 These lectures, I'm gonna give you four lectures on weak interactions. The outline, well let me tell you a little what the goals of this lecture series will be. Giovanni asked me to kind of start slow in these lectures and I thought it might be interesting instead of going into great theoretical detail to concentrate in these lectures on how we know what we know. So the main focus of today's lecture will be V minus A theory and basically the basic structure of the weak interactions. Tomorrow the lecture will be about precision electoral week. Wednesday will be a only slightly more theoretical lecture. It's about the Goldstone-Boson equivalence theorem and then on Thursday we'll talk about the properties of the Higgs Boson and what's known about the Higgs Boson already experimentally. So that's the outline. If you think the level of these lectures is too fast, please come up afterward and let me know. If you think the level of these lectures is too slow, then I'm gonna be around all week so just ask me anything and I'll see if I can answer it. I would like to just take the first minute to expand a little on one of the things that Michael Angelo talked about. It's a historical note. You should know that this beautiful theory of the charm content of the proton that he described to you first appeared in the PhD thesis of a fellow named Edward Whitten. I don't know what happened to him after that but this contribution was very important for phenomenology. Okay, so let's then begin a little, the sketch of the history of weak interactions. There's some very important dates in the theory of weak interactions. 1931 was really the starting point. Well, of course there's a date up here, 1896, when Becquerel first discovered radioactivity and radioactivity was a mystery for a long time. The physics community then was much smaller, things progressed much more slowly. It took a long time for people to realize that beta decay was very different from alpha decay. In alpha decay you have some process where a nucleus A will go to B and emit an alpha particle and B would be some specific nuclear resonance. In beta decay you would have A plus B going to a beta particle and electron but this electron, it took some time for people to realize, couldn't be explained by a series of resonances and it had to be a continuous spectrum. And so in 1931 there was the famous letter of Pauley which begins, dear radioactive ladies and gentlemen, please raise your hand if you've read that letter. Ah, very good. Those of you who haven't Googled it tonight, it's very amusing. In which Pauley postulated that there must be an extra unseen particle here, now the anti-neutrino. Very soon after that Fermi formulated the Four Fermion theory of beta decay and one could really get started trying to formalize how the weak interactions worked. But again, there wasn't a lot of progress until another very important development in 1956 when Lee and Yang pointed out that the weak interactions were actually a separate interaction from the nuclear strong interactions and one in which parity conservation, which was known to be a very good symmetry of nuclear physics, had never been tested. And they postulated that the weak interactions in fact violated parity and this was very quickly confirmed experimentally by many groups. Now theorists took that idea and ran with it. So if parity is violated, parity should be violated maximally. And so in just a couple of years, one had a very precise theory of weak interactions due to Feynman and Gelman and Marschak and Suderschein. And this is what I'm going to call the universal, V minus A theory. And so the first thing I'd like to do and I'm gonna spend about half the lecture on this is trying to tell you how we know that the universal V minus A theory is true. And then from that we'll work our way toward the standard model of weak interactions. So by the way, how many of you have read the chapter in surely you're joking Mr. Feynman about the Feynman-Gelman theory? Okay, fewer hands. So this is also something that you need to check out. Gelman never liked this account because it's mostly Feynman as are all the stories in that book. And one has to question the historical veracity of some of it, but it really is very amusing. And the crucial point is when they get a letter from, in the, how many of you by the way have read the Feynman-Gelman paper? Fewer. Okay, yeah, the history of physics is really interesting. And this is a very interesting paper because they had to go against the established experimental data on beta decay in order to postulate this theory. And they just came out bluntly and said the experiments were wrong. And indeed the experiments were wrong. In this account by Feynman that I described, he says they got a letter from Valentine to Legde at Chicago saying the fund, the FG theory of beta decay is no FG. And then a couple of days later got another letter from to Legde saying please excuse me, I left out the recoil correction and your theory works perfectly. So it's very interesting how in just a few years one got from just the notion of parity violation to a very complete theory of at least the charge-changing part of the weak interactions. So what is this universal V minus A theory? Today we would write it like this. So for GF over the square root of two, the square root of two is a relic of the era before parity violation was introduced. J plus mu J minus mu. Where very importantly these currents are currents of purely left-handed fermion fields. So J plus mu would be, let me use a chiral notation here. New bar electron, the left-handed piece is only and U bar sigma mu D, the left-handed piece is only plus the same thing for higher generations. By the way, in this lecture all the Kibibokobi-Ashi-Moscow mixing angles are going to be taken to be zero and the corrections to that will appear in Zoltan-Leggettius lectures. So you have this very simple form where you're