 First, an announcement about organizations. Those of you who need certificates, which confirms that you've been here. They're all printed by default, and you can pick them up from the info point. So just come up there, say your name, and we'll get the episode paper, which will say something useful. Let me start with summary of lecture number one. So we were talking basically about conditions under which a Fermi liquid, at low enough temperatures, will have a resistivity, which has a residual part, and the part coming from electron-electron interaction. And of course, soon we're going to be replacing this power law by something more generic, but we were talking about our Fermi liquids, and we specified three conditions. One, which is the canonical one, to have unclubs, but the two are caveats. First, your Fermi surface needs to be large enough, and second, your interactions has to be short-range enough. If either of these two conditions is not satisfied, unclub scattering is going to be suppressed. The second way is to have a multi-band system. If the bands are of the same polarity, let's say electrons, or let me call it uncompensated case, you still need some help. You need intra-band of relaxation. Let's say due to disorder, your resistivity will be bounded in between two limits, low temperature one and the high temperature one, both of which will come from a relaxation by a disorder in each of the bands. In between, the resistivity may show up some power law, which in the middle is going to be squared. If the system is uncompensated semi-metal, then with the same plot, you may add intra-band of relaxation, or you may not, but if you do, you will start with some residual resistivity, but there will be no upper bound on resistivity that can go in this model infinitely high. And of course, in reality, if we think about this version, this saturation is going to be masked by, let's say, funds. So this is the model in which you have only electron-electron, a friction between the bands, and the purity. And I started to talk about the third case, which is normal collisions, so normal in this context, meant not umklap on an esotropic thermosurface with disorder. So that's what I'm going to finish now. I think this is the last slide from the previous lecture, where I brought in glialen invariance again. Let me do it again, switch it on, switch it off. Oh, okay. Oh, theories can do something, sometimes. Okay, so again, if the system were glialen invariant, then a conservation of momentum would be identical to a conservation of the current. So what I would do, I would take the Boltzmann equation, I would integrate it by the current, which is velocity. In this way, I would get rid of the impurity term, sorry, I would get rid of the left electron term. The impurity term will survive, and that will determine the relationship between other current and the conductivity, sorry, and the field, I'm not used to this yet, okay. Which will give me the drude form of the conductivity, where there is no effect of electron interactions. Now, if I bring back the lattice, I still have momentum conservation, because that's the game I chose to play, eliminated on club, and I still have only normal collisions. They conserve the momentum, but for trivial reason, the velocity on the lattice is not equal to k over m. So when I take the Boltzmann equation and I multiply it by the velocity, the impurity term still gives me other current, but the left electron term doesn't vanish, because velocity is not the same as k over m. And being non-zero, it will give me the rate of change of the current due to electron-electron R collisions. And so that tells me that I cannot eliminate electron-electron interaction, it will R contribute. To understand how it will do it, one can look at two limits. One is the low temperature limit. So at low temperature, the scattering rate of electron-electron interactions is small. Whatever is the mechanism, it always scales as some power of temperature. So it means that electron-electron collisions are rare. Electron-impurity collisions are frequent, right? So I have two parts in the collision integral. One is small, the other one large. We'll do a perturbation theory in a small part of the collision integral, which means that I will solve first without taking into account the electron-electron interaction. And what's important that what I will get here is a distribution function, which by itself is going to be proportional to the dot product between the velocity and the electric field. Then I'll plug it back into the Boltzmann equation, find our correction, and that will be the correction to the current and to the conductivity coming from electron-electron interactions. This is kind of simple algebra. This gives you the result, which is long, but it's fairly easy to read. So this is your resistivity, it's some of two parts. This is the residual part, this is electron-electron. I can integrate over the energy and scale out a factor of a t squared. We are still in a fermi-liquid. The conservation laws come through this factor, which is the imbalance of the current in the initial and the final state. So V sub k plus V sub phi times the charge is the total current in the initial state. This is the total current in the final state. As long as this is nonzero, I will have a nonzero t squared term in resistivity. We can check immediately for a gradient invariant system. If I go back to the case, my velocity is k over m, you replace each of these velocities by k over m, p over m, and then because we have momentum or conservation, the entire term are vanishes. So in gradient invariant system, you do have t squared, but it's multiplied by zero. All right, but this is still the limit when we did the perturbation theory, which means that the t squared term in resistivity is small. What about the other limit of high temperatures? Okay, so now if you go to high temperatures, high temperature here means that the electron-electron scattering rate is higher than the electron-impurity one. So I have very frequent electron-electron R collisions, but by themselves, they cannot control the current. All they can do is to equilibrate the distribution function, which is going to be a fermi-deric distribution, but with a drift of the center of mass, and the center of mass is going to be moving in a way which is controlled by electron-impurity scattering. And that's why this limited resistivity is, again, saturates, and it saturates at a value which is controlled by electron-impurity scattering only. So explicitly, you will have two of limiting, so in this case, these are the conductivities, and how they differ, they differ in how we average the components of the fermi-velocity over the fermi-surface. If I have a sphere, it doesn't matter how I average, so in this case, the upper limit is going to be equal to the lower limit, and the space in which you have the temperature dependent variational resistivity will shrink back to a constant. Now, generically, if I take any fermi-surface, the variation between the upper limit and the lower limit is a number of order one, right? It is dimensionless, and I cannot make it two in the tropic. So in this case, you can have a contribution of electron-electron interaction, but it is limited by, say, number one. So in this sense, the two-man model, which we had here, it's also the same variety. I also have the lower limit and upper limit, but here I can play this masses of light-band and heavy-band, and if one is much larger than the other, I can make this upper-bound very high, and then I may not even see it. All right, so this is the summary of the scattering mechanisms without umclubs. So if you have a compensated semimetal without umclubs, that's the only way when the resistivity starts from the residual value and then takes off in the thermo-liquid as T squared. And the residual part and T squared part add up as prescribed by the Madison rule. If I have anything else that is either a multi-band metal but not are compensated, or just in the tropic thermo-surface, the resistivity varies in two limits, which by itself are controlled by a disorder, but differ depending on other parameters of the structure. Okay, so today is somewhat special case, which just tells you that you have to be careful, we had a couple of talks last week, which were on integrability, on exactly soluble model. So everyone knows that a system of particles in one D is integrable, because as you probably remember from your freshman physics course, if I have a ball or a cart on a rail and I run another cart into the stationary one, all they would do is to swap their velocities. The one which was moving would stop and the one which was at rest would start to move, but would be exactly at the same velocity. So this is the notion of total integrability because in this case, the number of conservation laws is equal to the number of unknowns. And that means that there will be no relaxation over the momentum or velocity in the system because if I prepare a bunch of these cards and I'll run one cart and then I'll look away and then I'll look back with exactly one cart, which will be moving at the same velocity at the initial one. And because I am kind of cart blind, I don't know which one was first or last, there is no relaxation here. So of course, we're not talking about one D here, we're talking about multi-dimensional systems, but if we're in two dimensions and we're looking at the processes which happen near the thermosurface, we force our fermions to move on the surface of the thermosphere, which in this case is a perimeter, a circumference. And so in this sense, we recover one dimensionality and it turns out that this kills the T squared term in the resistivity. And the next term when it starts from is actually T4 rather than T squared. And one can see it going back to this connection to the resistivity, which we derived in the general case. So this did not specify whether it was a two-dimensional or three-dimensional system. But I still need to, so let's say this is a two-dimensional system, so how do I read this formula? Let's say I fixed momentum transfer, which is controlled by mass scattering probability. I need to find the initial states and the final states, which belong to the thermosurface such that the energy and momentum is conserved. And it's actually pretty hard in 2D. And to see how hard it is, let's go a thermo-liquid demo. Let's go back to the Lerner invariant system, which is now a circle. And I will have the initial state K on the thermosurface. I will have another initial state P on the thermosurface. This is total momentum K plus P. Okay, and now I need to find two other vectors also on the thermosurface, which add up to the same sum. And the choices are very limited. One is that I can cheat a little bit and I will choose the initial configuration in such a way that total momentum is equal to zero. Okay, I can find any two other vectors, which also add up to zero, and this is so called Cooper channel. But the total current in the initial state is equal to zero. Total current in the final state is also equal to zero, so nothing happened. The other possibility I can do is swap of the momentum. K goes into P prime, P goes into K prime, but again, I will have a compensation between the various terms of this velocity imbalance. And finally, I can do nothing, which is forward scattering, so the initial and final states are kind side, but obviously this doesn't lead to relaxation. So the only processes which I can have on a two-dimensional thermosurface on a circle, at least, do not lead to current relaxation. So why so few choices? Well, because a circle is convex. And as any convex object, it can have only two self-intersection points. But then I will claim that the same is true, not only for a circle, but for any convex shape in 2D. So as long as my is my thermosurface is convex. I will have the same problem, and that means that my unpredictability will have zero in front of T squared term and the first non-managing term will be T4. So this is actually the long story which goes back to the name of Guruji in 1980s and then various groups of people ran into the same paradox. Okay, there is an easy way out of this. First of all, we can go to 3D. And in 3D, as long as we step away from a sphere, the manifold of the intersection between two thermosurfaces is aligned, not a set of points. And so in this sense, the system is always integrable. We can go multi-band as we did before. And again, if I move two circles simultaneously, I will have more than two intersection points. So even if I say that my band, my thermosurface in each band is a circle, two circles already take me out of this. Or I can look at a concave thermosurface in two dimensions and a concave figure has more than two tangents. That means more than two self-intersection points, right? So if we have a 10-band model, if it doesn't have a T prime term, then the correction to the resistivity is going to be T4. If it has a T prime term and T prime is larger than the certain value, then I'll recover T squared. All right. Now I'm going to do maybe a couple of formulas on the blank board. This is about Wiedemann-Franc's law, which is usually considered as a hallmark of the Fermi liquid behavior. In fact, it is not. It is a hallmark, but