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

Thank you very much again. Um so so let me uh start by rectifying something that I I said yesterday. Yeah, Austin noticed that that I made a mistake at some point. So um when we were talking about single fields draw roll and we estimated the effect of a cubic vertex. So we had our potential and we were expanding it. So at some point I showed you a bound on this guy. um that then fed into our estimate of FNL. Okay. And the way we got the bound on this was starting from um this quantity. Okay. That I call the third draw parameter. But in reality the third draw parameter is this guy. So which and if you recall the usual definition it's the same as this. Okay. So this is the quantity that should be let's say smaller than one and so the bound on this quantity becomes instead of pz to the 12 pz to the 1/2* eta okay so about 100 times or so smaller then in turns in turn this means that fnl which is of order uh let me which is of order 1 / pz to the 12 over apple should not should not be smaller than one but should be smaller than ata okay so this makes it uh well in the end the qualitative conclusion is similar because yesterday we said it's somewhere in between large case structure and 21 cm this puts it deep into 21 cm territory but uh well in the end it will be hard to see. Um all right so having said this we can uh we can move on to something new. So yesterday we ended the lecture talking about uh what other way so what what can enhance the signal. Okay so what simple particle physics scenarios can give you a large FNL and I mentioned two options. So one is a chemical potential And the other one is a time dependent mass for for particles coupled to the inflaton. And today we want to understand well on the one hand how generic these possibilities are. So if they required some work on the particle physics mold the building side or not and on the other we want to understand how the enhancement of the signal works. Okay. [snorts] Um, so we already mentioned some of these yesterday. So we can start with just sketching the basic intuition. So a chemical potential is the same as shifting the Hamiltonian by minus some constant times a charge. And so it makes it favorable to produce particles of a given charge which in turn so in in flat space in a thermal bath is going to change the number density of a massive particle to from being exponentially suppressed to potentially being unsuppressed. Okay, provided that the chemical potential is sufficiently large and we will see that even though the effect of a chemical potential is not identical between flat space and the sitter in the end this intuition will work out. So we're going to find that uh the propagators of massive particles instead so in the limit where the chemical potential is much bigger than the mass bigger than abble some components of the propagator will scale well not exactly like mu over but they will be enhanced okay they will scale more like m squ mu over Um yeah um sorry minus okay but they will be greatly enhanced compared e to the minus m / apple um and for what concerns the time baring mass we're going to find something similar although not not identical because we're going to see that the chemical potential in the sitter is in practice introducing a time dependent mass. Uh so we're going to find in the case of the time dependent mass something that looks like this. So imagine that the mass of some new particles particle depends on the inflat for example which is the easier way which you can get a time dependent mass during inflation. Then we can expand it and well we have some sort of zero to order mass plus some coupling five dot and the time variation. Okay. And in the limit in which this term is large enough compared to the zero order term we're going to have again that propagators instead of being suppressed by m over abble are suppressed by this quantity. And so this is going to allow us to leverage the fact that phy dot is bigger than apple during inflation as we said as we said yesterday. So in practice it's going to allow us to produce particles that are much heavier if their mass varies fast enough with time. Okay. In practice this is already the answer. Okay. But now let's get to it and also let's see how these two conditions might happen in a particle physics model. Um [clears throat] yeah so first of all we're going to talk about the chemical potential and well so if you want a more detailed discussion about this point I think a nice reference is this one. So this is done uh in the language of correlators and they explore several ways in which you can add a chemical potential and they try to be general. They do it for firmians for vectors for spin 2. Yes. >> Sorry. >> Can I put this one? You mean? Okay. Sure. Like that. All right. Yeah. So, um I'm going to repeat some something that I said yesterday uh answering a question so that we have the whole story, let's say, in one place. So well the first thing that that you might want to do is uh take a complex scalar with some charge Q under some symmetry and then the current is just going to be the usual one. So k dagger k dot minus k dot dagger kai. Okay. And then you can add to the lranjon this so sorry this is the zero component of the current times mu. So this is the charge and mu is some constant. It's our chemical potential. Okay. So if we add to the lran this quantity we're in practice shifting the kinetic term of kai. So going to get something like this. And as we said yesterday, this chemical potential doesn't really have a physical effect because well for several so there are several different ways to see it. One is that it's the same as this time component of a of a gauge potential which as you know you can gauge away. This is the first way to see it. The second way is that you can do a field definition and get rid of the chemical potential without doing anything else to the rest of the action. And the last uh way to see it is that it's just shifting the frequency. So it's shifting the zero point of the energy uh and uh well this is not uh not physical. Okay, at least the same story carries over. So, so this is the simp let's say the simplest thing you can do for a scalar. The same thing happens if you do the simplest thing that you can do for a firmian. Okay, let me again add some firmium sai with some charge Q and then I can add to the lranion again some constant times the zero component of the current. So, so the charge of S and the exact same thing happens. So now you've again uh shifted the kinetic term but for exactly the same reasons we said for the scalar this chemical potential is going to do nothing. So again this looks uh like the time component of a gauge potential and it can also be field redefined away. Okay. Uh sorry. Um yeah. So again if we look at the third reason why this does nothing, we might get a hint for how a chemical potential can actually do something instead. Okay. So in this case what we did is that we took the dispersion relation for these particles and we shifted omega. So by adding the chemical potential in this simple way we just shifted the energy okay by either a positive or negative sign depending if you look at positive or negative frequency modes. And this we just said doesn't have an effect. Okay. But there's something else we could do. So if we could engineer a model that