 So we will use, to compute perturbations in this case, we will use a rather commonly used gauge choice. We will fix our coordinates in the so-called zeta gauge or uniform density gauge, which is defined simply by using the scalar field as our clock. So I imagine here's the global picture of space time, time and space, and we're foliating the space time using surfaces of constant phi. So these surfaces, these space-like surface, they all have phi equals constant. In other words, the choice of gauge is such that the perturbations of phi are zero. Okay, we can always, we have two degrees of freedom for scalar diffeomorphisms. I've used one to fix the clock to be phi and I can fix the second amount of freedom that I have to fix the form of the metric on each of these slices. So Hij will be the spatial metric on each of these slices and we can use the remaining gauge freedom to set this spatial metric to be of the following form. So the scalar perturbation will be simply a conformal mode, which we call zeta. And then there will be tensor modes associated with gravitational waves and these we can write as e to the gamma ij where gamma is transverse and traceless. So gamma ii is zero and di gamma ij is zero. And meanwhile, the other degrees of freedom of the metric, namely the lapse function and the shift vector are fixed by the constraints of general relativity. So we've completely fixed the gauge with this particular gauge choice. Now it's true that this completely fixes the gauge, but in the more precise sense as follows, it fixes the gauge completely if you consider difumorphisms that fall off at infinity. If they fall off sufficiently fast at spatial infinity, then you cannot do anything other than what I wrote down. But there is room for gauge transformations that don't fall off at infinity and this is what we'll focus on. So let us indeed consider the residual symmetries of this zeta gauge story. So here I'm gonna follow a treatment of a following paper, Interbickler, Huey, and myself, 2013. And so we're just gonna proceed. We're gonna consider possible difumorphisms, so small difumorphisms, call them C mu of x and t that preserve this gauge. And these we will identify as symmetries and they will correspond to some constraints on our correlation functions or ward identities at the end of the day. So first of all, we wanna preserve the fact that phi is purely a function of time. We wanna keep the fact that we don't wanna change our foliation, phi should remain the clock. And so as a result, clearly the zeroth component, we don't wanna redefine time, so the zeroth component must be zero for this diff. So all we can do is to do spatial difumorphisms to change the coordinates along the slices. And so we're left with xi. So this is all that we have at our disposal, xi of x and t. And so the goal is the following. If we manage to find field variations, so specifically we wanna find field variations, delta zeta and delta gamma ij, if we can identify such field variations in such a way that these guys mimic a difumorphism, then this will be manifestly a symmetry of the theory. So in other words, what we wanna do is, we wanna identify some field variation of the spatial metric, which we wrote down. We want this guy to be a lead derivative with respect to xi of the metric. And I'll write down what the lead derivative is in case you've forgotten. So where the lead derivative of an arbitrary tensor tensor gij is given by xi k dk gij plus di xi k gij plus dj xi k gij. Sorry, this is supposed to be k, okay? So the idea is the following. A lead derivative of course is just what a difumorphism would do. So if we can find field variations in such a way that delta zeta and delta gamma exactly mimic how a lead derivative would act, then these will precisely correspond to, they will act exactly as a dif would. And since our action is assumed to be difumorphism invariant, this will be a symmetry of the action. So for every delta zeta and delta gamma that satisfy this property, we will have a symmetry of the theory. Now this is actually non-trivial to solve in particular because of gamma, okay? So gamma being transverse and traceless. The equation is non-linear in gamma. This is hard to solve. So we will do so perturbatively in gamma. So solve order by order in tensors, okay? So in other words, what we want. So I wanna expand everything in terms of tensors. So namely the dif itself, xi, this difumorphism will be a leading order difumorphism without tensors plus a correction that's linear in the tensor plus dot dot dot. And similarly, the variation of zeta itself, the field variation, the symmetry variation also can be thought of as a, will be perturbed in this fashion. And finally, even delta gamma itself, it may have a piece, it will have a piece generically which is 0th order in gamma plus et cetera. Now for this discussion, we're not gonna actually go beyond 0th order, okay? But I'm just at least outlining what one should do. And in fact, in the paper, that's what we do. You can compute these corrections order by order. For our purposes, we're just gonna stick to the 0th order in gamma. So to 0th order in gamma, let's expand both sides of this equation. So on the left-hand side, when I do the variation, so first of all, I'm working to 0th order, but I have to be careful because delta gamma, when I'm gonna take delta gamma, this, they can have a 0th order piece. So I don't wanna set gamma equals to zero from the get go. So we have the following. So let's do it. So by the chain rule, I get, I can vary the zeta piece. So I will get twice delta zeta e to the 2 zeta. And then here I'll have e to the 2 gamma, but now I, sorry, e to the gamma, but now I can set gamma to zero. So this will give me delta ij that comes from varying with respect to zeta. Now I vary with respect to gamma, and then I set gamma to zero. So this will give me plus e to the 2 zeta delta gamma ij. Notice I've dropped the subscripts, the superscripts gamma zero, gamma one, because it's all gonna be implicitly to 0th order and gamma, henceforth. So that's the left-hand side. And the right-hand side, we just have to compute this lead derivative. But the lead derivative, this guy for sure, since I'm working at 0th order and gamma, I can set gamma to zero from the get go. So here's what do you find when you turn the crank. You find e to the 2 zeta. If you just compute this lead derivative, it's really straightforward. You get twice ck dk zeta, plus di cj plus dj ci. Okay, does that make sense so far? Yes? Okay, so the factors of e to the 2 zeta as they cancel across the board, and this is our final statement. So, and then we can isolate delta zeta and delta gamma. So we can do this as follows. Let's take the trace of this expression using the fact that delta gamma, being a tensor mode, has