 Yeah, sounds good. Let me start sharing the screen in the meantime. Maybe just a second I will change the Wi-Fi just to make sure that it should work back in just a second. Okay, can you hear me? Yes. Okay, great. Yeah, I'm ready. Okay, perfect. Okay, then we can start the last lecture today with the fourth of Standard Model by Stefania please. Okay, thanks a lot. Yeah, so for today what we want to do is to finish the Higgs discussion that we started yesterday and then we want to talk about the clever physics. So that's the plan for today. Okay, finish Higgs discussion plus clever physics. Okay, so having said that, so yesterday maybe just a reminder. So the last thing that I showed you, let me actually share these slides. Just a second, please. Just a second, sorry. I wanted to show you these slides that we saw yesterday as the last thing, as a reminder. Here it is. So yesterday we concluded mentioning that the ones that we have the Standard Model Lagrangian, including electric symmetry braking, our Standard Model is highly predictive and it can tell us everything about the Higgs boson phenomenology, including the phenomenology that we studied at the LHC. And then if you remember, I was showing you these couple of slides. So the first slide was about Higgs production at LHC and the second one was the Higgs decays at LHC and we learned that there are many different ways for the Higgs 2DK ones that we fix its mass to 125 GV. So that's to connect to the class of yesterday and now the question is that we go back to the iPad. So the question is how to compare experimental results with Standard Model predictions. So you might have heard from the LHC collaborations that what we do is, one of the things that we do is to extract information on the Higgs couplings and we want to understand how do we do this. How do we do this? So at the LHC we measure rates for the Higgs particle and so we measure actually additional features of the Higgs boson but one of which are the several Higgs rates. And so let me mention one example. Suppose that the LHC we measure the rate for the Higgs produced, so proton-proton collisions going to a Higgs that then goes to gamma-gamma, so to two photons. So this is our measurement. Now if we have this measurement then we want to match this measurement with the prediction of our theory. So we have to impose or to look for a theory that does predict the same thing, so that predicts the cross-section times the branchination of the Higgs to gamma-gamma being the same as the experimental value. And this I mean, I write theory in the sense that this is a whatever theory that matches data. It can be the Standard Model or it can be beyond depending on data, depending on this experimental measurement. But then how to link this equation to the couplings of the Higgs boson because ultimately we want to understand how we accept information about the Higgs couplings. So let me play a little bit with this equation. And so I can normalize the right-hand side of this equation in a little bit different way. So this is simply equal to the Standard Model prediction. And then I write again the right hand side but with the denominator that is the Standard Model. And then the same for the branchy ratio. So the Higgs to gamma-gamma in the Standard Model. Now the reason I've done this is that we know how to do very precise predictions for this quantity as we have learned yesterday. And actually if you look at this ratio, what is this ratio? So suppose that I consider the Higgs production in glonfusion. So this diagram that we saw also yesterday. Then this ratio basically tell us what is this, let me call it effective coupling of the Higgs with glons normalized to the Standard Model prediction. So at the end of the day this ratio here is given by this effective coupling that I just mentioned that I call kappa g to the second power because you know the cross-section scales as the second power of the coupling. So this is a reduced coupling of the Higgs with glons. And then I can do the same thing to express this branchy ratio. Now I have to be a little bit more careful because the branchy ratio will depend also on the total width of the Higgs. So let me rewrite this expression in the Standard Model prediction. Here I have kappa g square. Then I have here the ratio of the width theory over Standard Model. And then here I have the ratio of the total width. So the total width of my theory whatever that is and and the one of the Standard Model. Then again this will be a reduced coupling square more specifically kappa gamma square. So the reduced coupling with with photons. And this is simply a ratio of width that I need to know. So finally I can write this as Standard Model times kappa g square kappa gamma square and the ratio of widths. And this as we said at the beginning this is the quantity that I measure so that I have to match with my measurements. So I do have, so DHC is giving me many measurements. And then what I can do is a global feat of data to extract all my reduced couplings, reduced couplings, kappa, as well as my total width. So that's the way that we are doing global feat of DHC measurements. And then let's see what is the status of the extraction of these parameters. But yeah let's see if there are questions on this procedure and otherwise we'll pass to the discussion of the experimental measurements of these couplings. Seems there are no questions so let's go to the slides. So what is the status of the measurements of these kappa parameters? I hope you can hear me well because I get the message that my internet is unstable. I hope it's fine. Okay great yeah in case let me know. Okay so here is the status. I'm taking this plot from the Atlas collaboration maybe I can put it full screen maybe. But so let's understand a little bit this plot. So as you can see here you have all these several kappas that are introduced in particular introduce this kappa g. So they reduce coupling with coulons and the one with photos. And so if all data are fitting well the standard model I expect to have this all this data point aligning here at 1. Because 1 would be the prediction of the standard model that I have a reduced coupling that is exactly equal to the coupling as predicted by the standard model. And indeed we see that all these data points are you know roughly aligned with this line at 1. Okay and then you see what level of uncertainties of experimental and theory uncertainties I have in the extraction of these reduced couplings. Okay. Now you might wonder what are these two different results because you see that I have one result here and one result here. So what is the difference? So the difference goes back to our equation that we saw on the iPad namely that you remember that we have this rate that did depend on the total width of the of the Higgs boson that I write down here. And in principle you know you can think about having a theory that you know predicts an enormous width so something that is much bigger than in the standard model and then you can compensate with having larger kappas. So you can you know shift everything up or down and you can still match measurements in this sense for this rate. So you need to do an assumption for fitting LHC data. So in the first set of data that you see here the assumption is that the the total width of the theory is as as predicted by the standard model once that you fix this reduced coupling kappa. So you don't have any what we would call exotic Higgs decay or exotic Higgs decay so the Higgs decay to any other extra particle. So this is the assumption that we do and then we get this fit of data. But then you can do also another assumption namely that you have some beyond the standard model decay of the Higgs so the the width of the theory is bigger than the one of the standard model. But then you impose that the reduced coupling of the Higgs with the W and the Z what is denoted here with kappa V is smaller or equal to one. Okay so this is another set of assumptions that you can do and then you get this result of the fit that is a little bit different than the first one. But bottom line is that we have to keep in mind that we need to make an assumption in such a way to do a fit of the kappa coupling the reduced coupling of the of the Higgs boson. Okay that's the message here. And then okay I'm showing you another plot here that basically tells you that in a quite nice visual way that the Higgs couplings they measure the Higgs couplings aligned with a line predicted by the standard model. So we have seen yesterday that the couplings of the Higgs are always proportional to the mass of the particle that couples to and and