taking explicitly only the left-handed parts of fermion fields and so this theory is actually maximally parity violating. The parity conjugate of that is the thing with only right-handed fields. It's a totally distinct operator. This structure is extremely important not only for all the things that we know about the weak interactions but as the week proceeds for all the mysteries of weak interactions and so I think there is some importance in understanding why this equation is true. And so the first thing that I'd like to do is to tell you four really striking pieces of experimental data that support this universal V minus A interaction. And so let's go through them one by one. The first one is obviously that, oh and let me just provide a little more notation. Maybe I should do that over here. So in these lectures I'm always going to talk about chiral fermions. Fermion masses will appear only every once in a while and so I'm going to use this basis of gamma matrices. It's the same one that appears all over my book. So the sigma bar is what appears down there. The Dirac Lagrangian in this notation, so that's d slash minus m psi reads psi dagger left i sigma dot del psi left plus i psi dagger right. That's a sigma bar, that's a sigma minus m. So the purely left-handed massless states, if the mass were zero, the purely left-handed parts of the fermion field and the purely right-handed parts of the fermion field would never talk to each other except as some of you know through anomalies. And the addition to the mass term requires the mixing of left and right which is eventually going to require that the Higgs boson comes in and intervenes here. So there are various things about this chiral projection which are going to now become important and we'll see how they work in the data. So let me now talk about the evidence for this structure of the weak interactions. Well the most obvious one is that if the weak interactions communicate only with left-handed electrons, then if you actually measure the polarization of electrons coming out of beta decay, those electrons should be preferentially left-handed and one can make that a little more precise. So let's remember that in the notation that I wrote there, an electron spinner is this, so the momentum parallel to the three axis is this for a left-handed electron and it's E minus P and E plus P downstairs for a right-handed electron. Now if I have a, if I'm coupling to what I called over there the left-handed component, that's the upper two components in this matrix. So that means that the polarization of an electron coming out of beta decay, which is the number of E-lefts minus the number of E-rights divided by the sum of those things, is the square of what you see in those entries. E plus P minus E minus P over the sum, which is P over E. And so there are lots of beta decay nuclei, but what you can see is as you increase the gap between the two levels, so the electrons come out more and more relativistic, they should come out left-handed or left-handed polarized with exactly this degree of polarization. Now here's what the data looks like. This is mid 1970s data, it may have improved since then, but already it's pretty impressive. So on the bottom axis you have V over C from zero to one, on the vertical axis you have the degree of left-handed polarization from zero to 100%, and you see it really works spectacularly well. So the number two piece of evidence for V minus A is something that when I first started studying elementary particle physics seemed to me a puzzle, but somehow the people around me all seem to understand it, but I think it is something that's really remarkable. In the universal V minus A theory, there of course are three kinds of neutrinos, so there's another term here with the muon, and the term with the electron is exactly the same as the term with the muon. But on the other hand, if you know a little about real elementary particles, you know that the branching ratio of a pion to E nu over the branching ratio of a pion to mu nu is extremely not equal to one. In fact, it's 1.23 times 10 to the minus four. So where did that come from? So that's again a consequence of the universal V minus A theory when you actually work out the details of it. The matrix element here is again, four GF over the square root of two. There's a J plus and a J minus. The J plus is the thing that, sorry, the J, sorry, how do I wanna do this? Let's put positives everywhere. The J minus is the thing that annihilates the pion, and this is a Goldstone boson type matrix element. It's proportional to P mu and F pi. There's a one over square root of two. That's the only thing we're gonna need about it. This quantity here is made up of lepton spinors, zero on one side and nu E plus on the other side. So it's going to be U bar sigma mu V, the neutrino here and the positron here. And so what we have to do is to dot P mu into this and evaluate it. The relevant spinors are for the neutrino, the square root of two times the energy of the neutrino. Let's get the neutrino moving in the positive three direction. For the electron, we have here V, the electron is, or the mu on let's say is massive. So we have the same structure. The square root of E minus P and E plus P, zero one, zero one. The electron is going backwards, so I've, please excuse me. The electron is going backward, oh, and there's a minus sign here. The electron is going backwards, so I've rotated the spinner by 180 degrees. And it's a down spinner because this is the V. It's, this is the whole that, this is the particle which when you eject it gives a hole which is the positron. Now P mu times sigma mu is equal to one. So now this is really easy to evaluate. It's just U dagger P dot sigma bar V is equal to the dot product of the upper components, which has entered a square root of two E and a square root of E minus P. So let's now go over here and just work out the mass dependence of that