 screen. Hello to everybody. So let's continue our morning session. And sorry, Natasha, sorry interrupt, but this display is, see, I don't even see my mouse. Then I minimize this. Then you should not minimize it. Go there and go again and hide loading, meeting controls. Okay, great. Okay, with that, we continue. And the next talk is by Shira Wu from Stanford and he will tell us about pair density wave order from electron repulsion. You have 40 minutes and thank you. Great, thanks. All right, thank you for the invitation. It's always great to be back and see old friends. And this work is done in collaboration with three very bright young scientists, in particular, Yiming Wu, who's a postdoc who's about to start at Stanford. And we should have a paper out reasonably soon. So let me start by reviewing some familiar notions. What I'm going to be discussing in my talk is a rare and exotic type of superconductivity that's related to something that was studied long ago, known as FFLO superconductivity. And remember, in a metal with inversion symmetry or time reversal symmetry, pairing in the clean limit involves antipodal points on the Fermi surface between k and minus k. And that's the ordinary superconducting state. And we can solve the linearized gap equations that understand basically everything. By contrast, the FFLO state involves pairing knots between k and minus k, but there is some net momentum, which I call q. That's the momentum of the center of mass of the Cooper pairs. And you can see the difference in the linearized gap equation that it involves fermions at k and minus k plus q. Okay, so there's two types of such pairing. There's the FF state in which delta is a plane wave. And since all observables must be gauge invariant, this state remains translationally invariant. It's just a phase. By contrast, the state of Larkin and of Chinikov has a modulated density, where the order parameter goes like cosine. So I'm going to be focusing on the Larkin of Chinikov state in my talk. So this state breaks, of course, the U1, as superconductors always do, and it also breaks translational symmetry. But it can also break time reversal symmetry, as you see in this picture here, where you imagine Zeeman splitting Fermi surface, so that breaks time reversal symmetry. And then you have, on top of that, this modulated superconducting state. So a related state to the Larkin of Chinikov state, so often Larkin of Chinikov state is accompanied by a uniform superconducting component. If there were no uniform component, this Larkin of Chinikov state has been named the pair density wave order. Yes, Pierce, you're right. You can think of the Larkin of Chinikov state as two FF states, e to the iqr and e to the minus iqr, and then you get a cosine out of this. Andre? So the question was, yes, the Larkin of Chinikov state also has no q equals zero, but in terms of mechanism, it usually came from taking a uniform superconductor, polarizing it, and then having both uniform and non-uniform components. Anyway, these are nomenclature, so we now call these things pair density waves. Okay, the beautiful thing is that not only are these rare and exotic, they exist in nature, as shown here in this very important result that hasn't been published yet in the group of Abe Pashupati. This is a europium-based iron superconductor, and you can see below transition temperature, you can look with your own eyes directly. If you see the gap as a function of position, it's oscillating in space. And from that, if you pour a transform, you can associate a brag-like wave vector associated with the superconducting state. And finally, if you look at the coherence peaks as a function of position, they have the spatially modulated form, which is like the Larkin of Chinikov, or pair density wave state. Other promising candidates include these recently studied and actively studied Kagome metals, cesium, vanadium, and antimony, perhaps UTE2. And of course, it's been invoked as being relevant in the cuprates. I don't know if that's true, I think that still remains controversial. Okay, so we're interested in the pairing mechanism of such states. So let me remind you that if you look at the pair susceptibility, which is defined here, B is a pair of fermions. If I look at the, in a translationally invariant system, it only depends on I minus J. If I look at the static piece and Fourier transform as a function of Q, Q would be the center of mass momentum of the pairs. It has the following feature that it's, first of all, a positive definite function. That's important. That'll be invoked in my talk. It's peaked at Q equals zero. The peak at Q equals zero is the celebrated logarithmic divergence of BCS. And shown here in the limit as Q goes to zero, it's just the density of states times some constants involving the cutoff and the temperature. So because the susceptibility is biggest at Q equals zero, that's why uniform superconductivity always wins. And if you want to have a mechanism for Larkin of Chinico of like pairing, you want to really build on this already large susceptibility at Q equals zero and maybe try to perturb the metal by, say, a Zeeman field if we're talking about singlet pairing. And so you usually have