 Here, call the boss. Okay, okay, okay, okay, let's continue. So it's a second lecture, and it's a blackboard, so really come closer, and here is what I will try to do. The story I will be talking about is against superconductivity, but now I will explore in more detail this feature that I discussed in the first lecture. Namely, that even if you start with repulsive interaction, then there is a way to convert repulsion into attraction, and the way how you do it, essentially, if we go back, say, to quadratic lattice and cooperate story. You want to increase interaction between fermions in one corner and fermions in the other corner. And more or less know what it means. It means that the system wants to develop something at large momentum transfer, and either looking at the phase diagram of the cooperator, just looking into this diagram, we know that this is magnetism. That the system wants to develop SDW order, or anti-framagnetic order, depending on the preference of words, with the momentum which is either equal or closer to pi pi. And the same story happens in iron-based superconductor where magnetic order is somewhat different. So the story is told so far as that interactions that tends to develop magnetic order pushes good component of pairing interaction, the ones that, in this particular case, favors d-wave, the one between patches. And eventually makes interaction attractive. And the story before, in previous lecture, was that A, how to properly catch this, and second, what is the interplay between superconductivity and other potential order, including that magnetism that was responsible for superconductivity in the first place. What I'm going to do today, sorry, in this lecture is somewhat different. Let's regress, and I will say, look, let's look at the phase diagram, say, for cup rates. Again, phase diagram is very complex. Let's suppose that it's only whole doping. We have magnetism. We have pseudo-gap phase. We have superconductivity and possibly charge order somewhere here in the phase diagram. But I want to look into this player for example, and say, well, we know that magnetism is there. So can it be that, by some reasons, magnetic fluctuations are strong, which means predominant interaction is between fermions, is not a small momentum transfer but a large momentum transfer. Those of you who know phase diagram of the cup rates also know that there is a charge order here. There is a region of charge order whose boundaries here are a little bit blurry. So you can play a different game. You can say, well, what about charge fluctuations? Promoting superconductivity. It's another possibility. And given this and similar examples in iron-based materials, I want to play a different game. And basically say, why don't we try to think from the very beginning, that effective pairing interaction between fermions, which I label like this. I take fermions with momentum K and minus K, P and minus P. And I want to put both K and P on the fermi surface. Originally, that was this G1 and G2 terms that we are talking about. If you have more than two possibilities for K and more than one possibility for placing K and P. For example, you can place K here and P here. This is momentum transfer pi pi. Or I can put K here and P here. This is momentum transfer zero. So what I want to do is to say by one reason or the other, this interaction can be viewed as effectively mediated by some boson, K minus K, P minus P. And bosonic propagator, which sits here is somehow enhanced at some particular momentum transfer. Again, with the risk of repeating. In this particular case, it will be enhancement at momentum pi pi. If you want to place a game with charge order, which is axle charge order with some non commensurate momentum, it will be increased on this incommensurate momentum and so on and so on. I will discuss in a few minutes one example when this can be crudely derived. But generally speaking, to be honest with you, nobody can derive this accurately. There is no small parameters that will allow you to go from a regional interaction and end up saying that we have effective boson mediated interaction. Which in some sense means that we replace phonon by some collective mode of electrons. And the story I'm going to talk about with you is the story of the pairing mediated by collective mode of electrons. Anticipating what I'll be talking about, the story will go this way. By itself, it looks that I'm just violating some principles but gaining something for free, namely. If I say that this is interaction peaked at momentum transfer pi pi, then I don't care about what happens with the interaction inside this page. I only care about interaction between patches because it's already by hand, by me, made large. Then it's not a question, definitely, there will be attraction and d-wave channel. So not surprisingly, on the first glance, it looks that if there is some kind of long range order that the system wants to develop as a function of some parameter, call it x. And suppose that this is long range order being magnetism or charge density wave or something. Because by construction, I put interaction attractive. And sort of large because I want close to transition. Then, inevitably, it looks that I should get superconductivity before I reach magnetic order because even before I reach magnetic order, my interaction is large and attractive. So this is what I essentially put by hand. The game I'm going to play with you is the game that this effective interaction necessarily becomes dynamical. And once it becomes dynamical, it has two properties. On one hand, by saying this interaction is wavy line. On one hand, it mediates pairing and produces strong attraction in particular channel. DS, D plus ID, whatever. On the other hand, I can take the same interaction and translate it into self energy. It's the same interaction. And it turns out that when interaction is dynamical, it produces your self energy if I take a fermion at frequency omega, just for sake of the argument, suppose that I'm at the Fermi surface, somewhere on the Fermi surface. It will produce me self energy, sigma of omega, which will be large. And you will see that at the critical point, the self energy will have non-farm liquid form. Now, if the self energy is non-farm liquid, what it means in practice? It means that fermions become incoherent, right? Incoherent fermions is the ones that are mostly diffusive rather than propagating. When we talk about superconductivity, we say, yeah, there should be attractive interaction. But we mean more than attractive interaction. We also, back in the head, means cooper logarithm. That no matter how small attractive interaction is, we always get superconductivity, right? Now, once fermions become incoherent, if you put the self energy into particle-particle bubble which normally should give you logarithm, you will not get logarithm. It will be non-singular at all. Which means, physically, that incoherence acts against superconductivity. You will see formulas later on which explicitly show that incoherence acts