does this, then we would definitely have an effect during inflation. One simple way to see it is that this is introducing a time dependent mass because K is equal to the moving wave number divided by the scale factor. And so this gives you terms that look like so you're introducing a new scale in the problem. So that that is essentially the time derivative of the mass. And then from the old story of parametric resonance during preheating you do expect something to happen. Um so in which way can we do it? Uh well so we said that we want to shift the momentum but the momentum is not a scalar. Okay, it's a it's a vector under rotations. And so we have to find a way to build an invariant term that we can put in the lranion. And well, just from this very simple discussion, we see that uh it's hard to imagine something like this for a scalar because we have nothing to dot k into to build out something that's invariant under rotations. Okay, so we expect that it will be possible to add a chemical potential that behaves like this only for a particle with spin where we can construct invariance of this type. And indeed if you try to do the exercise explicitly and uh just write down all possible operators that uh that uh are proportional to something like this in the nor relativistic limit you find well let's say two obvious examples one for a firmian and one for a vector in uh in four dimensions. So for the firmion you're going to find something like this and for the vector you're going to find the trans Simon's uh current and actually both these terms are well this shouldn't be unequal but so both these terms are already well known uh in inflation as leading to enhanced particle production. Okay, so this is one of those cases where um a large signal at the cosmological collider let's say can be mapped into a very old story of particle production during inflation. Um so now I want to take one of these couplings and do the exercise a bit more explicitly. Okay. So add this to the lranjon and see what happens to the mode functions of the firmians to the propagators and to the signal that we expect in the three-point function. Um well so first of all so what what you want to do during inflation if you want that coupling so that that current to lead to a chemical potential well is the simplest thing that you can imagine. So if you add this dimension five operator to the lranjon you are going to generate a chemical potential from phi dot okay uh and well I'm going to do it for for a bile fiance. So so far I've written down direct fiance but uh the story is exactly the same. So now I'm going to go to two components. So for me the two component guy will be this and whenever I'm going to use it again but not much the dra guy with four components will look like this. So almost the same but not identical. Um so you can just write u uh the lranjan for the two component spinner in the usual way where this d mu uh is just uh capitalized to remind you that you you are now in a quasi deer background. So you have to take care of the metric factors in the derivatives and then okay there is a mass And then there is our dimension five coupling that in two component notation looks like this. Now we can uh use the fr FRW for the metric as usual and we find it's also convenient to rescale the FMAN. Okay. So if we do that then the coarian derivative becomes a normal derivative. the mass gets a factor of the scale factor. And finally, well, I'm going to put uh sorry, I'm going to put the inflat on to its background and generate the chemical potential. So where now this guy is just five dot over lambda and we're going to ignore the special derivatives of the inflaton that don't change the story qualitatively. Okay. So at this point we have all the ingredients. So we can take this lranjan solve for the equations of motion of of pi and see what happens. Okay. So as usual we're going to de compose p sai in free modes and okay we're going to work in conformal time. I didn't say it explicitly but uh well we are using all the same uh tools as yesterday. So these are the two elicities that the two component firm can have u and this is the lawrence index. These are the usual mode functions. These are the FIA coefficients that then become creation and annulation operators. And okay, I guess you're familiar with the rest. So as usual we need two um different mode functions for this charge firmian. And now okay we can just uh plug this back into the equations of motion and sol solve for the mode functions. uh before doing it it's convenient to write SI and Kai in a slightly different way. So essentially they compos it into um elicit states. So we're going to write XI as U times H. And similarly we're going to write kai uh wait sorry here I yeah we're gonna write kai dagger okay as the so where they carry and Elicity state where these H's are defined by the fact that the elicity operator acting on them just gives S times the momentum. Okay. Okay. After having done this, we can write the equations of motion. not going to derive them but it's a simple exercise that I encourage you to do and okay we're going to label the different components by plus or minus depending on the elicity and for the equations of motion you can check that one gets where prime is again a derivative with respect to conformal time you just get this Of course, you can check if you're familiar with the let's say normal equation of motion without chemical potential that this becomes the same when you send new to zero. Oh, sorry. Okay, the equations look relatively simple, but it required some work to solve them. I'm not going to do it for lack of time, but you can look uh at the paper that I brought up somewhere, maybe deleted there. Okay. So if you want to see how to solve them, you can look at this. Um, and the solution are a series of special functions of which we're going to be so but but we're going to mainly be interested in the late time limit. So I'm going to write for you the solution and then we're going to start again talking about the parametrics and how the chemical potential is making a difference. So the solution for you looks like this with a mu chemical potential over two apple. So this is already a good sign. So you see that the chemical potential is enhancing the mode function. And then here you have this with tucker function. Uh so which is a function. Well let me call it k tild because my k and my kappa look the same. U to the 12 and 2 I k to where uh mu to the 12 is equal to square roo of m^ 2* + mu ^2 / h and k tilda is equal to -2 - i mu / h. Um it's not immediately clear to see it from here, but uh the chemical potential has two effects. On the one hand, well, it's giving you some exponentials that compensate for the exponential suppression coming from the mass. On the other, it's also shifting the mass. Okay? So, it's not only doing good to us. It's also increasing the suppression from the mass. As I said, it's not immediately clear to see it from here. If you don't know how this W function scales at late times, but we're going to see it in a second more explicitly. So, so as usual there are four times more propagator than it would be in flat space. So, so there's a total of eight propagators. To write all of them, I have to write the solution for each one of the U's, V's, etc., etc. As I've done so far, I'm going to give you just uh one