 Zelo prejdi vsih, da te ne zaino, je tudi je, da se pripravimo, tudi je pripravimo, kdo je ne nezakon, ker je tudi še zelo, zelo. Zato se nezapravimo in igračo, da tudi pripravimo ati ctp.it, vse vse te poste v pdf forminu, vse vse poste izprout, nekaj, hrbno, hrbno vse, nekaj. je to format do vse poste, in vse poste je začala, da je začala vse smr 335, je tukaj nekaj, nekaj vse poste, ker tukaj vse poste. Vse poste, nekaj vse poste, nekaj vse poste. If you have a poster printed already, I will tell you, so the issue is the following, I would like to have the poster session up on the terrace, where we also will have wine and cheese, but the weather forecast is uncertain. OK, so we're waiting until the last minute to see whether we will have it on the terrace or here, because there are poster boards all around this room. I would, of course, prefer to do it on the terrace, but we cannot commit by now. So I'm gonna tell you where to put the posters on Monday. Any questions regarding this? Oh, everybody who sent an abstract is accepted, by the way. There are no crazy people among you. I was afraid somebody would present a poster, like the theory of everything. I can explain why E is equal mc squared using cellular automata. Yes? You had a question? No? No question? OK, very good. So if there are no questions, then I leave the microphone to Professor Dalibar, and here we go. No, I think I will have enough space, I hope. OK, welcome back everybody. So let us continue with what I call the first lecture of this series of two. So I remind you where we were just before lunch. We were asking ourselves how can we describe an interacting 2D both gas, and I was proposing to use a classical field analysis, which means that I am assuming that all my particles share the same wave function, which I called little psi of x, y. I have frozen the vertical motion, the motion along z. So the real wave function in the 3D space is this psi of x, y times a fixed wave function along z, which is just the ground state of the harmonic oscillator confining the particle along z. And I explained to you that the energy functional of this classical field is a so-called Rospitejevski functional. It involves two terms. The first one is a kinetic energy term, which is proportional to grad psi modulo square, integrated over the whole space, x, y. And the second term is a psi 4 term, and it results from describing the interactions between particles as a contact term, a direct distribution, particular interactions only that are on the same point. And the factor in front of this interaction energy term, the one that characterizes the strength of the energy, is a dimensionless coefficient, which I denoted g tilde. And if I want to connect g tilde to real life, this g tilde actually is a ratio of two lengths within a factor scroll of 8 pi. On the numerator I have the 3D scattering lengths, which characterize how do particle interact in 3D. And in the denominator I have something which is the thickness of the gas, so to speak. This is the size of the ground state of the harmonic oscillator along z. But you don't need to know A and AOH separately to describe your 2D both gas. The only thing you need to know is the value of g tilde. And as I told you, g tilde is typically between 0.1 and 1. And to say it briefly, 0.1 means a weakly interacting gas, whereas above 1 this is a strongly interacting gas. But this will become clearer later and probably tomorrow. OK, so as I tried to explain you just before lunch, although we are doing a classical field analysis, we are not getting rid of all quantum physics and there is still something which is quantized, which is the circulation of the velocity because the velocity is simply h bar over n times the gradient of the phase of my classical field. I would like to start this second part of the lecture by showing you that indeed I still have some quantum features, so to speak, into this classical field approach. But now I would like to look at something different, which is the Bose-Einstein law, the one which is still written on the board here. And I would like to convince you that the Bose-Einstein law contains this classical field analysis. And for that I would like to come back to a simple way to justify the Bose-Einstein law. I mean if you are on a desert island and you have to explain for yourself to rederive the Bose-Einstein law, how can you do it in two lines? So first let's derive the Boltzmann law. And for that I just look at the processes, the collision processes that can occur in a gas if I have a gas at equilibrium. So I can have two incident particles with some energy associated to the state G1 and energy associated to G2, so E1 and E2, a collision between the partners and I get G3 and G4. And so I have a collision, G1 plus G2 gives G3 plus G4 if E1 plus E2 is equal to E3 plus E4. And of course the inverse process is also possible, that is I can play the movie back that G3 plus G4 gives G1 plus G2. And if I am at equilibrium I have a detailed balance condition which tells me classically that this process has the same rate as that one, therefore NG1, NG2 has to be equal to NG3 and G4. As soon as I take the four energies, such that. And if I assume, which is valid for our system, that NJ, the equation of a state J, depends only on its energy, J, this means that the function N of E must be such that any E1, any E2 equal any E3, any E4. So I don't know whether you know how to prove that. Well, no, excuse me, this means that. And the only function that allows this to be satisfied for any quadruplet of energy E1, E2, E3, E4 is a function which is an exponential of the energy. So let me simply prove that. So I am saying, telling you that I have any one, any two equal any three, any four. As soon as E1 plus E2 equal E3 plus E4. So let me call f of E the log of N of E. And then the function f of E, when I take the log of this, should be such that for any quadruplet E1, E2, E3, E4, satisfying