 Abbiamo già tutti i partecipanti ammessi o ancora no? Si, si, il meeting è aperto. Eh, è già aperto, ok. Quindi in a minute we will begin. Wait just one minute. Ok. Ciao, we start. Stefania, ok, so the last lecture today will be on standard model by Stefania Goree. I remind you that you can ask questions. I encourage you to ask questions by raising your hand rather than in the, you can also do it in the chat of course. But I'd rather have the hand raised because in the chat there are several comments and I may lose one question. All right, whenever you want, Stefania. Thanks. So welcome back everybody. And yeah, today as I was anticipating yesterday, the main topic will be the electric Lagrangian of the standard model. So we want to understand how that part of the standard model works. And before getting into this main topic, let me discuss briefly a little bit of leftovers from yesterday. So I wanted to introduce briefly natural units and dimensional analysis because this is something that we'll use later on, especially when we start discussing more the phenomenology of the standard model. So this is going to be our section 1.3. So again, natural units and dimensional analysis. So as you have probably already seen, so the idea is that in high energy physics there are some, sorry, some fundamental constants that we put equal to 1. So what we do is to impose to say that the speed of light is equal to 1 as well as the reduced plant constant. So each bar is equal to 1. Now, we know also that, you know, the speed of light, the dimensions in, if you use, for example, the international system of units, this is a length over time. And instead for each bar, here we have a energy times time or equivalently a mass times length squared times time to the minus 1. So from here we see that we have a relation between length, time and mass and we can express everything in terms of a power of energy. So basically from here we gather that the mass is nothing else than length to the minus 1 and also t to the minus 1. And this is also equal to energy. So when we try to estimate things in high energy physics, at the end of the day we want to have, we perform a lot of dimensional analysis to see how things scale as a power of the energy. So energy to some power is equal to our physical quantity. So that, okay? And so what is, you know, we can look everywhere in the web or in our books and see how what is the value for the speed of light and for the reduced plant constant in whatever units and then you get relations. So one relation that is quite useful that is used a lot of times in phenomenology is to remember actually that H bar C is equal to 200 MeV times a femto bar. Sorry, a femtometer, not a femto bar. And then also it's useful to remember that one EV is very roughly speaking equal to 10 to the 15 seconds to the minus 1 and also very roughly 10 to the 7 meters to the minus 1, okay? And this is, you know, coming directly from imposing that C is equal to 1 and H bar is equal to 1, okay? And then of course, you know, once that you have EV then you can express every form of energy in terms of seconds to the minus 1 for example you can express joules into these different types of units, okay? Now, for dimensional analysis, so this is natural units so dimensional analysis and we see an example of this of the power of dimensional analysis later when we will introduce the weak Lagrangian. It is very useful to see what several quantities including the fields that we introduced yesterday what is the dimension of these quantities. So first, the starting point if you want is to understand what is the dimension of the action. So notice that whenever I put to this parenthesis means that I want to investigate the dimension of some physical quantity. So what is the dimension of the action? Can anyone tell me? I need to unmute if someone wants to speak. No, we have an answer in the chat. Okay, great. Yeah, so this is dimensionless. That's why I put it, this is equal to 0 or if you want, you know, I can put, yeah, let's put this equal to 0, so dimensionless. And if it is not obvious, we can think that what is the action, right? Or what is the Planck constant? So the Planck constant is the elementary quantum of action and we impose that reduced Planck constant is equal to 1 so it doesn't have dimensions that's why the action is also dimensionless. And then from here we can extract everything if you want because we know that the action we have introduced it yesterday, at the very beginning, is simply the integral of the Lagrangian in four dimensions. And then from here we gather that the Lagrangian has dimension four so it's an energy to the fourth power. Energy to the fourth power. And then here let's play a little bit around with fields so what is the dimension of fields? So we can do the exercise for scalors. So we can remember, for example, that one term, one allowed term in our Lagrangian for a real scalar and what we have seen yesterday was this type, the mu phi, the mu phi, and okay, there was some factor of two for a real scalar, it doesn't matter. Then we know that the derivative is equal to this quantity here and therefore has the dimension of one, so this is one over the length and therefore energy. We know also that the Lagrangian has dimension four and then putting everything together, we get that the dimension of a scalar is also equal to one. And then, you know, is exactly the same exercise for fermions and gauge bosons. We have just to remember or to know how to write down some terms of the Lagrangian and then from the dimension of the Lagrangian we get the dimension of the field. So fermions, we had the term of the Lagrangian