 Our last speaker today will be Maurizio Fagotti, and he will talk about entanglement evolution and generalized dynamics. Maurizio, please. Maurizio, you are muted. So unmute yourself, Maurizio. Nothing. I'm mute. Okay, now you are good. So let's start from the beginning. Maurizio, please. Maybe it's working. Okay. A moment, huh? But it's not. I don't want to see this. Okay, anyway. So, also, okay, I would like to thank the organizer for organizing this virtual conference that brought together more people than we would have expected in a real conference, and also get to invite me to give a talk. So, and today I will talk about the simple application of generalized aerodynamics and it's about the time evolution of the entanglement entropy in the case of inhomogeneous systems. So this is the plan of the talk. So I will start by introducing first the ingredients that we need to know in order to understand the results. And then I will, I'm sorry, do you see also the small? So we see your big pointer at some point, yeah. Do you see a big pointer? Yeah, I see it. Unfortunately, I can't see it. So, okay, anyway. So, first I will introduce the ingredients, understand it, and then I will just go into the details of entanglement evolution. So the first ingredient is about bipartite entanglement. So what is it? And the idea is to imagine your system and then the idea is to split the system into parts that we can call A and B. A is the subsystem, B is the rest. Now, we can better understand the bipartite entanglement if the system is prepared in a pure state because in that case, we have an important result, which means the composition, which means that the state can always decompose as I'm showing here in terms of an orthogonal basis of the subsystem A and of the rest B. And the important result is that all the correlation, the quantum correlation between the subsystem and the rest are encoded in this coefficient Pn, which are nothing but the eigenvalues of the reduced density matrix of A, or also B if you don't consider the zero one. And, okay, in order to measure correlation, quantum correlation, you can use any function of the Pn, and let's say that the functional, the attractive more tension has been the entanglement entropy for several reasons. First, because it's a very nice properties, and then also because it can be connected with the statistical entropy and also with the Shannon information. So, but what do we know about the bipartite entanglement in quantum antibody systems and in particular in one dimensional systems, finch chains? So, well, yeah, I can give you just a general picture of it. So, if you consider a Hamiltonian with the local interactions, then in general, the bipartite entanglement entropy in the open energy state in the middle of the spectrum satisfy what is called the volume law. So, it's proportional to the subsystem length, okay? There are exceptions to this rule, and in particular, if you consider a very low energy energy states, then they are characterized by bipartite entanglement entry, which scales as the boundary of the subsystem. So, in particular, one dimension means that it's independent of the length of your subsystem. This is for non-critical systems. Now, if you consider a critical system, there are, the situation is a bit more complicated. And in particular, at low temperature, you can have a log breaking of the RLO, which means that even at low, even when the temperature is zero, you have the entanglement entropy scales as a function of the subsystem length, but now it's not linear, it's not extensive, but it scales as the logarithm of the subsystem length. And then also, if you consider integrable systems, you can find this kind of excited states even in the middle of the spectrum, okay? Now, consider this is just a general picture of what can happen, but there are also exceptions here. So, you can find also fancy behavior in the middle of the spectrum if you look for it. But in general, if you just pick a random state, this is what you find. Okay, so this is my introduction about bipartite entanglement. Then I did a very short introduction on quantum quenches, although Lorenzo already did it and I thank him. And okay, because indeed I would like to discuss the time evolution of the entanglement entropy, okay? After this we go to quantum quench. And in particular, so I want to consider time evolution, the unitary time evolution, which is generated by given Hamiltonian age. And so, depending on whether the initial state is a pure state of density metrics, then it's described as a Schrodinger equation or by the phenomena equation. This kind of very simple quantum dynamics was called quantum quench in a particular case by John Cardi. And the case is when you imagine that the initial state is for example, the ground state of the Hamiltonian prepared for a given parameter, given coupling constant. And then at the given time, one changes abruptly the coupling constant and follow the time evolution after this sudden change. This was called quantum quench. Now the typical examples of many body systems where people study these dynamics are spin-lactic system and quantum field theories. And I will talk here of spin-lactic systems, okay? In particular, I mean that the local Hilbert space has a finite dimension. And for spin one half, it means for example, as I mentioned too, okay? And well, one of the reasons why as Lorenzo said, why people study this problem is because of what, in the thermodynamic limit, you find that the local degrees of freedom seem to relapse to some stationary value. So many, many people trying to characterize this particular, the properties of the stationary, the stationary properties of the system. And okay, if you are more interested, you can find more details in this missed review. So, but okay, this is very general. And, but now I would like to, so my aim is to apply generalized aerodynamics to the time evolution of the entanglement entropy. So, I'm interested in that in homogeneous systems. And in particular, I had to also define GHG, what is generalized aerodynamics. So, we already had more than two talks