 Oggi vorrei parlare del sistema quantum, in particolare dei modelli fermi, che sono introdurati da Matteo già l'altro giorno, e vorrei discutere delle proprietà per equilibrio dei gas fermi e della quantità, che è la energia di correlazione, che sarà introdurata in un momento. Lo che vedete è basato in un lavoro con Nils Benedicter, in Milano, Pham Tham Nam, in Munich, Benjamin Schlein, in Zurich, e Robert Salinger, in Vienna. Ok. So, let me tell you what is the setting that I would like to consider. So, the setting is about an interacting and confined... Oh, sorry, I should probably turn this off. Sorry for that. Sorry. Right. So, N interacting and confined thermionic particles, where N is going to be a large number, eventually it will be taken to infinity, but the system will stay confined. Ok, so it's a mean field regime. So, we're talking about a quantum system, so the state of the system is specified by the wave function. Ok, and this is going to be an anti-symmetric wave function for a system confined in a box lambda, which I will take with periodic boundary conditions. Ok, so my lambda is going to be a three-dimensional torus. Right. And the side of my torus is going to be 2 pi for convenience. Ok, so you should think of that as being populated by N particles and the typical distance between two particles is N to the minus one third. So, it's a very high density regime. Ok, so anti-symmetric means, let me just remind you that if you look at the wave function and you permute, well, if you permute the arguments, if you permute the labels of the particles, what you do, what you get, well, you basically just pick up a sign, the sign of the permutation. Right, so that's my setting, so it's a very high density situation and I will consider a mean field Hamiltonian. It's going to be a self-adjoint operator on L2 of lambda N. Ok, of the type that Matteo already introduced, but now I will focus on the mean field regime. So, there's going to be a first part which encodes the kinetic energy of the particles which comes with a sort of semi-classical parameter epsilon squared. This is going to be part, I mean I will motivate that. For the moment let me just write it and then on top of it I will introduce a many bad interaction of mean field type. Ok, so it's a very interaction with the two-body potential which later I will give condition, I will specify what kind of potential I will focus on for the moment just think of B to be a bound potential. So, what about this parameter epsilon? Well, this parameter epsilon is an independent parameter it's n to the minus one-third, or order n to the minus one-third. Ok, so you see that with respect to the bosonic, I mean classical mean field or bosonic mean field, the difference is that we have to, we are taking into account this semi-classical parameter which makes the problem harder because it makes the spacing between the eigenvalues of the Laplacian smaller and smaller as I am accustomed to. Ok, so what we are focused, what we are interested in for the purpose of this thought is the ground state energy of course on the fermionic sector of the Hilbert space so en is going to be denoted, is going to denote the infimum overall wave function in L2 anti-symmetric lambda n which are normalized of the expectation value of hn over epsilon. Ok, so that's the quantity that I would like to discuss. I mean the point is not really, I mean the number per say is not that interesting physically but what's interesting is to get some amount to compute this number with high precision which is in fact what we are going to do, you have to develop a precise understanding of how the ground state looks like and that's actually the content of this talk. Alright, so that's about the model now in the remaining in the next minutes I would like to motivate why we have this epsilon here so the reason is that I mean what you want to do really is to be in a setting in which there is a balance between kinetic and potential energy so as usual I mean if you think of V to be a bounded potential if you look at the interaction energy over a reasonable class of states then you would expect this guy here of order n and so basically the role of this parameter epsilon is to make the kinetic energy also for the n ok so I want to kinetic energy and potential energy to have a chance to compute. Ok, so to understand that let me consider as an example the free Fermigas ok so or if you want a non-interacting problem and I would like to discuss the ground state of this non-interactive problem so the Eigen states of my Laplacian the Fermionic Eigen states of the Laplacian they are just anti-symmetric tensor product as we have seen from Mathieu's talk these are called later determinants ok so you have some of our permutations and you got the sign of the permutation and then you have the product of Eigen states of the Laplacian single particle ok so this fk denotes a plane wave so fkx is 1 over 2 pi to the power 3 over 2 that's correct l2 normalization e to the ikx and I want my wave function to be compatible with the periodic boundary conditions so I will take k to be a point in z3 and then another feature of this wave function is poly principle