 Okay, so our last speaker for the session is Nikolai Prokofiev, he's going to tell us about Bipolaron Superconductivity. Yeah, thanks Tigran and other organizers for let me speak about what we did recently. What we found something which I knew probably right after the high school, this was the long story about bipolaron superconductivity by Aleksandr Afmoot and many others. And what I found more recently is that not a single calculation was done well quantitatively, which means it was all talking, talking, talking, talking, but not a single number anywhere for the most relevant case when you are in the adiabatic regime with one frequency smaller than the bandwidth and then doing bipolarons with full numerics so you know the effective masses, you know the sizes and you understand how to calculate DC in the interacting with the gas. None of this was ever done even despite the numerous work about bipolaron superconductivity. So probably the first half an hour, I will simply try to create the field of how the problem should be approached and what kind of numbers you need to actually make quantitative predictions for how the bipolaron superconductivity will look like. And only at the very end for 15 minutes I will discuss a particular model and try to argue that you can reach transition temperature potentially, which is maybe four times higher than the highest possible TC you can get from the Ilyashberg state. Okay, so let me start. So first let me thank the collaborators, it's Chao Zhang, she was at UMass and now she moved to East China Normal University in Shanghai, Boris Svistanov also from UMass, Mona Birkiu from Vancouver, Jon Sos, Andrew Millis and David Reichman from Columbia, Jon moved to Stanford I think, the San Diego, sorry, Stanford. Barbara Capagross Sansonasi is at Clark University and Massimo Bernicelli from the University of Alberta. And most of the results are now on the archive or published in a couple of Fizreff papers. Now the outline of the talk will be as follows, the many body parts of the bipolaron superconductivity is more or less the easiest thing, contrary to everything else we have in the strongly coupled electronic systems. Yes, the coupling is strong, but the moment you are talking about bipolaron superconductivity the many body aspect of the problem is probably the easiest one. And that's the one which I will try to quantify first. Well, next once you have the many body physics under control, there's only probably a small exception that you have to be sure that when you create bipolarons they do not face separate. This will require very hard many body calculation. So assuming that face separation is not happening, that bipolarons behave like a bazoonic liquid, calculating TC is relatively easy. So next I will simply just discuss what happens for bipolarons is a very steep competition when they form between those bipolarons being extremely heavy or very, very large. And essentially you have to compromise between the two limits because either of them will result in a small value of TC. So you have to compromise and optimize the properties of bipolarons to have high TC so they are relatively compact and relatively light, but you are never happy with both of them. And finally, just while I'll discuss the very end a particular model when I will claim that you potentially can reach transition temperatures as high as a thermal frequency divided by five. But remember that in some high pressure hydrides, omega can reach a fraction of electron volts. So divide by five and decide whether this is high enough or not. So point number one, if you have bosons, because bipolarons will be bazoonic particles, if you have bosons, they don't need a mechanism to become superfluid. So no matter what you do with bosons, no matter how strong they are, they couple. They will bazaar condense and go superfluid. They will not bazaar condense, but they will go superfluid in two dimensions as well. And essentially there is very little you can do to prevent this from happening except creating a crystal or going to a phase separated state, which is a crystal again. So otherwise, it's very robust phenomenon. For example, if you take helium three, which is very strongly coupled state in terms of kind of many body physics, you have to do numerics to compute TC, you end up with the result which is on the point seven of the transition temperature in otherwise an ideal gas, even though we are in a strongly correlated liquid. Formally, you also have to remember that while we talk about bosons, almost always bosons are composite particles and even helium four is probably the good bound state of 14 different thermionic species. So those are the two aspects which we have to worry about bipolaron superconductivity as well. That first, what is the transition temperature when we form bosons, what will be the effectiveness of this boson and how large or not compact this boson looks like? But at some point we'll have to stop the density. So let me first kind of go through the first point. How robust is transition temperature in the interacting boson gas if you change the model or change the interactions? So this is the case for three-dimensional case, I know that while this is easiest, if I simply stay further away. So if I look at the three-dimensional case and I look at heart and soft spheres in the continuous space, then you say, okay, how much DC, sorry, how much DC will change if I change the density? Well, the density, I know, 10 minus 1, that's already more or less when the heart spheres will start touching. So you go all the way from extremely dilute gas, first transition temperature actually goes up if you increase interactions, it will rise, well, it's big on the plot, but just look at the vertical scale. It's at best maybe 6% and then it drops, dramatic drop by another 7%. So essentially, no matter what you do with heart spheres or soft spheres from dilute limit to pretty dense gas, almost creating a liquid, transition temperature stays exactly the same as the ideal gas within the 10% accuracy, if you like. So okay, yeah, I will allow you to probably go further and maybe even touching the floor, but this will be given the vertical scale maybe 0.91. So it's pretty robust. Well, this is about the short range interactions. So we have heart spheres, soft spheres, looks good, but you say, okay, well, what happens if I have coulomb interactions? And the answer is not much will change. So recently we did the calculation for the coulomb gas for different values of RS, lattice and continuum. So on the lattice, we use a Harvard U, for example, eight times T, strongly coupled, we can use a coulomb repulsion on the nearest neighbor side is also eight and then it decays as a coulomb law, three dimensional coulomb law. So this is again TC as a function of density, starting from half feeling, okay, 15% down and then going to the dilute case, 10% doping, it's close to one. Well, you go to the continuum and we can look at different values of RS. Well, for values of RS simply just remember what is RS. So if you introduce the bore radius as the unit of length, then you simply measure the interparticle distance in terms of RS. So RS is a dimensional parameter giving you the volume to particle. So if I go to very, very large values of RS, that's very, very dilute case because you have a huge volume to particle, like 25 cube, it's already big number, so it's close to 10 minus 3. And there's all the pi is probably 10 minus 4. So you change density here essentially from half feeling to 10 minus 4. And