All right. So once again, I showed yesterday the QCD Lagrangian. We eventually arrived at this one here with the various pieces and the representation in Feynman diagrams of what the various bits and pieces actually mean. And I pointed out yesterday that with respect to the QED Lagrangian, the QED that I assume you know and that also I saw Stefania presented to you yesterday, there are essentially three main new features in a sense. These new features are the existence of color, the existence of self-glue on couplings. And as I said, the need to use ghosts when quantizing the Lagrangian and when performing calculations. And all these three consequences are essentially a direct consequence of the non-nabilianity of the theory. And you find essentially the same kind of characteristics by the way in the SU2 part of the standard model which is also the electric standard model which is also a non-nabilian gauge theory. So what I'm actually saying here is actually you find it in electric theory even if with a slightly different terminology. So let's see a bit what we mean more in detail by these three elements. So let's start with color. So color as I was saying is essentially that color matrix or if you wish the generator of the group that you find in the Lagrangian and in the various Feynman rules. For instance, consider the Feynman rule for a gluon interacting with a quark, something like this one. So you have colors, let's say C and B here. You have a color A there and an index, a Lorentz index mu for your gluon and indices I and J for the spinors of the, spinors of the spinor index of the quark which is a fair one spin one half fermion. And so the vertex for this kind of interaction is something like TACB gamma mu JI. Now you do recognize of course the part with a gamma mu which is essentially what you would find in QED. And the new part of course of the Feynman vertex for this kind of interaction is the TA. TA is one of the A generators and we said in the fundamental representation it has, it is a three by three matrix with indices C and B in the fundamental representation. And it is these indices here that eventually will interact, will be contracted with those of the fermion and will give the transform the color for the fermion. So this is the first important difference. You will have to have this kind of things. The safe coupling, well, it's essentially not much more than what I said yesterday. If you write down the Lagrangian, you find out that you have this kind of interactions. Once again, self interaction of the bosons are a characteristic of non-abiliang age theories. And that's the same reason why the W and the Z in the electric theory also interact among themselves. Another way of saying this is that the gluons transport color charge whereas the photon does not transport electric charge and therefore the photon does not interact with itself. So when you write down the various processes, of course these kind of interactions will enter in your calculations. The ghost is a little more tricky. Perhaps it is best illustrated by a process like this one. Let's take, let's say a process, GG2QQ bar. This is a process that you can easily imagine happening in a hydro collision. So you extract from the gluons two, from the protons two gluons and these gluons interact and give a QQ bar. So how does this happen? Well, it happens through a number of diagrams. This is one, for instance. This is another across the one simply. So these are the two diagrams that you also find in QED. But then there is a third one thanks to the gluon vertex. So now let's say you calculate this. We are not going to do it in detail but let's say you calculate your amplitude through the sum of these three diagrams. And then of course in order to calculate the cross-section you need to sum over the colors and the polarizations, your matrix element squared. And so somewhere in the matrix element square there will be the polarizations of the gluons. Let me how do I write it? Let me write it. Well, let's write it this way. Perhaps it's better. Let's write this as mu nu and then epsilon mu star, epsilon nu star where these represent the polarizations of the gluons. So you would normally have to write down this kind of thing, so squared, sorry. Then what do you do when you square this? You find yourself in the square with terms like epsilon mu, epsilon mu prime and epsilon nu, epsilon nu prime. And this is what you have to sum over. Now, if you were doing QD you would know how to deal with this. In QD you would simply write that the sum over epsilon mu, epsilon star mu prime is simply minus g mu, mu prime. And the same thing for the other polarization sum. So in QD, this would be sufficient to obtain the right result because thanks to a particular word identity of QD the non-physical polarization states of the photon cancel, even if here you are summing also over the non-physical ones. In QCD, you are not so lucky. In QCD, you must do something different because if you use this particular, the tricky part here is that if you use this sum in QCD there's no special warning, at least not immediately that you're doing something wrong. You get something that looks sensible but it's simply the wrong result. So it's not like your calculation explodes because you're using the wrong polarization sum. It's just that it's wrong and it's wrong because you are including terms that are not physical. So what you in this state have to use is to use, oops, sorry, the sum over the physical polarization so epsilon mu, epsilon star mu prime and this becomes minus g mu mu prime, sorry, a little bit, but then there is another part which is plus k mu k bar mu prime plus k mu prime k bar mu divided by k dot k bar. And if k mu is simply the usual four vector, this is the gluon momentum, one of the two gluons momentum. This is just this one, which by the way, if you take the z direction assuming, of course the gluon is massless, suppose you choose specifically the z direction, this becomes simply k zero, k zero. And then k bar, you can be chosen as k zero minus k and therefore k zero minus k zero. With this choice, for reasons I have no time to explain now, with this choice, you do away with the unphysical degrees of freedom on the polarization and you get the right result. Needless to say, this complicates a bit your result, especially if you are calculating it by hand and not by machine algebra. Of course, if you're just throwing this, sorry. If you're just throwing this into mathematical, you may not care too much of another tool for symbolic manipulation. You may not care too much if you have just a single term in the polarization summaries, you have three. Of course, if you're doing this by hand, this will complicate a little bit your algebra, but eventually that's what you get. Okay, so this is one way and it is the typical way to include ghosts in leading, especially in three level calculations. In three level calculations, the preference usually goes to doing the polarization summaries this way. I should still point out that there is another way that is actually available also in three level calculations and it is to simply include an additional diagram with the ghost and the coupling of these ghosts with the