this is what is plotted here that's why we have a line. And we see that we have all these data points for the Higgs coupling to top WZ bottom tower and and the Neon. So the conclusion is that the standard model seems to be working with these measurements of of the Higgs rates. Okay but then I have a question for you here. So that yeah there is a question first yeah. Yeah hi sorry so can I ask you something about so this left diagram. So it looks as I said with this fit on the right hand side it's not much worse than the fit on the left hand side. I mean does that mean that the LLC cannot listen to English these two cases if there is BSM or not or do I understand something wrong? Yeah so I agree with you that they are not vastly different. I mean if you look a bit more into the detail there is some difference right. If you look I don't know for example this Kappa B you see that you have quite a bit of a larger uncertainty here than here and also the central value is a little bit shifted. So there is a bit of difference you know. For the second part of your question so to distinguish the two cases I think this wouldn't be the best way to distinguish the two cases but you would need to look for Higgs exotic decays so some new decay modes of the Higgs boson and then obviously if you discover something like that then this would tell you that this assumption is not correct. So there are ways to distinguish but it's different than the discussion that we are having now. You have to look for BSM decays. Okay thank you. Hello so I was wondering how you get the signs of the coupling if it scales by you know squares. That's a great question that I was avoiding to discuss but since you asked. So actually so you see here so for Kappa T there is also a solution that is negative around minus one and so this is performing not incredibly well in the sense that I think this is a two sigma yeah this is a two sigma interval instead of one sigma but still there is this solution because as you say in general measuring rates we have sensitivity to the square as we have seen in the iPad for the Higgs to gamma gamma rate. There are exceptions to these so there are processes that so there are rates that we measure that scale with the first power of the coupling but then you need to have some interference. So maybe I can draw for you if I am on diagram otherwise I'm afraid it's a little bit unclear and just a second let me share my iPad quickly. Yeah so they if you look at the Higgs single top production this is one of the cases so you have let's see what are the two diagrams you have a w down top and then you're ready to Higgs and here you have a still your top and this is a quark so it's Higgs top quark but then you have another diagram of this type top w and then you're ready to Higgs from here and this is still no Higgs top quark production but you see that since you have two diagrams this they will interfere and then you have something that also scale with the linear power of of kappa top so there are processes that are sensitive to the to the first power and therefore the sign but in general this is this is hard to do that you cannot get a signal sign from the LHC. That's right so this is hard to do and this is a sort of a you know a good example of something that especially in the future will be able to do for the other couplings like kappa v or you know the smaller couplings is super super challenging I don't yeah I'm not sure there are any any prospects but then what we do is to extract the the Higgs feats that I was showing you before is to assume that they are a big and unzero but yeah my doubt came from a paper which was looking at the limits on super Higgs coupling that was asymmetric on both sides it's not like so it's a less than the coupling less than b not like minus a plus you know so yeah I did not understand even then yeah yeah thanks sure mm-hmm okay so so the question that I had for you let's see if somebody can answer is so I showed you what is the present status of Higgs coupling measurements but then the question is what is missing in these plots because you know yes I released it all the Higgs couplings I predicted by the standard model then I showed you all the measurements and then I wonder is there something missing from from a plot like this or you know from a plot like this anyone let's think so if we look back well we cannot do it but if we look back at the video of yesterday what I was mentioning is that the Higgs couples to every massive fermion of the standard model and okay very good so I see in the chat same way you mentioned up down quark up down charm and strange quark and indeed this is something that is missing so you don't see any data point for the coupling of of the Higgs uh indeed with light-flavored quarks so for example the strange of the charm as well as you don't see information about the Higgs coupling to electrons uh reasoning that these measurements are super super challenging to be done um so the rates are small because the couplings are very small and then also some of the signatures from the experimental perspective are challenging so they are background they are overwhelmed by background and then another thing that is missing are the Higgs coupling because we learned yesterday that you know we can write down the Higgs potential and then what we find is that there are couplings like Higgs to the third power of fourth power um and uh but so far we don't have uh really evidence for these couplings at the DLHC and one way that the the LHC collaborations are looking for these couplings is to search for die Higgs production and the reason is that if you you know there are diagrams like these for die Higgs production that do involve a triple Higgs coupling so this coupling here between three Higgs bosons so testing die Higgs production we could extract information about this uh this coupling predicted by the standard model and therefore information about the Higgs Higgs scalar potential but then you see that the rate for this die Higgs process is very very small um and it's basically three orders of magnitude smaller than the single Higgs production so again this process is this process is pretty challenging for the for the LHC and maybe for the high luminosity LHC we'll have prospects to know a bit more and maybe to have some evidence for this process okay so that's um all I wanted to say about the Higgs phenomenology and so let's go back to my iPad and so what I wanted to to do now since also I think someone yes it was was asking me is to talk a little bit about custodial symmetry so this will be our section 3.6 custodial symmetry so basically you know broadly speaking we are done in introducing the standard model of grandeur and now we would like to understand you know aspects here and there either you know connected to symmetries or to phenomenology and this is what we are going to do for the rest of this course so what is custodial symmetry so the idea is to define a new global symmetry uh so let's define a symmetry that is an SU2 left times SU2 right and this actually this first SU2 left can be identified with the gauged SU2 of the weak interactions and then you see here I'm adding an additional SU2 symmetry that is global I don't need to I don't need to gauge it and and then what I can do is to see if the standard model of grandeur can be made invariant under the symmetry or if it is already invariant so is the standard model Lagrangian invariant so whenever I ask a question like this it's always useful to express the several fields that I am working with in terms of representations of the of the symmetry that I am talking about so first I want to introduce the Higgs field let me put it in bold face because this is a in a different representation if compared to the Higgs field that we have introduced yesterday in particular so this will be a by a doublet of SU2 left times SU2 right I will explain a little bit more what what that means and the way I define it is okay I put some normalization and and then here I've write these metrics where phi is the let me say the Higgs field of yesterday I mean that we introduced yesterday that I can write in SU2 components as phi plus and phi zero and that has you know the hypercharge plus one half so this is the field that we already know of but then also define a phi till the field that is simply given by taking the epsilon tensor so the anti-symmetric tensor and applying it to the complex conjugate of phi where phi is this guy here okay and then you know in components so this is given by phi zero star so the complex conjugate and then here I have a minus phi minus and then you can check that this