 So I'm going to talk about the work that I did at the end of my PhD with these people, Machete and Tremac that were my supervisors at that time, Tony Afien, who was a group leader at IGFOR and Ignacio and Gemma that were at MPQ at that time. The paper you can find it there. I have been told that this talk is recorded and so you can also find this talk in YouTube. It's not the first time that I give it and, well, here you can find what can happen when you put your talk on YouTube. Luckily for us, the reference comments were much more positive and, well, if Bob Lutter is in the audience, I would be happy to discuss whatever he has to say, okay? So I hope it has better reception. So this is the outline of what I'm going to tell you. I will first motivate the results and the question that we are investigating, right? I will present what is the main idea and the setting and then, well, for this talk I will have to discuss two optimizations. One of them is quantum. The other is classical and, well, the connection is through assigning a belly inequality to a physical Hamiltonian that I will also present. Then if you have some symmetries like translational invariance and everything is much nicer and depending on the time that I have left, I will discuss some examples and, well, conclusions and further outlook. Okay. So I think that many people in the audience may not be familiar with the so-called device independent approach or non-locality and, well, I will just give first of all a general introduction to the field and the way that I like to present this is that, well, if you think of quantum distribution, maybe this is the most mature quantum technology at the moment. I mean, there are some companies, right? I mean, there was in science last June, the new record in this, right? And if you look at why people claim that this is secure and so it's based on the theory, right? The theory makes claims such as that you have a Hilbert space in which you have qubits, for instance, right? Of course, if you go to the lab in this, I would not dare to call it C2, right? It's something that is much more complex and, well, clearly there are some mismatches between theory and implementation and, well, if you exploit the loopholes that appear in this mismatch, right, you will have someone that you can't call quantum hacker, right? This is also a real picture. And, well, so it's in 2010, one of these commercial devices was hacked, right? And the reason was not that the theory was incorrect, right? The mathematics were fine, but of course, I mean, there is a leap that you have to cover, right? And, well, then the new approach in the field was this so-called device-independent approach. So instead of looking at my laboratory with polarizers and beam splitters and things that are very difficult to characterize, I will just treat it as a black box, right? And this will have some knob in which I select some measurements to perform, and each measurement can give some output, right, like the spin is plus one or minus one, right? So it will have some outputs, and, well, I will make measurements on this box and I will get some statistics. So now the paradigm is a bit shifted, right? Because I'm interested, for instance, in what is the number of inputs and outputs of my box, right? And the question that I like to ask is whether the observed statistics are non-local, right? So rather than, for instance, if my quantum state is entangled, right, this would be a question about hyperspaces. So if the statistics are non-local, OK. So why consider bell correlations, right, or non-local correlations? So they are a resource for the device-independent quantum information processing tasks, right? As I argued, if you make less assumptions, you have more security, right? You don't need to work with keyword spaces, right? And, well, so these correlations have been actually measured in some remarkable experiments two years ago, right, in this so-called loophole free belt test. Of course, it's much more demanding to show non-locality than entanglement because you perform less assumptions, right? And therefore, it's harder to show this, right? Also at MPQ where I work, there have been some experiments like this, right? And with this, you can do, well, device-independent and whatever quantum information task from this list, like quantum distribution, randomness expansion, amplification, self-testing, many things. The connection with entanglement is that bell correlations are stronger than entanglement. So not all entangled states can display non-local correlations. But if you have no local correlations, the only way to generate them is from an entangled state, okay? And, well, if you look at the literature, so most of the studies and applications of bell correlations deal with small systems. And here, we try to pose the question in a bit different way, right? So we were interested in seeing if these bell correlations appear as a natural resource in low energy states or physical systems, okay? So the rest of my talk is going to deal with many body states, right? And we want to see if these correlations appear naturally in their ground state, for instance. This is a study problem because of the mathematical complexity that you have. It's also more experimentally demanding because you may need to measure more particles, and so you have a question? Okay. And, well, you know that the quantum description of multi-partite states grows exponentially with the system size, right? So you also have to overcome these difficulties, right? In previous works, we dealt with permutational invariant systems, right? And this had been shown in some experiments of hundreds of thousands of atoms in a thermal state of a B or on a BEC. So there has been some progress in that. And in this talk, I will focus on spin systems in one spatial dimension, okay? So not permutational invariant, but you have to think of one of these systems, okay? I will spend one slide on defining what is non-locality and so just to follow more easily the