other terms will again eliminate the unphysical degrees of freedom. And so you may write down, instead of the three diagrams that I've written down, you may write down, well, the three that you have again. So the same ones, this one, but then you add another one and this one here. And now you can use the simple sum epsilon mu, epsilon star, mu prime equals minus g mu, mu prime. If you have included in your amplitude, the ghosts, then you can go back to the usual covariant polarization sum because the unphysical degrees of freedom have already been taken care of. If instead of doing three level, you do loop calculations, you have no choice. If you are doing loops, while you don't have the freedom to choose the form of your polarization sum, in a sense it's already implicit in the terms you're including in. And so if you have a term like this one, for instance, the loop of a gluon, you must sum the corresponding ghost diagram to the amplitude. Once you have included these things, then your calculation is correct. Excuse me, my God, there's a question. Yes. My Julian Boyle. Yes, hello. So I'd like to ask what is the deeper reason why these ghost cancel or why you can take a simple form of the polarization in QED and why doesn't it not work in QCD? Well, because I'm not sure if there is a deep reason. The reason is that in QED, there is a word identity that takes care of that. And that particular word identity does not hold in QED, so you can't get away with it. In a sense, it's lucky that it works in QED because physical photons still have only two polarization states, whereas you are summing over four. What happens in QED is that thanks to this word identity, they cancel in the physical, in the square dump digit. And that does not happen in QCD. And so you have to take care of them explicitly, either by using only a physical polarization sum or by including the ghosts. I don't know if there is a more profound and deep reason for this, but that's what I can tell you. All right. Then there's another question by Padza. So my question was, can you modify gluon propagator such as to don't use costs in the calculation? Because I mean, numerous gluon propagator corresponds to polarization sums. I guess you may use a physical gauge. Yes, something like that should probably work. Probably, yeah. That's also an option. You may use a physical gauge. Yes, I think it should work. Then good luck with the calculations and the quantization as soon as you get out of covariant gauges because it gets much more messy, I think, but I think you can, yes. Thank you. Okay. So let's see the consequences of these three differences now. And let's start with the consequences of color and actually, we'll essentially stick to the consequences of color because, well, the ghosts, I've already told you what happens and then you have to do the calculation. And as far as the three and four gluon vertex, you just have to add the diagrams. But let's see what happens with color because, well, color has some, let's say phenomenological consequences, quantitative consequences. Let's start with, this is an interesting relation. And the relation is essentially the following. Let me just write it down. N is the number of colors, delta IK, delta LJ, plus two, TA IK, TA LJ, and there is an implicit sum over A. This can be represented in the following way. Oops, sorry. A quark line, GI, an anti-quark line, LK, this is equal to essentially one over N, a quark line that I write like this, JL, and another quark line that I write like this, IK, plus twice, this is JL, everyone, and it's IK. Okay. So this is essentially the statement of a tensor product, a composition. So this is the statement that three times three bar is equal to one plus eight, composition of the answers. Why is this useful? Well, it is useful because we can use this relation to calculate something else that we need. So for instance, what you can do, what you can think of doing in this particular relation, let me actually, let me copy it to the next slide because this way I will be able to use it. So let me change the slide and copy this relation. Okay. And now, consider doing something like contracting I equal to J. So essentially you contract this, and you contract this. So as I said, take I equal to J.  So you sum TA, IK, TA, LI, and therefore, you can rewrite it as TA, TA, LK equal one half N minus one over N delta LK. And so eventually N squared minus one over two N. Delta LK. So this is a useful relation because without going too much into group theoretic arguments, this is essentially the one of the Kazimir operators of the group SU3, quadratic operator, TA, TA. And so this is called the Kazimir of the fundamental representation of SU3, of SUN given in this form in fact. And it's called CF Kazimir of the fundamental. This CF sometimes is actually spoken of as a color factor, which is nice because it is also CF color factor. But technically it's Kazimir of the fundamental. But the reason why we also like to call it a color factor is the following one. Consider the emission of a gluon of a quark. Something like this. So IJ, this is a gluon of color A. We have seen before when we wrote down the Feynman rule for this particular vertex that this is proportional essentially to the matrix TA JI. Nevermind there is also a direct gamma matrix, of course, but nevermind that's not what we are interested in. And that part there will be identical to what you would have in QED. So that's not where we are interested in or we are interested in is the new effect of color. So this is what you have. And suppose you want to calculate the probability of gluon emission. So I want to calculate the probability of gluon emission of a quark. So exactly, of course, the probe is proportional to the matrix element squared of this thing we have just written, summed over, let's say, JA. Let's say we sum over the final state. Another option would be to sum over all states, I, J, A, and then divide by the number of states, color states of I, it doesn't matter, you can choose. You either fix the initial color or you also sum over the initial color and then you average over it, it doesn't matter. So this is what I want to calculate, what the probability is proportional to. Now this will be, we'll go like the sum over JA, TA, I, J, TA, J, I, which is the sum over A, TA, TA, I, I, which is nothing but CF according to the definition. We gave earlier delta I, I. So what does this tell us? This tells us that the probability of gluing emission of a quark is, and actually, sorry, since I'm not summing over I, that's simply CF. So if I had summed over I, I would have got a three and then I had to divide by three again so I would have got the same result. So this tells me that the probability of gluing emission is proportional to CF other than the usual coupling and all the other Lorentz, Lorentz stature. And so you see why this CF, this cousin of the fundamental can also be seen as a color factor. It's a color factor that enhances the emission of a gluing of a quark with respect to what you would have if you had emitted a photon of an electron. And of course, for N equal to three, CF is equal to four divided by three, which is not a huge enhancement over the simple G