vanishing trace. So this guy is zero. So if we use this fact, I'm gonna be left with a six delta zeta from the trace of this guy. And then on the right-hand side, sorry, there should have been a delta ij here. The delta ij. This will give me also six ck dk zeta. Zeta plus twice dk ck. Cancel the factors of six, makes it a third. And that's it. That's the variation of zeta. And then we can plug it back in to the equation and then solve for delta gamma. Can plug this in and then we can solve straightforwardly for delta gamma. Here's the answer. After substitution, you find that delta gamma will be equal to dixj plus djci minus 2 thirds. Because I have here, I have two. Sorry, but you gotta divide by two. So I'll have one minus a third. Anyway, it comes about to be minus 2 thirds. Yeah, it's two minus 2 thirds. So it gives me at the end of the day. Minus 2 thirds, dk ck delta ij. We're about to do that. That's right. We're about to do that right away. So so far, it seems arbitrary. And now we will derive the one equation that it must obey. Okay, very good. Exactly, that's right. That's right. So I'm looking, in other words, precisely. So if I just did a random generic diff on this, I would start messing up with this form. Indeed, I would pick up some didj of some scalar. So I don't want that. So I'm not gonna consider the most general diff. I'm considering a subset of diffs which preserve this form, meaning if I can associate a field redefinition of zeta and gamma for those diffs, then by definition, I'm staying in the same gauge. So indeed, I'm not generating anything else in the spatial metric. All I'm doing is redefining the fields that were already present. That's right. But you're right. Also it is true that I haven't said anything yet about xi itself, but we're about to say something right now. Very good. So the one constraint we have not imposed is that delta gamma remains transverse. So in other words, the shift in gamma, the shift in gamma should be such that gamma remains a tensor mode. And so our remaining constraint is transversality. We wanna take the divergence of the delta gamma variation and wanna impose that this is zero. So that the divergence of the shift of gamma be zero, so that indeed it remains an honest to goodness tensor fluctuation. And this you see imposes the following. So if you take the divergence of this equation, you find Laplacian of xi plus di, sorry, this should be j, excuse me. This is then di, dj, di, ci. That's the second term. And then the last term minus two thirds, the same thing, dj, di, ci. Okay, so in other words, our defimorphism xi is constrained to be satisfying the following equation that the Laplacian, the spatial Laplacian on xi plus one third divergence, gradient of its divergence must be zero. Okay, so that answers your question. So any diff which obeys this equation does a job. Okay, so now we're gonna study a little, yeah, yeah. No, like anything else in perturbation theory, it's probably asymptotically, yeah, asymptotic convergence. In any case, for our purposes, we think the small parameter in this expansion will be the, if you want the amplitude of tensor modes, which is very small, okay, so that's our parameter. So to that extent, these are all small, these are all small, you know, these are all small corrections, but it's an asymptotic series like anything else. Okay, now I wanna prove for completeness, let us convince ourselves that indeed, if I impose that xi vanishes at infinity, then I get no new solution. This will convince ourselves that for vanishing diffs at spatial infinity, our gauge was fixed completely. So indeed, let's suppose that xi, i, our diff does go to zero at spatial infinity, and I wanna convince you that in fact, the only solution in that case is trivial. So let's see, let's take the divergence of this equation. So if I hit this with a divergence, what I find, I'll find Laplacian on the divergence, plus same thing, a third of Laplacian of the divergence is zero, okay, so in other words, the Laplacian of the divergence is zero, and since now I'm assuming things fall off at infinity as usual, I can invert the Laplacian trivially. This implies that the divergence is zero, yes, but when you plug it back, sorry, when you plug it back to the equation, you're left with Laplacian of xi being zero, so this implies Laplacian of xi is zero, and again I invert the Laplacian trivially, and therefore xi is zero. Okay, so this confirms that indeed there's nothing new if I just focus on this that fall off at infinity. So non-trivial solutions, non-trivial diffs don't fall off. So first let's consider, let's first consider the case where only zeta is transformed to start with, so consider the case where, in other words, we wanna focus on the scalar sector of these diffs, so we're gonna impose that the tensor mode doesn't change whatsoever. So we're starting with gamma equals zero, we're not changing gamma, it remains zero, so these symmetries we're about to identify are symmetries of zeta itself, nothing else. So in this case, imposing that delta gamma is zero to remain in the scalar sector gives me directly an equation for xi, which is stronger than the one we derived by taking the divergence of the equation clearly, because I'm not taking the divergence, so we get a straight equation for xi, which is the following, di xi j plus d j xi is equal to two thirds d k c k delta i j. And of course, by taking the divergence of this, I'll reproduce this guy, but now it's a stronger statement. So who recognizes this equation? It's a well-known equation. Excuse me. It's a conformal-killing vector on R3, that's right. So this is the conformal-killing equation on Euclidean space, on Euclidean three space, and this makes sense intuitively that this is the answer we found because after all, if I'm setting the tensors to zero, our spatial metric is just this guy, so the only diffs that are left over that will keep me in this gauge will be conformal transformations on R3. Conformal transformations on R3 will pick out a conformal factor, which I can absorb just by shifting zeta appropriately. That's exactly what's going on. Okay, so that's obvious at posteriori, and so the solutions are dilation. Well, of course, there's a 10-parameter group. We have, let's just say that, to start with, we have rotations and translations, but those are isometries of the background, of our background, spatial translations, and rotations are isometries, and so they're not gonna do anything. We know how they act on correlation functions. Correlation functions would be strictly invariant under those two symmetries, so that's not what we're interested in. We're gonna be interested in the dilation and the special conformal transformation part. Okay, so let's write them down. So we have, for dilation, this guy is just a