 Ok, je to, da se počutim, da se dobro sega izgleda se izgleda se izgleda se izgleda. To je izgleda, ne? Tak. Tak. Tak. Ok. Ne zelo, da se ne zelo, da se prišlo, Nikolo? Zelo, da se ne zelo. Zelo. In zelo jaz označenje, ker sem čest, da je učinil, tako je zelo je začal na tjude. In ga je razstajala, da je advancesk, da načem v že, da bo vzela, da je označen, da je označen, da je označen, da je označen. Moj, da se kot pa, da je označen, da je označen. Predstavito, da sem to je želila, da sem bošnja, da sem tečnega. Slepšo, ki sem tukaj, da je. Nemo, da ga zelo, da se je. Zvom, da se bošnja v loči, da se je občal, da se je zelo. Selo se je zelo, da se je zelo, da se je zelo. V zelo se je natunal. Pistaj, da se je zelo, da se je zelo. Zelo se je zelo, da se je zelo. Tukaj, da se je zelo, da se je zelo. OK, a da sem si pozna, da se je komputila, in so in utopili zelo in nekaj je trefo ja... V odličnoj napisteva, nekaj nekaj je ta dobrovodnja,atorskri obdej, zelo igno, in je je v masternji, o da bo matov, zelo, patiš, če bo cah nato, če se... OK, se je to je現在. Nekaj je, nekaj je Nikola, nekaj je Vnikola, nekaj nekaj je, Trai kritične in izvrstke ideje, no, trai kritične in izvrstke z heading, in glasbe, ztega, nič doagnana vse je. In sve glasbe je v delovoh, ki so tudi z njenim sej, kot tudi nekako. Mi je neko što. Teh da sem spodljel, pomembnem po požeštih. Moj jo v deteženju, s kaj sem tukaj lahko vsej so počutiti, da res predstavimo na vseh prijev. Zato. To mi počutimo, jaz njih jaz sem tukaj neko počutil je z vsev. Tako, nekaj sem počutil, nekaj so počutil vse, nič. Dobro. Dobro za če nekaj vse. Zelo je jedna qeština? Ne. Ah, ne! OK. Spod njih imam alunje, da pa ne poželijo mi. Mi moraš da? V tom, da ja... Dobro. Pistim, da se ovega na dobro, njih imamo dobro, z delala vzifli avanje in z delala vzifli z našem vzivnjom. Ben sem zelo vzivnjom vzivnjom in je zdaj ne ovo, da je zelo vzivnjom vzivnje, ne bom zelo vzivnjom in in lobimo in Časovosti s n Profilice in skupaj. S njom núč flowers. To je jinama nr. A v nemajnico, da je in такой, da nerega bookeda, čas več nekaj zniknjih, ne bo decadn-več nr.no tega obeljana, in to je visi, da tega vso nekaj pojom, da nekaj gre tkem srednost in nekaj spesihov. izgleda se na stranje, ki je kratum in ki tablejna rekovača in tač, ki je alpha, kaj je autiliv, ki je, v my frema, dobro tako uniko skupov, in v fiziku. In je instotne, ja ne zelo lahko jaz bo potrebjemo o zač Official Quell, because jeel anglobi kako skupivo kratum in pa created. and simulate a spin Hamiltonian. So once again our focus is going to be on a spin Hamiltonian, like the one that I've been discussing yesterday, so the transverse fieldizing model. But in this case, but now we have that interactions decay as a power law, so Vij is now a power law of the distance between the two spins along x. There are various ways to describe long range interactions, so what is important to notice is what we were discussing yesterday that as a function of alpha one has two radically different situations. If alpha is larger than the dimension of the system and notice that in this case I'm focusing on the case of a one-dimensional system, so I'm talking about, you see here, a one-dimensional chain. Now in this case the dimension is one and in general one can say that one has two regimes for long range interaction. The regime where alpha is smaller than one and the regime where alpha is larger than one, so smaller than the dimension and larger than the dimension. The case when it's equal to the dimension I will not comment too much, it's a little bit in between the two, but more similar to alpha smaller than one. The difference between the two regimes is that in the regime alpha is smaller than one, one needs to introduce a rescaling in front of the energy. One needs to introduce a rescaling in front of the energy in such a way that the energy of the system is extensive. The cat rescaling Stevano has already been talking to you about and you see basically that also for general alpha so also when alpha is away from zero but it's still smaller than one you need to introduce such a rescaling but the rescaling has to be able as a power rule of the system sides. I also put some coefficient in here but these are not essential. What is crucial is this term here which rescales the internal energy. On the other hand when alpha is larger than one this is just a normalization coefficient so it doesn't modify the behavior of the model and sometimes or most of the times when alpha is larger than one so when alpha is larger than the dimension of the system we prefer to redefine it and define a new exponent sigma which is nothing but alpha minus the dimension and this is convenient for some reason that we're gonna talk about later. So basically all this normalization condition derive from the value of this submission here so if this submission this is the submission over all possible distances of the interaction energy of the system and this submission is the one that determines if the internal energy is convergent so the thermodynamics is extensive or if it's divergent. Ok, this is a bit mathematical I don't want to spend too much time about it just to show you that one can show or prove exactly even the divergence of that submission and one can basically prove exactly that the divergence of this submission is the one that I indicate here. So now just to look to make the formalism more mathematical to look it nicer I change alpha into s but this is not so important. So one can use the standards Euler-McClurin formula that converts a submission into integral and one can just say that the submission that we are talking about which is this one is equal to the integral of the same