Can you hear me? Yes? Good. So happy to see so many of you in this early Saturday morning. So let me say one thing before I start. I put some references on the website under the lecture yesterday. And I actually gave the three reference that I gave. So I gave some flavor physics about 10 years ago in several summer school that I wrote up. And this review is kind of a short review that I hope can get you basically all I'm covering in this three lecture. And then there's another much longer review that I wrote with a former student of mine who's now a professor at Riverside named Flip Tanedo. And it's called Just a Taste. He's an amazing writer. I mean, it's really fun to read. You should read it. It's like he writes in a very nice style. I'm also signed up on this, but all the writing is him. So you should really try to look at those lecture notes. I really like them. And the third one, there's actually a book that we are writing. So me and my advisor, his name is Yosinir. And we are writing this book. I know I told you to some of you. We are writing this book now for more than 10 years. And it's almost done. So I really hope that in two years it will be out. And there's actually a draft that it's kind of OK to read the draft. And so I send the link to the draft. And every time I update the draft, every month or two, when we make some progress, we are updating the draft. So feel free to look up there. There's, again, much more information there. And if you find any mistakes there, because it's definitely not ready, please just send me and say, hey, in the draft, and send me the day after the draft, the draft of whatever. June 14, there's a type on page 233. In equation this and this, there's a minus sign missing. So this is going to handle things. Good? Yes, so take a look. And hopefully you like it. So I can hardly talk. Can you hear me? I'm sorry. Let's see now how to do this. OK. So first I want to start with the, nice, huh? So the question is, we do this, I know it's early in the morning. But when you see this picture, what's kind of come to mind?  When you see this picture, I'm angry. What come to mind? Do you recognize there are two people here? I see you, where is, it didn't show up to his, and he knew it is here. Well, well, I cannot see with, oh. OK, OK, I was worried you didn't come to this. Amazing. So yes, of course, you know, you see, they are smiling, they're happy, they are here in a summer school in Trieste. OK, nice.  When you get into an ice cream shop, what the first thing that come to mind? I'm the only one. What flavor should I get? Very nice. So it's flavor physics, OK? And usually, do you take one flavor or you do more than one? What is the fundamental, very important difference in one flavor and two flavors? Ah, not symmetry. I know most of the time that the answer. What's the big difference in one flavor and two flavor? The color? Yeah, I want the color was kind of close but not. Flavor mixing, flavor mixing, very nice. And what happens if you are really angry, if you are really angry and you take three flavors, what would happen? Nice, here we go. So next time you go to a shop, try it. Say, can I have one flavor and try it? And say, you know, that's really nice. But let's see what's happened when I took two and then you start having the effect of flavor mixing. And the three, it's amazing. I mean, really, I remember this. OK, so now I want to move on and have one plot that I should have shown yesterday. So thank you for all of you who didn't help me with it. And I want to show the following picture. Let me kind of recall what we did yesterday. So what we did yesterday, we start talking about a standard model at one loop. And I make the following statement that when we look for electric precision measurement, we can actually look and test the row equal one relation and actually some other relations at the level of one loop at the level of 1%. And this is the data. So these two parameters are called S and T. This is a very amazing name. So it's Michael Peskin idea. He called them S, T, and U in order to make clear that you're very confused compared to S, T, and U. So this S, T, and U has to do with scattering. This is the momentum exchange, the T channel, the U channel. So he called them S, T, and U for whatever reason U is not so important. And S and T are two things that we measure. So this S and T are two things that you go out and you do all those very precise measurements of the masses, the scattering angle, atomic pi evaluation, all the things that I kind of touch upon. And this is, and the standard model is normalized such that the standard model prediction is at 0, 0. So you see the standard model prediction. There's a very, very small arrow. It sits here. And this is the data. So this is 1 sigma, 2 sigma, and 3 sigma of the experimental data. And you see this kind of gray line? That has to do with the mass of the Higgs. This mass of the Higgs is 125, which is the physical Higgs. And this one has to do with the, before the Higgs was discovered, when people didn't know what, and you change the mass of the Higgs up to 1 TV. So you can see, actually, just from this plot, you see how we were able to tell that the Higgs is lighter than the 1, 140 GV just from this precision. Do you see it? But now, what I want you to look at this picture, and it's so much, how the standard model is amazing. And you hear this statement so many times. I don't know how many times I said this. I said, the standard model is such an amazing theory. It explains everything. There's so many measurements that it can explain. And this one actually tells you that it explains everything at, wow, I cannot do it. No, it doesn't. OK, it can do it at one loop and see how precise it is. You see? Do you see how nice the standard prediction sits in the middle of this ellipse? And this measurement we're done in a huge number of data points, a lot, a lot of efforts come in order to make this plot. And we see that the standard model works. So I really want you to be impressed by this little picture that when you make the measurement of one loop effect in the standard model, they agree so well. Are you impressed? Thank you. That was the idea. Good. So what I want to do now, I want to go on. And I don't know how to do this, let's see. So what I want to do now, I want to actually start with one loop effect in flavor physics. And let me remind you what we did yesterday. So yesterday, we show some branching ratio on the blackboard. And we saw that flavor-changing neutral current in the standard model is much, much smaller than the charge current. So flavor-changing neutral currents are much, much smaller than flavor-changing charge currents. OK, this was the big statement that we make. And then we asked why the standard model is so small. And we say, because this one is one loop and this one is three level, so this one should be smaller. But I told you there's actually more to this statement. And I also mentioned the fact that historically, the absence of flavor-changing neutral current were used to actually measure the standard model. And we understand the rough structure of CKM from this idea. So to make the very big statement, the very big statement is flavor-changing neutral currents are a very important thing of building the standard model and very important thing when we try to find constraint of physics beyond the standard model. So now what I want to do, I want to actually kind of calculate some flavor-changing neutral current amplitude and see what are the suppression factors that come into the game. And what we're going to find out, that the suppression factor of FC and C, the suppression factors are one loop, which is a generic for any standard model. And then we have two of them that are unique to our standard model, to this standard model. We always have some CKM suppression. And this CKM suppression can be very significant. And we also have something called GIM suppression. And let me just kind of show off hand. How many people heard about the G mechanism? Ah, OK. Most of you. Very good. So I will try to get a little bit into this of the G mechanism and see how it's actually working. But the bottom line is the following. That flavor-changing neutral current are suppressed by these three factors. This one factor is generic to any standard model. And these two factors has to do with the specific structure of the CKM that we see. So let's look, in order to get it, let's do some simplification. Let's work in a two-generation standard model, where we only have the first two generation. So I only have the first two generation of the standard model. And I want to look at the decay. S going to D, neutrino, anti-neutrino, which to some approximation, you can think about it as the following. So it's S going to D with some Z star, meaning an off-shell Z. And I want to look into this process. Experimentally, people look for this process. It's an extremely interesting process. It's k pi going to pi ni bar, or k long going to pi zero neutrino, anti-neutrino. And there's a dedicated experiment at CERN, NA62, that look for this one. And there's few others that we look for this one. This one, we're almost there in terms of measuring it. Hopefully, in a year or two, it will be measured. Very interesting. And this one, we're a little bit farther out. But hopefully, we can also measure it. We're going to learn a lot from these two measurements. What is the amplitude that I want to write for S to D, Z star? There's actually several amplitudes. But let me just put one of them. There's an S, there's an D. I need here to have the Z star coming out. So I need something here to run in the loop. So what I can do, for example, I can do something like this. I can put a W and the Z coupled to this thing. So what runs in the loop? What I should put here in the loop? So this is a W. What is here inside the loop? That one. Which one? Up. So should I put here U? Let's do like this, U and U. Someone else want to give me one more diagram? Charm, very nice. And in principle, there's also top. But we are doing a two-generation, so we are not caring about the top. Now, very important thing about quantum mechanics. When you have two diagrams, what you have to do? You have to first add the diagram and then square them. You don't first square the diagram and then add. I know it sounds so trivial, but really remembering it. It's important. So we have to add the two diagrams first and then sum them up. So in general, what do we have here? We have here some CKM. The CKM that we have here, let's call it VIS star. And what I have here is VID, where I is either a U or a C. You see this CKM. And there's some loop function. There's some function that tells me something is going on in the loop. And this loop function should depend on the masses of the quarks over the mass of the W. What else it can be? And I want to say something very important about physics, that when you ask what it can be, and you just say, what else? And what else, usually in everyday conversation, sometimes I say, why? And I say, why not? It sounds very not nice. But in physics, it's really OK. Said, what else it can be? So if it's just the mass of the quarks or the mass of the W, that should be the answer. You agree with me? Yes? I know you heard it so many times. But others, you agree? Yes? What else? So it should be this. Good. So let's write the amplitude. So the amplitude is should be the sum of the i equal to u and c times v i s star v i d times some function of, let's call it x i, and let's define x i. So x i is defined to be mi squared over mw squared. We're OK with how I got this diagram, with this expression. And this function f is some loop function that you have to calculate. And actually, when you have to be a little careful, because this diagram by itself is not gauge invariant, you have to add other diagrams. The other diagrams where the z actually touch other places. There's also diagram, which is kind of a box diagram. You have to add all the diagrams in order to make sure that your result is gauge invariant. But at the end of the day, no matter what you add, it's always have the same structure, because you have the same C cam. So this function at the end is some gauge invariant function. OK? So now let me ask you the following thing. What would be the result if this function f was independent of x i? If this function f were just a constant, what would be the result? 