that looked like this as we have seen yesterday. And then from here again, we have a derivative, we have learned what is the dimension of the derivative and then putting things together we learned that the dimension of this fermion field is equal to three over two. And then the last one is a gauge boson. Again, we have the Lagrangian which is given by minus one fourth, F mu nu F mu nu, where F mu nu was given by the derivative minus this other derivative. And then again, we get that the dimension of a gauge boson is equal to one. Okay. So you might wonder why is this useful? Why is it useful? So there are several reasons. So these natural units and dimension analysis can allow you to perform estimates of physical quantities quite easily. So estimate of physical quantities. Justo anticipate what we'll see later. So what we'll do is to estimate actually the width. I will explain obviously much more later but I will estimate the width of a mion based on dimension analysis. So that's pretty useful if you don't want to enter into the, you know, tiny details of calculation but just to have a rough estimate of whatever physical quantity that you want to compute. And then also it allows you to understand what are the operators that we can put in Lagrangian. So what operator in Lagrangian. Because we mentioned that we want to have a, for example, a standard model Lagrangian but also if you think about a BSM of UV complete Lagrangian that is renormalizable. And this tells you that the dimension of each operator should be at most equal to four. So dimension smaller or equal to four. And then having the dimensions of the fields of course I'll show you what operator you can write down and what you cannot. Now, so we saw what are the dimensions of the fields and then we can just write down the dimensions of other physical quantities as for example cross sections. So if you take a cross section so this is an area. This is what a cross section is. You can think about it as the area of the target very broadly speaking. And so this is an area and therefore this is equal to minus two. So length to the second power or energy to the minus two. Then we have the lifetime of a particle that I denote with tau. This is time and therefore in terms of energy to some power we have energy to the minus one. And then finally with the width of the particle that is the inverse of the lifetime. And obviously I mean I'm writing a little bit sloppy but hopefully this notation doesn't confuse you. Well, I should probably put the parentheses in this way. And this is equal to one. And obviously we'll use this relation when we want to estimate the width of the neon later on today. So this is just a little bit of a reminder of, as I said, the natural units and dimensional analysis and then now we are ready to enter into the main topic and to start a discussion of the main topic of today namely the electric Lagrangian or the standard model. I don't know if there are any questions on this reminder. Otherwise we are ready to move. Okay, so since there are no questions. Let me start. Let's write here the goal of today. So we want to write and understand the electric Lagrangian of the standard model. And then for tomorrow we will break the electric symmetry and we'll speak quite a bit about the Higgs boson. That's the goal for today and tomorrow. So for doing so, let me first speak a little bit more about symmetries. We introduced some of them yesterday but I want to discuss with you a little bit more about symmetries. So in particular we are going to focus for now on a SUN group, a SUN lig group. So this is a special unitary group and this is a group of transformations among the components of a complex and dimensional vector. So if you take phi being a n dimensional vector a complex vector then SUN leaves the norm invariant. So invariant. So you will have that if you do a transformation on this field, a SUN transformation, this is simply equal to your starting point. Okay. Now the reason that I am mentioning SUN is that actually we have this type of symmetry to describe the electric interactions. So in particular we have to introduce what we call a gauge or a local symmetry. So this is a symmetry or let me say transformation that depends on space time, on x. And so if we focus on a local SUN symmetry and we apply to this vector phi that I introduced before what happens to phi. Pi will transform as ux phi or I can write this transformation explicitly this unitary transformation explicitly as the exponent of i alpha a of x t of a phi. I can see what these symbols mean. So first of all here as you see I have some parameter, some continuous parameter that does depend on x and that's why we're considering local transformations and these are what we call the generators of in this case SUN. To be a little bit more precise and this is a note for a little bit more for experts I should put here an R in the sense that these generators will depend on the particular representation of the group that I'm taking. So phi can transform in many different ways under the SUN group. We'll focus on the low dimensional representations of SUN but in principle you can write down as complicated representation as you want and then you will have different type of generators of this symmetry. And then what you see here also I'm putting an index A in order to tell us that indeed we'll have more than one generator in particular we'll have for SUN we'll have N squared minus one generators. So the index A will go from one to N squared minus one. And just to introduce a little bit of symbols because we'll use them later so these generators of SUN will obey the algebra namely some