Thank you and welcome back So I have some answers to some questions that I couldn't answer yesterday And I just want to encourage you to keep interrupting me seriously Just if you have any questions stop me yell at me.  That's great Just go ahead and ask questions whatever comes to your mind So someone was asking me about yesterday it came up in the discussion about the face convention independence or a dependance of Epsilon K and and and of course what was said here is absolutely correct that The any physical quantity is of course independent of face conventions But if you look at the standard expressions in the literature, so this is a review of Buras and collaborators Then then what you see can you hear me? Is this on? What what what you see is some expression That's obtained after using CKM uniterity And so Epsilon K in the standard expressions is related to imaginary part of M12 I'll define what M12 is in a moment and if you look at these expressions It looks like that guy that I enlarged over Bell and here one has used CKM uniterity Partly because uniterity is also required to make the standard model result finite and If you look at this result, then it's completely unclear how some refacing of the CKM phase Will drop out of the result but But but it does it's just it's just highly non-apparent There was another set of questions that we had a long discussion on about the renormalization of the CKM matrix so Actually, there are a bunch of papers on that in the literature. I just didn't remember the results. So this is one of them from 1998 and What these guys calculate is the running of the CKM element So how why up and why down the up and down you cover matrices run from the hundred gv scale to the gut scale 10 to the 15 gv and what you see is Actually, I should how is this I? I lost the paper Anyway, so it's mostly it turns out that not surprisingly It's mostly the elements which are in the search generation that that run more than the others and so okay So how empty runs that's obvious So for example, if you take VCB Then at the low energy scale it has a value about 0.04 and we'll talk about it a little bit And you see that if you run it up and that to the gut scale there are like 10% changes and there are similar IG effects for for a VUB etc and But these are yeah, so okay, that's that's that's all I wanted to say about that and Any questions There was another paper that I lost wait Well, it's It's this guy so This is yet another set of waters from 2003 and there you can see explicitly that So there's a lot of discussion in the literature about some scheme independence of how you do the renormalization of the CKM matrix So there are some subtleties that actually I didn't fully end up understanding last night But you see that you know in a first approximation You just do a universal rescaling of the VUB VCB and VTS VTD entries And this leaves the seat that the shape of the unitary triangle invariant I Think that we've said that they ask again variant would not run but it also runs that also has a scale dependence So if any of us wants to read more on this then there is a lot to learn for me as well as possibly for some of you So any questions? So we left it off and yesterday After an hour and a bit was sort of deriving how the unitarization of the up and down You cover matrices leads you to the CKM matrix and how the unitary triangle is a useful way to look at many constraints on CKM elements and check whether they are whether they are not consistent with one another. So switch off the laptop for some time and What I Okay, so so we'll go through some of the constraints and I have cheated some expressions that will be relevant for this talk I put up on the blackboard so that I don't need to waste time with it And I just want to say a few more words about sort of the Conceptual foundations of how I think about this this this flavor physics program So from a low energy effective field theory point of view We saw that any flavor changing interaction is mediated in the standard model by by the w boson and So from the point of view of Experiments at a few gv scale at the be quiet at at at mb or the d meson mass or whatever Well, you can do some experiments from some hadronic decays whether you look at something which is Neutral meson mixing. So for example this diagram mediates transitions between b0 and b0 by If you look at some penguin diagrams, which is another celebrated process b2 s gamma Well, you just look at ordinary some electronic decays like b2c Electron neutrino From the low energy point of view each of these things are just some local Operators which mediate some for a fermion interaction. So if I'm at 5 gv and I'm calculating some matrix element I really cannot resolve that this is a top quirk and this is a double u boson It's just some effective for a fermion operator Which in the standard model happened to look like? Sorry, so this is like b if I do bb by a mixing It's just this operator and it's coefficient. I can calculate in the case of b2 s gamma. It's again some Other local operator because all of this is like a point-like interaction when sort of the distance scales I can resolve is only One over mb these things happen at a much higher energy and therefore a much shorter distance scale So I can represent it as a local operator in the standard model the operator that's generated is sigma mu nu f mu nu the dipole operator and Because it flips chirality it necessarily comes with one over with an mb factor and it's suppressed by Some scale because these are dimension six operators and Similarly the semi-latonic decays It's just yet another operator that is a two quirk two lepton for a fermion operator Which in the standard model and of course these scales in the standard model are the weak scale some order one factor times mw and When so from this low energy point of view this program is asking the questions When you calculate do the standard model calculations? Can we see evidence for a non-standard model contributions to these operators or can we see evidence for operators? Which are forbidden by some symmetry in the standard model to be generated by new physics So for example in the standard model This is when all of these sort of this and that interaction is a left-handed current because the w only couples to left-hand left-handed Quirks and leptons in the presence of new physics you could get other operators which are Exactly zero in the standard model and you can ask the question whether there is any experimental evidence for that and So again when when when you interpret these measurements and you try to do the calculations most of the time the question Is how well can we calculate the coefficients of these higher dimension operators? and and how well can we compare it with the data and Figure out if there is some evidence for a new physics or a better reason and The reason that in particular for understanding CP violation that beam as on decays offer a very rich Playground is because so we saw yesterday or at least we didn't see it but we talked about it that If you think about that plot that I stupidly took off the screen then epsilon K CP violation in KK biomexing That's a important and strong constraint on the CKM parameters But we tangentially talked about it that the other CP violating quantity in the K on sector Which is called epsilon prime and it's related to direct CP violation that has so large Hadronic uncertainties from QCD that we at the moment don't know how to understand Quantitatively that we don't know how to plot that on that plane other than to say that it's order of magnitude