0, very nice. Why? Unitarity, very good. So I hope that everybody see when I take it away. This relation is just the unitary relation of the CKM matrix. And even if I have three generations, it will be the same. It's just the fact that two columns of the CKM matrix, when I sum them up, is 0. That's direct consequences of the unitary of the CKM. Good? So now let me do one other little thing and actually write it explicitly. So I use the unitarity of the CKM. And I say I do it in two generations, so it's a little easy. And in two generations, CKM, what I have, I have that VOS star VUD is equal to VCS star VCD with a minus sign. OK, that's the unitarity relation. And this is only two numbers. The sum of two number equal to 0 can be written like this. And since it can be written like this, I can actually put out this factor and what I get. Let me write it here. So that's when I can write as equal to VCS star VCD times f of xc minus f of xu. OK, I just use the CKM to write this expression. And here, clearly, you see that the result is 0 if these two functions are constant. So actually, you can see something a little more important. You see that this is 0 if x is equal to xu. Another way to say it, if they up in the term where they generate this function, it will be 0. So we learned that the amplitude must be proportional to the splitting in the mass of the uptight quarks. And I look for a decay that has to do with the down type. So I have an S going to D, which is an exchange of the downtight quarks. And the result is for the splitting in the uptype. And now I do one more approximation. And the approximation is the fact that x is very small because the mass of the quarks, the up in the term, are much smaller than the w. So this function, x, is very small. So I can expand f. So I can write that f of x is equal to 1 plus f prime at 0 plus x plus tata tata, Taylor expansion, right? And then based on this, I can say this is, roughly speaking, bcs star bcd times mc squared minus mu squared over mw squared with some number in front. This number is some f prime. And I would ask you what you expect this number to be. Then you would say, well, it's some order 1 number and you are correct. Yes, you are correct. Very nice. This is another 1 number. So we learn the following things. That because of the CKM unitarity, the amplitude is propellant to the splitting of those masses. And you may argue why it's squared, et cetera. So let's not really get into the details. There are a little more details here. But the point is that this is what we see. We see that the amplitude is propellant to the mass splitting because of CKM unitarity. We are good with this? Yes. What factor? Ah, yeah, yeah, yeah, yeah, yeah. Of course, yes. So there's a 1 over 16 pi squared. I meant this other one up to the loop factor. Yeah, you are totally right. Yeah, yeah, yeah. Absolutely. OK. So I want to actually stir a little bit of this reason. Actually, it's much better to do it your way, OK? So I will ask you, what is this factor? And you would say all the one, but you were wrong. No, no, I was. So that's some number of all the one over 16 pi squared, right? So now you agree that should be all the one? Good. Now you agree. You were correcting me. So now we can actually stir at this formula. And as we talked about formula, love to be stirred at. And what do we kind of see from this formula, OK? And I kind of hinted on this before that this formula actually tell us the three suppression factors that we have, OK? So one suppression factor is this. It's the loop function. So that's the one loop effect. This one is the CKM suppression. And you see that when I said that there's CKM suppression, what I mean that when you have flavor chain units are current, we must have some of diagonal term of the CKM. And since the CKM is very close to a unit matrix, I always have here something that is small. In this case, it's the VCD, which is of all the lambda. Actually, I really forgot to tell you the value of lambda. Lambda is about 0. Let me write it here. So lambda is about 0.22. And sine squared theta w is about 0.23, OK? So it's kind of easy to remember. These two numbers are roughly the same. But one of them is sine, and one of them is sine squared. So this one is suppressed by something of all the lambda. It may not be such a large number and such a small number. In this case, but in general, we can have here actually much more CKM suppression. And then the last term that has to do with the splitting in the masses of the uptight quarks, that's called the gym mechanism, OK? So this one is loop, this one is CKM, and this one is gym. And gyms stand off a glascio in a polos and Miami, OK? These are the name of those three guys who did it. So this glascio is the same glascio from the standard model, OK? Good. Any question on the gym mechanism? Yes. So here it's crucial, right? So here, that's why we get this very large suppression. So actually to a very good approximation, I can forget about the mass of the U. And you can actually write it like this, mc squared or mw squared. And this is the ratio. And let me ask the following. So what is the fundamental small number in the Lagrangian that is actually play here a role? It's the Yukawa coupling. It's the fact that the Yukawa is very small. So again, it's the same story. So let me say a few remarks about the gym mechanism, OK? One important remark is the following. When you do those calculations, you find that the result is finite, OK? And generically, if I come and I tell you, hey, go and do a one-loop calculation, you should expect the result to diverge, right? And this result is finite, OK? Someone give me a general argument why it must be finite, yes? Because the leading order is 0, very nice, thank you. So that's a general statement. You all know the three-level calculation are always finite. But this is actually a special case for a more general statement. The more general statement is the following. Always the first order that is not 0 must be finite, OK? So if the three-level is 0 for whatever reason, one-loop must be finite, OK? Technically, the reason for this is the following, because the three-level is 0. You cannot write a counter term for your calculation and the renormalization. And it has to do with some symmetries. I'm not going to get into the explanation. I just want you to hear this statement. And there are some cases in the standard model that the three-level one-loop and two-loop are 0. That's the case of the electric dipole moment of the neutron based on this. In this case, the three-loop is the first one that is non-zero and the three-loop is finite, OK? So always the first non-zero term is finite. So this one is finite. And actually, so this is very tightly related to the fact that we don't have SCNC at three-level, OK? So I would ask you the following question. What happens if I add something to the standard model such that I do have SCNC at three-level, OK? And we did it like in the homework last time. And then you find actually that this result diverged like it should. And if you want to do a really cool exercise, it is to actually totally do the same for B2S gamma. So instead of B2SZ, I can put a gamma in this vertex, OK? And everything that I was telling you would be the same. Everything I said will be the same, OK? However, what's happen if I do the same trick that I add something to the three-level? Because the three-level I can never have B2S photon coupling, then the B2S gamma is guaranteed to be finite no matter what model I have, OK? So you can actually do the calculation and see the differences. It's rather cool. And what else I wanted to say? And we talked about it yesterday. Let's see if you all remember it. It's linearly in the mass of the charm. Should we be surprised by the fact that it actually grows with the mass of the charm? It should violate the coupling theorem, right? So why I'm not really worried that it goes with the mass of the charm? You remember we talked about it yesterday. So let me remind you. Let me remind you. I see that people are not like, ah, of course, you talked about yesterday, but you talked about so many things. So naively, you expect when I have a particle that their effect will be smaller. And some of you we see. We see it's very nice with the W. If the W is heavier, the effect is smaller. But if the charm is heavier, the effect becomes larger. So how come that when I have a heavier quack, the effect becomes larger? That has to do with the fact that the yukawa. Because masses actually get their masses through the Higgs mechanism, the whole effect is proportional somewhat in the underlying theory. It's proportional to the yukawa interaction. And it's actually the case that the yukawa go up. And that's why we have dependent in the numerator of this. The same thing. It's exactly the same story as the electronic precision measurement that we saw yesterday. And let me say the last thing that I want to say on this is then what is the different suppression that we have for the different cases? So if I have kdk, so in kdk the gym is very important. So all of them have one loop. The ckm is not very important. It's only lambda. But the gym is very large. It's mc squared over mw squared. For ddk, which is charm, we have the one loop, of course. And the ckm is also not very important is of all the lambda, because it's the 2,1c. And the gym mechanism is much more important. It's actually ms squared over mw squared. So the gym suppression for charm dk is extremely important. And moreover, for charm dk, the leading order result is much bigger. Because for kdk, for ckm dk, the tree level result, the tree level dk is also for lambda. In charm dk, the leading order dk is going up to 1. So maybe I do the following game. Let me add here, let me add here, tree level. The tree level dk. So the tree level dk for ckm is s to d. So it's lambda. For d, it's c to s, which is all the 1. So for charm dk, the gym suppression is much more important. It gives you much smaller branching ratio. Because you take this diagram and compare it to something of other one. So in k on physics, the tree level and the ckm factor in the loop are the same. So the only thing that we have is the one loop and the gym. For charm physics, we have everything, the ckm suppression and the one loop and a very strong gym suppression. So we expect FCNC in charm to be extremely, extremely small. And so far, we did