squared, but nonetheless, where it gets more interesting is if you repeat the same thing and you try to calculate the probability of a gluing emission of another gluing. And in this case, this probability is proportional to the sum of this kind of diagram. And if you do the calculation, what you get is that this is eventually proportional to CA. What is CA? Well, CA is the casimir of the adjoint representation and it happens, the calculation is a bit longer, so I'm not actually doing it. It is given on the PESKID, for instance, on many, many quantum theory books. I did the calculation for CF, I'm not doing the one for CA, but the idea is roughly the same. So CA is equal to N. So we said four over three is not a huge enhancement with respect to simple colorless emission. Now N is equal to three, remember. N is equal to three in QED. So enhancing gluing emission of a gluing by a factor of three starts being a large number. So it starts being a significant factor. So you see that one of the consequences of color, of having all these color charges spread around not only over the quarks, but also over the gluons, is that then you get that this emission of gluon by a gluon can be a fairly large number. Okay, it's difficult here to gauge exactly what that CA will do. We'll see in a moment what that CA does, but for the moment, just keep in mind that one of the consequences of color is that there is an enhancement in radiation. Gluons like to emit other gluons, even more than quarks like to emit gluons. Okay, so now what are the macroscopic consequences of what we see now? What are the macroscopic differences of QED with respect to QCD? To QED, sort of QCD with respect to QED. Excuse me, there's a question which we're going to do now. Go ahead. Hi, Paulina. Yes, thank you. Concerning the probabilities, is there any normalization factor missing because there are bigger than one? In that sense, well, yeah, yeah, sure. I mean, this is just a factor that that enters a calculation with, I'm saying the probability is proportional to this number, but it could be proportional to many other numbers. There will be a coupling, there will be kinematical factors, there will be plenty of other things. So this is nowhere near a final result. I'm just saying that if you compare a result, this result with something you would have calculated in QED, other than all the other factors, you would find this number. So yes, sorry, it's not really appropriate. Okay, thank you. And next question by Max. Yeah, my question would be, so you just showed that this CF, in the case of N3, is bigger than one, so that's kind of an enhancement. So I think if you would choose N equals two, it seems that this would be smaller than one. So is there some intuition why it seems that for SU2, actually the probability would be less and due to this cosmic factor and only for N bigger than two, it's enhanced by just some mathematical. I think it's pretty much an accident. And the reason I say so is that that enhancement in the global emission of a quirk is not, it probably does not have a real important phenomenological consequences. What is actually real important, and we'll see in a moment why, is the CA. This is what changes some characteristics, but actually, but even in that case, I think that even for SU2, you have those same characteristics. So no, I would say that the little announcement or no announcement at all in the emission of a quirk is not relevant. What is more relevant is the existence of the self boson coupling and the casualty joint. So what are the macroscopic differences? What happens in QCD that does not happen in QED? Okay, so the two main macroscopic differences are probably one is confinement and the other is asymptotic freedom. So confinement is the statement that we never observe the fundamental degrees of freedom that are quarks and gluons. What we observe at large distances, so at large, sorry, distances. And remember, these large distances means larger than 10 to the minus 15 meters. So everything is relative to the microscopic world. The size of a proton is a large distance as far as QCD is concerned. So we only observe hadrons. This is confinement. The quarks and the gluons are confined at a scale which is less than the size of a proton, roughly. So this is a fact. If you look around you, you don't see a quark or a gluon unless you probe at larger distances and therefore at smaller, sorry, at larger synergies and therefore at smaller distances, the matter. I should stress that there is no proof of today, there is no proof that QCD predicts confinement. There are hints, we'll see some of these hints. There are some indirect evidence at best, I would say. The fact that QCD seems to describe very well everything else we've managed to describe with it. And we are just missing this proof of confinement. There are symmetries, there are plenty of things working with QCD, but theoretical analytical proof of confinement is still missing. And it is actually something that is potentially an open research field, of course. As I was saying yesterday, there is even a prize, there's a million dollar prize for the person who actually proves confinement. So this is one of the biggest differences, of course. The second statement is asymptotic freedom. The second difference is that if freedom is the statement that the coupling, alpha strong, has a function of the energy scale at which it is, well, measured or normalized, as you prefer, goes to zero for large energies. So the stronger you probe something in terms of energy, and remember, large energy means smaller distance. So the smaller distance you are probing, the smaller the coupling becomes. So strong interactions become in fact very, very weak at small distances, so let me, strong interactions become weak at small distances. And again, by small, I mean much less than 10 to the minus 15 meters, okay? So they are stronger where they need to be strong to keep a proton together, probably. Again, there is no for confinement, probably. Then as you go and probe smaller and smaller distances, they become weaker and weaker. And this weakness, for instance, is given by the, oops, this typical plot of alpha SQ as a function of Q. So a typical number could be, let's say, let's put it here, about 100 GV, it's all the 0.1, 0.1, perhaps. And then this is something that goes this way. This is how it behaves. So this is somewhere, let's say one GV, this can be a thousand GV, et cetera, et cetera. So this has been observed experimentally, unfortunately, since I cannot cut and paste, I cannot immediately show a plot, but this has been measured, repeatedly, over all this energy range. And it has been quite accurately tested that indeed this is the behavior of their running. This is called, because it is a function of the scale, of course, this is called a running coupling. And if you want to be very linguistically absurd, you may even add the running coupling constant. Of course, a constant that runs is not a constant anymore, but since it is sometimes of use to call it a coupling constant, then it can become a running coupling constant. But besides this small linguistic inconsistency, this is what happens. There was a question I see. Yes, hello. Yep. I would like to ask that, well, as we look at the running coupling of the QCD, we see how small energies gets, like goes to infinity. So I'm sorry, is this something that we, is something that hints to the confinement or no, we got no. So in what direction of my plot? In small energies. Small energy, okay, this is a good point. In fact, so let's put it this way. It is something that may suggest confinement because you say, oh, but then it's clearly here. So this thing is going up at small energies and therefore large distances. So of course, at some point, this will become big enough that it will confine everything because the coupling is very, the interaction is very strong. Okay, one could be tempted to draw such a conclusion, but in fact, it's wrong. The reason why it's wrong is because, and we'll see it in a moment, the prediction of this behavior here is actually perturbative. We obtain this curve and actually we then test it with the data using perturbative calculation.  