E1 plus E2 equal E3 plus E4, then the function itself should be such that f of E1 plus f of E2 equal f of E3 plus f of E4. And it's very easy to show, this I will not do it, but it's a simple mass, to show that the only type of function which satisfies this when I have this relation between the energies, are the function f of E of the type which are linear in E plus some constant B. These are the only ones. And again it's really not complicated to show. So if f of E is like that, then N of E which is exponential f is like exponential A of E plus B and this is nothing but the Boltzmann distribution. So it's a simple way to redefine Boltzmann distribution. What about the Boltzmann time law? Then you have to remember that when you are dealing with quantum particles, you don't have only collisions like I described, but you have actually stimulated emission process. The collisions like I described, E1, E2 gives E3, E4, and therefore any one, any two equal any three, any four correspond to spontaneous emission, that is, I have two particles colliding in one and two and I can go into the final state, three and four, and it was not stimulated. But if now I am dealing with bosons, I have some stimulated emission which is possible, which means that when I have one and two which are colliding, the rate at which I can produce three and four can be stimulated if I already have some particle in three and some particle in four. I have some announcement if I am dealing with bosons, if three and four is already populated. And as you know, this announcement is simply one plus N, the stimulated emission term. So the process one plus two gives three plus four as it occurs actually with the rate, if I have stimulated emission, N1 and two, the occupation of the initial state, one plus N3, one plus N4. And this has to be equal again to the reverse process, three plus four gives one plus two, so N3 and four, the occupation of the initial state for the reverse process, times one plus N1, one plus N2. OK, so what are the functions which satisfy this? It seems a bit complicated, but actually it's not. I just have to take this one plus NG1 and put it in the denominator here. Same thing for one plus NG2, same thing for one plus NG3 and therefore, and so on. And now I am left again with the function of E1 times the function of E2 as to be equal to the product of the function of E3 plus the function of E4, so it was exactly the same type of ration as here, except that instead of N, I have N over one plus N. So again, same thing, N over one plus N has to be the exponential of a linear function, exponential of Ae plus B and this, now it's a simple algebra to show that this N of E therefore is equal to one over something exponential minus Ae plus B minus one and this is nothing but the Bosnian chain statistics. OK, so this is the few lines way of recovering Bosnian chain statistics. If you understand that in a collision process you have not only spontaneous emission but you have stimulated emission. Spontaneous emission is this one. Stimulated emission is this N. OK? Any question at this stage? No? OK, so now if I try to look at this Bosnian chain law and understand what are the two extreme regime of this law that is the high energy regime and the low energy regime, the high energy regime is what allows you to recover from Bosnian chain Maxwell Boltzmann. If I take, say that the energy is very large here in the denominator the exponential of E over KT is very large so I can forget about the minus one and then I recover N of E equal exponential of U minus E over KT this is the regime of classical particles Maxwell Boltzmann. So this is the true Bosnian chain law in black and Maxwell Boltzmann is the dotted blue line that you have here. Now if you take the opposite regime the regime of low energies and I remind you this is the regime in which we are interested for BKT physics because I am looking at large wavelengths so small wave vectors, small energy then in this expression here the energy and mu are very small compared to KT so I can linearize this exponential and I get then something which is simply KT divided by E minus mu which is very large compared to one by hypothesis here and then I recover a population proportional to T which was what I got just before lunch for a classical field I told you that for a classical field the average of the modulus square of the amplitude of a mode was proportional to KT so this is what I recover here so I recover the regime of classical field with occupation number very large compared to one so when you say let's take the classical limit of a Bosnian chain law it doesn't mean much I mean the two extreme regimes are classical here is classical in the sense that you neglect stimulated emission you keep only spontaneous emission and then you have a classical regime with respect to particles whereas if you look at what is happening here you neglect spontaneous emission in front of stimulated emission and then you have a classical regime in terms of field so the Bosnian chain law actually is a way to connect these two so to speak classical regime a regime which is particle dominated here and here a regime which is field dominated Bosnian chain law actually takes into account both the particle and the field nature of the matter so again this classical regime that I'm going to use contains some quantum feature it contains this limit of the Bosnian chain law the low energy limit any question at this stage no ok so maybe I can skip that which is just another way of stating what I said so let's go back to this 2D Gospiteevski energy functional and let's try to see what we can say about it so I remind you what we have kinetic energy grad psi square interaction energy psi 4 and the wave function is normalized such the integral over whole space is equal to the number of particles that I've put in the system I had some questions during the break about that what I'm really