schematic phase diagrams like this where, oops, okay, there we go. Still works. Where you have, for a large part of the phase diagram, a uniform superconductor and just a sliver, say, if you apply a magnetic field in the regime right near the upper critical field, you might see an FFLO state. But what I want to do is ask, so this I would call an extrinsic mechanism. It relies on some other order that's about to condense. And this was the strategy I think also of Larkin and of Chinico. But what I want to ask is, is there a more intrinsic mechanism where you get naturally of spatially modulated superconductivity without the need to invoke a uniform component? Okay, and for that, because the susceptibility is not logarithmically divergent at Q away from zero, we need to go away from this tyranny of weak coupling. We need to get away from weak coupling. So in my talk, I have good news. What we have done is that we found robust low energy effective theories hosting the pair density wave order for a variety of lattice continuum systems in D greater than one. But there's always bad news. The bad news is that the low energy theory requires some special circumstances in order to emerge from microscopic models, solid with the say the Hubbard model. But I'm an optimistic person. I convert bad news into good news. The fact that it's so rare, this requires special circumstances, I think explains in part why these states are not found everywhere. It's a rare phase of matter. Okay, so what we study is a Fermi liquid, Fermi liquid with some dispersion, and an interaction, which is a BCS like interaction. And when this is positive, that's a repulsive BCS interaction, I'm going to study repulsive interactions. In Andre's talk from the other day, Andre's also studied repulsive interactions at weak coupling. He showed quite clearly that you have unconventional superconducting states, which are uniform states. But what I'm going to do is go away from the weak coupling limit, where we're going to look for non uniform states. How to solve, we don't have a small parameter. So we're going to we're going to solve using McDowell-Eliashberg theory. Okay, so that does not require being pinned to weak coupling. Yes. Because the logarithmic, oh, sorry. Yes. The question is, why do we need strong coupling? You know, strong coupling, I just mean, you don't, you cannot find it at weak coupling because the logarithmic physics that gives rise to your usual superconductivity is coming from Q equals zero. Away from Q equals zero, the log divergence of the pair susceptibility is cut off. So then you need some finite strength. That's not strong coupling, but it's intermediate coupling. Not, you know, weak or strong is words, but it's not a marginal instability of a Fermi liquid. Nesting is not relevant in this talk. Okay, good. So we're going to solve by McDowell-Eliashberg theory. And what I mean is this, we have a set of solutions for exact fermion propagator, inverse propagators and exact Boson propagator. So where's the Boson? The Boson will come from the BCS interactions by Hubbard-Stratanovich identities. From this, we will introduce a Boson, which is actually a collective mode of the fermion itself. It's an intrinsic property of the fermions. So that defines a bare Boson propagator. And we find the exact propagator for the Boson. And we find the exact inverse propagator for the fermion. And this is how we solve everything. And from the knowledge of the propagators, we extract the physics. So if we want to do thermodynamics, we look at the static pieces of the propagator. If we want to do dynamics, we look at the full propagators, et cetera. Notice for the experts, this is McDowell-Eliashberg theory in the particle channel, as you can see by the arrows. So we're not doing McDowell-Eliashberg theory in the particle whole channel, the way he would do, say, for the electron phonon problem or other such cases. We're going to start with a BCS interaction and then solve it beyond weak coupling using McDowell-Eliashberg. Yes, Andre. We will find self-consistent solutions. Yeah. It doesn't have to be at one loop. Oh, the question Andre asks is, is this just self-consistent one loop? In principle, these are fully, pardon? Okay. So maybe this is going to answer your question. If you wish to study a model in which these equations are the saddle point, take the large n action, which is a form of a bosonic piece, a fermionic piece, a bosonic piece, and a coupling. And what I'm going to do is take a large number of flavors. You've heard lots of talks on this this week. We have a large number of fermion flavors, number of boson flavors, and a coupling, which is in this random Yukawa form. And notice that this is a coupling in the particle particle channel. So the saddle point is what we solve. And if you say that's one loop self-consistent, then yes, I agree. Okay. Yeah, vertex corrections. They are n suppressed. Yeah, in the large n. At n equals one, we don't compute vertex corrections. The thing that I will allow for is the possibility of having either attractive or repulsive BCS interactions. So there's a piece that involves Psy dagger Psy dagger. There's a piece that involves Psy Psy, which out of laziness, I