Yes. In the region where alpha becomes larger than one, of course, your calculation is reliable anymore. So you can't say that you have predicted alpha becoming much larger than one. Yes, but no.  You can say it increases. Yes. But then you can't say anything what happens in this region here, for instance. You don't know. It keeps going up. It flattens. It keeps coming down. I don't know. Yes, yes. I don't know. So you cannot logically say that you have predicted confinement because you see that the coupling increases. Besides, you don't necessarily need an unfinished coupling for confinement. So you can't. You can't. I mean, the plot I've shown here is a plot of perturbative origin. And perturbation theory cannot tell you much about confinement. Yes, again. Let's see. Thank you. Okay. All right. Next question is by Abid. Yeah. Hello, sir. Yeah, my question is from running coupling constant. This is not my question, but someone asked by someone asked to me that if I will ask you as like a referee that when we work on a different quark sector from up down stage to top bottom sector. So normally we take alphas as a fixed value. And it is very difficult to take alphas different values for different quark sector. So my question is, although it is an approximation, when we'll take alphas as a fixed value, but as an approximation, it is good. So what is your comment? If suppose we are calculating a radiative decay with or any decay with, and there we are using alphas value as a fixed one fixed value. So in what kind of, in what kind of process? Yeah, it's a radiative process. Yeah, but it depends, let's see, in a radiative. So at what energy scale? Because that's a question. Yeah, the energy scale is not fixed because suppose I'm predicting with my model about the radiative decay widths of all baryons starting from a strange baryon to bottom baryon. So in that case, how I will choose alphas? It depends on the typical, indeed, it depends on the typical scale of your process. If it were, let me just, sorry, let me try to close my phone because just a second. Okay, so if you were calculating, for instance, the radiative corrections to the decay of a top quark, it would make sense, for instance, to use the mass of the top as the typical energy scale to put into lambda, into alpha strong. And so you would be choosing a value around here. If you are looking at the decay of a bottom quark, instead, then the typical scale is five GB and it would be around here. So indeed, yes, there would be quite a difference, but I don't see why you can, first of all, choose different values and where the problem would be. I mean, typically the rate, okay, we'll take along, it will take quite some time to explain how one chooses the value of alpha. Strongly as a function of the energy, but perhaps you can just say that you typically choose the scale that is the typical energy scale of your process. So if it's a bottom decay, you take five GB. If it's a top decay, you take 170 GB. And these will lead to different values of alpha. So I'm not understanding why you say to take the same values. Same value means I'm trying to calculate in the decay width of from strange to bottom variance. So now the thing is in a strange energy sector, alpha is different for the charm and bottom sector alpha is going to change. So if there are different energy scales in your calculation, in your process, indeed, it's problematic because whatever you choose, you will induce somewhat large logarithms. And so, I mean, there is no perfect solution. The perfect solution would be to be able to do the calculation to all orders in particular the QCD. Whenever you do a calculation truncated at a certain order, there will be some shortcomings in a sense. So I cannot give a general answer because it really depends on the kind of process that you are considering and also what kind of accuracy you are aiming for. Sometimes you could just content yourself with some value and it will be overall accurate enough for what you want to do. Some other times you may have to consider much more carefully what values of alpha you are considering and whether you have to resum some logs that may still appear in your calculation. So I cannot really answer generally the question, I think. Okay, another question about the choosing of alpha's value. If my working energy range is from, suppose 500 MAV to 5GV, then how I will choose the value? Means I will choose value from a 500 MAV energy scale or at a last 5GV energy scale or in between as I have to take the alpha's value. It's, well, again, it depends on the specifics of the process. Sometimes you may find that there is a certain value that allows you to include everything. Other times you will be forced to resum the logs that come out of the big energy difference. It depends on what, it depends exactly what kind of emission your alpha is describing and whether these are completely generic in the process or whether you have to provide different values. It's very difficult, again, to answer this question in a general way. Okay, thank you. You're welcome. All right, please go ahead. Okay, so I won't say anything anymore about confinement because as I said, it's just something that would need to be proved. Instead, I can say a few words more about asymptotic freedom because this is something that instead comes out, I mean, this is a prediction that comes out from perturbation theory. So I can actually predict asymptotic freedom in QCD using perturbation theory, assuming that somewhere it is small enough that I can apply perturbation theory. So typically less than one. So if I can calculate at this scale, then I can predict this kind of behavior. So how do you do this? Well, you write down a renormalization group equation for alpha and that's the typical for the coupling and that's the beta function as a function of G. So if this beta function is less than zero, then you have asymptotic freedom. So it all comes down to calculating this thing in a particular theory and see whether it comes out negative or positive. An interesting anecdote is that in the 50s, Landau calculated a beta function for QED. It found it to be negative and it was very, very excited for a while because then it has a very important consequences potentially and then after a couple of days, some one of these students found the mistake and so that goes back to QED having a positive beta function.  