looking at here is a kind of canonical ensemble but I imposed some temperature from the outside because I'm supposing interacting with some energy reservoir ok so the first thing I would like to mention about the analysis of this of this energy functional is that when we are at very low temperature or if you prefer when we are at large density so what I mean is large actually phase space density product of rho times lambda square which is rho over temperature because lambda is square root of t so large rho over t then the density fluctuation are suppressed and this is relatively easy to understand let's take simply the box of size l times l, the square box and let's look at the ground state of this box, so I start at t equals 0 then if I take a square box and I look at zero energy if I take periodic boundary condition then the ground state is just uniform gas, no kinetic energy and no density fluctuation so psi of r is simply square root of the density with a phase equal to 0 by convention so no e to the i theta this density is simply the number of particles divided by l square the area of the box and the interaction energy is therefore h bar square over 2m this g tilde and the product rho zero times m so now let's go to nonzero temperature what I claim is that if I am at relatively large phase then the density fluctuation will be very small and I can neglect them in first approximation here I have plotted this modulus of psi as a function of x, y and what I am saying is that the modulus of psi is nearly constant and the reason for that is that if I take a large phase density so rho zero times lambda square very large compared to 1 and the fluctuation have a very high energy cost I take the interaction energy which is written here I calculate it per particle so I divide by capital N and when I take this quantity h bar square over 2m g tilde rho zero and I compare it to kBT you see that what I see appearing is simply this h bar square over 2m divided by kBT which is nothing but lambda square here h bar square divided by m kBT times the density rho zero so what I get in the end with g tilde is g tilde times the phase phase density divided by 4pi and if I take the limit of large phase phase density d then it means that this is very large compared to 1 so the interaction energy per particle is very large compared to kT so it cost a lot to create a significant density fluctuation the particle are repelling themselves so strongly that I prefer to have a uniform density of course I will be left with phase fluctuation but the density fluctuation are very strongly reduced at large phase density so I concentrate now on phase fluctuation so let's look at the energy related to phase fluctuation so the interaction energy now is frozen because the interaction energy is just a modulus of psi to the 4 so the phase does not enter into the interaction energy so the only thing which will remain is kinetic energy this energy is therefore the gradient of psi and if I plug the fact that psi is the square root of density times e to the i theta and this does not fluctuate because it is fluctuation of frozen I am left only with the gradient of theta and I have kinetic energy term now which is h bar square over 2m the density times the gradient of the phase square d2r and this is the energy which is relevant for describing a both gas at low temperature or high phase phase density gradient of phase and I would like to emphasize that this type of energy where the energy depends only on the gradient of the phase of a field is typical of a superfluid what I see that the energy which is relevant in this system is the energy that I need to twist the phase at zero temperature the phase is uniform and the relevant energy in the system the relevant excitation are excitation that we are trying to twist the phase and when you have such an energy which is relevant for the system it is relatively easy to understand that this corresponds to a superfluid system and the best way to understand that is to think of a good test of superfluidity which is a so called rotating bucket experiment what is this experiment the rotating bucket experiment is just to put your system 2d of 3d in a bucket and rotate the bucket and see whether if you rotate the bucket at uniform speed whether the fluid which is inside the bucket will rotate together with the fluid or will stay at rest in the laboratory frame and to compare the two possibilities either the fluid stays at rest or the fluid rotates with the bucket you have to think of what happens if the fluid rotates if the fluid rotates it means that the fluid in the lab frame which is supposed to be an inertial frame will acquire a non-zero phase as it rotates it corresponds to twisted boundary condition for those of you who are doing numerical simulation superfluid which is at rest in a rotating frame is a superfluid which in the lab frame which is inertial has some twisted boundary condition the twist being proportional to the rotation frequency so saying that we have an energy like that means that it takes some energy to twist the boundary condition takes some energy to put the superfluid in motion and therefore if the rotation of the bucket is small enough the superfluid will prefer to stay at rest in the lab frame it will cost less energy than to twist the boundary condition so what I have done implicitly here by neglecting density fluctuation and writing my energy like that is I have said that I actually dealing with the superfluid with the superfluid density which for the moment is raw so it is a bit subtle here so I wanted to say it explicitly but this corresponds to a real experiment twisted boundary condition that you use in numerical simulation corresponds to the experiment of rotating the bucket in which your fluid is any questions at this stage you don't see any so let me continue so let's deal with this phase situation today the algebra will be relatively simple it will be