Raffaele D’Agnolo - 2/3 Beyond-the-Standard-Model meets Cosmological Correlators

All right. Thank you very much. So, we went ahead and uh finished the technical part this morning. So uh now we're going to start uh having some fun just one last step which okay it's well known from quantum filter in flat space. So we have uh we have built uh we have built our path integral that looks like this. with some uh boundary conditions at the final time that tell us that the minus and plus fields are the same at that time and some epsilon prescription that I'm including implicitly in the action and so from here we like to compute uh our correlators and we can do it as usual in perturbation theory so I will give you say a lightning review of perturbation theory. Um so if we have a party integral that looks like this where s0 is quadratic in the field. So it looks something like this with all some operator. Well, then we know how to do the fat integral. Okay, so we can go to the ukidian and and realize that this is a goian and uh and we know how to solve it. But we can do more. So this is not the only part integral that we know how to do it. We know how to do it also with a with a source. Okay? And when there is a sorts, we know how to do it just because we can complete the square and go back to an integral that's purely quadratic. So we can uh we can shift our integration variable where this DF is precisely the inverse of our quadratic operator. Okay. And so this is this is just a constant shift. You see this part doesn't depend on the field. So the integration measure stays the same. And the exponent instead becomes um something quadratic in the field. Well, say well let me sorry define this with one half since the scalar is real. Then there is a 1/2 here. And there is also a 1/2 here. So with this shift we went back to a quadratic key integral. So the part on phi just gives us uh if this is a scalar a determinant of our uh our operator. So times whatever is left. So e to the minus 12 d for x d for y jx df jy and I'm reminding you of this basic facts because uh well this is potentially an infinite factor the iapsilon is telling us let's say if we want to solve this inverse in 4 space what contour to pick to go back to physical space But uh in any case this part whenever we compute correlation function is canled by the normalization of uh of the function. So, so the correlation function is always uh defined as this guy divided by this. And so, okay, the this potential infinite factor never matters for us. All that matters is this guy. And I'm sure you all are familiar with with this. The only point I wanted to highlight is that if now you have a more complicated part integral that you don't know how to do, so your action is not quadratic anymore, you can always uh split it. Okay? So split your action into a quadratic part and an interaction term plus you can add your source term still. And now if you know only how to do the part of the integral without the interactions is not a problem because you can rewrite this as at least formally e to the i interaction of derivative with respect to j of the part integral that you know how to evaluated then at J equals zero. Okay. And you can use you can use this trick also to compute correlation function. So if here you have some powers of the field, you can also trade them for powers of derivatives. Um, of course, I mean, you're cheating when you're doing this because really the way in which this exponential is defined is uh is as a sum okay of uh of polomials in fi. So, so what you really have is something like this. So, so let's say you have some small coupling y in your interaction lrang. So, your interaction lrang is something like this. And so when you bring this piece outside and replace phi with a derivative with respect to j, you're inverting the sum and the integral. Okay, which is often fine if y is small enough but when you start going at large number of external legs you're going to have a problem. So technically you are not allowed to invert the sum and the integral and that uh comes back to bite you uh in some cases that are not going to concern us but but it's a fact that's useful to keep in mind anyway. So all this discussion it's well known from from uh initial courses on quantum field theory. Uh the point that I wanted to highlight is that uh now you can essentially write whatever you want if you have small enough couplings because otherwise you have to resum infinitely many derivatives as a bunch of derivatives acting on some part integral that you know how to do which is proportional to that uh stuff up there. Okay. So J B FJ so you can reduce computing correlation functions to computing this and you see immediately that this is giving you products of propagators. Okay. So this DF is as we said the inverse of whatever operator was in the quadratic part of the action. So doing these derivatives is just bringing down products of propagators. So writing fine man diagrams is just amnnemonic uh for keeping track of these derivatives and the products of the propagators. And of course okay if you have a more complicated theory with some interaction these derivatives come into two classes. Okay. So there is the type of derivative that is accounting for whatever operator you have here that you want to compute the correlation function for. And then you have also operators. So you have also derivatives which come with some power of the coupling and correspond to vertices. And so well a fine man diagram is nothing but products of vertices and propagators. Okay. So that's that's what we're going to do from uh from now on. Apologies to those for which this was already completely clear. But uh well [clears throat] I find uh useful sometimes to make this very quick connection to more basic facts of life. Okay. So the only difference for us compared to this very standard uh discussion is that we have a plus and a minus for each field. So we're going to have four propagators for each field. And uh well one possible type of notation is that uh when you have two plus scalars you call the propag you describe the propagator as a line between two black dots and when you have two minuses between two empty dots and everything else in between of course so you can have also the mixed propagators. Okay. And I'm going to call G uh the propagator momentum space. Okay. where this delta AB or well yeah let's call it delta is the propagator in position case and uh if you remember well the boundary condition that I just deleted so the fields are the same at the final time slice that I'm considering so the one where we make the observation and so whenever a propagator goes to uh the boundary at to at the final time it doesn't matter if it ends with a plus or a minus field. So I'm going to put a square. Okay. So in this case we're going to have just G+ and G minus. So let's see. So again this notation is from that paper I I wrote down before but I find it uh pretty good. Okay. So as I said all we need are propagators. Here they go. And vertices. So for vertices the only thing to keep in mind compared to flat space is that they come with powers of uh of the scale factor. Okay because in our in our action we have a determinant of the metric and okay we might also have other factors of the metric when we have derivatives. [clears throat] So each vertex comes from comes with an integral over times the determinant of g which is a to the 4. Okay. And with the coordinates that we've chosen. So in conformal time a to the 4 look sorry a looks like 1 / h times conformal time with a minus. If you have derivatives you see you have also the inverse metric. And so okay