not see any FCNC in charm. And actually, we do not expect to see any FCNC in charm with current level of precision. So if we will find FCNC in charm, it will be extremely interesting. And now we do it for b. And for b, the tree level is b going to s. So it's lambda squared. We have the one loop. The ckm here is also lambda squared. And the gym is empty squared over mw squared. Because then I need to do the full tree generation. And this is all the one. And the approximation I did is not correct. So this is actually like an all the one. And I don't have anything. So in b physics, the only suppression that we have is basically the one loop. So in b physics, FCNC are suppressed by one loop, which is 10 to the minus 2 at the amplitude level, which is roughly 10 to the minus 4 at the rate. And that's what we see experimentally. In charm physics, we have this extra important suppression. And we see this extra suppression in the data. And for d, we didn't see it yet. So we see kind of where we are on FCNC. And the point is, the important point to make is that it's far from trivial. So this pattern that we see, you see this is far from a trivial pattern. In a generic standard model, it would be all one loop. And that's it. But we see this very different pattern of different decay of k, d, and b. And that's what we see in the data. Within cd, the b is kind of the FCNC are kind of large. It's only one loop. The k on is suppressed by another 3 or 4 order of magnitude. And for the d, we didn't see it yet. Good. Any questions? Yes. No, no, no. So actually, this was just for the purpose of the discussion. And you can actually do the same with the photon and the gluon. In the gluon, it's a little harder to know, because then you end up with some mesons. And sometimes the mesons can also come from charge current. So it's actually very rare to understand and say, I only have an FCNC with the gluon. But when you have a gluon, you also have a charge current interaction that give you the same final state. So what we really usually care is about the z and the photon. And we also care about some other diagram that we call box diagram that I'm going to talk about next. So I just gave you the general picture. And of course, in any case, you have to make sure you know all the diagram and do all the calculation. But this kind of the big result that I presented here, it's correct. Yes. Yeah, yeah, yeah, yeah. So the point is because of the hierarchy, when I do MC squared minus MU squared, I can neglect the MU. When I do MT squared minus MC squared, I can neglect the C. When I do here MS squared minus MD squared, I neglect the MD. So that's why I kept only leading all the results. But you are totally right. It's always different. It's always proportional to the splitting in the sector. And you also have to do it a little more precise in the fact that there's three generations. And then the unitarity relation is a little more complicated. So there's everything. I don't think there was even one thing that I was telling you during this seven lecture that I kind of did from start to end. I just give you the idea. And then you can go and understand more going on. Yes. No, so when you calculate the one loop and it's finite, then the two loop can diverge. It's always the first order must be finite, and the higher order can be a divergent correction to the first one. OK, so let me move on. And what I want to talk about is another very important process that led us to deep understanding of the standard model. And that has to do with meson mixing and oscillation. So Andrea already talked about mixing and oscillation in neutrinos. And the quantum mechanics of oscillation is very similar in the case of neutrinos and the case of mesons. But there are some really cool subtleties that are actually different between those cases. And I want to do meson mixing and oscillation, so I want to touch on the differences and see why meson mixing are so important for understanding of the standard model. First, let me say what is the difference between mixing and oscillation? Anybody knows the difference between mixing and oscillation? What we usually mean by this? Yes? Yes, very nice. So mixing is just the fact that when you look at your Hamiltonian, you actually have two states that in two leading order are degenerate. And then when you put some perturbation into them, there is actually a mixing. There's actually a rotation from one basis to another. That's what we have. It's a rotation from one basis to another due to some perturbation. It's not a dynamical statement. And oscillation is the dynamics. It's a time-dependent experimental thing that we see time-dependent processes. That's what we call oscillation. Many times come together, and it's totally mixed up. But we want to understand the difference. So what is the idea of meson mixing? So the idea is this following. Let's consider, for the sake of the national, let's consider a b-meson. And a b-meson, what quarks make a b-meson? The b-meson is made out of the b-bar quark and a d. So that's a b0. And a b0 bar is made out of a b and a d-bar. And it's a little confusing, but you just have to remember that a b0 is made out of a b-bar quark. So the b-meson is made out of a b-bar quark. The charm is not the case. It's only for the b and the k that you have this opposite convention, that the meson made out of the antique quark. And for the charm, the meson is made out of the d-meson. Let me just write it. So this is actually made of cu-bar. And there's some historical reasons why this is the case. But it doesn't matter. It's just convention, so you have to remember this convention. So we have these two mesons. We have the b and the b-bar. And these two mesons, if I turn off the weak interaction, these two mesons are degenerate. And the reason that they are degenerate is because there's actually a flavor symmetry of QCD. And if I turn off the weak interaction, I mean I turn off the W exchange, then these two mesons are degenerate. However, they do have a different quantum number. This one is a b-bar, and this one is a b. So now when I turn on the weak interaction, or more generally, when I introduce breaking of the flavor symmetries, that's the more general statement, then these two could mix. There should be an amplitude that takes you from this one into this one. And when you have two states that are degenerate, and I turn on a perturbation that can mix them, then the eigenvalue of their system now becomes there's a splitting in the system. We've seen it so many times in quantum mechanics. You have a degenerate state. You put a perturbation that connects them, and you get a splitting. So that's what we have here. This splitting is extremely small. And what we find out experimentally is that the splitting is of all the 10 to the minus 15. So delta m over m is of all the 10 to the minus 15 for the current system is 10 to the minus 15 for the b's, whatever, 10 to the minus 12 or 13. But it's all the 10 to the minus and number that is larger than 10. And this is an extremely, extremely small number. And when you have extremely, extremely small numbers, then you think about oscillation. Because oscillation tells you that since this is a very, very small number, and I produce what we call a flavor eigenstate, then this flavor eigenstate starts actually oscillating due to the mass eigenstate. When del time is very, very large, then we can think about mass eigenstate. We can think about mass eigenstate when del time is very, very large. And I said, A, I produce a pi or I produce a d. Pi oscillation are not there. But when the mass difference is very, very small, I have to worry about the oscillation. So that's really the story here. And then we can actually measure the oscillation. And once we know this number, we can actually get sensitivity to the amplitude that move between those two. And the amplitude that move between those two is a flavor change in neutral current amplitude. And we get sensitivity to FCNC. So that's the story. So what I wanna do now, I wanna talk first a little bit about the mixing and then talk about oscillation. And then I'm going to discuss how we actually calculate the mixing amplitude in the standard model, okay? So the idea is as following. Let's say that I produce a B at T equals zero. So I have my B at T equals zero. It's a B-meson. So I assume that I produce a B-meson. And how do I know that I produce a B-meson? Because in QCD, the thing that I produce in QCD are those states, right? Because QCD conserves flavor quantum numbers. So I produce either this or this. So let's assume that I produce a B-meson at time equal T. And then when times go on, I start having a superposition of a B and a B bar because of the amplitude that makes them, okay? So in general, I define this state, B of T is a state that started at B at T equals zero. And it's not a B, it's an oscillating state, okay? So B at T, I can write it as a very general, at AT of B plus BT of B bar plus what else? So this is all I'm just saying that it's oscillate. Since it's oscillated, it becomes a superposition of a B and a B bar. Makes sense? And now I said there's more. What else it can be? When you look at the B that start to move on in time, what else it could be when times go on? Do you see why it can be a B or a B bar? It's like a neutrino. I said I start with an electron neutrino and after some time it can be a superposition of electron and muon neutrino, okay? And there you can also have a new tau, okay? There's three of them. Here I'm asking what else we can have here? A decay, right? So when I look at the time of a B, it could also decay and that's different than the oscillation treatment that we did with neutrino. In neutrino, we assume that the neutrinos are stable and we didn't consider decay. But in general, when I have a state that go forward in time, it could also decay. So let's write it something like this. C1 times F1, which will be the first one, first file I state, second final set, et cetera. In general, there's actually infinite of them because I have three body decay that have, you can think about them as infinite number of things. So basically, this one is an infinite sum of state that we can go on, okay? When we have infinity, we usually have to do something to get rid of the infinity. So what we can do? So here we do the following approximation, okay? We treat our state, we treat the B state not as a stable state with decay as a perturbation. So when we do quantum field theory, we think to leading order we have a stable state and the decay is a perturbation that I put on top of the state. And those state we call particles, okay? Here, I wanna do something a little different. I said, I cannot really think about my decay as a perturbation. I want to take the decay and bring it up into the leading order result. When I bring the decay up into the leading order result, then my particle is what we call a resonance. And a resonance particle have, actually it's because it could decay, he have non-unitary evolution. It sounds very bad, oh my God, non-unitary evolution, just don't do it. But what it just mean is that if I have a particle that decay, after some time the probability to find the particle is zero, unitary evolution tells me that it's always one. Non-unitary evolution tells me that I start with one and at some point it becomes smaller than one, okay? And that's the whole idea of decay. So I actually