against superconductivity. On the other hand, if somehow superconductivity wins and the system becomes superconducting, then feedback from superconductivity changes the form of bazonic propagator. Why? Because it's collective mode of electrons. If electrons become superconducting, this guy necessarily changes. And the changes are such that self energy gets smaller. There are zillions of experiments on cooperates which show how quasi-particle peak appears below TC. This all goes under peak-deep-humped storing cooperates. But message is simple. If you are below TC, system becomes more Fermi gas-like than it was before. And you hear probably balance. On one hand, tendency towards non-Fermi liquid tends to act against superconductivity. On the other hand, if superconductivity develops, it tries to recover Fermi liquid behavior. So the question that I will pose and will try to answer is who wins? Tendency towards non-Fermi liquid or tendency towards superconductivity? If superconductivity wins, you get this behavior, right? Before you reach any order, system will become superconductor. If tendency towards non-Fermi liquid wins, you will find the phase diagram which you can see in many books on quantum criticality. Namely, there is the order and there is the fan, there is the region when the system becomes as non-Fermi liquid, there is a Fermi liquid here, but superconductivity will not develop because it will be killed by non-Fermi liquid physics. And of course, you can have intermediate behavior where superconductivity is a minor, minor bump somewhere here than it's really, you have both. You have non-Fermi liquid at higher temperatures and superconductivity at lower temperatures. So in some sense, this is what I'll be talking about. And in some sense, you will find that surprisingly, it's very difficult to get this picture, almost impossible that it always goes like this by special reasons. Okay, let's do it. So this is basically really summary of what I'm going to talk about. So let me start with something that is probably the only example when you can do something more or less, yes and no, in a sense that in practical purposes, of course your question is right, but on the other hand, you can ask fundamental question. Can you see a naked quantum critical point not interpreted by superconductivity? Because whatever I put here, it will tell you that this point is unreachable, that it's under the umbrella of a superconducting state. And I think it's reasonable to ask a question, can you get naked quantum critical point? How to kill superconductivity? Oh yes, but I basically said, okay, let me put it more accurately to say what I mean. Suppose that it goes like this. So then there's no space for non-formal liquid physics at all. And of course everything in between is also possible. So we'll start trying to understand which is picture is closer to reality and can you change one picture into another by varying some parameter? Okay, let me start with one example and you will see basically on this example tendency towards this behavior. Namely, suppose that I want to consider just a system of fermions with some, just for sake of the argument, hybrid interaction. So what I do? I need to write down paving and direction. And let me write dashed line will be interaction u. Again, k minus k, p minus p and this is some repulsive interaction we start with. Well, first point which I want to make is that if you want to study superconductivity, this is not an object that you have to look at. Although my previous lecture I used hybrid u which is okay, but you need to be more careful with this. Namely, these two are fermions with spins. So let me put spins, alpha, beta, gamma, delta, generally all four components alpha, beta, gamma, delta can be either up or down. Interaction preserves spins. So this term will be u, delta, alpha, beta, delta, gamma, delta, meaning that spin is preserved along interaction line. I deliberately don't put which one is up, which one is down, it can be anything. But who said that I need to take out going fermions this way? I can easily, they are undistinguishable, right? So I can interchange them. If I interchange two fermions, I acquire extra minus sign. So physical vertex that we call gamma is always anti-symmetrized interaction. I mean that by itself interaction needs to be anti-symmetrized to measure any physical quantity. So what I have as a vertex is this guy, the original one, I will not repeat, oh no, let me do it. Alpha, beta, gamma, delta, just spin, and this will always be k minus k and p minus p. Minus, this guy, oops, dashed lines here. I interchange them, it will be delta, this one will be gamma, this one will be beta. I just interchange two fermions, and when I interchange two fermions, of course I have to change sign of the interaction, and I also change momentum. If this was k minus k, p minus p, this will be k minus k, but this will be minus p and p. If interaction is just hybrid U, we don't care much about changing momentum. If you want to keep interaction as momentum dependent, of course it matters. So what we get here is a combination of two terms. It will be U, delta, alpha, beta, delta, gamma, delta, minus delta, alpha, delta, delta, beta, gamma. This is your physical vertex, which includes Pauli principle for fermions. Why I'm doing this? Because there is wonderful relation for the second delta function. If I write down sigma, alpha, beta, sigma, gamma, delta, it's equal to minus delta, alpha, beta, delta, gamma, delta, plus twice delta, alpha, delta, delta, beta, gamma. This is identity, who remember? There is a name associated with identity. No, there is a specific name associated with this identity. I don't remember. Somehow some colleagues of mine who know all high energy literature that says that somebody has a name associated with it. Anyway, this identity, identity on which relates delta function and sigma matrices. Why I'm using it? Because this guy, the ones that I have here can be written in terms of this guy and sigma matrices. So let me do it. Continue. You will get U over two, delta, alpha, beta, delta, gamma, delta, minus sigma, alpha, beta, sigma, gamma, delta. And already at this stage, you find out that by doing this simple anti-symmetrization analysis, you find out that your effective interaction has two pieces. One piece contains delta functions in spin indices, and this is what we normally say interaction in charge channel because charge fluctuations are scalar. And this piece contains sigma matrices. This is something associated with spins. So I can equivalently write down my gamma S. I'll put it now like this direction. Alpha, beta, gamma, delta, there is one term which will be U over two, delta, alpha, beta, delta, gamma, delta. And the second piece will be with minus sign. Same, but in the vertices, I will get sigma matrices. Sigma, alpha, beta, sigma, gamma, delta. If this is alpha, beta, and this is gamma, delta. This is first step to splitting the physical interaction described by vertex functions into interaction and charge