 All right, so we're ready for the last lecture of the day with you. Well, okay, we can keep this nice pictures really nice, huh? Can you hear me? It's okay. Oh, now I've lost this one. Okay, so let's see how we survived the second talk of the day and let's go directly into where we were. But it didn't work. Shit. Okay, so every time I will come back and press. Nothing, also the thing. Okay, anyway, so let me remind you what we were ending last time. So last time we end with the following big picture and we say we understand all those aspects of symmetries and now we're ready to actually start doing model building and we understand how we do model building. We choose the coordinate, we put the symmetry and we write the Lagrangian and we go on. So what I want to do now, I want to start actually building models and what I'm going to do, I'm only going to present models that works. Actually, it's a big surprise. Okay. And in a way, if we had a lot of time, I would think that really the best way to do this kind of exercise is to actually give you 10 or 15 or 20 models that do not work. And then when you see the one model that works, you would like wow. But then instead what we are doing, I'm just going to give you the one that works and then you might be like, okay, of course it works. What can we do? And you will not appreciate how amazing the model that works actually is. Okay. But since we don't have much time, I'm going to give you just the model that works and my plan is to give you some model that do not work actually for tomorrow, for the homework and you will be, I hope you can actually do some calculation and see why it doesn't work and what is so nice about the one that do work. Okay. So let me start with a model called QED. It stands for Quantum Electrodynamics. And I'm sure many of you have seen QED before, but usually the way you study QED in quantum field theory is not from the principle of model building. It just said, hey, here's the Lagrangian. Let's start doing calculation. You do all those traces of gamma matrices. You calculate cross-section. It's really cool. But what I want to do now is to actually get it from a different approach. So the first model, just as I told you, here are the things. I tell you what the symmetry is. The symmetry is a U1, a local U1. And I want to emphasize, we call it U1 electromagnetism. But this EM that is there is just a label. It's just for us for bookkeeping. There's no fundamental meaning for this U1. The EM, what is physical, is just the fact that it is a U1. I only consider to start just a electron. So it's just a photon and electron. So the U1 gives me a photon, and then I give you two electron fields. I give you a left-handed electron and right-handed electron. And I tell you what are their charges minus 1. Why I want the charge to be minus 1? Just because somehow historically the electron has a charge of minus 1 rather than plus 1. OK? But we so got used to it, so that's what we are doing. And there's no scalars. And now that's it. You see how easy it is to do models? All you need to do is just to tell you for three lectures all the background, and then we just do the model. So now what is the next step to do? You have to write the most general Lagrangian, truncate the dimension 4, and start doing calculation and make prediction. So the question is what can we say about this Lagrangian? And I remind you there's all these four terms. Kinetic is a kinetic term for both the boson and the fermions. The L psi is the one that is mass through the fermions. The Yukawa is one that have fermions and scalars. And L phi is the one that have just scalars. OK? So to start with, out of these four, can you tell me some that are zero? L phi is zero? So L psi is not zero, actually. So we're actually allowed to write mass terms for the fermions. So L psi is the one that give mass terms for the fermions. We're going to write it in the next reference, but it's actually not zero. And we also know that the kinetic term can never be zero. So that's actually conclude for us that the two that are zero are the Yukawa and L phi. So why is there zero? Why L phi is zero? Because they're no scalar. Why do Yukawa is zero? Because no scalar. Very nice. So actually, it's kind of an easy case. We already solved 50% of the model. So when you grade your students, when they say 2 is zero, you already give them 50%. That's very nice. OK. So now let's actually write what we have. So the two are zero. And then we can write a master. And why I can write a master? Because this theory is a vectorial theory. The left-handed and right-handed fermions have the same charge under my symmetry. So I am allowed to write a master. OK. It's the M E L bar E, right? And by the way, maybe I didn't talk about it. You know the plus H C, what it stands for? It stands for plus Hermitian conjugate. And whenever I have fermions, because they are complex, I just write one term and I put plus Hermitian conjugate. So my Lagrangian is Hermitian. Many times we just don't write it. But when you don't write it, you have to know that it is implicit. Our Lagrangian is always Hermitian. So I have a master. And then I have a kinetic term. And what do I have in a kinetic term? Again, I write what I can. And I have the kinetic term for the fermions, for the gauge boson, which is the F mu nu F mu nu. And then I have the gauge boson for the kinetic term for the left-handed electron and the kinetic term for the right-handed electron. And we have the covariant derivative. And in the covariant derivative, we have the derivative, which is really the kinetic term. And then the extra term that is I E Q A is the coupling between the gauge boson and the fermions. Okay? Just like we talked in the morning today, where I make here a following distinguish. E and Q always appear as a product. So I could just call it some one number. So I write it as E times Q in order to get us ready for the non-Abelian case. And Q is the charge of the electron. It's just some number that I assign. Okay? That's a number that I assign. That's the input. And what is E? E is the coupling constant that emerge in the theory. So the difference is that Q is an input and E is a parameter. Of course, since they are both come together, I can always combine them into one number. Okay? But when we go into the non-Abelian case, I will not be able to do it. So I'm also not doing it here. Okay? Any question about this first model building exercise that we are doing? If not, so let's ask some question. How many parameters I have in this model? So you remember when we talked about model building, we say we write the Lagrangian, we truncate, and there's finite number of parameters. How many parameters this model has? Two. Can you tell me? Which are the two? M E and E. Okay? Kind of nice. Okay? So we have the coupling constant E is one parameter, and the other parameter is the mass of the electron. So all we need to do is to measure those two parameters and we're done. How do we measure the mass of the electron? I don't remember. Okay? But you know all the history, right? They put it in this oil drop and all this. But since then we measure it extremely precisely. So it's 511 kv plus, it's known to like 10 gd. And how do we measure E? So we usually, instead of measuring E, we measure some other combination that we call E squared over 4 pi. Let's go under the name of alpha. And it's probably the most famous number in the world, which is 1 over 137. And when I just arrived here, it's really true. And then Giovanni gave me, who gave me, you gave me the key. Where's the key? I'm so disorganized. And can you imagine what room did I get? 1 37. Okay? And it's not the first time it's ever happened to me. It's happened to me once before. It was really amazing. So you know those stores that you go and buy like sandwich and they give you a number, right? So in America there's some chains called Panera Bread. And I went to Panera Bread. They just opened it in Itaca where I live. And I went there and you know what number I got? 1 37. So I was so excited. I went the second day and you know what? I get 2 1 37 in a row. Anyway, so we all know about it. And actually we measure it extremely precise. And one thing I want to mention already now that this 1 37 is correct only at a very low scale. When we go to a higher scale, you actually change the thing. So that's it. We have a model. We make two measurements. We measure the mass of the electron and alpha. And then we can actually start making predictions. So we assume we're going to discuss some of the prediction. What are the Feynman rules of this model? Feynman rules that mean I want some vertices that give me some interaction. So what are the Feynman rules of this model? I look up and I said, oh, there's only one interaction term. This interaction term comes from the kinetic term. And it's actually just this. It's electron, electron with a photon. Or if you like, I can write it e left, e left, and e right, e right if you prefer. Or just write the electron as a Dirac field. And what I have here? Here I have the e times or minus e times gamma mu. Very easy. So I have my Feynman rules and then I know how to do any calculation that I want. And last question I want to ask is what's happened with P and Cp. So what's happened about parity? Is parity conserved or not? Yes, so parity is actually conserved in this system. Let me ask you the following question. Is parity is an accidental symmetry of this model or an impulse symmetry? Did I impose parity? No, so it is accidental. So do you think parity should be broken by high order