didn't write, but it has the same form. And the key is that if it's repulsive, there's this coupling is imaginary. If it's attractive, it's real. So we're going to study the repulsive cases. Okay. So the main point of my talk is that repulsive BCS couplings can lead to this pair density wave order. You get long range order when the repulsive forces are actually peaked at finite distance. This is kind of reminiscent of what Andrei was talking about in the frequency domain. But this is as a function of possession. We've solved models with long range order. And then I'm going to speculate on the relevance of this to reality. And we're going to see a phase diagram like this. There's a Fermi liquid, quantum critical point, a long range pair density wave order. There's three regimes in which we solve. In regime one, we just need to solve for the static. I told you the game is to solve for the exact propagators. Okay. For regime one, when we look at the finite temperature transitions, we only need the static propagators as a function of temperature and momentum, and the solution of the linearized gap equations. In region two, when we go into the ordered phase, we need again the static propagators. We can, of course, ignore the fermions. We can integrate out the fermions and look at only the bosonic degrees of freedom in sort of Ginsburg-Landau like theory. But in region two, we need to solve for nonlinearities as the order parameter develops below TC. And finally, in the quantum critical region, we cannot integrate out the fermions. We need to keep the fermions and the bosons and look for self consistent solutions of the dynamical propagators. It'd be a fair question to ask if you uncover your pair or your interaction purely in the pairing channel to tell us what would it presage for any of the normal state response functions as a function of the field. The only response function that enters is the pair susceptibility. So the particle hole. You have not asked the other thing. Correct. Correct. Good. Yeah. Yeah. Okay. Yes. In principle, I know all the physics, but I think what my interpretation of Chandrase question was, do I include interactions in the particle hole channel and look at the renormalization of those things? I'd just like to know what your model implies for normal state response function. Well, in principle, you can calculate anything. Yeah. As Andrei says, from the propagator. I can tell you the answer. We can, yes, we have thought about it. And I will show you some expressions, especially near the quantum critical point where this is important. But away from this, it's not so important. Okay. First, if an equal one is not large and what would instabilities could happen in strong coupling that will destabilize the pair density wave or that's hard to know? I don't know. I cannot speculate. I don't want to speculate. I will not speculate. Okay. So first let me tell you why the most reasonable thing you would do, short range BCS interactions fail, fail for the program. And so what I call that is, as again, an optimistic person. It's a fluctuating, but actually failed pair density is the correct approach. So as I said, a form of the bare interactions. Was there another question? The bare interactions that I wish to study determines, sorry, the interactions I wish to study determines that defines the bare Bosonic propagator. Okay. So for instance, if I were studying an exponential form of the VIJ, in Fourier space, the bare propagator has this Ornstein-Zernike form, which is just a constant plus Q squared. And then if I solve McDowell-Leashberg theory, I won't tell you the method. I mean, I won't bore you with details. I'll just tell you the answer. The fully dressed Bosonic propagator has this Ornstein form plus a self-energy coming from the pair field susceptibility at the static because I'm only looking at finite temperature response at this point. And what you need to know is that for small Q, the logarithmic divergence of the pair susceptibility is cut off by omega divide over B Fermi Q. So if you put this together, it's a positive number. This is a positive number. That's a positive number. Everything grows positively. But what you do find is that the solution of the inverse propagator has a minimum at finite Q. It's the minimum has gone away from Q equals zero because of this logarithmic piece of the pair susceptibility, because we're talking about repulsive interactions. Now the Q equals zero piece gets weaker and weaker it doesn't grow as it would with attractive interactions. But what you find is that it's peaked at finite Q. So that's encouraging. That means it would condense if we could just push this down. So for those of you who don't have intuition to think about inverse propagators, let me just say if I integrate out the fermions, the exact propagator, inverse propagator would be the coefficient of the quadratic term of Ginsburg-Landau theory. And we know that that has to vanish if we have a phase transition. And clearly it's positive. That means we're above the phase transition. Even if you go, if you recompute this even at t equals zero, it will always remain positive because, as I said, the pair susceptibility is a positive function. So we