channel where we have vertices which are scalars, and interaction and spin channel which we have vector vertices with spin indices there. So, you see, so far it's rewriting. The next step is to say, okay, let's now renormalize. And there is a bunch of diagrams that you can put in dressing this interaction. What I mean by dressing this interaction. I want to consider irreducible interaction in a particle-particle channel which means I don't allow any renormalization in the particle-particle channel. Just fix it. But nothing prevents me to renormalize interaction by, say, doing this, creating particle whole bubble or creating something like, sometimes it's called wine glass, or creating exchange interaction. External lines are always like this, and this is K minus K, P minus P, and I have to undissimitize again. Plus higher order, plus higher order, plus higher orders. So, this is how you dress your interaction. As I said, there is no well-known procedure how to do it, except for one case when internal quantity, which K and P here, and this is what we call polarization operator as function of K minus P, which generally depends on geometry of firm surface, of excitation, et cetera. If you allow me to play a game, and it's really playing a game, when this quantity as a function of K minus P rapidly decaying without even asking me a question how to justify it. Just to play a game and see something. I want to show you one thing. Then you can sum up infinite series of corrections every time saying that when I have this small momentum transfer, I take maximum. If it's large momentum transfer, I get nothing. And I'll tell you what the result will be without going into details of calculations. It turns out where is our guy, our guy's here, right? So let's erase what we don't need, and let's write down the result and then we'll see what we're talking about. So the result will be that at the end of the day, our vertex will be gamma, will be U over two, one divided by one plus U times polarization operator at zero because this is what I want to look. This is P of zero, and I take it as the largest one times, let's put the numerator, delta alpha beta delta gamma delta minus U over two, sigma alpha beta sigma gamma delta divided by one minus U P of zero. This is how people normally split interaction into interaction mediated by spin fluctuations and charge fluctuations. Because if, you know, this is summing up infinite series, what I suggest you to do, it's a fun problem for five minutes, calculate leading correction. You will find that in order to obtain leading correction, you need anti-symmetrization. Without anti-symmetrization, you get zero. Nothing. But one way or another, if this is the result, then we can say two cases. Case number one, suppose that U is positive, and I kept repeating this several times, positive U is good for magnetism. And suppose we come to a point when U times P of zero is approximately one. This is known as Stoner criteria for magnetism. Because I talk about zero, I used to say a word for magnetism, but I generally want to not specify too much. If this guy is close to one, obviously this interaction is enhanced, right? So now you have largest terms, then you can forget about this one and say this is what I need to consider. You end up with effective interaction. Let's put it here. Your case one, you have C in which you have effective interaction mediated by, because of the sigma matrices on both sides, mediated by spin fluctuations. You can expand this in small q and get a term with q square with standard Ornstein-Cernicke form of propagator. Namely, if you expand this, one minus U P of q will be approximately some correlation length difference between one and U P of zero plus q squared. So this guy will then will be one over psi minus two plus q squared. Notice minus sign here. So it looks that we get attraction, negative interaction, attractive. Wrong, but sign is right. It's just logic is wrong. Or you can think about another case when U is negative and when U is negative, now you have a different story. Now this one is non-singular, but now this one becomes singular, right? And now you have singular interaction mediated by charge fluctuations, right? So this is case A, case B, we have gamma, we have interaction, and this have delta function and delta function here. Very good. Yes. Oh yeah, sure. Yeah. Oh yes, because if you want to go to a case when U P of zero is larger than one, forget about this introduce, which means that if you like, it means that correlation length square becomes negative, which means that they haven't introduced long range order. So I always, of course, assume that U times P is still smaller than one. I'm just approaching transition. Yeah. Are we on one wavelength so far? Good. Okay. Sure, go ahead. So far no, so far no frequency. It's just simplest case. It will become functional frequency, no. So far I'm just doing one step at a time. So far there's just no frequency, static. All I want to tell you, and this is roughly, this called RPA, why is the word random face approximation, which in some sense means that you sum up ladder and bubble diagrams without any combining them together and writing something more sophisticated. Sorry, you can combine them together, but you cannot write anything more sophisticated than just separately ladders and bubbles, or just only ladders and bubbles. Nothing more stronger than, for example, you cannot put inside the renormalization particle, particle channel somewhere in between. Anyway, what this tells you? This tells you that this non-rigorous procedure tells us that cases, either when this is small or when this is small, when you can approximately write your interaction as mediated by soft boson that is about to be condensed at the critical point. Soft boson can be in charge channel here, soft boson can be in spin channel here, but in both cases this is soft boson. Attraction is guaranteed. At this level, what is even guaranteed is that when denominator is small enough, your interaction is pretty strong. So we are heading towards this direction, in fact. Strong paving near the critical point. Just one caveat, and if possible, I will not spend time on this, I'll just ask you to do simple exercise. Game is this, minus sign here, right? Let's go to positive view. Does it mean that we have attraction in spin channel? Minus sign normally means that we do have attraction, right? It turns out that the answer depends what kind of pairing you want to consider. If you want to talk about what we were talking about, this problem and other problems, then it's spin singlet pairing. Question, make sure that for spin singlet pairing, when the pair that you want or pair stays that you want to consider is spin singlet. When you combine it with interaction that contains sigma sigma, you get extra factor of minus three, minus. So you have extra minus sign when you play the game with spin indices. Minus converts this minus into plus and you cannot just get attraction out of nothing. You started with repulsive interaction, you