example because parametrically the unsuppressed one all scale the same. Some components of the propagators are suppressed, but we're going to care about the unsuppressed ones. And to build it, okay, we need we just we just need one more uh solution which is the one for u minus which now scales like this. Okay. And now we're going to construct the one propagator we're going to care about, which is D minus plus uh alpha beta dot. So you see there it's both the minus plus for the doubling of the fields in swinger calish in in the two sporial indices as usual and then this is going to depend on the momentum and the initial and final time as for the other propagators that we've seen so far and this is given by kai to one time sidagger time to Okay. And so now we're just going to write down the late time limit. So K to much smaller than one, which allows us to write this function in a way that is a bit more manageable. And this gives us e to the pi mtild where uh well actually there's no point in defining mild. So this is just new 12. Okay. And then there is gamma squared of 2 i mu to the 1 alpha apple uh and gamma of 1 + i mu - uh sorry well sorry actually Um well this is not new to the 1/2 but it's new to the 1/2*able. So let me just write new to the 1/2. Uh and so this is minus new to the 12 * all divided by apple and okay so this is a bit long but we're going to get to the physics in a second. Do not lose hope. So this is again move to the 12 times abble all over and okay then there is the elicity that we've chosen plus many other pieces plus new to the 1/2 that goes to new to the uh to minus new to the 1/2. Okay. So these other pieces are those with uh sorry this is H+ with other elicities. Okay. So this plus or minus er refers to the fields in swinger calish. Okay. So so we took up capsai minus and upsai plus okay but for each one of them we can choose the elicity. So this is this plus+ is the elicity and this other term that I didn't write are other choices for the elicities. Okay. But uh again all we care about is the parametric scaling overall which is well represented by this term. Um and we're going to consider the limit which is interesting for us. So the one that makes that gives the biggest signal which is the chemical potential dominating over everything. Okay. We're also going to take m much bigger than able to showcase the fact that even in this limit the chemical potential allows to have a detectable signal which normally wouldn't be the case. Okay. So we we can use uh the asytoic scaling of of the gamma function which goes like 1 / roo of y * eus / 2 absolute value of y for y going to infinity to expand this expression in this limit. Okay. And well, let me do it there. And you can check actually it's not too hard to check. Okay, you can do it almost by I by looking at the expression that this guy is going to scale as m over mu which will not matter too much. But what matters are the exponentials which look like this. So e to the pi square of m 2 + mu 2 over apple divided by so I'm doing I'm really keeping all the pieces so it's easy for you if you want to check to do the expansion yourself. So I complicated my life uh a bit by writing all the pieces but uh in the end uh this is relatively simple. So one can check from this expression that it just scales as e to the pi mu / h time e to the minus pi m new to the 12 which is the effect that I was telling you before. Okay. So this mu to the 12 essentially is just the shifted mass. So the fact that the chemical potential is also making our life a bit harder by making the mass larger but then it's compensating by adding this exponential enhancement into the number density of particles that carries over to the propagator but so this whole thing scales like this simpler product of exponentials and uh well more precisely if you simplify the expression you're going to get the following which you can then expand in the limit of interest to us to get what we discussed at the very beginning. So you're going to get this factor which in principle can be order one. Okay. If mu is sufficiently large there is no exponential suppression. This exponent goes to zero and you get that even for particles much heavier than abble you might be able to see something. Um I would love to do the same uh rough estimates of fnl as we did yesterday. Also in this case it turns out it's not so easy. Okay you cannot ignore the integrals as we did yesterday. I can sketch for you the reason why uh the reason why you would get the wrong result if we did what we did yesterday is uh at least intuitively what I was saying at the beginning that that when you add this chemical potential you are effectively adding uh a time dependent mass okay something that looks like this plus dot dot dot And uh [clears throat] a time dependent mass is doing something qualitatively interesting whenever m dot over m squared becomes larger than one. Okay. At this point you cannot neglect anymore the time variation when you do your calculation and you get some nonadiabatic particle production. This is let's say in the old fashioned language. Okay. And the region where this is true um in uh in momentum as uh so it's centered around k of order mu and as a wid of order k of order m. Okay. So when you do the integrals that we this that that always appear in these correlators you will find that they don't have support over the whole range but only in a relatively small region. So so you're going to get some factors some factor that is the volume of this region which is of order this. Okay, in practice, so the reason why this is the volume is that you're looking at a spherical shell of radius mu and width m. Okay, I realize this is very rough, but uh well, it's to give you an idea why the estimate fails. And if you want to see the actual calculation again I refer you to to this paper. Say my goal was just to show you qualitatively why the chemical potential is making a big difference. Okay, we saw it from the mode functions. If you prefer you can also do it via particle production. So really solve directly the equations of motion for s k instead of the mode functions and check what's the number density of pi. Okay. Um so once you realize that uh that this chemical potential can greatly enhance your signal then you can play all sorts of games and uh actually most of the papers that that claim to do BSM. Yes. >> Yes. So, so I'm uh yes, so so this this this goes back to to uh something I was talking about yesterday when we discussed the effect of derivative coupling. So you so in this in this game you have several scales. So the the biggest scale is the potential of the inflaton. Then you have phi dot. So and then you have able. Okay. So this is the hierarchy from the largest to the smallest and so uh if you forget about the inflaton but just look at the f of inflation with the goals of time translation you can in principle take lambda just a bit bigger than okay we don't we don't want to do that because it's very easy to find corners of this EFT that are very hard to UV complete where you might think you have a large signal but you don't. So in this lectures I didn't dare to venture in this territory. So minimally I'm asking lambda to be a little bit bigger than five dot to the 1/2. Um which is fine. Okay there there is nothing wrong with an EFT that looks like this. It's just an assumption on the