talk. So you have several players, Alice, Bob, Charlie, and so on. And each of them has their own black box. So this will be their laboratory. And they have some inputs, X1, X2, X3, right? And this is the measurement that they choose to perform. They don't know what's exactly happening inside the box, right? But they read out the measurement, right? So classical information goes in and classical information goes out, right? And at the end of the day, they repeat this many, many times and they can estimate these conditional probabilities of the outputs given the inputs. Right, and you can collect this huge table of numbers in a vector. And this you can represent in a point in some geometrical space. So now the question is that, well, depending on the assumptions that you make on the physical theory in which this experiment is carried, right? There is a feasibility region according to your physical theory. So if you have a local hidden variable model, right? This region looks kind of picture like that. So it's a polytope, right? So it has flat faces and a finite number of vertices. And well, they'll show that the correlations that arise from quantum theory are actually larger than this set, right? And you can also consider more general physical principles, like the nosing and the link principle and so on, right? So the operational way to interpret this set is so the LHV set, right? The set of classical correlations, well, local hidden variable model, is that the vertices are very special in this set. So they correspond to when everything that happens inside of each of these boxes is a deterministic program, okay? So given the input, the outcome is determined and only by the input of such of the same box, right? And then, well, this players could have agreed in the past to follow some strategy, right? This would be like analogous to the hidden variable. And here I have made an example, right? So if I call lambda my hidden variable, right? Ali's, Bob, and Charlie could have agreed on some probability distribution on lambda in the past. And they say, okay, when the value of the hidden variable is one, we are going to do something, right? And then when it's three, we are going to do some other thing. And in this way, you see that you mix different vertices, right? And then you can access any point inside of this region. Okay, then a belly inequality is simply a half space that contains the local polytope, right? So geometrically, you can think of belly inequalities, like that facets of this object. Okay, so let's go to the idea of our result, right? So we would like to start with a Hamiltonian, right? And from this Hamiltonian, we assume that we know how to compute its ground state energy, or at least to prepare a state of sufficiently low energy, right? I will illustrate that with Jordan Bigner or some numerical methods. So these are all known tools to achieve this result. And we will establish an equivalence from this Hamiltonian to so-called Bell Operator. The Bell Operator is when you have a belly inequality, you put quantum measurements in it, then you have an object that is in the Hilbert space, right? And then you can very easily identify that with the Hamiltonian, right? And then, well, you perform measurements on your system, and you get some expectation values. So this is the quantum value, right? And then we will connect that to the belly inequality. So it means that, well, Hamiltonian or a Bell Operator, if you choose the inequality coefficients appropriately, then you get some inequality, but it also works the other way, right? So from Bell inequality, if you make suitable observables, suitable measurements on it, then you generate different kinds of Bell operators or Hamiltonians, right? So this belly inequality is what's important about them is their classical bound, right? So the limit that they can achieve within the local hidden variable model theory. And with this, I will use a technique that is kind of underused, I would say, in the field of quantum information, but it's very general and maybe used in many ways. And, well, finally, by comparing these two values, you would see if you have Bell correlations, so, well, namely, if you violate this belly inequality. If you, on top of that, you have some symmetry like translational invariance, and everything is much nicer, right? For this side, we will have analytical expressions, and for the other, we will have an exponential speed up in the way to compute this classical bound. Okay. So to fix the setting, I will, as I said, I will consider spin one-half Hamiltonians and particles in one spatial dimension. If they have open or purely boundary conditions, it's not so important, so you can think of either of them. What is important is that their interaction range is short, right? So I will call that capital R, and you have to think of this as a constant, right? The system size is variable, but this R has to be a constant. So it could be a Hamiltonian that has this form, right? So, for instance, some sigma sets, right? And then some string operators like that, like some X, some Z in the middle, and some X or Y at the end. Why do I put such a particular form of a Hamiltonian is that, well, because this, I know how to compute the ground state energy easily and analytically, but the method is not restricted to that, as I will illustrate in the end of the talk. But I think that it may be, I mean, some people may be familiar here with the Jordan-Biener transformation, that is how you solve these kinds of systems, and then it's easier to follow. This is also XY model in a transverse magnetic field, would be a particular case of this Hamiltonian. Okay, so how do you find the ground state energy? So, in this case, you can do exact diagonalization. And, well, there is a mathematical trick that maps spin Hamiltonian to a fermion Hamiltonian, right? So, on this CI0 and CI1 are Majorana fermions, so they obey some canonical anti-commutation relationship, right? And then every Hamiltonian of this previous form that