function plus a reminder and the reminder can be shown to converge I don't want to prove it to you right now just you can basically show that this reminder is smaller than the integral of f prime alone than the modulus of the integral of f prime so the modulus of this is smaller than the modulus of the integral of f prime but obviously f prime converges because if s is between 0 and 1 f prime converges because it's the derivative with respect to x of 1 over x to the s so this guy converges and the only diverging part is the one that you find here and this part it's straightforward to do the integral of the x of 1 over x to the s is just go like x to the 1 minus s and so it diverges like n to the 1 minus s so basically it's this term is this integral term in the Euler McClurin formula that gives you the divergence that I have been talking about here okay there is one can do much better than what I have done down and one can prove that this summation goes like this where you also got the right coefficients and some corrections but this is not so interesting for us so what it's interesting for us is that the divergence is proportional to n to the 1 minus alpha so that we have to put in front of this guy i coefficient n to the alpha minus 1 which rescales the energy okay so I believe this is enough mathematics but this is kind of a basic mathematics you should be convinced that we know how to treat these divergences what I also want to convince you about is that this system has a quantum phase transition for any value of alpha and basically you should be able to have an intuition for that from the fact that we had a discussion yesterday about the fully connectedizing model so the case alpha equal to zero and we have explicitly made the calculation of the quantum phase transition so I have given you some details and hopefully you have been able today or maybe in the next few days to reproduce that calculation in details but also the case where alpha is equal to infinity so the case where alpha is equal to infinity it corresponds to the nearest neighbor case no so this is the case alpha equal to infinity when alpha is equal to infinity only sides that are at distance 1 between each other survive 1 because I have set to 1 the lattice spacing of my system and so you should know from your basic courses that also the nearest neighbor limit so anizing model with nearest neighbor interaction as a quantum phase transition between a symmetric phase where there are exponential correlation and no paramagnetic order and a ferromagnetic phase where there is ferromagnetization along the x-axis you have seen this stuff during your courses I already asked this question yesterday but I will re-ask for completeness is there anybody that doesn't know about the quantum phase transition in the transverse field design model apparently not so I believe that if you know that the nearest neighbor system has a quantum phase transition and yesterday we discussed the case of alpha equal to zero basically you will not be surprised to know that for any value between alpha equal to zero and alpha equal to infinity there is also a phase transition how can we see the properties of this phase transition well the first thing that we can try to do is is as a called also in premarkov transformation which some people call also I think quite rightfully a spin wave approximation so we try to transform our spins into bosons and this transformation is one of the very basic tool of many body quantum theory maybe you have already seen it around and it basically consists to approximate the spin as a bosonic state the idea somehow is that n spin as n states a boson as infinite many states no, because I can put an infinite number of bosons in the same site but into some approximation one should be able to reduce the dynamics of the boson as if it has only few states and if it has only few states then the dynamics should be the same as the spin well there are ways to make this analogy formal like considering not just bosons but arc or boson so boson which have a strong interaction and the physics of boson with a strong interaction so with an arc or constraint is basically the same as the spins but in our case we are not going to do that we are just going to reduce so we are going to employ really the austen premarkov transformation in bosons which have no constraints so this is the transformation you see here so basically you transform the spin ladder operator in bosonic ladder operator or if you wish you transform the sigma x and sigma y component of the one half spin into combination of bosonic ladder operator and the value of the spin around the polarization axis that now we choose to be z is going to be given just by the density of the bosons so if there is no boson on one side the spin is one half if there is one boson inside the spin is one half is minus one half and if there are more than one boson this state will be the wrong one that we don't want to consider so this is an approximation as I already said but it's a very powerful approximation because it allows to reduce this model that is not treatable analytically in the beginning into a model which is fully solvable because once I have replaced my spin operator into bosonic operators the Hamiltonian is now a quadratic boson Hamiltonian and it can be solved by a bogorubov transformation is there anybody here that doesn't know what it is a bogorubov transformation are you sure guys? sorry Stefano I couldn't hear you someone does not know but others maybe know I don't know maybe okay maybe you repeat to say something more okay fantastic somebody says he doesn't know what it is very good so well let me give you some words about the bogorubov