so the k's have to be all different otherwise you get identically 0 so k i has to be different from k j for i different from j and that's due to the anti-symmetry of the wave function if you want poly principle alright so computing the ground state in the absence of interaction amounts to find the choice of k that minimizes the energy of the state ok so let me do that so the energy of any such state so it's just kinetic energy let me forget for a moment about the parameter epsilon squared that's just simple computation somewhere k of or if you want some over i from 1 to n of k i squared where those k i's are those that appear in this later determinant ok so you want to minimize this quantity enforcing the constraints that k has to be a point in z3 and the k's have to be different so you can feel only one state every particle I mean you cannot have w occupied states so how do you do this well that's clear so let's look at ok so you should think of that as being z3 no plane because otherwise I would make a mess so what do you do you start to occupy states keeping somehow making sure that the energy is as small as possible right so you start from k equal to 0 and then you go on and you start to populate the states that have the smallest norm once you measure it from the origin ok so you get this 4 and then you keep going you get those guys you start to feel a lattice and you see that this thing here as a radical symmetry ok so you start to populate a ball which has radius kf this is called the fermi radius or fermi momentum which grows as n to the one third ok so clearly if you just give me any n particles then you might have that the ball is not completely filled there might be boundary terms and from now on I will always assume that n is chosen in such a way that the fermi ball is completely filled so the number of points in the fermi ball which is this ball here is equal to my number of particles and that's of course in general there might be boundary terms ok so in this setting then you can ok so if you just look at this quantity here for for k in the fermi ball then it's an exercise I mean it's not difficult to see that if you just sum those numbers what you get is something which is asymptotic to n to the 5 over 3 ok so probably principle it's making kinetic energy higher with respect to the bosonic case so that's of course something you can just somehow it's an insight that you get started from a special class of states but there are general inequalities that give you a lower bounds for the kinetic energy that are compatible with this this is called the leap theory in equality so this epsilon squared is there then to to turn this kinetic energy into something of order n alright so that's a non interacting problem now I would like to go to the interacting problem ok v non 0 and that's of course much more difficult because while in general the ground state of the system is not going to be uncorrelated but you will have to be able to take into account correlations among the particles ok so even in the mean field regime computing the ground state energy of this problem is a difficult task because you have to understand correlations among the n particles alright so that's what we want to do and so I mean of course that's a problem which is somehow it's been studied in physics and a popular approximation for this for this problem is the Hartree-Fock approximation which tells you basically that while it's based on the assumption or on the simplification that you replace the full fermionic the space of the fermionic wave functions by the space of uncorrelated fermionic states so you replace the two anti-symmetric by the space of the set of slater determinants ok where those f i's they are orthonormal functions in the two of single particles ok so these are the slater determinants that we have seen in Mathieu's talk these are fermionic uncorrelated states and the hope is that in this replacement the problem simplifies because if you don't have correlations among the particle then in this mean field regime you would expect every particle to experience a mean field due to everybody else that's the same mechanism that we have seen in Peter's talk yesterday so it's a large number type of thing ok so from a mathematical point of view an advantage of slater determinants is that they are sort of Gaussian states in the sense that they are k particle density matrices that have been introduced by Mathieu completely determined by the one particle density matrix completely determined by the one particle density matrix exactly as for a Gaussian field ok so in that case it would be the covariance so the one particle density matrix the general definition is just a trace over all particles except the first of the project over the state of the matrix of the project over the state psi n you can compute that for a slater determinant and what you get is a rank n projector over the sub space of the Hilbert space which is spent by the n orthonormal wave functions that enter in the slater determinant ok so it's a rank n orthogonal projection ok so that's a an interesting class of states where they're simpler than usual fermionic states because there are no correlations and you can compute the energy of such states ok so psi n here is a slater determinant you can compute the energy of those states and as Mathieu told us I mean in general the energy the interaction energy of a many body problem depends on the two particle density matrix but the two particle density matrix can be expressed in terms of the one particle density matrix for slater determinants so what you get is the final expression is this expression here so there is a first part which is exact is the kinetic energy plus a term which encodes the many body interaction and which is quadratic in the density matrix ok so there is a first part which is of density density type exactly as for a classical mean field or right so there would be just density density type of interaction and then you have to correct the fact that particles are not really uncorrelated but there is the anti-symmetry requirement and that's why you have to take into account these other extra terms so this is called the direct term and this is called the exchange term ok so this row by the way is row of x is the density of particle at x is just the diagonal of our density matrix ok so this is just you see it's a non-linear functional of the density matrix omega it's also called the Hartree-Fock energy function so that's a Hartree-Fock energy functional which plays a central role in Hartree-Fock theory ok so that's of course an easier quantity to consider than the full many body problem but still it's non trivial task to understand the minimization problem associated to the Hartree-Fock energy functional ok so let me define the Hartree-Fock runs at energy as the smallest energy that is later determinant can reach ok so since there is a one to one correspondence between those rank and orthogonal projections and this later determinants you can just minimize over rank and projections your Hartree-Fock energy functional ok so that's that defines the Hartree-Fock grand state energy ok so that's the same thing as minimizing over later determinants so that's the definition and clearly since we are minimizing over a smaller class of states with respect to the original problem that gives an upper bound on the many body energy ok so trivially the true many body grand state energy is less or equal than the Hartree-Fock grand state energy ok so that's so if you're able to compute this grand state energy at least you're able to come up with an upper bound for the original many body problem so of course much harder is to understand the lower bound of the many body energy in terms of Hartree-Fock and that's the thing I would like to focus on and in particular I would like to discuss the so-called correlation energy which as the name suggests is the energetic contribution due to the fact that fermionic states are not exactly the later determinants so the definition of the correlation energy is just the many body grand state energy minus the Hartree-Fock grand state energy ok so what I would like to discuss today so our goal is to is to understand the asymptotics of this guy as angostin field ok so that's what we would like to focus on so let me make a few remarks no sorry I have to find out oops let me make a few remarks about this Hartree-Fock problem so as I said in general it's a difficult minimization problem I mean it's finding out the exact grand state of the Hartree-Fock problem by itself a very difficult problem however in the present confined case it's actually not so difficult due to the fact that after all the Laplacian has a gap so you can exploit that so let me just give it to you as a remark in our setting so confined system with bounded potential and then later I will make more precise assumption on the potential the Hartree-Fock grand state is actually equal to the plane wave state in general this is not too so let me call it plane wave to say that this is the free Fermigas in general that's not true so when I say in general I mean in the thermodynamic limit where the spectral gap of the Laplacian is just not there anymore you might have symmetry breaking in the infinite volume limit and in fact there is a very nice paper of so in general not true there is a very nice paper of mathematically rigorous paper by Gontier Heinzel and Levine from 2018 where what they proved is that in general what I mean what they prove is that this plane wave state in general is not the grand state but is exponentially close energetically to the true grand state of the system exponentially close I mean exponentially close in the density ok let me not write down the precise estimate but that's just what I want to do to emphasize here ok now what about rigorous work on in fact not really computing the Hartree-Fock grand state energy but understanding the validity of Hartree-Fock theory starting from the many body problem ok so let me make here another remark rigorous work on Hartree-Fock on on validity of Hartree-Fock equilibrium ok so I will mention two words one is work by Falker Bach in 92 where what he did is to prove well he didn't really consider this setting I mean he considered the setting of large atoms which however can be recasted in terms of mean field problem by