computational relations so we'll have that the commutation between T, A, and T, B is equal to I, F, A, B, C, T, C where these F are structure constants. Stefania, can I read a question from the chat? This is a question from Ankita Kakoti you have written gauge or local symmetry. Gauge symmetry can be global too, right? What I mean with a gauge symmetry is a local symmetry that does depend on the space time. So it's not a global symmetry and these are two different things. Thanks on behalf. Ok, very good. Ok, so this is just a very generic introduction about SUN transformations local transformations and then we can be a little bit more specific I connect these to the letter week symmetry. So if you think about electromagnetism so what we know is that electromagnetism is related to a U1 local transformation that is a very special case and so it's a simple case just rotations and for a U1 if you have something that is transforming under this U1 local transformation then your generators are simply identity matrices. For phi that is charged under U1 so for example if you take an electron that has an electric charge the electron will transform in this way here where you have the exponent of your charge but here you have something trivial that is simply an identity matrix. So let's write it explicitly so for an electron or let's write for whatever fermion with a charge fermion with a charge q this fermion will transform in the following way e to the Iq we have discontinuous parameters let me put them here alpha x and then phi so this is a direct consequence of what we saw before but then we end up having a problem so let's recall the Lagrangian that we have seen yesterday for a fermion so yesterday we brought yesterday and we wrote that we have terms like this for a fermion but then we can see what happens to the Lagrangian if we do this U1 local transformation let me just for simplicity let me call this q of x just for a matter of simplicity to write it down so let's do a U1 transformation to the Lagrangian so here we'll get e to the minus Iq of x then we have I, gamma mu then we have e to the Iq of x the derivative applied to the field but then we have also second term so this guy here will transform as we have seen q of x psi but then we have this derivative that is acting on both of these terms both the field and this exponent depend on x so here I will have two terms and then here again I have psi e to the Iq of x ok but then you see what is the problem that even if you simplify this expression so there will be terms that go away so this exponent this exponent will go away and of course should not forget that I have also mass term but nothing happens to it but you see that we have this extra term so what this little exercise is telling us is that if we just take the Lagrangian that we have seen yesterday this Lagrangian is not invariant under a local yoan symmetry ok and I think someone asked me yes if I remember correctly but so here are the what we are trying to do is to write down a Lagrangian that is invariant under the symmetries that we are imposing and here I am imposing a local yoan symmetry so what I want to have is a Lagrangian that is invariant under yoan so there is something that I need to do on top of the Lagrangian that we introduced yesterday Stefania, someone asks if a q of x can be complex so this let's see no so these are real parameters and so I should have said these generators are actually Hermitian so the idea is that this u transformation is a unitary transformation and then ok excellent so we need to do something and what we do is so what to do is to replace this derivative with what we call the covariant derivative so basically we need to introduce gauge fields in order to have a Lagrangian that is symmetric so introduce gauge fields we put s in parentheses in this case we just have to introduce one gauge field that is the photon so the covariant derivative will be defined as the partial derivative plus iq a mu where this is the photon that we know and notice that just not to get confused this q here is simply this q here is the charge for example of the electron and but this is not the only thing we need to do what we also need to do is to specify the property of transformation of this photon under the local u1 and so the photon will transform in the following way as a mu minus the derivative of alpha of x but again alpha of x where the continuous parameters that we saw above here and so a little exercise so if you want at home we can check that the Lagrangian that is written as psi i gamma mu d mu minus m psi is now invariant under this u1 local symmetry and actually I can even add an additional term namely the kinetic term of the photon so this f mu nu and then again we can see that the full Lagrangian is invariant ok so we found we brought down all the possible terms of our sorry elettri so our em Lagrangian dettar invariant under this local u1 symmetry ok so this is our qed Lagrangian question from Gautam one of the q term in a mu field as in the gauge transformation I guess Gautam you want to speak to yourself hey Stefania I was wondering so in the transformation for the amu field under u1 for the photon we will not have a 1 hour q term for the oh very good it depends how it depends how you define it so using the definition that I have you don't have sometimes let's see because we have an IQ in the covariant derivative so I'm just assuming yeah but actually you can check that with the notation that I'm using you don't need it so if you consider the transformation I'm writing above here as e to the i alpha x then you need to put this q this 1 over q but it depends I mean you