think about my B not as a particle that could decay, but actually as something that decay and therefore in the propagator, if you like, I have the i gamma and the evolution is not unitary. So when I think about it, what I do, I erase this and I understand that my B have a non-unitary evolution and I must include the width into the story, okay? So then I write my Hamiltonian as m minus i over two gamma, okay? So what I did when I erase those final states, I changed my Hamiltonian. So for a particle, the Hamiltonian is just the mass. Yes, you know, for a free particle, the evolution goes by the mass. However, if instead of talking about a free particle, I talk about a particle that could decay, my Hamiltonian is m minus i gamma. This is the non-unitary part of my Hamiltonian and that has to do with the fact that I kind of ignore those decay and I think about the decay as a non-unitary thing that go outside of my system, okay? So that's my Hamiltonian right now. So now let's write this Hamiltonian. What is the size of this Hamiltonian? I mean, it's a matrix. What is the size of the matrix? What is the size of the matrix? Someone speak up. Two by two, because it's actually a matrix that connect my b and my b bar state. So it's a two by two matrix. So I can write my Hamiltonian as a two by two matrix and I can write it like this and write it as m11 minus i over two gamma 11, m12 minus i over two gamma 12, m21 minus i over two gamma 21, and m22 minus i over two gamma 22, okay? It's kind of nice when I have a finite size Hamiltonian so I can write it. So now I'm asking you the following question. On this Hamiltonian, what do you know about this Hamiltonian? Tell me anything you know about this Hamiltonian just from the mathematics before we do any calculation. What are some properties that we know generally on this Hamiltonian? Hermitian, why? Because Hamiltonian always has a mission, right? Well, no, why not?   You dare? Yes?  You stand on this. You've been in quantum mechanics somewhere and say Hamiltonians are a mission, right? I really like, you don't give up, yeah? And of course you are right, okay? So the whole point of this, this is an effective Hamiltonian. It's effective Hamiltonian different and the point is that it's not Hermitian because I take away the decay. The Hamiltonian would be Hermitian if I include the infinite number of things that it could decay. But since I only think the effective two by two Hamiltonian and I think about those decay as the rest of the Hamiltonian that I didn't include, this Hamiltonian is not Hermitian and I actually wrote it explicitly in the following four. M and gamma are a mission. So M is the Hermitian part of the Hamiltonian and gamma is the non-Hermitian part of the Hamiltonian. And I really hope that you have this intuition. The mass term is the one that is correspond to the unitary evolution of the state and that must be Hermitian and the gamma is exactly the decay of the state and the decay of the state is what take you outside unitarity. It's what tells you that the probability to find the particle is not one is actually e to the minus gamma t and e to the minus gamma t is exactly the non-Hermitian, the non-unitary part and non-Hermitian Hamiltonian corresponds to non-unitary evolution. Hermitian Hamiltonian, unitary evolution. That's why we always talk about unitary evolution say that probability is conserved. Particle cannot disappear. You remember that in quantum mechanics they say particle cannot disappear. Therefore, it must be psi squared, the integral of psi squared is always one. But what we have here, the particle do disappear. It does decay and particles that decay are not, the evolution is not unitary so the Hamiltonian doesn't have to be Hermitian. It's an effective Hamiltonian. The full Hamiltonian of course must be Hermitian because the full theory is unitary. But this state, the evolution of this state is non-unitary. Good with this, okay? Very deep non-unitary Hamiltonian. Okay, but M and gamma are so M and gamma are Hermitian so this is the anti-Hermitian part and this is the Hermitian part. Good. Can you tell me something about the relation between some parameters? Since it's Hermitian what can you tell me about M12 and M21? So we know something like M12 is equal to M21 star. The same with gamma, okay? What can we tell me the relation in M11 and M22? They're really nice. Yes, because it's Hermitian, very nice. What else you can tell me about M11 and M22? Anybody, yes? Sorry? Bigger than zero, so actually, yes. Very nice, good. What else you can tell me about them? I see you're getting there, you're getting there. Well, how do you think about this? Yes, yes, so you already can tell but that's not based on the mathematics, that's based on the observable and based on the data you will tell me something that M12 is much, much smaller than M11, okay? You are totally right, but I want one little extra thing that you are almost there. What, I want you to have this feeling, okay? It's always so important to develop those kind of feelings.  So yeah, we follow the algebra we get but what M12 means, what M12 means in terms of words? What is the word that goes with M12? M12 has to do with the amplitude of going from this state to this state. M12 is the coupling between this and this, yes? Maybe I should have been more explicit. This Hamiltonian is something like B, B bar, B, B bar, okay, that's the index things, okay? So one, one is BB, okay? So M12 has to do with the transition between one, one, between a B and a B bar. What M11 is, M11 is, is there, thank you, hi. And you know, you were not afraid, you talk loud, that's the way, it's the mass of the B. M22 is the mass of the B bar, okay? What do you know about the mass of the B and the mass of the B bar? They are the same because of which symmetry? CPT, very nice. A particle and antiparticle must have the same mass because of CPT and a B and a B bar has the same mass. So M11 and M22 must be the same. And let me emphasize that CPT tell you that the parameter in the Lagrangian of the mass is what has to be the same. It's not the physical mass that we get after diagonalization. So CPT actually guarantee that M11 is equal to M22, okay? And let me tell you one last thing that CP, not CPT. CP actually guarantee that M21 and gamma 21 is real and CP violation can make these two complex, okay? But that's what we know in general for this matrix, okay? Good. So let's us move on and after you stare at the matrix, I think I said something wrong in the past. You remember that I told you what you do when you see a matrix? You remember we talked about it and you see I say that diagonalized there, right? So I think I was wrong.  So you stare at the matrix for a little while, understand what she is doing or he doing or he doing and then only then you diagonalize the matrix, okay? So we did the staring part, we understand what the matrix is and now we have to diagonalize it, okay? So you go and diagonalize it. It's a little more complicated than a usual matrix because it's not a mission but I'm not going into the detail, you diagonalize it and you find the eigenvalues and the eigenvalues we call them BL and BH. BL and BH, BL stands for light and BH for heavy. So by definition BH is heavier than BL and it's given by B plus minus, just make the following notation. Plus minus Q of B bar, okay? And the normalization is such that P squared plus Q squared is equal to one, okay? And the overlap between those two eigenstates, BL and BH is equal to BH. BL is equal to P squared minus B bar. Q squared. So let me ask you the following question.  I'm soon going to discuss it and I'm telling you that they're propellant to this thing, P and Q and when you stare into this, hopefully it looks a little different than what you would have expected, okay? What it is that, yes, yeah, yeah. So actually I diagonalize, I must, what I should diagonalize? Should I diagonalize M separately and gamma separately and the diagonalize if M give me the mass and diagonalize if gamma give me the width or should I diagonalize the total H and then look for the real part and the imaginary part? What is the right procedure? So you see there's two procedure to do here. One procedure is I diagonalize M and the eigenvalues of M give me the mass and the eigenvalue of gamma give me the width or I diagonalize the full H and then I look at my result and the real part give me M and the imaginary part give me gamma which is the correct procedure. Should I diagonalize H or diagonalize M and gamma separately? H, why? Because you always diagonalize the Hamiltonian, okay? Yeah, you totally right, you totally right. So let me give you two answers, okay? One answer is that in physics we never care about all those crazy ideas that mathematicians care about and it's a, we don't have a proof to it but I didn't even see one counter example, okay? All those crazy things that you can study, like I don't know, like if you take some mass course and they study all those things what is the condition to diagonalize a matrix for half a semester? It's totally irrelevant, it's always get it, okay? And actually in our case, you can actually prove that this is indeed the case and the reason is that as always gamma must be smaller than M and when gamma is smaller than M then this matrix is diagonalizable, okay? So the point that should surprise you is the following thing. When I diagonalize a matrix the eigenvalue should be orthogonal, yes? You know that when I take a system and I have two eigenvalues and I do a basis rotation the eigenvalue should be orthogonal. And here I wrote this thing, okay? And I tell you that this is square minus q squared, okay? So I didn't tell you that p and q are one over square root of two. If there was one over square root of two that would be a orthogonal set. But if p and q are something else then this is not orthogonal, okay? Does it make sense that the eigenvalue of this matrix is non-diagonal, is not orthogonal? Can I have a system that when I diagonalize it the eigenvalue, the eigenvector are not diagonal? Actually that's right, but I do, I can still think about those as the propagating states for whatever size it is. What does it mean that I have some, why do I have, why in general those eigenvalue are not orthogonal? What is the fundamental thing that I did that led to this word result? Yes, very nice. It's the same thing that we did again and again is the fact that this Hamiltonian is not Hermitian. When you have non-Hermitian Hamiltonian and you diagonalize it, the eigenvalue doesn't have to be orthogonal. And let me tell you the following. They are not orthogonal only if I have Cp violation. When I have Cp, they are orthogonal. When I have Cp violations they can be non-autogonal, okay? Okay, let me go back to here. Yeah, so I said that the condition, I want to be a little more precise. If I impose Cp, then they must be orthogonal. I'm not going to prove it, I'm just stating it, okay? If I have Cp violation in general, they are not orthogonal. If I impose Cp, they must be orthogonal, okay? So you can actually do the calculation and under some assumption, then I'm not going to go everything, you find the delta M, delta M is of order 2M12. Okay, and delta gamma is of order 2G12. There's more going to it, it's just a mere two like this, okay? But that's the intuition, the intuition is that the splitting is propelled to the off-diagonal term, okay? And the