large atoms I mean that the nucleus has a large charge of the order of N so what Bach proved is that the correlation energy in this setting here would be a little o of N to the one third where N to the one third is the smallest contribution entering the Hartree-Fock energy which unfortunately I raised the exchange term so N to the one third is actually the size of the exchange term and what he proved is that this correlation energy is smaller than that so actually Hartree-Fock gives the leading approximation one question may I ask a bit late sorry so you made this remark that the Hartree-Fock energy is actually the so the minimizer do it's claim that what you say there is that there is a minimizer yeah and it's a plain way state yes in the confined case yes so the other work I'm mentioning here is a work by Graf and Solovey where in 92 94 where they extended this work to the case of Jellium so Jellium means Coulomb system in the thermodynamic limit ok so that's ok so that's what I wanted to say and now I would like to present it here ok let me take it here so this theorem is about the asymptotics of the correlation energy and it goes as follows so let me take v such that the Fourier transform of v is non-negative that's important and such that sum over k of 1 plus k v hat k is finite so we need v to be bounded and then to have a bounded derivative ok then the correlation energy is equal to an explicit thing in which I call e rpa rpa stands for random phase approximation up to some sub bleeding term so alpha is some explicit rate ok where this e rpa is a rather complicated expression that I will write once here ok so this e rpa is a highly non trivial function of the many body interaction ok, so let me write it here there's a 1 over pi it's an integral and there's a log I mean the explicit expression is not enlightening at all maybe let me put here there's a k naught then there is a function of lambda here which is 1 minus lambda arc tan of 1 over lambda and then here I close this parenthesis so this is integrated in d lambda minus a term which is linear in v which is this one, so this cancels the first order term in this expansion of the log ok so it's this complicated expression here and this k naught is just 3 over 4 pi to the one third ok, so this is complicated but explicit and in fact it corresponds to a I mean it has interpretation of an infinite resumption of perturbation theory certain class of diagrams so just a few remarks so the first proof of this result was given through matching upper and lower bounds in a work with benedicternam Schlein and Seieringer 2020 and 2021 this was actually under the assumption that v was small for small potentials the version representing here is actually joint work with Niels benedicternam Schlein and Seieringer in 2021 that's for this version in particular there is no restriction on the size of the potential and so that was not the first result about the correlation energy, it was a previous work that I had with Christian Heinzel and a master student Felix Rix in 2018 this was a sort of a rigorous second order result ok, so we are only able to compute the correlation energy at second order and let me also say that the same result has been obtained well the same time by Christian Heinzel and Nam in 21 with a different method so same result but different slightly different method so that's about some of the history of that now that's of course these are rigorous works as I said this number here captures an infinite resumation of Feynman graphs in the expansion for the ground energy which is called the random phase resumation or random phase approximation so the result agrees with the random phase approximation which was an approximation scheme introduced by introduced in physics by Bohm and Pines in the 50s and the last comment I would like to make is about the fact that so this besides agreeing with the random phase approximation this number, this complicated expression that you see there also agrees with the ground state energy of a quasi free bose gas non interacting bose gas so and that's not an accident in fact but that's because the method of the proof which in fact it's the interesting thing about this result is allows essentially to describe in a precise way excitations around this fermi ball in terms of an emergent non interacting bose gas so in a sense this simplification arises from the fact that those fluctuations around the fermi surface can be understood in terms of a non interacting problem that you can diagonalize and that's the ground state energy ok so that's a rigorous version of what in physics is called bosonization and that's applied to a three dimensional fermi gas ok so sorry say it again do you really mean non interacting verbally non interacting what is the role of the interaction potential so the interaction potential is there to essentially define the non interacting of the boson so this interaction potential for the fermions enters in the bosonic theory in defining the non interacting Hamiltonian so the non interacting Hamiltonian is not just a Laplacian it's something more complicated sort of can think of a strange external potential being there which is