rescaling by lambda, and the special conformal transformation is the following thing. So this is the infinitesimal diff, just 2v.xxi minus x squared bi, and we can trivially now compute how delta, how zeta transforms under these two transformations, and this will make contact with our discussion yesterday. So we find that, so by plugging in these diffs into the variation of zeta, you find that zeta under dilation, it's 1 plus x dot grad zeta, and under an sec, it's given by 2b dot x, plus 2b dot xxi minus x squared bi zeta. So now let's compare this with what we discussed yesterday. So yesterday we wrote that a transformation of a field, let's consider dilation. In that case, we were thinking of this as a spectator field. We wrote that under this conformal group, we had the following transformation. It was delta plus x dot grad phi, okay? So now I ask you to compare this expression with the one for zeta, and tell me the obvious differences. It's obvious, but I want you to say it's not me. Delta, so what is delta for zeta? Zero. So indeed, there is no term which is just zeta here, it's just x dot grad. So as a result from this immediately, and this will be important for later, the weight of zeta is zero, okay? Which makes sense, it has mass dimension zero, it's a massless field, so perfect. It has weight dimension zero. And what else do you notice? You notice the constant bit, okay? So you see that here there's a constant, not multiplied by zeta, whereas here it was completely multiplied by phi. So by definition, this is known as a nonlinear transformation. It's not linear because the answer is not proportional to phi. It's just a constant shift. So you see that zeta transforms nonlinearly, both under dilation and under SET. By the way, this jargon of particle physics is rather confusing if you're not used to it. It's still, it looks like a linear transformation, okay? But it's nonlinear in the sense that if you could think about the number of particles in each state, so here you start out with zero particles and now you have some particles, because you've given a VEV to zeta, okay? So that's a nonlinear. And a nonlinear transformation of this type is the hallmark of a spontaneously broken symmetry. So this tells us that dilation and special conformal transformations are spontaneously broken. And in fact, since zeta is the quantity that transforms nonlinearly under those transformations, we identify zeta as the Goldstone boson for that spontaneous breaking. So this is very nice. So D and K are spontaneously broken and zeta is the Goldstone. We identify zeta as the Goldstone or the dilaton for the spontaneous, for the symmetry breaking pattern, which is SO4 comma one, the group of conformal transformations on R3 being spontaneously broken down to spatial translation and spatial rotations which remain linearly realized and that's the inhomogeneous special orthogonal group in three dimensions, okay? So this is very nice. Yes, very good. Yeah, that's a good question. I was not gonna discuss this, but indeed. So the time dependence is completely arbitrary. Now, at the end of the day, this is not gonna make any difference for the consistency relations I'm gonna derive. But technically, we would like these diffs, as you know well, we would like these diffs to be the limit of a physical mode. And so you have to run through this adiabatic argument which you can and Paolo and friends did it for SCT and you find so dilation by itself demanding that it's an adiabatic mode forces lambda to be time independent for SCT it forces you to combine B must be time independent and moreover, you must combine it with a time dependent translation. And so that's true in general, there's always, but at the end of the day, the bottom line is there's always gonna be a one to one map between the symmetries we identify and the number of warded entities. It's just you have to take certainly near combinations. By the way, at first sight, it's a bit surprising. So we're breaking here four symmetries, it's potency breaking four symmetries, dilation and three special component transformations and yet we only have one goldstone. Now this is not a violation of Goldstone's theorem because we're breaking space-time symmetries. It's a well-known thing that with space-time symmetry breaking, you generally generically have fewer goldstones than broken symmetries. And you see already what's going on. So there's one goldstone, but of course there will still be four consistency relations that will ensue. And you can see it, you can see the physics here, although there's one goldstone, you see that by the symmetry transformation, you're probing different moments of the goldstone. Here just a constant and here a linear gradient. Before we go on, let's just, you can be more general and identify the most general diff so we can solve this equation. So noting that, so if I want to solve star, noting that we don't get anything for free, if we don't get anything new if things fall off at infinity, we just want things that grow. So let's just do a Taylor series around x equals zero to find the most general solution. So more generally, let ci be an infinite series, n equals zero to infinity of one over n plus one factorial m i l zero l n x l zero x l n. Okay, so a general polynomial. And you can see if you plug that in to this equation and solve order by order in x, it's clear that all you're demanding is some condition on the coefficient and it's just a trace condition. So when you work it out, this array of m, which by the way of course is already symmetric by construction in its last n plus one indices by virtue of the fact that it's multiplied by a bunch of x's. But then if you, by star, you find an extra condition which is that, so it's just a trace m i l l l two l n is minus a third l l i l two l n. Okay, so there's some trace condition on this array. Which is if you want, is the condition in order for you to stay in this gauge. Okay, so there's infinitely many symmetries that preserve this gauge, okay? And they're all of arbitrary order in x. And there's a lot of, physically it actually makes a lot of sense. You can do the counting and before Merdad asked me a question about this because we had some exchange about it. I'm gonna be, for the purpose of the lecture, I'm gonna brush over a subtlety having to do with tensor modes and adiabatic modes with the tensors. Please allow me to do that. But if you do the actual proper counting and you exchange a few emails with Merdad, you find the following answer. The physical picture at the end of the day is very beautiful. So imagine that you're a local observer in the universe. So here's your little region of space. And there is a long wavelength perturbation superseding you, okay? Which is a combination of zeta and gamma. So locally, in your little