is said this actually has hypercharge minus one half now let's not go at confused so this is not any field in the sense that I'm not adding any new degree of freedom so in terms of degrees of freedom these are still the standard model Higgs degrees of freedom I'm just taking a sort of a complicated coordinate of this Higgs field okay and then this this object here will be a two times two matrix of fields okay I can write it explicitly just for completeness so this will be a phi zero star minus phi minus phi plus phi zero okay and then I I can ask how this this field transforms under SU2 times SU2 and so the way that it transforms is to take two of the factors that we introduced two days ago namely one on the left with some coefficients omega l i then these are the generators so the fundamental of SU2 so this would be my transformation for SU2 left and then I have the corresponding for SU2 right where in general the coefficients are different in the sense that this coefficient here on the right on the left is different from the coefficient on the on the right okay maybe max let me finish here the Lagrangian of the Higgs and then I will take your question and then what I can do and I can check at home that this is correct is to rewrite the Lagrangian of the Higgs boson and what I can see is that this is given by the trace of the covariant derivative of this phi dagger current derivative of phi minus mu square trace of phi dagger phi plus the lambda coupling trace by dagger phi square okay so that's the way that I can write down my Higgs Lagrangian in a SU2 times SU2 invariant form yes max I just want to be sure so the relation between phi minus and phi plus is just that so so like phi plus plus adjoint is phi minus or or is it something else yeah that's correct yeah that's right that's right yeah yeah okay thank you is another question um yeah so just to be sure like the w r are not spaced independent right correct yeah this is just the global yeah that's correct yeah yeah in principle I mean for the for for this discussion even this this left here could be you know independent on x but then at the end of the day we want to identify this symmetry with SU2 or the standard model and therefore it has some x dependence but but definitely this is a global symmetry so no x dependence on omega r okay um excellent so um this Lagrangian here is um I mean it's it's easy to check and I would suggest you to to think about it at home to see that this is indeed the SU2 times SU2 invariant so basically if I do these transformations here uh nothing changes in my Lagrangian um uh but then um the Higgs gets a web so we have a lattice symmetry breaking and that the minimum this bi doublet representation can be expressed in the following manner v 0 0 v and uh and actually um um what one can see is that this vacuum configuration is not anymore invariant under a generic SU2 left times SU2 right symmetry SU2 right um symmetry um again one can see it starting from this type of transformation that I brought down here in all generality uh so the the symmetry that I introduced is broken even though it was a good symmetry before of breaking a Latruic symmetry for the Higgs Sector um so it's broken and then one might wonder what I'm breaking the symmetry to and I'm breaking it to the um um v so this is the diagonal SU2 meaning that you know I still do these type of transformations I'm putting again the arrow here but now they are not generic in the sense that my parameters of transformation um omega l is equal to omega r okay so this is the the diagonal SU2 transformation and this is actually the symmetry that we call custodial symmetry okay that is preserved by the vacuum of the Higgs yeah there is another question um yes so um regarding my prior question like how can you guarantee this if the omega right r space I'm independent of the omega l if you want to consider it or identify it with an SU2 l symmetry is in general because it's a gauge symmetry space independent or do you just consider this SU2 l to be like some other global symmetry and you take this chiral symmetry like as a whole leather discussion um yeah so at the end of the day also the SU2 custolia will be broken um at this level I think you can just think about having also the SU2 left that is a global symmetry and then you don't have to worry about about this um I agree with you that uh yeah when when once that one considers this as a gauge symmetry one has to be a bit careful but again at the end of the day right now we'll learn that also the custodial symmetry is broken so yeah okay so you can just like consider the SU2 l as a completely new single field in that sense yeah just an SU2 yeah yeah okay thank you yeah um um yeah only that you see that I mean if you look at this by doublet um this I mean the way that we introduce the Higgs field is that this was indeed the SU2 left gauge symmetry um but yeah um okay Vangelis you have a question um yes thank you um we have considered so far the the Higgs potential the Higgs sector and what about the the other part of the southern model yeah yeah I'm getting there yes so so there's a connection right I mean because um because of the Higgs sector to be involved in that means that this other model itself could be involved under this uh absolutely yes yes yeah yeah uh let's let's get okay so indeed what about the um the other parts of the standard model Lagrangian um so I'm I'm telling you the conclusion um in principle you know we can do the same exercise of taking each part of the standard model Lagrangian and try to write it down in terms of representations of SU2 times SU2 uh you know just an example and then I would suggest you to do it at all more concerned references where it is done more explicitly for example the W of the standard model is a SU2 left triplet and SU2 right singlet so we have to understand and know how all the fields uh what is the representation of all fields of the standard model under this SU2 times SU2 write down the Lagrangian in terms of these representations and then see if there is some if the Lagrangian is invariant or not and the conclusion um is that if I write down the hypercharge interactions so the u1 y that we introduced two days ago um these interactions do break the custodial symmetry okay um and indeed you know intuitively you can understand it because you know this W i where i is equal to 1 2 3 is not anymore um you know they generate in the sense that the masses uh will be different once that you introduce the hypercharge because the hypercharge will mix with the third component of the isospina as we said so there will be some breaking there of this diagonal SU2 sorry and and also there is there are other interactions that are breaking the symmetry namely the yukawa interactions and in particular um you know you will have a bigger breaking of course from uh from the top yukawa that is a the larger yukawa interaction okay um so this to say that uh basically the the lesson that we learned here is that we have the custodial symmetry that is valid in the Higgs sector of the standard model but then is broken and so you can think about it as as an approximate global symmetry of the standard model that is broken by this type of of interactions now this also connect to what we introduced yesterday that was the raw parameter that was defined as the ratio between the mass of the W mass of the Z and the Weinberg angle here um now at the three level this is equal to one three level um um deviations from one deviations from one indicate uh breaking of the custodial symmetry um so one can show that indeed any even you know beyond the standard model if you have any effect on this raw parameter this effect is breaking farther this SU2 diagonal and and in fact one can show that if you go beyond the um the three level you have loop effects we call delta raw um that uh uh that are coming from the yukawa coupling um that as we learned are breaking the custodial symmetry and in particular you will have an effect that goes like a Fermi constant mass of the top square that is basically you know the yukawa top square I mean it is proportional to that and then you have here some factor um in particular you have the the loop factor okay um and then again if you go beyond the standard model every effect will be proportional to a custodial symmetry breaking uh diagram okay um so that's uh all I wanted to say about uh the custodial symmetry and I brought to these sources an example of an approximate global symmetry uh of the standard model that is accidental so as you have seen I I didn't impose anything by hand it just comes from ones that I I I have my standard model