terms? Yes. Do you have an idea what is the full high energy theory that actually give you parity violation for high order term? The weak interaction, right? So we know that high order term break parity and they actually come, they end up here from their weak interaction. Anyway, so now I want to go and do my, so I hope this is simple. But if not, please ask me a question. Even if it's simple and you want to ask me a question. Yes. Do I, why? Or why m cannot be complex actually? Let me rephrase. So he asked why m cannot be complex. And let me rephrase the question. And I said, you know that m is complex. Why you don't care about it? Okay? m is totally a complex number. There's no reason for it not to be a complex number. Right? So why I don't care about the fact that m is a complex number? Yes. So the answer is somewhat technical that we can actually absorb the face of it. And either tomorrow or on Friday, I plan to teach you how to count. It's very, very nice. And we're going to count until 18 for another reason. And then we're actually going to see exactly what is physical and what is not. So at this stage, we just say, oh, it's easy to see that this is not physical because we can rotate it. Another way to think about it is that actually what is really physical is always the mass squared. That's what really appeared in the dispersion relation, is the mass squared. Okay? But you are totally right and we have to be a little worried about it. Any more questions? Yes. Oh, it has to be dimension six. Okay? It's go under the name of the Euler Heisenberg Lagrangian. And this Euler, it's a very interesting story. So if you never heard of the Euler Heisenberg Lagrangian, it's a dimension six term. And it's how come that Euler and Heisenberg work together? So Euler was 200 years before Heisenberg. Anybody could solve the puzzle of what is the Euler Heisenberg Lagrangian? And for years, I was sure that it's how somehow Euler's got the, you know, Heisenberg learned something from Euler. Anybody know? So there's actually a second guy named Euler who was actually a student of Heisenberg. And I mean, I guess I don't know if it's luck or unlucky to be called Euler and you do some fun stuff. Anyway, so there's another Euler. It's called the Euler Heisenberg Lagrangian and it's basically just a dimension six operator. Okay? So if you don't have more questions, I want to ask you some questions and let's do our usual I'll be quiet for two minutes and you'll be talking to each other kind of exercise. And I want to ask you the following questions, okay? So, so far I gave you the model with one generation. Now I want to add a two generation. So instead of having just one set of EL and E right, so before it was E left and E right with a charge minus one. And now I actually put an index I on top of it and I go from one to two. Okay? So instead of having one left-handed and one right-handed, now I'm telling you I have two left-handed and two right-handed. And I'm asking you to write the Lagrangian, basically what I did before and modify it, tell me how many parameters in this model and what are the accidental symmetries of the model? Is the question clear? Start working, start talking. It was so quiet. You need to talk to your neighbors, make friends. By now you all know each other or almost, right? So these are the questions. Let me write the questions here and then I go over. So what is L? The number of parameters and the accidental symmetry. Okay? And let me go back one. Okay? So this is the case for one generation and I will ask you to modify it. And in order to save time, if something doesn't change, you just say it's the same. Just write the things that are different and tell me how many parameters there are and what is the accidental symmetry? Question? You done? This is the master. Good. Yeah, yeah, that's okay. But what is the number of parameters? How many parameters? How do you know? It looks to me. Why am I? You should write everything that is allowed. Write everything that is allowed. What else is allowed? I really want you to write. Okay. So someone who is willing to tell me what is the Lagrangian now, what's happened to the kinetic term? So the kinetic term, we just add also kinetic term for the other things. What happened to the master? I'm interested in a master. What's happened to the master now that I add two generations? What it will be? Someone? Yes? We have a math matrix, very, very nice. So what I have now, I have M ij, e left bar i, e right j. And I want to emphasize the fact that this is a matrix and how many parameters there are in this matrix? Four? Maybe it's complex. So maybe it's eight. So let's do this. So is the mass itself should be a machine? Yes? Or the only the Lagrangian. So when the mass should be a machine, if you have the same vector on both sides, if you have two different vectors, it shouldn't be a machine, right? It's only a machine when you have seen on the two sides. So actually this math matrix, when you write it, there's no, it's the plus emission conjugate that makes a mission. But let me, let me not go there for a second, because what we want to do when we see a matrix, okay? When you see a matrix, what was your initial reaction? Diagonalize. It's very good, right? So if you go in the street in the kind of a down and suddenly a matrix come in front of it, diagonalize it, right? We're all good in this. So let's diagonalize it. And after we diagonalize this matrix, we get here two eigenvalues and we can actually choose the basis that they are real. So we choose two real things. And let me ask the following question and I'm not going to answer. Can I always diagonalize any matrix that I want? Can I diagonalize all the matrices that appear in physics? Only mathematicians care about matrices that we cannot diagonalize? Is it correct? You should say of course. I mean, do we ever see in a real world matrices that we cannot diagonalize? Anyway, so basically all the matrices that we care about as physicists are nice. They don't have all those crazy property that mathematicians spend so much time on them, okay? So we basically can diagonalize this matrix. One thing that we have to worry, a little worry about is that if I diagonalize this matrix maybe some other matrix become non-diagonal. I don't have to worry about it yet, but hopefully by tomorrow we're actually going to discuss a lot what's happened in this case. And you know it's the kind of CKM. Okay, so after I diagonalize this and I get two eigenvalues and in general these two eigenvalues are different. How many parameters do I have? How many parameters do I have in this theory? Three. So the answer, the correct answer is three because I have two masses, the two eigenvalues and one coupling constant. Yeah? So actually in this case the mass eigenstates are also the interaction eigenstates. And the reason is that when you diagonalize this matrix, actually nothing happened to the kinetic term. And the reason that nothing happened to the kinetic term because it's popular to the unit matrix, so you actually end up with three parameters, okay? What are the accidental symmetries of this theory? So now I have a theory that I have two fermions, each of them have its own mass. What is the accidental symmetry of this theory? So we already talked about parity as an accidental symmetry but there's one extra accidental symmetry in this model. Extremely interesting accidental symmetry. Yes? Which one? A U2 symmetry. So a U2 symmetry would be an accidental symmetry if I take the masses to be actually degenerate. So if the masses were degenerate then we actually have a U2 symmetry, okay? And if the masses were zero then we have U2 squared symmetry. But what is the symmetry that I have in this where I make the assumption and let me just call one mass eigenstate, the mass of the electron and the other one, the mass of the muon. So they are not degenerate. So what is the accidental symmetry? What? Which one? Which number? Lefton number. Thank you. I didn't hear this lefton number. So what we see after we diagonalize this matrix, we see that we have a mass eigenstate and a kinetic term for the electron and a mass eigenstate and a kinetic term for the muon. So the electron can rotate and the muon can rotate and they don't talk to each other. So we actually have what we call lepton number and we have U1 electron times U1 muon. And I impose only U1 electromagnetism so I impose only one U1 and I end up with U1 squared. So if you like one combination of U1 is accidental and I'm not going to discuss too much into this, it just already we see that lepton number is an accidental symmetry of QED. And of course we can do the same with three eigenvalues and we get also the tau and we will be done. So I was just telling you about QED and there's actually a lot of experimental tests of QED and I'm not going to mention many of them, I just kind of tell you that QED is doing extremely, extremely well. Let me mention one far from trivial experimental test is the fact that the QED predicts one of our potential between charged particles. It's so obvious that we never actually mentioned this because of course we know that that's why we built this model like this. But it's actually a very good test of the model but that's how we built it. And then the non-trivial test of this model, the most famous one is of course the G-2 and you actually probably heard about the fact that we measured the anomalous magnetic moment to very, very high precision to 10 decimal places and it's basically agree with QED and it's so precise that we actually need also a weak correction to