I showed becomes quadratic, and this is the key point to find the ground state energy in this way. After you make all the substitutions, right, you find that you have a Hamiltonian of this form, but that is quadratic in terms of fermions. This matrix of coefficients is also real and anti-symmetric, and you can exploit these properties, because real anti-symmetric matrices, they admit what is called a Williamson eigen decomposition. So, it means that there is an orthogonal matrix that almost diagonalizes them, right? You have these two times two blocks, and it's a block diagonal. And then, well, this actually means that you have a new family of Majorana fermions, and, finally, your Hamiltonian looks like that. So, a sum of n terms, right? And each of these operators mutually commutes, so this means that you can recover the spectrum of your Hamiltonian from these numbers, right? The eigenvalues of these operators are plus minus 1. So, you see that to find the ground state energy is very easy, because you just have to take these numbers with plus 1 eigenvalue for this, and with minus 1 for the rest, all right? There are some subtleties that are like a parity in post-superselection rule, but I'm not going to go into these details. The point here is that you can compute that very efficiently with the system size, because you see that this is not exponential in n, it's just linear. Okay, so now we have to assign a very inequality to the Hamiltonian to ask what is the non-locality of its lower energy states. And, well, the key is that we want a Bell operator that resembles as much as possible as the original Hamiltonian, otherwise it's no use to find this ground state energy, right? So, we want something that is just a Hamiltonian, but shifted by a value. So, to go from a very inequality to a Bell operator, you have to choose some measurements, right? So, it means that in the block sphere, Alice has to do some measurements, and now it's the moment that you have to choose them. And the most general thing that you can do here is to take measurements in the x-y plane and one measurement in the z direction, and then you can, so, making many measurements here in this plane, and another one in the z direction, you can actually obtain Bell operators that have this form, and then the very inequality looks like that, right? You would have one, if you have a single operator here, you have to measure a many-body correlator. If you have a one-body operator here, one-body correlator and so on, but now if you know how to compute the classical bound of this inequality, then you can hope to answer the question. Okay, so how do you compute the classical bound of the inequality? So, the inequality, I will write it in this form, I plus the classical bound greater equal than zero, so if the expectation value of this is negative, it will be violated, and there will be no local correlations. And, well, this is the definition, right? The optimum over all the local hidden variable models of the inequality, right? So, geometrically, it means that I have a hyperplane in some direction in this space, and I just have to push it up to the edge of this set, right, this is the optimization. So, Fein's theorem, this old result, tells you that this is enough to optimize over local deterministic strategies, right, over the vertices. And, well, so how many local deterministic strategies do you have in a many-body system? Well, this depends on how many measurements the parties can perform, right? So, for instance, if you can perform three measurements, and on the number of parties that you have, right? Alice, Bob, and so on. So, you could say, okay, the deterministic outcome for the first measurement of Alice is going to be black or plus one or whatever, right? And then for the other two, it could be like this, right? And then you fill this table, right? And this is the description of one of these vertices. So, you'll see that this is very bad news because, well, you have a number of vertices that is exponential in the system size, right? Also in the number of measurements. But, well, our very inequalities actually are built in a kind of 1D geometry, right? They contain short-range interaction terms, right? So, it means like low-order correlators. We are not exploiting this at all, right? So, with this technique dynamic programming, you can actually find this efficiently. Okay, so this optimization that you would do with linear programming in the general case, and you have to put as many constraints as vertices so it's impossible for many-body inequalities. It's actually, well, doable and extremely efficient for 1D. So, to apply this dynamic programming, you need some ingredients like recurrence relation. You need to be able to compute and store intermediate sub-solutions and you order them, right? This is just a very abstract way of defining it, right? Because it's a very general technique. And, well, you get normally a polynomial scaling and a constructive method of one of these vertices, right? And so it's much better than the brute force approach, right? And, well, to convince you that this is a very general technique, right? If you go to some, I mean, this website, they make coding competitions. And if you look at the kind of problems that they put and the technique that you use to solve them, right? Dynamic programming is a third, which is quite remarkable, right? So, it's something that's useful beyond this talk. Okay, so for this particular case, so we want to express the value inequality as a sum of smaller ones, right? And then the optimization is, well, you have some sub-solution at some point, right? So, you have filled part of the local deterministic strategy that is optimal, right? And, well, you have something beyond that you really don't care because you want to optimize this part in the middle. And the inequality is made of short range interaction terms, right? So, whatever you put here will not affect what the solution is going to look like afterwards, right? So, I did not put the