fail to capture the pair density wave order in this way. Okay. Yeah. Yeah. Everything else is just, so Andre says, what if you go to a more accurate order in Q? Yes. Yes. Yes. Oh, the speed of the speed. Yes. I see. So do you get the, so the question is, do you get this Q squared term coming from the pair susceptibility or do you have to put it in by hand? I think you have to put it in by hand. Yes. I don't think it comes. Yes. Okay. So now we want to do something to this model to obtain long range order. And the strategy is to do this. Instead of choosing a, what physics would say, you should choose something that, a BCS repulsion that decreases with position of pairs of tunneling. If you had, however, for whatever strange reason, a non-monotonic BCS interaction, no problem. And intuitively, it's simple why this is the case, because if I Fourier transform this, I now get already from this model attraction at finite Q, right? If I Fourier transform this, there's also already attraction at finite Q. So, but it's still not an instability of the, of the Fermi liquid. We need to increase the interactions beyond some threshold. And then if we do, we will obtain a condensation at finite Q. It's very simple. And when it condenses, we will have long range order. So to make things even simpler, because I like models that I can solve exactly and analytically, consider a case where the pair hopping is a delta function. So it's zero. And then it's some number, it's a delta function at, at r equals r zero, and then zero everywhere else. So it's a spherical shell, spherical cow, spherical shell approximation. This Fourier transform, of course, is probably, as you all know, is a Bessel function. And it has this oscillating pieces. And then when you solve this in the large N, so this defines some bare propagator, and we find the exact propagator. What you find is that the exact propagator is this Bessel function, but plus a piece that depends on the pair susceptibility. But now the important thing is that the sign of the interaction matters. So in the, in the regions where it's repulsive in momentum space, the inverse propagator grows. That means the Landau-Ginsberg coefficient gets bigger. But in the regions where it's attractive, it weakens. And so the Landau-Ginsberg coefficient gets smaller, so we're approaching a critical point in the sense. So if you then solve this model, this is the inverse, this is the solutions for, as you vary the strength of the interaction. So what I'm plotting here is the inverse propagator as a function of q times r zero. You can see that at different temperatures near q equals zero, first of all, as you lower the temperature, the inverse propagator grows. That means the coefficient of the q equals zero piece is increasing. You're getting further away from condensing the q equals zero boson. But by contrast, at finite q, where the oscillations in momentum space leads to attraction, you get closer to condensation as you lower the temperature. But you need to increase the coupling. It's not an instability. Beyond some threshold effect, you will get pairing at finite momentum. And it's trivial in some sense. Inverse exact static bosonic greens function. Yes, exactly. So we can proceed in this way and find at different temperatures, where's the temperature dependence coming from? The temperature dependence comes from the pair susceptibility. It has some temperature dependence. And so we can obtain a phase diagram by looking at the solutions of the static propagators at different temperatures. And we have a phase boundary between the normal state and the pair density wave state. And the dashed line corresponds to regimes where we cannot compute because at low temperatures, this is a numerical solution of the self-consistent equations. And at low temperatures, you have to keep lots of lots of months of borrow frequency. There is a question. Yeah. From Yuri Gehrin. Recently, it was shown by Yusbashyan and Ben Shuler in PRB that McDowell-Iliashburg theory loses validity at finite value of lambda critical between 3 and 3.7 of the coupling, regardless of the underlying model of Newtonian. How it is consistent with your approach? This is particle channel. I don't think they analyze this. And anyway, what, if you wish, take the story that this is the solutions in some large N theory. And that's true for all coupling. Probably I didn't answer the question to that person's satisfaction, but let me move on. Okay. So then we can look at the temperature dependence below the phase transition in the pair density wave order regime. And we then have to solve the nonlinear equations and look at self-consistency with the inclusion of the expectation value at finite q of the boson. And what we see again is indication that the transition is continuous. The order parameter develops smoothly. Again, we have finite size effects. So I don't, I don't, I draw that with the dashed line. I want to show you what we actually compute. We can also determine the ordering wave vector of the pair density wave by looking at the location of in momentum space where the inverse propagator is minimum is a minimum. Actually, it has to be zero for the critical point. And also at the, at the