written down in terms of Fourier transform of what we call it so not real space but yeah all right ok so this approach of course this kind of methods were developed in the condensed matter physics community and in this mean field regime one can actually use them to prove theorems rigorous version of them let me also say that this bosonization techniques really started from a mathematical physics paper namely from the exact solution of the Laftinger model by Mattis and Lieb ok, so then maybe in the remaining part of the talk I would like to tell you something about the proof which goes through a fox space description of the problem and it's based on a matching upper and lower bounds for the ground serenade so let me introduce the fermionic fox space exactly as Serena did but this time the you start from L2 antisymmetric and not from L2 symmetric ok, so that means that the general element in the fox space is an infinite sequence of fermionic wave function which have arbitrary number of particles sort of a grand canonical picture ok and in this fox space the Hamiltonian also can also be lifted so you can introduce a fox space Hamiltonian as the direct sum of Hamiltonian's acting on states with a given little n number of particles and this fox space Hamiltonian has a form very similar to what we've seen in Serena's talk so there is a kinetic part and there is a many body part which is quartic in this creation and annihilation operators so there is a sum of 3 momenta here the full transform of the potential and then you have a quartic operator in the fermions that describes the scattering among our fermions ok, so Serena gave a nice interpretation of this many body interaction let me just say that now this fox space operators a, k, a, q they don't commute but they rather anti commute ok, so this curly brackets is the anti commutator same for the star version and the only non trivial anti commutator is a, k, a, k ok so you can use those guys to build slater determinants so if you apply n of them over what is called the vacuum so the vector in fox space describes a case in which you have no particle at all you get a slater determinant with plane waves f, k, 1, f, k ok, so this vacuum vector has a 1 in the first component it's a vector that describes a case in which you have no particles ok so ok, so what do we do next while we introduce a, no maybe, well I will just continue with those two so I would like to efficiently compare my ground state energy with the energy of the of the free Fermi gas which is my Hartree-Fock ground state and to do that I will introduce the notion of fermionic Bogolubov transformation so these are unitary operators on the fox space so Bogolubov transformation these are unitary operators in the fox space that preserve the anti commutation relations ok, which so it has the following feature so the existence of this guy can be proved so it has two important properties the first one is that if you act with this fermionic Bogolubov transformation on the vacuum you get the free Fermi gas so you can understand the free Fermi gas as a unitary conjugation of the vacuum so this K i's that really form the the Fermi ball and the other property is that those operators they introduce a so-called particle all transformation so if you take the operator that annihilates a plane wave with momentum K and you conjugate it with R what you get is again a K if K does not belong to the Fermi ball or a star K if K belongs to the Fermi ball ok, so these are useful algebraic properties why are they useful well as a consequence of the first property the first property implies that you can compute the Hartree-Fock energy as the expectation value over this unitary conjugation of the vacuum of your Fox space Hamiltonian ok, that's clear because this guy is precisely the free Fermi gas and that's an n particle vector in the Fox space so that's just rewriting of the energy of the free Fermi gas the second property is tells you something about the fact that if you take any vector in the Fox space not necessary are omega by using the unitarity of R what you can do is to rewrite the energy of Psi as R star Psi R star H R R star so that's just one because it's unitary Psi so now what you do next is to use the algebra that I gave you I mean this simple relation to transform this operator it's just an algebraic computation there is nothing deep about it and you can bring the result into normal order meaning that you first write the you first start it from the less you first write the creation operator and then the annihilation operator so you have to anti-commute things around if you do that you get a constant term which is precisely the vacuum expectation value which is the energy of the free Fermi gas plus something let me call this vector here Psi plus the average of what I call the correlation Hamiltonian over this vector Psi I'm sorry I'm sorry in your first line you say that the first concept then the Hartree-Fock energy is the matrix element of interacting Hamiltonian with the ground state with the non-interacting ground state yes so as I said in this special confined case the Hartree-Fock ground state is the free Fermi gas so I'm just rewriting