actual formula. I think that, pictorically, this is easier to follow, right? So, you would like to, at each, there will be n steps, right? At each step, you would optimize these squares. They may depend on something that is a little bit beyond because the inequality is short range, right? But not on more things, right? So, you have a recurrence relation and you define this by, well, seeing what are all the possible values here of these triangles. Because this is a finite number of values. You can compute all of them and then you have this recurrence, well-defined, right? And then you find a classical bound as the end product of this optimization, right? So, here, what I think it's non-trivial is that the overall complexity of this is linear whereas in the general case, this would be exponential in the number of parties, right? So, you have an exponential speed-up in computing this value. So, if you have two national invariants, now things become much more nice. So, in the case of the classical bound, you would have here the same link everywhere. So, it means that, well, pictorically, I think that it's better that I describe the figure, right? So, you would have several sides with end parties, right? And then you can define a function that depends on, okay. So, that would allow you to remove the even sides, right? So, you would say f of x and y would be the minimum over z, that is the one that is in color of that link plus that link, right? And then this is a well-defined function if you compute it for every x and y, which can take a finite amount of values. And then you can go to the next row, right? And then you can keep repeating that, right? And then you achieve exponential speed-up because you remove half of the parties every time. And, well, not everything is a power of two in life, but then you can apply just the previous scheme, but you only have a logarithmic number of terms left, right? So, you compute in log n steps the classical bound, right? So, it's a doubly exponential gain with respect to the most general case. Well, if the r is greater than one, you can also do it, it's a bit more tricky. I don't have time to explain that. Okay, so, in the quantum case, this H matrix of coefficients has the additional property that is block-circulant, right? And then you can hope to find the grand state energy analytically even, you don't have to run any numerics. And the thing is that block-circulant matrices are diagonalized with discrete Fourier transform, right? So, this orthogonal matrix that was diagonalizing H in the general case, now you know which one it is, right? And then, okay, depending on the parity of the system, you have to do a couple of fixes, but basically both of them work, right? So, you have a matrix that its elements would look like this, right? So, it would be the same element on the diagonal. And then, well, okay, long story short, means that you have an analytical solution for the epsilons, right? That would be the Williamson eigenvalues, right? So, you put all these formulas here inside of this, and then you can compute analytically what is the grand state energy, okay? So, maybe I illustrate that with an example. I think that, okay, maybe the most interesting one, I put many of them, but I would like, okay. So, if we consider x, x, z model, and inequality in quantum information that is called jizan's, well, he calls it elegant inequality, the inequality looks like this. So, you have two parties, Alice and Bob, right? And you have these kind of coefficients defined in the inequality. So, Alice has four measurements, Bob has three measurements, right? And then, here, I added this delta to then make the mapping to the x, x, z model, right? And, well, Alice and Bob performed the measurements in these directions in the block sphere. So, they are quite symmetric, right? But the point is that the Bell operator, when you perform these optimal measurements for jizan's inequality, it looks like this, right? So, sigma x, sigma x, plus sigma y, sigma y, plus this delta, so this is why I added it, sigma z, sigma z. And so, this is very close to the x, x, z model, right? If I would make some of these inequalities, right? I would have, well, actually, something that would be exactly the x, x, z model. So, here, you see that the, well, the equivalence between the Bell inequality and the Hamiltonian is, can be non-obvious, because here, we have a Bell operator that is permutationary invariant. If I exchange Alice and Bob, nothing changes. But the Bell inequality is not symmetric in any way, right? Because, I mean, not even, it has the same number of measurements at each side, right? So, but, well, nevertheless, we found this correspondence. And then, well, you can build up a toy model, like using this jizan's inequality here at these links with these weights. You can add an epsilon parameter, right? So, some weights are stronger than others. And then, you have that the classical bound looks like, well, you can find it analytically, and it kind of phase diagram, I would say, but this is for the classical bound, right? So, depending on the epsilon and the delta that you put in the inequality, right? You have, well, several values, several kind of inequalities, depending on the parameter region that you have. So, now, how do you compute the ground-state energy of this Hamiltonian? So, in this case, you cannot use the Jordan-Biener transformation because this would not work in this model, but you can run some numerics on it, okay? So, this would be the Hamiltonian that I am interested in studying, right? So, it's Hamiltonian exact jet model. So, if you use matrix product states, or the MRG, or well, any tensor network method, so here we used this one, you have a result that looks like this, right? So, this corresponds to a different number of parties. This would be the parameters of the, that you put in the model and the inequality, right? So, delta and epsilon, and you see that there is a region in the middle that is non-loga, right? So, for any value