environment, you can write the spatial metric Hij. You can start writing it as a Taylor series. So Hij, I'll write it as bar to be the value at the origin of your patch plus dk Hij xk plus a half dk dL Hij xk xL plus dot dot dot. So this is bar again. You do this Taylor series. And now you know, of course, that you can do coordinate transformation to simplify this metric. In particular, you can use coordinate freedom to set the constant mode to zero. You can make this delta ij. There's enough freedom to do that. And that corresponds precisely, that freedom corresponds precisely with a number of these nonlinearly realized symmetries for dilation. So in particular, if there's a constant conformal mode, you can bring that to zero simply by shifting zeta suitably. Similarly, you can also set the value of gamma locally to zero. You can also set the linear gradient to zero. That's just the equivalence principle. You can set this guy to zero, which corresponds to removing a constant acceleration. That's what this SET is doing by removing the linear gradient and zeta. And similarly, you can do that for gamma as well. So the first two terms, you can make trivial. Now, as you know, at quadratic order, you cannot set that to zero. That's physical. Part of it is physical. It gives you, at best, if you want, you can go to Riemann normal coordinates. This will give you 1 third, Ricci, whatever, i, k, j, l, x, k, x, l. But if you count how many components, this is the Riemann tensor in three dimensions, how many components it has, it has six. And if you compare that to the number of components at quadratic order, there are more components at quadratic order. There are 12, if you do the counting right. And the difference between the number of components, which is 12 minus six, is six. And that's precisely the number of symmetries you have at quadratic order. You see? So the number of symmetries we have at each order is precisely the difference between arbitrary components in this Taylor expansion and the physical components, which you can write in terms of Riemann and derivatives of Riemann at higher order. So it makes beautiful sense. At lowest order, it's just the equivalence principle. And at higher order, it's just the redundancy of these different coefficients. Okay, so now that we've identified the symmetries, we can write down consistency relations associated with them. Questions so far just about the symmetries are a bit of caffeine. By the way, I told this to Alberto, and caffeine powder is very dangerous. It's being sold on Amazon, if there's an article in New York Times. If you buy caffeine powder, they sell you 100 grams of it in the package. It's not regulated, you can buy it. But it's like almost as lethal as heroin. If you take a teaspoon of it, you die, okay? You have to take at the milligram level to measure it. It's like heroin, okay? So I'm just mentioning this. It's amazing in the U.S. We're worried about many things, but somehow caffeine powder is totally fine. Okay, yeah, somehow a teaspoon is like 25 cups of coffee. And the package is like 4,000 Starbucks coffee mugs. Okay. Ward identities. So here, let me give some credit. First to us. Sorry, it doesn't mean that way, but it just happened. Okay, so our paper, that's the one I gave earlier. And then there are many related papers of people in the audience, in particular, Kohlberger, we, and Alberto, 793. There's a paper by Bauman and Green. And then there's a paper by Paolo. And it's very nice because these papers, they all get the same result using very different methods. So they're all complimentary and worth reading. I'm going to focus on the method we followed for the only reason that it's a bit cumbersome, but at the end of the day, it's easily applicable to many different relations. Okay, so that's what I like about it. So I'm going to focus on that method. So let us start our starting point to write constraints on our correlation functions is a triviality. It's just a statement that the charge, which generates the symmetry transformations, the charge commuted with the field. So let's just call it some arbitrary operator O, which will be a product of zetas and gammas at the end of the day. By definition of Q being the charge, this is just given by the symmetry variation of this operator. It's a triviality, a tautology. That's what the charge does. And we're just going to take that statement and take its expectation value relative to the in vacuum. So to compute an in-in correlation function like we do in cosmology. Okay, so this is our starting point. And again, it's a tautology. And indeed when we worked on this, we were getting one equals one for a year. At some point after a year, you managed to massage this to something non-trivial. But literally for a year, we were wondering whether we were just doing one equals one. At the end of the day, you get a non-trivial statement. Okay, so first of all, just to stay an obvious point, if the symmetry were unbroken, if we were talking about linearly realized symmetries, if the symmetry is linearly realized, then the statement is obvious. By definition, if the symmetry is unbroken, it's a symmetry of the vacuum. And so the charge acting on the vacuum would vanish by definition of the symmetry being unbroken. And as a result, the left-hand side would vanish identically. And so therefore, the statement that if a symmetry is linearly realized, it just tells you that your correlation function is invariant under the symmetry, okay? That's like what we did yesterday on the rest. So 4, 1, 1, our correlators were invariant under that symmetry. If, on the other hand, the symmetry is non-linearly realized, which is what we're considering today, if it's non-linearly realized, in fact, what you get in that case, as we'll see, is a soft pion theorem, analogous to what you find in QCD, okay? So this is what we want to do. We want to compute a soft pion theorem for our symmetries. And it will suffice to simply focus on dilation, just to illustrate the method, and then once you've done it, the method can be applied to anything. In fact, it's easy to do it for all these x to the n transformations. So let's do it. So first, let's consider... Right, so we're going to do the dilation as an example. And so for dilation, the next following we have that the symmetry transformation of zeta is 1 plus x dot grad zeta. The right-hand side is easy. So the right hand... Oh, sorry, moreover, we're going to focus on, for simplicity, we're going to take O, this product of n fields, just for simplicity to make our life easier a little bit. I'm going to imagine this is a product of zetas, an equal time product of zetas. All right, here we go. So the right-hand side is easy, because we did it yesterday. So the right-hand side, that was your homework, although I don't