again, you see the transition temperature barely changes within if I just draw the line somewhere in the middle plus minus 10%. So coulomb interactions are also within the same picture. So if you have interacting bose gas, short range, soft spheres, hard spheres, coulomb interactions, strong, weak, you are still good if you estimate TCS if it's an ideal bose gas. Now 2D is slightly more subtle in the sense that while ideal bose gas in 2D has no transition temperature. But it happens to be that the dependence on interaction, so this is the scattering length in 2D, we'll say for hard spheres, the dependence on interactions only is through double log divided by 2 pi. And just believe me that double log divided by 2 pi can never be large, no matter what you do. So unless of course your density is one boson for the solar system, then probably you feel this number. Otherwise it's pretty much kind of almost static and nothing changes. Well, for example, this is again numerical calculation for hard disks. Well, this point is they almost touch. Maybe I know three times more intensity and they will touch, but you see again the scale of changes from 0.80 into 0.22 when you change the density by three orders of magnitude. Of course, there will be Wigner crystal at very low density, there will be Wigner crystal. That's what I say, okay, if it's a super fluid, don't worry, it's ideal gas TCTCTC and then boom, Wigner crystal in very strong cathodic. Oh, this is a separate calculation. So if you do two dimensional gas, okay, let's go to the pooling case. Just to finish my story. So the end of the story is if I have pooling gas in 2D, again on a lattice with strong hybrid U, strong pooling interactions, so you do continue for very different values of RS, again, covering essentially all densities from high to low transition temperature barely changes relative to say, you know, some baseline like 1.2 plus minus 10% for the entire range of densities and different coupling constants. And that's the way you read that, well, okay, going further in RS beyond 50, so 50 is the last point. Once you go beyond 50, you start seeing Wigner crystal. And that's why kind of I'm not doing anything here for the following reason. A direct flow transition between the super fluid and the Wigner crystal is forbidden. So you cannot even determine the phase boundary because it's intermediate kind of range of coupling strengths when you have bubble phases. Because in the long range system, you cannot produce any change in density unless creating a negative surface tension. So on some scale, you will start creating a staircase of different bubble phases. And numerically, this is essentially impossible to study with huge finite size effects, because it's a staircase of different transitions. Okay, so because first order directly is forbidden. And it is a kind of very strong change from liquid to crystal. Yeah. This is a coolant gas of bosons. This is continuous coolant gas of bosons. This is latest gas of bosons coolant. Again, it's on site coupling is eight nearest neighbor coupling is eight. And starting from this point, it's one or it's a divided by R, where R is measured in terms of latest constant. Again, the point is, all those plots are simply telling you if you want to estimate what is the transition temperature of the interacting two dimensional boson gas, no matter how it's coupled, just say it's 1.2 and divided by M, and you will not be mistaken by more than 10%. In 3d and just say it's 3.2 and two thirds divided by M and you will not be mistaken by more than 10%. So that's the end of first story, which is saying the many body part. Once you form bipolarons and their bosons, in a sufficiently dilute limit when you don't break bosons apart, calculating the Tc for the system is very easy. Just essentially use ideal gas expressions, which may be slightly changing the number a little bit because this is not ideal gas expression, but say 1.2 and divided by M means to this type estimate, you are accurate. So this part is under precise control. So with all the calculations, we see that we can understand what will be Tc in the interacting system. So next, we have to go back to properties of bipolarons. Okay, so I hope this message simply says, okay, well, we don't have to worry about what is Tc once you form bosons. Well, it would be, whether it's sensitive to the dispersion of bosons. Since I'm talking about more or less dilute systems, not that much in a sense that I don't feel the entire bandwidth. But this is the latest calculation. And for example, the last points are close to half feeling. So at half feeling probably you feel the entire dispersion relation because it's half feeling and transition temperature is, you know, half of the bandwidth. The answer is barely. You are talking about one or two percent. So if I calculate, for example, take tight binding model and calculate transition temperature for the tight binding model for the ideal guess, or case squared over double m, you will get roughly the same numbers. So it's a question about multiple. Well, okay, well, then if you ask the ground, whether if you change the dispersion relation to the more at latest, then everything is completely different. Now, of course, we are talking about simple parabolic behavior or simple band with simple minimum, and the minimum is at K equals zero. If you create a minimum on the circle, it's a totally different story. We don't even have bosons. Yeah, in this sense, yeah, this is a good point that yes, sometimes even the bed dispersion is such that you completely change with it. Well, because it's already liquid on the edge of forming a crystal. And starting from those 70% of ideal TC just drops to a crystal. Again, I would say just given everything in the strongly coupled remuneric systems numerically, 70% is already not bad. But I'm saying if I am slightly diluted, not liquid, which is about to form a crystal, you'll be sure about 10%. This is simply removing one trouble in the discussion. So don't worry about TC once you have bosons. Okay, so that's more or less the outcome of this discussion. So if you want to estimate transition temperature in the interacting boson gas, just use this type of expression. And those are the numbers which I suggest for you to use within 10% accuracy. Okay, of course, depending on circumstances, it can be slightly higher, can be slightly lower, but not by much. Next, probably what you have to remember is that, well, strictly speaking, when bosons form, they form as bound states of fermions, and those may have finite size. And strictly speaking, everything I discuss here, what is the transition temperature in the boson gas at the given density, is written under the assumption that I don't break bosons apart, which means it's written for densities when bosons are not allowed to overlap. If they strongly overlap, then probably those estimates are no longer accurate as I claim. So this is only done for the sufficiently dilute case, which means I can use this type of expressions for densities all the way to the point when say, particle density times the volume of the bosonic of the fermionic pair, which is forming a boson becomes 41. Going further, probably those estimates are no longer very reliable, but that's how you get to the estimates of what is the highest Tc by the bipolar on mechanism is possible, because I understand, of course, the higher the density, the better Tc, no problem. Keep increasing the density, but at some point I have to face density, which is so high that it's either