the average of my Hamiltonian over the free Fermi gas in this fancy Fox space language but then I'm saying ok in general I might have any state on the Fox space and I want to compare it with that so how do I do that while I just use unitarity to extract what I want and then I have to show that this is sort of small ok so then the task reduces to a study of this correlation Hamiltonian which I'm not going to write down because it really involves many terms I'm just going to tell you what are the two most important terms la rappresentazione Hamiltonian è, come ho detto, una parte di molte termi è quello di conjugare l'original Fox space Hamiltonian con questo operatore r quindi in termi di questo tipo puoi ripetire l'energia di correlazione come l'enfimum di Rolks psi che dimenticano la conjugazione della restriczione in particolare del Fox space con R star perché poi è tutto quello che stiamo facendo di l'energia di Rolks psi quindi questo correlation Hamiltonian è l'esterno di qualcosa e questo qualcosa si può scrivere, è una parte di molte termi i due mostri importanti, sono un tipo di termi di energia e un termo quartico e poi ci sono i termi più piccoli che non posso scrivere quindi questi due ragazzi sono uno termo che arriva dalla conjugazione di l'energia che è una energia relativata e questo ek è una quantità positiva, è solo il valore di epsilon squared k squared quindi è l'energia kinetica di l'energia con momentum k con rispetto all'esterno di mu quindi mu è epsilon squared e l'esterno di momentum squared è una quantità in ordano 1 quindi è un operatore positivo e poi hai questo q che è quartico che arriva dalla conjugazione della interazione di molte termi che ha la seguina in termini di la seguina di b operatori quindi è il sum per p hai la interazione di molte termi e poi hai qualcosa che sembra quadratico in termini di quei operatori bp b-p plus la misione di conjugazione più 2 b star p quindi cosa sono questi b ragazzi in termini di quadratico perché tutt'ora si deve essere quartico quindi per una buona risposta quindi questi b operatori sono molto importanti creano le excitazioni particole quindi b star è un creatore particolo è diventato da un po' di punti in termini in termini di p che k plus p non è una termina di b a k star a star k plus p ok, quindi cosa sta succedendo pensiamo di avere un stato in il foc spazio che ha un po' di articoli ok, quindi quello che avete aspettato è che dopo conjugare con r star, la maggior parte del stato è trascogliata il fatto che si trascoglie e se avete un b star su questo quello che hai è il sum per p ok exactly the same thing where now you add the two particles ok, one is here it has momentum k plus p outside of the fermi ball and one is inside ok so that is this important constraint which means that if p is fixed these two points have to be closed because the radius of this guy is n to the one third so these are excitations around the fermi surface and they behave as almost bosonic operator told us last time it's a very important thing that you could think pairs of fermions as behaving as a boson and that's precisely what we want to do here so this b operator behave as a boson in the sense that well if you look at the commutation relation for this guy they are bosonic so they commute ok and if you look at the anti commutative of b star p with b q what you get let me be very informal here you get a constant term which is of order there is a Kronecker delta in p and q and then there is epsilon n so that's n to the two thirds then there is p plus something which is small but small means on states with few particles I'm not going to be more precise let me just say that if the state here where you are acting has few particles so much less than n to the two thirds then this guy wins and you can think of that as being an error term so if you normalize these operators by this n to the two thirds then those guys really behave as bosons ok these are almost bosonic operators so if everything was quadratic in this almost bosonic operator then you would be left with almost quadratic, almost bosonic, Hamiltonian and quadratic operators can be diagonalized and you would be done essentially just one detail is that actually your commutator is the other way around it could be yes so maybe there is a minus sign here yeah so you say almost bosonic but epsilon n is not all the one it's n to the two thirds when I say almost bosonic I mean that you have to normalize them ok ok the error is much less than the two thirds when you normalize that's a bleeding with respect to one ok so I would like to get to a point right so if everything was quadratic in those guys then we would be done we would essentially use the Bogle-Luboff transformation that Serena told us and this problem could be exactly diagonalized you could find a spectrum and you could compute this correlation Hamiltonian as the ground set energy of this bosonic problem the problem is that it's not quadratic in the B's ok so this is quadratic in the fermions not in the bosons so there's no in this try here seems a little hopeless but what you could do is to compute just to probe how far this is from being a quadratic operator to compute