agrangian um so there are questions um yes I have a couple of questions actually so so the the first one is so when you write the hyper charge interactions in the break this s2 uh in the organa so you mean this so you mean the net terms of the fermions and the x and in particular the b mu part there right interaction between the b mu gauge fields and the fermions and the x this is the one which which which breaks the symmetry right yeah so not necessarily the fermions even if you just focus on the gauge part of your lagrangian um you can see that uh that part uh so if you write down the covariant derivative of the hicks that involves all the all the fields or the gauge fields you can see that that part is already breaking uh um the cursor symmetry okay and then so um so the so the so the symmetry there's not this s2 left cross as to right but only the the diagonal part after uh uh spontaneous symmetry breaking is that what is called uh typically yes I think in the past I found the reference that was calling the full s2 times s2 the custodial symmetry I mean at the end of the day it's a matter of name um we have learned that uh you know as long as we know what we are breaking and what not this is name but I think typically yeah in in in in reference is typically the oops the diagonal one is uh is called custodial yeah okay and then one last question could you please explain so why is the symmetry called custodial symmetry because I thought that so I think that I heard that there's a very particular reason for that because it's somehow yeah whatever some value to become in this case probably one or something but yeah maybe you could comment about yeah so honestly I don't know the origin of the name um it is true that I mean this the symmetry is something that is the symmetry that is um indicating that this row parameter that is defined here will be close to one because you know if the symmetry is exact you will get the row parameter being exactly quite one and all the effect of symmetry breaking will uh will appear as a breaking of that relation um yeah okay thank you yes so sorry again um so like I have two questions first this s2 l s2 right symmetry is this like can you also call this a chiral symmetry or would you confuse people by doing that um I think it would be confusing because usually the chiral symmetry what we mean is okay something else uh yes it's sort of a bigger type of chiral symmetry if you want but I wouldn't yeah it's yeah okay okay so um and my second question like because you like this does not care about like qcd and su3 at all like you you can you can define this like without having gluons and that kind of stuff um what are actually like the pseudo number goes to bosons which come out of this theory because I mean you break a global symmetry and you have three degrees of freedom and they become massive so like what do you define as the pseudo number goes to bosons of this breaking oh I see uh very good I should think about it maybe you should take the soft line yeah okay uh but um yeah yeah yeah okay um yeah there are many more questions you guys like the consider symmetry apparently um is next um okay hello uh thank you for taking my question um I what I have a question regarding this whole problem but but is it actually and what does it tell me I mean it tells uh so far I see it like it's already deviated from one because standard model itself breaking uh historical symmetry so correct yeah what can I learn about it for bsm stuff when it's already deviated from one yes so in general what we have learned is that the deviations from one is predicted by the standard model are small in general so I I brought down explicitly you know the expression for the breaking coming from the yukawa couplings and um we see that we have a loop suppression no so if we put numbers this is something like order 0.01 roughly speaking okay uh now if you go beyond the standard model and then actually there are uh you know measurements from left that are constraining this row parameter or some of you might have heard about the t parameter that are you know closely related uh and uh so what we learned is that the bsm cannot introduce two large deviations from one because indeed the custodial symmetry it seems to be a good approximate symmetry of of nature so whatever you introduce in your bsm you can the bsm cannot be too large contributing to to raw um so maybe yeah I can I can I can mention an example so suppose you know we work with effective theories and already at the level of operators you can write down operators that are breaking the symmetry the the custodial symmetry so you can write down an operator I mean this is a little bit for the expert expert but since you asked you can write down an operator like this um uh this is a dimension six operator so you will have a some new physical scale square are um at the denominator this is breaking the custodial symmetry so it tells you that you know this scale here should be pretty large uh because otherwise the contribution to the row parameter would be too large to be in agreement with data um or you know this is an example from effective theories you can ask it's in question for more uv complete models like typically you know um I don't know if you have hicks extended models so if you have a hicks triplet representation so I see two uh of s2 left I mean the gauge the s2 typically introduce quite a bit of breaking of the custodial symmetry I mean there are tons of examples just that you know in general you have to be a bit careful when you study bsm models that they don't produce too large of additional breaking of the custodial symmetry okay and a second question regarding this row parameter when I calculated so I searched for deviations uh from the uh relation the set in the w boson and when I calculated for bsm models I look for loops uh or uh three level correct or corrections to this ratio so if I want to calculate the row parameter for bsm model right that's a way I mean I think that is quite involved the way of doing it it is simpler if you if you consider the t parameter t parameter that is row minus one and the t parameter maybe I've write a disacquestion and then we should take it offline because it's a long discussion actually but so typically what you can do is to define the the the vacuum polarization amplitudes of the several gauge bosons like this w mu three and then some pi w mu three minus let me write this expression then uh w mu minus and then basically the t parameter is uh given by the the difference between the three three minus the w w computed at q square equal to zero over m w maybe there is some normalization I don't remember um uh and then you need basically to compute these vacuum polarizations um um uh amplitudes in your model computed t parameter and this is basically row that's the simplest way but yeah probably I mean if you have more questions about this we should take it offline otherwise it's a one hour discussion yeah okay other questions otherwise why don't we um so I have a proposal why don't we take now if I mean a break and then uh we keep going with the class and then at the end if you have more questions on the custodial symmetry we can discuss then otherwise I'm afraid that we have just one hour discussion on custodial symmetry this is a little bit uh okay so let's take maybe five minutes break till 57 uh we can yeah okay so we're yeah okay should we start again Giovanni? Yes okay yeah and then yeah again I wanted to repeat but please don't forget your question about custodial symmetry and we discuss more um during discussion session later um um so um uh before passing to flower physics uh um what I wanted to mention is uh uh so we presented the electro with theory uh you know with symmetries and everything and uh and we mentioned that this is a pretty uh very predictive um uh because it only depends on uh on a few couplings what we call now the the coupling strength uh g and g prime and then eventually either the vacuum expectation value of the hicks or some mass say the mass of the I don't know of the z boson okay and then once that you specify these parameters everything is fixed so all the sorry phenomenology of um of the gauge bosons is fixed uh so in principle um you can compute from the electric theory everything about the the z boson uh phenomenology you know all the widths um of of the z to whatever to headrons to left ones and so on um you can uh compute also um some additional observable like uh uh symmetries so there is the famous uh forward um forward the backward symmetry of the of the bottom cork uh that is defined uh in this way uh sorry this is nf plus