deal with repulsive interaction as it is. So you again need to play this game that sign of order parameter here and here will be different. So we cannot get attraction. However, if you want to consider another set of problems which similar to what Aaron Kapitulnik was talking in his lecture, p-wave pairing or f-wave pairing, any spin triplet pairing, then for spin triplet, combining spin symmetric combination with the sigma matrices, you will get plus one. So minus sign, remain minus sign, you get attraction. Which translates into phrase like this, that magnetic fluctuations promote attraction in p-wave pairing channel. And I think until recently, this was the main theory for pairing in helium-3. Now as usual situation is a little bit more murky, whether it's mod physics there or tendency towards mod physics or just for magnetic fluctuations, people are saying different things. But at least there is definitely attraction. One remark about this as I showed in this very oversimplificated story. This is just like you need to change in sign of you and then you will get attraction. But this is how people consider two things. A, pairing by pneumatic fluctuations. For this, you need to take interaction which is not hubbored to you. You take u of q and take d-wave component of this u of q and say by some reason the d-wave component of u is attractive. This is how you obtain interaction which describe pairing by pneumatic fluctuations. Another example, pairing by charge fluctuations, it's also here. Final remark about this, I put p of zero everywhere. If your story is maximum interaction at pi-pi, well place the game when you're replacing by q here and place the game when p of q as a function of q, this is pi-pi. Suppose that it goes like this. It's phenomenology. I don't know how to derive this and I doubt that anyone knows how to derive this. Regulus. But if experiment shows that you are close to anti-frameditism, then it's reasonable to assume that this guy will be maximized at pi-pi and then close to anti-frameditic transition to still get this interaction but now you expand around a different point. Okay. Saying all this, I want to introduce a model. The model will be, it's called Fermion boson model. I probably told you all possible justifications for this model, but I guess the key justification is really closeness between superconductivity and some kind of order, either nematic or spin or charge. What this model tells you, it's formulated in Hamiltonian language, not Lagrangian language, but Hamiltonian language, means dynamics is not there yet. And this model consists of fermions with some Fermi surface and some dispersions around Fermi surface. So HF term is just sum over k, epsilon k, ck dagger ck. Let's put in this as alpha, alpha and sum over alpha means that, of course you hope without changing speed. So conventional quadratic form on fermions. Next term is a boson term. And this boson term is described by sum over q, chi q inverse b q dagger b q. Either a scalar or vector in terms of spin, it will be vector in terms of charge, it will be scalar. Just bosons, which are described, so propagator of fermions is solid line, that's green function of fermions. Wavy line is chi for bosons. We take what we do physically. We take collective mode of electron as a separate entity. We can say we call it boson and treat it separately from electrons at this level. How is this chi is made from fermions? But the idea is somehow separation of energies. It is assumed, again, in words, not in deeds, that largest contribution to chi come from high energy fermions. So if I gain, I'm not doing Rg, I'm just in words integrating over high energies. So assuming that I somehow managed to integrate over high energies, I put upper limit in energy integration, suppose it's energy units. And I put upper limit at some lambda and consider everything here. And I assume that everything here is integrated out, I get some bosonic propagator. And the last term here is fermion-boson coupling and this fermion-boson coupling is equal to sum over K and Q, C dagger K plus Q, CK and BQ. If it's charge, alpha, alpha, alpha, if it's spin, alpha, sigma, alpha, beta, beta, and this is a vector boson, so it has three components. One way or another, this is called fermion-boson model. You probably have better memories. Somebody wrote this model during the talk. I don't remember who, but formula was there. I remember that somebody wrote this fermion-boson model in this case. The question is, right, we have fermions, we have, and sorry, what I forgot to write down is a coupling constant here. So we have interaction between bosons and fermions, which diagrammatically, let's write down interaction, is like this, fermion coming in, fermion going out, and then there is a boson here. Of course, I can write down, I have to be more accurately as B with BQ dagger to be accurately here. There is a boson, and in the vertex, either I put spin matrix or I put delta function if it's charge. I probably will use, will work with spin simply because not to say every time, spin or charge and just to minimize the number of words. So we have a vertex, you have chi. So obviously you have effective in pairing interaction. When you put another sigma here, this is very similar to what was on the blackboard. Okay, so chi, we need to make assumption about what is chi. We assume that chi, let's put it here. So epsilon k to first approximation will be just Vf, k minus kf if we extend around the Fermi surface. For chi inverse, I will assume that it's psi minus two. Some scale which tells you how close you are to a bosonic transition, now let's call it magnetic transition, and let's say q minus capital Q square where capital Q is pi pi, if it's pi pi transition. If it's Q equal to zero transition, then it's simply Q squared. So this is the model reformulated in Hamiltonian language. It gives so far static interaction between fermions, attractive interaction in a proper channel, we know this. And then, I guess the first thing to do about this model is to say that there is one interesting feature about this model, that, yes, double counting in what sense? I can, I can, but let me be careful with this. Maybe I put again that on energy, oops, on energy scales, bandwidth will be somewhere here. I want to treat this as a low energy model. So I put some upper cutoff lambda and pretend that everything here is integrated out. So if I find any correction in the theory, which will depend on the upper limit, I will be in trouble. Then it will be double counting, because then it means that it comes from the same energies which I counted once. But if I found corrections, which are completely sitting in this region, then there is no double counting, right? Okay. And there was no snapping staged, but I'm going to talk exactly about this. Because one thing is pairing interaction. Another is that we also, when we look at this, we can say, look, what about this piece? When boson, wavy line, transforms into a pair of fermions and they recombine then into bosons. You saw this