UV. I mean it's telling you that at lambda there's a bunch of states that modify the derity couplings of five but do not touch the potential. If you really wanted to be completely conservative and forget about uh the structure of the UV and not be careful at all about the UV completion, then you could take lambda bigger than V2 the one quarter. But but uh uh yeah, also in this example I'm uh kind of focusing on this regime and and then I'm being generous and taking lambda very close to five dot to the 1/2. Yeah. So, so in the end my my chemical potential uh if you remember was phi dot over lambda and so it's going to be of order um sorry what am I saying? Yeah, it's going to be of order five dot to the 1/2 more or less. Okay. And that's why I'm allowed to to take this limit where it's much bigger than able because from the power spectrum I know that this can be of order 60. Yeah. So I'm cheating a little bit by taking it this really order one. Um okay. So, so, so once you um once you realize that uh that this chemical potential makes your life so much easier, then yeah, you can play all sorts of games and the vast majority of the BSM papers at the cosmological collates I found are literally just doing this. Okay, they're adding they're adding this coupling to the lranion for some standard model firmian and then they're saying okay now you can see all sorts of stuff um well I can mention a couple of uh of interesting uh um ideas along these lines but maybe before before doing that I should say that even if I cannot offer you an explicit estimate of FNL In this scenario that we just discussed in this limit of the FT you can easily get to FNL of order one or even a little bit bigger. Okay. So um without doing any violence to your model without doing any weird model building just by adding a sufficiently large chemical potential. uh so yeah so this is the starting point of various uh ideas one in this paper is that you might be able to see at the cosmological collider if the X has a deep minimum okay so so in the standard model maybe you you know maybe you don't that um well so the the X boson has some potential that looks like is and is squirty coupling of course runs with energy and there is a scale where it crosses zero and it actually stays close to zero for a while okay the scale it's it's around 10 to the 10 GV and so if you add this effect to the potential this effect of running to the potential, you're going to get some sort of field dependent value of the quartic. And if you plot this potential, so not just the three-level one, but the loop corrected one, you're going to get that it has a sort of shallow minimum near the origin, then a small potential buyer, then it goes down to a deep minimum close to the scale. Okay. So here here everything is of the order of the weak scale. So this is let's call it 100 GV to the fourth. The height of this barrier is also further 100 GV roughly. So the typical scales of the standard model but this this this deep minimum is at much larger scales. Okay. And uh the idea in this paper is that okay you can imagine that at during inflation you're probing these large scales. So you might have able big enough that uh it made this potential buyer disappear and the eggs roll to the minimum. If this is the case, well, as as you probably know, uh in the standard model, firmians get their mass from yukawa couplings to the x of this type. So, this is for an up quark. This is for a down quark. This is for uh electrons uh where I'm using again two component spinner notation. Okay? So you have to think of this as some sort of SI left direct firmion and this one is upsai dagger right and so uh the masses of the standard model firmians depend on the x vacuum expectation value through this coupling. Okay. So if you end up in this minimum the masses of the firmians become much larger than we observe them to be today. Okay. And something that I did not tell you at all, but I hope someone else told you in the first lectures. If you're able to measure these cosmological corridor signals and you're even able to look at their momentum dependence, you you can in principle um measure the mass and the spin of the particle responsible for say the nonzero value of delta fi cq of the correlation function by looking at some oscillations. Okay, so it is in principle possible if the signal is big enough to not only uh see an oceanianity but also end up concluding that it comes from the exchange of a firmian of a given mass for example. Okay. Uh and so well the whole idea behind this paper is just that if you add this kind of interaction for the standard model firmians you might be able to produce a signal large enough in the three-point function of the inflaton uh that you will be able to tell that you ended up in this minimum for the X. Okay. Um yeah. Um [snorts] similarly, well another idea along these lines is that uh if you add all possible higher dimensional interactions up to dimension six of the nutrinos to the inflaton and to the standard model, you might see nutrino signals at the cosmological collider. Uh this is discussed in this paper. which actually is the one where these guys took all the calculations of the mod functions of the firmians from. Okay. So, so this is actually the the real source at least within the BSM community that I could find of this old story of the enhancement from the chemical potential at least at least the through the mod functions. Okay, the fact that the chemical potential leads to an enhanced signal is some sort of old well-known fact. Uh and then okay, I could list many others but but the basic idea is always the same. You add this chemical potential. It allows you to produce particles that are much heavier than able and enhance the signal and maybe you see them. Um, okay. So, I think we can uh we can move on to to the time varying mass. Um, and as I mentioned at the beginning, the story is almost the same. Okay, because we've repeated over and over that a chemical potential is effectively giving you a time dependent mass. Okay. So, uh instead of doing the exercise through the mode functions, I'm going to go back to the uh Bolio uh picture that we had yesterday for the thermalb. Okay. So, you see more or less the same thing but in two different languages. Um and this one is in a so there was a reference I wanted to tell you about but I forgot to write it in the notes. So too bad. But uh yeah there is a nice paper by Senator Eva Silverstein and collaborators uh of 2016 where they discuss the story of the time dependent mass uh in way more detail that I'm going to do. uh but the basic idea is uh that you want to produce a mass that depends on phi dot and this in this way again you have access to this larger scale during inflation and you might be able to produce either heavier particles compared to what you would normally do just from the temperature in the heater or produce more lighter particles. Okay. So that's that's the basic idea and you can realize it again I mean in in the simplest possible way. So you can add a scalar with some coupling to the inflaton that give it a inflaton dependent mass and then well what you're going to do is again uh write the fa components of your scalar are and okay you can go the usual route of solving for the mod function. Um if you ignore abble expansion the equation of motion looks particularly simple something