transformation so it's very important this is also so we are using very standard tools of many body quantum mechanics okay obviously this is a school of statistical physics this is also statistical physics as we will see in a moment but I understand that maybe some of you they do never encounter this concept before what I just wanted to say is that if you plan or are planning in the future to work in the field of quantum physics and many body quantum physics it really important that you go and attend some courses on these topics or also revise these topics independently on some books and if you write me I can suggest one some so basically the bogorubov transformation it's a 2 by 2 transformation so what's the idea the idea as you see sorry so I start here as I said I took the spin operators and I mapped them into bosons now these are bosonic fields these are bosonic operators and so these represent bosons that are up on the lattice but this is not a pure Newtonian it also has this anomalous term you see this term is a bit strange because it doesn't conserve it doesn't conserve the number of bosons you can basically create two bosons from the vacuum without paying without destroying anyone so this non-numbered conserving operator may seem a bit strange but we know that in many body quantum physics it's not so anomalous it appears already in BCS theory of superconductors and it appears also here because the idea is that if you have a many body system you are always in the grand canonical ensemble so many body quantum theories is almost always describing the grand canonical ensemble where the number of particles doesn't need to be conserved in some sense so you can have Hamiltonian where there is this kind of operators but what do you do in that case so in that case is pretty simple so first of all since the interaction is translational invariant you do a Fourier transform you do a Fourier transform this is standard tool I guess you know it and you go to momentum space since the interaction is translational invariant in momentum space you don't have any coupling between the sides but the anomalous term so the non density conserving term it shifts its couples different commenta so let's try to break down this Hamiltonian let's try to break down this Hamiltonian so this Hamiltonian it's a sum of n independent Hamiltonian or n half independent Hamiltonian k lives on its own now you see there is no coupling between the momentum k's so you can consider this h as a sum of h of one Hamiltonian for each k there is no scattering between plane waves or between Fourier modes however there are modes that are coupled and the modes that are coupled are just the modes that have k and minus k so let me repeat it once again to just maybe could be boring for someone but for somebody that haven't heard about that it's important I went from real space I went from real space to momentum space in momentum space I have n modes as many modes as are decided in the lattice these modes are positive and negative k because of the system being symmetric and all these momentum modes are independent from each other but the coupling that emerges due to the pairing term and this coupling couples states which have momentum k which states that have momentum minus k so basically you have n half Hamiltonians which represent the Hamiltonian of modes with k and minus k coupled to each other now we want to remove this coupling because we want to really to diagonalize the Hamiltonian we want the Hamiltonian to be diagonal like this one this one is not diagonal in momentum space because it still couples momenta which have different values basically each k is opposite we want to go to an Hamiltonian where there is which is completely diagonal where you have some quasi particles which are fully non-interacting how can we do that? well since these are basically two-state system we can do just a volume of transformation which is nothing but a 2 by 2 matrix so it's a transformation which it's a 2 by 2 matrix and it helps us redefining the bosonic operator as a linear combination of creation and annihilation at k and minus k so you see it here I take my old operator and I express it as a combination of a new bosonic operator and a new both creation operator but at minus k if you do this in metric space it's going to be evident to you it's just a rotation in the spin one half space so it's just a rotation in the space of 2 by 2 matrix in the space of 2 components vector and it's a rotation in the space of 2 components vector because basically 2 states and so in order to resolve that coupling we only need to make a rotation in the space of 2 components vector which is the space of which is the space of spin one half vector which is s1 but this is ok I don't want to bother you with group theory and s1 symmetry of the spin operator what it is important is that you understand technically how it is done and technically is just introducing new bosonic operator b which are linear combination of the old here I am writing only the old as a linear combination of the new but if it's a linear combination I can always invert it to make it another linear combination and then I have to determine which are the u and v which make the Hamiltonian diagonal I don't want to do this explicitly now you can do it yourself it's very simple just plug this expression into here the dagger is obvious you plug the expression for this ak and this dagger into the Hamiltonian and you can then assume that all the terms which are not diagonal so that are not of the form b dagger b or b dagger vanish and if you do that you will immediately find this u and v v so I can do this bogolub of transformation and I can define new bosonic quasi particle