crystal or I start breaking bosons apart and I create a fermionic soup. Okay, so remember this. And second, we are not in a continuum in condensed matter, we are living on the lattice, which means that the effective mass of a bosonic bound state will be strongly renormalized relative to the bare particle. So the bare particle mass or bare polar mass can be m already renormalized, but the bipolar mass is not double m, because we don't have Galilean invariance, and this can be orders and orders of magnitude stronger than double m, and this has nothing to do with E equals mc3. Okay, now you can ask the next question. You say, okay, what if I try to squash bosons a little bit, so I create a fermionic soup. Not much is known about this particular setup, except one problem which we know for sure how well it works. It's called unitary fermi gas. We know properties from the cold atoms, but also from the calculations which are reliable within the arobus for the unitary fermi gas. That's how the unitary fermi gas behaves. So you have transition temperature, which you measure in terms of fermi energies. The estimate for fermi energy is roughly the same as my ideal bosic gas temperature, up to a number, roughly close to one fifth. So that's what we know. So if we have very strong interactive interaction between the fermions, they create very compact bosons, and then the transition temperature is the same as for the ideal bosic gas, slightly going up, because this is universal, we can interact in bosic gas. The moment you kind of create finite density temperature, it goes a little bit up, then it reaches a maximum and goes down, and it was reliably computed at unitary to be, I know, one, six, five. Again, the difference here is not dramatic, maybe 20%, between the unitary limit, when bosons already completely squash each other, everybody wants to make a bound state with everyone. So this is already beyond the overlap. So up to overlap, the transition temperature is the same as ideal gas. That's the outcome from the unitary fermi gas, but well then start to the overlapping transition temperature slightly dropped, but not by much. So which means estimating the possibility for having the high STC by bringing bosonic pairs to the edge of almost touching is probably still not spoiling the estimate for how high PC can be. Probably the other point I would like to say that, well, there is nothing unusual or you would not have to pick out if it happens that transition temperature is higher than the binding energy of bosons. Looks kind of a little bit strange, you say, okay, yeah, for example, somewhere here, you have a piece which says, okay, well, there is no overlap formally. I'm just plugging numbers for unitary fermi gas, as is. It says, okay, well, you already have no overlap from very simple criterion density times the volume per particle equals one. So that's what I call no overlap, because partially there is overlap, but that's the criterion. Well, here already transition temperature is higher than the binding energy. Of course, if unitary, the binding energy goes to zero, but transition temperature is still 0.16%. In this sense, I say, okay, yeah, it's not so bad if I simply allow bosonic density to go all the way up to say touching or getting to the dashed line when the density is roughly inverse volume per pair. So this will be my highest estimate for TC given whatever system I am trying to solve. That's the many body physics. Of course, this is in continuum and this is the only fully solved strongly coupled fermionic system I am aware of. On the latest, many things can happen different from continuum. For example, I can estimate TC as going like this curve and I reach the point of touching and then you say, okay, what will happen to transition temperature after that? We don't know. It can slightly go up maybe by 50% or it will probably drop down. And if I start approaching say half-pilling, then it's in some calculation this was observed. It's quite possible that you can create even the insulating state. It can be anti-permagnetic mode state or some other structure which is not super fluid at all. So which means if I go to density all the way to half-pilling, nothing is reliable. So which means all of this which I am trying kind of to discuss with you now is a sufficiently dilute system which means low density of fermions definitely far away from half-pilling when I'm not responsible for what will happen in this system. So essentially everything is up to no overlap, what happens beyond that, that's questionable and requires a separate calculation. Okay, now just brief introduction of what are polarons. It's kind of trivial. So, a polaron is essentially a particle which is modified by its coupling to the environment. So that's the notion of the polaron. Okay, here it's written like particle Hamiltonian, phonon Hamiltonian and behind the image you have the interaction Hamiltonian. Well the interaction Hamiltonian the most standard one would be that electron density couples to the displacement of atoms maybe on the same side, maybe on different sites which means you have a whole range of different coupling constants. Well but the simplest model would be simply to write down the local coupling that there is a local coupling between the electron density and the displacement of some phonon mode. Typically people consider the optical mode and this is known as Hallstein model. For the generic model with all kind of couplings probably I can even make an image. It's an electron but if I just inject this electron it will push atoms around but if electron has to move it has to drag the entire latest deformation with it and that's what makes it heavy and standard more or less picture. Okay, when I talk about bipolarons I literally mean that while you have one electron coupled to the latest it will change its energy but maybe if I have two electrons also coupled to the latest energetically it's favorable for them to form a bound state. So this is a bipolaron. Once you form bipolarons they will buzzer condense according to the transition temperature which I showed to you but why I need to know properties of bipolarons I need to know the effective mass and I need to know their size because their size will stop me from using two high densities I cannot allow them to overlap because beyond that I cannot control that. Okay so that's essentially the the root of estimating what is the highest possible PCI can get from the mechanism that's how you have to proceed. We already know all the numbers in this formula in 2d and 3d. Well now we simply go to the highest density possible and we estimate it from the density times the volume per pair equals one and that's the final expressions. While you estimate volume as pi r squared in two dimensions or four pi divided by three r cubed in three dimensions so for this I simply have to plug in this expression or this expression and those quantities are computable numerically this is just mean square size of the bipolaron bound state and I need the effective mass. So those are the estimates which I will be plotting well maybe plus or minus 11 maybe 0.5 but okay with 10 percent accuracy those will be the numbers which I will be plotting when I estimate the highest possible PC by the bipolaron mechanism. Again the main body part is under full control but the final estimates I would say not control with 10 percent accuracy because they estimate I simply stop the density at the point when the density times volume equals one. This is not precisely control but this is a