this in order to agree with the experiment. There's many, many more scattering, etc. and I'm not going to get into the details. And because this theory is actually perturbative and we have a small alpha, it's kind of small and we can do perturbative calculation, we can do very, very high precision calculations and we can do very precise experiments and when we do all this we actually see that we are already there and we start seeing deviations from QED and we start caring about the weak interaction. So that's all I wanted to say about QED. Any more discussion about QED? Yes. Oh, why do we assume that we have the same coupling to both generation? So let me say the following. We don't really assume it should be all out come output. So what is really the ingredients that guarantee this that they do have the same coupling and that's also go under the name of lepton universality. Lepton universality is the fact that the electron, the muon and the tau couple with the same strength to the photon. So what is the thing, what was the ingredient that I put in as an input that guarantee that the output is that they have the same coupling? What was the input? So I could have one u1 and I can actually make them couple very differently. There was some other thing that I put in that guaranteed that they couple the same. Yes? Yes, so it's actually the fact that they do have the same charge. So the coupling is defined by the charge. The charge, we talked about it at the end of the morning lecture. The charge is what tells you the strength of the coupling. If you look at the Feynman diagram, it's the charge that appear here. And since I choose the two generations to have the same charge, then I guarantee that they have the same coupling. It's the input. I put an input that they have the same charge, okay? And of course it would be really interesting to ask the question, what's happened when they don't? And then this model doesn't describe nature. And I will, hopefully as I said, tomorrow we're going to have one of these homework exercises, a model that doesn't work. And in this model, actually I play with the charges and it's not going to work. Any more on QED? What do you think I'm going to do now? Nobody gets it. Thank you. Yes, it's kind of an obvious thing. QCD is just like QED. It's just non-Abelian versus Abelian. So all I need to do is just instead of writing Q, I write some matrix and everything follows, okay? So let me talk about QCD. And I want to define it because I really love to finally do model building because I know the algorithm. So I provide the input. So what is the input? I provide a symmetry. The symmetry is SU3 and I call it C4 color, so it's SU3 color. And I provide six left-handed fermions and six right-handed fermions. And these fermions are called quarks, okay? And there's no scalars in this model. So we do the same story that we did before. And I provide, I say that the left-handed and right-handed quarks are fundamental or triplet under the SU3, okay? So is this theory vectorial? Yes, because you see that the left and right are transformed the same under the symmetry. Okay, so then we ask the following, the same question. Let me write the Lagrangian. I have the four terms in my Lagrangian. And obviously the Yukawa and the scalar potential is zero because there's no scalars. And the kinetic term is the same as with this QED. The only difference is that instead of calling it A, we call it G for gluon. And there's another index because it's a non-Abelian symmetry. And basically it's the same. What you see is the Lagrangian looks almost the same as the Lagrangian of QED, okay? So let me ask the following question. How many parameters? What are the Feynman rules? P and C and accidental symmetry. So how many parameters? Seven. We love this number seven, right? It's always the number that come out. We actually, for example, you give seven, right? But actually it happened to be the right answer. Kind of interesting, right? So why seven? Because we have one coupling constant and the coupling constant we call G strong. The coupling, so instead of the, instead of G or Q, we call it G strong. And then we can also define alpha strong. It's defined as G strong squared over four pi in analog to alpha, which is the coupling constant square over four pi. And the six masses that we have, the six mass of the six quarks, okay? What are the Feynman rules? So the Feynman rules are basically the same as we have here. And instead of E, I put here a Q. Q left, say. And instead of the photon, I put a gluon. And here come a very big difference. I don't know why, but the gluon is like this. Very nice, huh? And now it's very different, right? So that's a gluon. And what appears here in the coupling, before it was just the charge, and here there should be some matrix. Which matrix we want to put here? The Gell-Mann matrix is because they are the one that corresponds to the three of SU three, okay? So it's basically the same as Q, as Qd in this level. It's very simple. We just make some small changes, okay? What about P and Cp? And it's a vectorial theory, so it should conserve P. About Cp, at the level that I wrote to you, it's conserved Cp. But there's actually some subtlety going on here, which I'm not going to discuss to great details. It's called a strong Cp phase. And it's, in principle, it's there. It has to do with a very complicated non-perturbative stuff. What are the accidental symmetry of this theory? What is the accidental symmetry of this theory? So when I write the Lagrangian, I will just see that after I diagonalize my mass matrix, I have, for each quark, I have a master and a kinetic term. Master and kinetic term, and all the interaction give you the same I. It's the same I for the interaction. So what is the accidental symmetry of this theory? One, six, six what? You want to the six. So the accidental symmetry is kind of you want to the six, and part of one, one kind of part of it is part of the SU3. But it does tell us that each quark have its own conserved quantum number, and basically tell us that in Qcd, quarks number cannot change. So in Qd, electrons stay electron and muon stay muon, and in Qcd, a uptight quark stay an uptight quark. And we give them name, and what the names that we give those quarks, we give them the name according to the mass. So according to the mass, we give them the name like this, U, D, S, C, B and T. Okay? And that stands for up, down, strange, charm, bottom and top, or beauty and true. But nobody really. Beauty we do use. Did anybody ever use truths for the top? No. But for bottom, we do use beauty. When you have this, you have the beauty conference, so you really want to go, right? If you say bottom conference, maybe you don't want to go. Question? Already solved question. Oh, you do? Yeah, instead of top? Nice. I didn't know. That's really good. So you say, oh, what you are doing, I say, I do true physics. And you do false physics. I never realized this. Okay. Really cool. Okay. So it looks almost the same, and I already conclude important stuff that Qcd cannot make quarks decay. And then there's one last little thing that it's actually kind of different, and this last little difference actually makes the phenomenology totally different, okay? And that has to do with running coupling constant. Okay? And let me ask the following question. How many of you heard about running coupling constant? Okay. Almost all of you. That's good. So let me discuss it because you know it. So what is really going on when we say that we have running coupling constant? What does it really mean? And what we mean is basically that in many cases, our three level Lagrangian may not be a good enough prescriptions because sometimes the higher order effects become important and significant compared to my experiment precision. Okay? And one way to actually take care about higher order correction is to start writing the diagram and calculate them. So for example, I can write this diagram. This is so nice. So before I have a three level and now I have one loop and one loop diagrams are higher order compared to three level diagram. So I need to calculate them and then how I actually start getting the correction to the scattering process that I have. Now if the coupling constant become bigger and bigger and stronger and stronger, eventually I need to calculate higher order terms. Okay? So now it's turned out that instead of doing all this calculation doing Feynman diagrams, there's an equivalent prescription which sounds very kind of out of the blue but is actually a really cool one and said I can actually think of my coupling constant as if they depend on the energy. Okay? So instead of calling my alpha s and alpha s, I call it alpha s as a function of the energy that is actually floating in the theory. Okay? And we know how to calculate it but at the end of the day, there's nothing about this, the running coupling constant more than just say I have to include higher order effects into my interaction. The interaction, the three level interaction is not enough to describe the physics because I have high enough precision. Okay? So that's the idea of running coupling constant. It basically tells us that the coupling itself is not really physical and I kind of already talked about it a little bit. What is physical is only experiment. Okay? I do one experiment and then I can make prediction for another experiment. The coupling constant are just a mathematical tool in the middle of the calculation. Okay? So then it's turned out that that's actually give us a very good description instead of thinking