mixing angle, what is the mixing angle, by the way? What is the mixing angle of this matrix? Look at this matrix, what is the mixing angle? Let's forget about gamma, let's say gamma is zero. What is the mixing angle? If gamma is zero, then it's a usual rotation, right? When gamma is zero, I have an emission matrix, the diagonal, we know everything. If gamma is zero, what is the mixing angle? What is the mixing that, the angle that it corresponds to the diagonalization of this matrix? So you have 45 degree rotation. So it's the fact that the eigenvalue is has to do with B plus minus B bar, okay? So this one, sin theta, theta is equal to 45 degree. Okay, and delta m is m12, and delta gamma is gamma12. And we also defined this two parameter, x is delta m over gamma, and y is two delta gamma over gamma. I find this extremely interesting that when people do flavor all those formalisms, they have like this amazing name, like P, Q, X, Y. It's so, it's tell you nothing, okay? So I just want you to remember, x is delta m over gamma, and y is delta gamma over gamma. What is the intuition of x? Let me ask you, now, okay, now I want to move to the time dependent, and when I move to the time dependent, I start having this oscillation, right? So what is the importance? What is the physics of x? X is delta m over gamma. So when do you think oscillation will be important? When x is smaller, when x is large? When x goes to zero, what happened? When x goes to zero, there's no oscillation. If there's no delta m, there's no oscillation. When x goes to infinity, what's happened? Then the oscillation are extremely fast, and I can really think about things as mass eigenstate. And when x is of all the one, then I really care about oscillation. And the intuition is that x, tell me, there's two timescale in this question. Is the oscillation and the decay? So if the oscillation timescale is much longer than the decay, there's no time to oscillate. I decay before I oscillate.  And when x is all the one, when I have only one timescale in the system, in the case, then I really care about oscillation and I can actually probe this system. Okay? So let me talk a little bit about the time evolution. So again, I'm not going to derive it for you, but the final result is the following. I want to define one more parameter. They turn out to be very important. And this parameter is the parameter lambda. So I define lambda, so, so, okay. Let me back up for a second. So the way we do oscillation experiment is the following. And here we are also different from neutrino oscillation. In neutrino oscillation, I put my source here, I put my detector here, and I know how long the oscillation happens. When I have decay, who tell me how long I can oscillate? It's not me, it's nature. So nature tell me, okay, sorry, you have to stop your experiment right now because I decided to decay. When I decay, you stop oscillating, right? So in neutrino oscillation, it's kind of nice. You say how long you want to do. In this kind of experiment, it's the decay that tell me that I stop oscillating, right? So what I really care about is that I start at t equals zero and I measure something when something decay, okay? So I always, the time that I'm there is from production until the decay, okay? That's the oscillation. So the result is actually very sensitive to what I could decay to. And in a way, the way you think about decay, decay is a quantum, you can think nature provide measurements for us. In neutrino oscillation, you have to work really hard to measure different things. If you want to measure charge current, you build you whatever, this amazing chlorine thing or this super-camio-canda, hyper-camio-canda, all those amazing things.  In B-physics, you don't have to do anything.  When it's decay into leptons, it's one kind of measurement. When it's decay into, say, K plus K minus, it's another kind of measurement. It's the final state that decide what you are actually really measuring, okay? So I want to define the following thing. I want to define AF. AF is the decay amplitude of some final state F. So the amplitude is the amplitude of B going to F, okay? And the oscillation formula would depend on what is the final state. Depend on the final state, every time I measure something else. You get the intuition. If I measure different final state, I can actually get different sensitivities. So for this, I define this parameter lambda F, which is Q over P A bar F over A. And this parameter is physical. A, Q, and P, all of them are not physical in the sense that I can rescale them through a phase. And this one is physical. And because it's physical, it's going to enter into the things that we measure, okay? So what we are doing next, we are not looking into the state that I produce. So I produce, say, a B and I see how the B decay to something after. And it depends what is this lambda. So I want to consider two cases. Of course, there's a general formalism. And first I want to consider the case that lambda F is zero. What is the case of lambda F zero? Lambda F is zero is when the B cannot decay into the final state F, but the B bar could decay. So A bar of F is A bar of B bar going to F, okay? And when lambda is zero, when lambda is zero, I say this one is zero. It's down to the B bar to F, and this one is not. In practice, this is the case when I have semi-laptonic decay. In semi-laptonic decay, lambda is zero. And the intuition is that semi-laptonic decay tells you the flavor of your state. So if I have a B in semi-laptonic decay, for example, a B decay into an L plus and a B bar decay into an L minus. And I was very careful here not to make mistake. And I said B decay into an L plus. And I didn't tell you if it's a B meson or a B quark. And I don't remember, okay? But then I don't make mistake, okay? So the B decay into an L plus and the B bar decay into an L minus. So I know when I see an L plus, I know that I have a B. And when I have an L minus, I know that I have a B bar, okay? So this is what we call flavor tag. A semi-laptonic decay of flavor tag because it tells me the flavor of my meson. It's measured, so I have a state that oscillates between B and B bar and it's measured the flavor. It's not either a B or a B bar. Makes sense? And in this case, the time evolution is like this. Lambda B is proportional to one plus cosine delta MT over two, okay? And lambda of B bar is proportional to one minus cosine delta MT over two. One thing that we expect, you expect that the argument is delta MT because it's the ET. And what I really make the assumption here that the B is not boosted and when the B is not boosted then E is equal to M and delta is equal to delta M. And sometimes we have to care more about the boost. In neutrino oscillation, the boost is extremely important. In B physics it's not so important so that's why I have delta M and not delta M squared over E like in neutrino oscillation. So that has to do with the oscillation and you see that the T equals zero. At T equals zero the cosine is one and you expect if I start with a B it should be B and you see that this is the case. At T equals zero this is one and this is zero. So at T equals zero I know that I have a B and not a B bar and then when times go on I oscillate it in B and a B bar and the amplitude is 45 degrees. So at time point I can have full, the B is totally moving to a B bar, okay? So I can go out and produce Bs and start looking at the time dependent and measure the oscillation and get delta M, okay? And let me tell you the following. When menu measure oscillation, usually the result is extremely precise. It's much, much simpler to measure oscillation than measure a decay, okay? So measure oscillation going to give you very precise number and we indeed know this delta M to a very high precision, okay? And we know that X for the B is about 0.7, okay? And we know it more precise, okay? It's kind of surprising that it's 0.7 we didn't expect it to be all the one but that's what it's at end. And let me talk about one other case and the other case is when I have lambda equal to one. So I take the case that lambda F is equal to one and that's the case that F is equal to F bar. F is equal to F bar is when I have a CP final state. For example, I can think about F as a D plus D minus. A D plus D minus is the final state that is the same so it's a CP eigenstate. So in this case that lambda is equal to one, the result is that lambda B is one minus imaginary lambda F sinus delta Mt and lambda of B bar is one plus imaginary lambda F and sinus delta Mt. And here the result is a little bit more surprising. So you expect that when I have a CP final state, if there's no CP violation, both the B and a B bar should decay in the same way into the same final state, that's what CP tells us. So you see that if there was no CP violation, there's no imaginary part. If all the imaginary part were zero, then we have just one and one. So this should be like a factor of two here or something. But CP violation tell me if they are not the same, we see CP violation and CP violation is probably up to the imaginary part of lambda and that's the intuition that we should have. Imaginary part always related to CP violation. Okay, any question on this? Let me just tell you some data that people got. So when we go out and measure it and we measure it and we find out you use this to measure the oscillation and we find the following thing. We find that X, let me do here a little table, X and Y, and here we have K, D, B and B sub S and we find the following thing. We find that for K, X is for the one, Y is for the one. For D we find 10 to the minus two, 10 to the minus two. For B we find here all the one and here we find less than 10 to the minus three and for B sub S here we find 20 and here we find 10 to the minus one. So we find that actually the different meson have very different values. So they tell us that yes, we are, we could really, the standard model must explain this kind of very different structure and of course it does and that's far from a trivial test of the standard model and another very surprising fact is that actually all of these numbers are somewhat all the one, okay? In particular the X's and there's a priori no reason for it to be the case but actually two of those are, this one is 0.97, this one is 0.7, 0.7 to something like this, okay? So we actually are sensitive to oscillation. Okay, so what I wanna do next is I want to actually calculate this in the standard model so I'm not going to really calculate it, I'm just going to show you the result. So what I need to do, I need to find an amplitude that moved me from a B into a B bar. So I start here with a B, so this will be a B bar D and I want to end with a D bar and a B, okay? It was a, yes. So I want to do something like this and the diagram is this diagram and this diagram has a name, it's called the box diagram. Actually, this diagram that I had before, this diagram is called the penguin diagram, okay? And I hope that you can see that this diagram looks like a box, it looks like a box, somewhat.  