reasonable estimate maybe it's twice as high maybe it's twice as low just so be careful with that because I'm pushing everything to the point of overlap. Okay so that's the main body part now I need to understand okay and finish the job is computing the effective mass and r square average by simply looking at the r square average using the ground state wave function which is behind the main. Okay so that's how it looks like but now we're just looking at the expression you say oh the following will happen for sure suppose I have a very shallow bound state very shallow bound state probably will be relatively light because the bipolaron mass by Galilean evidence now will be twice the polar mass it's relatively light but huge if it's very large in size the moment the bipolaron forms it's huge in size so I cannot push its density too high so transition temperatures will be relatively low well but then you increase coupling and the size of the cloud will shrink so I can go to higher density you go to higher density but then the effective mass will start increasing and at some point the increase in the effective mass will be a much more important effect than making the bipolaron compact of course ultimately you reach say the one or two three latest spacings and you cannot push the size of the bipolaron further but the effective mass will keep increasing so you'll go through maximum which means we have to look at the optimal by optimizing effective mass times the size of the bipolaron squared and that's what will determine the best possible chance for high tc in your system okay so this is the framework for the discussion okay now I want to discuss a little bit kind of the other side of the story because bipolarons form when if an electron form on coupling is relatively strong if it's weak forget about them you do the mcdowell ashberg theory and so the question is okay how you go from one picture to the other when you have to change mcdowell ashberg theory to the bipolaron picture okay and I will argue the following so if you increase the coupling constant in the mcdowell ashberg theory beyond one the following three things potentially can happen first this can be the latest instability you completely restructure the latest and then do the calculation from scratch because you will change the phonon spectrum you will change the electronic structure you will change the electron form coupling constants and maybe it will go back to less than one so that's one second maybe you will simply drive some insulating phase because I was showing to you that well at half feeling many things can happen including an insulator well the other one which I can easily argue is very generic because the first two are latest dependent density dependent many things depend whether one or two happens but three is trivial to compute and I can immediately claim whether this will happen or not as a variational argument so bipolaron collapse I can compute properties of a variational state and you cannot fight with variational state I simply present the energy of this variational state and it's better if it's better than the mcdowell ashberg theory forget about mcdowell ashberg theory probably it's a very strong first order phase transition which mcdowell ashberg theory will not see at the level of one diagram because you neglect vertex corrections but if you go to vertex corrections of order 20 suddenly you will discover there is a much better state but nobody does it but variational I can immediately establish that well that's the case so that's the generic argument and I'll kind of try to convince you that well okay by going through the generic argument I can show you that well you cannot cross in mcdowell ashberg theory coupling constant lambda bigger than one which I define in terms of bare Hamiltonian parameters because sometimes coupling constant people define it through the normalized parameters which themselves are changing as you compute more and more in terms of bare parameters you cannot cross lambda equals one and I'll do this argument well it's actually pretty old if you look at the references which I hear this is roughly when I still was in the high school so that's how all these arguments so let me show you even though sometimes I hear from people that well if you do mcdowell ashberg theory for lambda much bigger than one transition temperature goes a square root of lambda all of this is fiction mcdowell ashberg theory does not exist for lambda bigger than one so let's see the proof it's the proof for the most realistic case when you have phonon frequency much smaller than the bandwidth this is what's called a diabetic limit because atoms are much slower than electrical which energy bandwidth yeah you'll see not from energy bandwidth so that's the standard definition of the coupling constant it's this is coupling between the displacement and density dimensionless displacement times density and it's divided by the bandwidth and the phonon frequency so that's the standard definition of the coupling constant sometimes people use different numbers but that's the number which I will use so let's start doing the calculation I will argue that okay lambda equals one will be the important one but lambda equal one is when the coupling constant is somewhere in between the large bandwidth and small phonon frequency so it's square root of the two now let's do the calculation so if I'm sitting in the band and I can do a second order perturbation theory using electron phonon coupling just compute for example the energy for zero momentum state second order perturbation theory well but this integral goes all over the bandwidth there is no fair manager I'm talking about just one electron and you can easily estimate okay well the energy will be a minus w divided by 2 that's half the bandwidth and some correction which is second order perturbation theory g square typical denominator is the bandwidth and maybe some number well but remember that okay lambda one will be the important one which means g squared divided by w well g square divided by w if lambda is one is probably on the scale of phonon frequency which is much smaller than the half bandwidth so this term is still dominating this is small correction as expected for the perturbation theory being valid it's okay so good let's do the other calculation simply stop the electron and not don't allow this electron to move well this is trivial you just complete the squares in immediately tell me how much I gain in energy if I have a localized electron not moving if the electron is not moving you simply gain energy which is minus g square divided by phonon frequency it's standard complete the squares well if it moves if it's moves if this will move I'm just completing the squares here for you just to be sure if this electron will move a little bit you'll find that well like actually exponentially heavy in this particular case just because the overlap between the phonon clouds when you try to move the electron in the lattice in this particular case for lambda equal close to one will be already exponentially heavy but this can only lower your energy so if I'm talking about variational state this is already good enough I can make it slightly better but not the big deal okay so this is simply taking care of the overlap that the localized electron can actually move because the system is translation invariant you compute the overlap between the two clouds and the answer is exponential minus g square divided by phonon frequency squared I'm sorry about the menu but there is