about higher loops effects and all this. I always do calculation as three level. Okay? I always think about a three level coupling and instead of thinking about it as some G strong, I think about G strong as a function of energy and to a very good approximation, it's capture all this one loop and two loop and whatever loops that I care about. Okay? So that's kind of the very simple explanation of running coupling constant and I would love to actually discuss with those of you who are interested about somewhat deeper aspects and how we actually deal with those infinities that happen in the calculation and why they are not really kind of we don't have to worry about them. Anyway, so at the end of the day, what I want to tell you about alpha strong is the following story. So when I look at electromagnetism, in electromagnetism, when I actually come and do an experiment in higher and higher energy, the coupling constant become stronger and stronger. Okay? And when I do calculate when I do calculation at the mass of the Z, alpha instead of being 1 over 137, it become 1 over 128. Okay? It's a little bit bigger, not by much, but a little bit stronger. When we do the thing with the strong interaction, because, and that's again, I'm not going to get into the details, but because that the glue on have self interaction. So remember this one thing that I say, oh, there's a difference that in QED, the photon do not interact with itself, but in QCD because it's non-Abelian, the glue on do interact with themselves. Okay? So this self interaction of the glue on turn out make a really cool interesting thing that is actually make the fact that when you go to higher and higher energy, the coupling constant becomes smaller. So in QED, when I don't have self interaction and I go to higher energy, the coupling becomes larger. In QCD, when I go to higher and higher energy, the coupling become actually smaller. Okay? So far so good, but now let's look at the other way around and say, what's happened when I go to lower and lower energy? So when I do in QED and I go to lower and lower energy, the coupling become smaller. In QCD, when I go to lower and lower energy, the coupling become larger. Until at some point, boom, it's become too large. What does it mean it become too large? It's become non-perturbative. And what's happened when things become non-perturbative? What can we do? What we do when we don't have a perturbative expansion? What do we do? Nice. I usually cry. So we kind of get stuck, okay? Because we have this, you know, our amazing tool of perturbativity is kind of gone, okay? And what is turned out? It's turned out that when you look at the strong interaction, somewhere roughly around few hundred MEV, the coupling constant become too strong and we don't know what's happened, okay? So the way we understand it, and it is in a way it's an open question in theoretical physics, it's actually to prove that that's what's happened. But what's happened is at some point, we just cannot actually think about quarks and gluons as fundamental fields, because quarks and gluons and fundamental fields are there only when I have a perturbative coupling constant. When the coupling constant is not perturbative, I cannot think about quarks and gluons anymore. And what we found out in nature is the phenomena called confinement. Confinement is the phenomena that those quarks, when I start taking them apart, taking them apart, maybe I go to the infrared, I go to low energy, I start to take them apart, and I just cannot do it anymore. Why? Because the system prefer to actually pop up a QQ bar from the vacuum. So the idea is as following, if I could do this experiment, I take two quarks, a quark and an anti-quark, and start putting them around. At one point, instead of keep putting them around, the energy, the system, prefer to actually make another QQ bar from the vacuum. So I have QQ bar, I start taking them apart, QQ bar, until at some point, the system will do something like this, where this QQ bar, I said, pop up from the vacuum. It's a very, very hand-wavy argument, and like all hand-waving arguments, it's kind of correct, but it's not precise. So what's happened is that when I have the quarks at high energy, eventually those quarks, when I go to low energy, must come together with something that is a color singlet. Okay? So what we have, we have the following really interesting phenomena, and I want to emphasize it. It looks so, so similar to QAD, except this one little thing that I have, the gluon self-coupling, and because the gluon self-coupling are there, it's turned out that the running, instead of being positive, has become negative, and because the running become negative, suddenly I have this amazing difference that at low energy, I cannot talk about quarks and gluon anymore, and I have to talk about some other objects, and these other objects, I do not have a fundamental filter to describe, and these other objects are called hydrons, and these hydrons are divided into bions and mesons. Bions have spin half, or spin half integer spin, and mesons have spin as an integer spin, mesons are bosons, and bions are fermions. And the point is that at the fundamental level, we don't know what to do, okay? And then we have to do a lot, a lot, a lot of things that are not very precise, and you write models, and you use some approximate symmetries, and you kind of going around, and there's a lot, a lot of fun going on, but our fundamental tool of perturbativity only works at the UV. So I can only do perturbative calculation of QCD at very high energies, okay? Good. So let me talk about some experimental test of QCD, and we did it at the UV, and we actually see that the proton is made out of quarks. How do we see that the proton made out of quarks? We take an electron, scatter it off the proton, and when the energy is high enough, we start seeing kind of a scattering of three different particles, okay? And we see the running of alpha s, so this phenomena that I was telling you about alpha s, you can see, you can actually measure the coupling constant at the z pole, and at the z pole, you get alpha s equal to 0.11, and you can also measure alpha strong at the tau, and tau decay, and then you get 0.34, okay? It's a factor of three different, and you also measure it at the b, and then the b is alpha strong is 0.22, okay? So you really see the difference, and you measure alpha strong at different processes, and you get a different number, and this is a number I totally agree with our calculation, okay? There's a huge number of other tests of QCD, of perturbative QCD that I'm not going to get into, but the point at the end of the day is the following. We understand QCD as a theory in the UV, and when it comes to the infrared to understand QCD at the infrared, it becomes much, much more complicated, and I'm not going to get into much details, and hopefully tomorrow or on Friday, we're going to talk about sometimes how we can actually get some idea of what we do in the QCD in the infrared. Any questions? I'm sure you have, but we'll go on then. Okay, now we can actually move to the weak interaction and the leptonic standard model. So what I'm going to do instead of jumping directly into the full standard model, because of QCD and QCD is a little bit complicated, I want to start first with a model that only talk about the leptonic on the leptons and then we go on. And just kind of a little of history, when Weinberg wrote his famous paper that explained the weak interaction, the title of the paper was a model for leptons. So when Weinberg actually understood the standard model, he only cared about leptons, and only later on the quarks were kind of put into the game, okay? So I want to do just the leptonic standard model. So what I'm going to do, I'm going to tell you the model and we're going to write the Lagrangian and keep going like we did before. Good. So what is the symmetry of the standard model? The symmetry is an SU2 L cross U1Y, L stands for left and Y stands for hypercharge. How come Y stands for hypercharge? Because the second letter in hypercharge is Y, no? I don't know. So I don't know why it's called Y, U1Y. And I give three fermions generations. They are the famous three generation. And in each generation, I have actually two different things. It's L, which is a doublet under SU2 and have hypercharge of minus a half, and E, which is a right-ended field, which is a singlet under SU2 and have a hypercharge of minus one, okay? And there's one, only one scalar, and this scalar, it's called the Higgs boson or the Higgs field, and it's actually a doublet under the SU2 and have a hypercharge of a half. And because of this, we can actually have spontaneous symmetry breaking that correspond to SU2 cross U1 down to U1 electromagnetism. So before I go on, I want to talk a little bit about names, okay? So I was telling you that when we write a model, I just tell you the symmetry, I tell you the representation, and I write the model in a Lagrangian, and then I have to measure the parameters, okay? The model that I gave you, depending on the value of the parameters, either could have spontaneous symmetry breaking or not. It depends on the value of this, on the coefficient of the phi squared in my theory, okay? So this standard model that I defined it so far, it actually could be the standard model that we know with spontaneous symmetry breaking and the one without it, okay? Now, the name, when we say the standard model, we mean the one that we have spontaneous symmetry breaking, okay? So far, I don't know if I have spontaneous symmetry