This renormalization group equation is usually rewritten more commonly not so much for the coupling but for this combination here, alpha S. And so, and then it becomes something like mu R squared the alpha S over the mu R squared not to the square here equal, that's the beta function alpha S and it is usually parametrized in the following way. And this is the perpetuity expansion of the beta function. Sorry, there is a minus here, plus, et cetera, et cetera, et cetera. And today we know this to quite high perpetuity accuracy I think at least three or four powers of alpha. So with, again, with B zero larger than zero, again asymptotic freedom because it is now defined with this minus here of course. So this is just a choice. So the result there you can easily solve this renormalization group equation and you get for alpha S a behavior of this kind, alpha S mu equal alpha S mu zero divided by one plus B zero alpha S log mu squared divided by mu zero squared. And you easily see that this gives a behavior like this if B zero is positive. So this is mu, this is alpha S mu, and this is what you get. Note that, so this could be the coupling of QCD. Note that for QED it's more something like this. QED has a positive beta function and therefore it evolves in exactly the other direction. And actually, well, in QED you can even extrapolate to zero one over 137 at some point, of course. So, and also remember that this is the small distance region and this is the larger distance region. So QED and QCD have the opposite behavior in terms of the running of the coupling. And again, we'll see in a moment that this opposite behavior of QCD comes exactly from the color part, if you wish, and the self-coupling of the gluons. Let me see how much I have before we can take a break. Yeah, let me, can I do two more minutes perhaps? Yes, I will. Okay, so how do you calculate the running coupling? So let's start with QED. You want to calculate the renormalization of this interaction between a photon and a electron current. And so the diagrams that you calculate are typically this one, that's the leading one, that's re-level, and then you will have a first loop and then you will have the self-energy correction and sorry, the vertex correction and then you will have one self-energy and you will have the other self-energy. This is the whole set of diagrams. It happens in QED, again, due to a word identity, this thing cancel, the vertex correction, the ultraviolet, of course, it is the ultraviolet divergences of this loop that drives the renormalization of the coupling. It happens that due to a word identity in QED, the ultraviolet divergence of the vertex, this one here is equal and opposite into the ultraviolet divergence of the self-energy. So these diagrams here cancel, you're only left with this one, which automatically obviously lead to universality of the renormalized coupling because the fermion never enter. So you calculate this diagram here, it's a single loop that you have to calculate and then the beta function of QED is quickly found to be one over three over pi alpha squared electromagnetic plus the higher orders and this is, of course, larger than zero. And this explains the behavior of the coupling that as I was saying earlier in QED, the coupling decreases going to, going to large distances, the coupling decreases. It is the screening effect of the vacuum. You know, I've probably seen this when you take two charges, the electric charges apart, between them you start forming out of the quantum vacuum, electron, positron pairs, and these screen the charge, the electric charge so that at large distances, the value of the electric charge is a smaller than at the small distances. It is what I had, what you had here. This is at large distances, the value of the QED coupling is less than what you have at small distances. As you pull two charges apart, two electric charges apart, the value of the strength of the interaction decreases. This is QED. Now, what happens in QED? In QED, what, yes, that is a question. Yes, so I was wondering because QED also has a lambda pole like at some like absurd high number. Like, does that have any physical implications or consequences or do you just say, okay, we cut off our theory like long before that and we don't care? I don't know, I won't speculate, I don't know. I don't think it has any practical consequences but not in every day's QED, as far as I know. But there could be some very special situations but I'm really no expert about that. But it is true, I mean, of course, yes, I agree with you, of course, that the thing is there and one should think about it. Okay, so QED, in QED again, you wanna do something similar. Now, let's say you have a balloon and you wanna measure its interaction with quarks. What diagrams you have to deal with? Well, it happens that some of them are quite similar. Of course, these are essentially the same diagrams. As you can see, I'm just replacing the photon by a balloon. These are the abelian diagrams. If QED is stopped here, it would be just a replica of QED. Of course, we know it's not the case because then you have this kind of diagrams too. And as you can see here, in these non abelian diagrams, these are the non abelian diagrams. I'm sorry, let's do this, non abelian diagrams. And I call them non abelian because of course, you see they have the characteristics that come out directly from the non abelian part of the QCD Lagrangian, the triple gloom vertex, and the ghosts. These things you won't find in QED, of course. So these things lead to a calculation for the QCD beta function, which is the following. Beta QCD, which is equal to minus, and already this tells you that it will be negative, 11CA minus four NFTF divided by 12 pi alpha S squared plus, et cetera, et cetera. And you see that this will be negative four NF, less than 16. NF is the number of flavors, is the number of quark types that can go around that loop. By the way, if in this formula, you take CA equal to zero and NF equal to one and TF is equal to one half, this is just a renormalization of the color matrices, you see that you get back this result here with the right sign. So this part here is QED like once you choose the proper parameter. This part here is QED like, and the fact that it's negative comes exclusively out of this term here. This term here is proportional to a CA. The color in the casimir of the adjoint is a typical signal that this term here is generated by non-Abelian diagrams. This one's here, okay? Almost, here there is also CA, but let's forget about it for a moment. It's not relevant. So this thing is negative. It is negative as long as there are fewer than 16 flavors. Well, we know six flavors, quark, six quark types, up, down, change, strange, charm, top, bottom. And so, well, we predict in perturity QCD with NF less than 16, we predict asymptotic freedom because then eventually the behavior of the coupling will be exactly the one that we showed before here. So this is what you predict. And as soon as your beta function is negative. I think this could be a good time to stop and