can shuffle q's around as long as you are self consistent is all fine but it's true in some books you find that here they have 1 over q not with the notation that I'm using I think the notation I'm using is self consistent ok ok very good this is obviously the simplest example but this is a very important example because it is describing our QED and then notice that because of this local symmetry we have introduced an interaction term between two fermions and one gauge boson then we can generalize this to the weak interactions so this was electromagnetism and then we can do the same I mean a similar thing for weak interactions this will correspond to a SU2 local gauge symmetry and then we have to play the same game basically just that we have to be quite more careful because this is a non-habilian gauge symmetry meaning that the generators are not commuting non-habilian but in terms of sorry before you go to the nabilian would you like to answer another couple of questions one from Mandeep who asks why we want our Lagrangian to be invariant in terms of their local gauge transformations yeah so this is the way I presented is that we are imposing a U1 local symmetry U1 local symmetry and then I want to have a system in Lagrangian that is invariant if I do those type of transformations now at the end of the day what we are trying to do is to describe the interactions between fermions that have an electric charge and the photon and then this is well described by Lagrangian that has this U1 local symmetry but that's the goal describe the interactions between fermions and photons yeah and there's another one from Ankita and I'm sure can we define coupling constant in this context maybe Ankita you want to hello ma'am ma'am as you have written when you introduced the gauge fields that is the covariant derivative is equal to del mu plus i q a r that is the photon field so can we define ma'am the coupling constant in this context yeah that's right this q will be my coupling constant so you can think if you think about an electron here you will have just the electric charge and this is telling you indeed how strong is the interaction between your electron and the photon yeah that's right does it answer to your question okay very good for the weak interactions these are well described by an SU2 local symmetry and then we have to understand what are the generators TA we have learned at the beginning that we have n squared minus 1 generator for SUn so in this case we expect to have 3 generators for the for this SU2 and and these are actually for the fundamental representation of the group these are the Pauli matrix sigma of A and actually typically people use a different normalization so there is one over two here it doesn't really matter too much and then we do the same thing so we can check that the Lagrangian wouldn't be invariant under SU2 if you don't do anything so you have to introduce a covariant derivative and that is defined as the partial derivative plus IG and this again is my coupling constant as we were discussing before for electromagnetism A mu and let's write explicitly and then TA times side and now the these gauge bosons here so we have three gauge bosons because notice that here we have an index A that goes from 1 to 3 so these are to be seen tomorrow they will correspond to well let's write them as they will generate the W and the Z boson of the standard model not exactly we'll see all the story tomorrow and then how do they transform under the SU2 so let me define Mu as MuATA this combination and this transforms as U MuU to the minus 1 plus I over G derivative U to the minus 1 where U was the thing that we were defining above namely this thing here let's go second above so this was the U transformation ok so and that's that is what is these are the transformations and the covariant derivative that allow us to write down Lagrangian that is invariant and the SU2 and then again our weak Lagrangian looks like before just that we have a different covariant derivative d Mu minus M psi again at home you can check that is invariant under SU2 again at home and you might wonder what about the kinetic term of these gauge bosons so I would like also for the weak gauge bosons to write down something like this F mu nu and this works only if I define F mu nu in a bit different manner if compared to the Abelian case so they so what I need to do so this would be you know my F and the field strength in the case of electromagnetism but now I need to add this extra term that is the commutator between a nu and a nu and in this way again this term here you can check that is invariant under my SU2 SU2 transformations qui question whether G can have a group label can be dependent on the generator very good no, so this is a universal capping constant that is defined in my field and so for as far as we know so far you know the the fermions that we have discovered in nature and then also actually the Higgs boson itself are in fundamental representation of SU2 and basically everybody interacts in the same manner with the gauge bosons once that you write this week Lagrangian essay as I wrote it down so this is just a universal coupling and actually even on top of that everybody is in fundamental representation I wonder if he refers to the different group factors you'll have later on different group factors not sure what you mean oh the fact that then you have eventually different capping constants in the standard model strong oh sure so this is obviously only the weak coupling constant that is different from the Q from the electric charge that is describing electromagnetism that is different from the strong coupling constant