the simple way of counting is that for each derivative you have an inverse power of a. So let's say if you had a vertex with two derivatives it would come with an integral over d to a squ instead of a to the 4th. Okay and so on. So for each derivative you get one more power of a at the denominator. And with this we are ready to answer some physical questions. Okay. So the first question that that I want to ask even if uh yeah so the first question that I want to ask is what can we say about single field inflation So we have our action from before. So this shows you also what's the relation between my M plank and my G Newton. And well the first obvious question to ask here is can we say anything about this from observations made today? and uh well let's expand it and start checking. Okay, so the quadratic ter so so the the mass goes into the twooint function. So we're not going to care about the mass in some sense. We've already measured it. So let's start with uh the threepoint vertex. Okay, there are of course also other terms where here I've expanded around the classical value of the background field. So this is the field driving inflation. It has some classical background value that depends on time and well I've expanded around it for its fluctuations delta fi. So naturally we expect this to leave an imprint into the threepoint function. So something that looks like this. Uh this means that these three points are taken say at the end of inflation. So whatever we see today and this vertex at the center can be either a black dot or a white dot. Okay. Now, okay, we have everything we need to compute this diagram, the corresponding delta fi uh correlation function and see if the FNL that comes out can be observed or not. Before doing it, let me make uh a simpler estimate. So this uh this threepoint function sorry this this three-point vertex is affecting slow roll. So before we were asking that this epsilon and data parameter be much smaller than one and we need them to be small for some time. So we need also the derivative of aa to be smaller than one. Okay. And if we want to construct a dimensionless quantity, we we can estimate the derivative of eta divided by abble. So this object is well the derivative with respect to time of what I wrote down before. So m plank squared over v the second derivative of the potential. Okay? because I'm putting a sim equal because this relation between ETA ata and the potential already assumes slow roll so it's approximate and then we have able here and so this is going to give uh well the derivative can act either on v or on the second derivative but we are interested to understand what's the maximal size of this guy compatible with observations so we're just going to care about the piece where the derivative acts on V module of finetuning. Okay, so we don't want the sum of the two pieces to be accidentally small. We're going to assume it's that the two pieces that the derivative can give you are roughly of the same size. Okay, and so [clears throat] in this case we get m plank squar over vi dot times three derivatives of v and plus okay also the derivative acting of v. But as I said, we're assuming that this piece and this other piece do not cancel each other accidentally. And so we want each one individually to be smaller than one. Okay? So we're going to ask that this whole thing be smaller than one. Um and this gives let's see this gives that the third derivative of the potential be smaller than V H over M plank squared that if you remember the definition of this Laural parameters uh oh sorry I forgot five dot Okay, over a fi dot. So if you remember the definition of the slural parameters, this is just h the square root of epsilon m plank which is also the square root of what we call pz and so this is order 10 to the minus 4. Okay. Uh sorry let let me actually divide this by apple. You're going to see why uh this is the relevant quantity in a moment. So we're finding that this third of the potential divided by apple is tiny. Okay, this is already a bad sign, but it's not enough to conclude that it's impossible to see this vertex in the threepoint function. So let's actually compute the threepoint function or rather estimate it as brutally as we can and see uh what comes out of it. So to estimate it, we need the vertex. Okay, but we know so the vertex has no derivatives. So it's an integral over d to of a to the 4th times whatever value of the potential has in the classical background times three propagators. Okay, that I'm just going to call G cube. And we're going to take this inflaton to be approximately massless compared to abble sorry what am I saying? Not not the inflaton. So the fluctuations are massless. Okay. So the fluctuations are are massless. You you can see it if you expand the action. And so we want to um we want the propagator for a massive a masses field in uh in the sitter approximately. So in this laural background I'm going to write for you the solution. Okay. I'm sure that uh you've seen uh already a few times how to get the solution. So this is what you get if you solve for the mode functions of a massless scalar in our background. And for us all that matters are the dimensions of factors. So we're going to take the this scales like h squ over k cq. Okay. So in practice I could have even avoided saying massless or massive because all we care about is the scaling. All the rest for our purposes combines in the integral to give you some factors that can also be large or small but uh we're going to consider order one for this lecture. So we just want to estimate the prefactor to get a feeling for the size. Okay. And so uh maybe we need this. Okay. So let me me use the blackboard behind Okay. So now we want to estimate this guy. We're going to factor out the delta function. We said it comes from this diagram. This is what the diagram gives. suppressing all the dependence on time and momentum. And so roughly what what are we going to get? So this is factoring out all the dimensionful quantities. All we need to do now is just uh define yet another dimensionless variable to really get all the dimensions out. >> The what? Oh yes, sorry. That's right. There is the our friend here the the the coupling. Yes. Um and okay, so this gives us another K cube. So there's a k to the 4th at the numerator and a 1 / k from the integration measure. So we're going to get indeed the vertex time k cub / k to the 9* h to the 6 times some integral that we're going to completely ignore. Okay, very good. So to get uh fn we have to use what we had derived before. So we're going to multiply by k to the 6 to get rid of the dimensionful piece. So I didn't say it but uh let's say you can interpret this k as being the largest of the three momenta. Okay. So, so that uh inside the integral then you're left with just uh dimensionless numbers of order one. Um of course I mean if you actually want to do the calculation is much more complicated than that. There are some plus and minuses that make everything convergent when you include everything. But as as we said, as I said many times, the spirit of these lectures is just to get a feeling for the size of these objects. So we're going to forget all the sublies unless you ask. Uh and okay, so let's see. Let's see what's the size of this object. So we still have a power spectrum to the 1/2 and able cube. Okay, from what we had derived before. And so if we plug if we plug that in, we're going to get this. Okay. So we put it together with Oh, okay. With that thing that is now up in the sky. So you see that the that the relevant parameter is really v the third derivative of the potential over apple which should