like this. Okay, where the fact that this frequency is time dependent takes into account what we were saying before. So we're going to expand the mass uh and have some uh some time dependent piece and well at this point you have several ways in which you can find the result. you can just go ahead and do what we did for the propagator or uh well as I as I did yesterday you can um do a bulb transformation. So let me uh yeah so one way to do it is solve this equation using the following answers. So you're going to do this. So sum coefficient times e to the minus i integral d to prime mega to prime to divided by k of to plus k star e to the i beta prime. K prime divided by sorry this is yeah K of T to okay so these answers will solve this equation but it will work well only when particle production is adiabatic so actually away from the interesting regime where the time derivative of this frequency is much bigger than the frequency squared but that's uh that's enough for us so if you are able to solve the equation using the answers then you can also reshuffle the effect of uh particle production from the mode function to the state. Okay, exactly as we did yesterday. So you can define so let me call this annulation operators in operators and you can define some out operators which are just um the bogio transformed in operators. Uh note that for this to work when you solve the equation you have to impose this condition in such a way that uh the commutation relations of our out operators remain canonical. But if you do that then you can repeat the exact same exercise that uh we went through yesterday. So you can uh move the effect of this time dependent mass from the mode function to the state and uh I find this uh uh useful because it's really recasting this this result uh of correlators in terms of some thing that looks a little bit like a thermal bat or or particle production and it gives you a different perspective as you were saying. So uh you can imagine that you started in the vacuum state of the inoperators at early times before particle production started and you can ask what does this state look like to an observer uh in uh that that sees the out operators instead. Okay. So we're going to define some new state n which is just some unitary transformation applied to the vacuum of the in operators. And if you want this transformation to bring you uh essentially in the same place where you were okay so to to keep uh what you're doing all the same you're just moving particle production from V to the state. Then as we said yesterday, you can use this relation to define S. And you can solve these equations to find an explicit form from S that looks like a squeezing operator. And finally well it's very easy just even using only this part to show that uh the number operator so a dagger A for the out operators in this state gives you the same result as the number operator of the in the original state. Okay. Well, let's do it for each fa mode which is nothing but the number densities of particles in the K mode which you can show now using the second relation is just beta absolute value squared. Okay. Uh sorry in so in the in so if you use the in operators the mode function looks like this. If you use the out operators the mode function looks much simpler. Let me call it out. It's just e to the minus i t dt. Sorry to v to prime okay to prime divided by square root of omega k of toao. Okay. So you see you reshuffled as promised the effect of particle production from the mode function to the creation operator. So in the out basis let's say you have a very simple mode function and a complicated squeeze state. In the in basis you have the vacuum but a complicated mode function. Okay. Uh I did it twice because I realized that yesterday was going quickly. So I hope that repeating it clarified some points. uh you can go ahead and do explicitly the calculation for a time varying mass which I'm not going to do it but I'm going to give you the result. So if you go ahead and do the calculation, you're going to get that this beta k^ squ is e to the minus pi mass + k^ 2 over gi dot. Okay. Where uh m I think I defined it. Yeah. So m is defined here. So it's the leading well it's not the leading term in the mass because we're going to go into a regime where this one is big but it's the mass that Kai would have if the inflaton at zero time derivative and then okay you can just compute the total number density of k by integrating over all momenta uh and not forgetting that there is a scale factor relating co moving with physical momentum And if you do that, you get a very suggestive form that really looks like a thermal bath. So from here you already start seeing that you produce in an appreciable amount particles that are not at the scale but are at this larger scale. Um to conclude this discussion uh I want to give you uh yet another way to do the calculation. So at this point you have all the ingredients to do the calculation in the standard way. So you can uh you can just compute the delta fi cube correlation function from all the machinery we saw yesterday uh by using uh either the in or out mode functions uh to write down the propagators for kai. Then check how they feed into the vertices with phi that will depend on the detail model that we did not specify. But you know how to do the calculation at this stage and you can choose okay you can choose to do it with the mod function in the in basis and sandwich your delta fs between the vacuum or you can do it with the mod functions in the out bas basis and use this more complicated state to compute the correlation function. But there is another way to do it which is again the oldfashioned way which is realizing that uh um well whether you do it in the in or out basis it doesn't matter but k square let's say out between our end state as a expectation value that is different from zero. Okay. And if you put it this back into the lranjen, it acts as a source for fi. Right. So we have our fi dependent k mass which now we can expand for five around its background and well we're going to get the usual sorry this is minus the usual mass for kai but then we're also going to get a term that looks like a sorts for the fluctuations of the inflaton. Well, this sorts is just the expectation of Kai in the relevant state times a derivative to fi of the kai mass evaluated at the background. And uh in principle you can start uh from here instead of doing the the the calculation that we sketched yesterday and just solve for fi uh using greens functions. Okay. As usual, when you have a source, your fi is going to be the convolution of the greens function with um your source with some uh okay with with as usual don't forget the scale factor. Okay. So if you do this and then plug it back into delta 5 cube, you're going to get the exact same result as if you do it with all the machinery that we discussed in the past two lectures. Okay, so so this kind of closes completely the loop. Okay, yesterday I I told you how to go in the particle production picture uh up to here, but then you still needed to use the same machinery to compute the correlation function. If you don't want to and really want to go back to the 90s, you can also instead do this. Okay. Really do it fully particle production style. Yes. >> Sorry. Like >> which one? Sorry. >> This one the end out bracket by >> this