breaking, I have to go and measure it, but I still call it the standard model where I know that they're going to be spontaneous symmetry breaking, okay? I don't have to, but the standard model is only the one with spontaneous symmetry breaking. Good. So let's say, keep going, and we have four terms, and let me ask the following question. Which one is zero? L psi? Yeah, you keep trying until you get it right. That's good. Yeah, in this one, you're totally correct this time. And how do you know it? Because it's a chiral theory, actually. Look how chiral it is. Do you see how chiral it is? Both this, you two, and the U1 are different. You see, it's two minus a half, one minus less. It's a chiral theory. A chiral theory cannot have much time for the fermions. And indeed, that's mean that L psi is zero, okay? For the kinetic term, I'm going to talk about next, in the next transparency, and then we have a Yukawa interaction. It's new. We didn't have it in QED and QCD. And the Yukawa interaction is something like Y, Y is a coupling, and it's a matrix, and then I have L bar E times phi. Okay? And you can see that this is okay. Okay? It's have a doublet, singlet, doublet, which can be a singlet, and it's have a hypercharge. The L bar is a minus a half. The E r is one, and the phi is a my, my, sorry. Ah, let's go back. Check. Ah, the L bar is plus half. The E is no. L bar is a plus half. E is a minus one, and phi is plus half. So half minus one plus half equals zero. Okay? It should have been easier for me, but it wasn't. But then you see that is invariant. Now you see. Okay? And then I write the scalar potential. So I have the Yukawa interaction, and I have the scalar potential, which is nothing but the famous Mexican head that we discussed. Okay? It's just x squared plus an x to the four. Right? You see that is x squared and x to the four. And I choose my, the science such that I want lambda and mu squared to be positive, and then it's, it's this kind of, the minimum is not at the origin. Okay? So let's start discussing differences and see what, what come out from this theory. Okay? So let's start with the kinetic term and the SU2 cross U1. So we have four gauge boson degrees of freedom. Why four? Because in SU3, in SU2 I have the, the gauge boson has, has to be an adjoint of SU3. It has to be a vector. So there's three, and the U1 have one degrees of freedom. And this is the covariant derivative. And the covariant derivative has two terms. One term that has to do, to do with the coupling into the W boson, and it's perpendicular to the matrices T. And one that is perpendicular to the coupling to the U1 gauge boson, which is B. And the coupling is perpendicular to Y. Okay? So Y is the number that I import, that's the input. And the T's are the matrices that are there, depend on the representation. So if I have something that is a singlet, this T is zero. So there's no coupling. The singlet of SU2 is not coupled to the W. And if I have a doublet of SU2, then these matrices are just the Pauli matrices, after a factor of two. Okay? So I actually know how to write these things. So what I'm asking you to do is to actually write the covariant derivative for those fields. So write the covariant derivative. You know, the usual thing that we are doing, I do it twice today, right? And write the covariant derivative explicitly for L and to N4E. Do you understand the thing? So I gave you the model. And in the model, I tell you what is L. L is a doublet. Ah, shit. There's one extra number that should not be there. It's the one from the SU3, which I forgot to take away when I was... Anyway, L is just a doublet and a minus a half. And E is just a singlet and minus one. And I want you to write the covariant derivative for them. Okay? Not so clear. Okay. So let me do the following. Let me write the covariant derivative of one field, and then you will do the other one. Okay? So what is the covariant derivative? I'll do the following. Ah. Okay. So how do I write the covariant derivative of the L field? I said the covariant derivative in general, I have to ask what is the charge, what is the representation of the SU2? And since the representation is a doublet of SU2, then in front of the W, instead of T, you see I have TA on the top, for the L, instead of the TA, I get sigma over 2. So you see that the T is replaced by sigma? Yes. And for the Y, for the B, for the B gauge boson, which is the gauge boson of the U1, instead of Y, I have the hyper charge, which is minus a half. Now I do the same for the E. And for the E, it's a singlet of SU2. Since it's a singlet of SU2, instead of the T, I have zero. So you see there's nothing there. And instead of the Y, I have minus one, and you see that I have minus one. Okay? So now, let's do this exercise. Now that I hope that you see those, can you write the covariant derivative of the scalar? Of phi, which is one, two, and it's a doublet and a half. I don't know how many times I look into this transparency, and I didn't see that I kept the SU3 index. But I should take it away. But do you see? Don't worry about the first index. It has the index and the SU3 color. It's a singlet, so we don't care about it. Can you write it down, please? Go right down now, the covariant derivative of the Higgs field. Can we? Please, please. If you don't have a paper, I do. Anybody need a paper? Because Shachar gave me some. Anybody need some paper? Look for the nice logo. Yes? Write the covariant derivative for the Higgs field. Okay? I really hope that you understand how this kind of algorithm works, okay? So I know what is the charges, I know what the representation, and then I write the covariant derivative, and that's what happened in the kinetic term for the fermions and for the scalars. Okay. So let's do it together. d mu is equal to d mu phi is equal to d mu. What should I have here? What is the coupling to the SU2? Let me write it. It should be plus i g w a mu, and now it should be, what should be the TA that I have? I'm missing you guys. Sigma, how do I write sigma? Sigma, do I need something else? Oh, oh. Now you'll start talking. Good. And now let's talk what about the coupling to the b mu. What is the value of the hypercharge? What is the value of y? It's plus, so I have plus i g prime b mu over 2 times i. Okay? Good. There's no more field, so I cannot do more exercise, but I can actually invent some fields, and I can still give you an exercise for tomorrow. Okay? Maybe we'll do this. Okay? So we understand how we do it. We understand how we get the covariant derivative, and once we have the covariant derivative, we know how to write our kinetic term, which is just the covariant derivative instead of the standard derivative. Okay? So we actually done with writing these things. And the importance of the kinetic term and the covariant derivative is that it actually gives me the gauge interaction. It's actually the thing that provides us with the coupling between the gauge boson and our fields. So now let's talk about spontaneous symmetry breaking in the standard model and the leptonic standard model. So we look at the scalar potential, and when we look at the scalar potential, we look and it's basically just phi to the 4 minus phi squared. And if the coupling are positive, then I have breaking. And as I say before, we don't really know. We have to measure, and the way we measure it is by the fact that actually the electron is massive. And just by noting that the electron is massive, we know that we have spontaneous symmetry breaking. So we know that mu squared is indeed positive, that's indeed spontaneous symmetry breaking. So what we have? We have actually four degrees of freedom for the phi field. So the symmetry, it's an SU2 cross U1, and we have a degenerate vacuum. The degenerate vacuum is just like the one that we have in the U1 case. So in the U1 case, we just have this Mexican head. Here it's a little more complicated. It's a four dimensional space. Four dimensional space are a little hard to imagine, but still it's a four dimensional space. And all I know, I know the magnitude of the vacuum expectation value. I know the magnitude of the minimum. I just don't know the direction of the minimum. Okay? Just like the case, you remember that we put the magnetic field in the z direction. So we have to choose the direction of the minimum, and we have to choose the direction of the minimum in my field space. So I have my field phi, and my field phi have four degrees of freedom. Why do I have four degrees of freedom? Because it's a complex scalar, and it's a doublet. Okay? It's a complex scalar and a doublet. So a doublet means there's two degrees of freedom, and each degrees of freedom is a complex number, so it's two. So I have total of four degrees of freedom. So I have to choose one direction where I want to put the vev. Okay? Anybody know which direction you want to choose? Okay, let me ask it a little different. Do you think you care which direction you choose? No, you can choose any direction as long as you choose. Okay? Okay? You cannot just say, I don't want to choose, then you don't break the symmetry. You have to choose the direction, and this is totally arbitrary. Okay? So the direction that we choose is to put the minimum in the real component. We have a real and imaginary component. We put it to put in the real component, and we choose to put it in the lowest component. Okay? So I can write my field phi. I can write phi as phi one plus i phi two and phi three plus i phi four. Okay? You see explicitly this is the top component and the bottom component because it's a doublet, and