zero, it has to have a, it has to have a minimum as you approach from the normal state because the propagators must be positive for stability of the theory for the Glendale-Ginsberg free energy. And so you can determine the ordering wave vector as a function of the coupling constant, the dimensionless coupling constant G squared. And what I want to say is that it changes with G squared. So you might have thought that, you know, in this, in this problem that I'm solving, the only scale is k Fermi. And so you might have thought that the pair density wave is going to be at 2kf. But I want to stress that this is not true. It's sensitive to the interactions. And so 2kf is not a characteristic scale in this problem. Also near this point of ordering, the exact propagator can be, can the static propagator has this parabolic form as a function of momentum q minus capital Q squared, where capital Q is the ordering wave vector. And this is just obtained by looking at the second derivative of the inverse propagator. So we can do all of these things. Now, everything I told you was for this direct delta function with this Bessel function problem, we can also now go back and, you know, as you all know from electromagnetism, you can superpose delta functions and get a realistic potential. So when you do that, you can study a potential like this, which is again, non-monotonic as a function of position. And you get the qualitatively same. I don't want to repeat the whole exercise, but you get similar results. Yes. Yeah, correct. Yeah, the question is, in the model in which there's only one length scale, r0, how is the q related to r0? So you can see that q from this, from this answer here, q r0 is characteristic, q, the characteristic value of q is one over r0, as you said, but it changes with g squared, changes in a modest way. Okay, so we find similar things for a non-monotonic BCS coupling, but also with armed with this intuition, we can go and study in this McDowell-Eliashberg theory, lattice models, say where you have a Hubbard model on a square lattice with nearest neighbor hopping, and with nearest neighbor repulsive pair hopping. Okay, now what we find is again, in this McDowell solution, where we obtain propagators, when u is less than four times the nearest neighbor pair hopping, and when it's bigger than twice the hopping, you get a pair density wave solutions again. And now the ordering wave vector of this pair density wave is close to pi pi, but it depends on the ratio of rule. And we have not looked at this in full detail, but we know that it's not exactly pi pi. Okay, how am I doing with time? I think I would be done fairly soon, actually. Okay, perfect. If there are more questions during the talk, feel free to raise them because I'm wrapping up soon. Okay, yeah, please. You mean at zero temperature? Yeah, so, okay, thank you. The question was, can I tell you a bit more about the phase transition between normal state and pair density wave state at zero temperature or finite temperature? I haven't gotten to the zero temperature part yet. I'm about to do that. But the finite temperature, it's completely conventional, continuous transition like a Ginsburg-Landau mean field like transition would predict. Okay, so you have some coefficient of a quadratic term, it's at finite momentum, and that coefficient vanishes. And then there's a quadratic term, which is required for stability. Yeah, yeah, Andre. So the question was, is it a generic statement that when you have this non-monotonic BCS repulsion that the system, can you explain? I mean, the logarithm is only at zero momentum. Yeah. I don't want to take time for me to talk about zero temperature, but you understand the question. Yes, yes. And so the question is, is it really a extremely generic statement of the same level as con-lattenger did with non-S wave? I think it's, I think it still requires that you need to lower the Q equals zero piece, sorry, the short distance piece relative to the finite distance piece. And I don't think that's always generically guaranteed. It requires special circumstances at the bare level. Yeah, other questions? All right, so now I'm going to go to the zero temperature problem where I go from Fermi liquid to finite Q superconductivity across a quantum critical point. And again, let me solve this delta function potential. The somewhat curious, maybe not important, but curious aspect of this model is that the critical phenomena of this problem occurs on a ring in momentum space, as opposed to bosons usually condensed at points in momentum space. But this case, if you look at the solutions, the contours of the exact inverse propagator, the static limit, it has this condensation, the zero occurs at a finite momentum radius. And it's because this is a rotationally invariant problem. There's no characteristic direction in momentum space. So it must be a ring. So if you study this, it turns out that the Fermi surface, you can take a point on the Fermi surface, scatter with any of these momentum, and go to another point in the Fermi surface. So the entire Fermi surface is a hot region. Think of the anti-ferromagnetic critical point where you have hot spots, but now we have hot regions. Okay? If we