so this is specific for the weak interactions um hello Mr please hello ma'am ma'am what is the physical significance of adding this commutator term in this F mu nu for SU2 local invariance oh very good so basically this you might wonder how so you have this field say mu and they as you can see from here they transform in a weird manner if you want under SU2 and the reason being that SU2 is non abilient so you are dealing with generators that are not commuting and therefore what you learn is that also your field strength will have a term that is coming from this that is directly connected to the fact that the generators are not commuting now in terms of physical interpretation actually this is something I wanted to comment on because it's very very important so you see that if you plug in this F mu nu in the Lagrangian contrary to the electromagnetism what you learn is that you will have extra terms in the sense that you will have interactions between three gauge bosons and then also between four gauge bosons ok so these are new interactions arising in a non abilient gauge symmetry and this is obviously very important phenomenologically this is something that we can check in our experiments and it was found to be true and this is of course not happening between photons so photons don't auto interact so in this sense this additional term has very important phenomenological consequences and only thanks to this term you can write down this term in a SU2 in abilient way thank you ma'am and there's another two questions maybe we can take them if they are short otherwise we can defer them Manuel yeah thank you again regarding this universal coupling constant is the reason why we take chi in universal coupling constant just due to phenomenology so we know that z and w boson interact the same way or is there a theoretical reason why it has to be universal so that they cannot say c boson and w boson a couple differently yeah so they so let's go back maybe to our let's see one second where did I write it down um yeah so if you consider a SUN let's see yeah so we know let's see yeah so what we are doing is I wrote it down explicitly here but for a u1 but it's the same also for the others so what we are doing is to to take let's see how do I explain this yeah so the bottom line is that we know that actually everybody is interacting the same way with w1, 2 and 3 all the weak interactions and so this is a universal coupling that we are plugging in this SU two transformations um yeah and so the idea is I thought I wrote it down sorry the yeah it was here so the idea is that you are factorizing out from here a universal coupling and that is the same no matter what what a you are taking yeah and this is phenomenologically something that has been checked very accurately and found to be true yeah ok thank you and this is the last question by thank you the last is also related it's just that as in the u1 you see that the charge can depend on the fermion or on the particle that it's collecting but here we have a charge that it's equal for all the fermions because as I saw like when you write the f the field strength term we add the charge on there so it seems that we can't do a charge specific for each fermion yeah that's true it has to be universal then because of that yeah as you can see I'm not specifying so this is the same G for whatever fermion that is charged under this SU2 gauge symmetry yeah right thank you yeah of course I mean we should not be confused so tomorrow we'll discuss quite a bit about it and then it will turn out that the interactions with the W boson and the Z boson is different from the fact that we are breaking the last week symmetry so but we'll learn all about it tomorrow so for now we have only this universal coupling G with everybody yeah so I think this is actually a good moment a good time for taking a break so maybe you can take a 5 minutes break all right perfect then we will reconvene at 55 okay okay we are ready to resume but before can I ask to the technical assistants if they are listening to make me co-host again because I lost the connection for a moment otherwise I cannot unmute people okay great thank you very much it's fine so we can continue does let me see if there are questions okay no so these are more of comments so please if you want to ask questions I recommend that you raise your hand so I can tell them from comments having said that please Stefania great so now it's time to start our chapter 2 so here a little bit more into the details of the electric week Lagrangian and so let's start with a little bit of introduction of why the week interactions so why week interactions and this is if you want a little bit of a historical perspective and so the reason that historically we introduced the week interactions is that we had quite a bit of data that we didn't know how to explain in the context of electromagnetism so several phenomena could not be explained by electromagnetism and and the classic example is beta decay so the beta decay of atomic nuclei so beta decay you can write down as you take a some whatever element x with some mass number a and atomic number c and this will decay to a different element x prime with one more proton so z plus 1 and then emitting an electron now actually we'll comment a little bit on the fact later on that we need to conserve the letter number so actually there is not only an electron coming out but we're sending a neutrino and so this is a process that we had plenty of evidence for you know with different type of nuclei at the elementary level I mean at more elementary level this can be written down as a neutron going to a proton and an electron