But there's not the name that everyone has called penguins, penguins, penguins, penguins. And this one is the box diagram, okay? So let's look at the box diagram. This one is a B and this one is a B bar, okay? And you see that this diagram connects those two. So the calculation of this diagram going to give me M12 and M12 going to give me delta M. So by measuring delta M, I'm actually measuring this diagram, okay? Before I go on, let me say something that we have in quantum mechanics, we always learn the following thing, you take the amplitude, you square it and you get the decay rate. You never care about the amplitude, you only care about amplitude squared, okay? And I was telling you right now that when I measure delta M, I'm sensitive to the amplitude, not the amplitude squared, okay? That's a very important difference. I'm not sensitive to some decay rate which is amplitude squared, I'm sensitive to linear order in the amplitude. You see, it's a little bit surprising. How come I have a linear sensitivity to the amplitude rather than quadratic? When I have a decay rate, it's the amplitude squared. Think about it, I'm not going to kind of get into the details, I just want to emphasize that that's what's going on. When you have oscillation experiment, you are sensitive to the amplitude, not to the amplitude squared, as you do when you actually care about decay. So I am sensitive to this diagram so I need to calculate this diagram. And the calculation of this diagram is very similar to the FC and C example that I did to you before. You see there's actually, I have to take to care about the UC and T running in the loop. So I have nine diagrams because this loop, this one and this one is independent. I have nine diagrams, you write it down, you use unitarity, you get all the G mechanism and at the end of the day because of the G mechanism, which diagram is kind of the more important one? Because of the G mechanism, which diagram out of the nine is the more important? The top one, do you see? Because it's probably not to the mass of the cloud. So the top diagram is point to empty squared, the TC diagram is M top time MC. So we have to care about the top diagram. You have to be a little more careful of what's going on because it's not gauge invariant, but somehow in the correct interpretation is the TT diagram is the more important one. And it's proportional to some CKM and the CKM that therefore we care about. So this diagram is roughly proportional to VTB, VTD star time VTD star VTB, which is equal to VTB squared, VTD star squared. And I like to emphasize that there's no typo here. This is VTB squared. It's not VTB absolute value squared. It's VTB squared. And this one is VTD star squared. It's not VTD absolute value squared. This one twice appear without a star and this one twice appear with a star. This diagram do have a phase. The phase of the CKM do enter into this diagram. And before I go on, let me say that just doing this box diagram calculation, we kind of understand this structure. And the reason that we see this structure is, for example, the box diagram for charm is very, very small because the gene mechanism for charm is much more significant. So that's why for charm, X is very small. For B sub S, the CKM factor is larger than for B sub D. And therefore for B sub X is larger. For B sub S, I replace the D by S and I have VTS instead of VTD and again a factor of lambda. How much is lambda? Roughly, one over lambda is about 20. So this is this 20 that I see here, okay? So this rough structure is totally explained in the standard model in this picture and it's because the gene mechanism and the CKM separation that we know this specific value of our standard model can actually make it into here, okay? Good. Any question on this? Okay, so the last thing I wanna talk about is CP violation. The reason we care so much about CP violation is because in CP violation, we are probing the structure of the CKM. So when we do flavor physics, we care so much about the CKM and we care about FCNCs, for example, and we also care about CP violation because then we can get sensitivity to the one phase of the CKM. Another way to say it, that we get sensitivity to the angle of the unitarity triangle. So remember we have this unitarity triangle. You remember what the size is?  And by measuring CP violation, you get sensitivity to those angles and then we can start actually test the standard model, test the pictures of the standard model. And in order to see CP violation, what we have to do, the definition of CP violation is that you have to see some decay. So I define ACP in the CP asymmetry. It defined to be some decay of some state, A to B minus A bar to B bar, okay? Over the sum. That's the definition of ACP. In the CP limit, A to B is the same as A bar to B bar. And this result must be proposed somehow to some phase of the, in the CKM or some, to leading order to some phase in the unitarity triangle, okay? And there's actually quite a lot going on into the CP violation and I don't have too much to talk, too much time to talk about it. But what I wanna do is, let me erase. No, I don't want to erase this. I don't. There's nothing I left to erase. I need another blackboard. So in order to see CP violation, it's really non-trivial to see CP violation, okay? And the reason is the following, okay? CP violation must be an interference phenomena, okay? CP violation is an interference phenomena. And the intuition is the following. CP violation must be proportional to some phase in my Lagrangian, yes? And a phase in my Lagrangian, in order to see phases in physics, you have to do interference. That makes sense because if I do not have interference and I take whatever amplitude I have and put them absolute value and square it, I lose all information of the phase. The only way I can get sensitivity to phases is when I have interference, okay? So CP violation must be an interference phenomena. So it's making it kind of hard to see CP violation because you have to look for interference and in order that the effect should be large, you better have two amplitude that are large because if I have a tiny amplitude that interferes with the big one, the interference effect is very, very small. Makes sense, right? So in order to see CP violation, you have to look for cases where you have large interference between two diagrams. And these two diagrams must have a phase difference between them, okay? And it's turned out, and I'm not going to get into the full derivation, that not only that we need one phase different, we need another phase different between them, you need two of them. So let me define the two phases. And one phase that we usually denote by phi, we call it a weak phase. And that's a very confusing name, but everybody use it. And what we should have called it is a CP odd phase. CP odd phase. What is a CP odd phase? A CP odd phase is a phase that change sign under a CP transformation. And those are the phase that appear in my Lagrangian. Delta Km, the phase in the CKM is a weak phase. It's a phase that change sign under a CP. And then I have another phase that we call it delta, a very bad notation because the CKM phase is also called delta, but delta is a weak phase. But that's the notation that people are using. Delta is a strong phase. Strong phase. And it's a very bad notation, but that's what everybody use. And the strong phase, you can think about it as a CP even phase. What is a CP even phase? A CP even phase is a phase that does not change sign under CP, okay? Can you think about a phase that does not change sign under CP? What kind of a phase do not change sign when I apply CP on it? So the phase in the Lagrangian must change sign when I apply CP. What kind of phases do not change sign under CP? You know? Anybody? The strong phase, the definition of a strong phase. The definition of a strong phase and it doesn't change sign under CP. So I'm asking you from your intuition of quantum mechanics. Can you think about a phase that do not change sign under CP? And I give you a hint. Until I was telling you that there is actually a weak phase, all the phases that you know are strong phases. And you do know about phases in quantum mechanics. You've been playing with phases of quantum mechanics very, very long time, right? And those are strong, yes? The time evolution phase. The time evolution phase does not change sign under CP because both the particle and the anti-particle go forward in time. So it's always e to the minus iet. Yes? So strong phase is the phase that come from time evolution. Weak phase is the fundamental phase of my Lagrangian, OK? And it's turned out and I'm not getting into the details. And unfortunately, I do not have a real hand-wavy argument why it is. If I had, I, of course, would have tell you. It turned out that in order to see CP evaluation, you must have two amplitudes that have a relative weak phase. That's obvious because we need to see an interferon that's provenant to the phase. But they also must have a different strong phase. So somehow I need two amplitudes that not only that they have a different weak phase, also their time evolution is different. Make sense? So it makes it somewhat hard to see CP evaluation because you need something that both have two amplitudes that interfere. They have different weak phase. And on top of it, they have to have different time evolution. Yes? So let's make it hard to look for CP evaluation. And it turned out that in order to have two kind of time evolution, you can have two things. One thing that you can produce some strong final state. And then QCD kind of go and do something in their time evolution. And when I have QCD, time evolution is really, really not simple. Our usual e to the minus IET is simple when I have a free particle. When I have QCD and strongly interaction, the e to the minus HT, it's not trivial. And I don't know how to calculate it. So if there's some QCD effect, this strong phase is something that I don't know how to calculate. It's still very interesting and very important and sometimes we can calculate. But in general, I hope you understand. When then I have strong interaction, it's not easy to see it. It's not the simple e to the minus ET. When I do know how to calculate the strong phase, when I have oscillation, in oscillation, the strong phase is trivial. It's just the e to the minus delta ET. So one way to look for CP evaluation, it's not the only way. It's actually to look for a case that I have oscillation. And the intuition is, let's erase this. The intuition is that the following two amplitudes occurs. I start with a b. And the b could decay into some sign of state f. The second amplitude is that the b first go to a b bar. And then the b bar decay into a sign of state f. So first, you see that I have two amplitudes. This one is, I said, decay. And this one, I said, first oscillate and then decay. So there's two amplitudes. These two amplitudes could be different in the weak phase. Why? Because this one is profiler to AF. And this one is profiler to BB bar mixing. BB bar mixing is q over p times b bar. So this one is proportional to this. And this one is profiler to this. So it turns out that the interference between these two diagrams is profiler to lambda f. I'm not going through the algebra. And you are really encouraged to go through this algebra. But I hope that you see the intuition. The intuition that I have interfered between this pass and this pass. And this pass is q over p because I have BB bar mixing. And where the strong phase is coming, the strong phase come from the time evolution of this thing. So if I have no time evolution, there should be no CP violation. When I start having time evolution, I start to see CP violation. So what I actually do experimentally, experimentally, I do