then I will... Let's take as more break of five minutes. Yeah, if there is a question, I don't know. Maybe we can take them when we are back because maybe there will be other questions. Okay, good. See you in five minutes. Okay. Okay. Let's assume. Start again. Okay, so we essentially arrived at this expression with also... Maybe we should take the questions first, sorry. Oh, sorry. I had forgotten. Yes. So there was a question by Max. Yeah, so thank you. So I had a question about those monobillion diagrams which you drew because why is that not also this diagram where you have this aquatic self-interaction of the unglued one? Because, because, because how do you fit it in this particular final state at this order? I've only drawn the diagrams that contribute to this final state at this order. For instance, this one. That's G-square though. It's one order higher. Right, this is only G-right. I think that's the reason why. No, but no way to say, but this is also... Oh, right. No, no, right, yeah. I mean, if I have forgotten some diagram, I have forgotten some diagram. Of course, the right result will include all the diagrams that should be there. I was just thinking about maybe there was some reason which I've missed and... That could be the reason, the power of the coupling that now you make me doubt that I think that's the reason. Okay, thank you. All right, next one by Dong. Okay, thank you, Professor. So I have a question about you said that when we look plus all of these dioramas, the diversion will cancel out each other. But as I remember, we must have to plus some, you know, like some shock for the shock for the emission dioramas and it gives you shock to emission dioramas to cancel the diversion. Which one here? You mean here? Yes, yes. Mm-hmm. So in your dioramas, they don't have like the... They only look dioramas but they don't have the shock for the emission dioramas. Oh, no, no, here you don't have emissions, sure. Here you only have the virtual diagrams because you're looking at the ultraviolet. You're looking at this particular final state here and so you're only looking at the corrections for this final state. So there are no real corrections here involved in the dioramas. Oh, okay, okay. Ah, thank you. All right. Next one is Manuel. Well, hello, thank you. I have a question regarding the number of fermions. So at one point, one would reach energies which possible fourth family would have, then the QCD behavior would change then. Yes. Let's correct it, okay. Thank you. Yes. Okay. So let's go back to this equation which is essentially the result that we found. Actually, our result was more like the value of B zero on the form of the expression is given by the termization group equation. The interesting thing here is that one can rewrite this in another form which is the following, which is interesting. B zero log of mu squared over lambda squared. This is called dimensional transmutation meaning that you are trading in a dimensional parameter like alphas mu zero for a dimensionful one, which is this lambda here. And this lambda is such that, well, you can see it here and this is essentially defined as the place where the expression diverges. You can see, so it's alpha S lambda goes to infinity. And this is related to one plus B zero alpha S log of mu squared over lambda squared being equal to zero. So of course, then the denominator of this expression here goes to zero and alpha diverges. And therefore you can solve this for lambda and you get lambda equal to mu zero minus one over two B zero alpha S mu zero to get an expression. So the reason why this is useful is because lambda essentially is where perturbation theory breaks down. And of course, it breaks down in a very dramatic fashion. The coupling diverges. So not only it's larger than one, one could be somewhere here. One, okay, of course then in all this region, of course, in all this region and perhaps even a little more perturbation theory would not be reliable anymore because the coupling is of order one. But more dramatically at lambda you're sure that your perturbation theory is not reliable because the coupling has exposed, the coupling as predicted by perturbation theory. So in a sense, perturbation theory via lambda is a signaling to you where it breaks down, which is a very useful thing, of course, to know. So it is interesting to look at this and actually try to estimate this thing from data perhaps because since we can measure the value of lambda of alpha strong from the data, perhaps we can also use this to estimate the value of lambda and see if we get what we expect to get. Let me first tell you what I expect to get. Because of all the arguments that I've given in the past few hours, I do expect perturbation theory in QCD to break down at a scale of the order of the proton radius. I do not expect QCD to be a perturbative theory at distances larger than the proton radius. But remember, a proton radius means 10 to the minus 15 meters, you convert it in energies. It's something like 200 MeV. So I do expect, I do fully expect this lambda to be able to be of the order of 200 MeV. And let's see what we get. And we can use the lab data to actually estimate it. Lab has measured alpha strong at an energy scale which is very perturbative, 91 GV. So it's at least an order of magnitude above the size of, sorry, it's actually, yeah, two orders of magnitude actually above the size of a proton. And this is of the order of 1.12. So quite small indeed. It's not as small as alpha electromagnetic, but it is reasonably small for perturbation theory. Now, the value of B0 with NF equal to five, and I say NF equal to five because up to the Zebozo mass, I encounter five flavors of quarks up to the bottom. The top is larger, it's heavier, but I have at least five. So if I take the results that I have calculated earlier, this is the value and this is 0.6. And so I can now estimate lambda using this expression here. I can plug in Mu0, 91 GV. I can plug in B0, which is this number here, and I can plug in alpha S at Mu0, which is 0.12. And so I can get a number. And if you do this exercise, you get lambda of the order of 100 MeV, which is indeed quite a success because we are getting precisely the typical hydronic scale that we were expecting. We get something which is of the order of 100 MeV, which is of the order of the proton radius, which is of the order of the pion mass. So all the hadron scales, you might have dreamed of there you find them. So this in a sense, this is some sort of internal validation of QCD, perhaps not the most rigorous one, but if you think about it, you take QCD, you take it to the perturbative phase that you do discover to exist because you can actually find that it has asymptotic freedom thanks to the fact that it's a non-nabillion-gauge theory and then also experimentally, you find that there is a certain point where the value of the coupling is small enough to allow for perturbation, remember. The simple asymptotic freedom would not have guaranteed that in the energy ranges that we can access, the coupling where necessary is small enough. This is something that we have to check and we have to measure. It happens that at energies of the order of a few tens of hundreds of