mb where basically these are the number of events so we are talking about the z decaying to bb bar so we study um this decay and what we are looking at especially at yeah we're looking at this observable at t plus and minus machines and and then if you produce a bottom cork um sorry i exchange the signs so this is a minus here plus here so if you produce a bottom cork in this direction as i showed in this cartoon this would be my forward direction and then i count the number of events of this type um if it is this direction here this will be counted as a backward uh event so you can do also these measurements of our symmetries and everything is predicted by the standard modeling you can match with your experiments and indeed there is tons of data on on the z boson phenomenology and most of it if you want was coming uh uh from the 90s um so data uh from lap that was the the collider before the lhc uh at cern as well as the slc that was the corresponding one at slack and um uh put in together all this data uh what we were able to do is to see if the electric theory of the standard model uh was matching our data and and and and the conclusion is that everything works so in this slide just to complete you know the the discussion of our standard model agrangian i wanted to show you that indeed things work so let me show this uh my slides again so this is the set of observables that we were measuring at lap mainly at lap and slc you see you know the the forward backward the symmetry that i was justifying then we have the several partial widths of the z boson combined in some clever way and and then we can do you know global fits all this data um and what we see is that the standard model is performing fairly well with a large value large p value um you can see that you know there is a the famous uh uh small anomaly in the forward the backward the symmetry with the bottom court that i just defined but this is uh you know relatively small and overall the fit is performing pretty well so the conclusion is that um the our unified electric theory of the standard model is a theory that work uh surprisingly well um and now we can you know be sure about that after now after the hicks boson discovery because actually you know you can compute the loop corrections to these observables and some of the loop corrections will depend on the mass of the hicks and having the mass of the hicks to be 125 gv as we found uh at lhc is telling us that everything fits uh fits together okay that is pretty remarkable okay especially considering the small amount of free parameters that the standard model grungeon has for this part of the of the Lagrangian okay so this is uh what i wanted to say a lot electric interactions um and um so if there are no questions um what i'm now planning to do is um to start the discussion about flower physics uh um so yeah we have to go back to the ipad a bit of a switch is today um so this will be our uh chapter four that is uh flower physics first uh we will focus on uh quark uh flower physics and then also yeah tomorrow uh we will talk a little bit about uh uh lepton flower physics and then we'll discuss about neutrinos that's the the flow um of the uh of the coming discussion um so 4.1 um so what i want to do here is to introduce uh uh okay quark masses and this we already know but what we don't know what we have not discussed together is the ckm matrix so we'll discuss this a little bit and um and then we'll introduce uh uh pre-rechange unit occurrence um okay so excellent um so let's go back to um our uh representations fermion representations of the gauge group of the standard model now i'm putting also su3 we didn't really talk about su3 but we don't have to forget about it um and and then we saw that we have in terms of fermion representations we have of the group we have the following representations you're right uh and then okay let me put also leptons for for completeness here it is so these are the five representations that we know of the gauge group of the standard model in the fermion sector but then here what i want to do is to introduce an index uh and this will be my flavor index i equal to one two three and this is for each representation because we know that quarks and leptons come in three flavor uh uh for example here uh well let's write let's write these so for for the right-handed down quarks we'll have a triplet um um under under flavor so we'll have the d right s right and b right and then similarly for all the other uh representations okay um now let's let's take some piece of the standard model Lagrangian that we already discussed in particular i want to write again the interactions of the fermions with gauge bosons coming from the covariant derivatives so we have that in the Lagrangian we wrote down now i'm putting the sum over the several flavors uh and actually over the several representations as well then we have gamma mu d mu psi i now if you remember this expression for the covariant derivative this is something that we discussed on Tuesday i think this covariant derivative doesn't depend on flavor on the flavor of the fermion that i'm considering but it only depends on you know the quantum numbers um under the the gauge group of the standard model okay um so this is what we call uh uh flavor uh universality someone asked about this i think on monday or tuesday uh meaning that you know each flavor each each generation of fermions interact with gauge bosons in the same way uh so from the gauge point of view um the different flavors are the same particle they really behave in the same manner okay um so this flavor universality actually can be connected to a again an approximate global symmetry uh for those of you who like to um think about symmetries um uh so we can write that the uh the gauge interactions so the interactions that we wrote here um of the standard model uh have a large uh uh global uh sorry global uh uh flavor symmetry uh can anyone tell me what this global flavor symmetry is just in case we're writing the chat i was i will tell you um um uh s u3 uh almost so indeed it will be a you know in general u3 because three i have three different flavors but then basically i have a u3 uh freedom of rotating for each representation of the fermions so at the end of the day i will have a u3 to the fifth power so one each uh representation you can think about you know taking this Lagrangian here and then for each representation for each of the five uh do a u3 rotation and what you will see is that your Lagrangian is invariant okay uh so these are rotations in uh flavor space uh sometimes you you see these presented in terms of an s u3 to the fifth power times uh uh u1s um it doesn't matter so it's the same type of of symmetry okay now so this is for again uh for gauge interactions um but then we know that flavor universality is uh is not a property of nature because we know that the several quarks and leptons have a very different have different very different masses so at the end of the day there is some some aspect of the theory that is breaking this large flavor symmetry and in fact uh the symmetry is broken at least uh you know most of it uh by uh the yukawa interactions okay and uh so let's write them down again um and contrary to before what i want to do is also to specify the flavor indexes so we saw that we could write down three yukawa terms for the uh for the fermions of the standard model uh so we have q left the Higgs field uh u right plus d q left five d right uh five e right emission conjugate and now um so this is what we already saw and now we want to put flavor indexes so here i put i and j i j i and j and then this will be uh matrices right three by three matrices in flavor space in all generality and then i will have this i j i j and i j okay um um so this is uh uh you know if i write down the yukawa couplings in uh uh using flavor eigenstates um so this for example i mean every every fermion is in flavor is a flavor eigenstate uh but then what we want to do is to write our Lagrangian in terms of mass eigenstates right so at the end of the day since you know these terms will be related to the masses of the fermions as you have seen once that we replace this phi with its vacuum expectation value um um so to find the mass eigenstates i just need to diagonalize this this matrix is okay um so um i need to take four unitary uh matrices for uh diagonalizing the cork sector so in particular um what i will have is that the the mass that's right here the mass matrix of corks that is diagonal so once that i go in mass uh eigenstates um this is given uh by the mass uh matrix in uh flavor space but then i have uh two