nothing I can do but so what's written here maybe I can exit the the show mode oh it's good sorry oh what did you do sorry now this is good enough just leave it okay so that's the estimate for the effective copying of this localized state which is fully deforming the lattice so it's this exponential our job is done so this is the energy done by perturbation theory in the band this is when I simply stop the electron and don't allow this electron to move the energy is minus g square divided by all you immediately notice the denominator is omega here denominator is the bandwidth and remember that one is much bigger than the other so you just compare the two forget about corrections they are very small so you simply compare mine minus w divided by two and here I have g square divided by phonon frequency so this one the right hand side one will win the moment my coupling constant is one okay now we can see everything so that's the simple argument which says if I have polyrons just one polar on at the band at the bottom of the band will decide to undergo a self-localization called self-localization transition at lambda equal one now we have a Fermi liquid you will start building additional energy because you have to place other electrons on top of the first one and the Fermi energy will go even up so which means if I have a many body states with many electrons it will collapse to localize polyrons even earlier than lambda equals one because localized state is exact I will just place localized electron on different sides and I'm done and it's variational state and for each electron I gain energy g square divided by omega so I definitely win if you have more electrons I win more so this will happen even earlier okay well that's roughly how this will look like for one particle it's a very light branch barely normalize but once I reach lambda one there is a very sharp cross so when I go to extremely heavy branch with exponentially large mass because remember my g square divided by omega squared if I use it in terms of lambda lambda one is the transition point and that's the number which is multiplying it's the bandwidth divided by photon frequency which is supposed to be much bigger than one because it's a diabetic limit so you need to have exponentially heavy particles the moment you transition from the band behavior slightly to normalize to strongly normalize now I will repeat the same argument for bipolarons it's very easy so if I have two polarons far away in the band well the combined energy is minus the bandwidth with small correction but if I localize two electrons spin up and spin down on one on the same side I gain four times the previous energy because now density has doubled and the phonon displacement is doubled so the energy gain is the square of this I gain four times okay so now the transition will happen at even smaller value of lambda which means my polaron collapse which I discussed before at the variational level bipolaron collapse will happen even earlier so we have finite density of polarons at lambda roughly one half in the halstain model the state will definitely go somewhere but it's not a familiar liquid type behavior so for lambda equals one half I already know that the variational state of localized bipolarons is already better in energy you can estimate the bipolaron effective mass and again the same story it's exponentially heavy I yeah no no it forms for lambda bigger than one half its variational argument says okay it will form for lambda bigger than one half and there is no dimensionality in this argument it's variational so in 3d I probably definitely need some critical value of lambda to form a bipolaron but here I'm just simply saying it will definitely form just don't even worry it will definitely form for lambda bigger than one half yep I cannot say that I for sure understood the question I'm saying this is variational argument which is saying it's one half but if you do numerics two numerics probably it will be 0.463 but you can do it numerically because you can figure out okay when they bounce state forms I'm saying at the variational level I prove that one half already tells you it's better but the exact solution probably will say it will happen a little bit earlier because variational energy is slightly higher than what we want but it simply says that going to lambda much bigger than one makes no sense for daily arbitrary yep well this is happening because I'm gaining more energy if I localize the two particles together because if I have one particle localized this is g square divided by omega times two because they're staying far away from each other but if I localize them on the same side I gain four times g square divided by omega so placing double the density will give you four times high energy if they are giving them far away this is twice as bad and it's very tight state because I'm gaining and this gaining energy is already competing with the bandwidth so it's a very good tight binding state yeah yeah this is extremely heavy crossover because everything here the change from light to heavy based on the exponential factors it's very very very very narrowed crossover in this region you understand that if I have one polaron so sorry okay because there is actually a theorem which says if I have latest model electron phonon coupling energy has to be analytic function of the coupling function which means it's not a voided crossing because you say okay why I'm saying it's a crossover because exact crossing is not possible there is no symmetry which says okay you cannot make a transition from light to heavy diagrammatically I know what happens lowest of the diagram lowest of the diagram lowest of the diagram and suddenly in this particular parameter range if I just do numerics you'll see the diagrams of 41 are competing with diagrams of 4040 they're fully reducible diagrams of 4040 and if you tune properly you can see that yeah with equal probability your particle wants to be either in the perturbative regime or fully dressed very heavy and it's extremely narrow transition between the two nothing it's a large bit theory lowest of the diagram and you are done and all corrections are small second order even smaller third order even smaller 15th order almost negligible and suddenly fourth 40th order is competing as much better yeah yeah I'll go through this it's a variational state whatever you do better is better in energy you have to understand the power of variational argument it's undeniable and it's killing you because obviously you say okay suppose I have hubbored you I have to increase the energy I'll do it don't worry I'm just this will happen no no because now you say okay the Hamiltonian is different from what we say let's add hubbored you how this will behave this will be actually hurting you because if you have hubbored you you will delay the formation of the bipolar and so when it forms it happens with an even bigger coupling constant because I have to overcome the hubbored you and then it will be even heavier but then you say okay how can I compare with the band theory and then good luck solve me the hubbored model because it's easy for me kind of to argue for bipolarons with you but fully interacting hubbored you many bodies state of electrons nobody actually can do it so actually I'm not progressing anyway no the question is from Andrey it's the same question we already discussed with him but he's doing it for your sake whether I have to distinguish between the bare coupling constant which I define here in terms of bare parameters