these terms to cancel. So let's try and work this out pretty carefully. So there's a minus IE for the photon diagram and IE which comes from this vertex two times the square root of two times P momentum in the initial state times the right-handed polarization vector of P so that's just this vertex here. Two times the square root of two times the square root of two E electron is what should appear there. And then there's a minus I over S and then there's the Young Mills vertex which is epsilon minus dot epsilon plus K minus K plus plus epsilon minus star minus Q minus K minus dot epsilon plus plus epsilon plus star Q plus K plus dotted into epsilon minus. So that's the full structure of the vertex that you get from Young Mills theory. Remember it has three terms which have basically the structure. And now what I'd like to do is to evaluate this expression using the polarization vectors for extremely boosted longitudinal W bosons. So now I'm going to consider these W bosons as longitudinally polarized and I'm going to put in epsilon zero minus is equal to K minus over Mw and epsilon zero plus is equal to K plus mu over Mw. And so for this I'm going to get as before two S over Mw squared. So that's the thing that I had before which was dramatically enhanced. And actually when you work it all out the result of all of those diagrams is also proportional to this structure. It's S over, sorry not two S, S over two Mw. It turns out that these terms are equal to minus double that. So it turns out to be minus S over Mw squared but still with this funny enhancement which yes is unitarity violating. However what's interesting is the structure of this diagram. Remember this is now an E right. So that diagram has the following structure. It has a one over Q squared minus Mz squared, sorry one over in this terminology S minus Mz squared and now you have to get all the factors right in this expression. There's a minus Sw squared for the, divided by Cw squared, Sw squared for, sorry divided by Cwsw. So this is the Z coupling of the right-handed electron. And then if you ask what this vertex is it has a factor Cw over Sw. And all of this is very cool because what you see is that all these factors cancel. There is a leftover minus sign and so these two expressions cancel asymptotically in S. And so at the end of the day what you end up with is something which is of order Mz squared over S squared for that expression, for that expression plus higher orders. And so interestingly this S squared cancels the additional factor of S here cancels the numerator factor of S here and we get something with the usual dependence. And when you work out all the details what you find is that this is equal to minus i E squared two E of the electron and extra factor of root two, epsilon right of P dot k minus minus k plus which is now exactly the amplitude for producing a scalar particle, sorry, times one over S. So that's now the amplitude for producing a scalar particle. And then there's an extra factor that let me just, oh I'm sorry, I left out something. These Mw's I didn't keep. So there, oh yes there's an Mw squared. So there's an Mz squared over Mw squared which factor then becomes one over cosine squared theta w. Good. Now, why is that a good thing? Well let's think about the process that this is related to by Goldstein-Boson equivalence. It would be the following process where I take a right-handed electron annihilated on a left-handed positron. There's a gamma plus z star and I produce a pair of Goldstein-Boson's pi minus pi plus. In the high energy limit, I can actually even evaluate this more simply by remembering that I can change the basis in the high energy limit to Su2 and U1. And so I can equally well write this as linear combinations of A3 and B. But the right-handed electron doesn't couple to A3, it only couples to B. So in fact the diagram simplifies to one with a single exchange which would be an exchange of a U1 gauge boson which it doesn't matter whether it's massive or massless because we're at very high energy. And then I'd just like you to remember that both of these couplings are one over E over CW which is G prime. And that actually explains the complete structure that we have here. The QED scalar production, one over S, and the one over CW squared is just the product of the extra factors in these couplings. So how that works, it's like magic. But that it has to work, that's guaranteed by gauge invariance. And in the end what you find is that the sum of these two diagrams for a right-handed electron is just exactly equal asymptotically to this diagram which is the production instead of W's of the Goldstone bosons that the W's were supposed to have eaten when they became massive. Okay, now what happens when you do this with the electron left? Well now this cancellation doesn't occur because there's an extra term here, a half minus sine squared theta W. So that a half term doesn't cancel out. But in the notes it's explained that that extra half is canceled by this diagram. And so in the end it works out again that the unitarity violating terms all magically cancel. Now unfortunately the simple approximation that I used here is not sufficient in the left-handed electron case to compute the next term which would verify that even for a left-handed electron, so for a left-handed electron we would wanna show the following. That the set of three diagrams is proportional to a three plus B, left-handed electron and pi minus pi plus. But let me just say that if you use the exact