these b and b dagger are bosons so they satisfy the same commutation relation as the a's and the a's were bosons so they satisfy the user commutation relation and in terms of the new operator b the Hamiltonian is diagonal with a certain energy spectrum omega k ok please, I have a question ok ok I wish you know if I wish you know if the transformation is unique I mean for an operator ak you have a unique bk and b-k basically basically there are two which are equivalent so that is the one that I am showing you and this is another one in which I also exchange the role of the creation and annihilation operator ok so basically there is one and there is the one that instead of giving me b dagger b gives me bb dagger but that one I can always bring it back to the first one just by swapping the two operator and adding a constant so it doesn't really matter ok so it's not unique but it's like if it were because it's the one that's not unique is just well, these are bosons and not fermions but in some sense it's unique thank you then my mathematical physicist friend they made me a long argument on how this transformation is simple etiquette so it's obvious that it's unique and the equivalent one for fermions state is unitary so it's not exactly unique but for the sake of all practical physical application the volume of transformation it's always it's unique both for bosons and for fermions so we have this transformation and we have this transformation and we have the spectrum of the system so now I solved my model I solved my model and my model is is diagonal and they we call this bogoljub of quasi-particle so the bidager beer we call them bogoljub of quasi-particle but I don't like so much to call them quasi-particles because they have not any finite lifetime so this model it's exactly solvable and the bidager beer are modes which live forever but people call them quasi-particles because somehow they have in mind that the spin-wave approximation is just an approximation and so it's expected to break down at some point and so this is why they call it quasi-particles but for the sake of being precise these are not quasi-particles they are modes of the Hamiltonian which make the Hamiltonian perfectly diagonal and so they have infinite lifetime what is interesting for us is what is the dispersion of the mode why is it this interesting for us well because I have been telling you that this system has a quantum phase transition as a quantum critical point for certain values of h now when the ratio of v over h reaches a certain value but close to this quantum critical point the properties of the quantum critical point they always depend from the supposed continuum limit not continuous but continuum the continuum limit so they only depend on the small momentum modes I guess this you know no in phase transition even in classical phase transition to say that what is responsible for driving the transition are critical fluctuations which have long length scales and if they have long length scales they have small momenta so it's interesting now to look at the expansion of this spectrum so these are the modes of the system asking what happens to the modes which have low momenta the one that are the most interesting from the point of view of the macroscopic physics and when I look at these modes their energy depends on this vk which is nothing but the Fourier transform of the interaction potential that was 1 over r to the alpha and so now I have to do the Fourier transform of 1 over r to the alpha which is nothing but this sum with the cosine kl do you understand this no? Do you see the Fourier transform of the interaction potential? Is there anybody that doesn't see the Fourier transform of the interaction potential in this formula here? Well, I guess that everybody sees it no, it's easy so this was the interaction potential so this was the shape of the interaction potential I want to do the Fourier transform the interaction potential is transversional invariant sum only over the distance between the sides that now itself and I have to put a cos kl on top so when I do this Fourier transform well, you can do some mathematics and show up that basically this is related, this is equal to the sum of the poly logarithm of this exponential of the momentum this is not so interesting as a redefinition of this Fourier transform based on the definition of the poly logarithm what is very interesting that when I go into k equal to zero limit so when I go a slow momenta this function is non-analytic so this function is non-analytic it has a momentum term in its Taylor expansion which is not nor odd power of the momentum it is a k to the sigma with sigma a real number which is equal to alpha minus d and since I am in one dimension in this case it's alpha minus one so I hope you are appreciating what's the strong effect of long range interaction in this case so yesterday we have been talking about interactions which have alpha less than d but now we see that also interactions which have alpha larger than d they can do something pretty deep and profound on your system they can generate a non-analytic momentum term into the low energy theory I don't want to talk for now of the case where alpha is smaller than d but different from zero we will see tomorrow also the case alpha is equal to zero so I am not going to talk about this for now for the moment let's focus on alpha larger than one when alpha is larger than one you see exactly what I was saying a minute ago the low energy theory contains a fractional we call this a fractional momentum term in the sense of momentum term which is non-analytic in k but this is deep this deeply changes the this deeply changes the physics of the system because