and anti-neutrino or also a down-cork going to an up-cork in an electron and anti-neutrino and obviously I mean you cannot write down any final diagram that can generate this process just using interactions with photons so we needed something extra on top of electromagnetism and then the second example that is quite important is actually muon decay so we knew that muons were decaying and then we have this decay mold into an electron and then two neutrinos I mean an anti-neutrino and the neutrino for the muons so once when you have a lot of data that you cannot explain then you have to invent some new physics and the first proposal was coming from Fermi in the 30s and the idea was to write down X term in the Lagrangian so the Fermi Lagrangian let me call it LF so the Lagrangian that he proposed was this type where you have some coupling constant coupling constant and then he wrote in interaction term like this where these are our gamma matrix here is a proton and the neutron and then you have here an electron and the neutrino so you see that this is a sort of a four fermion operator and it has the same same structure of the electromagnetic interactions in the sense that you have a vector coupling for example between the electron and the neutrino you have simply a gamma matrix and then obviously you can do something similar for muon decays so you will have again some coupling constant and then your muon neutrino and then an electron and your neutrino for muon decays so if you think about it in a fermion diagrammatic way you have these four fermions coming together so you have a very very short range force so contrary to electromagnetism that you have a long range force ok and the strength of this interaction is parameterized by this coupling constant G and actually we can see what are the dimensions of this coupling constant so let's write it down here so what are the dimensions so we can use the tools that we saw at the beginning of this glass to understand the dimension of this coupling constant can anyone maybe write it or tell me see so what is the dimension of this G ok minus 2 very good yeah so the dimension is minus 2 so 1 over mass square and I was asking this because actually here is a little exercise that we can do using dimension analysis to see how to give an estimate of these quantities as for example the width of the neon that is decaying so using dimensional analysis what we expect is that the width for a neon to decay to an electron and the two neutrinos we scale will be proportionate to G square so the coupling constant square this will give us 1 over the mass to the fourth power we know that the width is equal to the energy in terms of dimensions and then we need to put here a mass to the fifth power right and we expect that to be the mass of the neon and actually yeah this give us a good estimate of the width if you do it correctly actually you get quite a bit of suppression compared to this because you have some powers of pi and you have a big number here in front but ok I mean the dimension analysis give us the right scaling at the leading order for this width and actually interestingly enough is that this coupling G the measurement of this coupling G is extracted from the measurement of the neon lifetime so measurement of neon lifetime is telling us the is giving us a determination of G elimination of G ok ok per le persone che sono usate a lavorare con una constanza che è GF questo è basicamente il G che sto parlando da lì solo per completare ok questo è fermi in order to explain this data for better decay but then there were two problems on this so the first problem so first problem of fermi interactions so the first problem is that so take the Lagrangian for the neon decay as an example so we brought it down again as this way and and then what you can do is to split your fermions in terms of left handed and right handed fermions so for example E you can write it down as P left E plus P right E where these are the left handed and right handed projectors that we introduced yesterday and then you can do the same thing also for all the other fermions so at the end of the day if you do this exercise you will find several terms in this Lagrangian so you will find a term like like this where you have a P left here and another P left here but then you have also a term where you have both P right and then the other combinations right so you have all the combinations of left and right fermions and this is again coming from the fact that the structure the Lorentz structure that Fermi guest was the same structure as for the electromagnetism so this vector interaction just with a gamma matrix if you want but then there were studies of a set decay of several nuclei as for example cobalt by Wu in 57 so the cobalt was completeness was decaying to nickel and then an electron and an antineutrino and basically studying the direction of this electron it was found out this direction had they preferred so the electrons were having a preferred direction and this was actually not consistent with the vector structure of these interactions so not consistent with let me write for to be brief with this gamma mu type of interactions ok so it turned out so the conclusion of all these measurements was that only right handed anti neutrinos and left handed electrons were participating to the process that's why yesterday I was quite a bit speaking about carality and so on and here we see that indeed to describe this data for better decay we had some special combinations of electrons and neutrinos so at the end of the day if we want to write down properly this would be for me Lagrangian we should write it as in the