see any copies on my desk, the homework was simply to compute that the variation of O... Actually, we did this one, but the K1 was supposed to be U. Delta D of this guy is equal to exactly what we computed yesterday, but now we use the fact that zeta has vanishing weight. So the answer for this, you just set all the deltas to 0 in what we found yesterday. So we had 3n-1 plus sum from a equals 1 to n of ka.ddka acting on the correlation function. And remember, this factor of minus 1 was because we were stripping off the momentum-conserving delta function, and so this is the answer. So that's the right-hand side, and now we want to extract something non-trivial about the left-hand side. So for the left-hand side, we have the charge. And so the charge, by definition, is such that if I commute it with zeta, I get the field transformation of zeta, which as we said is just a constant plus x dot grad. And, okay, so let me say the following. So you see that there are two parts of the transformation. One is the nonlinear shift, and the other one is the linear transformation. I'm going to split the charge accordingly. So let's split the charge as follows. Let's write q as q naught, the free part, plus an interacting part. Okay? And let me say y, and I claim that the free part, q naught, so with q naught commuted with zeta, it's equal to minus i. Okay? And the reason I call this guy the free part and this the interacting part is let's think about it. Imagine you take your action, and you expand it out in powers of zeta. You have the kinetic term for zeta, and then you have a bunch of interactions, cubic, quartic, et cetera, schematically of this form. By definition, the lowest order piece of the charge will be a symmetry of the lowest order piece of the action. Okay? And indeed this is shift invariant, invariant under the shift symmetry. But now if I compute, if I look at, if I ask, is it invariant under this part, it's not. Because what will happen is that the linear shift of this piece, if I transform under this guy, I will get a term here which will be which will be quadratic. Right? And this will cancel the lowest order transformation on the cubic term, which will give me something quadratic as well. You see? So as a result, this symmetry, this, this, this transformation can only be a symmetry of interactions, of a theory that includes interactions where the lowest, whereas the lowest order piece is by itself a symmetry of the free theory. Does that make sense? Yes? Okay, so this is the, we're going to call this the free charge. It's the symmetry of the non-interacting theory. Okay, so what is the charge? Well, the charge must be such that when I commute it with zeta, I get a constant. So clearly it's just the conjugate momentum to zeta. So indeed we can write down q0 explicitly. q0 is just the integral d3x of the conjugate momentum of zeta because clearly if I commute this conjugate momentum with zeta, I get a delta function, do the integral, I get 1. Okay? So that's what you need. And equivalently I can think of this as the limit if I were to put an e to the i q dot x here, I can think of this guy as the limit as q goes to 0 of the Fourier transform of the conjugate momentum. So this is equivalently the limit as q goes to 0 of pi zeta of q. It's a zero momentum limit of the conjugate momentum. Now I want to use basic concepts from scattering theory. So if you remember in quantum mechanics when you do scattering theory, you usually do it off a central potential and you define asymptotic states as follows. You define asymptotic states states which far away from the interaction potential asymptote to free trajectories. Okay? So if this is the trajectory this will be your in state psi in, psi out and you define these in states basically by taking the state of the system evaluating to the far past using the full time evolution operator and bringing it back home with the free time evolution operator. That's how you define these in states and similarly for the out guys. Except for us we're doing in in so really our picture is the other way around so we want to think of it as some initial surface ti and we want to define our in state to come in to do some scattering and bring it back but the story is the same. We want to think of that near time, the initial time here you'll have almost free evolution both here and here and then the definition of our omega states are interacting vacuum will be exactly the same as what you do in scattering theory. Okay? So in other words are interacting vacuum so omega the in vacuum let me just write it. So this in vacuum is related to the vacuum of the free theory which will choose to be the bunch Davies vacuum if you want that's the definition of bunch Davies vacuum. As usual via a moiler operator okay so this is moiler operator it's defined so so first of all this guy is the bunch Davies vacuum by definition and this omega is by definition is this moiler operator okay so this is the moiler operator. Okay so just the intuition is obvious what you're doing is you take a vacuum state okay at time t equals zero that's omega you bring it back to the infinite past via the full time evolution operator and you take it back home with the free time evolution operator and that defines for you the interacting vacuum state okay just like in scattering theory perfect similarly we can define we will demand we will demand that our operators in particular the charge I should have said that as well so when we do scattering theory we demand that the potential is localized or in other words that far away from it you have a free theory similarly here we're going to demand that at initial times our interactions turn off that's what we do in the in-in so in the far past these will be free interactions so if I bring any operator so I'm thinking Heisenberg picture if I bring any operator in Heisenberg picture back to the past I should get the free theory operator so that's true in particular of q so we have what's known as an intertwining relation for q namely that omega the Moeller operator q so that's just taking q and bringing it to the past okay this guy will be just the free charge does that make sense yes no yes no okay so let's just to be clear what this statement implies is the following if you want if I were to unpack it in terms of u and u0 it tells the following statement it says that u dagger minus infinity 0 q u minus infinity 0 is equal to u0 dagger q0 u0 minus infinity 0 that q you bring it backwards in time to minus infinity it agrees to the operator q0 brought back to minus infinity according to the free time evolution operator okay it's exactly the same statement as saying I'm turning off interactions in the far past okay if you open Weinberg's book in the chapter on s matrix he exactly assumes this for in and out states