be smaller than pz to the 1/2 and so finally we get that this is roughly smaller than one okay which is roughly what's the current constraint from the CMBB on this third slural parameter exactly how much smaller than one okay depends on your model. And so, okay, this question lands somewhere between large case structure and 21 cm line depending on the exact shape of this function. uh probably it lies closer to 21 cm line given the the number I was quoting to you at the at the beginning on uh on this factor of 40 that this guy is looking at this scalar tint function of the CMBB found. So the FNL you can bound in the CMBB for this kind of uh function if you really take into account the dependence on the three momenta is order 40. Okay. So one is pretty challenging. Yes. >> Sorry. I just How do you know that this is roughly? >> Yeah. From from the estimate we did up there. Uh let me get it down if I can. So to preserve slow roll long enough. So let's say current CMBB data if you want tell you that this ETA dot over H should be roughly smaller than one and then this data dot over H is proportional to the third derivative of the potential and then if you manipulate it a bit you you'll find that this being smaller than one implies third derivative of the potential over H being smaller than this PZ to the 1/2 which is exactly what we find uh at the end in the three point function. Okay, so if you're an optimist, you might say we can tell what's the inflat on potential at least the three point vertex at some point before we retire. If you're a pessimist, maybe we'll never know. But uh well let let me give you so okay this is not the only question that we can ask about single field inflation there is more okay there is more and it but it's something equally generic so here we asked a very generic question so what's the first term in the potential if I assume that the fluctuations are small will we ever be able to tell and I let you decide on whether we ever be able to tell based on this estimate. Okay, but there is not just this. Okay, as we said the data are consistent with a slow roll scenario where the potential of this fi is very flat. So the most natural option from a model building perspective is that this fi has some shift symmetry or some other symmetry that makes its potential small. So it's totally natural to expect that its leading interaction instead of being uh the third derivative of the potential it's actually a derivative interaction and the first such term that you can write if you have only phi is something that looks like this and as I told you at the beginning this gives us some extra hope because phi dot so because the typical scale the typical scale of quantum fluctuations during inflation is abble. So, so the estimate that we did before for the twooint function of zed scaling like h to the 4th over five dot squared. So if you translate it to delta phi it gives you something that scales like abble squared. Okay this from the relation we we had mentioned before that zed is delta ph dot. Okay. So if you translate what we had before into delta fi you get this which is essentially telling you that if you leave a scalar alone on a flat potential in a deserter background it will do some random walk taking steps of order abble okay so the typical scale of quantum fluctuations is abble and that's what went into this three point function but now if you have a derivative vertex you have access to this other scale Okay, that as we said before can be quite a bit bigger than abble. So maybe it's easier to see derivative interactions than it is to see the uh 5 cube vertex in the potential. Additionally, derivative interactions are more informative because if you see something like this, you're not only learning something about the inflaton, but you're also learning something about a scale where something else is happening. Okay, where maybe this inflaton potential is UV completed. So this is interesting for these two reasons. Better chance at being seen and more information on high energy physics. And um actually this is the estimate that tells you that even if the scale of inflation is bounded to say order 10 to the 14, you might still be able to see particles that are 10 to the 16 GV. Actually, we're going to see this better. I don't know if at the end of today or or tomorrow but you have to keep in mind that uh in inflations there's more than one scale in your effective theory and one is uh is a little bit bigger than than the others. Okay. So um let's now uh do the estimate for how this contributes again to our threepoint function. And well, as you can imagine, from now on, we're always practically going to do the same exercise, but declined into different types of questions. So now we have our fourpoint vertex. We're going to send three legs to the observer and we're going to set one leg to the background. So this means that five is on its classical trajectory. Okay, then this again goes like a to the 4th but well you have three derivatives so divided by a cube uh times f dot over m to the 4th coming from the vertex times three more derivatives acting on the three propagators. Okay. And now, okay, we just turn the crank. Okay. We again, we again define our word Z K to this time. Um, this time we this guys give gives us no K's, but this one gives us a K cube. And then we have again a square over K cube from the propagators. This guy gives us a k cube as I said and then we have phi dot over m 4th um which ends up giving us h to the 5th over k to the 6* 5 dot / m to the 4th. Okay, if from this estimate you don't get k to the 6, you should worry but all the rest can be anything. Uh and finally well we again want to estimate FNL as we did before. We plug this result in here and we're going to get h 2 over Z to the 12 dot M to the 4th. But well, if you use the definition of PZ, you're going to get PH dot squared over M to the 4th. Okay. And here comes again well how generous we want to be in this case not to observation but to our effective field theory. Okay. What m what's the smallest m we take we we can take without breaking the theory. Okay. If we were doing the effective theory of inflation. So forgetting about phi but uh only focusing on the gold stone of time translation of broken time translation invariance. We could take m just a little bit above. Okay. But here we have already in UV completed our effective theory of inflation adding this uh scalar field and so we are not allowed to do that. We have two options. So we can be conservative and take m bigger than the biggest scale in the problem which is the potential of phi evaluated on the background or we can be generous and take m bigger than just five dot to the one alpha. Okay, this is the bare minimum you need for your effective theory that contains the d to the 4th vertex to make sense. Okay, so this is saying that at the scale m there are some new states that in induce the relative interactions of phi but don't bother the potential. Okay, instead okay the with this you can sleep tight okay your effective theory is totally fine you don't need to bother build anything at the scale m and again okay so depending of what nature has in store for us whether it's been generous to us or not this could be roughly order one if this is the case or much much smaller than one if this is the case okay so again we are in a region where it's hard to think that the CNB will say something but next generation experiments might and again the answer on whether we can see this derative couplings of the of the inflaton or not is uh left to your optimism. Okay. Um so this is all I wanted to say about single field inflation which