thing. The rate for pi plus to E plus mu is the square of the matrix element that I wrote there. So E minus P times the energy of the neutrino there's of course the usual one over two M pi, one over eight pi for phase space. There's also a two momentum over M pi for phase space. And this factor is another P is the same thing as the energy of the neutrino. The process is two-body kinematics into a mass less than a massive particle. So that means that the energy of the electron is M pi squared minus M E squared over two M pi. And the momentum of the electron is M pi squared. Sorry, this is plus, this is minus M E squared over two M pi. So E minus P is just the mass of the electron squared. And so this now turns out to be the mass of the electron squared. And then these various factors of P give you factors of M E squared over M pi squared. There's one from here and there's one from phase space. If I wanna know what the relative branching ratios are for electrons and muons, I would just take the ratio here for electrons and muons. And if you just work this out with no higher order corrections, you get 1.28 I believe times 10 to the minus four. So the basic principle I think one can state more simply. The decay we were talking about has a pion which has no spin. It has a neutrino which is always left-handed. So the electron, the positron spin has to go this way. The positron also has to be left-handed. But the left-handed positron is the antiparticle of the right-handed electron, which it doesn't couple in V minus A theory except via the mass. And so this process has a natural mass suppression. When you square it, you get this M squared factor here. If the weak interactions weren't precisely chiral, you'd get something different here and something much larger. But this is apparently the way the world works. Okay, number three is the theory of mutiK. And actually, mutiK is one of the places where parity violation appeared very early in experiments. As soon as people had the idea to look for parity violation, the relevant experimenters, Garwin and Letterman and Tlegdi and Friedman, realized that if parity violation were large and weak interactions, then you could just see it as an order one effect in mutiK. And indeed it's there and the calculation is very kind of very fun to do. So let's talk about that. MutiK to an electron, an electron anti-neutrino and a muon neutrino. The matrix element is again, 4G firmly over root two, U dagger for the E, sigma mu and V for the electron anti-neutrino, U dagger for the muon neutrino, sigma and a U for the muon itself. Yes, I'm sorry? Yes? Right, because if it was equal, then we wouldn't have that problem. Absolutely, and that's in this formula. It's this factor here. So this number, it could even be smaller if it were not for that. Sorry? Oh, please excuse me. Is this too small? Okay. Okay. There are lots of ways of getting the square of this. One rather simple and effective way is to use the Fierce identity. Sigma alpha beta sigma bar mu gamma delta is two epsilon alpha beta, epsilon gamma delta. So now, what this, sorry, please excuse me, alpha gamma and epsilon beta delta. That's the relevant Fierce identity. So what that does is to split this expression into two scalar factors. U dagger E alpha, epsilon alpha gamma together with the other U dagger V of the new E bar beta, epsilon beta delta U, delta mu. Now each of these factors is a Lorentz scalar. So you can go to a preferred frame to evaluate it very easily. For example, if you evaluate this in the muon frame, then it is more or less the square root of the mass of the muon, which is the normalization factor for this, the square root of two E new bar, and that's it. And this one, the easiest way to do that is to go to a frame where the electron and the muon neutrino are back to back. And in that frame, this thing evaluates to four E electron, E neutrino. And if you look at these things, what they are are the square root of two P E dot P new. And this, the square root of P new bar dot P mu. And now we're done. So the matrix element squared is just four, the square of this two times two P E dot P mu, times new mu times P new dot P new bar. Now, to get all this straight, you have to do three-body kinematics, which is the other kind of very elementary kind of relativistic kinematics that should be just second hand to all of you. Here are the basic formulae for three-body totally massless kinematics. You define some quantities Xi, which are two Q dot PI over Q squared. The three X's, because it's P one dot P two dot plus P three in the final state add up to two. You notice that P one, so this is, please excuse me, Q goes to P one plus P two plus P three. P one plus P two squared is Q minus P three squared is Q squared minus two Q dot P three plus P three squared is equal to one minus X three. And then here, there would be an M squared if this particle were massive. Right now I'm going to take it massless. Well, now we're pretty much all set. So we can just write the total, and one more important formula. Three-body phase space is then just a simple integral over these X's. X one, X two, Q squared over 128 PI cubed. So now we're all set. We can write this as M U squared over 128 PI cubed. The integral over the energy fraction of the electron, the integral over the energy fraction of the anti-neutrino and then just pick up the factors from this expression. This one is an X nu, this one, that dot that gives us a one minus X nu, nu bar. And then if you just do this integral, this is from the phase space region is over a region that looks like this. X electron and X anti-neutrino. So we're going to do this integral from zero to one and this integral from one to one minus X E. If you do this integral, what you get is X E squared over two minus X E cubed over three. And so we get now the following structure. D gamma in terms of the fraction of its maximum value that the electron can have is