unitary transformations one on the left and one uh on the right q right dagger so these are the two uh unitary uh matrices that i use um and then of course i have to do that both in the up sector in the down sector okay to diagonalize everybody and uh and then in terms of mass eigenstates what i will have is that the q li the mass eigenstate is related to the flavor eigenstate using this um q left uh matrix uh q l j and then similarly for the right-handed corks li mass is equal to q right i j q right j and these are again flavor um eigenstates um but then we have to remember that um um so u left and and d left belong to the same uh representation of the gauge group of the standard model um so what i can do is to write down these rotations in a uh more um uh i mean in a su2 fashion um so namely what i want to do is to write these transformations so this will be now what we call the q l i um so these are again uh flavor eigenstates and these are related to the mass eigenstates in the following manner so i have u left j mass eigenstate mass eigenstate actually this is right here okay and then here i have to put the u matrix is so i will have here u left j um uh sorry let me write in front of the parenthesis so u um um q left dagger i j so this is pretty you know simple math this is q left uh so let's see so this is uh u u left u d left dagger j k and then here i should put actually u u left um i j okay so this is just the same transformation as above and i just wanted to write it in a more su2 evident way let's say okay but the reason i do that is that you see that from here it's evident that what i'm doing is a different transformation on the up component and on the down component of of my su2 doublet and in this combination here that you see um this is what we call the ckm matrix that is giving you really a mismatch of rotation between the left the up component and the down component of this su2 doublet okay and uh yeah so this is coming from the authors uh that first uh um uh predicted the i mean uh mentioned this uh this matrix namely uh cavippo uh kobayashi and muskawa um and the reason that this is a an important matrix is that then it will enter in the uh physical coupling so the uh quarks with gauge bosons once that we write down um this this covariant derivative uh covariant derivatives here in terms of mass against states um so what are the um the physical uh couplings of quarks uh and gauge bosons um uh let's specify the the couplings with the w that are the most interesting ones from this point of view so we saw that in flavor against states in the flavor against state basis i could write them down in this way uh where you have uh w plus you left i gamma mu d left i plus emission conjugate this again in flavor eigen states and uh and then you do a rotation to uh mass eigen states and what you find is g over square root of 2 w mu plus um you left i mass gamma mu and then here you have your ckm matrix the elements ij and then here you have dlj mass plus emission conjugate um so in terms of you know we can represent this in terms of a uh Faman diagram so you will have a w and then here you have um you left i dlj where all these couplings are different from from zero no matter what i and j you are choosing and the final rule for this will be something like g over square root of 2 and then the v ckm uh ij okay so what we see is that we have a flavor violating uh couplings of the w boson now these are the only um flavor um violating uh uh couplings of the standard model um in fact uh you know at home you can check uh uh you know um from the Lagrangian point of view that if you consider the coupling for example of the z boson these couplings will be diagonal in flavor space uh so the index here the flavor index has to be the same uh the same if you write down the coupling with downcorks the same if you put a photon here um you don't have any flavor violation in this this type of couplings and the same if you put a hex boson here okay so everything is is diagonal uh in in flavor space um then you might wonder what happens if i want to consider a process that doesn't change the electric charge uh can i have a flavor violation um so this brings us to our section 4.2 uh that is flavor changing neutral currents i will abbreviate with fc and c's okay um now obviously the w medates uh charge currents uh we have uh no the w has a charge and then it will mediate you know the it will it will interact with quarks with a different electric charge and then here again let me write it down what i said earlier is do we have uh flavor yeah do we have flavor changing neutral currents in the standard model um now based on what i was telling you here um the answer is no at three level right so i cannot write any uh three level uh couplings that are flavor violating and then also that are not changing the electric charge uh but we do have them yes uh beyond uh the three level so ones that you start considering uh loop diagrams um and let me give you an example because this will also uh allow me to show you that indeed the flavor changing neutral currents in the standard model are very very suppressed and the example is the classic example of uh uh uh keon um anti-keon uh mixing uh so first of all what is a keon uh so a keon is a bound state of a down quark and the anti uh strange quark and then we have a anti-keon that is the uh the bar s and these two states are actually mixing to form uh my second state and um obviously if we pass from a k to a um k bar so we'll have d s bar and s d bar um um so what what we are doing is to change flavor by uh two units okay and then also we see that uh you know the the the current is a neutral current because we are not changing the electric charge so we already know that um to write down a contribution to this process to this physical process we need a loop diagram okay at least a one loop if not more and indeed there are loop diagrams that are contributing to this process uh but i will write down here um so we need the w boson because otherwise we cannot change flavor and indeed here i have two w bosons so in such a way that the electric the the charge doesn't change and so here i start with the keon so i start with the strange and then anti down um and here i end up with an anti-keon uh so we will have a down here and an anti-strange here um so here i have a w boson so that means that here running in the loop i will have up quarks um and in all generality i can put whatever up quark i like so whatever index i so i can put an apple charm on the top and the same for this other leg but in general i and j are different um i is different from j in all generality that can be the same of course um okay um now from from what we learned above um so let me put an arrow here so the interaction of the w boson will scale like the corresponding ckm element uh so here and here and here and here we'll have ckm elements and um um so i can estimate the amplitude of this process uh how how it will scale so the amplitude will scale like we'll have obviously one loop suppression that i put here uh then i will have a sum because i have all possible combinations of levels i and j um i will have uh uh ckm elements so i will have a v is vid star so just to make sure so this will be uh the the first one that is this thing here then we have this and so on and so forth and then indeed we have the corresponding ones v j s v j d so these are nothing else than the seven elements of the ckm metrics and and then we'll have some loop function that will depend on the masses of the fermions involved in the loop so we'll have a loop function of mi and mj okay um now um suppose that um so this is uh what i get is to me for for for for this one diagram um that is leading to ccm mixing um so the question is uh is this small or not we do expect this transition to um to be a rare transition because we have a loop suppression um so from that point of view uh we know that it cannot be that large but then this is not the only suppression of the of this process because uh as we will learn at this point i think tomorrow um so the off diagonal terms of the ckm so v ij where i is different from j are also small so you can think uh for now about today ckm uh metrics as a matrix with a form like this where these other elements are uh quite small um so um for uh you know k-1 anti-k-1 mixing or any uh you know processing while being a flow change intercurrent we expect to have extra suppression on top of the loop coming from uh small off diagonal ckm elements um but then also we have a third uh source of suppression of these processes and this is what we call the gym uh mechanism that i will