diagram in Young-Galai's talk many times. It doesn't matter, in his case, there was a photon in, photon out, but there's still particle whole bubble. That's called particle whole bubble because direction of errors. One error goes this way, another is this way, which in practice means what's particle whole. That when you start calculating this, you find that one particle should be above Fermi level and another is below, otherwise it equal to zero, it gets zero. You can ask what is this? What you get out of this? And out of this, you have two terms. One term accounts for double counting. If you want to calculate, particularly in anti-farmagnetic case, static correction, correction to this guy, because this is self-energy for a boson. If you want to calculate static piece, it will be some integral which will be sitting at the upper limit, you either forget about it, I already counted it. But there will be dynamical piece. And if this guy has dynamical frequency omega, then what you get out of here, let's do it in real frequencies, will be i omega times g square. Remember there's a coupling g and g here. And this is land-out dumping for large momentum transfer. Remember I put here momentum transfer capital Q this while land-out dumping has just linear and omega term. Why I'm saying that it's not double counting because you can check yourself, it's a nice exercise, that if I put some frequency omega inside this region, somewhere here, then energy of both fermions, both, not one, but both, are here. Neither energy can be larger than omega. You can check this. If I'm playing the game with Q equal to zero transition, say pneumatic transition, then form of land-out dumping is something different. It's i omega divided by Q times g square. And in both case there's some number which depends on fermion energies, et cetera. So I just want to show the functional form of this. This is important, why? Because for example, look at this, omega over Q, I have to compare it with, let me be here currently, chi Q minus one is equal to xi minus two plus Q square if Q is much smaller than one. And this is when Q is approximately capital Q. So this is pi pi and this is what happens if I consider transition with Q equal to zero. Keep in mind always that when I say Q equal to zero, I mean pneumatic transition. So this is dynamics. And this doesn't just finish the phrase and then I say, sorry? This is big Q equal to zero. So here either I put Q approximately capital Q or Q which tends to zero. Either way, yeah, okay. If I'm confusing you with notation, just raise a hand and you know how it works here. It actually works here. There was a lecture here where a lecture gave everyone a piece of chalk. So yes, if you don't like what you see, if you won't take it. Anyway, let's move on quickly. By the way, what I'm doing time wise. How many? 20, so only 20. Okay, so we need to go a little bit. Sorry? Okay, we need to go a little bit faster. So what do we have here? We now have dynamical problem, right? And with this dynamical problem, we can say, okay, let's take this guy. I'm doing something that justified a posteriori in fact. Let's take this and add to bosonic propagator. So now my bosonic propagator acquires self energy which makes bosonic propagator dynamical. So what I have, again, let's consider one particular case. Let's consider this case just to do something. My chi will be psi minus two plus q square plus i, actually minus i omega divided by q. There is a g square here and some number which depends on other parameters. Very good. Dynamical bosonic propagator. So far nothing special, so what was it again? It's still when omega is equal to zero and q goes to zero, it's still very large because of this chi minus one is small. I don't change repulsion into attraction, it's still attractive interaction. So far I'm still alluding to this picture. Now, the story keeps going on that now I can calculate fermionic self energy. Why dynamics is important? Because even direction is static then self energy doesn't depend on omega. You can just shift frequency and integrate over it. You need dynamical interaction to get self energy being frequency dependent. Before I tell you what the answer is, let's very quickly ask our self what you expected the Fermi liquid for self energy. Fermi liquid is not a Fermi gas, fermions are interacting. So, but there are particular forms of the self energy in a Fermi liquid. Let's state it and then compare. So, Fermi liquid, self energy is sigma prime of omega. Again, let me put k on the Fermi surface in order not to deal with k minus kfps. Self energy sigma prime is proportional to omega. Sigma double prime of omega is proportional to omega square, right? This is your Fermi liquid. Real part of self energy, just master normalization, imaginary part give you omega square and the rest actually follows. This much more robust than t square and resistivity and I will not take bread from Professor Maslow who will tell you that omega square and self energy and t square and resistivity are generally completely different things. So, let's see what we have here. If xi is not equal to zero, xi inverse is not equal to zero. You substitute the sky. You calculate self energy. It's a quick procedure. You get Fermi liquid. So, sigma prime will be equal to omega times something. Sigma double prime will be equal to omega square times something. So, apparently not a big deal. We have Fermi liquid, everything is fine. The question is what is something when xi goes to infinity? Xi minus one goes to zero. The result depends on dimension. If we are in three dimensions, then this term is logarithmic. This is log xi. This goes as omega square over xi cube. Which tells you something interesting going on even here but looking at this term, yeah, it's only logarithmic. Interesting, but this is what is eventually will be called marginal Fermi liquid theory. If we are in two dimensions, I need space and I'm doing this badly. So, let me quickly go here. So, same in dimension equal to two. So, dimension equal to two here and I get self energy sigma prime goes as omega over omega times xi. I'm sorry, omega square times xi cube, sorry. It is big. And sigma double prime goes as omega square times xi to fourth power. Now you have more serious stuff. Your self energy, particularly mass renormalization d sigma over du omega goes to infinity and not as logarithmic as power. So, now combining these two terms, you can say what if I'm right at the critical point when my xi is equal to infinity. We can still do calculations. You get finite results. What are these results? You find out that in three dimensions, this is where marginal Fermi liquid is in all its glory. Sigma prime will be proportional to omega log omega. Sigma double prime will be proportional to omega. This is not Fermi liquid. Compare these forms with these forms. Why it's called marginal Fermi liquid? Because if you compare this guy with this one, they differ only by logarithm, right? If I go to dimension three plus epsilon, this will be eventually omega and this