but now where he takes the Hamiltonian or the momenta or all kinds of operators obeying these kinds of relations okay so that's your homework tonight you open Weinberg page whatsoever you read s matrix story it's exactly that okay so now with this simple these simple statements we can get a long way so we have the following so in particular this intertwining relation if you want this intertwining relation implies the following equivalently q times omega minus is equal to q is equal to omega the model operator times q0 so in other words when I commute this model operator I transform the operator into its free its free partner and so that's it now we can apply this to our identity on the left hand side of our identity we have the charge acting on the vacuum we have q acting on omega and so here we go q acting on omega is q times by definition of the in vacuum times the model operator acting on the bunch davies and then I can commute the model operator past the charge metamorphosing it into the free charge in the process so this simply becomes omega minus infinity q0 acting on the free vacuum okay so this is simply simple but it's amazing because now we've turned the problem this rather complicated problem of acting with a full charge on the fully interacting vacuum we translated it to a problem of acting with a free charge on the free theory vacuum and that we can all do so we just have to do that and then we're done so let's do it we just have to compute the free charge which is here let's leave it the free charge acting on the bunch davies vacuum ah very nice it's the same so if you want this usual I epsilon deforming the contour business is equivalent to turning off interactions in the far past because it basically it takes it maps e to the minus i kt it adds to it a damping part like so so it's precisely the I epsilon trick is exactly a way it's exactly equivalent to this exactly equivalent in a way that nobody has made precise but anyway it is exactly a way of shutting things off in the past but it's true nobody has quite in the literature made this very precise but indeed it's exactly correct so it's just that's by definition it's by definition the bunch davies vacuum when you do that absolutely okay yeah I should say this is yeah if you get me going on this I can keep going in non-equilibrium theory people have considered problems like this but usually they consider problems where you have a Hamiltonian which splits into a free Hamiltonian which is time independent plus an interaction piece which is time dependent that's the type of problem people are interested in non-equilibrium dynamics but then by hand they can then turn this guy off with an exponential okay adiabatically which is what as you said the complication in cosmology is that even the free part is itself time dependent okay so anyways I couldn't find any precise analog in the literature in non-equilibrium dynamics but at the end of the day okay so let us now compute this free charge acting on the free vacuum and this is quite remarkable it sounds very simple but this is again where we kept getting one equals one it's very nice actually because the simplest way to do this I think is to actually use the Schrodinger picture of field theory which people rarely use but for this purpose it's actually exactly what the doctor orders so we're going to actually think in terms of the Schrodinger picture so in the Schrodinger picture as you know our operators are time independent so in particular in other words I'm thinking Schrodinger picture in the Schrodinger picture we have that our field operators zeta is actually time so it's a pure function of x and the eigenstates of this guy are also time independent they're just basic states and so actually we're going to consider eigenstates of the free zeta operator actually in Schrodinger picture I think it doesn't matter whether they're free the operators are free or not it doesn't matter so we're going to define this guy as the eigenstate under zeta like so we're on the left hand side as an operator on the right hand side as a function and we're going to insert a complete set of states of these guys it's exactly what you do in quantum mechanics so we want to compute this charge so q0 acting on the vacuum is equal to so it's equal to this pi it's the momentum acting on the free vacuum so we have limit as q goes to 0 of pi the conjugate momentum acting on the free theory vacuum this is exactly like in quantum mechanics you want to find how p the momentum operator acts on a particular state and it's often convenient to go in the real space representation in which p is just d dx so that's exactly what we do we insert a complete set of states so it's a continuous basis so we have integral over c0 let me erase the charge at the bottom we have the integral so complete set of states so c0 c0 pi sorry I had my limit let's put it outside limit as q goes to 0 pi zeta of q of not ok so this is indeed the real space or configuration space representation of how p acts on a state so it's just given by d dx on the wave function in this case d dx zeta and by the way this is not a path integral it's a good old ordinary integral just over all possible configurations of zeta so now the limit q goes to 0 integral d zeta zeta not minus i functional derivative with respect to zeta of minus q if you work it out it's minus q instead of q acting on the wave functional the real space wave function the real space the configuration space representation of the bunch devi's vacuum does it make sense it's just quantum mechanics on steroids ok yeah sorry what u q yes it's just an operator on the Hilbert space it's a charge operator here's the transformation when commuted with zeta yeah yeah in fact it's not it's like saying the analog there's the precise analog in quantum mechanics so here we're shifting zeta by a constant that's like the analog of shifting the position operator by a constant and you know that in quantum mechanics the generator of translation is just p since p commuted with x is i and similarly here the generator here the charge is the conjugate momentum zeta and etc the analog is perfect except that zeta is not just yeah it's a field but otherwise the techniques are the same so to compute this quantity we just need to have the wave functional representation of the bunch devi's vacuum and that's easy because the bunch devi's vacuum is just a Gaussian so now we plug in the bunch devi's vacuum it's just a Gaussian so we can represent it easily it's a Gaussian because it's the ground state of the harmonic oscillator and we know the harmonic oscillator has ground state which is a Gaussian so no surprise here it's just a Gaussian as well so it's an exponential except that I have a sum over all possible oscillators so I integrate over d3k zeta k and the