is well You would expect it to be so any any questions so So what's special about be physics and there are some Both on the theoretical side and on the experimental side it has sort of it's a lucky coincidence of of several nice features So on the theory side we said we talked yesterday about So in the in the chaos system all of the CP violating quantities that that's so epsilon and epsilon prime They are very small in in the B system. You can have large CP violation and we'll see examples If you look at flavor changing neutral currents in general and be be by a mixing So flavor changing neutral currents that I will abbreviate as FCNC's they do not have Suppression due to small you cover coupling so there is no gym suppression and there is also no suppression by small CKM elements because because of this approximate diagonal structure of the of the CKM matrix VTB is Extremely close to unity So so many of the suppressions that apply for a K on CP violation don't occur for a BCP violation And and and we'll get to it tomorrow that some of the hadronic physics Some of the QCD physics is tractable namely one one can do many first principles calculations for hadronic physics in B decays Essentially as a consequence of the be quick mass being an order of magnitude larger than This typical scale of hadronic interactions about 500 MV or so and at the same time the As I said it's a kind of a lucky coincidence that experimentally there is it's a fortunate Coincidence that VCB is small and therefore the be lifetime Is is fairly large it's something like 1.5 picosecond So just to indicate this so VCB in magnitude is about 0.04 You know that if you look at the lifetime of a B meson or a D meson it should The total decay bits goes like the fifth spiral of the mass of the particle So a B meson which is three times as heavy as a D meson should be much shorter lived And that's not the case because VCB turned out to be Substantially smaller than for example So VCB is substantially smaller than Vcs sorry VCD that is the Kabibbo angle and and why that's important is Because the long B meson lifetime allows you to study this process is experimentally So this is a plot from 25 years ago from the Olaf experiment at lab And I just like to show it in in in talks because it's just such a beautiful example that you know Do you have a 10 meter big detector and already in the early 90s or late 80s? There was a technology to resolve This B meson traveling a few millimeters in this 10 meter detector before it decays to some particle and You know you have something very quantum mechanical happening and you can draw this classical picture because of the long lifetimes You can reconstruct the B decay and then the subsequent D decay So with all these tracks coming from displaced vertices and that's because then that essentially boils down to this fact that VCB is as small as it is another experimentally Useful or a critical element of was that there is this Upsilon resonance So this is a BB bar resonance at like 10 point some GE V which is Barely above the mass threshold of decaying into a B zero B zero by a pale So we'll see that by doing E plus E minus collisions at a particular center of mass energy You can produce essentially B mesons in a very well controlled environment and study their decays In a way that is impossible for K mesons and D mesons, which you always produce together with other type of head runs and The third thing which was kind of important for this program at least in a plus E minus colliders It's less so for the LHC That in the B sub D sector so not for B sub S but B by D mesons the mass difference happens to be Comparable to the total lifetime So the delta M over gamma is some number of where there are one I don't remember if it's point seven or whatever it is But what and what this tells you is that so we'll build We'll come in a moment to be via oscillation. So if you produce an initial B meson then the two mass eigenstates have a very small mass splitting and you have oscillations between these two states and That characteristic timescale of the oscillation is comparable to the lifetime. So that allows you to study it experimentally quite nicely any questions So Unfortunately, I have to tell you a little bit about the formalism how Of neutral meson mixing in general so I'll use the B mesons as an example so we're Probably you will know because otherwise you should have asked me a B zero meson is defined to be a state which is Don't ask me why it's a B by D Whereas the B by zero Is the state which is composed of a B and the D by There are some historical reasons for that. It's maximally confusing and so these two states the This box diagrams we talked about yesterday for a KK by mixing so the analogous diagrams in the big case so be can emit a W and turn into an up chime where top work and become an S eventually and And Sorry Let's do that. So these are W's so that is Second-weather whatever that is second-weather electroweak processes which give you non-zero transitions from an initial B zero to a B zero by or vice versa and The elsewhere you end up with so that the so these are the flavor eigenstates which have well-defined flavor properties and And as we'll see for how the particles decay in big interactions These are the convenient states to describe what happens however, the Propagating degrees of freedom sort of the the eigenvalues The eigenstates of the full Hamiltonian are not these states by but states Which are usually labeled as be heavy and be light for just the heavier and the lighter mass eigenstates because you have to call them something and These are usually defined by some linear combination of B zero and B zero by With some coefficients, which are usually called P and Q. This is just notation. So these are the mass eigenstates And these are the guys which if you look at their time evolution They have trivial time evolution in the sense that it just looks like whatever e to the Minus I am either heavy or light So they are the two mass eigenvalues and the two different lifetimes So there's so there's a phase factor that you pick up in the time evolution Which depends on the mass and there's a lifetime which tells you how quickly these states decay times the states at time zero and and if you want to understand this time or whatever what what how you get the eigenvalues of the How you get the mass eigenvalues and the lifetime eigenvalues you have to solve a Schrodinger equation Which describes the time evolution of these states so I DDT So there's this two components state, which is B zero and B by zero So the complication is that it's not a Hermitian operator which describes their time evolution because there is a piece Which just describes the mixing and so M is a Hermitian matrix But it also decays so there is another Hermitian matrix gamma which comes with an factor i and That's what describes the the decay of these states. So both of these are these are two by two Hermitian matrices and And that's it So if you if I so if I can calculate the entries of this matrix M and gamma Then I can diagonalize this matrix and that will tell me what I p and q that defines the linear combinations that gives you the mass eigenstates of this of this of this B zero B zero by a system that oscillates and decays and and and and and and and that's the System that we want to study So an important point is that the CPT theorem Tells you that the diagonal entries of this matrices M11 so M11 has to be the same as M22 Because in the absence of mixing this would be just the masses of B zero and B zero by a and and in the absence Of