proportional to this expression here. X E squared times one minus two thirds X E. If you look at that function, it has a very distinctive shape. It's quadratic for small values and it actually has zero derivative at the endpoint. So it's, I'm sorry, just like that. And when you go and measure this, wow, that's exactly what it looks like. There's a little rounding you see at the top that's due to photon emission from the outgoing electron, the radiative correction. But the basic shape of that is given exactly by the function that I presented here. And again, everything depends on all the particles being left-handed projected. It's again a prediction of V minus A theory. Now there's one more very interesting prediction of V minus A theory that I think I just have to be a little more sketchy about, but I'm hoping that, oh, sorry. There's one more thing I have to tell you about this which is totally amazing. Let's now think about the kinematics that happens at this end where the electron has its maximum value. So in that case, the muon is here. The electron is emitted with its maximum value. So Xe equal to one. The two neutrinos have to recoil against it. The mu nu, according to V minus A theory, is left-handed. The nu E is right-handed. And the electron, if I can ignore the electron mass, and that's a good approximation here, is also left-handed. So this can only conserve angular momentum if the muon spin points this way. And so there's an interesting observation that for at Xe equals one, d gamma d cosine theta is proportional to one minus cosine theta. That is, the electron is always emitted in the direction opposite from muon spin. Now, it's possible to test this rather sensitively. And this was done in the 80s in a beautiful experiment at triumph by a triumph Berkeley group. You create pions, you stop them, the pions make muons. And you bring the muons over here and you stop the muons in some magnet. Muons from pion decay by the argument that we made over there are all right-handed, essentially 100% right-handed polarized. So the muons coming in here are all right-handed muons. And you know exactly the spin of the muon at the time that it stops. Now, you let the muon decay to an electron, and let's say forward here, maybe off to the side, I'm not sure where they put it. You put some kind of counter that measures the presence of electrons. And now you put the, you turn on the magnetic field, the muon spin processes. And so as a function of time, you should see an oscillation of the number of electrons that you see here with complete extinction when the muon spin points to the electron detector. Well, they didn't see complete extinction because of course there's some depolarization when the muon stop, there are some depolarization when you transport the muons. But they saw actually 98% extinction with the simple argument that the muon who spin points to the electron counter can never emit an electron at the full energy in that direction. So again, V minus A at work, everywhere that we study high energy weak interactions. Now, there's one more set of formula that I'd like to write here, which are very important. And so maybe it's worth my just taking the time to at least sketch out how this works. And that's deep in elastic neutrino scattering. Now, I guess Michelangelo, he's interested in much more high energy versions of applications of QCD. But of course the first place that quarks were discovered was in deep in elastic electron proton scattering. And there are some very beautiful formulae for that that were developed in the 1960s, which it's worth recalling. So there are two, the process that we're interested in is this. A lepton, in this case an electron comes in. It scatters by a virtual photon. It disrupts the proton. Both the initial and the final electron momentum are measured. And so the momentum Q that goes here is known. The process depends on two crucial variables. One is called Y, which is two, let's call this momentum K and K prime. Two K dot Q over two K dot P. So Y is the fraction of the total incoming energy of the electron, which is transferred to the adoronic system. The other one is called X, and X is, so Q squared is space like, we can write it like this, this Q squared over two K dot Q. And what, I'm sorry, I shouldn't do this without consulting my notes. That's much better. The proton momentum is P. Two P dot Q over two P dot K. If I evaluate this in the proton rest frame, it's Q zero over K zero, which is exactly the energy transferred from here to here. This formula should be written like that. And this has a very interesting interpretation, first written down by Feynman, where you think about making a part on description, a factorized description, as Michelangelo talked about of this process. So then here you have a single quark with momentum fraction C. Q comes in. A single quark goes out. The momentum of that quark is, please excuse me, this. This is two C P dot Q minus Q squared, but the quark is massless, so we can solve for C, and it's exactly this expression that I wrote here, X. So this is the fraction of the electron's energy, which is transferred to the hydronic system, and X is the fraction of the proton's energy that's carried by the quark that was struck. So these are two very useful variables which live between zero and one in this high energy limit that I'm describing. Okay, so now in terms of those variables, one can write the cross-section by basically just putting together this picture and the quantum electrodynamics cross-sections, and it takes a very simple form. d sigma dx dy is the sum of all the quarks and anti-quarks that you can hit, they're charged squared, the part on distribution function that Michelangelo talked about, and then there's a quantum electrodynamics cross-section, two pi alpha squared s over q to the fourth and a one plus one minus y squared. The part of this that I'd like to concentrate on is this factor one