variance of this guy is the power spectrum or 1 over the power spectrum which you can check if you compute the two-point function using this wave functional I tell you it's just like quantum mechanics so just compute the two-point expectation value the variance should match the power spectrum and you'll find it's exactly 1 over p that you need here to match the correct answer s, 1, 2, 3 no no yeah that's right what I'm aiming at here is just to find how I'm just getting to the point where I can show how this charge acts on the vacuum at the end of the day so that's just the point of this exercise no it's not zero that's the point very good so we're done so we can compute now this functional derivative on this state since it's just a Gaussian you can see what happens I'm just going to bring down a power of 1 over p times zeta and when you put all the right factors yeah so dot dot dot that's easy and so here's the answer that the charge q naught acting on naught right so you do that completeness relation and here's what you find this is for tonight in the privacy of your room you find the limit q naught goes to zero 1 over 2 p zeta of q times the free field zeta naught q acting on the bunch Davies vacuum and now it's an operator statement and so I can think of it in any picture I want so let's think again back now in terms of the Heisenberg let's think back in terms of the Heisenberg picture we revert back to the Heisenberg picture now so if you want what I mean by zeta naught here so I should say this guy now depends on time in the Heisenberg picture and what I mean by zeta naught is the field which evolves in time according to the free Hamiltonian very nice so we're done except for one technical assumption by the way let me make a comment what's very nice about this derivation although it looks a little bit technical is that it's very clear where the initial state comes in the initial state came in right here so you want to modify bunch Davies because you feel like it you could just add extra terms you could add a zeta cube term and so on and so forth this will directly descend into your variation and so on and so forth so the assumption about the initial state is manifest in this way of doing it now the final assumption which is really which is really a necessary condition for this consistency relation is the fact that zeta must go to a constant at long wavelengths so this is where the assumption comes in so if the growing mode for zeta is a constant so in other words if the limit as q goes to 0 of zeta of q is a constant so if the growing mode is a constant and similarly for the free the free guy so limit as q goes to 0 of the free operator so in the long wavelength limit it tends to a constant then since these guys in that limit are just constant operator clearly they don't care how I time evolve them and so in particular this implies that they obey just like the charge they obey an intertwining relation so namely the limit as q goes to 0 of omega so if I bring this guy past the if I bring a more operator past zeta I turn it into the free zeta so this is where the assumption of constancy of zeta comes in ok so let's recap where we were essentially were done so we started out on the left hand side with q acting on the in vacuum we turned this into the free charge acting on the free vacuum ok so now let's write the result the result is we have the moiler operator acting on this object over here q not on the free vacuum so let's write this out we have limit as q goes to 0 1 over 2p zeta q zeta of q and this whole thing acting on the free vacuum sorry this was the free field and then we can undo bring back the moiler operator through use the fact that zeta is conserved as we just showed when I bring it through it turns zeta not into honest to goodness zeta and then omega will act on so not and make it the interacting theory vacuum so that's the last step this quantity is equal to simply the limit as q goes to 0 1 over 2 p zeta of q times the full zeta acting on the full vacuum of the theory and so that's it the only thing we haven't done is to take a commutator but that's easy so I was a little bit delinquent there was an imaginary part here which I dropped the imaginary part cancels yes madad absolutely thank you I was zooming out yes absolutely so here also there should be a limit as well thanks so this should be zeta not and then the last disclaimer is that here there should have been also a phase bit from the wave functional but this cancels out it would carry on all the way here there would be an extra correction but remember at the end of the day we're taking the commutator q with 0 and then in that case that imaginary part drops out okay so at the end of the day you get the following statement at the end of the day we have and we're done we have the statement that the left hand side which was this product this expectation value of q commuted with an arbitrary product of zetas this guy therefore we've shown is the limit as q goes to 0 oh and by the way when you take the commutator also the factor of a half becomes 1 so it just becomes 1 over p zeta of q zeta omega zeta of q which depends k1 kn very nice and then right hand side we know what it is this guy is equal to the variation of o so it's this variation of o which we wrote down earlier it was minus 3n-1 minus ka sum dot d-d-k-a on o yeah yes prime prime and that's it so as advocated in this case because it's a nonlinearly realized symmetry what we have is not that the correlator is invariant but instead the variation of the correlator is related to inserting a soft mode of the goldstone this is exactly what happens for example with the soft pion theorems having a soft pion soft momentum pion gets related to a symmetry transformation of the correlation function without the soft mode let's see if the computer works that's probably the trickiest part let's see laptop I don't know oh it's there maybe it's just warming up it's coming won't you spoil the joke no there's no joke this time okay so I'll just show you some results that come out of this this is our conclusion of what we did on the board you can generalize this to all the other transformations we found so in particular we found that we had delta gamma and delta gamma we found an infinite number of these transformations that went as x to the n and the transformation goes as x to the n on the left hand side this will correspond to many derivatives the same number of derivatives acting on the soft mode so for arbitrary n this is what you see this is the structure of the consistency relations so it's n derivatives with respect to q of these correlation functions and now in general they involve a soft zeta mode and a soft tensor mode and they get related to the corresponding symmetry variation of the operator O and again based on