very close to what we have done for the crystal of Mr. Payers in the first hour this morning I will suppose that the variation of the phase are relatively soft this set of arc fluctuates on the sample but I assume that I don't have any phase defect for the moment no vortices, vortex is a phase defect which is very localized since I told you yesterday and I will say it later the phase around the vortex rotates by 2 pi irrespective of the size of the contour I take around the vortex so vortex is a very strong phase defect so let's neglect for the moment the vortices I suppose that the phase varies gently over the sample so I assume that I can Fourier expand my phase, theta like with Fourier component which are labeled as cq here at thermal equilibrium I know that the cq modulo square will be proportional to kT the same algebra as the one I did this morning for the vibration of the crystal and therefore exactly like when I calculated this morning the variance of uj minus u0 in the crystal here if I calculate the variance of the phase theta of r minus theta of 0 in my both gas that is I'm comparing the phase that I have here and the phase that I have there and I'm looking at how these the difference between these two phases fluctuate I will get something not redoing the algebra but it's really the same will vary like the log of the distance r divided by a length scale which is the thermal wavelength again which is written on this board so here we are simply recovering the result by pihors and the factor in front of the log which is important is simply 2 over the phase phase density so I'm skipping some algebra but nothing mysterious in this algebra and now let's ask ourselves what does it mean the order parameter of the system the order parameter of both gas told us this morning again is simply the wave function of the psi that I'm taking so what does it mean for the one body current function if you want j1 this psi of r psi star of 0 how does it vary with r if the phase of this varies like this so I want to calculate rho 0 again I have frozen the density fluctuation so the density is uniform in the system and I'm left only with the friction of the phase so I have to calculate the thermal average of e to the i theta of r minus theta of 0 and there because the theta of r is a sum of Gaussian fields the cq which is here I can use a well known result for Gaussian field which is that the average of e to the i u where u is a Gaussian variable is simply the exponential of minus of u divided by 2 so I have to take now the exponential when I want to calculate this I want to take the exponential of this 2 over d times the logarithm so what does it give the exponential of a log give of course the function itself with some power and the power will be this 2 over d which is here 2 over d so this is what I have written here the one body correlation function calculated from this formula is equal to 1 over r to the power alpha where alpha is equal to 1 over d the phase phase density so what we get here is a result which is notably different from what we had for the ideal both gas for the ideal both gas I remind you that I told you here because I have interactions because these interactions have frozen the density fluctuation and I'm left only with phase fluctuation I have a decay of g1 which is much smoother which is algebraic and this is why people speak about quasi long range order and so here on this graph I'm comparing typical situation if I take a phase phase density of 10 typical of what we can achieve in a laboratory with cold atoms if I plot this g1 function with respect to the distance r divided by the thermal wavelengths lambda t so typical thermal wavelengths when you are at temperature in the order of 10 to 100 nano kelvin is a fraction of micrometer so here I'm looking over distances of typically 100 micrometers which again corresponds to what we can do experimentally here you have the result for the ideal gas this is this exponential decrease that we had seen this morning and this is the algebraic decay that we get for the interacting gas for this is still a result which is compatible with Mermin-Wagner theorem g1 still tends to zero when r tends to infinity but as you see it tends to zero very very slowly you have to go very very far in order to see g1 equal to zero so for any practical purpose actually you will see some phase coherence between the edge of your sample and the center any question at this stage don't be shy excuse me I use plane wave exactly so if I come back here I use plane wave to expand the phase yes exactly yes here this is valid only if as I said soft variation of the phase no vortices exactly when the vortices are here this type of expansion is not valid anymore and we will have to then we will lose the algebraic decay as soon as vortices at least three vortices can be present we will lose algebraic decay and this will be the Kosterlitz-Stowless foundation but for the moment I have low temperature and at low temperature we will see later tomorrow that it's reasonable to assume that there are no free vortics in the sample and since there are no free vortics this expansion of the phase is legitimate yes exactly yes a good point yes that's true I mean the interaction is hidden somewhere yes I will have a slide on that later but I'm not sure it will answer fully your question you need interaction to freeze the density fluctuation of course but once you have frozen them as you see the interaction energy is constant the g tilde is here it enters into the energy but it's a threshold for the energy it's a constant and it does not enter anymore as a function here there are some universality aspects in this 2d both fluid because precisely of your remark yes that's a very good point yes for the interaction yes absolutely interactions in 2d when you want to describe really in a quantum way are a bit complicated because you have some logs again you have seen many logs