I realize it's not a lot but it's already giving you an idea of where we stand. Okay so this is the leading derative of interaction in the simplest model. This is the leading non-derivative interaction in the simplest model. They all give uh uh signals that are smaller than what we can see today in the CNB but not that but if we're lucky they're not that small that we will never see them. So it is conceivable that uh this program of measuring the by spectrum the threepoint function will tell us something about the model of the universe even in this minimal case where there is just the inflaton and nothing else. Okay. So so let's say that that today yes I want to be optimistic. Okay. All right. So we're going to go back to uh inflation at the very end tomorrow uh where we're going to drop this single field part. But for today uh now I want to start asking a different question which is going to take a much longer time to answer because uh well when you add more fields and more BSM freedom obviously it becomes harder to make generic statements. But uh I think I will still be able to tell you some general lessons. And the question is what can we learn about particle physics? So, so if there are some particles say with masses in the neighborhood of one of our two scales that couple to the inflaton. Can we see them or not? You can take a guess if you want but the answer is that there is no right answer. It depends again. So if they have certain kind of interactions, we're going to see them. If not, we're not going to see them. Uh if you want the the interesting part of this very open-ended answer is that typically we're going to see them in all the cases that were already well known from very old-fashioned methodologies in inflation where you had a lot of particle production. Okay. So if you're familiar with chemical potentials or preheating parametric resonance during preheating when you produce an exponentially large number of particles those are the cases where you're going to see them in this in these correlators and I'm going to make an effort to go back and forth between the language of correlators and the more traditional one let's say and all the in all other cases it's not impossible but it's way harder. So we're going to look at examples in both realms. So uh cases where you give these particles special couplings that enhance a lot of reproductions and cases where you just put in generic couplings and see what happens. Okay. And of course the most interesting case is well when you make no effort and you just assume that these particles have generic couplings to the inflaton. Um and to gain to gain some intuition um on this case I want to step back for a second from computing correlators and actually go closer to to some of the stuff that that Dio was telling you before. Okay. Well, so far we have kind of in a in a pretty cavalier way we just said that these were our coordinates and we completely forgot about observables. Okay, so we just assume that we had access to the all the sitter manifold. But I mean as you know the scale factor during inflation is increasing exponentially and this amounts to having an horizon. Okay, so exactly the same horizon that was talking about before. So each observer is carrying with himself an horizon and it does not have access to the old manifold. Okay, it will never see these global coordinates. Um and this as an interesting uh physical consequence. Okay. So we we can make a change of coordinates and go to uh a gauge where this horizon is manifest. Okay. So you you just see it immediately from the metric. I'm not going to tell you what's the change of coordinates because okay it's it's a bit everywhere uh in textbooks and in the literature and I hope that some of the previous lecturers told you what it is but in the end you get with this metric you see immediately that this goes to zero for some value of the radial coordinate and uh you know that there is an horizon there what's interesting for us so from the point of view of particles floating around in the sitter is that uh if you assume that you started your life in the vacuum state of uh the the global coordinates let me call it I don't know omega global okay then what an observer for which these are the natural coordinates so so a guy in the static patch what this guy sees is something that looks like a squeeze state with a thermal spectrum Okay, I'm going to call it the vacuum of the static observer. And uh well, one of the many standard ways to see this is to do a bou transformation. So, so imagine that you uh have uh done you started your life in these global coordinates. you've computed the mod functions of your fields and your propagators and that's exactly what we did. Okay, so I didn't tell you super explicitly but when I wrote the propagators that's what I did. I started in global coordinates quantize the fields and solve for it their mod functions. Okay. So, I'm going to have some fields that look like this with some annulation operators in global coordinates and some mode functions in global coordinates. Okay. So now I can repeat this exercise but I want to solve the equation of motion now in static coordinates and let's assume that I know exactly what's this function okay you can actually solve the equations and find these functions then you can look for a transformation that turns one into the other. So a transformation that takes the annulation and creation operators in global coordinates and relates them to the static ones. So if you fix the norms of these coefficients alpha and beta to respect this relation, you can show that the commutation relations between the operators are preserved. Okay. So you started with AK global AK dagger sorry AK global dagger K prime being proportional to a delta function of K minus K prime and if you do this transformation and impose this normalization condition you're going to get exactly the same for the static one. So you start with some good creation operators and you end with some good creation and annulation operators. Um so the the interesting fact about this is that as usual I mean you can see these uh these transformations in quantum mechanics either as acting on on the operators or acting on the states. So you can do something completely equivalent which is just saying that your global vacuum is related to your uh static vacuum. Sorry, let's do the opposite. Your static vacuum is related to the your global vacuum via some operator s and the two vacua are defined as those anilated by the respective anilation operators. Okay. And [clears throat] if you want this uh s to do exactly this, so to move you from the solution in the static patch to the solution in the global coordinates or vice versa. It's easy to show that uh this guy should just be defined by this relation. Oh, sorry. And one can solve okay one can solve all this and find that the operator looks like the following. So notice that maybe I don't know if this stuff is familiar to you but it was very familiar to cosmologists a few decades ago. So this is the static dagger and this is the static dagger. Okay, if any of you uh has done any quantum information course, this is exactly the squeezing operator. That's why I was calling it a squeeze state. Uh okay. So, so we are almost there. I promise that the punch line is coming. So, now we've seen that uh um you can phrase the physics by saying that if you start so that that if you start your life let's say in the vacuum of the global coordinates a static observer is not going to see a state that looks like the vacuum. Okay. So if really you assume that the universe is in this state, you as an observer today with this