were on the lattice by contrast, you would not have this feature. The condensation would occur at discrete momentum, maybe related by some rotational symmetry, 90 degree rotations. And then you would see the problem of Fermi surface hot spots. Okay, now I know some of you must be seeing this ring of condensation or critical phenomena, and you must be thinking, what about fluctuation driven first order transitions a la Brasovsky? Well, the beauty of the large n limit, the McDowell-Aliashberg theory, is that it suppresses all of that, all of these important physical effects. And so the transition remains continuous. Yeah, yeah. Something else. So on this picture, where is the Fermi wave vector? Oh, it can be something completely different. It's here, as was asked, Q depends on R0, the range. So it can be different than KF, arbitrarily different. Physics does not depend on the ratio between capital Q and KF. Correct. Well, Q has to be small compared to KF, in the regime where we're computing. But beyond that, there's no dependence. Usually, the condensation of those photons happens into some even single quantum mechanical state. So can we associate the single wave function with this ring? Because it seems like there are many states characterizing this. But yeah, but remember if what happens is that, okay, so, okay, the question was, what was the question? I was going to interpret it in my own way. Unfortunately, it was a single, single, single particle wave function where all photons condense. Yeah. So now you have a ring which seems to consist of many single particle wave functions. Yes, yes. So that, can you associate to the condensate a single wave function out of this ring? Yes. Or how does it? So the question is, can you associate with the condensate a single wave function or a linear combination of infinite number? The answer is the critical phenomenon occurs on a ring. That doesn't mean the condensation occurs on the ring. So because we have, in Ginsburg-Landau theory, there will be an important terms. And that will choose among this degenerate manifold some discrete subset. And then you'll be safe. Five minutes. Perfect. Okay. So we can solve, again, in this MIGDAL approach, the propagators at the quantum critical point. Now we cannot get rid of the fermions as I've been doing before to look at just the bosonic problem. We have to keep the fermions at play. And then if, in the limit where the ordering vector q is small compared to k-fermie, we can obtain analytically the bosonic and fermionic propagators. And the bosonic propagator takes this form here. I've defined gamma for you. It's the second derivative of the inverse propagator at capital Q. And then there's the Landau-Damping form, which looks similar to, you can calculate this quite easily. It looks quite similar to the Landau-Damped form of the anti-ferromagnetic order parameter with the denominator capital Q. Then the inverse fermion green function has a self-energy which can also be computed. It has this omega, mod omega to the power one-half form. And the thing is this is a self-consistent result, because if you go back and compute, recompute the bosonic propagator, the Landau-Damping does not change once you include this self-energy. And then you can feed back and everything is self-consistent. Okay. In the lattice model, this form of self-energy would still apply, but only at the hot spots. So this is very similar to the anti-ferromagnetic quantum critical point. And so we can understand the quantum critical. So that's the second part of this transition. Now it turns out we're in two dimensions, and the dynamical exponent determined by the bosonic Landau-Damping is two. So what you have here is a problem where you're near at the upper critical dimension. And so the critical exponents up to logarithmic corrections to scaling laws will be mean field exponents. So we know that z is two, and we know that nu is a half for mean field theory. And so the line of thermal transitions terminating at the quantum critical point comes in with slope one, up to logarithmic corrections, because nu z is one. Okay, I'm done with the analysis. Now let me speculate. Okay, based on what I've done. Okay, so now everything I say, everything I've said thus far is not controversial, but now everything I will say is controversial. Well, the first thing is not controversial. It's that I've shown you that unique, if you have finite repulsive BCS couplings, which are non-monotonic, you get pair density rate. Great. Unfortunately, this is not common in solids. Okay. Usually, more over BCS interactions, hopping of pairs on a lattice are microscopically known to be small. Okay, but that's microscopic physics. At the level of a low energy effective theory, anything's possible. For instance, let me speculate where you can get this non-monotonic behavior. Suppose I have screen coulomb interaction and strong electron phonon coupling, local electron phonon coupling. As you know, in the strong coupling limit, you get an instantaneous non-retarded attraction mediated by the phonon. If that short distance attraction can be sizable, you can be in a regime where this effective interaction is non-monotonic. I think this might explain in part