now it means that in the continuum limit and sorry this is a misper it should be continuum in the continuum limit this system becomes a gas of bosons where the dispersional relation instead of being k-square is k-sigma so a gas of boson between analytic dispersional relation so I'm sorry this I have to ask does everybody know what I mean when I say I go into the continuum limit or this is something that you hear here for the first time there is no shame guys so if somebody doesn't know what I mean when I say continuum limit just say it and I will try to explain it a bit more so you have the chat also in the world nobody is writing in the chat and you don't dare to raise your hand you can you can put it in the chat okay I guess that the world continuum limit is somehow famous and it's the same in some sense this is not really related to quantum physics now this is exactly the same case that will happen in classical physics you can take a classically critical model and that's on the lattice like the ison model and do the relation to the continuum limit and you find that the ison the classical ison model is the same as a pi to the fourth theory in the continuum limit and this is somehow the same that's happening here I've taken my quantum ison model and I have done some manipulation to map into a gas of bosons and when these are still bosons on the lattice so this Hamiltonian still represents bosons that are hopping on the lattice but then I made a further step I make an expansion of my Hamiltonian as low momentum and when I do an expansion of the Hamiltonian at a small momentum I get a gas in the continuum so I get a gas that lives in the continuum space without the lattice but this gas now has a non analytic dispersion relation so as a dispersion relation which is not the standard k square over 2m that you know or maybe you also know the phonon dispersion relation that is k simply k, linear dispersion relation no the gas of bosons hopping on the lattice in the continuum looks like a non analytic gas with k to the sigma and then you can play the game of studying the properties of this boson gas in the continuum with non analytic dispersion relation and I don't know if you ever try to do this this is some time given in exercise in statistical mechanics in quantum statistical physics course so when you study bosons and condensation you have studied I guess bosons and condensation in the standard 3 and 2 dimensional cases now you can study bosons and condensation in fractional dispersion relation and it turns out that when you study bosons and condensation in fractional dispersion relation it's like studying bosons and condensation in fractional dimensions and this you see very easily if you try to compute the density of state the density of states is how many states I have this is standard calculation in when one study the bosons and condensation you count how many states you have a fixed energy so you have to integrate over the entire momentum space the omega k and the omega k it's now non analytic it goes like k to the sigma rather than k to the square and so when you do the computation you find that this goes the density of state goes like the energy to the d over sigma minus 1 notice that in the standard case of bosons and condensation you will have d over 2 minus 1 you know this formula density of states goes like energy to dimension 2 minus 1 you know this as usual if somebody doesn't know he can speak up now we have generalization of this formula the standard formula for bosons and condensation it goes beyond the case of local interactions and for long range interactions you have d over sigma minus 1 as be replaced by sigma but this is not surprising because the standard case of a boson gas will be k to the square no the kinetic energy of a boson gas is k square over 2m but now with long range interaction we have k sigma over 2m sorry I call cdc is 1 over 2m now we have this k sigma and this tells you also something else that the case sigma equal to 2 it's special and it is indeed special and you see also from this formula because when sigma is smaller than 2 this is the correct expansion when sigma is larger than 2 when sigma is larger than 2 this is not anymore the correct expansion here I indicated that I have a k to the square term and so when I do the expansion at low moment that when sigma is larger than 2 this term is sub leading with respect to this k to the square and it should be replaced by the k to the square and so it means that long range interactions they only modify the low energy behavior if sigma is less than 2 which is tant amount to say that alpha is less than 3 and this can be somehow sorry I am not skipping that I want to just to show you so this somehow tells you already something that sigma equal to 2 is a special case and then for sigma larger than 2 the system behaves exactly as if it were short range as if it were nearest neighbor and only for sigma less than 2 you have some real long range so you have some contribution from the long range interactions to the low energy theory and then when sigma is smaller than 0 we go back to the case that we treated yesterday the case of strong long range interactions ok now I hope I have convinced you that the that this system that the ising model can be related the physics of the ising model with long range interactions can be related to the physics of a Bose gas with a non-analytic dispersion relation and at the same so also the existence of the phase transition can be somehow related to the possibility for the gas to to condense the ferromagnetic phase transition from the spin in the language of this both in the Bose gas language can be seen as a condensation and I don't want to do