explain now uh so this is coming from glascio uh heliopolis and maiani so what is the g mechanism um so the the g mechanism is telling us that these processes will also be suppressed by um some ratio of masses like the quark mass square over the the mass of the w square and then of course if you take a chum quark in the loop this will be a large extra suppression uh but we would like to understand how how we get to this result you know why we have this additional mass suppression of this flow change intercurrent uh so question is why um so to understand that uh so this is a general mechanism whenever you have flow change intercurrents involved so a process that is involved being a flow change intercurrents but let's see how it works for this kind of mixing as an example uh so for that let's go back to our amplitude that we brought down here um and let's suppose that all quark masses are the same this is of course a an assumption that is actually pretty readily broken because the top mass is much larger than the others but let's make this assumption for for a second to see what we get for our amplitude uh so for the amplitude so the the the loop function will be factorized so we'll have a loop function of some mass m since all the masses are the same and then we have the sum over the sabersikian elements b j s b j d star um so i want to understand what is the sum here so i can split it in two let me write down just the sum in i and then you know i will have also the sum for j um um so i can write this explicitly um because i would like to convince you that actually this combination is equal to zero um so if we write this explicitly what we have is the first is a v us v ud plus v c s v c d plus v t s v t d um why is this equal to zero um the reason being that uh the vc cam is unitary it's a unitary matrix as we could have observed you know when i define it so when i define it here you see that this is a product of these two unitary matrices that we introduced in terms of uh you know to rotate to my second states and if it is unitary you know one of the properties of the unitary matrices uh is that if you take this type of combination of elements you get zero okay um so we have learned that in the limiting which all quark masses are the same i don't have this flavor transition okay um so for this reason the only terms in a that are different from zero terms that are different from zero need to be proportional to this combination of mass of the quark square over mass of the w square or to be more precise i need to have dependence on on difference of masses um but you know i will have always this dependence on on a quark mass uh over the mass of the w and this tells us that indeed we have this additional suppression that we call uh the gene suppression of this flavor change in the current uh processes okay um so bottom line is that what we have learned is that because of these three uh suppressions the flavor changing uh neutral currents in the uh standard model are let me write them tiny so in general they're very very small uh much smaller than what i would expect just using a loop suppression question uh huh um yes so um what do you mean with um the quark masses are these the masses of the constituents of the kaons or once you're running in the loop yeah no these that are running in the loop yeah these guys here um but where's the suppression if i take the top quark that's a very good question so um um that's because i don't enter in the tiny details of this calculation but so basically um obviously you will have a contribution coming from having top and top here uh no and then because of gene uh uh you expect to have this scaling like um so the top square over the mass of the w square and then you say ah there is no extra suppression that is true but actually this diagram here will be very very suppressed by ckm so if you look at the the combination of the ckm elements that are entering there you will have no um uh bts btd uh square and this this combination here of ckm elements is very very small much smaller than what you would get for uh you know if you are in the loop the charm so again you will expect the mass of the charm square mass of the w square and then here we have vcs that is much bigger than vts and bcd so this part is much bigger but then you have a much bigger gene suppression and this is typically what happens uh uh when we say that the process is very very gene suppressed is uh what we what we observe here that you will have a top contribution that doesn't have mass suppression but is very very suppressed by ckm and then all the other contributions have this mass suppression um that that comes from gene okay um yeah okay so um this is uh yeah I wanted to say about flavor change in total currents and I know again um uh we have to keep in mind that this is uh so flavor change in total currents that are predicted by the standard model are um uh are well tested so we have a lot of tests of flavor violating processes at our experiments um the the you know the leading uh experiments in this field um nowadays are the lhcb collaboration and uh bell two collaboration in terms of big experiments and then we have uh smaller experiments for example for measuring chaos and so on and so forth okay and and then um so overall there is good agreement with the standard model predictions but there is uh there are a few anomalies here and there that depending on time tomorrow I might I might mention okay um so this is uh yeah all I wanted to mention and I think yeah I should stop here and take questions and then tomorrow we'll finish a little bit uh this uh uh flavored part connecting a little bit more with data so with uh with measurements okay very good thank you let's then open the q and a session so there were a question from before hello can you hear me yes I can can you write an effective lagrangian include the Higgs plus minus and Higgs plus minus are coupled the the fermion and I think something like the city violation effective lagrangian that is written for a fermion Higgs coupling um sorry can you repeat the question so you want to write down a lagrangian of of what sorry I I didn't hear what you can write an effective lagrangian include Higgs plus minus and the Higgs plus minus are coupled to fermions something like the city violation effective lagrangian that is written for a fermion Higgs coupling um yeah so Higgs plus minus so a charge sticks you want to um you're you're you're mentioning right or um yeah so definitely I mean if you go beyond the standard model there are theories with charge sticks bosons I'm not 100% sure if you're referring to that but please don't make no no if you are referring to something else but yeah if you go beyond the standard model you can have the appearance of charged Higgs bosons so Higgs bosons with with an electric charge and they you know the classic example is a tweak starboard model where you have this charge sticks state that is not a goldstone boson so it's a a physical particle and then you can you can couple these two to fermions so you can write down couplings like like this um you know you can have a I don't know a top left then here you have a b right and then you have some you have a coupling here so you can write down this this type of of couplings yeah and then I mean in models like this they can have also new sources of cp valuation and so on definitely so in a generic tweak starboard model you have or you can have also additional sources of cp valuation we'll mention cp valuation tomorrow so far we have not yet introduced it but it is true that you can have new sources I don't know if that answers to your question yes thank you okay very good next is Mandip hello can you hear me yes so my question is related to the ck matrix you may have mentioned it somewhere but maybe I missed uh so ck matrix is something that is related to the rotation of up and down quarks right right yeah so my question is why the rotation matrix is different for up and down quarks um yeah so in general um um so let's see um you need to diagonalize this you cover up and you cover down uh so this is what you need to do in order to go to mass against it and then to do so um what you need to have are four unitary transformations for unitary matrices um one for the up and sorry two for the up and two for the down because if you have only three you cannot diagonalize both of them basically um um um so yeah what I was writing here in a bit compact form but basically you will have no yeah the u u left u d left and then you have u u right and u d right and these are the you know the four matrices that you need to diagonalize the system but