some recent DMRG simulations done by the group of Steve Kibelson and his students, where they found pair density wave order in some Hubbard-Holtstein models at strong coupling. Maybe this physics is somewhat reminiscent of what they found. The other thing I want to say is that I alluded to the fact that pair density wave order has been observed in these Kagome metals, the CC of vanadium antimony. Now, near the van Hoog filling is where they're seen. My friend, Ronnie Tomali, made an important observation a few years ago that when you look at fermions near the van Hoog singularity, you have a hexagonal Brie-Wan zone, and you can approximate the fermions dominantly coming from points, patches, near the van Hoog singularity. They have this property known as what he called sub-lattice interference that the wave functions all belong to one out of the three members of the real space sub-lattice of the Kagome lattice. What that means is that if you look at the states here, they all belong to the A sub-lattice, B sub-lattice, and C sub-lattice. Now, what we want to do is we want to look at, in this patch-like theory, the renormalization starting with sub-Hobbard interaction, the renormalization of the BCS repulsion. Maybe, as Andre mentioned in many examples in his talk, the inter-patch BCS coupling can be bigger than the intra-patch, and then you would be exactly in this regime where the pair density wave order can form. I think I should stop here. This is the summary in the form of a picture, and thank you for all the questions and your attention. Thank you very much for the talk. More questions? So for the pair density wave, do you need non-monotroponic interaction or just any interaction that gives a negative, when you Fourier transform to momentum space, a negative component? The latter. So a real-world question about Kagome's. If you're going to have the PDW physics that people are talking about in their Kagome's, does that come along with a fundamentally unconventional order parameter that should, for instance, be disorder-sensitive? Not necessarily. It can come along with a uniform S-wave component. So here, everything I was doing was a sort of, the pairs consisted of fermions on the same site. So it would be, well, I mean, the pair density wave, of course, is very sensitive to disorder, but if it were accompanied with a uniform S-wave component, there's no problem. Perhaps I could ask you to push your speculation even further, please. There are a number of experiments, particularly, for example, Shamis Davis, where you see pair density waves with their standing tunneling microscopy, but they also see superconductivity. Can you comment on whether the pair density wave is just a spectator or an active participant there? I told you I was going to push you. Probably a spectator, I would say. I mean, it's this notion that it's the mother of all facts I don't necessarily subscribe to. So it might be a small fraction of something that's on top of a uniform component. That's my speculation. Of course, I could be wrong, so that's what I think. And I guess, following up on my previous questions since I have the microphone, when I asked you about the transition between Fermi-Liquid to pair density wave, I guess my real question was, what would be the observable characteristic of that transition? Observable characteristic of the Fermi-Liquid to pair density wave transition. Finite temperature or zero? Yes, finite temperature. Well, so you would see, if you do X-ray scattering, you would see a bragg peak that would grow at below Tc, and the bragg peak, the bragg gets sharper as you lower the temperature. Just like a charge density wave order. But how would you distinguish it from the charge density? Because of the resistance. The resistance is zero. Okay, so you have to do transport as well as... The pairs have a translational, so the question Chandra said was pairs don't couple to X-rays, but with every pair at finite Q, there's a accompanying charge density wave component, which is always present, just by symmetry, twice the Q. Yes, so you would see the ordering, you would see the bragg peaks twice the Q. Particular as the last example, but it's one of the four points. In your case, and in the case that many citations leads to non-S-wave pairing from initially repulsive interaction, you start with initially repulsion and then dress up pairing interaction. In your case, you basically, you add corrections in a particle-particle channel. But you have Q-square, as you said, coming from somewhere. On the other hand, if you add randomization in a particle-hole channel, you get, you did it by yourself, you get effective attraction in non-S-wave channels. At Q equals zero. At Q equal to zero, yes. So question is, there are two competitors, right? Yes. So it's just one possibility or the other. Yes, yes. If you don't include this Q-squared piece, you could, but you know, in the large end limit, this cone-luttinger, all these diagrams are one over N suppressed. You made it, you made it. I did it by hand, but of course, as N goes to one, what I envision is that what I call Fermi liquid in the left side of the phase diagram here will maybe be some unconventional superconducting