the entire calculation here I hope so all of you some notion of Bose Einstein condensation have you studied Bose Einstein condensation in the past if somebody hasn't studied Bose Einstein condensation speak up as you very good so if you know about I haven't said one student does not ok, very good so you don't have notions about Bose Einstein condensation so let me see how to explain in a nutshell in a nutshell well in a nutshell Bose Einstein condensation of course there is when you have a Bose gas a gas of quantum particles which are boson what physically what it may happens that these are quantum particles so they have wave function when you lower the temperature at a certain point when you lower the temperature the particles become feather so they become more and more fat because the wave function the debris wavelength grows this is standard from quantum mechanics course as now we always say as the debris wavelength grows there is more and more overlap between the wave function of the particles in the gas and there are more quantum effects now it exists a critical temperature below which the quantum overlap between the wave function is so big is so strong that the particles that there is an establishment of coherence in between the particles so the particles tend to form a condensate so a coherent state where the particle lose their identity have behave like a very coherent homogeneous fluid now this is theoretically this is justified by the fact that the bosons can stay in the same state so bosons can stay in the same state and if the temperature becomes so low they all start to have zero momentum because this is a gas which only has kinetic energy it's a gas that only has kinetic energy so the minimum energy state is the state with k equal to zero so all the particles want to go in the k equal to zero state and when there are too many particles in the k equal to zero state you cannot use any more a statistical mechanics description because the statistical mechanics description somehow relies on the fact of the particles being to the energy being continuous and the particles being well distributed according to some distribution into the entire energy levels into all the energy levels but if all the particles occupy the same momentum state they all have momentum zero in the continuous representation of the gas so it means that the standard statistical mechanics description fails one way to see that the standard statistical mechanics description fails is that when you count the number of particles in the system you get the wrong result and this is basically what I'm trying to show here I put just this formula and I assume that most of you have seen the same formula in the case of nearest neighbor interaction but let me try to break down this formula, this is the number of particles in the gas at a fixed at a certain momentum at a certain chemical potential value mu so according to the standard notion of statistical mechanics from a chemical potential I can obtain the number of particles that are in my system and this calculation is easily done is just the summation over all possible energy states so this is the density of states that enters here that counts the energy states of the probability for a particle to occupy that energy state and the probability for a boson to occupy that energy states is given by the boson's distribution this is nothing but 1 over exponential of beta d energy minus 1 so this function here the 1 over this is the boson's distribution that you probably have seen many times this term counts it gives you how many energy levels there are at a certain energy and the integral over all the energy should give me the total number of particles in my gas this would be true if there were a statistical mechanics description that worked but this statistical mechanics description fails to work when the gas becomes a quantum of boson's and condensate because when the gas is in a condensate it means that you cannot describe it as an homogeneous distribution function because all the states are at zero energy and this you see from this fact when you do this when you do this integration and you do it at a small temperature because we are thinking about something that happens at a very small temperature when you do this integration and you approximate it at a small temperature you get something like this now you see that this guy vanishes as the temperature goes to zero but we expect our gas to have a fixed density no, we want our gas to have a fixed density and so this as the temperature goes to zero the vanishing of this term has to be compensated by a divergence of this term but this term does not diverge as long as d over sigma is larger than 1 so this sorry, please I don't know if you are using a mouth because when you are pointing a term we can't see it sorry, you cannot see it ok, that's very bad there is a red dot on the screen, I don't know if this is the mouse so if there is a red dot now I'm pointing at d on the second line no, it points to the integral it's moving ok, then this point I don't know what it is I'm sorry guys I thought I had made it in such a way that it was visible but ok you see now the mouse ok, now we see the mouse now you see it ok and also the red dot moved I don't know if it's this red dot is not visible for me so I don't know ok, very good so let me redo this with the mouse so I was saying this guy is the number of particles of the gas at a fixed chemical potential we have how to calculate it is the sum or the integral because we are assuming that we can treat this gas as a constant so this so I'm not telling it would be that I don't see the as you see so go on Nikolaj, sorry so this, the number of particles in