expression for the longitudinal polarization states and you get a very big piece of paper you can work it out and it works perfectly. So I encourage you to do that but it's not a five minute exercise. It's a serious exercise to get the next term. But at least the cancellation you can see very clearly of this term against the effect of this diagram using this approximation. Okay, well once again there's, we can ask the question what does experiment have to say about this? So actually there is experimental data relevant to this question because at LEP the energy of the accelerator was increased to 200 GeV, actually in the end to 208 GeV and they actually observed this process E plus E minus to W pairs. Actually here is an event where what you see are two jets from 1W. This is interpreted as a pion from a towel, maybe a row from a towel with the other W decaying to towel new. And let's see on the next slide I think I have a four jet event from W pair production in which both Ws decay electronically. So LEP two observed a large sample of E plus E minus to W pair events over a range of energy and they were able to actually take the sum of these diagrams actually even the one loop electrolyte radiative corrections were provided and compare that to the rates measured in their experiment. The result is really interesting and here it is. So the things with error bars are the data. The blue curve is the standard model prediction which incorporates this so-called unitarity cancellation. So the standard model prediction goes to if there were only S wave it would peak and go down again. The fact that this diagram has higher partial waves causes the cross-section to slowly increase logarithmically. So that's what you see with the blue curve. And these things YFSWW and Raccoon are programs that incorporate the one loop electrolyte corrections. But if you computed this at the tree level the curve you would get would be very similar. On the other hand, if you leave out this diagram you get the purple curve here. And let me just, yeah if you leave out this diagram you get the purple curve which is wildly discrepant from the data and is heading upward in a direction that clearly violates S channel unitarity. And if you leave out both of these diagrams you get the blue curve which is even worse. So one can say then that the cancellation which seems extremely weird and magical when you actually do it with Feynman diagrams is verified experimentally. If you are a theorist you would say well that cancellation had to happen because of the Goldstone-Boson equivalence theorem. Whatever mess you get over there at high energy it has to turn into this or this. So it's interesting, this Goldstone-Boson equivalence theorem allows you to intuit when you get a large effect from this longitudinal W polarization vector and when those effects have to cancel. You just replace the longitudinal Ws by the Goldstone bosons that they ate. Think about that process. All of a sudden everything becomes much simpler and you can see your way through it. And let's just say there are many examples of high energy physics and as we go to higher energies at the LHC there'll be more and more where this kind of thinking is going to be just tremendously valuable to you. Okay, I have another 15 minutes and I'd like to discuss one more topic which is relevant to this and which is quite interesting for LHC physics. And that is, I'd like to discuss the Parthons splitting functions for the W. Now Michelangelo in his lectures talked about the Parthons splitting functions for quarks. Namely a quark can, if you have some hard reaction that involves a quark the quark can from the initial or final state radiate a gluon and what he showed you is that when the gluon becomes collinear with the quark there's a logarithmic enhancement of the cross-section and that's given by these D-Glap or Alter Aliperisi splitting functions. There's some alpha S over pi, D P T squared over P T squared, D Z where Z is the momentum fraction of the quark carried away by the gluon, one plus Z squared over Z, there's a color factor of four thirds. And so one gets an expression like this for the probability that a quark will radiate a gluon on the way in or actually the same probability that the quark will radiate a gluon on the way out as the first stage of forming the jets that Michelangelo was talking to you about in this morning's lecture. Now, if you have a really energetic quark then that quark can radiate a W boson and then go off and participate in some hard process. Actually, another thing that we're gonna be very interested in is that instead of the quark going off to participate in a hard process the W can go participate in a hard process. And in fact, if you have another quark over here so this is let's say an up quark which turns into a down quark and a down quark which turns into an up quark you can have W plus W minus that you radiate off the quarks these things can then collide and give you a W plus W minus scattering process. This process is called W fusion. It's extremely important in Higgs phenomenology because one of the things that you can have here as we'll discuss in the next lecture is W plus W minus coming together and creating through this vertex a Higgs boson. So one would really like to know what is the parton distribution of a W in the proton in the language that Michelangelo was using and that'll be generated by the splitting of a W and almost collinear W from the quark. Well now there's an interesting question that one can