GV, indeed the coupling is as small enough, it is essentially this plot where I have this plot. Yeah, this plot here at energies of order of 100 GV, indeed the coupling is small enough so that perturbation theory can work. So we have a region where we can apply perturbation theory, we can measure and when you take this extrapolation and this measurement and extrapolate back, well, you find that indeed this perturbation theory breaks down precisely where you would have expected it to break down if QCD is the right theory of the strong interruptions. Again, it's not proof, but at least it tells you that the perturbative phase of perturbative QCD does not push into where it cannot work if I want QCD to be this theory of strong interruptions. At a scale of the order of 100 MeV, you want QCD to be non-perturbative in order to be able to predict confinement and all these kind of things. So it is one test, perhaps not the most stringent one, but it is one test that QCD can be the right theory of strong interaction. It is not killed by itself by trying to predict things in the non-perturbative region. Of course it is not a proof of confinement, as I already said earlier, but it is something that goes in that direction. Let me also point out that both this expression here and this particular value of lambda are purely perturbative results. And not only that, they are perturbative results at first order in perturbation theory. Of course, you can keep calculating your beta function at higher orders, and this will change both this expression here and the value of lambda. Lambda is not a physical observable. Lambda is a perturbative quantity that depends on the order of perturbation theory and it depends on the renormalization scheme that you have chosen. So for instance, suppose you calculate the renormalization group equation to higher orders. And so, so let's say beta QCD. So we have minus B0 alpha S squared. Suppose you also calculate B1 alpha S cubed. You do the calculation and it has been done already immediately in the 70s, I think, or something like this. E1 equal 17 CA squared minus five CA and F minus three NFCF, I think, divided by 24 pi squared. And then you write an expression for alpha S which is one over B0 log mu squared over lambda squared. One minus B1 over B0 squared log, log mu squared over lambda squared divided by log mu squared over lambda squared. Note that this is not an exact result. There is no closed form expression for the solution of the RGE beyond leading order. So this expression here is not the exact solution of this one but it is a solution that is accurate at that even perturbative order. There are other expressions that are possible. This is a conventional choice, is the one that is most often used, probably. And also as a result, also the value of lambda, again, is largely conventional but it will still be of the order of 200, 100, 200. So it's always there. There is one more thing I wanna say about alpha S. This particular result that you have here, this one, it was calculated for the first time in 72 or 73, I think, by Gross, Bilsek and Politzer who for this got the Nobel Prize about maybe 15, no, about 10 years, 20 years ago, 2000, 2001, something like this. So the thing I wanna mention and the reason I wanna bring it up in a school like this is that Bilsek and Politzer were PhD students at the time. And if you remember what Celine Boehm said the first day of her lecture, pebbles and silk also were PhD students when they did and they wrote very important papers. So this is to point out to you that if you have a chance of making an important discovery or something important, chances are you make it when you are young or when you are PhD students and you first enter in a new field with new ideas and things like this. When you're older, you may do other important things. You may do management, you may do teaching, you may do plenty of other things. But there is some evidence that many discoveries even if they are awarded a Nobel Prize 30 or 40 years later many important discoveries are made by young people that start with new ideas. So this in order to actually encourage you to work on new problems and put in new ideas in the field because it's not when you're 50 that you typically have the new ideas. So this is one evidence. I'm sure there are many others of people who have made their discoveries when they were very young. Even I think Higgs, I don't think he was much older than a young postdoc or a young professor when he brought down the Higgs mechanism with the Higgs balls and so on and so forth. It's not because you are awarded a Nobel Prize when you are 80 that you made your discovery when you were 80. Typically it's very young people. So again, take this as an encouragement to actually think about these things right now. Okay, questions? I'm closing the chapter on Alpha S. Yes. I have a question by Paulina. Let me remind you Matthew that the lecture should last until 3.30, right? Yeah. And then we move to the Q&A part. Yes. So I will just sketch probably the next, I will just sketch the next topic and we'll continue tomorrow. We won't take five or 10 minutes because there were a lot of questions. So Paulina. Yes, I have a question concerning the WIC theory. There is also a non-Abelian-Gage theory and what happens there with the Kaplan constant? The same. Okay. So there is a synthetic freedom in electric theory also? Well, there isn't, so there is a trend of evolution of the coupling. You don't talk about a synthetic freedom there because you don't expect to have a confinement phase but the running of the coupling is similar. Okay. Thank you. A synthetic freedom, but it's a Higgs set, no? Low energies, low Higgs, they get a mass, no? Okay. Thank you. Okay. No other questions? Okay. No, I don't see anything else. Okay. So let's start calculating processes now. We have gone through all the characteristics of QCD and now we wanna start using QCD to calculate the processes that actually happen at colliders. And the easiest process to look at, of course, is E plus or E minus two hadrons. So you take an E plus or E minus collider, you collide them and you might expect not to have to do with hadrons in an E plus or E minus collider, but in fact, nothing prevents the photon to then go to a QQ bar pair since they are electrically charged. And then this thing through confinement will eventually evolve into hadrons. Again, this is the non perturbative part of the process that we will not deal with. We will just assume that the sigma for E plus E minus two hadrons is similar to the sigma for E plus E minus two QQ bar because why not? Where else can the QQ bar go if not into hadrons? They will have to go somewhere. They will have to addronize. And of course, more technically, this is actually true modular, the so-called power corrections. So you can show that the confinement part actually produces corrections to the perturbative result that are actually very small when, so lambda is the lambda we just mentioned, the typical adronic scale, root S is the center of mass energy of the collision. And so you see that this is a particularly simple and very inclusive process. As I was saying, the Q and the Q bar can do nothing else but to hadronize