mixing obviously this would be the correct mass eigenstates and then because of CPT theorem They would have to have the same mass and similarly gamma 1 1 has to be the same as gamma 2 2 and The point is that Yes, I can't hear you Sorry, let me come closer The CVT theorem tells M11 equal M22 and the gamma is the same It's not as specific for the B meson it is for any neutral Neutral meson so that I'm using B as a label the same formalism can be used for D mesons for B sub S mesons For a KK by mixing it's just that the parameters would have different values in these different systems But the fact that the fact that the two diagonal entries in the mass matrix and the decay with matrix are the same That's true for all of them. Just they are numerical values are different in In the four different neutral meson systems so So so so the key question is what are these off diagonal entries of in M12 and gamma 1 2 that will come into Describing the physics of these of these things and and so gamma 1 2 Comes from real intermediate states So so things that which are common final decay products of B 0 and B 0 by so in the case of the B system You draw the same kind of box diagrams but now you can only have the up and the charm quirk in the intermediate state and certainly The up and charm can be on shell In the decay of a beam as on Whereas whereas M1 to That the leading contribution to the BB by a transition will come from the top work because The top work will have the least amount of suppression by CKM elements and also because the top work will have the Will not have this gym suppression. So as we said yesterday for the K on case Then when I when I mean the top where contribution That's really some contribution which comes like which enters as m top square minus m times square where I'm up square over Mw square But this is not a suppression for the top contribution. So so it turns out that the top country completely dominates M1 to and Because now this box diagram is it has two W's and two top quarks Each of which is much heavier than the external 5g the Analogy scale that I have as maximal momentum entering these Quark lines this has to be It has to be an extremely good approximation that this can be represented by a local operator and this can be calculated To hire Either adjust this order or including higher order Qcd corrections using perturbed if Qcd because this loop momenta I offer the hundred gv that is relevant for this problem and not So the gv scale does not matter for For what goes on in this loop? It only matters at the end of the day for evaluating the matrix element of this and we'll get to that in in a few minutes so so because So the point is that gamma 1 2 comes from these box diagrams With up and chime quarks, whereas M1 2 is comes from the box diagrams with the top quark Because of the gym mechanism M1 2 of the gamma 1 2 I Should have written it the other way around gamma 1 2 of the M1 2 Absolute value has to be of order Some scale of whether Mb square over Mw square just from the fact that gamma 1 2 Does not have the top contribution here so in general diagonalizing these matrices Gives you some fairly complicated solutions, but if I Can use the approximation? That gamma 1 2 is much less than M1 2 Then one can actually write down the solution of this eigenvalue equation in a fairly simple way so What one finds is the delta M Which is defined to be the heavy state minus the light state. So this is by definition positive Is given by twice the absolute value of M1 2 So basically you calculate this box diagram and its matrix element between B and B by step B 0 and B 0 by states and You think that's a complex number in general and you take the absolute value multiplied by 2 and that's the mass difference and Delta gamma is Is minus 2 times real value real part of M1 2 gamma 1 2 star divided by M1 2 I can never remember any of this and that's also the same as twice gamma 1 2 times the phase of gamma 1 2 over M1 2 and Why am I why I write down all of this will become clear? soon enough so And you can write down that what is Q and P and that's also going to play an important role So Q over P is given by Minus M1 2 star over M1 2 times 1 minus a half Imaginary part of gamma 1 2 over M1 2 and someone should have yelled at me that I didn't define what delta gamma was and for unfortunate reasons Whereas delta M is defined as M heavy minus M light the logical definition would be to define delta gamma as gamma heavy Minus gamma light, but it is usually defined as gamma light minus gamma heavy Just to make things confusing and so the reason for this is because it's with this definition The delta so in the in the B sub D system delta gamma is actually a tiny quantity And it's not yet measured experimentally in the B sub S system Delta gamma is much bigger than in the B sub D system and this definition is chosen such that in the standard model You expect delta gamma to be positive in the B sub S system And that's now confirmed by LHCB measurements in any case the any of these signs by themselves Entirely convention dependent, but the relative signs between delta M and delta gamma are physical and that can be used to test our understanding of the short distance physics so one immediate consequence of of of of being of Of M12 being much bigger than gamma 1 2 in the standard model model in both the B sub D and B sub S systems That you see that in general I Have to calculate this box diagram and evaluate its matrix element between B zero and B zero by estates There will be some associated hadronic uncertainties in that in the calculation of that matrix element which is done either using Historically it was done using various models and for the last five ten years It's essentially there is no competition to doing it using lattice QCD methods and actually that's a Very well understood. Let this QCD calculation How to calculate the matrix elements of this four fermion operators between B zero and B zero by estates There are still some hadronic uncertainties in calculating delta M in the standard model of whether something like B25 percent for the calculation of delta gamma because it relates to Physical intermediate states it's much more susceptible to long-distance Hadronic physics and the calculations of delta gamma even though there is a formalism to calculate it in a From first principles QCD the calculation of delta gamma has larger uncertainties than delta M But you see from this expression right away that if m1 2 is sure distance dominated Then q over p in magnet and then gamma 1 2 we said is much smaller than m1 2 So this the second term is tiny Then q over p is going to be a Complex number whose magnitude is unity because it's just m1 2 style m1 2 and It's phase is going to be determined entirely by short distance physics. So That's going to be extremely important for for for understanding CP violation in the B system that Even though delta M and delta gamma has substantial uncertainties We can understand the phase Associated with BB biomexing From short distance physics without knowing essentially anything about Non-perturbative QCD interactions to a better than a percent level any questions So that was a plot of just looking at time dependent So okay, so this is a beautiful measurement by LHCB So I said that in the B sub D system that characteristic timescale of oscillations and decay is comparable That's very different in the B sub S system the oscillations happen much much more rapidly than the decay itself and What's on this plot is the LHCB measurement of? of Essentially this oscillation so they can identify an initial B