one you mean? Or uh let me see. So I might I might have inverted them indeed. Uh so this is s dagger omega in. So this is a dagger a in which is a dagger out. How did I define? So s dagger. Okay. This goes away out and this is s uh no I think it's uh I think it should be okay. The in defined by the in operator >> sorry >> the in vacuum is defined by the in operator. >> Yeah. Yeah. Yeah. Uh yeah. Sorry. Sorry. You're you're right. Yes. I actually I think I >> I have inverted I have inverted the S and the S dagger. Okay. Yes. That's that's right. [snorts] Um yes. Sorry about this. Um so now let's uh let's try to draw some conclusions. Okay. Yes. >> I was just wondering what happened to the >> M. >> Oh, this guy. >> Uh oh, sorry. This is an M squ. So that's uh so yeah. Yeah. So so indeed it's carrying through. So uh the point is that instead of going like e to the minus m over abble it's probing this bigger scale. So indeed it is it is suppressed if you make m heavier than the kinetic energy of the inflaton but uh yeah but you're gaining this factor of 60 which is not bad. Um okay so um let's see so what what what is the take-home message? Um well we've seen that uh maybe I should write our lessons down although they are lessons that we've drawn from a few examples. So nothing uh is guaranteeing that someone very clever cannot find something better than we found. But uh I would say that in general so for generic couplings it's not so easy to see particles coupled to the inflaton. Okay. So you need to be in some sense uh well not only lucky because they have to have a mass comparable to able but also for most couplings the signal is going to be small. So, so roughly this is my take-home message that if you have so, so this this is in answer to the question, what can we learn about particle physics? Well, let me call it high energy physics so I can abbreviate it. Okay, what can we learn? So, gener typically you can probe them at 21 cmters, but it's it's not guaranteed that you'll be able to probe them before. Okay. However, uh there are some couplings coming from parity odd interactions. which lead to large signals. Okay, you can ask yourself how likely it is that these couplings are there. I mean the answer is that uh if we our generation had understood anything at all about nature much of our our troubles would would not exist. Okay. So I I don't I don't uh consider myself able to understand what nature is thinking after well all sorts of things. Okay. So after flavor after the proton didn't decay after George I predicted it after not finding any explanation for the X mass or the CC values anywhere. So there's nothing wrong with these couplings. They might be there. How likely it is it's a question beyond our capability to answer at the moment. Um [snorts] there is so another generic way to enhance the signal is to have m dot big enough that m dot is different from zero is generic in inflation if you couple some particle to the inflat because of the background. So this is generic but then if this is time variation is big enough then you can again have large signals and these were both scenarios that were known well let's say since the 90s at least or or uh as ways of enhancing uh particle productions during inflation. [snorts] Um so let's say these are not much of general lessons. uh we have seen essentially that uh we can probe uh um both the structure of inflation so the potential of the inflaton and higher dimensional operators with derivatives if we can get to FNL essentially smaller than one and the same remains true for other particles coupled to the inflaton and there are some special cases which do not require enormous model building gymnastics that can give you much bigger signals. Okay, so um you can let's say get to order one in these cases. Um so I still have some time I guess, right? Okay, >> right. Yeah, maybe I'm not even going to need all of them. Uh may maybe I'm lying, but okay. [laughter] So now I want to change gears and kind of go back to the first question we asked which was what can we learn about inflation only. Well, yesterday we asked the question with this important caveat in the middle. Today we're going to ask the question by dropping it. Um maybe before getting there a disclaimer about this. Okay, the so this this list of conclusions comes from my imperfect knowledge of the literature. So as I was saying, it might be that one of you comes up with a clever idea and that's one more case where the signal is big or well I just missed a paper that already existed. So don't take it as a comprehensive list of conclusions. Uh but okay having said this we can move on uh to the second question. Um and there um I'm going to [snorts] give you just uh one example but uh which kind of gives you an idea that the moment uh you're willingly to modify inflation at order one then you can really get huge signals that you might even see in the CMB. Okay the problem well I mean to some extent we already discussed this yesterday. Okay if you have massless particles floating around during inflation going generate huge signals. Now if you want the whole game behind answering this question is how can I have a master's party consulting around during inflation that doesn't screw up completely the twooint function. Um and okay I'm going to show you one example. There are others which for which uh I I have not seen explicit models but the way they were described to me uh make made my art uh I mean like made made me feel some pain because they they they are they kind of fall into this class of ideas that are also used a lot for primordial black holes where you just change the inflat potential however you want and forget about tuning. Okay. So if you change the infotton potential however you want in such a way that you make the laural parameters very big but only in a small field range then you can get big signals. Okay but typically okay if you have a potential with that has inside the three scales that are very different from each other you're tuning usually. Okay. So I instead of talking about that those that class of examples uh I want to talk about another option which is if we imagine that there is one field which again I'm going to call phi which drives the expansion of the universe. Okay. So this guy is the inflaton and dominates abol. Okay. So abble roughly scales like square root of the potential of this five. But then we're going to have another field still a scalar that we're going to call curve aton. And this guy dominates uh the fluctuations. Okay, this seems uh hard to do and very dangerous. In reality, it's not so hard to do. Uh and you can easily fit observations with this model. Uh I'm going to for simplicity assume that the two scalarss are decoupled. Okay, in general it's it's not uh super easy to do it in a technically natural way because there are always terms that are neutral under all symmetries that can couple the two scalarss. One way in which you can imagine that this happens is if these two scalars for example are localized in an extra dimension and are very far away from each other. Okay. So, so their wave functions have very small overlap as a consequence of maybe the fact that this extra dimension there is some warping. So you have some ads profile and effectively these two scalars don't know about each other because of that. Okay. But but this is just a technical