about that. So, well, you, I think you already know what I'm talking about. Yeah, I would like to give you just a different perspective, slightly different, because in the end, the questions are the same. And okay, my point of view about this, this theory is that you have your initial state, which let's call it consider density matrix, zero, the time of all this under the Hamiltonian. Now, we already know that if the initial state is stationary, then this time evolution is trivial. And in particular, we can describe the state using, for example, the root densities that Lorenzo also introduced before. So, there is a clever way to represent the state in other way, in the other sense. Now, my point of view about GHG is that the generalized aerodynamics should be the most clever, the clever way to parametrize the state in such a way that you can write a dynamically question for the root densities. That's somehow decoupled from the other decrees of freedom. Now, this is not necessarily possible, okay? But nevertheless, okay, the aim could be just to reduce as much as possible the degrees of freedom in this time evolution, okay? Now, any particular, okay, we know for sure that if one considered the limit of low inhomogeneity, then this is possible. And it was shown that when you try to expand this function as a series expansion in terms of the inhomogeneity of the state, then you find that the first order, you find just a functional of the root densities. So, the time evolution, the dynamical equation for the root densities is closed, okay? And now this is, okay, this is a, if you want, I like to call this first order, generalized aerodynamics, because there is just the first derivative respect to the position. And, okay, what I want to stress is that this is a classical equation. So, there is no H bar above in this equation. And this, as you will see in a moment, is very important for what I'm going to show. And, okay, but before I go in there, I want to give you an explicit example of what I mean with functional, with a system of this functional that allowed to simplify the time evolution. So, what does it mean to have a clever representation? Parameterization. And the example is non-interacting spin chains as always, okay? Because if you study the time evolution of a slated determinant, so a state that has, for which the Vix theorem can be applied, then what you find is that Vix theorem is applied at any time. And there is a way to parameterize the density matrix, the time t, as a functional of two fields. One field is the density, and the other field is just some auxiliary field that somehow describe the off-diagonal matrix element of the density matrix. And analogously, you can parameterize the Hamiltonian as a functional of the expression energy, okay? So, yeah, I'm not writing what does it mean explicitly, but I can tell you the result. There is a way to parameterize state and Hamiltonian such that when you write the Schrodinger equation for the state, then the Schrodinger equation can be decoupled between the two degrees of freedom. So you have in particular any question for the density, which as you can see depends only on the density, and you have any question for the auxiliary field, completely decoupled equations. Now here, maybe for whom or you who are familiar with the big description of phase, phase formulation of quantum mechanics, so you can see that these are, their density here appear like the bigger function in free system, is the same. And it actually satisfy the same equation. And this operator here, this star product, yeah, in Moyer star product, is the equivalent of the product of operators in quantum mechanics. But in this phase space formulation that where operator has function of position and momentum, that it becomes this kind of star product, okay? It's just a different way to write things. And, but what is nice here that you can decouple the two degrees of freedom, and in particular you can, okay, what I would like to call the first equation, generalized adoranami equation, at any order, because it's just a function of the density. It doesn't depend on anything else about that. Okay, but now let's try to use this result to our problem, okay? The problem of the time evolution of the entanglement entropy. And first of all, okay, and for this, I would like to introduce a semi-classical picture. Now, if we, now we have seen, I'll show you before that if you expand this generalized adoranami set, the first order, you obtain just a continuity equation, okay? For the, and moreover, this order, it is possible to reinterpret the root density as a density of particles. Indeed, as far as I can see, at this order, the root density is semi-definite positive, so it shouldn't be negative. This is actually not true if you can see all the orders, but at least at this order, I think this is correct. So you can always reinterpret it as a density. And so in other words, we can try to represent the state as a collection of particles that move according to this equation. Now, what is very important that I tried to stress before is that at this order, this equation is classical. There is no H bar. So this means that the, you cannot create quantum correlations by the equation. So whatever all the quantumness of the state is at the initial time. So if you know how the particles are entangled at the initial time, that you know how they are entangled at any time. The only thing that you should do is to evolve to the trajectory of the particles. Trajectors that can be very complicated because generally they depend on the state itself if there is interaction, okay? And so from those, as you can see, the if one is able to trace back the problem of the time evolution of the bipartite entanglement to the entanglement particles, then we are done because the entanglement particles are simple. Time evolve is in a very, very simple way. So the question is how can we do this? So it's, let's consider for a moment just non-attracting systems, which have slightly differences with respect to generic integrable systems. Okay, in this case, moreover, okay, we can also write the