zero where B zero by Sorry, this is B sub S or B sub S by meson at The production and I will show you how that's done at least qualitatively and what and and you can study as a function of the elapsed time from the from from from from the production of this B sub S or B sub S by meson But at the time of the decay it decays as the particle that was produced Whether it decays as as as the as the anti particle as whether an initial be zero the case as B zero B sub S zero or B sub S zero by and You see that the challenge in the case of B sub S was that this oscillation happens so frequently Compared to the lifetime that essentially at the time of the decay It's a random chance whether an initial B sub S meson decays as if it was still B sub S Or whether it decays as B sub S by so there's almost perfect mixing and because of this very rapid oscillations it it We knew for a long time that B sub S oscillated But the problem in the B sub S case was to resolve these oscillations by having very very fine time resolution to be able to To get to this measurement that was first done at at CDF at the Tevatron and then now much more precisely at LHCB so So once you know how to calculate delta M and delta gamma So so we said before that if you were interested in the decay and the production of of a B meson from weak interactions Then it's the flavor eigen states which are the natural states to Described production and decay. However, the time evolution is simplest in terms of the mass eigen states and you can Just work it out that if you produce an initial B zero a B zero bar, you decompose it as B heavy and B light You time evolve it with this pure exponential form Until time T and then you project it back onto the flavor eigen states And so I wrote down the equations that govern this time dependence. So an initial B zero After some time T has some probability to still be a B zero or become a B zero bar and vice versa an initial B zero bar after time T can either still look like a B zero bar Or it may have oscillated into a B zero and these functions. So all of what I wrote down is valid in the limit Well, we are the lifetime difference is actually set to zero Otherwise these expressions are more complicated and so these functions g plus and g minus heaven sort of they pick up a phase over time and of course there is a term describing the decay and they look like cosine or i Times sine delta m t over two and why I bother writing putting it on the table will become clear in a moment So, how can you study CP violation? in some sense the simplest form of CP violation is CP violation in the decay of some particle and So, let's just take an example if you look at Some process like b goes to k pi Which at the quark level comes from? b goes to u u by s There are several contributions in general to some b decay. So in this case You can there's a there's a contribution which it just comes from An ordinary foil-thrummy interaction well b goes to a u u by s But you also have this so-called penguin diagrams well This is sort of schematic drawing because I should really talk about some effective operators, but never mind What's relevant? It's just that there are several contributions. So if I want to write down what is the amplitude? Well, say in this case b goes to some final state f let let that be k pi Then in general there are several contributions to this decay amplitude So there is some sum sorry There so there are some sub amplitudes the matrix elements of each of these contributions Let me use a different index than I So there are two different kind of phases that can occur in this hadronic matrix elements some of them Which are called strong phases relate to some hadronic rescheduling effects and some other phases Which are called weak phases that come from the Lagrangian So in the standard model it comes from the complex phase in the CKM matrix in the presence of new physics There could be new sources of of these weak phases and the point is that They are distinguishing properties that one of them. So the strong phases which come from QCD rescheduling effects RCP even Because strong interaction is symmetric under CP once we have neglected tether QCD which we know that is negligible for a whole of flavor of physics and These other type of phases which come from complex parameters in the Lagrangian RCP odd So what this means is that if I write down the Opposite so whatever the CP conjugate process, which is B by goes to F bar Then there will be the same type of contributions But some of these and some of these phases will be the same that come from QCD rescheduling effects and some of these phases will change sign and Once you have more than one contribution to a decay Then you can have situations that These strong phases and these weak phases are non trivial Relative in in the two contributions and you can end up in a situation that if you just take the the decay rate, which is the Whatever the absolute value square of these amplitudes then So if the decay rate for a B2 decay to some final state is not equal to the decay rate for The CP conjugate process then obviously that's a manifestation of CP violation and And so this type of CP violation always comes from interference between different contributions to a decay Their characteristic property is that in no cases that I know of can we compute? Well enough the hadronic matrix elements that we can understand this kind of phenomena At a precision level from first principles So this is the type of CP violation that is epsilon prime in the K on sector and also in the B system in particular in this B2k pi type decays It was discovered by Bobois and Bell that Whereas this type of CP violation in the K on sector is at the 10 to the minus 5 level in the B system It's really an order of one effect. So So for example in in this B2k pi type decays there Is something like a 10% direct CP violation of this type, which is experimentally observed and on the one hand That by itself is is quite exciting because it tells you that in the standard model in the quark sector CP violation in some sense is not a small effect. It's really an order of one effect Which was however? For decades it looked like a small Effect in the K-on sector because in the K-on sector it is sort of masked by the gym suppression and the small CKM effect. So there's some particular Suppression factors in the K-on system which makes CP violation tiny in the K sector But in the B sector in general it's a large effect and and sort of the exceptions is very small and not when it's large Yes Yes, so in the standard model this phase is always related to the CKM phase and in the presence of new physics There could be other phases which are unremovable complex parameters in the Lagrangian that would Change sign for a process and its CP conjugate But in the standard model phi is always related to the one phase in the CKM matrix So all of these so all of the direct CP violation phenomena, and I like See that I don't even know how many of these have been observed with the more than five sigma significance All of them somehow connects in the standard model to the CKM phase The only problem is that we can't calculate the hadronic phases This the CP even phases and the magnitudes of the hadronic amplitudes well enough to make very precise tests of the Standard model by doing this type of measurements There is one particularly Interesting case that sometimes some of you may have heard about which is sometimes called the k pi puzzle which actually relates to the difference between two of these CP violation measurements in different B to k pi final