ask. What is the polarization of the W which is radiated from a light quark a U quark or a D quark? Now certainly if you think about very high energy you ignore all the masses a W aside from the fact that it's SU2 rather than SU3 would be just like a gluon. You would radiate a transversely polarized guy. It would probably have the same kind of distribution function that you would have for a gluon. The problem with that is that the Higgs boson which is after all a Higgs boson couples mainly to longitudinally polarized Ws. Because these are the things that in the high energy limit are equivalent to the pieces of the Higgs sector that the W absorb to become massive. And so a very interesting question is is it possible for an up quark which is a very light quark to radiate a longitudinal W which is then present in the structure of the proton and available for a collision with a longitudinal W coming from the other side. Naively you would say this is zero because you use the Goldstein-Boson equivalence theorem. What appears here is the up quark Eucala coupling and that's suppressed by something like M up over V which is a number which is 10 to the minus five and then you square it. However, it turns out that this is not right. And this was clarified in another beautiful 1985 paper by Sally Dawson. And basically the reason it's not right is because the W emission process should really be viewed in the closer to the rest frame of the W where the polarization states are more difficult to distinguish. And so let me just sketch that calculation for you. And in the end we'll come to the splitting functions for the various polarizations of W's that come from emission from a quark line. I think Michelangelo is like a real QCD calculator and so he's used to using very automatic methods to generate QCD amplitudes and these are very much like QCD amplitudes. I'm more of an amateur in this field so I'm just gonna use these stupid helicity methods but maybe it's a little easier to understand. So let's sketch the calculation in the following way. I'd like to consider an incoming quark with momentum P, a W boson going off with momentum Q and so this is let's say an up quark, this is a W plus, a down quark going off with momentum K. The W is going to be viewed as approximately collinear with the up quark to try and give some kind of enhancement of the production rate. So let me make approximations which are relevant to that situation. So let's now write the momentum vector as follows, P which is the on-shell momentum of the initial quark is EEQ carries away a momentum fraction Z. So Z is between zero and one. It'll have some transverse momentum. I'm gonna make that in the one plane so this component will be zero. This component here will be ZE and there'll be some correction to that which I'm gonna write in a moment. K by energy conservation has one minus Z here. By momentum conservation it'll have the opposite PT. Here I have a one minus Z, but it's gotta be on-shell. So the first correction to that will be PT squared divided by two times one minus Z times E. So now to order PT squared, this momentum is on-shell at zero mass. And now for reasons of energy conservation, I have to write the opposite here, namely plus PT squared over two times one minus Z times E. And this guy is off-shell, but please notice in this diagram that guy is off-shell because it's in an immediate state of some more complicated Feynman diagram. Okay, now what is the propagator in this Feynman diagram? It's one over Q squared minus Mw squared. So that's equal to one over minus PT squared. And then minus PT squared, then the square of this four vector gets another term, it's minus PT squared over one minus Z and minus Mw squared. So it's space-like as the kinematics requires. And when you put it all together, it is PT squared plus, sorry, over one minus Z plus Mw squared. Sorry, did I get that right? Oh, sorry, Z times this puts a Z here. Yeah, now I got it right. Okay, very good. So now we're in excellent shape. Now I've got all the kinematics straightened out, now I just have to compute the amplitudes and that's pretty straightforward. Let's see. Time is running out, maybe I just need to sketch the calculation of these amplitudes, the details are all in the notes. The matrix element is Ig over root two, the u for k, sigma mu, these are all left-handed particles, u for p dotted into epsilon star of q. Now for these elements, I just take the relevant polarization vectors. So u of p is a left-handed quark polarization vector, two e, two p, actually times zero one. u of k is two p times one minus Z, that's the energy of that particle, times a rotated polarization vector. So it's one here, but up here it's pt divided by two times one minus Z times e. You rotate through a half angle, so that's the origin of this factor two. And then the polarization vectors for the w are one over the square root of two, zero one i, zero. But now you have to again rotate the w into its correct orientation. So that will give here a factor of pt over Z times e. And sorry, it's minus here q over mw, pt over mw, zero, Z e over mw. So that's the rotated, I guess I left out a Z here. That's the rotated longitudinal polarization vector. Okay, so now put this, this, this into here and crunch a little and outcome the following results. The matrix elements are i, g, the various cases are left, right, and zero. They're in a different order, please excuse me. Here you get the square root of one minus Z. And here it turns out you only have to