first correction if you want, which is a quantum correction and well, quite interestingly, I would say the first correction as a third derivative with respect to the position. And this is clear because, okay, there is no diffusion in interacting systems. So we don't expect a second derivative. But okay, now we are just neglecting this term and we consider just the generalized of the dynamics at the first order. The difference in non-attracting system is that the velocity is now independent of a state if you consider homogeneous Hamiltonian. And so what happens is that the trajectory has just straight lines, okay? And now, okay, so it's kind of simple to perform an analytic calculation of the position of the particles at any time. And this is actually what was done in this old work by Pasquale, Calabres and John Cardi when they provide this semi-classical picture of the time evolution of the entire intent of this, exactly using the time evolution of particles. So here I'm just telling you that if you, now, instead of starting from their point of view, you just start from generalized aerodynamics, this becomes very, very natural to consider a particle time evolution, then the fact that they don't increase the entanglement. They can, especially in the free case where also when it's clear that when they cross, they don't care about each other, okay? But actually this is true also in the interacting case. Again, because the generalized aerodynamics is classic. Okay, now, well, recently then the situation considered by Pasquale, Calabres and John Cardi were a bit generalized. In particular, they focus on what happens when the initial particles are entangled in pairs. And more recently it was realized that there are states for which the particles are entangled in multiplets. And so, well, they consider also this case. But you see that from this perspective, there is no really big change. So what we have to do is just knowing the entanglement initial state, we have to follow the trajectory of the particles. And then clearly, for example, in this picture, if you look at the plane on the bottom, then you see that the contribution to the entanglement of the subsystem given by the particles given by just by the three particles, okay? Because they are, this is when the, because clearly the fact that they are entangled create quantum correlation between the subsystem and the rest. Because one is outside, the subsystem and the two of them are inside, okay? And in particular, okay, one of the hypothesis of the, of the semi-casual picture for entanglement entry was the low entanglement of the initial state, which in this context means simply that the disentangled particles are always close close to one another at the initial time. This doesn't mean that we remain close to one another because they are evolving with different velocities. So when you, when you wait for a sufficient long times, they will be very, very far away from each other. But it's, it's important to in order to find a very, to be able to obtain the result, actually one has to assume that the, the entanglement in the initial state is very low. So they are entangled only, only if they are close by. Okay, so more quantitatively what do I mean with reducing the special entanglement special entanglement into the entanglement particles? Well, I mean the following. So the picture that I showed you before when you reinterpreted this state as a collection of entangled particles can be written as I wrote, okay, the upper left corner like a tensor product of density matrices of this sets of entangled particles, okay? Now, since we are assuming that the state is low entangled we can actually somehow confuse locally the density matrix with itself. I mean, so we can create a multiple copies of the same operator, same density matrix everywhere in the chain. And why I am doing this because I would like to treat each small cell separately, okay? So the particle are defined in a particular cell when they are created and they are not entangled to the other particles. You did hear I put a tensor product of the various density matrices. Clear, okay, this is not our original problem. This is different. So I am clearly ignoring quantum correlations and also creating some quantum correlation in fact, somehow. But they, but we really don't expect that they will contribute at the leading board when we consider the time evolution. Leading order I mean the term of the entanglement which is proportional to the volume. Now, and okay, so then once this particle are defined we just have to follow the time evolution using first order ggd, okay? Which becomes just a dynamical equation for the semi-classical particles. Okay, so what is then the entanglement between the subsystem and the rest? Well, we just have to trace out all the particles that are not in the subsystem, okay? This is what we have to do. So in the schematic, let's imagine that the time t we have this kind of situation here. So there are several particle here, a, b, c, d, e, f, g. Now as you can see a, b, c are entangled, f is entangled with g and d and e are entangled. Now, when you trace out the degrees of freedom of the particle outside the subsystem you have to trace over f, c, d and e. So this means that you have to compute the reduced density matrix of rho a, b, c. The reduced density matrix of a, b in the density matrix described by rho a, b, c and reduced density matrix rho g corresponding to the state, the part of the state fg. But then, okay, there is the other part, the other two particles outside the subsystem. So when you trace them out, since the state is purifying one. So this is, you have just to consider all the contribution from particles that are partially inside and partially outside, okay? Now, in particular if you apply this picture at late times, what do we find? We find that if we wait sufficiently long times, only one particle of the multiplet will be inside the subsystem because all the other will be because they have a slightly different velocity. So if you wait sufficiently long times, there will be always one inside and one in the other outside. And