horizon are going to see a funny state which is the vacuum times the squeezing operator. And um if you compute what's the expectation value of uh sorry uh maybe the opposite okay so you can use all these results to compute what's the expectation value of the number operator so a dagger s on the global vacuum. So you're saying I know that my vacuum should be bunch Davis. So I get it uh by doing analytic continuation into space from global coordinates and I want to see what a real observer sees. Okay, how many particles is an observer in the static patch seeing in this state? And what he's seeing is beta k absolute value squared which you can solve for. Okay, I didn't give you the explicit expression of this uh mode functions but you can in principle work it out. And if you do it, you're gonna find a therma spectrum. So in this sense you can think as you can think of an observer in the sitter space or quasi deater as for us during slow roll as being immersed in a thermal bath with temperature h over 2 pi. Okay. So once you make this observation a lot of the things that come out of cosmogal correlators make immediate sense of course. Okay. So for example uh if you have a particle with mass much bigger than the temperature in a thermal b you expect that its number density be roughly m * t^ 3 * e to the minus m over t. And indeed, if you want to compute the exchange of a massive particles in these correlators, you're always going to find that this e to the minus m / h suppressions for massive particles. Similarly, for a massless particle instead, you're just going to find that the number density scales as t cube, which is the only scale, which is of order cube. And indeed when we were checking what a propagator of a massless particle was doing in these functions it was scaling like some the appropriate power of apple. Okay. So because it's the only scale you have that's it. Um okay so if you want this is the very basic reason why uh it's not immediately easy to see particles in these correlation functions because if they have a mass they will be boltzman suppressed. If they don't have a mass you might say okay yes >> this one Yes. Why are the spatial momentum the same when the spial slices >> the spatial moment this ah uh yes yes you are right I glossed over a big sap which is there is I mean the reason is just that uh these coordinates are describing the same space okay only this is describing a portion of the space and this is describing the all So in reality I'm I'm not Yes. So I cannot I'm not allowed to really do this. What I should do is I solve for the mode functions here. I solve for the mode functions here. And then in a patch that overlaps I find a third set of coordinates and then I match the two solutions using this third set of coordinates. So, so this this thanks for the question because I mean this was all very compressed into into uh into this rough uh formula. Um all right. >> Yes. >> You define the data coordinates. Can we show using transformation from global to that? So I mean that is the same or >> I mean I think you can also do indirectly from ER going to the Ukidian sphere that also gives you the bunch I'm not yeah I'm not I haven't tried to do what you're saying so I'm not sure but uh um all right so okay so so but this should immediately tell you that okay then why don't we just think about massless particles okay so if we think about massless particles. Since there are no scales, we expect this to just go like ab to the 6. Okay, that's it. and uh and then well if you plug this back into our FNL estimate so D5 cube so again well I I I often forget the prime okay but I'm always talking about correlation functions with the delta function factored out Um so divide by pz to the 12 / by cube and then here you get uh sorry k to the 6 here you get something huge. Okay, why this cannot be the case? I mean it's very simple. It's because from just the twooint function, we already know that there are not masses particle floating around besides the inflaton. Okay. Otherwise, uh all our nice CNB spectra with the multiples that are fitted perfectly with 12 decimal digit. Okay, 12 maybe generation. I mean they they all get completely screwed. Okay. So it's not that it's hard to see generic particles at the cosmological collider. Okay. It's that we've already excluded them essentially. Okay. And now only the art stuff remains. Um all right. So uh this this gives you some some general intuition. But uh I think it's useful to uh also look at some concrete examples and go more into the details. Okay. So let's just add some particles with some couplings to the inflaton that are consistent with observations and let's see what happens. Okay. Uh how much time do we have? >> 35 minutes. >> Okay. All right. So, uh down. Yes. So, we're again going to start simple and we're going to add some scalar. Okay. Some scalar with generic interactions to the inflat. some trinear coupling um some quadratic some quartic so here I'm adding a complex scalar but it doesn't really matter okay of course we're going to give it some mass because we just said that uh otherwise we're going to screw the power spectrum and so again we want to compute the correction to the three point function of phi coming from exchanging the scalar kai >> [snorts] >> And the leading thing that we can uh draw is something that looks like this. Here there is a lambda and here there's some kai. These are f. And again this means that this can be either black or white dots. Okay. And here there is a mu and okay you you're not going to be surprised if again I go through the usual machinery. So now I have this lambda coupling this mu coupling. I have integration over loop momentum. Then I have one vertex. I have another vertex and I have three massless propagators and I have two massive propagators that I'm going to call D. Okay, so going to call D the propagator of Kai in momentum space. I've been a bit all over the place with notation, but from now on I'm going to stick with that. Uh >> by the time >> No, this is mu. >> It's just some coupling. Um um yes. So what's the massive propagator in pract? Well, I'm going to write it down for you. One of them. Ah, yes. Before I didn't say I didn't say why I only wrote one of the four propagators, but you didn't complain. So I guess you know that they're all connected by complex conjugations, inverting the time, some functions. So parametrically they all scale the same way. That's why I wrote only one and that's why I'm writing only one also for the massive guy. Okay. So the one for the massive guy looks slightly more uh imposing but in the end it scaling is exactly the same. Okay. And I'm going to define this new in a second. These are ankle functions. I'm sure you've seen them in some previous lectures. And this new is defined as the dimension of space over 4 minus the mass over a squ. Notice that this is just the dimension of space. So for us it's three okay not of space time. So we see that when the mass is much bigger than double this factor is giving us the Bzman suppression that we were mentioning before. Okay. But other than that, these two functions are dimensionless and well, they blow up near the origin, but then the combination of plus and minuses in all the examples that we're going to look at makes the integrals finite. So, we're going to ignore them. So, all we care about again are the dimensionful pieces. But again, if we define our Z variables as before with the largest well with the only momentum at our disposal in this case, we're going to see that this D scales roughly like H2 over K cube. So in the exact same way as the massless one and if we are unlucky there