states so between as in k plus Pi zero and k plus pi minus and if any of you know about it and want to ask me about it Then we should talk afterwards because so that's an interesting case. Well by taking the difference of two channels One would hope that some of these hadronic physics can be understood better because you are only interested in understanding difference of two things and and I just want to say that that it is possible for many of these Effects that you know there are like hundreds of papers little written in the literature Discussing whether there could be some new physics hidden in these measurements and from my point of view All of that is inconclusive at the moment but there is only a possibility in the next 10 years to understand this better theoretically and and and and thereby Obviously the sensitivity to possible new physics contributions would increase Any questions so to to go towards more Understandable things Yes please sir, but When you are Comparing the magnitude for gamma 1 2 and m 1 2 for B systems So I didn't quite understand how you concluded that the gamma 1 2 will be much less than m 1 2 there So if you can explain I understand that if you can calculate the box diagram you just extract out the imaginary part and Call it gamma 1 2. So why that part is small? I mean, what is the difference between K system and B system? The magnitude relative magnitude of gamma 1 2 and m 1 2 so I should say it differently I think if you want me to make a strictly correct statement that what I should say is That I'm calculating all contributions that Contribute to the matrix element of the full weak Hamiltonian to any power between B0 and B0 by and I just decompose it as the Hermitian and the anti-Hermitian part. Yeah. Yeah, that's fine, but you have now the m 1 2 So So loosely speaking each of these intermediate quark contributions is proportional to the quark mass quail Over mw square. Right. So m 1 2 is dominated So therefore m 1 2 will be dominated by the top for a contribution But the top quark cannot contribute to an on-shell intermediate state and therefore gamma 1 2 only receives Contributions from up and chime in these box diagrams. That's fine But in when there is a top quark, there is a complex quantities in VTB and VTD So what about them? They will contribute to the imaginary part of the amplitude No, no, no, but they will so so m 1 2 and and m 2 1 are The complex conjugates of each other, right? So if I if I if I now into a change and I look at the so this is B0 by to B0 The other one is B0 to B0 by and you expect to have a complex conjugates So m 1 2 is not a real number m 1 2 is still complex So there's nothing wrong with m 1 2 having the complex VTD and VTB CKM elements in it the important thing is that That m 1 2 is m 2 1s complex conjugate, right or something like that Okay, okay, but but m 1 2 by itself is not a real number. It's a complex number. Does that Okay, let me ask you later. We can talk about it later where so So CP violation in mixing and we are getting towards more interesting topics so So if CP was a conserved if CP was conserved by the weak interactions then the CP eigen states and the mass eigen states would be the same and Since this guy is the comp is the CP conjugate of one another up to a possible convention dependent phase you would expect that the CP eigen states should be again up to just some trivial phase just Whatever b 0 plus minus b 0 bar and you would expect if the CP eigen states were the same as The mass eigen states that the absolute value of P and Q would have to be the same So this is what I meant before that there could be some phase factor, which is Entirely a convention dependent thing right for any external particle I can define a phase that That that that is an arbitrary convention So I by defining the phase of b 0 to be whatever I Wanted to be I could change the phase of P and I could change the phase of Q but if the CP eigen state so if if the CP eigen states equal to the mass eigen states then you would have Q absolute value equal to P absolute value and and that's it and Both in the so so so in But but but it is possible that this is not the case if if if so if you if you have a situation That Q over P absolute value is not equal to one Then that would tell you for example That the that the be heavy and the be light states the two physical states so you can just look at P be zero Plus minus Q be zero by so if I take the scalar product of these states That's just P square minus Q square right and so If the CP violation in mixing that is P over Q not equal to one in absolute value Then it's a funny situation that the physical states are No longer or orthogonal to one another And again This is sort of another way to see that CP violation is really intrinsically a quantum mechanical phenomenon that there's really no classical Analog that you diagonalize the Hamiltonian you find what the mass eigen states are and in the presence of CP violation It could happen That the mass eigen states are not orthogonal to one another And this is actually something that's experimentally observed to happen in the case system this is related to the real part of epsilon K and The experimental search is for this effect in the B and B sub S systems So you may have heard that a few years ago. There was an anomaly from the D zero experiment so what is plotted up there is Is sort of this CP violation in B B biomexing on the horizontal axis and CP violation in B sub S B sub S biomexing on the vertical axis The standard model is that red dot with extremely small uncertainties and I will define for you in a moment what this quantity ASL is and The D zero experiment at the Tevatron measure the particular linear combination of these two quantities And found what is quoted as about a three point some sigma discrepancy from the standard model expectation The E plus E minus B factories Bobo and Bell have measured This band they only produce B sub D mesons and the late CB and I think D zero as well. They measured just this directly just the CP violation in The B sub S case you see that this type of CP violation except for this anomaly if whatever this Intriguing measurement by D zero it has not been observed So this type of CP violation has only been seen in the in the in the chaos system but I Will explain in a moment that the standard model prediction in this case is extremely precise and You see that at least from one experiment. There is sort of a three sigma tension with the expectations so what is this quantity a sub SL it is called a sub SL because Historically people looked at it as measuring the CPA symmetry in some electronic decays So the definition of ASL T For an E plus C minus B factory like what Bobo and Bell was was defined To be the time-dependent rate of an initial B zero after time T to decay to a positive lepton and anything else Minus the time-dependent rate for an initial B zero bar after time T to decay into L minus X Divided by the sum of these two things and you might ask what the hell I'm talking about So let's think of for a moment. What is okay? Let's let's do this So what is how can a be so a B by a zero contains a B quack and Normally, it's dominant decay once it decays semi-leptonically would be with a virtual W So it would decay to a chime quack and a positive charged lepton and a new trino, right? No, I'm wrong. What what am I? Sorry Did I totally screw up my notes? Well a big quack if it decays to a chime. It's certainly an L minus so Correct me if I'm saying nonsense. I think my notes must be wrong So the dominant decay of a B has to be this guy, right? So