and fully renormalized parameters at the level of one or two electrons of course I can use bare parameters and there is nothing to renormalize further because it's zero density of electrons if it's a many body state with finite density of electrons sometimes the following is happening in the literature people define the coupling constant not in terms of the bare Hamiltonian but in terms of renormalized phonon frequency so when the softening of the phonon mode happens you can redefine the coupling constant by putting the renormalized phonon frequency in denominator and this renormalized coupling can be made bigger so this is definitely something which you have to watch what people call the effective coupling constant but there is also the question of double counting then because if you allow the electron phonon coupling to change the phonon spectrum the question is how much went to the phonon spectrum in the first place because usually when people form the phonon spectrum screening of coulomb interactions is already taken care of because otherwise for example you don't have acoustic phonons unless you screen coulomb interactions in the crystal you have plasma frequencies so some part of the electron ion interaction is already part of the phonon spectrum so once you start doing diagrams on top with the electron phonon interaction the question is how much has to be added or subtracted for example rpa was included in the phonon spectrum already and that's how we formulate the problem which we tried to solve this is typical for mig-w Albert calculations but then if you try to renormalize the phonon spectrum you have to compute the polarization operator and subtract the rpa expression which is already included in the formation of the phonon spectrum and this is becoming kind of slightly tricky but yes you have to pay attention in the finite density case to the difference between the coupling constant defined in terms of renormalized spectrum relative to the coupling constant which I defined in terms of original Hamiltonian parameters okay I'm talking about original Hamiltonian parameters in my case is zero then scanily probably if I don't finish that's not important as long as the discussion is going I'm happy okay so essentially we have a couple of lessons so electron phonon interaction may lead to formation of bipolarons if it's strong enough perturbation theory fails when the coupling constant in terms of bare Hamiltonian parameters is bigger than one for sure maybe even much earlier but bipolarons can be extremely heavy because remember the moment I form bipolarons it's immediately the rush to claim tc because we are done with many body physics but we have to know the effective mass and if it's exponentially heavy our tc is not going to be high probably when mig-w Albert theory drops to bipolarons tc actually drops to almost zero so that's the danger okay and kind of jumping a little bit ahead so if you do the health strain model the answer is yes that's precisely what will happen so if you do mig-w Albert theory you can reach certain values of tc but once you start forming bipolarons and the firm liquid theory collapses to the dc type of theory tc actually drops because for health strain model the heaviness of the effective mass is the most important to that okay so this is the end of the previous decades of studies of bipolaron mechanism of high tc as far as i see for the model which was always discussed in the literature it simply doesn't exist if you do numerics okay uh i will forget about macmilliform you are now the change of the story was that well okay health strain coupling is not the only one you can imagine happening in materials so let's imagine a different model at this point i'm not a material scientist i'm imagining a model so the model is that well okay the interaction is happening not through the particle density but through the particle current in a very simple sense i am saying that the hopping of the particle between the nearest neighbor sites depends on some form and displacement or atomic displacement happening somewhere on the bottom okay so that's the hematoma you can imagine two things for this to happen for example this was the original thinking of my teachers that well okay the electron can hop from side i to side j and there is some for example couple of atoms which are sitting on the tunneling path well if they move away it's much easier for the electron to to to hop if they are close to each other and closer to the tunneling pass the potential barrier goes up and then the hopping matrix element is smaller so literally saying okay depending on the position of those two atoms the hopping matrix element is some exponential this is the tunneling action which depends on the position of black atoms and then you simply linearize this and write down the model as one plus gx instead of this exponent and that's how you produce this type of coupling we found that this model is probably not the right one to think about this because in this model you can never go to strong coupling face with it when gx is bigger than one then who told you that you can do the Taylor series expansion and now we know this is definitely not possible but there are other cases for example suppose you have the case when electron can tunnel from i to j using two different passes and they almost compensate each other to very small value now if there is another atom which is modifying one of the passes in the superposition you can change the superposition significance that that's how you can still rely on the Taylor series expansion for small displacements but at the same time change the superposition even from positive to negative depending on the position of the atom so that's roughly the picture which we had from one of the collaborators David Richmond who told us okay this is maybe the realistic one so the electron can tunnel but there are different ways you can for example tunnel directly or through some other atom but there is another pass to go say through the intermediate atom and if this atom moves up or down this path is changing its amplitude and if there is another one which is opposite in sign they can compensate each other and then small changes in one can produce relatively large changes in the active hopping okay so that's the other model it's simply kind of a picture justifying how this type of Hamiltonian can be written and taken to the strong coupling case now what are the properties I will be pretty quick and say okay we do Monte Carlo and this particular model is sign free we can do one electron we can do two electrons we can take care of the harbor view we can take care of coom repulsion between those two electrons and we solve it exactly for any coupling constant for the electron funnel coupling why well because expansion in terms of path integral for the particle if tunneling is negative it's a sign positive coupling constant to phonons always comes in pairs so it's just where no sign problem from that and you keep in the path integral you keep all the potential interactions how about you coom interactions in the exponent you're not doing the expansion so this representation is sign free and you can solve it exactly so which means we can tell you all the properties of bipolarons in numerically exactly up to the arrow bars how they form the bound state what are the effective masses and how big they are so the problem is completely solvable because it's sign free so that's the Monte Carlo which we did we can also do it for not only for linear coupling but for