is also the exponential suppression. Okay, so roughly for us a massive propagator is just that essentially it's a massless one with a price to pay. Uh and so okay now we have all the tools we can also estimate this diagram. So we have our lambda uh mu we have uh a k to the to the 6 upstairs uh plus h 2 over k cube to the 5th. Okay, if you want, we can also put the Boltsman suppression. Uh, I probably forgotten some factors here. Uh, let me see six kum. Yeah. Then there is also the well uh I'm going to assume that the loop integral is mostly supported at the physical momentum. Okay, which is typically the case if the integral is finite which gives me another k cube. Okay, that's uh a dangerous estimate to make but in the end it's going to give us the right result. Okay. So overall we're going to get uh ah sorry I forgot that each one of these a comes with an abble. So there is also an abble to the 8 at the denominator. Okay. So finally we're going to get lambda mu over k to the 6 ab squared and the exponential suppression uh plus okay if want to be honest there is also a loop factor. Okay, when you integrate over the solid angle, it cancels some of the two pies in the 2 pi cube and you're left with roughly 16 pi squ. So, how big is this? So our fnl is again k to the 6 times the threepoint function of phi over p z to the 12 cube. We take this and we get lambda mu over 12 loop factor squared sorry loop factor over albon. Okay. So a priori we don't know what's the hierarchy between these couplings and whatever it's at the denominator. Uh sorry. Well there let's let's not forget this. Okay. However, okay, let's imagine for the moment that um that Kai started his life close to but slightly below. Okay. So that this can be taken to be order one. Very good. But the inflaton is moving and is changing the mass of Kai through these two couplings. So we we want that mu * delta 5 and lambda* delta fi squared where delta fi is the amount by which phi moved during inflation remain smaller than abble squared. Okay otherwise this exponential suppression might [clears throat] become important. Okay well sorry I called it mkai but then I forgot to put the kai back. Okay. Uh so uh so this gives us an upper bound here. Okay. So this should be smaller than um able cube over 16 pi squared by z to the 12 over delta 5 cube. Okay. Very good. And now you're going to see that once more Zurole is killing us because what is Zurole telling us? It's telling us something the number about the number of efforts of inflation. So the integral over delta PH of H over dot. Okay. So this is telling us roughly that delta phi is approximately the number of eals times pi dot over. Okay, assuming uh that uh oh we're we're approximating the integral as this product is consistent with the fact that we're assuming that these parameters vary very little during slow on. So abble stays roughly constant and f dot stays roughly constant. Um so if we plug this back here we're going to get that this upper bound is roughly so and and again we recall the definition of pz in terms of ph dot and h we're got we're going to get pz over number of eolds cube. Okay, if I put here the 60 needed to explain the CMBB, this gives me a tiny number. Okay, which is completely impossible to see at anything that you can uh that you can imagine. Um, so what is what is the lesson here? Well, the lesson here is that scalars don't want to be light the moment you couple them. that uh the hierarchy problems also always comes back to bite your ass and so uh there is nothing to do. So if you want to see something in uh in this correlation functions you need large couplings but if you have large couplings and enough rolling to explain the CMBB whatever you couple to the inflaton will become too heavy to be seen um within this decal bat. Okay. Um, so this is just an example and it already suggests uh how you could uh instead have larger signals. But if you want the take-home message is that uh if you just write something generic, you're going to encounter this problem. Okay. So if it was massless and generic, it would be fine to give a huge signal, but this is already excluded from the twooint function. If it's massive engineering, you have to be careful and typically whatever you do, you're going to go back either you're killed by the small couplings or the exponential suppression. Uh, however, well, there are many things that we can try. So, we can do the exact same as we did before. There doesn't need to be a coupling like this. Maybe the leading coupling is a derivative coupling and there we benefit again from fi dot. Or maybe you can invent a clever system to couple these uh these um particles but still protect their mass. And I mean if it was firmian you wouldn't need to be that clever. Okay, the mass is already protected by the car symmetry. So um but okay let's let's go a step at a time. Let's check what happens if you have some generic scalar coupled to the inflat via derivative couplings. Okay. So maybe we can put a sort of checklist here. So what we've seen is that scalar with generic couplings It's not so visible. Let's say massive scalar massless anything. with generic couplings is too visible. It's already excluded. Now let's check massive scalar with derivative couplings. Okay, which are also natural and in some sense generic. As I was saying before, you might imagine there is some shift symmetry that is protecting the inflaton potential. And in that case, now what you expect to be the leading interaction looks like this. I'm also going to add a quarterty coupling which is what going to is what's going to make our signal in the B spectrum non zero and well a mass for kai as before. So I think that by now I could call any of you to the blackboard to do my job. Uh I'm tempted I have to say but uh I'll do it. So let's again estimate this uh this pi cube. Now the diagram looks like this where this is a phi dot over m times some ve of k0 over m^ 2 okay I'm assuming here that kai as some web okay that doesn't change appreciably during inflation otherwise I would have no contribution solutions to the threepoint function. So again I'm not being completely generic. Then here I have my lambda kai vertex with the ve and then in all the other legs I'm mixing kai and phi. Okay so these solid lines are deltafi. These dash lines are kai. They mix through this interaction which gives a contribution to the mixing of this order and then through this interaction kai is generating a contribution to the triple function. Okay. So we can just estimate this as we did before. Okay. So now we have four vertices and just write da to the four. uh one vertex comes with four powers of the scale factor. The other three each come with a scale factor cube because one of the derivatives is acting on the fluctuations, the other one on the background. Uh each one of these three vertices gives me a phi dot kite 0 / m all cube. Uh and finally I have a lambda kai kai 0 from this vertex uh and my propagators. Okay. So I have one two three k propagators and one two three uh five propagators and three derivatives. Okay, by now you know the drill. Each one of these is going to give you one over uh k over abble. Okay, each one of these is giving you one over k. Uh each one of these is giving you k and each one of the propagators is giving a squared over k cub. Okay, so we can put it all together and we're going to get well not surprisingly one over k to the six. This I can just write down without even computing it. And then I do k0 / m cub and lambda k0. And if you check all the ables, they all cancel except for one. So this is my delta fi cube and again I'm going to translate it