it's B to C L minus an anti-neutrino So what we are interested here is exactly the opposite we want to have some asymmetry Which is the measure of okay, so there's the spectator quack of the B sort of the Of any so this is a B by a meson So we want to see what is the probability of a B by a meson Oscillating into a B zero before it decays so a B zero contains a big quack No, sorry a B zero contains a B by quack Again, this is so easy. Okay, I'm always confusing myself so a B zero by goes to an L minus Dominantly so I'm asking what is the probability that the B zero by when it decays it decays in the Oscillated state so as a B zero and Then it would produce an L plus Right, so Therefore, I'm conjecturing that my signs are totally wrong So what I'm asking is what is the probability that I made a B zero at time zero and after a time T It decays After having oscillated Well, what is the probability that a B zero by? decays after a time T From this oscillated part of its wave function So that I am measuring the piece Which is not the same flavor as the as the time of production, but the opposite flavor So you see that the difference between So that asymmetry that I wrote down there will Just get contributions from this piece and that piece And so all the exponential time dependence will drop out when I take the absolute value squares And what this will give me so That's too far away So B zero, what do I want? I want so it's going to be q over p squared For the absolute value square of this This other piece will be p over q square And there will be some universal exponential dependence which drops out from the ratio So this is just going to be q over p square plus p over q square and Since empirically We know that you see that this asl quantity Deviates from the standard model at more. Sorry. It's it's it's magnitude is less than a percent So p over q is to a very good approximation equal to one And therefore the denominator to a very good approximation is two And so I can write this as minus one So this is going to be the same as q over p minus one, right? Plus higher order terms which are extremely small And So so the reason that these are interesting measurements is because we said that in the standard model q over p minus one comes from The imaginary part of gamma one two over m one two We said that gamma one two over m one two Is suppressed As mb square over mw square. So that's why we expect it to be tiny In the standard model, there is another effect that If they up and the charm quark will degenerate in mass Then you would expect To be an additional suppression By m charm square minus m up square Over mb square So this quantity in the standard model is really extremely tiny Which is why the standard model prediction is just a dot on that plot with extremely small It's not that the uncertainties are tiny as a fractional uncertainty But the whole effect is so small that that's why it looks like a super precise prediction Which on this scale it is But the point is that if there are new physics contributions to bb by mixing Then This suppression factor could easily be eliminated by having a new physics contribution to bb by mixing And that's how you could get a large effect due to new physics in these kind of measurements So as you see from that plot this story right now is inconclusive We don't know whether there is or whether there isn't A hint from that green ellipse that that does not agree with the standard model point And it's just something that's going to be interesting to To see how things develop with more precise measurements with lhcb And and and the future upgrade of the bell experiment Any questions? So the case where one can really get precise information about About the standard model or new physics is actually the third type of cp violation Which is cp violation In the interference in some decay With and without mixing and I will explain in a moment what I mean so The picture of unusually draws is that if you have some final state f and we'll use a particular state j psi k short as an example because it's a It's it was a particularly important process for For the b factories and because it's also one of the cleanest example of this type of cp violation that it's tractable without much hadronic uncertainty So So in particular if f Let's just take it for simplicity to be a cp eigenstate final state which this guy is So this so j psi is a cc bi meson Is the light is is is is the Yeah, it's a cc bi meson and k short is a as is the Linear combination of s by d and d bar s Which is approximately cp even so So when we look at such a process then the in general the two different decay paths b zero can decay directly into this final state or b And b bi can mix with each other and then you can have this decay and The asymmetry that one measures experimentally is again similar to to this that you are looking at Producing a b zero meson at time zero and after time t you look at its decay rate to a particular final state so in this case You look at a similar rate difference just not for for a semi-laptonic final state But the cp eigenstate final state divided by the same things by the sum of the same so The difference divided by the sum and It would probably take me More time than I have to work out so so you can work that rate out So let me just make some notation. So a I'm going to call the af to be So a and a bi that are the amplitudes for b zero and b zero bi to decay to this final state And using the formula which are at the opposite end of the on the black of the blackboard You can easily work out that if you produce initially a b zero after time t It will it's decay to that final state will have a term that comes with g plus of t Times a because it's b zero decaying into the final state And there will be another term which is this other function of time times a bar and likewise for b zero bar after time t There will be a term which is some function of the ellipse time times the Amplitude a that I defined there And and another term times a bar and you take the absolute value square And that will give you an expression for this Decay rate difference and The final expression is going to be It's going to have a particularly simple. Well, it's not that simple I'll so there's there's a quantity lambda that one always defines for this decay. So lambda is defined as q over p times this a bi over a And it's convenient because one can write this This time-dependent cpa symmetry in a very simple form in terms of lambda And it really takes 10 minutes 5 10 It's like four or five lines to work out take the absolute value squares and work out this difference So there will be a term which goes like sine delta mt and there will be a term that goes like cosine delta mt and that's because Whatever g plus and g minus absolute value square has this one plus or minus cosine delta mt The cross term g plus conjugate times g minus will have a sine delta mt dependence And when you work through it you get an expression like this and And the reason this is interesting is because We saw on the previous plot that That empirically we know that q over p Absolute value minus one is less than something like 0.01 and so if we find Some final state f That a bi over a so if there is no direct cp violation if also a bi over a absolute value Is approximately equal to one Then you see that That will imply that lambda absolute value is going to be one. So this second term is going to vanish And the first term so once lambda absolute value is one It's just going to be imaginary part of lambda times sine delta mt And the imaginary part of lambda if