each of them is a complex field, so I have phi one plus i phi two. Phi three plus i phi four. Okay? And I choose to put the wave here. I put the minimum in this direction. So I have four dimensional space, and I choose to do it here. All the physics that I'm going to talk about, of course, is irrelevant. I could choose any other thing. It's just that once I choose this, that's the standard notation, and in all books, you will find the same notation which is the notation that we choose this one. It's exactly the same story that we choose the vector, the magnetic field in the z direction. Okay? Now, after I do this, I actually ask what is the spontaneous symmetry breaking, and the spontaneous symmetry breaking is SU two cross U one down to U one, and you can ask yourself, how do I know it? And the answer is that there's actually some formal way that you can do it, and if you've never done it, I can actually explain to you how you do it. It's kind of cool, but at this stage, I'm just telling you, I kind of tell you that we know that this is what happened. And remember that we say the same story with when I put the vector, I break the SO three down to SO two. Here, I break the SU two cross U one down to U one. So I start with four, I have four gauge bosons, and since I break three of them, I should have three massive and one massless. We talked about the fact that I can have spontaneous symmetry breaking that is only partial. So this is the case here. I have partial spontaneous symmetry breaking. My full symmetry is four dimension. It's SU three cross U one, SU two cross U one, and the unbroken part is U one. But the unbroken part is not the U one that I have. It's U one which is some combination of the original U one and some of the SU two that I have. So there's some combination of those that is unbroken. Okay. And we call the reminding symmetry electromagnetism. Why we call it electromagnetism? Because we want this model to actually reproduce QED. So QED should be the low energy part of this model. Okay? Why? Because we want to describe nature. Kind of a good reason. Okay? And when we actually do this breaking and out of these four degrees of freedom, three of them are become, give masses to the other, and I have one massless scalar. So we didn't discuss it to great detail before. So I'm just telling you that at the end of the day, the spectrum is such that I have three massive gauge bosons, one massless out of the four that I started with, and out of those four degrees of freedom, only one of them is physical and it's called the Higgs boson. Okay? So it's actually a very, very, very detailed prediction already at this stage. It's a prediction that I have one scalar degrees of freedom in this model, and I have three massive, massive gauge boson and one massless gauge boson. Okay? Okay? So where is QED? So we just told you that I want the U one that is unbroken to be QED. And then I identified the unbroken generator. So the unbroken generator is a combination of T3 plus Y. What do I mean that the unbroken generation is T3 plus Y? If I take any field that I, if I have a doublet, say I have a doublet. What is the T3 of the top component of the doublet? Okay, let me say what is T3. T3 is the eigenvalue when I apply my sigma 3 on it. Then let me, let me back up a little bit. Let me back up a little bit. You remember when we did, say, spin in quantum mechanics, and you say when I do spin, there's one operator that is diagonal, and we always choose this operator to be sigma z. You remember? Why we choose it sigma z? Just because, of course. You choose whatever you like, and we always choose it to be sigma z. So here we do the same. And when I have a state, I know I can identify my state with the eigenvalue of the operator sigma 3. So if it's a spin up, and I measure the, when I apply sigma 3 on a spin up, what would I get? What is the eigenvalue of sigma 3 on spin up? One or plus half. What is the eigenvalue of sigma 3 when I apply it on a, or sigma 3 over 2? When I apply it, when I have the, the spin operator, when I apply no spin down, I get minus 1 or minus a half, okay? So T3 is the equivalent of sigma z. Is the diagonal operator? I should have explained it. I didn't explain it good enough. So T3 is the diagonal operator that I use in the SU2, okay? So now when I apply a T3 on a top component, what do I get? I get plus a half. When I apply T3 on a bottom component, what do I have? Minus a half, okay? And then I'm telling you that I define electric charge, electric charge is defined to be the combination of the eigenvalue of the T3 plus the eigenvalue of y. And what is the eigenvalue of y is what I gave you. If it's an L, then what is the eigenvalue, what is y of L is minus a half, okay? And now I can actually do these little cool things. And I said if I have a doublet, I can start actually telling the different electric charge of the different component of the doublet. And of course, once the symmetry is breaking, I can start telling the difference between the top and bottom. Because when the symmetry was a good symmetry, I cannot tell the difference between them because they were symmetric. Now I break the symmetry so I should be able to tell the difference between the top component and the bottom component. And in particular, I can actually calculate the electric charge. So let's see what is the electric charge of the top component. So the electric charge of the top component for the phi, for example. So what is the electric charge? Let me write it here. So the electric charge is T3 plus y, okay? So let's look at the higgs. So here for the phi, phi. So for phi I have y is equal to a half. And what is the T3 of the top component? Here the T3 is a half. And here the T3 is minus a half, okay? So what is the electric charge of the top component? What is the Q of the top component? It's half plus a half, so it's equal to one. And what is the electric charge of the bottom component? It's half plus minus a half is equal to zero. And I can do the same for the fermion field that I have, the electric, the doublet of the electrons. So the doublet for the left for L, for LL, y was equal to minus a half, okay? So for the top component I have T3 minus a half, which is equal to zero. And for the bottom component T3 minus a half is equal to minus one, because here T3 is half and here T3, okay? And how would I call the top component that is the fermion with a zero charge? Let's call it a neutrino. We just learned from Andre that it's used to call a neutron. And then we call it a neutrino. So we call it a neutrino. And now we call the bottom component the doublet charge of a minus one. We call it an electron, okay? And of course at this point there are just names, but the whole idea is that these names correspond to the particles that we do see in nature. And we already see kind of nice thing that this one is a left-handed massless fermion, which is the neutrino. The neutrino is a massless fermion. And don't worry about the fact that it's a very, very small mass, because of course this very, very small mass must come from things that we neglected, right? So this is the fact that we have massless left-handed fermion that we call the neutrino. And here we have something, and soon we want to see where it's going to mass from, but it's something that has electric charge minus one, and it's soon going to be understood that this is the electron, the physical electron, okay? So far I just kind of play with the Lagrangian. And when I play with the Lagrangian and talk about spontaneous symmetry breaking, I kind of see where I am. I kind of see that after the breaking I have one unbroken generator that I identify with electromagnetism, and I kind of see what are the charges of those things. So now let's see that this really looks like the photon, okay? But before I do this I actually want to understand how we get masses. So, so far there was no masses for the fermions. Why there was no masses for the fermions? Because the theory is good. It looks all of you look like so tired because it's the end of the day. I know, I know I'm too. But you understand it's kind of so I cannot, but I do have a, I do get mass after spontaneous symmetry breaking. So you remember we talked about it that when I have spontaneous symmetry breaking, Yukawa terms can give mass to the fermion. And that's exactly what I have here. I start with the Yukawa interaction, and the Yukawa interaction is just the L, L bar E phi. And after the phi acquires the wave, then it becomes a number, then I actually get the mass for the fermion. You see up there, instead of a phi, I put a number, and then the mass is just the number, the wave times the Yukawa coupling. So actually, we see that the electron acquires a wave. However, it's only happened to the electron because only the electron has Yukawa coupling. So only the down component of this L, only the down component of this L is coupled to the E right. And the up component has nothing to, to couple to. So the up components stay massless, and the down component become massive, and we call it the electric field. So all we are doing is that actually I write the Yukawa interaction. I just follow the procedure that I was explaining you. And I find that I have one that the down component coupled to the E right and become a massive Dirac fermion that I call the electron, and the top component is the, it stay massless, and it have no electric charge, and that's the one that I identify with the Neutrino, okay? And we have three of them, and when I have three of them, we already discussed it. I diagonalized it, and I