work to first order in pt, so that's all I'm going to do. pt over Z times one minus Z. Here you get the square root of one minus Z times pt over Z times one minus Z, and there's an extra one minus Z that appears in the numerator. And down here you get minus one over the square root of two, the square root of one minus Z, times mw divided by Z. So it turns out that you get non-zero results for all three entries. This one, when you combine it with phase space and square it, will give you something analogous to this factor one that appears here. This one, the right-handed longitudinal has an extra factor of one minus Z. So it'll give the one minus Z squared here. So for a left-handed quark radiating a gluon, this is the left-handed gluon, this is the right-handed gluon. And then there's going to be also a contribution from the longitudinal polarization state. And now let me just try and write more explicitly what this works out to be. So let's consider a process which is the one I was discussing before where the quark comes in, the quark goes out. There's a virtual w here that collides with something x on this side and turns into y. And by suitably manipulating the cross-section, you can then get the following formula. Sigma of qx goes to qy is approximately in this collinear limit equal to the integral dz. The integral d2pt, there is a pt squared because here there are pt squared. Let me leave that aside for a moment. There's a numerical factor and there's the denominator which I wrote over there. pt squared divided by one minus z plus mw squared squared. Then what comes here are the squares of these amplitudes. There will be basically a pt squared and a one and some factors of z. There will be a pt squared times one minus z squared and some factors of z. And for the longitudinal part, there'll be an mw squared divided by z squared. So now when you simplify this expression, the following thing is gonna happen. Over here, we had a logarithmically divergent integral pt squared over pt squared. So what you get is a logarithmic distribution of pts for the transverse parts. And you see that that's also forming over here. It's dpt squared times pt squared over asymptotically pt to the fourth. And so once again, this is a logarithmic distribution of the integrals. But for the case of the longitudinally polarized, we only get mw squared. So the pt is limited, the integral is finite. And so that eventually is the end of the story. What we're going to get is, oh, and sorry, all of this times the cross section for x for w plus x going to y. And so when you simplify this and clean up all the numbers, which is done in the notes, what you find is that this is equal to the integral dz of a w parton distribution in the proton times the cross section for w at the momentum zp plus x going to y, where the parton distributions are given by the following expressions. Fw of z is equal to alpha w over four pi for the left-handed case, the integral dpt squared times pt squared over pt squared plus one minus z times mw squared. And then the rest of it is the alter eye per easy form, one over z. Here alpha w over four pi dpt squared pt squared, sorry, squared over pt squared plus one minus z mw squared times an extra one minus z squared over z. So that's exactly analogous to what we found for the gluon case. And then for the longitudinal case a somewhat different structure, alpha w over eight pi times the integral dpt squared times mw squared divided by this denominator times one minus z squared, it turns out over z. Well, very interesting. These two expressions have logarithmically divergent distribution. So there's a logarithmic distribution of pts that goes out as far as you like. When the pt gets really large, we're in the regime where the Goldstone-Boson equivalence theorem applies. You can't radiate a longitudinal w Boson out there. But when the pt is small and comparable to mw, you're in the situation where it's like you're in the w rest frame and you get a non-zero result. And in fact, you can actually do this integral and what you find is a distribution of longitudinal memento, fw of z is alpha w over eight pi times one minus z over z. So that's the longitudinal distribution of longitudinal in the sense of the beam axis, longitudinal in the sense of polarization states with w's in the proton. It's not enhanced, but it's order alpha w. So it's plenty big. And then one has some fuel to provide w fusion reactions at a high energy proton collider like the LHC. The transverse momentum associated with this function is a transverse momentum, which is strictly of order mw because as I say, this is a convergent integral for large pt. And so using this function or maybe some more exact computation, one can then work out the rates of w fusion processes and try and use this to compute the rates of reactions. What's especially interesting is that this is the longitudinal w, which is the equivalent of the Goldstone boson eaten by the w. So this is literally a Higgs sector reaction. Two particles from the Higgs sector coming together in a high energy collision to form presumably anything else that exists in the Higgs sector. And so this is a tremendous tool to be exploited at the LHC and at higher energy proton colliders. We just raised the energy of the LHC, so now we can see how many of these reactions we can collect. Certainly they already are shown to produce Higgs bosons and maybe at higher energy, they'll produce surprises as well. So that's the end of today's lecture. Thank you very much.