more, okay, then, and this is true also in the integral case. I'm not using actually that the system is non-integral here. This, the integral b will change the trajectory in a complicated way, the interaction, sorry. But it won't change the fact that in the end, they will be, you won't have more than one particle inside the subsystem. And moreover now, at this point, we can also use the fact that, well, the, what Lorenzo described in the previous talk, but then what's okay? I mentioned before, local accession. So the fact that late times, the state becomes locally prevent to a stationary state. Now, so if we think a bit careful of what we are doing, then you realize that the, when you trace out the external degrees of freedom, given a particle, then you are not doing anything else but what they did, young and young, when they started the thermodynamics on the Liblinian model, and they finally showed the entropy, the young, young entropy, so-called young, young entropy, starting exactly from this point of view. So we fix a particle with a momentum k, and then somehow you had to find all the possible way to create a particular density of particles, raw of k, and this is actually the meaning of doing the trace over all the external degrees of freedom. So from this actually is kind of natural to assume that our entanglement between two, between one particle and the rest, we don't know if there are more multiplats here. One particle at the rest, the particle of the of the multiplat is just given by the density, okay, if you can call it like this, the density of the young, young entropy. And in particular, if you use this result, you find that, okay, trivially that in the infinite time limit, you have that your entanglement entropy becomes proportional to the volume of the subsystem and the proportionality constant is the young, young entropy, okay. And, but okay, we can clearly apply this formalism also at finite time. Now, the problem is that at finite time, we this, when we consider this correspondence at infinite time, this can be used at any time only if the particle are entangled in pair. Because if I have a more complicated structure, and if I have a, I am at the finite time, then I will to deal also with the situation where there are for example, two particles inside the subsystem, one outside, and I don't know how to treat it. Because here I just consider the case when there is one inside and the rest outside. So the, so this restricts us a bit on the choice of the initial state. Because if we want to be predictive now, especially in the attracting case, we have to assume that initial state consists of pairs, okay. So that now we can just use the previous result and say that the entanglement between the pair, the entropy associated to the pair is simply given by the density of the young, young entropy, okay. Then, okay, what we have to do to obtain the time evolution of the entropy at the time t, we just have to evolve the particle. So we move the particles according to generalized aerodynamics, first-order generalized aerodynamics, and then we use the same result. So at every time, even at late times, the entanglement between the pair will be given by the young, young entropy computing at the initial time. If you consider the particle that started in a particular position, okay. We did this, and if you are interested in the final expression, you can find it in our paper. I don't think that this is the place where to show it, but I can just show that it seems to work, okay. All this formalism. In particular, these are results for the half-chain entropy. So the entanglement entropy of the statistic consists of half of the chain, okay. And for the X, X, Z model. And in particular here, we can see the initial state, that's called domain wall states, or no. Before it was called bipartite state, I don't know. So, well, we can see this particular state. So I left just a near state, and I'm sorry, right, is a fair amount of the state in a tilde position. And then we check our theory for different angle and different values of delta. And the numerical data are consistent with the analytic prediction, although, okay, it's very difficult to say whether this is actually exactly correct or not. Because at least from numerics, it's impossible, okay. And okay, okay, with this, I finished. So this is a very brief summary. I tried to re-obtain, in a sense, all the known results about the entanglement entropuses started from generalized agronomics. And in particular, we have seen that the first order GHD supports the semi-gascals picture of the time elusion of the entanglement entrop. So in a sense, we understand a bit better why it works. It was working since it always worked. And moreover, okay, using this theory is possible to make prediction even in the presence of interactions. Okay, I can't hide anymore that there is a problem. It's not true that everything is solved. And in particular, there is a subtlety related to how I describe things, how I present the things. In particular, I told you that you can just write down a density matrix to find an equivalence between density matrices. The problem is that if this were a case, if also in the interrupting case, we could do something like that, then we would have immediately an expression for the time elusion of the re-entropuses, for example. We could immediately find it. And because there would be some relation between the phonamine entropy and the re-entropes. Unfortunately, this doesn't seem to work, okay? And okay, the reason is kind of clear, but nevertheless, it's still an open problem that we didn't solve. And with this, I thank you for your attention. Thanks a lot Maurizio. So I think we are a bit late. So we have time only for very quick questions. So are there some really pressing ones? Yes, can I ask something? Please, Paola. Yes, so thank you Maurizio for the talk. So I would like to ask you, so I mean, I'm more familiar with the entanglement from the point of view of conformal field theory. So I guess that in the way, I mean, I don't know very well how much you can go, you can get explicit expression in