this case. I look for the time evolution into some CP final state. When I do the evolution into CP final state, is this minus this? And this minus this is for you to see the one cancel and this one have the same. When I subtract, they have the same sign. So I have imaginary lambda f times sine delta mt. Since I know delta m from the measurement we just described, what I actually want to measure, I measure the difference between them. And the amplitude of the oscillation, give me imaginary lambda f. And what is imaginary lambda f? Imaginary lambda f is directly the phase difference between this amplitude, which is the BB bar mixing amplitude, and the decay amplitude. So if I have, for example, f going to d plus d minus, it's this diagram. We're going to see CD. And what is the second factor on this one? This is vcb star. And here is vcd. I'm not going through the full details. I just want you to give you the general feeling. The general feeling is that using this experiment that I actually look for the time-dependent of a decay into a Cp final state, you get sensitivity to the phase difference between BB bar mixing and the decay. And what is the phase difference roughly between this one and this one? Very roughly speaking, this part is proven to the t side of the unitary triangle. You see I have here vtb, vtd. And this one is proven to the t side of the unitary triangle. Yes? You see it? Because I have vcb, vcd. That's the c side of the unitary triangle, OK? So when I do b going to d plus d minus, which angle of the unitary triangle I get sensitivity to? You have three options? No, four. You can also say zero. Which what do I get? Beta. Do you see it? Only one person said beta. Anybody else? Second, do I see a second? Beta, beta. Beta is the angle between the t and the c side, yes? So you see that the interference is must be proven to this side, to this side, OK? So actually, when you do algebra, you find out that the amplitude is exactly sinus to beta, sine, sine, not sine. Sine to beta, nice. I'm too tired. Sine to beta. So the amplitude of this is sine to beta. So people go measure it, and they get a very precise measurement of sine to beta. And the best mode to do is actually, instead of d plus d minus, the best one to do is psi k short. And you might hear about the fact that psi k short is the golden mode for CP, OK? So let me summarize. I'm not going into more details. Actually, many measurements of CP validation were done. I gave you kind of the simple story. But at the end of the day, we made many, many of them. There's some adjoining uncertainty that I didn't discuss, but that's always a big deal how we go over adjoining uncertainty. But the fundamental physics is this diagram, is this interference pattern. And we are able to measure beta, very precise alpha to a very good precision. And gamma, we are at good precision. I spend so much time on gamma. I mean, that's by far my favorite angle, OK? If you ask you, what is your favorite angle? Gamma. I spend, I have more than 10 papers on it. I really, really liked, OK? I love to tell you about those angles, gamma. So we are at gamma. So right now, beta, we know it's 1%. Alpha, it's about 5%. And gamma is about 7%. But in 20 years, this will be the best, OK? That's why we are so excited about this one, OK? So let me conclude and say, when I put it all together, we have the following picture. So here it is. So this one was the electroic precision. And some of you came late. So if you came late, you should get impressed by this. By now, there's the standard old bell and there's the experiment. So that was the electroic precision. And then we did the same program that took whatever, 30 years, hundreds of papers and all this. And we get this paper, this plot. What is this plot? This is the unitary triangle, you see? Alpha, beta, gamma, you see? And all those different colors are different measurements. This is the gamma measurement that I love so much that I was telling you about. This line is the sinus 2 beta measurement that I was telling you about. This one has to do with the BB bar mixing. That has to do with some other thing. That has to do with CP variation in k-ons that I didn't have time to talk about. But the bottom line, what I really want to tell you is this. This is, you see this? I cannot even, let me try like this. I cannot, I'm getting too old, I have oscillation in my head.  And actually, it's really cool you see this. This picture is from 2016. And you see the same picture in 1996. And in 1996, the area was something like this. All these were the area. And the bottom line, what I want you to be impressed is that this is like a huge number of measurements that depend on four parameters of the CKM. And they all agree, and there's no contradiction. That is, we learn the same story. At one loop, the standard model explained the data extremely well. And I gave you two, two examples, not only one. Say, hey, you give me one example. You shouldn't say it, because it's one loop. Should be impressed with one example. But I gave you two. One, two. Two examples that the standard model works to the present level at one loop. So we actually do calculation at one loop. We do measurements that are precise better than below 1% precision. Very complicated measurement. And they agree, totally, totally agree. There's few facts, few experiments that are at the order of few sigma. And they could be anomalies, whatever. We should care about them. I think we talked about it already, OK? But the bottom line is this, OK? For this, when I told you that I want to learn how to do the, anybody know how to play the trumpet? Anybody have a trumpet by chance? No? OK. I should have actually, I could download the trumpet. Ta-da, ta-da, ta-da, ta-da, ta-da. You're impressed, right? Without the trumpet even. You're impressed, OK? And that's really where I want to end my lectures, OK? Where you are so impressed, OK? Like, wow. And most of us, so let me kind of conclude this thing. And I'm not going to talk about the BSM, because you're going to have several lectures about BSM. But this is the status of the standard model. The standard model, when we say the standard model is doing extremely well, this is all what I was talking in the last few lectures is this really, really impressive result. This is what we are saying. The standard model works extremely well. And in our generation, it's a little bit of, I really wish I could see some deviation. Oh, it's again, agree with the standard model. It's so boring, the standard model. It just, it's like, he's so good, he just can solve everything. That's the standard model, is this person, right? No matter what you give to him, it can solve, right? It's just so impressive. And you should not say, oh, it's another standard model measurement. It's like, wow, again, the standard model works. And we as community were able to actually construct this model. And you construct this model based on this amazingly simple axiom and model. Remember, in my very first lecture in the second transparency, I said, we have a very nice solution that has to do with simple axioms. And that's what we are talking about. We're talking about the fact that we have huge amount of measurement, extremely precise measurement, and they all agree based on this really simple, simple axiom. Tell me the symmetry, tell me the fields, write them all down again, and truncate it, do some measurements, measure the parameters, and make predictions. And that's what we see. So I hope you appreciate the standard model a little more than a week ago. Just a little more that you can appreciate it. And I think it's a good time to stop. Eight minutes. OK, done.


Welcome to Morbules 14, last time we had studied quite a bit about some examples and so on of various notions like open set, closed sets all right, interior, boundary, no air dance and so on right. So today we shall start taking them up one by one. The foremost one is about closed sets which are just nothing but complements of open sets right. So they will be as important as open sets after all. So let us start putting various properties in one single place as many as possible. It is not possible to list all of them exclusively okay but it is possible to do quite a bit of good job here. Start with any topological space. Let A and B denote any two open subsets, any two subsets of x sorry. Then we have a number of property 1, 2, 3, 4, 5, 6 up to 10 I have listed here. They are all easy properties. Let us go one by one, go through one by one. The first one is the closure of empty set is empty. So that is obvious thing why because closure is defined as whenever you take a point all points which have a pouring property, whenever you take a point and a neighborhood of that it should intersect the set but the set is empty. Everything is intersection is empty there. Therefore no point will satisfy this property. Therefore closure is empty. A is contained inside a bar because a neighborhood of a point of A will always intersect it. So that is also easy. The third one if A is contained inside A sorry, A intersection B is empty and B is open then A bar intersection B is empty. Once again take a point which is in B I want to show that it is not in A bar. Then A bar is in B will be empty right. So take a point in B then B is an open set. So it is a neighborhood of that point and we have assumed that B intersection A is empty. Therefore by the very definition of closure the point is not a point of A bar. So every point of B to take is not a point of A bar. So it is empty. So you see all these things are very easy one by one but when you club them together they may become more difficult that is why we are going through them slowly okay. The third fourth one is A bar itself is closed. Showing that A bar is closed what I should do take a point in the complement namely I should show that the complement of A bar is open okay. Take a point in the complement in the complement means what it is not a point of A bar. That means the negation or that is no closure point what is negation there exists an open neighborhood of X which does not intersect A right. So that is a neighborhood if something is in the complement means right. So whole complement see all those points will have the same property. Once X belongs to an open set U U intersection A is empty all the points of U have that property for the same reason the entire of U must be in the complement therefore we have shown that the complement is open okay. Similarly now if A is closed then A is equal to A bar. A is already contained inside A bar I must show that A bar is contained inside A what is the meaning of that every closure point is inside A or conversely what I should show A bar is contained that means any point which is not in A is not in the closure that is what I have to show right. It is not in A because A is closed not in A means what it is in A complement but A complement is now open clearly A complement intersection A is empty therefore we have found a neighborhood of every point inside the complement of A what is that neighborhood 8 A complement itself which does not intersect A therefore they are not points of A bar.  