have three electron-like particles that we call the electron, the muon, and the tau, and each of them come together with its own Neutrino. So I have three Neutrinos. So I already made a huge progress, and I know you are not too impressed by this huge progress because I really didn't give you enough model that do not work, okay? And if I give you many models that do not work, you would be very impressed. So I want you to be impressed even though I didn't give you. Are you impressed? Maybe you are not impressed because you saw it so many times before, but sometimes you get used to these kind of things, but you should still be very impressed, because we put all those inputs and you see what we got. We got really nontrivial prediction. We got a prediction that we have massless neutral fermion that is only left-handed, okay? Really nontrivial, and a massive fermion that have a charge minus one under electromagnetism, okay? Which is what we see in nature, okay? So it's already a lot of progress, but we are not done yet. We're going to see that these are actually coupled the way we like them to couple and all of those kind of things, so that's what we're going to see next, okay? No, I cannot be impressed with you. It's really impressive. I mean, I don't know. You are like, everybody see it, yeah? We heard it many times. The standard model is so impressive, really, you know? Maybe next time that's what I should do. I have seven lectures, six lectures I give you wrong models, okay? And then by the seven, yes, yes? So the point, so the question is, I choose to be the VEV in this direction. The question is what's happening if I choose it in that direction? How would I would see, actually, that I have a massless fermion that doesn't have a charge? Is that the question? No, sorry. Yeah, yeah, so that's what I was saying. So the point is that it's following. No matter which direction you choose the VEV, there's always a U1 that is unbroken. And it's always this U1 that is unbroken is what we call electromagnetism. And it's the same story with the magnetic field in the hydrogen atom. So second hydrogen atom. And we choose the magnetic field in the Z direction. And then you see, wow, it's amazing. It's actually have splitting in the M quantum number, right? What's happening if I choose it in the X? It's not, it's not giving splitting in the M quantum number in the Z direction. Now it's giving splitting in the M quantum number in the X direction. But the physics is totally the same. I would just call instead of walking in the basis that the Z is the basis that I quantize around, I quantize around the X axis. Here it would be totally the same. No matter which direction I choose my VEV, I always know that I do have a massless gauge boson. We're going to see it soon. And I always know that I have one particle that the charge under the unbroken one, which I call here, I call it T3 plus Y. But if it would be in another direction, let's say I choose the top component, it will be minus T3. And if I choose the imaginary part of the top direction, it will be minus I T3 or something like this, okay? So the definition of Q would be different, depend on which direction I choose. But the physics will be the same. Okay, so let's go on. And let's talk about the accidental symmetry of this model. And this model have a U1 cubed accidental symmetry. And dimension five actually breaks it completely. And total left-to-number is the sum of all those. And I'm just mentioning it because when Andrei going to start discussing neutrino masses and left-to-number violation, I want you to remember that the dimension four, this is a symmetry. So we cannot have neutrino mass. The mass of the neutrino is exactly zero in this model. However, it's exactly zero in this model at dimension four. At dimension five, we do have mass for the neutrinos. And that's what we're going to see when Andrei we're going to discuss neutrino masses, okay? So I just mention it and he will get into the details. So now I want to talk about the gauge boson masses. And the point is the following. I was telling you that I know before I do any calculate, I actually did the calculation and many people did it before me. I would just copy what they did. But I did it myself also because I had to write a homework for my student. So I did check it. And what you find out that I have SU2 cross U1 down to U1. That is, I know that I have three generators are broken and one is unbroken. So I know that I should have three massive gauge bosons and one that is massless. I know that what I should do. So now I actually can calculate and see the mass metrics for the gauge bosons. So what I do? I do the following things. I write the covariant derivative. I write the kinetic term for the Higgs field. And from the kinetic term for the Higgs field, I get masses for the gauge bosons. And which kind of did it in the morning, I kind of show it to you. And what I will ask you to do and this will be tomorrow in the session of the homework, I ask you to actually do it, okay? And by doing it, I mean you take this d mu phi. So d mu phi is the matrix because phi is a 2 by 2 vector. And the d mu is actually a 2 by 2 matrix. So it's a 2 by 2 matrix times a vector. And then you multiply it together. So you have something long to calculate. And at the end of the day, you give you some numbers and you open the parentheses and make sure all the indices are right. And then what you're going to find out is some matrix. And then you look at this matrix and you diagonalize this matrix. And three hours should be enough. Actually, I really hope that you will do it in half an hour. So you have time for more questions. And you just do it, okay? And maybe then you will be impressed, okay? Because then you really kind of start calculating things. And you find out, and I said, you will do it. You find out the following thing. You find out that the mass of the w is given by quarter G squared, V squared, V is the wave. mz squared is whatever it is. And ma is 0. So what we find out, we have four degrees of freedom. Three of them are massive. The w plus minus are two degrees of freedom. The z is the third. And a is the fourth one. And actually, we can also see what are their charge under u1, the unbroken u1. And the w plus minus have charge plus and minus. The z have charge zero. And that's why it's called z, I think. z for zero. Is it correct? Yeah. Okay? At least it's a good guess. Okay? And ma, which is the photon, is massless. So we see really cool stuff coming out. We see that we have a massless gauge boson. And we have three massive gauge bosons. Okay? And what we end up seeing is that we have kind of a mixing. Again, we're just a basis choice. We just move from one basis to the other, like we're always doing quantum mechanics. So what we see is that the w plus minus are just 45 degree combination of the w1 and w2. But the a and the z are kind of cool. The a and the z are rotation between the w3 and the b. Okay? They are the two neutral components. And the a and the z are the mass eigenstate of those, the b and the w3. And the mass and the mixing angle is some non-trivial. It's not just some square root of two or something. It's some new parameter that we call sinus theta w. And theta w stands for either the weak angle or the one bearing angle because this model was first by one bearing. So it's nice that the w, the same letter, also can be weak and also one bearing. And this is just the basis of theta w is just the angle that rotate you from the a, the basis into the, or the w3, b basis into the a and z. A and z are mass eigenstate and w3 and b are what we call interaction eigenstate. They are the eigenstate of my unbroken generators. Okay? And this theta w is just an arbitrary number and it has to do with the parameters of my model. It turns out that this tangent theta w is g prime over g. g prime is the coupling of the u1 and g is the coupling of the su2. Okay? So now come the really cool part. Actually everything is cool. Now it's even more cool. Okay? So we can actually measure theta from interactions and I'm going to, I'm going to ask you to actually measure it. And you know, when I say I measure things as a theorist, you know what I mean? What does it mean to measure things for a theorist? Okay? Measure things for a theorist means open the PDG. Okay? So I ask you to actually measure it tomorrow. Okay? You're going to open the PDG and find the values of this theta. And the value of this theta comes from interaction. You look for interaction strength and you measure the value of this parameter theta. Okay? And then you have this amazing relation. It's called the rho equal 1 relation. And the relation is that the mass of the w squared over the mass of the z squared times cosine squared theta w is equal to 1. And this relation tells us the following. You measure the mass of the w. You measure the mass of the z. You measure theta w from interactions. Okay? You put it all together and you find that this is, that it is satisfied. Okay? And this is really the signal of spontaneous symmetry breaking. So you remember in the morning I was telling you about spontaneous symmetry breaking and I told you that spontaneous symmetry breaking give you a relation between parameters. You remember? And I told you to remember this moment because it's very important. So now I want you to recall this moment and that's what you see, what you see. We see a relation between parameters and they are far, far, far from trivial. You see a relation between the masses of the gauge bosons and the coupling constant of these particles. So