already but there are some logs which appear in the scattering amplitude and so on so at the class I'm indeed working a single vertex within classical field analysis yes so this is a remark maybe which will be related to the question that you had just one minute ago about where are the interactions hidden so in what I've said they are completely, completely hidden actually if you want to put them back in a heuristic way you should come back to the formula I gave you when I gave you this expression for the kinetic energy gradient of the phase theta this expression when you write it like that assumes that the whole gas is superfluid in the sense that when I twist my boundary condition I'm saying that the energy that it costs is directly proportional to the density of the whole gas this is a bit too much idealized in the sense that I've used for this a classical field approach but I assume that my density fluctuations are frozen and all these assumptions are valid only at a relatively large distance scale larger than typically either the healing length or the thermal wavelength so what people do usually in order to account for the fact that this is really a Hamiltonian which is valid only for large distance but not valid for a short distance or large wave vector they take into account the physics at short distance in a heuristic way by saying let's not put here rho but let's put the superfluid density rho s which is a phenomenological density at this stage but which is quantity which is smaller than rho and this is a way to absorb here all the short distance physics in this rho s and to concentrate on the long distance physics so now back to your question where are the interactions the interactions actually are contained in the fact that I want to replace rho by rho s in order to put all the short distance physics under the carpet but still remember that it's present and the rho s will not always be the same depending on the strength of the interaction rho s is a very strong rho s it's very close to rho whereas if the interactions are weak, rho s will be notably reduced with respect to rho it's not a precise answer but it tells you a way to recover some effect of interactions exactly, this is exactly a renormalization absolutely, yes so just a teaser for tomorrow as has been noticed my plane wave expansion is valid only in the absence of vortices so the question I would like to show very briefly here is what happens when density fluctuation becomes significant and what is the biggest density fluctuation that can occur it's when the density the fluctuation is so large that the density still becomes zero at a given point when I have zeros of the classical field psi then I have a big density fluctuation because the density goes from a small value to zero, it cannot go beyond zero it has to be positive so I just want here I will address vortices tomorrow but I just want here to tell you what is a zero of a classical field immediately show up when you look at the big density fluctuation and the zeros of the field so the classical field psi of xy as I said is a complex field so it has a real part and an imaginary part saying that I take a zero of this classical field means that the real part has to be zero and the imaginary part has to be zero so let's look at what it means so if I take the real part the real part is the real quantity by definition so it can be positive in some position of space and it can be negative in other position of space and it will take it will be zero along the line which will fluctuate in time because of thermal fluctuation if I take the imaginary part same thing it will be zero along the line and it may be negative say on the left and positive on the right and the zero of the field is a place where both the real part and the imaginary part are zero the line where the real part is zero in red the line where the imaginary part is zero in blue at the intersection of this line of these two lines I have a place where the density is zero because both real and imaginary parts are zero now let's look at the phase around this in this place here the real part is positive and the imaginary part is positive so if I plot the phase on a trigonometric circle I have a phase which is somewhere in this quadrant here in the first quadrant both real and imaginary parts are positive if I go here now here the imaginary part has become negative but the real part is still positive so my phase is here now my phase in this part both real and imaginary are negative so my phase is here on the trigonometric circle and now it's here and now you immediately see that when you rotate around this point you get a vortex that is you have a phase winding of 2 pi so it is plus 2 pi in this case it would be minus 2 pi so region of positive and negative here but taking into account big density fluctuation means taking into account the zero of the field and the zero of the field of simple zero is simply a vortex in this context so vortices naturally will show up when we look at big density fluctuation is that okay? and just to continue with that again to finish the teaser for tomorrow why are vortices going to change the picture of having a superfluid in quasi long range order in the phase so let's suppose that I have no density fluctuation so if I have no big density fluctuation no vortices so I have two points A and B and between these two points A and B there exist some weak coherence in phase this quasi long range order between A and B which as we see will vary like a power law with a distance between A and B suppose now that vortices can be present that is I can have some randomly I can put one vortex somewhere between A and B so vortex as I said is a zero of the density so here I have put one vortex what does it mean for the phase of the wave function it means that here the phase was phi putting a vortex in between because of the phase of the associated to the vortex rotates by 2 pi around it it