The sixth one if A is contained inside B A bar is contained inside B bar this is also as easy because starting with a point in A bar and a neighborhood of that point by the very definition it intersects A but B is larger so it will intersect B also therefore the point must be inside B bar okay so that is the property sixth here. Now A bar bar the closure of the closure is the closure itself A bar bar is A bar that is the statement 7 here how do you do this one you can go on taking point wise and so on but already you have done enough work here so this is a corollary consequence of that namely look at what we have done A is contained inside A bar here therefore when you take the closure what does it give you this sixth one it says that A bar is contained inside A bar bar okay one way you hold but the fifth one says the closure of the closure is already closed so it is it is equal to A bar so therefore the two are equal okay so you can use this one A bar is closed then A is equal to A bar so A bar bar is A bar that is directly the seventh one okay says that A bar is the smallest closed subset containing A first of all A bar contains A right here so it is a closed subset containing A now you take any B which contains A bar then you take bar again that A double bar which A bar itself will be contained inside B bar but B bar is B because I take any closed subset which contains A bar so A bar is the smallest closed subset so that is the seventh one the ninth one and tenth one are quite useful especially this one says what is happening to the union under closure so this is like you can say that the two operations of taking union in closure they commute each other first you take the take the union and then take the closure the same thing as first take closures of both of them and then take the union but in the ninth tenth one it only says he's contained in here you have to be cautious equality is not assured here it may happen but it is not true in general that is the meaning of this okay let us work out nine so first of all A is contained in the union therefore by sixth show A bar is contained inside the bar of A and B similarly B bar is contained inside so one way is clear namely right hand side is contained in the left hand side okay you have to show left hand side is contained in the right hand side for that you have to work a little harder so let me this time to use this all these things I have written down carefully you can go through them okay I have already done but the ninth one let us go to a ninth one now okay namely one way we have already done namely A bar is contained in the union because A is contained inside this one so bar will be contained inside a bar similarly that so union is contained inside union of the bars is contained inside bar of the union so we want to do the other way so start the point in the closure of the union we want to show that X is either in A bar or in B bar that is the meaning of union if X is in A bar no problem we are done so assume X is not in A bar then we must show that X is in B bar okay what is the meaning of X is not in A bar this means that there is an open subset V such that X is inside V V intersection A is empty okay now use this fact to show that X is in B bar okay how take any open set which contains X put W equal to U intersection B okay then W is an open set and X belongs to W now use the fact that A union B X is inside A union B bar that means that this W will intersect A union B okay if you intersect A union B what is A union B intersection VW it is A intersection W union B intersection W one of them must be empty one of them must be non-empty because it is empty it is non-empty okay but A intersection W is already in A intersection B we started with A intersection V equal to empty here so one of them is empty the other one must be non-empty what is the other one B intersection W but now B intersection W is contained inside B intersection U which is bigger therefore this is also non-empty so that is what we wanted to show starting with any open set containing X we wanted to show that U intersection B is non-empty so that that will prove that X is in B bar so what we have done assume that X is in the closure of the union assume that it is not in one of them closure of one of them then it is in the closure of the other okay so that proves right actually sir one question please here A union B is subset of A bar union B bar yes if I take both sides closer then right side is A bar union B bar is closure union of closures what I have started A is contained inside union B this is done other side other inclusion other inclusion A union B is subset of A bar union B bar I take both sides closer but you don't know you have used your A bar union B bar is closure of A bar is closure B bar is closure union of closure is closure union is closure yeah then bar will be equal to that also can be used sir yeah yeah you are sure thank you sir yeah you can do that way also very good yeah okay so let us come to the intersection take a point in the intersection if U is a neighborhood of X then we have U intersection A intersection B is non-empty because X is inside the closure but if this intersection is non-empty both U intersection A and U intersection B must be non-empty therefore for every neighborhood of X both intersection non-empty means what X is in both A bar as well as B bar so the closure of the intersection is contained in intersection of the closure okay but you have this you have this equality inclusion map here okay by repeated application of this you can always get finite union right actually 9 to 1 let's go to 9 to 1 here it's the equality you can take finite union okay there will be equality but if you take infinite union will this work equality usually that is what is going to be problem because just now as one of you pointed out infinite union of closed section may not be closed so you can't come back okay it may happen that they are equal but in general that is not true right so if you take arbitrary union the closure of the union definitely contains the union of the closure but if you take first separately closures and then take the union it will be contained in that solve but may not be equal may not be equal okay we are plenty of examples for example here is an example take each singleton where the point varies over the q okay that is a r is just one point a r is singleton r then you take this collection the union of all this a r is just q right and what is the closure of each point points are closed inside the metric space so their closures are they are itself so union closure is just q right but first you take the union that is q then take the flow here is with the whole of r okay so here I am taking each a single ton rational number this closure is that single rational number itself when you come here it is the whole of q its closure we know is the whole of r okay we know from elementary real analysis that q is dense in r all right similarly the tenth one here okay what is the tenth one a intersection b bar is containing equality may not be true here even when there are only two points and two sets for that what do I do I take a equal to q and b equal to complement of q rational numbers irrational numbers intersection is empty here so closure is also empty but what is the closure of a it is the whole of r what is the closure of b it is the whole of r so intersection is r so this is empty and that is the whole of r so that gives you that equality need not hold all right so that is roughly you know familiar to be close subsets okay so let us now go to something some more examples here about closed sets a hyperplane in l hyperplane l in k power n you can talk of r n or c n does not matter to think of r n no problem then every coefficient set retry will be real numbers that is all so hyperplane is given by a linear equations like this okay this is actually if you put r that is also called hyperplane this is called hyper surface okay equal to you can put any some real number or complex number here that will be hyperplane equal to r is the hyper surface hyper subspace hyper means what just one dimension lower surface subspace is linear subspace okay so if you write f as f of x 1 x n is this linear map a 1 x 1 a plus a n x n etc this is nothing but a linear map from k n to k okay we have seen that such a linear map is continuous because adding two continuous functions is continuous scalar multiplication by the continuous function is continuous you use it iteratively but you know 0 is a closed set inside k so inverse image of a closed set is closed under a continuous map because inverse image of an open set is open and a continuous map so this is what we have already seen right therefore what it emits to saying that each such l is a closed subset of k n easy examples now you can use finite intersection of closed sets is closed to get that if you have finitely many equations like this the common zeros will be also closed so those are all the vector subspaces of k n okay take any vector subspace of k n so suppose it is of dimension say m then you can write it as n minus m equations common solutions of n minus m equations only finitely many of them anyway so they are intersections of four dimension one subspaces okay so they will be also closed okay these are elementary uh examples but they have to be properly understood same explanation will be we will hold for these are half spaces like you can say x1 is positive x1 is greater than or equal to 0 or y1 is greater than or equal to 0 and so on so here a1 x1 an xn is greater than or equal to r greater than or equal to r that means what it is the inverse image of the closed interval r to infinity open inverse image will be a closed set so such hyper planes are also closed so intersections of these hyper planes a hyper half spaces is also closed okay so for example you can take all points which are bigger than or equal to uh x coordinate is bigger than equal to 0 and y coordinate also bigger than or equal to 0 bigger than equal to 0 both of them are closed sets intersection will be closed set what is it it is the closed first quadrant right so like this you can construct lot of them moreover you can go to other things like look at the closed disk drxd in any metric space d is little d is metric metric drxd set of all points y such that distance between x and y less than or equal to r right similarly the sphere what is sr it is those points where in the distance between x and y is equal to r okay so those things should be closed also why this time you have to use this fact namely fixing any y or fixing any x whichever whichever one you want to fix fix why x going to dx y this is a continuous function on the topology of the space namely tau d to the real numbers okay it is a continuous function from x to zero infinity what is the topology on this one it is the metric topology here okay so these things we have seen that they are continuous function therefore closed balls the spheres they are all closed subsets okay any questions so let us consolidate these things let us have any question if you have any questions you should ask now okay hello sir yeah so in in the hyper plane equation instead of zero it can be replaced by some real number yeah so definition this is actually this is hyper plane fine but this is actually a passing through origin when they pass through origin they are called hyper hyper subspaces okay okay so i wanted to concentrate on one thing instead of floating r first so if you can float r singleton r is also closed the same argument holds so that i have used in the second part if you are going to put r it is fine okay so let us stop here until next time thank you