our thinking of the equivalence principle the leading order so q0 and q which was the constant part and the linear gradient part they're completely fixed so the if you think of this guy as a Taylor series in q the constant bit and the leading order and the first order in q they're completely fixed in terms of the lower point function but at higher order it's only partially fixed some just partial information about the Taylor coefficients these are physical statements in the sense that there was a physical assumption that went into them and in the physical assumption was that zeta was constant outside the horizon these guys can be violated in fact you can test for example when we try to measure fnl we're precisely trying to check the validity of these consistency relations and one thing I want to stress which a priori is a little bit surprising but notice that we didn't assume anything about inflation we didn't assume slow roll this is exact statement arbitrarily far from slow roll so this is kind of remarkable because it basically says so somehow our symmetry breaking pattern remember our symmetry breaking pattern was SO4 com01 broken to ISO3 and this guy it's linearly realized exactly at the de Sitter point and you might expect that only for small departures from de Sitter but in fact arbitrarily large departures the symmetry breaking pattern so it's like de Sitter is the point where the symmetry gets restored the exact point where symmetry is restored and the deviation from it is described by this symmetry breaking pattern oh sorry the clicker doesn't work I'm moving arbitrarily far but for no good reason let's continue this is how they look, this was the schematic form this is actually how they look so when you put all the indices and stuff and of course as particular cases you reproduce things that people knew Paolo and Matthias in particular had derived using other methods but then you get other state new statements so in particular at n equals 2 corresponding to a quadratic shift so here's one I check this at some point and it works so left hand side matches right hand side it's very nice and that's as much as I could do so then I hire young people they did some amazing stuff so they did you can do so here's the schematic form of the of the consistency relation and the right hand side so this is for arbitrary sound speed the power spectrum of Zeta goes as 1 over epsilon times Cs let's just check the parametric dependence just for this discussion on the left hand side you find that up to order Q squared this is the dependence so Zeta starts at order Q squared in this case a 3-point function and gamma has beautifully the right parametric dependence you do the check it works out but then at order Q squared there's a paradox it's not much of a paradox but it's a bit non-trivial so you see that at order Q squared you see that the answer for Zeta goes as 1 over Cs cubed and the answer for gamma also goes as 1 over Cs cubed so for the identity to work out these two contributions the 1 over Cs cubed must cancel leaving you with the 1 over Cs and indeed that's exactly what happens so somehow the two of them and so Lasha and Jumpu they did all these checks so all the 3 to 2 checks up to Q to the 3 so that's very nice so it works then you can do also another thing you can do is to do multiple soft limits and this was work with my former student Austin Joyce and Marco and so this is the intuition comes from Pions that if you take multiple soft limits that's another way to probe higher Q dependence so if you have two legs you're differentiating with respect to Q so for example here's our general result at the end of the day so if I take two soft limits two soft limits with respect to Q1 and Q2 now I'm going to get you're basically getting things at order Q squared so two there is with respect to Q in other words and you basically get it's basically this structure where either two legs are acting on the hard legs and you can also derive this alternatively and this was what Alberto and collaborators followed you can work this out instead of thinking in terms of operator formalism you can work it out in a path integral fashion as a Slavnov-Taylor identity so here's something that you've seen in your quantum field theory course if you in E&M so after all all these symmetries are difumorphisms so they're really originating from a gauge symmetry and you know that in E&M gauge symmetry relates vertices so in particular the three-point vertex involving a photon and fermions is related to the fermion propagator when you contract with the four momentum of the photon so this follows directly from the gauge invariance of the theory it's known as the word Takahashi identity and so you can do the same thing in gravity okay so here's the answer you get at the end of the day you get something about the vertex functional varying with respect to zeta and with gamma being related to something on the right hand side and then you can take this sort of master identity and vary it a bunch of times with respect to zeta or gamma to get the corresponding to get the corresponding identities so here it is, in other words this is at the level of vertices as opposed to being correlation functions so here it is for the three-point function you see something very similar to what we've derived you have the three-point vertex for zeta and three-point vertex involving a tensor it's related to the two-point function of zeta okay and this statement is exact in Q, there's no expansion but at the end of the day to turn it into a statement about correlation functions you have to do the translations you turn vertices into correlators sorry I'm going fast but it's just to give you a flavor in fact I ran out of slides anyways, sorry it's here so you do sorry I thought I didn't have the punchline thankfully I do so you convert to lines going from vertices to lines and this is the final statement which is exactly what we got so that's all I had to say and in this Slavnot-Taylor way of doing things what is nice is that the assumptions about zeta being constant you can make it even more precise let me just say a quick word about that so when you turn when you turn this identity for vertices in terms when you try to solve it for the vertices you find there's an ambiguous part AIJ which is a physical part this is the part which corresponds to the physical violation of the identity and at the end of the day the assumption that zeta is constant outside the horizon amounts to an analyticity property of this guy so as long as it starts at q squared you can keep going and you get the result we derived okay I know this was fast this was just to give you a flavor of these consistency relations but what we did today for dilation it's very easy to do it for these arbitrary transformations okay thank you