the commutators with the B and what you get let me just tell you what the result is so this Ip ok, let me just write the things out k plus p does not belong to the fermi-bol and k belongs to the fermi-bol and what you have is k plus p a k and then here it is an retrieval function of k which is e k plus p plus e k that's just twice epsilon squared k dot p plus epsilon squared p squared so forget about this term, that's of order epsilon because k is of order epsilon a is of order epsilon to the minus 1 and so you see that if this guy was a constant that somehow this would look good because this is precisely a B operator and then this would suggest that on certain class of states H not behave as a bosonic operator however that's not a constant ok so then the trick is to localize so we would like to replace k dot p by a constant omega dot p and this would drop out of the sum if you can do that this goes out of the sum this replaces a B and then you have sort of a bosonic problem so just in a nutshell how does this go well this goes by a suitable localization technique on the fermi surface so you take your fermi surface your fermi-bol and you take composite in patches so these patches they are away from each other in a precise construction that I don't have time to review so there is also north and south pole and so on every patch is called B alpha so let me say that this B alpha it has an extension N to the 130 divided by N to the 1 alpha this is B alpha and it has a center which I call omega alpha so M is the number of patches ok so this introduces a localization on the fermi surface and this omega you have to think of it as the center of a patch ok so you introduce a new localized operators they are basically like the B except that now you only look at momenta which are inside a patch so there is a normalization forget about that then you have to sum over K let me call IP this this set here so K is in IP and then you also want this to intersect a patch so one thing that you can tell is that these patches are actually 3 dimensional objects so they extend a little bit inside and a little bit outside the fermi ball by order 1 ok so you have a K plus p a K in terms of those guys then the moral is that in terms of those guys this commutator looks indeed like a bosonic I mean the H naught essentially satisfies a bosonic relation in terms of those guys the conclusion and I will soon stop is that the commutation the commutator of H naught minus a certain bosonic kinetic energy with let's say c alpha p is essentially 0 on states with few particles so up to error terms that's really 0 and this db is a quadratic operator in the bosons sum over alpha and p of some bosonic dispersion relation which I'm going to write here so the bosonic dispersion relation would be epsilon times the length of p and then you have the scalar product of the normalized center of the patch with the normalized vector p times c star alpha p c alpha so that's the idea so this is the localization technique that allows to think of this correlation Hamiltonian in terms of a almost bosonic problem which is now quadratic in this operator c so what we do at the end is to study this almost bosonic problem and of course we have a lot of insight on this almost bosonic problem by the exact bosonic problem which is exactly solvable ok so in particular and I will stop here for the upper bound so let the exact bosonic problem so that would be db plus qb so this qb would be q after replacing the b by the sum over alpha of n alpha c alpha ok so if you look at this operator here and you treat those guys as being exactly bosonic then the ground state would be given by something very explicit would be a Bogolubov transformation of the vacuum in the ground state of the exact bosonic problem would be a vector that may have me call it phi that would just be, or maybe let me call it psi which would be a Bogolubov transformation of the vacuum where this Bogolubov transformation is just a unitary operator generated by something quadratic in the bosons so there is a certain Bogolubov kernel it's an m times i matrix then you have to make everything unitary so you remove the mission conjugate anyway it's something explicit forget about the explicit form and that would be the exact bosonic ground state so what you can do for the fermionic problem is to get inspired by that and take this as a trial state by taking the c as being fermionic as built by a fermionic operator ok so for the upper bound you have to take a trial state which in the end I mean which would be just an almost bosonic Bogolubov transformation and the vacuum and if you take this trial state you evaluate the energy of this guide and you see that leading order this gives a random phase approximation energy plus error terms of course you have to prove a matching lower bound and I'm not going to be able to tell you anything about it but to do that essentially well you can imagine what you really have to do is to use the fact that I mean instead of having the vacuum here can be written in this way where here you don't have the vacuum you have something which instead has few particles inside and so essentially what you have to do is to