surprising will be omega square at very low frequencies. It's just window will be very narrow. But if you want to do the same in two dimensions, what you get? Sigma prime will be omega 2 third and sigma double prime will be omega 2 third. And this is what's called non-Fermi liquid. Why non-Fermi liquid? Because A, real and imaginary part of the cell margin of the same order. And B, remember what was the physical meaning of a Fermi liquid? Physical meaning of a Fermi liquid was that you approach zero frequency and then progressively imaginary part of self-energy becomes smaller and smaller than the real part, which means that particles at smaller and smaller energies live longer, longer and longer. And finally, at the Fermi surface, they live infinite amount of time, right? This is the property of a Fermi liquid. You break this property here. Your imaginary part of self-energy will never be smaller than sigma prime and this at small frequencies is definitely much larger than Bayer omega with what you compare. So the first approximation can forget about Bayer omega and deal with this. Now, yes, go ahead. Oh, you take one limit was omega goes to zero psi finite. Another limit was psi minus one is zero omega finite. So there are just two different limits of the argument under the log. Notice that with imaginary part of self-energy is really huge change. Now, because I realize that time is running and I want to talk about superconductivity, let me now do this. I will include self-energy into consideration and I will now look for superconductivity in the presence of the self-energy which by itself will destroy Fermi liquid and ask a question what we get for this. Let's see. Let's quickly move on. Yes. So the next step is that I need to explain to you how to calculate the L.A.R. brick style which means I need to assume that fermions are slow compared to, sorry, fermions are, sorry, bosons are slow compared to fermions. Let me take five seconds and say what means. When people talk about L.A.R. brick theories of superconductivity or of electron-electron interaction they take one issue to make calculations rigorous. They assume that fermions are fast excitations compared to bosons. Let's check this. So what we have for fermions, fermions, Green function is equal to one over omega plus sigma of omega minus VF K minus KF. Let's take two-dimensional case because it's more interesting that three dimensions. Let's forget about this omega. Then typical suppose I take given frequency. For a given frequency what I get? I get VF K minus KF is proportional to omega 2 thirds. This is my scaling of fermionic propagator. For bosons I get one over, let's put psi to infinity, Q square plus or minus i omega over Q so far. It doesn't matter at this point, I just do scaling. So you see immediately that my typical Q scales as omega one-third. So yeah, yeah, yeah. He said 20 minutes and I start rushing. Point is, this is a very good point. I didn't tell you about self-consistency of this procedure. When we calculated omega over Q, we calculated bubble and this bubble was made out of three fermions, right? And we get omega over Q. Now I said, look, it's not like this, there is a self-energy. And self-energy is larger than omega. So what you get if you address these functions? If you include self-energy into propagators, the answer, omega over Q. Nothing changes, even coefficient will stay the same. And the reasoning is that frequency-dependent self-energy does not change density of states which remains the same as it was. And then when you calculate land-out dumping, there's a convolution of two densities of states. So this is a part of the story which is a substantial part of the story. Frequency-dependent self-energy does not affect land-out dumping. Just does not completely, it's exactly the same result as it was before. In superconducting state, of course it does, but not in the normal state. So this answer, we can check the answer, but I claim that it's like this. And as such, you still can work with this part. Very good. So just looking at this, one typical momentum, scales as omega one-third, one scales and other scales as omega two-third, for a given frequency, actually which one is larger? For a given small frequency, this one is larger, right? This one is larger means that essentially if you divide this by Fermi velocity, then essentially it means that fermions are much faster than bosons. And because fermions are much faster than bosons, there's a particular way how you can write equations for the self-energy and for the bearing. And let me just maybe emphasize the main things. So we can get rid of this. So what I want to do is to write down two equations. One will be for the self-energy, another will be for the quantity which is called pairing vertex. Pairing vertex is just like anomalous green function. I want to probe that the system develops pairing condensate and I call this quantity phi of omega. If this frequency omega, this is minus omega. And equations are phi of omega is equal to pi t sum over omega prime phi omega prime square root of omega prime plus sigma of omega prime square plus phi omega prime square and then chi omega minus omega prime. I'll understand what it means. Sigma of omega is equal to pi t sum over omega prime omega prime plus sigma of omega prime, same square root. Omega prime plus sigma of omega prime squared plus phi of omega prime squared and same local interaction omega minus omega prime. There are two notations here which needs to be, first of all, I made a logical step. Before I tell you everything in real frequencies. I can write this equation in real frequencies but it turns out that it's easier and more simple to write equation that's what's called Motsubara frequencies. So in writing this equation, I switched from real frequencies to Motsubara frequencies which is as you know pi t to m plus one. So summation here is a summation over discrete variables. At zero temperature of course summation goes into integration. Second, there is a quantity which is called chi L. Chi L of omega minus omega prime is you take your chi which is now a function of q and omega and integrate along the Fermi surface. In two dimensions it's just integration along the Fermi surface. In three dimensions there will be integration q dq because there are two components along the Fermi surface. It requires one lecture to justify how to derive this. So let's just take this because this is what you get and I will explain one thing right away. But I want to show you one quick thing here. So we said that we'll deal with this chi. If I convert it into Motsubara frequencies it will just give me mod omega here. It's just converting from real frequencies to Motsubara. Just make sure that it will be modulus with by all rules of conversion. If you integrate this guy, if you take this guy to zero, suppose that you are at the critical point and integrate along the Fermi surface you will get this proportional to one over omega one-third. So let's make it omega minus omega prime to power one-third. If I wanted