the gas is the integral over all possible energies of the number of states as a certain energy which is the density of states multiplied by the probability for the particles to occupy the state which is the Bose-Einstein distribution so this is the Bose-Einstein distribution 1 over the exponential minus 1 is the Bose-Einstein distribution so this integral should give the total number of particles of the gas at a certain chemical potential as if I go to the slow temperature limit to the small temperature limit to the limit of cold gas I can approximate this integral as it is written here on the right and you see that there is a pre-factor that vanishes as a power of the temperature so if I want my gas to have the same number of particles because we want our construction to be at a fixed average number of particles the vanishing of this temperature term has to be counterbalanced by a divergence of this term here of this integral but this integral does not diverge as long as d over sigma is larger than 1 you see the limit of this is the small temperature limit because in the low temperature limit the chemical potential also vanishes so this guy vanishes in the slow temperature limit this guy tends to decrease proportionally to the temperature and if this is happening it means that the particles are entering in the Bose Einstein condensate state it means that I am not anymore able to see them in this description because this description assumes that the particles are somehow smoothly distributed along all the energy levels but this is not the case because all the particles are condensing in the zero energy state and the zero energy state is not accounted in this integral or is not accounted properly by this integral so this is the standard study of Bose Einstein condensation the fact now that Bose Einstein condensation will occur between in dimension larger than 2 because normally you will have that this is not sigma, this is 2 and you will have that Bose Einstein condensation of course for d larger than 2 but this is not the case you see this is not the case now with long range interactions the 2 has been replaced by a sigma and this basically means that for this kind of gas Bose Einstein condensation can occur at any value of d as long as sigma is small enough and this basically means and this is also another way to say that a gas with a fractional dispersion relation a gas with a fractional dispersion relation is equivalent to a gas with a local interact so a gas it's equivalent to a gas with analytic dispersion relation so with the standard kinetic term k to the square but which has an effective fractional dimension ok and this effective fractional dimension I reported here is 2d over sigma and this is exactly very easy to show you can take any calculation that you have done on your Bose Einstein condensate states and you have to replace the dimension d with this number sorry with this value and you will immediately see that this calculation that these expressions are transformed into the equivalent one for long range interactions well let's a simple example for example the specific heat you know that the specific heat is the alpha the number of degrees of freedom in a local gas in a local non interacting gas we always say that the energy is 3 alpha and kbt no 3 alpha and kbt because we are talking about gases that are in three dimension and the standard result for the equation of state is that the total energy is 3 alpha and kbt now here I have said kb to 1 so for me the standard equation the equation of state of a gas is that the energy is 3 alpha and t and this means that the specific heat that is nothing but the derivative of the energy with respect to the temperature has to be 3 alpha m but in a long range interacting state instead of having 3 alpha n which is I have to replace the dimension with the effective dimension that is 2d over sigma and I get the result that is in here so you see the presence of long range couplings modifies my effective theory and it modifies it in such a way that I get the effective fractional dimension because I can take the standard formula specific heat is equal to d alpha n replace d with the effective dimension 2d over sigma and I will get the results of a gas with a non-analytic dispersion relation well I hope that you got this point please make some question tell me what you did I guess this is a bit of a convoluted matter because I started from anising model I mapped into a quantum gas but the special thing of this quantum gas is that it has a non-analytic dispersion relation and then I try to show you that the gas with ananalytic dispersion relation it's equivalent to a gas with ananalytic dispersion relation but which lives in a fractional dimension is this picture clear at least in its history well let's make a 5 minute stop here in the previous slides you had a quantity which is not defined c so this is the dimension of the integral because you have the where sorry in here first line last formula sorry, this is the c here it's 1 over 2 m basically it's just a coefficient sorry I always speak about critical phenomena so I never look at coefficients it's the speed of sound or the mass of these particles depending on how you want to see so it's not really speed of sound because it would be k square so it's only if it is k square the sound normally is linear so if sigma is 1 this would be the standard definition for the speed of sound so if this were a phonon gas this would be ck but since it's k to the sigma if this were a square the c would be 1 over 2 m the mass of the particle the speed of it in between these two so we can make a break 5 minutes 5 minutes break you can behave like bosons it's a popular task