means that the phase phi will become phi plus pi so putting vortices randomly will completely scramble the phase ratio between these two points because the phase can jump between phi and phi plus pi if the vortex is just put on the segment which is between these two points and actually it will fluctuate between phi and nothing and anything else if the vortex is not strictly on the line so as soon as I have isolated vortex v vortices with density RoV I expect that I will have no phase ordering on a distance which is typically inter vortex distance because if I take any two points separated by this distance I have one chance out of two to have one vortex in the middle so vortices are really something bad if you want to maintain long range or quasi long range coherence so that's why it's important to understand that I have neglected them so far and when I will put them all the conclusion will be changed is it okay? yes yes this is already topological defects so you are right there is a kind of unique picture grand picture here the vortices are objects that you can see immediately without having single atom resolution so it's a difference with respect to what Emmanuel was discussing where he was addressing single excitation with single particle whole or doublon was a single particle effect here vortex may involve thousands of particles the number of particles that are missing in the vortex core can be relatively big but you are right seeing from a sufficient large distance the two phenomena looks very much the same it's physics of topological defects and phase condition related to that no more comment okay so I don't think I want to go yes I don't think I want to go further now on the analysis of the classical field tomorrow we will discuss vortices and release a BKT transition and I will give you some examples with real atoms how this can be addressed in the remaining 20 minutes I would rather like to come back to the ideal gas and show you some particularity of the ideal gas that you may find in the literature I wanted to show them so that you wouldn't be surprised if you see them in the books and also since it is the end of the two-hour lecture I wanted something relatively simple to finish the lecture okay so I want to discuss about back to ideal both gas harmonically trapped system and discuss this Bosnian condensate of photons that was found first in the group of martin Weitz in Bonn which has been seen now in several groups so for the moment I have discussed only my 2dBos gas in a uniform box in the XY plane let's suppose now that we have in the XY plane a harmonic potential you remember I have a harmonic potential along z to confine the particle but now I add another harmonic potential in the XY plane much softer I don't want the particle to accumulate but I have a harmonic confinement in the XY plane now what is known is that you recover Bosnian condensation in this case even in the thermodynamic limit although you are in 2d but it's not a uniform system so it's not a violation of Mermin-Weigner theorem and the way to understand this is to look again at the convergence of the integral I have written this morning which I remind you what it is it's the Bosnian low in the denominator and the density of state in the numerator and I told you this morning that if I take a uniform 2d system the density of state was constant and because it was constant this integral was not converging but now if you take a not a uniform system in 2d but to take a harmonic oscillator the density of state here is proportional to e and this is relatively easy to understand if you take a 2d harmonic oscillator I can always look at the eigenstate of this 2d harmonic oscillator I am looking at the eigenstate along x and the eigenstate along y and so a typical eigenstate of the full motion in 2d requires the knowledge of both nx and ny the degree of excitation of the x oscillator and y excitation and therefore if you look at the eigenstate in energy this is what you get the ground state is non degenerate n equals 0 both for x and y the first excited state is degenerate twice because you can excite x and leave y in the ground state or you can excite y and leave x in the exact state by 1.2 then the second excited state is degenerate 3 times 4 times and so on so you see that the degeneracy increases with the energy and it decreases linearly 1, 2, 3, 4 so this gives you if you take a kind of thermodynamic limit d of e to e and then the integral converges which means that if you take a harmonic potential in 2d you have in the thermodynamic limit which corresponds to going to a very, very low frequency in the xy plane you have an upper bound on the number of atom you can put in all excited states which survives to the thermodynamic limit which is simply y2 over 6 times kbt over hbar omega to the square so it seems that actually as soon as you put a harmonic potential you recover both and shine condensation in 2d like you had in 3d actually this is not true this condensate that you get in 2d in a harmonic potential is singular and a good way to see its singularity is to try to calculate the density at the center of the harmonic potential and in order to do that a simple way to do that is to take again the Bose-Einstein law and take a semi-classical version of this Bose-Einstein law that is when you have the energy here you replace it by the sum of kinetic p2 over 2m plus potential energy treating p and r as classical variable so this semi-classical approximation which is valid when temperature is large compared to hbar omega the quantum of oscillation in the xy plane this semi-classical approximation allows you to write a kind of phase-space density for your system and now since you look for Bose-Einstein condensation you take the limit of the chemical potential mu tending to zero you integrate over the momentum to get the density rho of r and you get this expression for the density you can do all the