plus one minus y squared. This thing here comes from the scattering of left-handed electrons off of left-handed quarks and right-handed electrons off of right-handed quarks. This factor comes from the scattering of left-handed electrons off of right-handed quarks and right-handed electrons off of left-handed quarks. It's gotta have that structure for some simple angular momentum reasons. If I bring in an electron which is left-handed against in the center of mass, in the part on center of mass frame, a quark which is right-handed, the angular momentum are both pointed that way. If I wanna completely transfer the energy of this guy to this guy, I can do that by backward scattering, but for massless particles, which is a very good approximation here, helicity is conserved. So that means that this guy is still right-handed and this guy is still left-handed. Now, the angular momentum is just wrong. And so, for this process, it has to vanish when y is equal to one, when you have complete energy transfer, whereas for this process, it's not constrained. It might as well be one. If you do the explicit calculation, that's what you get. Okay, well now we can apply this to the weak interactions. And the application of the weak interactions is very interesting. We can look at neutrino deep and elastic scattering. A neutrino hits some nucleus, it interacts through the universal v-a interaction with up and down quarks, let's say that come out of a proton. If it's a neutrino, it turns into a mu-minus. And so, what do the formulae look like? I put this slide in here, it's quite interesting. This is a relatively modern neutrino experiment. These neutrino experiments go back to the 1970s and even to the 1960s. The first big deep and elastic neutrino scattering experiments were done in the early 70s. The ones at Fermilab were quite amazing. You shoot a proton beam into a target, you make pions. The pions are given something like 100 meters of airspace to decay. Then Fermilab's big, it's on a prairie. You go down there a mile, more than a kilometer, you dig a hole and you put a large amount of iron plates submerged in liquid scintillator into that hole. And you get pictures which more or less look like the top where nothing comes in, having these are the invisible particles that penetrate a mile of Earth. They all of a sudden create a muon recognized as an extremely penetrating particle that penetrates tens of meters of iron. And then if you magnetize the iron, you can measure the momentum of the muon. And so then people could start to measure these cross sections. So let me write the analogous formulae for these neutrino scattering cross sections. This would be something like this. There are four processes that we can think about with just first generation objects. A neutrino with a down quark can turn into a mu minus and an up quark. A neutrino with an anti up can turn into a mu minus and an anti down. A neutrino can never turn into a mu plus. That would violate leptome number. But a new bar could make a mu plus or on an anti down it could make a mu plus. And if you now put this into the V minus A theory, you get some very simple cross sections. G Fermi times S over pi times X F D, oh sorry, this is for neutrino proton scattering, plus X F U bar. And then please notice everything here is left handed. So there's a one. But here this guy has to be right handed. So there's a one minus Y squared. And for anti neutrinos, exactly the opposite. The quarks come with one minus Y squareds and the anti quarks come with ones. So that's a rather striking prediction. If you measure this variable Y, it's just totally different between neutrinos and anti neutrinos. If you ignore the anti quarks, which at low energy, Q squared of order 100, at least the quarks are very dominant, you should see characteristic distributions which are flat and which are one minus Y squared. So here's the data as given by the CDHS experiment done at CERN around 1980. And it really works. The top set of, the top curve is for neutrino scattering, not off protons, but off of I believe an iron target. It's an isoscalar, an approximately isoscalar target. The bottom curve is the anti neutrinos scattering. The little deviations from the law of one and one minus Y squared is readily explained by the small amount of anti quarks in the nucleus. So there we go. Each of these four things is very characteristic of this idea that the weak interactions violate parity maximally. They couple to left-handed particles, but they never couple to right-handed particles, at least the charge changing part of the weak interaction. Sorry? So everywhere here I've been ignoring the mass of the neutrino. Now if neutrinos are myerana, then what happens is that there's a mixing between the two states of the neutrino, the left-handed neutrino and the right-handed anti-neutrino plays the role of the mass term that I wrote for the Dirac equation. But a very high-energy neutrino is just approximately one or the other. So if the mass of the neutrino is sub EV, but its energy is 100 GEV, you can just ignore that mixing. It's left-handed and there's some tiny component of the right-handed state mixed in proportional to the mass squared divided by the energy squared. Sorry? Well, I think it's really semantics. It's what you call it. So what I call a neutrino is what's produced in the decay of a, let's say, 100 GEV pion. Okay? And now you might say if it's got a myerana mass, I don't know whether that's a neutrino or an anti-neutrino. But that object must be right-handed and it must be, actually, have the quantum numbers of the neutrino until, no, I think I'm just gonna stop there. That object must be, so I have an extremely relativistic pion. It generates an extremely relativistic neutrino. That guy must be left-handed and it must be a neutrino. Please excuse me. I'm not answering your question clearly. So I guess what I'm saying is you have to think about what