then what you learn is that indeed these two matrices here are in general different and and because of this difference we don't know that u left and d left are connected by su2 and since they are different this is basically telling you that you have this breaking of su2 and this is um uh you know what leads to our ckm matrix this is defined here that is telling us really how much is the mismatch of rotation in the in the up sector in the down sector left handed the sectors sector yeah okay thank you yeah I don't know if there are other questions on the custodial symmetry um that I was yeah I cut it a little bit earlier because um yeah uh those diverging a little bit but I would be happy to discuss more about the custodial symmetry as well now if people are still interested or you can send me an email so I'm getting a few emails of you and then I'm planning to reply to all of them uh in the next coming days so um you will hear from me okay this one from Giovanni hi um thank you for the lecture uh the question was already asked just maybe if we could talk about the go oh no oh sorry the goldstone bosons you you think later that's what you say right uh yeah that's right yeah yeah we can talk more offline about this it's a little bit uh yeah of yeah yeah hi thanks for the lecture so uh I'm just wondering why when you diagonalize oh sorry when you transform the favor oh sorry the mismatch is in favor space to the mismatch is in the mess uh mess uh mess space uh why the unitary matrix uh on the left and on the right are different so why do you ql and you qr okay I mean the the reason is that uh you know these mass matrices are connecting as we know no different fields um uh so if you look if you look at this term in the Lagrangian for example no uh this you cover this you cover term you see that you are putting together a left-handed field and the right-handed field and as we have learned the standard model um I mean the the the the the fermions of the standard model the left and the right are different in the sense that they uh transform differently under the um uh the standard model gauge symmetry so at the end of the day these are two different type of fields so I'm what I'm doing is uh I'm sorry um a rotation on the on the left for the left-handed uh quarks and the rotation on the right for the right-handed quarks is what I'm doing because they are uh yeah as I said the different fields I see okay so uh the mess matrix in the favor space is off diagonal right correct so in yeah in in general I can write these as being a um a generic three-by-three matrix in in favor space yeah okay thanks next question is by Marwan thank you teacher thank you professor so my question is there is some news for the massless or a trashy problem uh what problem sorry again the massless problem massless problem uh for for the for no for for the Higgs boson um more precision for more precision for the Higgs boson um can can can can you ask again uh what you have in mind maybe uh uh so what's your question again sorry news and some some news or some some new article to some new research for for this problem yeah but I didn't understand what problem you are mentioning uh can you uh massless problem or a trashy problem ah sorry okay okay very good sorry I didn't understand yeah so um the the hierarchy problem is um so in general I use the problem of understanding why the Higgs boson mass so I will mention it a little bit tomorrow just to mention it but the problem is that the mass of the Higgs boson that we have measured is vastly different from any other you know new any other physics scale that we can think of including you know the God scale or the Planck scale and the hierarchy problem I mean said in a very um you know rough way is the problem of understanding why there is uh the the two scales uh I mean the scale associated to the Higgs uh the Higgs mass and and other scales like the Planck scale are so incredibly different uh from the qft point of view this um you know the the mass of the Higgs is very very sensitive whenever you have an additional energy scale in the in the in the problem so you can think about having the mass of the Higgs that would be pulled either to the God scale or and and it's it's hard to keep it you know stable where where it is where where we have measured it so this is in general the the hierarchy problem said in a in a very intuitive way if you want um and this uh uh this this problem might call for uh physics beyond the standard model yeah yes I I would like to go back to this uh custodial symmetry well the this symmetry is uh respected only by the Higgs sector right correct well the sq2 left cross sq2 right uh is respected by only the Higgs sector while the other part of the SM Lagrangian is not respected right of course I mean also you know sq2 interactions per se uh so if you take the gauge sector so the gauge interactions uh and you forget about you one you have just an sq2 obviously this is also invariant right um um yes that's why we said that a hybrid charge like to break this correct that's right that's right yeah that well first of all I want to clear the terminology and when we say we have an accident symmetry I mean we think that there's a cement that's a part of the theory that respects that symmetry and the other part does not this right the okay no no sorry so what what we mean with accidental is that so as you have seen I am not imposing any um I'm not working on top of the standard model Lagrangian to make it symmetric right so this the fact that the Higgs part of the standard model Lagrangian is invariant comes out uh accidentally I mean I you know um that is invariant um now what I was mentioning is that this is a symmetry um that is only approximate um um because you know is the symmetry of some part of the Lagrangian and then there will be some breaking effects that in turn we expect them to be small because you know we know that the hyper charge interactions so the the G prime is relatively small and also we know that the uh you know the breaking that is coming from the yukawa is giving a correction to the row parameter that is a loop level correction so it's also small so in general this is an approximate symmetry um in that sense yes yes we see and well if this custodial symmetry does well connect with the this row parameter and this um well in some sense maybe point out to the theories that go beyond our model I mean I am I'm just asking if some we have some theory for beyond our model physics and we have we don't have such a custodial symmetry maybe that breaking of this custodial symmetry would affect the row parameter to be way over one so this is some kind of constraining restriction about what we're doing beyond our model this yeah absolutely absolutely yeah absolutely um yeah so I was uh yeah I think someone else also was asking but um so whenever you go beyond the standard model this is a constraint that you are to take care of uh so you know that your theory beyond the standard model shouldn't break too too badly the custodial symmetry um if if if you are introducing large breaking of the custodial symmetry then you are you are you end up in trouble with with experimental data um yeah and then I was making I mean I was mentioning a couple of examples very very briefly but you know from the gft point of view there are operators that you can write down that are explicitly breaking in the custodial symmetry also you know you can think about the you'll be complete more you'll be complete models that are introduced in large breaking of the custodial symmetry and the typical example are you know models with an extended the Higgs sector where you have Higgs triplet representation so as you to uh so where you have a new Higgs boson that is a triplet this is also a lot of times introduced in large breaking of the custodial symmetry so there are really um um you know in general uh you have constraints on BSM yeah and when you mentioned about the effective theories this lambda that would be like so suppressed that the custodial symmetry does not break so much right yeah that's right so I don't remember exactly the you know the the exact bound on this lambda if you write down this operator is probably I would say a few TV but it is true I mean you want to have this lambda being large enough that your extra breaking of the custodial symmetry is not too large mm-hmm yeah okay well thank you any other questions maybe we're getting tired so if there is no more question maybe we can finish here for today and thanks Stefania for nice lectures and answering questions and uh we resume tomorrow same time very good thank you very much thank you