prove that actually this number of particles is very small and you have to keep track of course you have to use a lot positivity of the interaction to bound quite a few error terms from below ok so that part I just I won't have time to tell you anything about it so I think I will just stop here and I thank you for your attention you're a little bit late so you are perfectly in time and we have time for questions is it possible to tell in short time which is the new ingredient that I love you to go from small potential to large potential yeah so I mean it's at the end so ok it's really technically in the sense that it boils down to more precise control of so there are several ingredients but one important one is to understand come controllare questi ragazzi da sotto. Quindi, l'unico bambino è già certo per un grande potenziale. Non abbiamo questo potenziale di potenziale, per l'unico bambino, il modo in cui le cose erano raccontate, erano usando molto il potenziale di potenziale di potenziale per controllare alcuni di questi error termini con questa energetica kinetica. E essenzialmente, cosa che dobbiamo fare ora, beh, essenzialmente, siamo capaci di avvenire con più dei migliori estimati per questi error termini qui e per esplorare in un modo più efficiente il modo di positività del potenziale. In un modo in cui non abbiamo costruito completamente questa energetica kinetica e, in realtà, il lavoro di questa energetica kinetica, cioè, abbiamo bisogno di controllare alcuni termini di positività, ma non usiamo più questa energetica kinetica, ma vi piaceremo di usare la positività del spectrum di spazio bosonico. Quindi, abbiamo sacrificato il spectrum di spazio bosonico in cui, nei primi paperi, abbiamo solo abbandonato da sotto per zero. Ora, abbiamo usato la positività di questo ragazzo per controllare, in realtà, alcuni dei error termini non bosonici. Questo è, ovviamente, quello che abbiamo fatto. Vedi che potete provare il potenziale? Quindi, ovviamente, ci sono due questioni naturali. All'anno, siamo solo abbandoni per considerare il potenziale con uno derivativo. Ci sono due problemi. Uno è di quanto fare questo per Coulomb. E' un problema molto interessante e molto difficile, e questo è molto difficile, e non sappiamo come fare questo. Il bambino d'acqua potrebbe essere... Ovviamente, il bambino d'acqua potrebbe essere più semplice, ma già che è abbastanza difficile. Ma, da mio punto di vista, io credo che cosa che trovo davvero più interessante è di poter considerare il limito tramodernamico, anche per un potenziale abbandonato. Quindi, tutto questo è fatto per un sistema confinato. E molti di nostri estimativi, quando si intende a un caso in cui un volume scega all'infinità, non scelano l'infinità con il volume, quindi bisogna riaggiare tutto e fare cose localmente e provare a metterne cose insieme. Nel momento, non sappiamo come fare questo. Con le necessità tecniche, c'è l'integrazione fisica di positività della tramodernamica? Beh, è relato con la repulsione dei particoli, cioè che è una notionale. Se cambiate l'area reale, il reale è un po' di tramodernamico, che non è così... Sì, sull'esercito di... C'è un potenziale che soddisfa... Potenzialmente non cercherà se è positiva o definitiva. No, no, in realtà, no. C'è un potenziale che vuoi, è che quando si inclusiva questo potenziale, l'energia sceglie. E, ovviamente, con la positività della tramodernamica, puoi assurre che questo succede, ma magari la storia è tutt'altro un'altra classe di tramodernamica. L'esercito di positività nella tramodernamica può essere generalizzato, ma, a l'altro, è vero per il potenziale di tramodernamica, che è un potenziale reale e fisicamente relevante. Quindi, in un senso, sì, è vero, è un po' di tramodernamica, ma è anche quello che il potenziale fisico e relevante soddisfa. Quindi, sì. Poi potete anche dire qualcosa di questo? O potrebbe essere l'esercito della tramodernamica? È una buona questione, infatti. Quindi, come è, ora abbiamo... Sangue, le tramodernamica, la tramodernamica finirà in basso? E' un'altra questione. Quindi, la quale ti bisogna direCome è l'esercito del potenziale di tramodernamica? Il potenziale di tramodernamica è, ma, se non ci sono due di queste piacute, che facciamo così? Si, e' un'altra questione. non riesco a raccogliere come questa metrizia densa sembra. Quindi se mi chiedi cosa sono le feature di questa metrizia densa comparedo uno dei gas frittimi, non lo conosco davvero. Lo che posso dire è che in una storia dimensionale, che è più o meno diverso da questo, il rilascio di questa exetazione bosonistica è per smutare il profilo in spazio momentumale. Quindi è una cosa cruciale che Mattis ha scoperto che con queste exetazioni bosonistiche stavano scoperto la discontinuità dell'occupazione. Quindi la superficie di fermi era desaparece, in un senso avevano qualcosa che, a un livello fermi, aveva l'infinito derivativo, ma tutto l'altro era smutato. Quindi, ovviamente, la cosa naturale è per provare a vedere se algo simile a 3D succede. Credo che ci sono argomenti che credono che questo non succede, ma non lo so, ma è una questione molto interessante. Ok, no other questions. Vi salte via il macchello. Thanks.