the experiment is. So you have a pion that comes in and it emits something, okay, which now you go some kilometers downstream and that thing has an interaction. So if you, so this particle here, it's created, the thing that's created is the new left before you added the myerana mass, the unmixed new left. And it basically just doesn't have time to oscillate to anything but the new left here. Now, if you did this at very low energy, you would find that there's some mixture of the neutrino and anti-neutrino states, but those low energies would be of order the mass of the neutrino. At relativistic energies, the mass correction is very small. It's proportional to the square of the neutrino mass. So what you wanna call this thing, I don't care, but what a relativistic pion produces is to a very good approximation, a massless left-handed neutrino, okay? Okay, so I think now basically all we have to do is to try and turn this theory into a more fundamental theory. And probably most of you know how that goes. Once people had the universal V minus A theory, it's a current-current interaction, it's natural to pull those two currents apart and put a mediating spin one particle here, the W particle. This necessarily is a particle which is spin one and it's massive. And so in the 1960s, people went around in circles trying to figure out how to make those two ideas consistent. And the answer, as we all know today, is the Higgs mechanism. That is, you have to start with a Young-Mills theory which contains this object, then spontaneously break its symmetry to give that particle mass. And so maybe it's good just to recall a few formulae of that structure. So the first thing we have to do is to try and write this object broken apart in two. So we need some structure that looks like this. The U dagger, we need to introduce something that I'll call Q which is the U D left doublet and L which is the new E left doublet. We need to write J plus as Q dagger, sigma mu, some kind of SU2 raising matrix, Q and L dagger, sigma mu, again an SU2 raising matrix L. And from that, we need to have, if we go to Young-Mills theory, we need to have this Young-Mills gauge group include SU2. So as probably you've heard when you've studied this in various places, there are various choices that you can make. The first, we now have to introduce something that's gonna break the SU2 so we can introduce a Higgs boson, a scalar field, in the isospin one representation of SU2. Now that's a very interesting theory called the Georgia-Glaschall model. It has the property that SU2 is broken to U1 because isospin one is like a vector in the three-dimensional rotation space. If I have three axes here, this symmetry is broken, this symmetry is broken, but this symmetry is not broken. So this theory then leads to two massive, one massless Young-Mills boson. It would be actually a great theory of unified weak and electromagnetic interactions. And in fact, Georgia and Glaschall proposed it for that purpose. But it doesn't work for various reasons. The most important of which is that this Higgs is not suitable to do the other thing that the Higgs boson is supposed to do, which is to give mass to the quarks and leptons. The next thing you can try to do is to introduce a Higgs and the isospin a half representation, but this completely breaks SU2. So then finally, what the right solution, the solution chosen by Glaschall, Salam and Weinberg is to enlarge the symmetry to SU2 cross U1. Now you have a covariant derivative acting on a fermion field that looks like this. D mu psi and a set of SU2 gauge bosons and a set of U1 gauge bosons. And the U1 charge here is called hypercharge. Now I have to assign this not only an isospin but also a hypercharge. And the choice of hypercharge a half is particularly felicitous, because let's now give this an expectation value, which we can always rotate into some form that looks like that. It's a doublet. Let's rotate its expectation value into the bottom component. If I now act on this with a general SU2 and U1 transformation, what appears here is this factor a half. If I choose alpha equals beta, alpha three equals beta, this is a phase e to the i minus alpha over two plus i beta over two. And if I choose alpha equals beta, this expectation value is left invariant. So with this choice, so the choice of the gauge group SU2 cross U1 and with the structure of the expectation value, there is one symmetry left unbroken. And so then we're going to get three massive and one massless gauge boson. Well, at this point, you can take this structure and work out both the mass matrix and the interactions of quarks and leptons. And maybe I should probably say this is a standard exercise. So let me tell you what the results are. The first thing that you find is the mass matrix of the gauge bosons. And it turns out to have this form. There's a v squared over four that appears here, a G squared, G squared and G squared from the SU2 part. So if you only had SU2, you would have complete breaking. If you only had this, I just spent a half Higgs boson and only SU2, not U1, you would have complete breaking of the gauge symmetry and the structure of masses where all three masses were equal. Down here, you have G prime. This matrix has to have a zero eigenvalue. So the only thing you can put in here is G G prime. And the only things that can mix are the neutral components of the gauge bosons. And so out of that diagonalizing this, you get the W bosons. W plus or minus is A1 minus or plus I A2 over the square root of two. The Z boson as G A3 minus G prime B, which I didn't write. And the massless gauge boson G prime A3 plus G times B. So from this matrix, these get masses. Mw is Gv over two. Mz is G squared plus G prime square to the one half V over two. And Ma is equal to zero. It's, the relation between these two masses is quite interesting. And it's something to keep an eye on as you go into kind of more advanced theories