lambda absolute value is one It's just going to be a phase And that's going to be the phase difference between So there's the direct decay which is described by the amplitude a there is the b zero buyer with the a buyer so this measurement in Particular final states where a bi over a is close to one. It's just measuring A phase difference between these two different decay paths from a b zero to f and b zero oscillating to b zero buyer goes to f and Everything about the hadronic physics how b mesons mix how these decays happen Is going to drop out from this time dependent cpa symmetries and you essentially measuring Some parameter in the lograngian some weak phase without having to know anything about hadronic physics so Actually how you do this expiry these these measurements experimentally that's an interesting thing And this was highly non-trivial. This is why these dedicated machines the e plus e minus b factories were built. So what happens experimentally As we said at the beginning that in e plus e minus collision There is a particular energy the mass of the this four s resonance That if you tune the beam energies to that mass then you are producing b zero and b zero buyer and There is another complication that in order to have a long enough decay time For the b zero and the b zero buyer you had to build this asymmetric e plus e minus b factories So that both the b and the b buyer are boosted in the same direction so what one measures experimentally Is that so you are producing a b zero and b zero buyer This measurement relies crucially on the fact that this is a quantum correlated state so if at a time t One of the so b zero and b zero buyer oscillate as you could see over there as we talked about it And as long as both As long as neither of the two mesons have decayed if at a given time one of them is b zero The other one has to be b zero buyer and vice versa So the way these experiments were done is that you reconstruct both of the b decays in one case you look at Some decay which tells you for example by looking at an energetic lepton that which so you look at one decay which Identifies the flavor of one of the bees And then you look at the other b decay into your favorite cp eigenstate final state like jpegs i k show it and but and and and basically You can directly measure experimentally this time difference between these two decays And that's the time that occurs here because When you flavor tag one of the bees That's where the clock starts for the for the other one because if you know that at a time t one of them was a b zero Then at the same time the other one was a b zero buyer and vice versa And that's how you have access to these decay rates as a function of the proper time of these particles The situation in some sense is simpler at lhcb So when you do the same kind of measurement at headroom colliders, there are some other serious complications, but But but just this so people call this flavor tagging because you need to tag the flavor to tell you what Is the initial state at time zero in at the at lhcb In some sense the situation is simpler because you are producing bb by And the hedronization of the two b hedrons is independent from one another. So The tie so so so so so that the two b So that so so one of the bees can for example Hadronize into a b sub d or a b sub s meson and then it oscillates with its given time dependence But the other b can go into any form of b hedron. It can be a charged b meson and not a neutral b meson It can go it can become a b baryon and so So this is not a quantum correlated state and therefore this time dependence in some sense is simpler you So you have different experimental techniques to study the same type of observables And so just to give you an example. So in the case of j psi k short Which is a particularly clean example of this So that's a plot from a bar bar paper So what you see? So in this case the second term is to a very good approximation not present So the red and the blue curves Are telling you That just the decay rates Of an initial b zero and an initial b zero biotag As a function of proper time on the horizontal axis to decaying into j psi k short And the lower part of the plot shows this asymmetry Where the oscillation Period is given by the mass difference and sort of the magnitude How big these oscillations are is So imaginary part of lambda in this case is Whatever it's just a sign of the argument of lambda, right? I'm saying something totally trivial So so that's so that's what the measurements do and so what So let me take five more minutes just to explain what this really measures in terms of ckm parameters and then I will So I should have said here there before that q. Okay, so So what is lambda for b to j psi k short? What I what what what is being measured here? um, let me Erase this So if you look at how this decay can happen then they're again So called tree and penguin diagrams, so you can have a b goes to c C bar s This is a d bar So people usually write this amplitude as vcb vc s star And I'm just going to call it symbolically a t which is some complex number And there are also so-called penguin contributions which is So this is b To s with the w and this is cc by and Again here you have up chime and top in this in in in in in the In in these loop diagrams and so there will be a piece which is vub v u s star Times sorry.  This is the What am I saying? So vub is here v us is there So each of these entries are sort of the scalar products of the second and the third Column of the ckm matrix right vtb is here vts is here So because of uniterity, I can write vtb vts as minus vcb vcs minus vub v us Right, this is just the sum of the three is zero So I can rewrite this whole thing as vcb vcs times these combinations and the point is that vcb vcs is of where the the cabiol angle Squared because vcs is very near one and vcb is of where the lambda cabiol squared This term is of where the lambda cabiol to the four because vub is lambda cabiol three and v us is Is just the cabiol angle itself So even without knowing anything about these hydronic matrix elements that I called p and t one would expect this second term to be Much smaller at most a few percent of the first term in addition There is so this is coming from a loop diagram So historically people expected this penguin amplitudes to be suppressed by one over 16 pi square compared to the three diagrams Empirically that suppression seems to be less than one over 16 pi square but still the absolute value of this Combination of hydronic matrix elements is expected to be smaller than this So So why why is that important because what I was okay, so sorry Let me backstep for a moment. So what I'm writing down here is the amplitude for So so this is an initial b zero by goes to j psi k short if I Wrote down the amplitude for an initial b zero goes to j psi k short Then again this t and p terms contain only strong phases coming from hydronic physics And one would get the complex conjugate of these ckm elements And the and and and and once This amplitude is dominated by terms with just one ckm structure That is enough to ensure That a by over a The ratio of the complex conjugate of this Sorry of the cp conjugate of this amplitude divided by itself Is going to be So that's direct cp violation cannot happen Plus small corrections And it also tells me that I can calculate what is the what is the phase of a by over a Because it will essentially be Okay, so the argument of a by over a Is just going to be vcb vcs star divided by vcb star vcs because So this was a bar and a Which is this amplitude Would