algebra, it's analytic you get a log of 1 minus a Gaussian function of r this is how the density vary in the trap when you have a Bose condensation and you see that when you go to the trap center you can again linearize the exponential and you get that rho of r varies like minus log of r squared and so this is the quantity which is diverging in r equals zero and this is why I said that this condensation is singular you get indeed a Bose-Einstein condensation but in the thermodynamic limit this Bose-Einstein condensation corresponds to an infinite density in zero it's still okay because the density is infinite when you take the integral of the density to get the number of particles you get something finite because the integral of a log even though the log is diverging in r equals zero the divergence is strong enough to make the integral diverging so there is no mathematical problem here but there is a physical problem an infinite density is not possible as soon as the particles interact even very very small it is not so relevant but it is relevant for non-interacting particles if I have strictly non-interacting particles well, I can accommodate for that at least within this semi-calcical approximation so one question is what are the non-interacting particles that I can have at my disposal and a good candidate for a non-interacting particle are photons in principle do not interact with each other at least at low energy so one question that we can ask is it possible to do this experiment a 2D both gas in a harmonic potential with non-interacting particles so in order to do that we have to solve a few problems we have to give a mass to our photons we have to trap them harmonically in 2D and we have to give them a non-zero chemical potential for those of you who have remember their course on black body radiation is not trivial in principle people always say that the chemical potential for photons is zero so how is all this possible so let's start with giving a mass to photons how is that possible so the experiment that was done in Bonn in the group of Martin Weiss was done in a Fabry-Perot cavity so Fabry-Perot cavities are just two mirrors in front of each other and these two mirrors are very close to each other so it is just a few wavelengths of the photon the photon that will be used will be in the range 500 nanometers yellow photons, say yellow or green and you take this L here such that it is only 7 wavelengths over 2 times the index of refraction and zero so it means that the photon actually will occupy only one what is called longitudinal mode of the cavity the photon will because they are selected in this range here they will occupy only the mode n equals 7 the longitudinal mode n equals 7 if they wanted to occupy n equals 6 they would have two low wavelengths and if it was n equals 8 it would be a two large wavelengths so the selection in wavelengths is done by taking a very narrow cavity so that what is called the free spectral range of the cavity is very large and all photons will accumulate in the same n of the cavity n equals 7 so it is a way of freezing the longitudinal degree of motion along the cavity so now how are these photons injected in the cavity so the cavity is filled with a die which has this index of refraction and zero and you pump this die with a laser beam so you excite the molecule of the die and then the molecule fluores and the fluores sends photon of these molecules should enter the mode of the cavity so we are interested in the photons which correspond to the fluorescence of the die molecules when they are pumped by the external laser which is represented here ok, so this is the experiment why do these photons have a mass in 2D well, the answer is the following look at this diagram here, so I have my photons which are bouncing back and forth between these mirrors with this between the wave vector of the photon kz and the length of the cavity such that kz is equal to 7 pi times divided by the length of the cavity l this is for a photon which is at normal incidence on the mirror but now let's suppose that all photons are not at normal incidence some photons have make a small angle respect to the optical axis I say small because I want to use paraxial optics but still it's an angle which is not zero then the wave vector of this photon kz square which again is quantized by this formula plus some k perp square corresponding to the small angle of the photons so k perp is assumed to be small compared to kz this is the so-called paraxial approximation so I can make an expansion of this square root here and I get that this energy is equal to h bar 0 kz 1 plus a small quantity k perp square divided by 2 kz square and this is exactly what I need to give a mass to photon I have a dispersion relation e of k, which is now quadratic with respect to k perp so with respect to the motion in the perpendicular direction perpendicular to the optical axis I have now this quadratic dispersion which means that I have a mass for the photon, an effective mass and this effective mass you can read it from this formula, it's simply h bar n 0 kz over c and when you do the numerics you find that it's a mass which is very, very small which is 10 to the 5 times smaller than the mass of an electron so we have given a mass to the photon but it's a very, very light mass which is good because it means that it will be easy to both condense a light object has a very long derivative length so it's easy to condense and I have also some energy of set h bar omega 0 corresponding to the photon which have no k perp so which are only providing along the axis so we have fulfilled the first item which is to give a mass to the photon second item is to have a harmonic trap for the photon harmonic trap means that I don't want the same energy if I am on axis here or if I have off axis at a distance r from the optical axis of the cavity and this is provided because I take not plane mirror for this fabric