of these parameters, well, the ground-state energy of this Hamiltonian has bell correlations, okay? So, you can certify, for instance, that the ground-state is entangled even without knowing what state that you have prepared, right? Without making assumptions on, for instance, if it's formed of qubits, right? Or what measurements you are actually performing on the state. And this seems to be robust as N grows, right? Because you see that this region doesn't seem to shrink. This is just an example. You can do it with many other very inequalities, right? But I have to skip them in the interest of time. So, let me go to the conclusions and, well, so to apply this method, if you want to show non-locality in a many body system, basically what we did, or what you would need to do is to solve three problems at the same time, right? The first is that you should be able to solve efficiently the combinatorial optimization to find a classical bound, right? And for this, I briefly explained that you can do it with dynamic programming, right? But if you find another technique that is efficient, this would also work. Then you have to show that the inequality is actually non-trivial. So that quantum physics can violate the value of this inequality. Otherwise, you are not going to show anything, right? So, in our case, because we have made this correspondence between the inequality and the Hamiltonian, this is reduced to finding the ground state energy of the Hamiltonian or a sufficiently low energy, right? It doesn't need to be the ground state. And you can do it, well, for some models exactly with the Jordan-Biener transformation, but for more general models, you can do it numerically and that's fine. And then on top of that, you would want to make it experimentally accessible, right? So if you have to show this non-locality by making many measurements or very complicated measurements that require single particle addressing or they are very difficult to do in the lab, then nobody could actually do it, right? So here, we think that it's advantageous that with a few body correlators or two body, right? You can actually measure the ground state energy because your Hamiltonian is a sum of local terms, right? But, well, you can also use other methods to estimate the ground state energy, right? If you have a quantum computer, you choose the phase estimation, for instance, but it depends on if you're happy with this hypothesis. Then, if you have a symmetry that is natural to the system, so we have here discussed 1D system, so it's natural to consider the relationally invariant. Invariance, well, in the quantum case, we have closed formulas and in the classical case, we have exponential speed improvement, right? So this is a tool set that allows you to study non-locality and physically relevant systems, right? So you need spin systems, one spatial dimension and short-range interactions. And here, short-range interactions, it's, I think it's funny because this is what limits mostly the classical optimization, but not the quantum one, right? Or at least, well, for some Hamiltonians. So let me discuss further directions that one could take from here. So we started studying this because, well, contrary to the permutationally invariant case, there is no definite restriction. So we thought that the inequalities would be more robust, right? So if you have a permutationally invariant operator or a permutationally invariant state, then you know that if you look at the reduced states of its subsystem, all of them are equal because it's permutationally invariant, but they go to a separable state with some speed, with a system size, this is what the definite tells you. This doesn't happen in one of the systems and, well, this makes this more robust in this sense. So this work I have presented that you can see the Hamiltonian as a particular realization of a value inequality. They have skipped the details, but I hope that this idea is clear. But one can also put the problem in a different way, right? So now you give me a Hamiltonian and I will try to find what is the inequality that best reveals the new local properties of its low energy spectrum, right? Because so the ground state energy of the Hamiltonian, I need to find it only once or the energy of the system. I need to measure it only once, but if I change the value inequality, if I make sure that also changing the measurements I keep getting the same Hamiltonian, I only have to do the classical optimization, right? And this I can do, I mean, not in the lab, right? And in the fully translational invariant case, so there is a phenomenon that is called monogamy of correlations. And it's still unclear to us how this affects non-locality. So monogamy of correlations is like similar to monogamy of entanglement. So if you have three parties, Alice, Bob and Charlie, and they measure, for instance, value inequality between Alice and Bob and the same between Alice and Charlie, you would see that, for instance, the most basic ones, CHSH, if Alice and Bob violate the value inequality, Alice and Charlie cannot violate at the same time, right? So there is some constraint that affects this when you make a many-body system. And well, we have been thinking also how to generalize this to more spatial dimensions, and well, there are some tools about code of extensions and semi-definite programming that we can discuss later. And then also I think that one of the referees pointed out a very interesting question that is to study the persistence of non-locality, right? So what happens if you miss some of the parties from your system, right? If you trace out some parties from your system, can you still show these results or they are not robust to particle losses, right? So for instance, if you had a state like G8Z, you know that if you lose one qubit, then this is indistinguishable from a classical mixture, right? So how robust are these inequalities to this kind of phenomenon? So with this, I would like to finish and thank you very much for your attention.