to do the same for anti-farmagnetic case when instead of omega over q I will get simply omega then this is pneumatic, two equal to zero case. And if I want to do it for anti-farmagnetic case I get omega minus omega prime to power one-half. This is anti-farmagnetic case. Now, what does the two equation mean? It means this is equation for the pairing and I took into consideration that we have attraction so all d-wave is already here. Notice there is a plus sign. So there is, we expect to have solution for this equation. If I neglect self-energy, just neglect it completely and want to solve for Tc then all I have to know about that my interaction as a function of frequency transfer is singular. So I get singularly strong interaction. I will get large Tc. But at the same time, the same interaction give rise to sigma of omega. And if you calculate this in assuming that phi is equal to zero if phi equal to zero you will get sigma of omega which scales as omega two-third as this pneumatic case and omega one-half in anti-farmagnetic case. This I didn't discuss with you, this I did. You go back to what you get. Same omega two-third that we had before. So we get non-formal liquid. And now see balance. If the system wants to have very large self-energy the self-energy is in denominator for the equation for phi so it tends to reduce tendency towards paving. If there is a paving phi is in denominator in the expression for sigma. So this phi tends to reduce tendency towards non-formal liquid. So the question is who is going to win? At this point probably I really need to, let me give me estimate. No, no, I'm not the boss. You are the boss. Come on. When I started. No, no, no, no, no, no, that's wrong. No, no, no, we ended late. No, no, no, we ended late and there was five minutes. No, no, no, no, no, no, no, no. Okay, seriously, who remember when we started? Okay, let me take 10. But promise not more than 10. Okay, here is that. I'm approaching interesting part of the story that's what I want to take. So let me tell you what is really interesting here. What is interesting is the following. So we have, we need to solve this equation in fact. Let's assume that we first of all want to understand whether or not we have superconductivity. For this we need to equation on TC, right? What happens at TC? We just start developing superconductivity. So our order parameter is infinitesimally small, right? If it's infinitesimally small, then I can do it in a blackboard very quickly. I get rid of this part, and I get omega plus sigma modulus. Omega prime plus sigma of omega prime modulus. Here if I eliminate this guy, this just give me sine function. It will give me sine of omega prime. By the way, this will give me exactly the self energy that I'm talking about. So I want to understand, self energy is still here in denominator. So I want to understand what will be the solution whether I ever get a solution of this equation. So here is what I will do. I will follow what normally is done in this situation and I assume that it will be also told to you at Prokofiev's lecture. I will look at this, an eigenvalue equation. Really, phi of sum, they're all discrete frequencies. Let's write it again. Omega is equal to pi t to n plus one. We dealing with discrete Matsubara frequencies. So we can easily write it like this. Phi n, here is phi m. Omega m, sigma of omega m and omega n minus omega m. Let's put it phi omega m here and phi of omega m here. This is not an equation. This is just, well it's equation but this I solve instantly, right? I take some form of chi, substitute and get result. So basically what we need, we need to solve for this. For a given form of chi, whatever the theory tells us, substitute it here and then solve for the, yeah. This is a well-defined procedure and this time is running. I will just tell you what the results are and when situation becomes out of control. So we want to solve this as eigenfunction, eigenvalue problem. I want to write here epsilon and as usual, solve this as a matrix equation. Find eigenfunctions, eigenvalues and I am interested in when epsilon crosses one, right? This is where I get the solution. So let's try to see what we should expect. Let me generally, without going every time into q equal to zero, q equal to pi and some other problems, generally assume that my chi L as a function of let's say frequency omega is equal to some constant divided by mod omega all together to the power of gamma. This goes under name gamma models. Gamma one-half is anti-phromagnetic case, gamma two-third is, gamma one-third is pneumatic case. Gamma equal to one is what people now want for more physics. Gamma equal to two is electron phonon interaction as the limit when the bifrequency goes to zero. Gamma point seven is what Alexei Tsverik once deal with his model of pseudo-gap. So there are a number of different gammas. Here's what I want to do. I want to write epsilon as a function of temperature and ask what are the options and then we'll tell you what the result is. What are the options? One option for, let's look at, remember we're interested, we are solving this as eigenvalue problem interested when epsilon equal to one because this is our problem. For epsilon equal to one we get our Tc, right? What happens with epsilon? Suppose that this is one and this is zero. Suppose that we go to very high temperatures. You don't expect superconductivity at very high temperatures, right? You expect that at very high temperatures interactions will not give you anything. So you expect that your epsilon is very small at high temperatures, right? And I'm talking about largest because I'm interested in the largest. No, what I'm saying is this, okay. What I'm saying is this, you want to solve it through. It's infinite matrix so there's infinite number of epsilon. I'm looking at largest positive epsilon and largest positive epsilon because if there will be largest I need to see whether epsilon crosses one or not. That's all I'm interested in. The question is whether it happens. I'll just give you two choices. Choice number one, the largest positive epsilon will go like this. Where is Tc? Here, right? This is our Tc. It can be large, it can be small, it can be anything but it's finite, right? Choice number two. Same story, epsilon is a function of temperature. It goes like this. And all others are here. All other solutions are over here. Then we never cross one, right? This is the story where we don't have superconductivity. Yes? In principle, yes, yes. But so far I'm just playing with pictures, right? I'm just saying what to expect, what can happen. And really, because I'm running out of time, I just want to say that the way how it was done first, there's one point which you can analyze, this one. This one or this one? Why is this is best point to analyze? Because it's T equal to zero. Instead of dealing with summation here and summation here, you just can do integration. If you do integration, then you know what the result is. Your self energy, KL says omega to power of one minus gamma, you substitute