of weak interactions and the Higgs mechanism. Conventionally, we write G over the square root of G squared plus G prime square is the cosine of the weak mixing angle. G prime over that same thing as the sine of the weak mixing angle. In the rest of these lectures, I'm gonna call these Cw and SW. These relations imply that Mw is equal to Mz times the cosine of the weak mixing angle. And as we'll discuss in the next lecture, this relation is really very well satisfied experimentally. It's something that it seems you have to hang on to if you try to generalize whatever Higgs structure you have in the standard model. So it's interesting to look at this matrix and see where it comes from. Basically, one way to obtain that matrix is you just compute it in the standard model. You do as you usually do with the Higgs mechanism. You write down d mu phi squared. You put in for phi the Higgs vacuum expectation value. You just work everything out. But on the other hand, the matrix has a logic to it which is number one, there must be a zero eigenvalue so that controls these two matrix elements. And the other thing is, there must be a symmetry which relates these three matrix elements in the limit where you turn off U1 and you only have pure SU2. The relation here is an unbroken SU2 symmetry that appears in the limit where you ignore U1. This relation then follows from an unbroken SU2 symmetry in the limit where G prime is equal to zero where you turn off the U1 interaction. And this symmetry is called custodial symmetry. The simple Higgs that you have in the standard model has custodial symmetry because the potential for the Higgs field is a function of phi squared. So phi squared is phi plus squared plus phi zero squared. But if you were to write phi, let's say as phi one plus i phi two, phi zero plus i phi three, if you were to write phi in terms of four real fields in this way, this would now turn into phi one squared plus phi two squared plus phi three squared plus phi four squared, phi zero squared, which is SO4 invariant. So accidentally as it were, this potential for the Higgs field of the standard model is actually invariant under an SO4 symmetry. SO4 is SU2 cross SU2. One of these is the one that's gauged and spontaneously broken. The other remains exact. It's explicitly broken when you turn on the U1 interaction. Now custodial symmetry is, this isn't the only way you get custodial symmetry. This is what happens in the standard model. But in principle, there can be many more complicated models of the Higgs sector. The first one was Technicolor, New Strong Interactions at 1TEV, which automatically have custodial symmetry and which protect this relation. And I think when you hear Marcus Ludi talk about models of electroweak symmetry breaking beyond the standard model, he'll try to arrange that the models that he has have custodial symmetry by essentially strategies that become weirder and weirder as the models become more and more complex. Okay, now I'd like to talk just a little about the structure of the couplings of the Z. It's a kind of standard exercise in the formulation of the standard model to work out what the couplings of the standard model gauge bosons are to various fermions. Let's just call this field F. So this is D mu I G A mu A. If F is an iso-double it, it's got a sigma here. If not, it has a zero. Here, there's the U1 boson in the hypercharge. Now what you want to do is to introduce the mass eigenstates. So here, what we're going to get is A1 sigma one plus A2 sigma two over two is minus IG W plus sigma plus over the square root of two plus W minus sigma minus over the square root of two. And from that, we're going to get the weak interaction coupling that the weak interaction coupling has here the value of G whatever the SU2 coupling was divided by the square root of two. For the Z and photon, it's a little more complicated. So what we have to do is to plug in the mass eigenstates into the A3 and B terms here. And let me do that as follows. So this is A3 times the third component of isospin or zero, this could be zero, minus IG prime B and the hypercharge. If you write this in terms of the mass eigenstates, you get minus IG times G times CW times I3, please excuse me, minus G prime SW times Y and then for A, GSW times I3 plus G prime CW times Y. Now, from the definitions over there, you see that these quantities are equal and they're equal to GG prime over G squared plus G prime squared. So whatever is the overall coefficient of this is a common factor, which I'm just going to now call E, the electric charge. So the bottom line of this can now be written in the usual form of the photon coupling to fermions, EAMU times Q, if we identify the electric charge as I3 plus Y. Now we can rearrange the top line. I think what I'd like to do is to note that this is proportional to SW squared. So let's add and subtract something proportional to SW squared, we would get Z times G over CW times CW squared plus SW squared I3 minus SW squared I3 plus Y. And now this breaks down into a very simple formula that the Z couples to a pre-factor, which is this, which is the same thing as E over CWSW, with the definition down there, times I3 minus SW squared times Q. And this quantity here I'm going to call the Z charge. And tomorrow we're going to go in great detail in what we know experimentally about the values of these Z charges. These Z charges are predicted in the standard model. The left-handed fermions have I3 equals plus or minus one, plus or minus a half. The right-handed fermions have I3 equals zero and then whatever the electric charge is. So the standard model has a definite prediction and tomorrow we'll talk about how strongly those predictions are tested. The first test came historically in deep and elastic neutrinos scattering. And maybe it's worth just writing those formulae because they're at least historically quite interesting. So you remember that I told