state. And they might be a phase transition from one Q equals zero superconducting state to a Q finite superconducting state. And maybe that's first order, because that's what Landau would tell us. But that's the speculation. I don't see any other questions. Excuse me. The fourth floor would like to ask some questions. I've had my hand up for 10 minutes. Sorry. I didn't see that. Okay. A number of questions. First of all, what about triplet pair density waves? Did you think about that? Yeah. Yeah. I did think about that. And quite similar things can be arranged in the triplet case. And actually, there might be regimes where the pair hopping can be somewhat more natural than the ones I used, because they were onsite Cooper pairs. If you have bond Cooper pairs, as you must in the triplet channel, then similar mechanisms can play. And follow up. There's some overlap between these ideas and those of Amperian pairing. And did you look at any, did you examine that possibility, the possibility that that there might be models that stabilize pairing at 2KF? This would be interesting because it would somehow mix particles and holes with the same velocity rather than opposite velocities. Did you think about that? Yes. I thought a lot about this. Amperian pairing is an appealing scenario, but I don't think it was found in any calculational framework. So basically, I did think about this. And we never found Amperian pairing. This is, I mean, it's the right country. It's France, but it's Coulomb, not Ampere. That's what I would say. Second of all, if you do study more exotic metals like Fermi surface coupled to some U1 gauge field, then you can have a situation in which there's a repulsive BCS coupling, like the ones I'm discussing here. But then the system has an IR fixed point, which is a non-Fermi liquid. And then you can have Amperian pairing solutions. But again, we never could find them at least in the way we were doing these calculations. Thanks very much. Sure. It would be great if they did occur though. I agree with you. Yes. So you did all this without bringing in any effect of nesting, but if I understand correctly, then the Q, the capital Q that you have corresponds to a very large distance over which the interaction is modulated. Could you possibly be able to derive that from incommensurate charge density or some nesting phenomenon? I don't think so. I mean, I think this is not a weak coupling phenomena where I have to couple points exactly on the Fermi surface. Once you go away from weak coupling, you can have some window about the Fermi surface coupling to another window. So it's not tied to any nesting effects. And that's what we find in this model as well. Did I correctly understand that you have a Yukawa coupling between bosonic modes? Yes, between a boson and a fermion. So how much does it depend on all these results on this averaging? If you would compute just for a simple G tensor, I would try to look into that. What do you mean by simple G tensor? That it is coupling between fermions and bosonic modes. If you don't randomize on them, but you choose a simple form for... Well, then, okay, so you could do that, but then you're not solving exactly McDowell-Eliashberg equations. So I would say I only invoke this because I want to, for those who insist on having a model, this is the model whose saddle point is McDowell-Eliashberg. But let me just say, I'm thinking about McDowell-Eliashberg. Yeah, just I wonder, I mean, for real material in the end, it will be kind of something fixed. Yeah, so there are issues with this random Yukawa model. It's very multi-critical, and there's other subtleties that I didn't want to get across here. So let me just say that for the finite temperature transitions, this model is perfectly adequate. And so since the physics is robust, you could probably do other large n expansions and find similar behavior. But the quantum critical point is subtle. There's... I think Andrei has it. Are we done? Of course. I want to make a very small comment. You used to work McDowell-Eliashberg many times today. Yes, yes. You are at finite temperature. You deal with static poles. Yes. McDowell-Eliashberg is a... But a quantum critical point, you... No, no, no. McDowell-Eliashberg has the idea that you have fast electron and slow boson. Here you have a static boson. So it's definitely not McDowell-Eliashberg in this respect. So I would prefer... I suggest to call it self-consistent one loop or large n something, but not something that has a concept of one excitation slower than the other. Because you... at finite temperature, zero temperature is a different story. Yes. But at finite temperature, since you deal with static boson, then there is no such thing as comparing of those... Well, at finite temperature, there's also a... There can be smooth analytic dynamical pieces that don't matter. Right. They don't matter. This is what I mean. Yeah, they don't matter. In this respect, it's not McDowell-Eliashberg because it doesn't use this concept of frequency. Yes, of one excitation much faster than the other. Yeah, yeah. I agree. That's a good point. Thank you. Thank you. So let's thanks again. And lunchtime, so we recommend it to talk. Do I just turn it off?