assumption made for simplicity. Okay. So as I said we're going to imagine that inflation is driven by fi and so is dominated by fi and also this laural parameters are determined by fi. Okay. Um but we're going to ch choose these lower parameters such that the curvature perturbation generated by fi is very small is much much smaller than what you see in the cm of or let's say 10 to the minus 4. So if there was only fine the universe you would see no power spectrum in the CMB. Um and then we're going to imagine that uh well phi rolls down its potential for a while then it arrives at a at the minimum starts oscillating and decays to say standard model particles. Okay. So standard end of inflation type of scenario. The only important point is that while this is happening sigma is still rolling down it's very flat potential. So at the moment in time say this is some time I don't know well what we call toa final the end of inflation which we set close to zero in our previous calculations. Fi is decaying but sigma is still rolling down its potential. Okay. Um and so well at this point but but even before okay so at at all times in this in this model you're going to have two uh contributions to to our curvature which okay you can you can easily compute if you want from the formulas we we uh discussed yesterday and it looks like this. Okay. So at at every point there is there are two contributions to zed. One from sigma and one from the inflaton. What I've written here is really valid at this time when the inflaton has already completely decayed to radiation. Uh and this f sigma is this. So I'm saying that this is valid only at this time because to get these factors of three and these factors of four I had to take derivatives of the scale factor for an energy density in sigma that is that is red shifting like matter. I'm going to show you in a second why and an energy density from phi that is red shifting like radiation. So row sigma is decreasing like the scale factor cube and row radiation is decreasing like the scale factor to the fourth. Okay. Um if you haven't if you never seen this before uh the reason why I'm saying that the energy density of sigma is red shifting like matter is actually the the consequence of just solving the equation of motion. So if you if you compute the equation of motion for sigma in an expanding universe you're going to get as usual uh this okay and now take now but now maybe you guys I mean given your interest you guys tend to think about more of the case of when this is the inflaton and abble is dominated by the potential okay but now you have to instead think of the case where Abel doesn't know anything about sigma. Sigma is a small subleading contribution to the energy density and abel is dominated by radiation. Okay, so now we are again at this time where the infatonas decay to radiation. If you solve this equation, okay, it's easy. So it's just a dumped harmonic oscillator and you're going to get that sigma goes like um a to the so scale factor to the minus 3 alps times well let's say that the potential is dominated by the mass okay so this is just m^ 2 sigma times some oscillations okay and then if you plug this back into the potential you have an energy density that is red shifting like 1 / a cube. Okay, so this is for for those of you that have not taught a lot about axons and dark matter, this is the reason why axons can be dark matter. Okay, so a very light scalar obsoar set free into a radiation dominated universe behaves like matter. Okay, by by set free I mean so so this of course depends on initial conditions. Okay, so you have to imagine that the sigma was initially displaced from his minimum. So it has some non-zero amplitude and then you get these oscillations okay or that it had some initial velocity otherwise this I mean a solution a perfectly fine solution to this equation is also just sigma equals zero okay so you get something non zero that red shift like matter if you have some initial displacement but it's only natural to expect some initial displacement because we're going to imagine that the universe is leaving at energy scales that are much bigger than the sigma potential so even just abble flux fluctuations of sigma. So the usual fact that a massless field gets fluctuations of order abble it's going to move sigma far away from its minimum. Okay, because we're going to assume well one assumptions that I didn't say but we're going to make is that abble is much bigger than the mass of sigma and yes just one second. Uh so one more thing that we're going to assume that I didn't say we're also going to assume that sigma has only the mass in the potential. Okay. So it's approximately shift symmetric and the only breaking of the symmetry is soft by the mass. Again this for simplicity. Yeah. Sorry. >> Yeah. Sorry. In that picture there was a flat potential and the sigma was also rolling but now it's oscillating a bit. >> Ah ah sorry. Yes. Yes you're right. Uh indeed. So, so uh this this case where it's oscillating, it's where uh it started this place from its minimum and it it had the time to uh go back and and forth. But um for for us we're going to imagine that the time elapsed as is not enough that he has reached this minimum and starting oscillating. So effectively we are we are expanding this cosine. Okay. And we only see the one. So we are in the slow roll regime indeed. Yes. Sorry. Um so so at this time and for a while longer sigma is so let's say its minimum is somewhere very far away and so it's still getting there. So it it is it is oscillating but it doesn't know yet that it is oscillating. It's still in the first uh part of the first swing. Let's say um okay and so um yes so so okay you can you can solve the equation of motion in this regime where the the time is much the time elapse is much much shorter than one over the mass. So, it's still slowly rolling. And if you do that and and plug and plug back the the solution into the potential to compute the energy density, you can also compute zed and you're going to get this where sigma c is again the classical value of sigma that it has at this point. We're imagine that is slowly rolling. So it's not moving a lot. The reason why uh this whole expression that contained radiation reduced to this is that we're actually evaluating it a bit later. So we waited for enough time that the energy density in radiation became negligible compared to the energy density in matter because it was red shifting faster. So if you wait long enough and sigma continues to roll uh slowly then the curvature perturbations will be dominated by sigma as I said on the scale of abble sigma is a massless field so if you compute the power spectrum of the curvature perturbations coming from sigma you're going to get the same result that we got for the inflaton on the first day so sorry yesterday the Only difference is that uh uh because uh yeah so the the energy density depends on where sigma is in the potential rather than on its derivative down here at the denominator. You're not going to get sigma dot but you're going to get just the value of the field. Okay. So you can do the calculation and convince yourself of that. And so this result is telling us that we can fit the CMBB no problem provided that H over sigma C is of order 10 to the minus 4. Okay. So this is what the CMBB is telling us. uh if we want to be completely