following way so you have again a gamma alpha but here you put the left projector and there you have a muon and then the same thing for an electron of course I mean I should always add the emission coordinate but so this is something that is much much more consistent with data and as you see we have only a left handed muon or neutrino here participating and correspondingly this guy here is right handed for the anti neutrino so in this sense you might have heard that we say that the weak interactions are v-a type of interactions so weak interactions v-a right because this is a gamma alpha and here you have minus gamma 5 over 2 so this is the structure for our weak interactions and what is interesting is and this is something that we could check at home again that these interactions here break parity so they are not invariant under parity transformations ok so again this is a four fermion type of interactions but now instead of having simply this gamma mu gamma mu type of structure you have a gamma mu p left and then another gamma mu p left ok now we are still not 100% satisfied this was the first problem and the second problem being that this is a an effective interaction so you see that so if you look at just the operator that I brought down so this operator here so this operator has dimension equal to 6 that's why we said G had dimension minus 2 so this is a non-renormalizable operator and actually is very badly behaved in the UV for certain processes so one can show that if you take some other process like the scattering of an electron and a neutrino to give for example a muon and another neutrino these again we receive a contribution from the operator that we brought above but then you can compute a cross section using that fatical Lagrangian so the cross section for this process and what you are going to see is that the cross section will be proportional to this coupling strength G square but now you have times S where S is the center of mass energy and you see that in principle you can imagine doing this experiment at the higher and higher energies and at a certain point in the limit this goes to infinity so in this sense this theory is not well behaved and what we say for the expert is that we are breaking perturbative unitarity so we don't like this type of Lagrangian and but interestingly enough this is a sort of a low energy description of our weak interactions so they if you want the correct description of the effective Fermi Lagrangian is indeed using the Lagrangian that we saw earlier today for the weak interactions where you have this SU2 local symmetry and then you can write down using Dalpa Lagrangia in the final rules that you extract from there you can write down a final diagram like this for example for myon decay so for myon decay you have a myon coming in you have anti-neutrino you have another anti-neutrino here in the electron and here you are putting your W boson and interestingly enough if you integrate out the W boson so if you think about it as being heavy compared to any other energy of the system you got your effective interaction that I denote with a dot with G so the coupling constant that we saw above that will be basically equal to the G of SU2 over the mass of the W square and actually the range of the weak interactions is short so this is a very important difference in the electron magnetism and this tell us that actually the W boson is has to be heavy ok so at the end of the day so what did we learn here we learned that the you know the SU2 gauge symmetry that we use that we introduced above is the you know the local symmetry that is allowing us to describe the weak interactions but also we have learned that only some of the fermions are participating to these weak interactions namely the left handed fermions or correspondingly the right handed anti fermions ok so now we can really write down the weak Lagrangian a little bit more explicitly compared to before namely we'll have a sum over all fermions of the standard model then we have this i, gamma mu, d mu so the covariant derivative that we introduced before and and then we have the part for the several gauge bosons namely we have one minus one over four sorry, f mu nu f mu nu but now you know this is actually exactly the same Lagrangian that we brought above but the difference being that we have learned that this d mu plus i, g I think we call it again so a mu a t a upside i and we learned that these generators so t a upside i are equal to zero for right handed fermions otherwise these are equal to our Pauli matrixes 2 for left handed fermions if I may interrupt the question from Mandeep related to what you said before what do you mean by integrating out the W boson yeah so it is true I didn't really explain this point so the idea is so say that I give you this five-man diagram to compute and then you can use your five-man rules to compute the width in this case for the neutrino decay right now what we can do is to match this width to the width that we would obtain from the fermi Lagrangian in the sense that you can do this calculation but you can also calculate the width simply starting from this for fermion Lagrangian and then you get two different answers now when you want to match the two you say that basically the physics the two the physics that is described is the same and then you compare the two results and then you get this sort of matching condition now whenever you write down a Lagrangian of this type the idea is that the interaction is super super super super short range that's why you can write down this point interaction so this four fermion and short range means that you have something like the W boson