just have the complex conjugate here and the second term is negligible So So what is lambda lambda is q over p times a by over a So q over p if you look there it comes from box diagrams with a top quirk So It's weak phase is given by vtb vtd star Divided by vtb star Vtd right because it's just you are picking up four ckm elements at each of the vertices of the box diagram a by over a We just said it there that it's Going to be vcb vcd star over vcb star Vcs And if you don't want me to cheat I should tell you that there is an additional subtlety that You see that b to j b to cc bar s So this is only giving you either so in this case you are getting a k bar zero Whereas in the cp conjugate process you are getting a k zero so in order for So it's really you're looking at b to j psi k zero and b bar goes to j psi k bar zero In order for these final states to interfere It is crucial that k zero and k zero by a mix And what you measure experimentally is not the k zero k zero by a final state But the k short final state which is an almost equal cp even mixture of k zero and k zero by that's why interference can take place At all otherwise if there was no k zero k zero by a mixing you couldn't have interference here And So there's an additional Term So what so what this calculation tells you Is really a bar for a j psi k zero bar j psi k zero But what we are interested in is a bar for a psi k short over a psi k short And so this will have an additional piece from kk by mixing which And since kk by mixing is dominated by the box diagram with the chime quack The additional piece so i'm just writing down what was there before And and the kk by mixing gives an additional phase Which is just the chime box diagram to a good approximation So in particular that guy goes away So that's well the next so okay, so if so so that's Because of kk by mixing Instead of vcs style over vcs. I have here vcd style over vcd and if you look at The unitary triangle on the top left corner of the box then you see That this term the phase of this term is just better that So we have vtb vtd style over vcd vcd style That's the term that I encircled there and the other piece is just the complex conjugate of it And if you are careful with signs Then you get that this is nothing else, but minus e to the minus 2i better so the Bottom line is that this time-dependent cpa symmetry measures A phase In the standard model Lagrangian the phase of those ckm elements that are defined there And the hadronic uncertainties Are at the percent level or below Without having had to conclude compute any hadronic physics at all so I Wanted to cover a little bit more but uh It's uh I will continue from here tomorrow so So what I what I wanted to say is that This time-dependent cpa symmetry measurements. So this was one example how it gives you Information on weak phases without hadronic uncertainties I will very quickly tomorrow show you a few other examples how Similar measurements for other final states Again can give you other theoretically clean measurements of phases in the Lagrangian either cross checks on this measurement or Other measurements which are sensitive to possible new physics affecting BB by mixing that could show up as a discrepancy between Measurements which in the standard model relate to the same phase I will show you tomorrow how that can be used to constrain new physics in BB by mixing and And that will probably take just the first 15 minutes of tomorrow and then we'll dive into a little bit of Discussion about heavy-quack effective theory And probably my goal for the end of tomorrow is to explain Some ingredients of heavy-quack effective theory and heavy-quack symmetry which will allow you to appreciate this plot which is Which is which which I will explain tomorrow. So this relates to some semi-laptonic Decay rate measurements in in beam as on decays by boboa and lhcb And bell which right now seems to deviate from the standard model in a very interesting way So thank you and sorry for running out of time and I'll take questions if you have any Okay, thank you


Lakers Warriors game one. Let's put some pressure on some people, Chandler. Who feels it more, LeBron or Steph to advance? I mean, this is a big legacy series for both of these guys. I think it's more pressure for the Warriors. I think the Lakers already exceeded expectations by getting here, by beating the Kings, by making the moves they did to even getting the playoffs. I feel like their season has already been successful. And I know they're not gonna say that. I know they're gonna want to obviously survive in advance and keep going. But we're looking at the end of an era, possibly with the Warriors, if they lose this series. And this could be the last time you see Steph, Dreymon and Clay together. So I think the pressure is for them. I think they know they're the better team. It's crazy that the sixth seed and they got home court advantage on to win the last time that's happened. But it just shows how competitive the Western Conference has been all year long and it's anybody's game. But it's exciting. It's just two of the greatest players of all time that can really just put a stamp on their legacy by whoever moves forward. Yeah, I think the most pressure is on LeBron. I don't really see Steph Curry really getting nervous. I don't really see Clay Tynes really getting nervous. You know, just judging from, you know, how some of the guys acted on a bench and for the Warriors, I don't think they see the bigger picture to really get nervous. So I think this is all a huge part of LeBron's legacy. Like we are you before, you know, you have Steph and LeBron, two of the best players of the era of the generation. And I think this kind of, one of the last few battles in our head that really solidifies who's really running what. So I see all pressure on the Lakers. Pressure is a funny word, but I'll throw a different guy out there. Anthony Davis.  Anthony Davis can put his stamp on whatever his legacy is here. He's a guy we all constantly go just like Kauai and Paul George. If he's healthy, if he's healthy, we look at him in that way. If he dominates this series, so I say after what we just watched happen, the Montessonus, and he helps push the Lakers to the win against the reigning champs, against Steph at this peak that he's at. And now we're looking at Anthony Davis different. So I think it's a big series for him. I think it's a good matchup for him too. Steph Curry, LeBron James, both have a chance to get five rings this season. We think Chandler, legacy's on the line. Can't wait to argue that. That's what I'm saying. It's a huge stamp of approval for whoever gets going through. Obviously they're both are gonna be first ballot. They both have had a hell of a career, but this is huge series for both of them. Just where they are in their career, what their season's been, kind of the up and down rollercoaster. Both these teams have been on. It's unbelievable to think that one of these teams, everything they've been through is gonna go to the conference finals and it has a lot to do with LeBron and Steph. Chandler, thank you for joining us on your own show today. We appreciate it. Tomorrow it's a boat or nothing. Next time he'll tell us exactly what Joe came know of or just text us in the thread. We won't. We won't say. Yeah, I know. Get him a drink. By load management, almost done. By the way, between this outfit last night, James Harden's 45 pointer and the Met Gala. Fit or break is chef's kiss today. That's coming up next.  She knowin' all over the map cause she make it clap. With a short song on the sky.