you, we wrote earlier in the lecture the formula for deep and elastic neutrino and anti-neutrino scattering. Let me write those formulae assuming they're equal numbers of U and D quarks. It makes the situation a little simpler. So there's a quark distribution. There's a GFS over pi. Then the thing is that the quark distribution for neutrinos has a one and the anti-quark distribution has a one minus Y squared. And then this is flipped from here to here when you go to anti-neutrinos. For the Z exchange, there's a similar process. A neutrino comes in, it goes out as a neutrino. It hits a quark and then shoves out the quark and disrupts the nucleus. So this is what's called deep and elastic neutral current neutrinos scattering. It corresponds to the lower picture in this plot where something comes in from miles away, initiates an interaction. The interaction is purely hydronic. There's no muons and then just goes out the other side. So the neutrino formula there is a little more complicated. It's EF squared S over pi because now there are contributions both from the neutral current interactions with the left-handed neutrinos and the neutral current interactions with the right-handed neutrinos. So what we're going to see is something like this. From the left-handed neutrinos, the Z coupling to the left-handed neutrinos, sorry, the Z coupling to the left-handed up quark and the Z coupling to the left-handed down quark and those come with a one and for the anti-quarks, oh, please excuse me and the Z coupling to the right-handed quarks which has no I3 factor and those come with a one minus Y squared. And for the anti-quarks, now the reverse. These factors here with a one minus Y and these factors here with a one but proportional to the anti-quark distributions. Now these formulae get quite complicated and so how do you try to make sense of that? Well, this is where a simple theory can make a lot of progress and the person here who had the very good idea was Chris Lou Allen Smith. So what he said was let's concentrate on the following variable, little r which is the ratio of neutral current to charge current neutrino interactions. Sorry, which is, please excuse me, the ratio of anti-neutrino to neutrino charge current interactions and for an isoscalar target, this is according to the formula up there very simple, just FQ times one minus Y squared plus FQ bar over FQ plus FQ bar times one minus Y squared. And then if you want, you could average this over the acceptance of your detector and just measure that number in your experiment. Then this quantity captures most of the complexity of the formulae that I wrote there. So in the limit of isospin symmetry, actually you can just eliminate r from those formulae and get the following very beautiful relations. That new charge current over neutral current over charge current is one minus SW squared plus five ninths SW to the fourth times one plus r and the similar quantity for anti-neutrinos is one minus SW squared plus five ninths SW to the fourth times one plus one over r. And once again, r is a quantity whose value is about 0.25 that you could actually go measure in your experiment. So then you get two rather simple relations with r are constant to evaluate against the parameter sine squared theta w. Now, this was back in the dark ages when people didn't know what the value of sine squared theta w was, so you could determine it in this kind of measurement. You just write the parametric curve that corresponds to these two equations for your measured value of r and it looks something like a figure that I drew on this plot, a figure that in the 1970s was popularly known as Weinberg's knows and this is the circa 1980 CERN data. So what you see is that already from the neutrino experiments we were focusing on values of sine squared theta about 0.23, quite close actually to where it's come out with more accurate data that I'll describe tomorrow. So this really was something very interesting. It now allows us to fill in the last missing parameter in the weak interaction theory. The basic parameters of the weak interaction theory are g, g prime and the vacuum expectation value. We wanna trade these for things that we can observe directly. G fermi is something we can observe directly. E and now SW squared as for the moment extracted from neutrino experiments on this curve. And as we'll talk in more detail tomorrow, those give you now values for all of the quantities that we've provided including values for the W mass and the Z mass. In particular, if you put the value of 0.23 here you get Mw about 80 GeV and Mz about 90 GeV and you can actually go look for these things in Hadron-Hadron collisions. So just to conclude the lecture here is not the first generation of Hadron-Hadron collisions but some LHC data. This is the distribution of electron pair masses measured by CMS. You see at just about 90 GeV there's a beautiful peak. This is what Michelangelo described this morning as the Drellian process producing the Z and that peak would be the resonance associated with the Z boson. There's some beautiful events associated now measured at LHC with these processes. So this is for example an event with an electron which you see very clearly as a track and some associated electromagnetic energy deposition. We call it against nothing. This has, when you analyze it a little further nice properties of W production. And here's another event from Atlas, a very beautiful electron positron candidate with a mass within errors just on the Z mass that we were talking about before. So now we've got a first picture of what the weak interaction theory should look like. I've written for you a lot of formulae and so in tomorrow's lecture what I'd like to do is to explain how those formulae can be tested kind of at the next level of precision by exquisite analysis of the Z and W in a further succession of colliders. So let me stop here and we'll pick that up next time.