that is very very very heavy otherwise you cannot write down a four four point interaction in this case so in this sense you are integrating it out with the idea of having a W boson that is much heavier than any other physical scale that is involved in the process in the process ok and this I mean is a base in terms of the Fermi theory but this is something that you do every time you have to deal with effectively theory where in effectively theory you have you know operators with a dimension that is bigger than four and then you do this type of you can do this type of matching I hope it explains a little bit the thing but I agree it's a little bit rushed I mean one can talk a lot about what it means to write articles and write down effectively theory but hopefully clarifies a little bit this point are there other questions not for the time being ok great so this is what we have learned more doing this sort of Fermi exercise compared to before and then I think to complete today the discussion of these electric interactions we can write down what are the standard model fermion representations so we have learned that to describe the electric interactions electric we have this we have this SU2 times U1 group and actually and this connects to the class of tomorrow so this is the SU2 the weak interactions but this is not the U1 of electromagnetism but this is called hyper charge for now you can think about it as some different U1 and we see tomorrow that this is actually connected to the electromagnetism that is not exactly the same thing but obviously I can apply the same rules of the U1 gauge symmetry that we have learned earlier today and this will be associated to a covariant derivative let's write it again let me use W this time it's easier plus I again I have another universal coupling called G prime and then I have B mu so this is the gauge boson for the hyper charge gauge boson and these are the several hyper charges of the several fermions and and and then let's see what are the several representations that we know about so you have quarks, you have up quarks that are right handed and so they transform as a singlet under SU2 because we have learned that they don't communicate to the weak interactions so you see here that the generators will be simply quite to zero there is no transformation, it's a trivial transformation and then the hyper charge will be equal to 2 third then we have down quarks, 1 minus 1 third, again they don't transform under SU2 and then we have left handed quarks these are doublets under SU2 so this is the fundamental presentation of SU2 transforming with polymetrics then 1 over 6 and then it's useful especially for the discussion of tomorrow to think about hitting two components as you left and you left please don't get confused so I don't mean this being vile spin also so these are two different fields and I'm writing explicitly the SU2 components ok so these are the representations that we know of for quarks and then we have leptons we'll have er 1 minus 1 and then we have the left handed leptons the third doublet and then minus 1 half and then again I can write down the two components as the neutrinos and the charge leptons and then in principle we could put some indexes here in I everywhere meaning that I equal to 1, 2, 3 and this is an index of flower because we know that we have three generations for each quark and leptons of the standard model and the question is so several of you might do some research on sterarn neutrinos we don't know if they exist I would say most probably but if they exist this sterarn neutrinos never done as an R I this will be simply a singlet representation so they are singlet under S2 and also they don't have any hyper charge ok obviously I mean these tell us basically everything about the standard model before electroesimic braking for completeness I should mention that we should not forget that we have also another group that is SU3 of color for the strong interactions so I want to discuss this during these classes but it's if you understand SU2 SU3 is very similar broadly speaking very broadly speaking and you can play the same game or covariant derivative and so on and then what you learn is that the quarks are in the fundamental representation of SU3 so these are three plates under SU3 that's why we know that you have different color three different colors for the quarks and so and then during this school you will have lectures about QCD by Kacciari so you will learn a lot more and to connect to what we will see tomorrow you might wonder what about this hyper charge and I can tell you already that what we will learn tomorrow is that the electric charge that we are familiar with is a combination of this hyper charge so why do you see here plus the third component of the isospine so the third component of SU2 so this is basically all I wanted to say about the electric interactions and also one message that I wanted to give from here is that we have learned that all these interactions are simply described so the electric interactions are fully described by two coupling constants and if we use this hyper charge coupling here what we have are the G and G prime coupling constants once we know that we know basically everything about the electric Lagrangian of the standard model since we have specified what representations we are plugging in ok and tomorrow we want to understand the origin of this relation here so why we introduce hyper charge and then we want to understand the electric symmetry breaking and a little bit of phenomenology of the Higgs boson ok so we are done and now let's start with more questions right, thank you very much Stefania