if this model were just an arbitrary model there was no reason whatsoever for the masses of the particle to be related to the coupling. They're totally unrelated things. However, because it's come from spontaneous symmetry breaking we have this relation and that's a prediction of the model. So the deptonic standard model predict that the masses of the w and the z bosons would be actually related to the coupling constant of the u1 and the su2. Okay? So we can actually go and check it. Okay? And in order to check it as I said I will ask you to do it tomorrow is to open your PDG and actually take the w and z decay to leptons and we're going to discuss soon either today or or tomorrow how you can actually get sinus to how you can get sinus theta w from the decay of the z into electron and neutrino. So you're going to measure it from the PDG and you're going to plug it back in and you're going to find that you actually this is satisfied. Okay? And hopefully when you do it you will be more impressed than now. Okay? That I cannot really get you impressed because I really want you to be impressed by this and somehow I'm failing so I still have you know a few more lectures to get to but if not I will keep giving you homework until you be impressed but I'm really really impressed just stop giving me homework and then I will know that. And actually there's huge amount of data and let me just say the following. So this relation the row equal one relation it's a very very very fundamental test of the standard model and we're able to measure it to extremely high precision and we can measure it to such a high precision that we are actually sensitive to loop effect and we actually have a whole program to look for loop effect. It's go under the name of electroic precision measurements and using this electroic precision measurements we actually test the standard model to even higher precision than this three level result that I was telling you. So let me talk now about the interaction that we have in this model and we have charge current interaction and how do I see this charge current interaction basically is from what I was asking you to do. I just take my Lagrangian diagonalize it put it back into the kinetic term and then I read off the interactions. Okay? Everything in a way once I put in the input you can write a computer program that speed out everything. It can speed out the Feynman rules and actually this program exists. You tell the program your symmetries and your fields it's write the Lagrangian for you it's do the symmetry breaking for you it's speed the Lagrangian it's write the Feynman rules it's actually calculate everything the only thing that it doesn't do for you is to write the paper. Okay? But you really can know and you can actually get all those things and what we see we see that we have this interaction that give you muon decay so we have this interaction that the muon decay into electron and to neutrinos and we know that the coupling is going to to G Fermi the famous G Fermi that Andre was talking about and we can actually understand what this G Fermi is is G squared over Mw squared and I can make prediction. I have an interaction and I can make prediction. One really cool part about this interaction is that the only the left-handed component of my fermion contributes so this is a very good one because as we know that only the left-handed part of the neutrino is there and that's what we see we see only the left-handed part and that's the statement that in the weak interaction only coupled to left-handed currents. Okay? And we can also do the same for the neutral current and the neutral current has to do with the z exchange so we call it neutral kind because the z is a neutral particle and you do the same and you find that the z exchange actually coupled both left-handed and right-handed things so unlike the W that only coupled to left-handed the z goes on coupled to both left-handed and right-handed fermions however with different strengths so it's violate parity also but it's not coupled only to the left-handed so the statement that we say the left-handed that the weak interaction only coupled to left-handed currents is only correct for charge current for neutral currents also right-handed fermions do participate in the weak interactions through neutral currents okay so we just see it now let me ask you the following questions the coupling of the z boson to fermions is brought out to a e over sinus theta cosine theta you see up there while the coupling of the photon is brought out to e and therefore the coupling that appears in the vertex of the z is actually larger than the coupling that appeared with the derse of the photon to the fermions okay so if I look at the Feynman diagram the z coupling to a pair of fermions is stronger than the coupling of the of the photon to the pair of fermions so why we call it the weak interaction if the coupling is actually stronger than the coupling of the of electromagnetism why we call it the weak interaction because it's a short range because of the mass very nice because it's short range according to mass at the infrared it's much much much weaker at large distances at very very very short distances the weak interaction is stronger than electromagnetism so if I have scattering process at one TV the z boson exchange is more important than the photon exchange only at very low energy photon exchange are much much much much more important than the z exchange okay good before I conclude let me also say something about the Higgs boson I didn't talk about the interaction of the Higgs boson the Higgs boson is is a fermion and when you actually write out the interaction you find that the interaction of the Higgs is proportional to the mass of the fermion and why is it why the interaction of the Higgs is proportional to the mass of the fermion because it's come from the same coupling it's come from the Yukawa coupling the Yukawa coupling is the one that gives you the mass of the fermion and the interaction to the Higgs boson so we have a very nice prediction the prediction is that the Higgs boson coupling Proposal to the mass of the fermion. So the Higgs boson should couple to the tau much stronger than its couple to the muon, by a factor of 17. And to the muon, by a factor of about 200 more than to the electron. But we found the Higgs boson not too long ago. It's about seven years ago that we found the Higgs boson. And we are actually still looking to see if this prediction is satisfied. So it looks as if it's OK, but we are not yet there to make any conclusion that the Higgs indeed couple like this. So before I conclude, I want to give you one more question. And I will ask you to do it tomorrow. But I really like you to think about it until tomorrow. It's kind of an interesting question. So you see that the Z boson and the photon looks very much very similar. The only difference between them is that the Z have mass and the photon is massless. But they are just neutral boson, neutral spin 1 bosons. And we know that the photon give you the photon, the column force. And I was making the big deal about the fact that we have the column interaction as a prediction of QED. Can I get the same from a Z exchange? Can I get a force due to Z exchange? Just like we have a Coulomb force due to the photon, can I get the force due to Z exchange? That I will call it the weak force. But the weak force, not in the sense that it's really macroscopic force between things. And if I do it, and I'm telling you, yes, you can do it, do you see that I can actually see it in atomic physics? And if you never thought about this question, I really want you to think, how can I see the effect of the Z potential? So when you do atomic physics, you always only talk about exchange of photons. But now I tell you that the Z, and somehow everything we did in atomic physics our whole life, we don't care about Z exchange. Why? Or maybe I should start caring about it. And if I should start caring about it, can you think at the principle level how I can actually see the effect of the Z? And let me ask you even another question. If I do have an interaction due to Z exchange, so the photon can give me bound state of electron positron. But the neutrino is not coupled to the photon. And since the neutrino is not coupled to the photon, I cannot have a neutrino anti-neutrino bound state. But the neutrino do couple to the Z. And since the neutrino do couple to the Z, and the Z can make some force, can I get a neutrino anti-neutrino bound state? Okay. I want you to think about this question, and I will, as I said, tomorrow we're going to discuss it. But there are some kind of question that we are having there. So let me just conclude, and I would say that the electro week model is rather involved. It's much, as you could tell, it's much more complicated than the QED and QCD. It's a very, very interesting prediction, really, really far from trivial prediction. In particular, the row equal one prediction is really cool, and it's really confirmed, and you are going to actually do the calculation tomorrow. It predicts left-on universality that is confirmed to a very, very high precision. And again, open your PDG. Just open your PDG. It's amazing seeing the PDG, okay? Open it. Go to the website, do Google PDG. Go and check how the W decay to left-ons, and you're going to find amazing result that is decay the same to electron, muon, and tau, okay? It's really, really far from a trivial prediction, and there's many, many more. So just on time, I'm done, and tomorrow we actually start going to do flavor physics.