it into fnl in the usual way. Okay, which finally gives me phi dot k0 over m cube lambda kai kite 0 over this guy. And note that well I I again uh dropped a factor from the Bzman suppression which is that should be there. All right. And now we want to estimate it and we're going to play the same game as before. Okay. So the first thing that we're going to check is what are the corrections to the mass of Kai and well first of all surprise surprise what I said before was wrong. Okay, the same thing that you can leverage to get a big signal. So phy dot is also coming back to bite you because it's correcting the mass of kai. So derivatives are not harmless during inflation as they are uh in flat space. they can uh also correct the potential and indeed the mass of k squar receives corrections of order phi dot squar over abble squared from this term and lambda k * k 0 squar from this other term and so we want both to be smaller than abble squared which we can plug back here to get an easier estimate to read for fn so we are trading I dot / M for abble and we're trading lambda kai k0 squar again for abble to saturate the bound and not get a big exponential suppression. So this f and l should be roughly smaller than again we're going to use powers of i dot and double to reconstruct powers of p of zed. We're going to get p of zed over lambda kai. Okay. So this is actually this is interesting and uh quite counterintuitive for me. So you're going to find that if you make this coupling very small you're getting a big signal. I still have to figure out the parametrics of this. All I'm going to say is uh well first of all obviously you make you have to make this coupling really tidy. Okay. Um if you want a big signal you are allowed to. Okay. So so I don't know if somehow I I find that uh some people already know this stuff even and some people don't even if it's everyone should know it. Yes. >> When you when you wrote down the estimate for the for that diagram. >> Uh-huh. >> You wrote down in detail and then a four >> should not be in d a four. All of that. >> Yeah. So this one this a to the four comes on this vertex and then each one of the other three gets an a cube because they have a derivative. So so the fi dot already has the a inside. So, so you have to remove one a per derivative that acts on the propagator. >> And then is it is it m squ? >> Oh, yes. This is m squ. Yes. Um everywhere. Um yes. So now I'm going to go back to like a basic fact of quantum filter theory which I I I find that it's it's less well known than it should be which I I learned from uh from the very nice BSM lectures of Ratati that he calls BSM for millennials because he doesn't know that I'm a millennial and and you guys are one generation removed from me. So he meant BSM for for Gen Z probably. uh but uh but uh aside from the title those are those are excellent lectures and one of the many questions that they answer in a nice way is is it natural to see a massive scalar and nothing else. Okay. So of course the answer is yes. Okay. So if you if you if you if you saw that this was the lranion of nature you wouldn't be bothered at all. Okay. It's a free scalar. there is no interaction that can correct this with loops and make it uh unstable under quantum corrections. But why? Okay, so what's the symmetry that's making it stable? It's not a shift symmetry because the mass breaks it. Okay, and uh and the symmetry is is uh it's easier to see in momentum space. Okay, so if you if you go to for space for the action, you're going to get something like this. Okay. And now you see that uh if you if you do this with alpha of minus p= minus alpha of p. This is a symmetry. Okay. You see it immediately. This is this is a symmetry with infinitely many generators because here you can put any function you want. So you can expand it into powers of momentum. The first term is translations but the other terms are just products of momenta or if you want products of derivatives. So this is a funny algebra where all the generators commute but you still have infinitely many generators. Um and the moment you add interactions the symmetry is broken. You can see it in many ways. Okay, you can just do it like add a a five to the let's say a k to the fourth term and check what happens. Okay. Uh notice well here I was using a complex scalar here I move to a real scalar but it doesn't really matter. Okay. Uh so you can just go to momentum space uh for this vertex and check that it's not a symmetry but it's actually way simpler than that. Okay the reason why it's a symmetry for the twooint function sorry for the twooint vertices but nothing else is just that the twooint vertx in momentum space has a delta function that looks like this. Okay. And so uh whenever you do this rotation if the function is odd the two phases cancel. Okay. But the moment you go to more momenta, you're going to have some delta functions that look like this. Okay? Times e to the i alpha p1 e to the i alpha p2 e to the i alpha p3. So some combination is going to cancel but never all of it. I mean so the delta function is just telling you this is alpha of minus p1 minus p2 but this is not enough to cancel the other two phases okay that that's it okay it's somewhat a trivial fact after you see it okay but but before you see it if you just stare at this lran the first thing to that comes to mind to justify why it can sit there in isolation is not going to momentum space okay so it's it's a one little fact about quantum field theory which is telling you that technically you can take this lambda as small as you want okay because as you send it to zero you are restoring this symmetry so it's technically natural to make this signal huge if you want then I let you decide how generic this is and how easy it is to UV complete but uh but in principle the couplings of the inflat on two scalers are potentially observable. Um, all right. So, we are not yet at the end of this chapter on what we can learn about particle physics, but we can already start to draw some conclusions. Okay. So what we learned from all these examples, all these rough estimates is that indeed this correlation function can say something about the new particles coming from the inflaton, but not always. Okay. So if you just throw at the cosmological collider, the first model that comes to mind probably so certainly you're not going to see it in the CNB you might see it in 21 cm line but uh well that depends on uh on what happens next uh in this old program. Um so um all right so so this is like a first let's say preliminary conclusion but the all these story with the temperature in the heater the thermal bath blah blah blah suggests that uh there are generic scenarios in which the signals might actually be very big okay much bigger than what we've estimated uh so are with gener in generic models. Okay. And uh I'm just going to going to state what these cases are and then I think we can call it a day because uh we are all pretty exhausted and I don't think it makes sense to start a whole new chapter like two minutes or whatever it is from from the end. Okay. So what are these cases? It's what I was alluding to before. Okay, case number one is a chemical potential. What does it mean to add a chemical potential where you're shifting the amonium by adding something like this? Okay, where this is a charge. So a nonzero value of mu is making it energetically favorable to produce particles of a fixed charge. And so again, if we use the intuition with the thermal bat, you're going to get a distribution that changes like this. And you see merely that if mu is much bigger than the mass, you might be in a situation