here, this summation transforms into one half integration over du omega prime. And you can ask, now I want to solve integral equation, not a sum. Statement, this integral equation has been solved exactly for any value of gamma, whatever value of gamma you want, yes please. Why not? What's wrong? Sorry? I'm solving equation for eigenvalue. This is what I'm trying to understand. If I get a TC, this behavior, then there is definitely TC, finite TC. If I get a TC, this behavior, there is no TC. Unless there's a crazy behavior then sign changes more than once, right? This is what I'm trying to do. And why it was done? Because contrary to equation at finite temperature, we can only ask your friend computer to solve. At zero temperature, you can ask your hand and equation can be solved. There is exact solution of this equation at T equal to zero. And this exact solution of the equation tells you that well, it depends. If you have, so this is the solution at T equal to zero and it tells you interesting thing. By the way, before I tell you what the solution is, detour for 30 seconds. BCS theory. What is BCS here? BCS would be when this guy is a constant. This would be BCS, right? If this is constant, we go back to Fermi-Liquid. This guy is also linear in omega. So I get logarithm, right? I'm at T equal to zero. There is no lower limit of logarithm. I get infinity. So at BCS, I know the result. This is BCS, epsilon as a function of T crosses one at TC and then goes to infinity at T goes to zero, right? Solution here. And this is the last thing that we discussed. Gamma between zero and one. Remember this exact solution. You get, basically you get this. Let's not write it again. This is gamma between zero and one. One thing is definitely different from BCS. It doesn't, epsilon doesn't go to infinity when T goes to zero. Why is that? Because now your interaction converges at large frequencies. So there's no divergences. There's no Cooper logarithm in the normal sense. So there is a threshold. But we are definitely above the threshold. There is a finite TC. At small gamma, you know, we even have analytical formula for TC, but who cares? It's finite TC. Gamma larger than one. It's like this only, one more careful. It's frozen here, so it always goes to zero. So you may, if you look at this, and this will be my answer before the last phrase. You're gonna say a lie. This is interesting story. If gamma is small, which means that interaction is almost like in a Fermi liquid. Self energy is, yes, it has some omega to power of one minus gamma, but it's almost like in a Fermi liquid. Continuity from Fermi liquid, of course, exists. From infinity with gamma equal to zero, you go down to finite value, smaller, smaller, smaller as gamma goes to one. Who wins? Superconductivity wins over non-Fermi liquid, right? However, if you have pretty strong self energy in this situation, it turns out that non-Fermi liquid wins over superconductivity, right? How I obtain this formulas? I said, I know this, I know this, I know this, I know this. I connected, let me put it more accurately, it should be dashed line. I connected what I know at high temperatures, which what I know at zero temperature. Seems absolutely normal, right? Okay, I am out of time, so let me tell you the answer. The soul is nonsense. This is the answer. It turns out that if you ask your friend computer to actually don't do integral, go back, do Pt sum over omega prime, solve this as eigenvalue problem, plot epsilon as a function of temperature for any value of gamma. You get a result, quite interesting, that let's look at this case. What happens is that, let's look here, this will be even better here. You get something interesting. You get this point, it's right, it's okay point, it exists, but the actual solution bypasses this point and goes up, which means Tc exists for any value of gamma. Tc exists, and if you plot Tc as a function of gamma, it gives you some curve which saturates its large gamma. So it never goes to zero. And let me say, so what the hell is going on? At one hand, at zero temperature, there is a clearly critical value of epsilon, as I said, exact solution. And this critical value goes down to one, and then you lose everything. And at the same time, Tc bypasses this point. So for the one hand, this is I guess the first example of isolated quantum critical points. There is no line coming off this point. Second, last phrase for me, why this happens? This happens because of very special reason. I told you that at the end of the day, it will be very special reason why superconductivity survives. I told you that self-energy scales as omega to power of one minus gamma. This is non-ferming liquid. And at zero temperature, this is the form that you get. If you are at finite Matsubara frequency, this guy scales as, I guess it scales as m to power of one minus gamma times T to power of one minus gamma, right? So if I divide it by omega, it will be singular. And zero, if m equal to zero minus one. There are two Matsubara frequencies. They are called first Matsubara frequencies for which self-energy no matter how strong at zero temperature just vanishes. You may ask why there are special reason why it follows from the formulas that I wrote to you. If you look carefully at the formulas, you find that this is exactly what happens. But story is that at finite temperatures, and you see this, you can only see when you are a discrete set of Matsubara frequencies, you find that there are two frequencies for which self-energy just zero, three fermions. Interaction between these two guys, it's still finite and large. So it turns out that final result, that there are two different system regimes that there is another crossover line here associated with what I had before with this line. When we are here, it's a conventional superconductivity. Namely, I can get it from T equal to zero solution, et cetera, which means all frequencies participate. This is a special superconductivity which is induced by first two Matsubara frequencies, which means that if I take, how can check this? I ask my friend computer to calculate TC by just eliminating two Matsubara frequencies. I still have 10 to 20 Matsubara frequencies. Eliminate two, you get this line. Nothing happens here. So this is all, so this is a statement that yes, superconductivity always wins over non-formal liquid, but here it wins by a very special reason and I wanted to have a little bit more time, which I don't, to say that this account, you may ask what it means for experiments. It does mean something for experiments. This is the situation when density of states never, when the gap in the density of states never closes. So maybe it's the last, very last picture. If you are here and you go down this way and you look at the density of states starting from this point, you start seeing that it goes like this, like that, like that, but the position of the gap never closes. So there are two qualitative different pictures. One you have here, another if you have here, he is already standing, so I'm done. Okay.