non-linear coupling as well and that's the result so it's more or less the competition which I was telling you before so if you have bound states this is binding energy of course it increases as you increase the coupling constant lambda well you see that while this is for phonon frequency three times roughly three times smaller than hotting given then the bandwidth is eight so it's 20 times smaller than the bandwidth so relatively good adiabatic parameter so if you increase hubbard u of course the binding energy goes down so this is binding energy for u equals 0 u equal 8 u equal 12 so binding energy is dropped well then you look at the sizes of course when the binding energy drops the state gets more and more delocalized this is say for lambda equals 0.5 somewhere here in the middle so it's much more delocalized and then you compute effective masses and that's precisely what we see on one hand if i increase the coupling constant bipolarons become much more compact so i can go to higher density but at the same time the effective mass will explode they become extremely heavy so i have to work and find the proper optimum in terms of coupling constant and hubbard u in such a way that i have to make sure that the product of effective mass times the size squared is the best possible place there is one piece which is kind of slightly strange at the beginning that hubbard u is helping as wave superconductivity just if somebody says okay hubbard u is hurting as wave superconductivity just don't believe them a very simple reason i'm giving you the mechanism suppose i have a state somewhere for u equals 0 but it's kind of too compact because it's compact it's very heavy and then tc is small because it's so heavy you'll give me hubbard u hubbard u will make the pair extended it will increase in size but if the bipolaron's state is getting bigger it's much much later so for very simple reason you just start pushing electrons away from each other it's kind of much easier for them kind of to move around and the decrease in the effective mass is the most important effect at this point and you sorry you increase the hubbard u and the transition temperature goes up and you can do it for different adiabatic parameters from formal frequencies i know 0.2 popping i know three times smaller two times smaller depending on different coupling constants you see okay if i fix coupling constant i can tune the hubbard u of course this was kind of for the design the q equal 8 for example will happen to be the optimum simply because it will optimize the size and effective mass of the bipolar mechanism so essentially hubbard u is not hurting you depending on the circumstances it can be actually to your advantage to produce aspects superconductive is relatively high tc because of the hubbard u otherwise it would be too heavy and this is more or less the final plot which summarizes the entire discussion that in this particular model now why this model is different from the holstein one holstein model gains energy locally i need two electrons on the same side to gain the most but hubbard u is on the same side so i am directly competing between phonon energy and hubbard u on the same side so to overcome the hubbard u i have to make a much bigger displacement in the phonon field in the bond polaron i gain energy from the phonon field by moving so electrons can easily move on the placket enjoy energy from the hopping that's how they gain phonon energy and at the same time they weigh each other so i'm much more tolerant to the hubbard u at this point and that's the summary so essentially by optimizing parameters for lambda different value here u was simply fixed to be eight we simply look okay what is the highest possible transition temperature if i do say migdal eliasberg theory this is migdal eliasberg theory when the coupling is of the bond polaron type and the other curve is migdal eliasberg theory when the coupling is the holstein type and you see okay this is curve number one and this is curve number two but they never cross phonon frequency divided by 20 here i am using the bare coupling constant lambda but you may see okay why it goes through the maximum it goes down instead of increasing its square root of lambda that's because i'm plotting tc as a fraction divided by the original phonon frequency omega so formally in terms of renormalized phonon frequency probably tc it goes up but renormalized phonon frequency it drops down so in absolute terms tc goes down okay so going to strong coupling limit in this migdal eliasberg theory we actually see the drop and it's actually a relatively small number if i do the holstein model without any hubbard hubbard u was there it's still kind of more or less at the same limit but typically it's smaller but if we do bond polarons we can increase this number roughly by a factor of five four so which means according to our calculations for bond polarons we are in a much better place in terms of having compact bosons relatively light they are not very light but relatively light so we can reach transition temperatures which are phonon frequency divided by five so it's a much better case to have let's call it non-locally like in phonon coupling it's to your advantage kind of in terms of the bipolar mechanism of supernative okay i'll probably stop at this so bond polarons can be materially compacted light and possibly reach transition temperatures the bifrequency of the phonon frequency divided by five again in some materials with hydrogen modes probably this can be i know even at the scale of 100 kelvin i'm not that optimistic but it's a dream material so formally in terms of engineering what would you want of course you want the higher omega is the better but we also see that typically the maximum we can reach depends on the ratio between omega and t and if this ratio is very very small usually it's extremely hard that's the argument which i started in the first place in the extreme limit of omega much smaller than t effective masses are just exponentially heavy and there is very little you can do about it this will not help you so having this ratio relatively large 0.5 is okay 0.3 is okay the higher the better so which means probably if the phonon frequency is fixed you need the material which looks like a tight binding model with smaller values of t so it can optimize the value of t okay so that's how the whether you have hubbard u you don't have to worry much hubbard u you can tolerate even relatively high hubbard recently we did also calculations how much of all of this will change if i also include long-range coulomb repulsion between the bipolarons in terms of many body physics we already discussed it nothing will change but in terms of the binding states how heavy how compact this will change and the transition temperature does drop by a factor of 2 or 3 if you include relatively strong coulomb repulsion long range obviously because i need much stronger electron phonon coupling to overcome the repulsion coming from the coulomb piece and this makes my bipolarons heavier but still definitely competing with the mcdowler okay yeah that's the question the numerics that you were showing that was for two dimensions right now we have three this one was two dimensions three dimensions have even higher value obviously high ratio but not much so this ratio for example was 0.2 i think in 3d if i don't include coulomb repulsion it can go to 0.25 not a big difference but there you were saying that you need a critical value of lambda yes and still beyond this critical value of