where even if the particle is much heavier than the temperature, much heavier than apple, you might still see it with no suppression. So all these games we played so far, if you have a chemical potential, are not valid anymore. And you can get you can see a signal in the cosmological collider and be consistent with Laurel inflation with much bigger couplings to the inflator. And well tomorrow we're going to ask the question how generic is it to have a chemical potential? Where does it come from? And do the calculation a bit more precisely. The second option is is almost the same. Okay. And and it's parametric resonance. where your particle gets some time dependence on the mass and the time dependence is important enough that this parameter becomes bigger than one. Okay, so it's well known that in this under these conditions you can have an exponential growth in the number of particles. So you expect the signal to be enhanced and we're going to see that these two cases are almost the same. Okay, you're just effectively changing the dispersion relations of this particle and greatly enhancing their number density which makes them much easier to see. Uh and again to some extent we're going to comment on how generic this situation is and uh what signals you can expect. uh after that uh uh so this will allow us again to go a bit back and forth between uh the new language of correlators and the old language of particle production because these are two somewhat standard cases for particle production let's say uh and finally uh after that we're going to ask again about inflation. So what happens if instead of a single field you have more than one maybe one is driving the expansion the other one is generating the power spectrum and there again you can have big signals and then we're going to conclude and try to get some sort of uh overarching message out of all these examples. All right so thank you very much. [applause] All right very nice. Uh we have time for a couple questions. Uh I think my concentrated my concentration flipper a bit I missed the the the usefulness of that discussion about the thermal path. I kind of >> Yeah. So I mean the idea was just to give you another perspective on why if you compute uh these propagators for massive particles they get this e to the minus m over h let's say b let's call it boltsman suppression. So if you let's say if you open uh your any of this paper on cosmological correlators they will just solve the equations of motion for the mode function compute the propagator and find this factor. Okay. But I think and I mean the reason is that the story I told you about uh um a static observer seeing a thermal bat is kind of standard is very well known. So so they kind of assume that you know it but but I wanted to tell you explicitly. Okay. So it's just a simple physical intuition behind the form of these propagators. That's it. >> Yes. You just talk about the uh copings to scalas. Would there be anything intricating if I couple to formulas and also technically would there be difficulties because we are trying to introduce formulas to a curve space. So um so in this case there's not going to be any huge difficulty in introducing firmians uh for uh yeah so you essentially in this case you don't have a boundary where the firm can bounce off and change kality as it happens in some other spaces and creates a problem. So that problem will not be there. What's interesting about firmians? Well, essentially is the story with the chemical potential. Okay, so u we're going to see it tomorrow, but uh for scalers it's impossible to add a chemical potential that does something useful and it it it in the end boils down to Laurens invariance because a chemical potential that actually does what I promised is shifting the spatial momentum. Okay. So you can think of it as a shift in the special momentum which means that if you want to build a scalar so a lawren so an interaction in the lranion that gives you this chemical potential you need to dot this vector into another vector and so you need some non-zero spin. So you are dotting the momentum into into a spin. I I realize that for now it's very vague and unclear but tomorrow I'm going to write down everything. So I would say that the big qualitative difference between firmians and scalarss and also vectors and scalarss is that for vectors and fmians you can turn on a chemical potential that enhances the signal and for scalers you cannot. So so this is the main difference as far as the size of these signals go. Uh I'm not aware of any particular difficulty for the furbian. So you you at least at the level of computing this uh um three level correlator. So you can still solve for their propagator and do what we did so far. Yes. >> And the we computed uh in what channel I mean what shape of the >> yeah well I I kind of completely gloss over it because uh it it there is some amount of model dependence. Okay. The typical signal that is advertised for this uh for this particle for doing let's say particle physics during inflation. So the cosmological collider is to go to the squeeze limit where one of the three momenta is soft. The reason is that if you do that then u the analytic dependence on the momenta of these three point functions give you some oscillations that allow you to in principle extract the mass and the spin of the particle. Okay. So, so it's it's an FNL with a shape that is not one of the standard ones. It's not the local one, not the equilateral one. Um, it's it's one uh I mean, let's say it's a special one. And uh I wrote the a paper this morning uh which now I don't know where it is anymore. Uh let's see. So in this paper, in this paper they uh compute explicitly the old shape for some of the signals. I don't know if ex exactly the signals I I told you about but similar ones with scalarss and uh they do the analysis of plank data and they get a bound on the corresponding FNL of order 40 at one sigma. So that's that's uh I probably there are more than this paper but okay this is the one I I found right? Oh yeah. >> Even Yeah, even that. So you you can you can add a chemical potential but it's not going to give you uh an enhanced number of particles. Uh the simple way uh to see it uh let's see. So, well, I wanted to do this tomorrow, but but let me uh give you a a preview. Okay, since you both uh since you both asked um so let's say you you add a chemical potential for a complex scalar. Okay. Okay, this this is what a chemical potential for a complex scalar uh looks like. Okay, so we're going to take the lranjen and add this. Okay. So, effectively what you're going to get is that you shift the time derivative in the kinetic term by I mu. Okay. And okay, you still have the special derivative. Now, okay, there are two ways in which you can see that this chemical potential does nothing. Okay. One is that it's the same as the zero component of a gauge field that you can always gauge away. The other one is that you can do this field red definition and it disappears. So yeah uh yeah yet yet another way to see it is that if this chemical potential you see in fa space it amounts to shifting the frequencies. Okay but the zero of the energy is not a physical thing. Okay, so you're just essentially re re lababeling your tower of states by putting this chemical potential. Uh and so what you really would like would be some sort of vector version of this chemical potential that instead shifts the momentum and and to do that you need another vector to create a lawren scalar in the lranion. All right, great. Uh, good. It's a long day. We're all tired. Let's call it here. But come down and ask. He's not done. [music]