consistent with CNB observations, we have also to compute the um the tilt and this uh I'm going to give the result without proof is - 2 epsilon from the inflator plus 2/3 m^2 / h^ 2 and so this implies that this m 2 over h 2 be of order of 10us three again if you want to fit the CMBB properly. Okay. So, uh yeah, I I didn't go through many of the calculations for lack of time, but they are relatively simple. I think they're a good exercise to do once you have the picture in mind. Um and at this point, we can check what happens to the by spectrum in uh in this scenario. So first of all we're going to compute the by spectrum of sigma and not of the inflat because it's sigma that's dominating the quantum fluctuations. Uh so let's check one of the examples that okay maybe I should not cover this board. Uh let me let me leave it there and delete that one. So let's now uh consider one of the scenarios that we considered before but in the in this different cosmological model. So let's take this sigma to the 4th over m to the 4th and check the effect on the threepoint function of sigma. Okay. Well, we can do the exact same exercise as yesterday. We're going to have a vertex. uh there are three derivatives so only one power of the scale factor survives and we're going to put one sigma to the background. So this is the vertex. Okay. So this is the usual diagram that we drew yesterday. So we put one leg to the background and three legs to infinity. So to us observers um and these other three legs give three derivatives acting on three propagators. I remind you that we can just estimate. So we can just use the scaling of the propagator and that we can rescale all conformal times by the momentum to get something dimensionless and this gives us uh the following estimate. So we have 1 / abble from this scale factor. We have abble to the 6 from the propagators, k to the 9 from the propagators and k cube from scale factors integration measure and derivatives. And finally, okay, we have sigma dot over m to the 4 from the vertex. And so we can again plug this in into our formula for estimating FNL which is just K to the 6 delta sigma cube. Well again this is delta sigma Q prime. So I factored out the delta function divided by pz to the 12 uh times able cube. And if you do that, you're going to get roughly sigma dot sigma c h over m to the 4th. So what can this? So what's the maximal size that this can be? Let's see it there. So first of all to get the final estimate we used uh this result. So the fact that P pz to the 12 is H over sigma C in this model. Uh then we can use the fact that sigma is effectively massless to estimate this kinetic energy to be of order squared. So this is a consequence of the fact that the mass of sigma is much smaller than abble during inflation. Uh and finally uh we want to choose m okay this time since the signal is very big as it is we're going to be conservative and take the cutoff to be bigger than the biggest scale in the problem. So the potent the inflaton potential to the sorry the the sigma potential to the 1/4 uh and this is of order 10^ the 3/4 time. The reason being that the sigma potential is of order m^2 times sigma c^ 2. And uh we said that sigma c is of order um is of order * 10^ the 4 from the power spectrum and that m squared is of order 10 theus 3 squared from the tilt. Okay. Uh and finally, uh well, finally, that's it. So, we can just take all these numbers and plug them back into here. And we're going to get the biggest number so far. Okay? Or a resounding 10. Okay? as as usual in BSM when you get a fantastic result is because you've cheated. Uh but but we didn't cheat that much. Okay. So the reason why it's so big is that we took sigma C much bigger than the cutoff. um which in principle it's fine as long as gravity is not dynamical which is the case in in all we've done so far. Um the EFT is still under control because we took m bigger than the potential. Note that uh the potential for sigma is much smaller than the potential for the inflat. Okay, but the two sectors are decoupled. That's why we are allowed to UV complete our uh theory for sigma at a much lower scale than where fi lives. Okay? Because these two guys do not talk to each other. So this is allowed. There's nothing wrong with it and it makes the effective theory for sigma consistent. You might be worried about large field excursions and you you would be right. So if we take m of order m plank then this becomes a problem. But if we take m much smaller than m plank in principle everything should be fine. Okay. So gravity doesn't like to be coupled to field excursions much bigger than m plank. Whenever you try something goes terribly wrong. Typically you're starting to decompactify some well when I say gravity I mean string theory. Okay. So when you do it, you you you start decompactifying extra dimensions. You get a bunch of kk states coming down. Your effective theory completely breaks down. But if you're happy with taking these uh scales or much smaller than mplank, even if we've cheated a little bit, we're still within the regime of validity of the f. Okay. So uh I think it's time for me to conclude. Um well in these lectures we've seen um a quick and dirty way to estimate cosmological correlators and get an idea of their size and then we use this technique to ask all sorts of questions about what you can learn about inflation and particle physics looking at these cosmological correlators. The bottom line is that uh this is the only thing that exists that even in principle allows us to probe scales much above 10 to the 10 10 to the 12 GB. Okay. Uh but there is no free lunch. Okay. So uh you're not going to be sensitive to any possible theory. So if you typically write the first thing that comes to mind a scalar couple to the infant with a cortic coupling well either you get to fnl so for their 10 to the minus two or so or you're typically not going to see anything. So you can be an optimist and believe in our experimental colleagues or just simply forget that there are humans as many theorists do and that they're going to ruin their lives even just to get to one. Okay. and and just say if I wait long enough we're going to see all this and we're going to know what's the inflaton potential looks like. We're going to know if it has derivative couplings. We're going to know if there are other particles coupled to the inflaton or uh you can be a pessimist and think that they will never get to subtract the foregrounds and maybe we're going to see nothing. Okay, I'm not gonna answer this question for you. I've given you all the numbers and I let you decide for yourselves. So, so the question we asked at the beginning was what happens if the sitter and BSM make a baby? Well, it happens that it inherits some of the amazing features of the sitter meaning that you can probe very high energies that you cannot do in any other way. But it also inherits some of the heartbreaking features of BSM which which is that there essentially there is no free lunch. Okay, the moment you really get close to the data and try to build a model that really works is natural and checks all the boxes, it's not so easy uh to see at least uh uh in the CMB. So well overall I would say that for my standards we are ending on a positive note. So well let's go out there and measure these correlators. Thank you very [applause] much.