the interacting case for the entanglement. For example, the asymptotics and this kind of stuff. But I was wondering whether, so if in this picture, you could imagine that the leading contribution is given by the fastest particles as in the usual way of thinking from the conformal field theory point of view. And in that case, you could think of following just the fastest particles, okay? And that would correspond in a CFT language to a CFT in carved background. So can you imagine of calculating? Yeah, okay, qualitatively, I think that you can find, see. So you see, the way I understood what you do from the deep point of view is like every particle has its own dispersion because it moves in a different way. And so each one in my way of thinking would correspond to a different carved background. But probably there is one which contributes more, which is not... No, okay, this is not true. No, there is not really particles that contributes more. Now, and especially if you imagine this kind of settings where they started with the domain wall case, then you really have to take into account all the other particles. They are, there is some democracy, right? Not really. You can find the qualitatively similar behavior. But if you want to be more quantitative, even if you now imagine to perturb it and you add the curvature or something like this, like Lorenzo did. So there are cases where you can actually reproduce the results like low temperature, but I'm not sure that you can find something accurate in general. So I... Even in the homogeneous case... Yes, yes indeed. You find something qualitatively similar, but then... Yeah, well, I've said that in the homogeneous case, you do find, I mean, like the main growth, at least, and where... Okay, for example, something you might not find if you can see, if you simplify the problem so much, for example, you can't see the effect of the interaction or the particle scatter before reaching the subsystem or something like that. If there is only one velocity, you are missing something. Which can be... You would not capture the leading behavior you're saying. It depends on what you mean with the leading, qualitatively, the qualitative behavior, possibly, although with some exception, the leading, no. The leading, I mean, the temporal portion to the volume. Yeah, that's... Or to the time, no. You want it. Yeah, you probably don't get even the slope, right? Because... No, no, you don't get anything. No, no, no. Maybe just a simple comment. If you look at the in homogeneous ground state, described by the left, and you do a small branch on top of it, you will excite only the particles at the local family velocity. So we'll have some particles moving this in homogeneous background or velocity background. So it's like you're neglecting a safety part. You can see that only the excitation. And the excitation will move with the velocity that you plug into the left. I think that this is the closest example to what you could have in mind. But of course, we are not... How many longer, I guess? But also we think that... But also we think that... As soon as you have a finite density of visor, you have these curvature effects in general. Yeah, yeah, of course. Like as Lorenzo was talking about. The behavior just in the homogeneous velocity. And if you want to go further on top of it, of course, curvature and so on and so forth. But if you want to measure a single velocity in homogeneous, you can play this trick. Can I ask a very quick question? Please, go ahead, yeah. So Manitza, the correction with the cubic derivative, does it produce entropy or not? That's okay, that is... I don't know. I didn't ever use the third order question to compute the entanglement entropy. But I mean, if you remove it... I am pretty sure that it has an effect. It should be a boundary effect, so you don't see it at the end order. Okay. Without doubt, the cubic term plays an important role at the icon. Indeed, you can see the KPC University class using the third order term. Yeah, but the total entropy is conserved by zero over the GHT, right? Yes. But if you add that term, is it still conserved? You mean that the young entropy, now you are talking about? Yeah, yes. I don't think so, right? You mean integrated over the whole body? Yes, I... So you will satisfy it with the same equation because it's just a linear equation in raw. With more derivatives. So when you consider derivative of... Ah, no, just a moment. No, no, no, it won't be... It will be... No, it will satisfy a different equation, sorry, because you have a function of a row. So it will be something different. So the question is whether it has a total derivative in space. So I mean... Okay, I don't know if actually it really contributes. I think... The diffusive correction increased the total entropy and this is because of this... What do you think? That's a very quick argument. It's time reversal invariance, though it won't. Right, that's true. But what I really think is that this can be really important if you consider really at an icon. When you consider time evolution, there is a moment when the subsystem is around the icon, then you can really see the effect of this third order derivative. Also on the entanglemental entropy. Yes, I think so. I think so that you should see something. You can see something. But... Can you just say something? Yeah, please. Very quick, Jerome, yes. So thanks, Mauricio, for the talk. So this term will also affect the density in the sense. So it will make it either negative or larger than one in places. Yes, true. So that you cannot interpret it as a... Yeah, indeed, indeed. Okay, this is a very important point, indeed. Because when you go beyond the first order, then you can't interpret any more raw as the density of particles. So all the pictures somehow doesn't work anymore. You have to... Because it could also be possible that you will have to reinterpret the density as something different, something beyond raw. So it's... Yes, you're right, it's not... Okay, so I would suggest to move any further discussion to the discussion session with Vincenzo. And I thank everyone for...