fabric perp cavity but curved mirror so taking curved mirror like this means that the distance L of 0 is larger than the distance L of r L of r again I can do a quadratic expansion L of r varies gravatically with respect to the distance so it's L0 minus r squared divided by the radius of curvature of the mirror and if you remember that the energy of a photon which was going straight was inversely proportional to 1 over L now since L itself depends on r it means that this energy h bar omega 0 will depend on r because of this quadratic variation so I have this h bar omega 0 of r which h bar omega 0 remind you it's this term the 0th order term in the expansion with respect to k so I have h bar omega 0 which depends on r squared and since I have a mass for my photon I can now derive a oscillation frequency omega and this oscillation frequency is 40 gigahertz in this experiment so summary of this analysis but thanks to this setup I have an energy for a photon at a distance r with a wave vector k perp which is quadratic in k perp and quadratic in r this is exactly a harmonic oscillator a 2D harmonic oscillator so it's a nice way and I think it's nice even for theorists because it tells you how you can achieve in the lab some Hamiltonian that you have on your paper it's a nice way to implement a 2D harmonic oscillator out of particle which in principle have no trapping potential and no mass by restricting to 2D this is what you get and so now we can see what happens in this experiment the photons as I said are limited by the fluorescence of the molecule of the dye these molecules have plenty of vibrational level both in the excited state and the ground state the photons have a distribution which correspond to a thermal distribution with the same temperature as the temperature of the molecules so the molecules have a temperature which is room temperature 300K and indeed when you look at the spectral distribution of the photons so this is the signal the fluorescent signal in terms of lambda you see that when you pump gently the system is described by a Bose-Einstein law with some chemical potential I will come back to this in a moment and the temperature which is the temperature of the dye molecule 300K and when you increase the pumping you put more and more particle in the system by cranking up the laser which pumps the molecule of the dye then at above a threshold you see a peak appearing close to the energy corresponding to the thermal state of photon accumulation of photon in the lowest mode of the KBT here and the condensation is expected as I told you before when the number of particles is proportional to KBT over h bar omega to the square I had a pi square over 6 before now I have a pi square over 3 because the photons have two polarization so I have twice as many photons in the absence of spin and the number is typically 60,000 in this experiment and there is a good agreement between theory and experiment so just one word here I am giving a chemical potential to the photon because I am using this Bose-Einstein law and I am saying nu is nonzero and it is only when nu approaches zero that there is a condensation why is it that we can give a nonzero chemical potential to our photons whereas I am sure you have been taught in your statistical physics lecture that the chemical potential for photons is zero in black body radiation is the following when you study black body radiation you have all the spectrum of light which is inside an oven and you may have a blue photon which arrives on the wall stick to the wall creates excitation in the wall and then the wall will emit two red photons as long as you conserve energy it is okay so the number of photons is not conserved in such a process you have one photon becoming two photons so this is what is written here you have one photon, excitation of the wall, two photons and therefore when you write the balance between this process you must have mu equal to mu because you have one can give two and therefore mu has to be equal to zero in the experiment that I just described there is a conservation law for the total excitation number that is here in this experiment in good approximation we have one photon plus one molecule in this round state give one molecule inside that state and vice versa give one photon plus one molecule in grand state so we have a conservation law associated with rotating wave approximation and because we have a conservation law there is a Lagrange parameter associated to this conservation law and this is precisely the chemical potential so we have in this experiment a conservation law and this is why there is a chemical potential which is not zero and this is why this Bosan-Schein condensation of photons could be seen ok so I will take question in one minute but I just summarized that the lecture will be over so what do we have seen in this first lecture we have used the modeling of the 2D gas in terms of the classical field and we have characterized the dynamics and the thermo-liquorium of the system our treatment was valid and when the phase theta could be fully expanded expanded in terms of plane wave and as we have seen all vortices have been excluded up to this point otherwise the fully expansion wounded have been valid this has led to quasi long range order in the presence of interaction with g1 decaying algebraically with distance and this alpha here is also to the inverse of the phase space density so this is dramatically different from the ideal gas case and the open question the one I would like to address later well tomorrow is what happens in the presence of stronger phase and density fluctuations and how can we connect this result to the high temperature case how can we connect this algebraic decay to the high temperature case where we know that in principle we should recover exponential decay and this is where the BKT transition and with this I stop for today and thank you for attention and expect questions