lambda well those lambdas are also not small so strictly speaking you can wonder where the bipolarons in 2d always form for very small values of lambda yes but they are outrageously large in size and completely relevant for tc so this is some way here they're too big well so you still go to relatively large values of lambda in 3d you need a critical value once they form you can still kind of have tc on the scale of 0.25 thank you cool sorry about dumb question i i guess there is no under some theorem for bipolaron mechanism so disorder will be okay this is a different story which i didn't show bosons don't care about disorder at all essentially whenever we studied how transition temperature in the boson system interacting with the system and interacting with the system will depend on disorder whatever is happening you go to the your regular limit essentially disorder is such that scattering lengths is interparticle distance or atomic distance tc will barely change by one or two percent you really have to go almost to be localization so that at a single particle level you want to localize everything and beyond that and only then tc will probably start suffering a little bit but bosons are living close to the bottom of your band so it seemed to be very easy to localize well no but then you have disorder which is kind of relatively strong and probably supporting bound states for them yeah no in this sense i agree if they're heavy well probably yes i agree so i i have to remember that while they can be so heavy that even weak potentials can localize them in this sense yes but i have to be close to the localization transition if you let me kind of say something else sometimes you see the following outrageously strong design don't have it for the latest so we we did calculations for interacting bosons with very strong disorder and try to understand okay how you modify properties of the super fluid the following will happen you have a very strong potential you say oh i will localize the boson good for you it goes there the second one will go to the potential well where this previous boson is sitting but they repulsive it's more shallow the bosons will simply feel your potential and the moment you have enough bosons if you have more impurities than bosons yes what can i do you didn't even dope my system sorry but if the system is doped in the sense that you'll okay you have less impurities than the number of bipolarons they will not care about your strong disorder at all they will fill out all the bound states to the point when the chemical potential is getting delocalized kind of above zero and they will go super good they just outrageously good in screening in this sense yes there is no boson that is only positive but this is a theorem for example there is no boson metal because if you have say conventional dispersion relation conventional repulsive interactions it's a sign positive representation for the density matrix which means whatever you compute your density matrix is always positive for if it has no sign consolation and then you can show that well the metal is not possible because you have always positive contributions to super fluid density yeah so it's it goes super fluid outrageously well unless your impurities will simply kill all the bottom yeah so when you talk about this bond hostile model uh in the presence of repulsive interaction either Hubbard or colon interaction yeah did you look for pairing in non-swave channel uh we didn't so we didn't for probably very simple reason because uh singlet for first we are not suffering from hubbard u because we can avoid two particles sitting on the same side because we gain energy by hopping in the high in the hudstain model you're right this was done before even before us in the hudstain model if you have very strong hubbard u you will trigger the transition that now triplet by polaron is better than the singlet one because you kill singlet because uh the repulsion is uh kind of give you high energy for the singlet then you have it for the triplet because triplet will avoid hubbard u anyway so yeah that's the case so for the hudstain model with local coupling to phonons yes triplet can be involved with the singlet for sufficiently large u so this is happening but again it's triplet bc so somewhat related question so if i add magnetic field now to this model that you've shown the results for right uh so that to go effectively imagined it will be polarized leave me so what did you do so i don't know but i mean no no my intuition would be since i well some some of the calculations were done for u equals infinity so given this for u equal infinity i can still have bound states and they're still relatively light and everything i would say that probably i will not suffer that much if you polarize everything if you polarize everything okay i'll have a triplet state but they can avoid each other anyway when u is infinity so i would say probably some of the numbers will survive but we did not didn't do the calculation so i would not say it will be the same so maybe the answers will change by effect of two or three i think yes because infinite u is still doable and it's still roughly the same properties well you have seen some of the pictures they were u equal 12 not not a big deal question from andrey chubukov is the difference with eliashberg theory the same for u equal to zero i'm not sure i understand because there is no eliashberg theory for non-zero u there is only eliashberg theory for u equals zero when you don't care about electron-electron interactions direct i'm not sure i understand the question because this migdal eliashberg theory is written it was done for u equal zero so probably that's how i have to interpret in this plot u equals eight was done for bipolar mechanisms that's when u equal eight was included when we did migdal eliashberg theory it means we didn't care about you at all it was u equals zero so which means for very large hybrid repulsion we have s-wave superconductive is much higher temperature than you have for u equals zero in migdal eliashberg u equals eight is only for bipolar on mechanisms for migdal eliashberg theory there is no u it's u equals zero so about phase separation you said that you're not going to mention it but in the hosting case there is a tendency towards phase separation at a large couple yeah it's possible we did not discuss it yeah is but is there a reason why that should be less of an issue in the uh definitely not yeah definitely not an issue if we have a long range coolant repulsion because only the micro structure quality crystal is possible and yes our numbers are still kind of relatively good maybe not as good as i was showing but still good enough compared to migdal eliashberg if you have long range coolant repulsion then phase separation is completely low whether you have phase separation for some of the coupling parameters which we started i cannot answer because we didn't solve kind of this type of say four bipolar on two bipolarons or four bipolarons just to see whether they want to form a bigger cluster we didn't do this calculation the calculation was done at the bipolaron level so we cannot answer this question so there is a possibility but in the coolant case it's rolled out so if you want to talk about realistic materials probably no any more questions so let's thank you for that we have uh according to the yeah um is there a discussion session or we just keep it we just had it but if there are any questions to max or to paola that urgent right now not then then let's