into hadrons. And so this is what is often called a non perturbative correction and also it has the form of a power correction. And it is a particularly small power correction. You see lambda over root S to the power of four because indeed it is a very inclusive quantity. And therefore, since you're not going to look at details, you are not so much interested about the details of this hadronization process that takes place here. And so calculating just the perturbative part give you already a pretty good prediction of this process. And it is precisely through this process here and this prediction here to certain order perturbation theory that Lepp was able, for instance, to measure alpha strong at the scale of the Z mass. So keep in mind that there are corrections but by and large for all practical purposes, you can consider that sigma plus and minus two hadrons is actually the same thing as sigma two Q bar plus perturbative corrections. Yes, there are a couple of hands. First, please go. Yeah, I can move this to the Q and A section if maybe it gets long and you want to go on. Could you sketch how I could get this correction? Like at least how I know this is the right order. Again, if this is long, maybe we can move this to the Q and A. No, it's even longer than the Q and A. No, I cannot really easily sketch it right now. I should better look it up before I can answer that. I mean, I have some ideas but they won't actually teach you much because they're not in a form to be pedagogical. What do you prefer, Matteo, to finish and then we get all the questions? Or if perhaps there, I won't be able to do much more. So if there are questions, perhaps... Okay, so I'll start taking questions then, no? Okay, so, Banjeris. Yes, thank you. Maybe in some future experiments, I don't know these number two corrections will be played a much more bigger role in the calculations. So, on one hand, the higher the energy, the smaller the corrections are. On the other hand, there are other observables that have bigger number two corrections. And the more exclusive you want to become, perhaps because you want to observe a certain particular observables, the more sensitive you may become to number two of the corrections. So, yes, I mean, there are definitely measurements that do have a non-negligible sensitivity to number two of the corrections. It's already the case and it will likely be the case in other colliders. Absolutely. This is a very special case. This is perhaps the easiest process you can think of and it is the process that has the smallest power correction that you can think of. In other cases, corrections are much more significant. Yes, so, one process, building like a positron electron collider, it's like these corrections should be small in... Yeah, well, you see the number, I mean, already at lap with the center of mass energy of 100 GV, this was less than one over 100 to the power of four. Yes. So it was a very, very small number in this particular case. But again, there are other observables where this hadronic correction is actually larger. So... Thank you. Yes, I see. Thank you. All right, next question is by Tim Moy. Hello. Hi. Yes, sir. Why we are considering this alpha s at mass of digits? Considering this alpha? Sorry, say it again. Why we are considering the alpha s at mass of the z at this process? At the center of mass energy, I will come to that. You mean why we are taking alpha at that particular scale? At the mass of the z? Okay, yeah. Okay, so it is actually not at the mass of the z in a sense. It's that it is at the center of mass energy of the process, which is this one. Then it is the process that has been made at the center of mass at the mass of the z. And the reason why will be in the next lecture. Okay, thank you. Okay. Next question, I think it was Surui first. Hello. Hi. Yeah, so I have a question regarding the behavior of the alpha s at low energy. So if we use non-perdibutive quciety, so are we being able to predict the exact behavior of this alpha s at low energy? So if you use non-perdibutive quciety? In principle, yes. But then you have to be able to actually calculate. You can do it using lattice. Lattice does give you some insight on the behavior of the coupling as small energy. So yes, potentially, yes. But not with perturbative techniques, of course. So is it like exactly can we predict or some, like using lattice, exactly do we, like are we sure that it is the correct mean? Lattice does give some fairly good predictions for things like the masses of the hard ones and these kinds of things that are typical non-perdibutive quantities. So yeah, I would say there is quite a lot of evidence that what we calculate with QCD be it perturbatively like I'm doing here for non-perdibutively via lattice or even some models actually does work pretty well. Okay, okay, thank you. All right, all right. Next question is by Manuel, please. Oh, thank you. My question is again, on this correction. How can we calculate the correction on a hard-organization process when we don't know how confinement works? So how can I say? One way of actually getting an idea how these things work is to look at what happens at the perturbative part, meaning when you do the perturbative calculation, you know that this cannot be the end of the story because you try to calculate a quantity that is expressed in terms of observables, but you express your asymptotic states in terms of objects that are not observables. So you know that you are missing something. And so one way of addressing the problem that you're mentioning is looking at the ambiguities that still existing in perturbation theory because you're using the wrong asymptotic states. In a sense again, perturbation theory can tell you that something is missing and this is something that is missing points to this kind of missing piece. This is one way of looking at it. It is perturbation theory itself that the same way that perturbation theory at small energies tells you, hey, look, I'm failing. My coupling is diverging. You can't use me here anymore. The same perturbation theory through another kind of failure or something if we're putting quarter, tells you that there must be another piece. And then you can also get some insight on the parametric behavior of this other piece. And this is how you can get this power to the fourth behavior. So that's one way. Has this correction been tested? Not here, I think. I guess it's too small. I suspect, I don't know. I don't know if it, well, I don't know if it has been tested here in this particular process, but there are other processes where the power is lower and therefore there are bigger power corrections. And yes, then they have been seen experimentally and checked, absolutely. What you don't know, what you can't calculate, of course, is the coefficient in front of this thing. But you can check the behavior, sure. That has been checked. Thank you very much. All right, next one. Okay. Next one is by Marwan, please. Marwanne, now you can talk, I think. You or she is not, I don't know, not working or... Well, so on. You're activated the microphone. You can write your question in the chat if your mic is not working.


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