yes. Right, so without unclubs, yeah, without unclubs. Oh, the previous statements were without unclubs. What I'm going to say now about Wiedemann-Franc's, it doesn't matter. What it is really a hallmark of is elastic scattering. Right, so namely, if you say that you are collision integral and I will not even specify what I'm interacting with, contains a combination of distribution function, which are correspond to elastic scattering process, which means that I have here a delta function which conserves energy. Here is some scattering probability. This is a sum over k prime. So with this collision integral, if I solve two problems, I will solve for the charge conductivity and I will solve for the thermal conductivity. Then I will find that there is a relation between any a component of the conductivity tensor. It can be on the lattice. It doesn't have to be in a gradient invariant case. And the, sorry, I need to take thermal conductivity. Divided by temperature, divide by the charge, arc conductivity and there is universal number multiplied by a ratio of fundamental constants. Now, when does it work? I invite you to think about history for a second and recall the original date on Wiedemann-Franc's law. Date is actually 1853. That's when this law was discovered experimentally. Now think about cooling techniques in 1853. No dilution fridge, no de-mag, no nitrogen. You have room and you have ice. That's it. Actually, there's a table in Ashkirchen-Mürmen which gives you the value of what is called at the Lorentz ratio and for two metals and at room and ice, right? Why does it work? We said elastic scattering or iso-elastic scattering. Usually elastic scattering is associated with scattering at impurities. But this is not all. There is one range of temperatures where in the conventional metal, scattering is also elastic. So let's say this is my aluminum, right? Which is not a superconducting, women copper beta, which is not oscillating at all. And then there are two regimes in here. One, when you have disorder controlling your resistivity. And this is definitely the case where the women France should hold and that does, as we learned actually only a fairly recently. But there is another range when the resistivity scales linearly with temperature. And at the very beginning of the first lecture, I told you that this is coming from Reckon-Fornen scattering, right? Women France worked, roughly speaking here, back in 1853. So this is scattering at Fonens outcome. It mimics elastic scattering. Well, it does because if you think about temperatures larger than at the Dubai temperature, the wave numbers of Fonens, which are being excited, are on the scale of the reciprocal of lattice spacing. So that's very short range. In the good metal, this is of the same order as your Fermi momentum, which means that every our collision of an electron with a phonon, with this wavelength is very effective in changing at the momentum of an electron, right? Energy-wise, it is a different story. Imagine that this is my Fermi energy. And I have an electron which typically is at energy T, RKB is equal to one, either above or below a Fermi energy. So now I want to bring this electron down back to the Fermi energy, and I can only use phonons. But the maximum energy of the phonon in proper units is at the Dubai temperature. That means that I can only change the energy of electron by a small fraction. Then I have to do it in many steps. I will go down, and then maybe I'll go up. Down and up, down and up, down and up. I will have a diffusion process along the energy axis. And takes a long time, right? Which means that in this case, the energy, the rate of energy relaxation is going to be much, the time of energy relaxation is going to be much longer than the time of momentum relaxation. And in this sense, the scattering is quasi-elastic. So that's why we recover Wiedemann-Franc's here at high temperature and also at low temperature. In between, where we have let's say T squared plus T5, a T5 scaling of resistivity, Wiedemann-Franc's is violated for a trivial reason because scattering is inelastic, right? So violation of Wiedemann-Franc does not mean that we have a non-thermaliquid. I will just show you our plot of Wiedemann-Franc's a violation in copper, right? Likewise, the system which obeys Wiedemann-Franc at very low temperatures and we'll also show an example does not necessarily mean that you have a thermal liquid at finite temperature. All it means that you have elastic scattering. All right, so this is data for copper, but actually it's from fairly recent experiment, at least on a historical scale. This is Louis Tellefer's group. And you need to develop very sensitive probes to measure thermal conductivity at low temperatures and the way it is done in this experiment as well as many other experiments coming from the same group that instead of measuring chi sigma xx, that is conductivity along the x-axis when at the current and at the field I applied in the same direction and the same for the thermal conductivity when the temperature gradient and the thermal current I applied in the same direction, you actually measure sigma xy and the whole analog of the charge conductivity which is whole conductivity, whole thermal conductivity, why? Or because in this way you get rid of phonons. Phonons also conduct heat but they don't couple to the magnetic field. And so if you do a transverse conductivity for the charge and for heat, then you get much better sensitivity. So this is copper, just plain old copper. This is as low as the end. You see, so pi squared over three in proper unions with KB and E equal to one is the what is called Zimmerfeld value of this ratio. And you see that you do recover, you do get somewhere in the vicinity of the inertial bound. You also see that you are closer to the inertial bound in Lxy than in Lxx, L without index is Lxx. Then the Lorentz ratio, the other ratio of thermal and charge conductivity drops down, you have a minimum. And then it goes back up. Okay, so this evaluation is a trivial evaluation due to elasticity of the process. Now, this is a system and it shows markedly non-fermal liquid behavior. This is one of the strands in ruthonite and there are many cousins in this family. So this particular cousin has a magnetic field driven quantum critical point, which is at the critical field of 7.9 Tesla. The nature of this point is not quite known, but it is suspected to be of the pneumatic origin. And the telling feature of the non-fermal liquid behavior is the linear scaling of the resistivity at this field. Okay, now you measure it as it was done also in Louis-Italiafer's lab, you measure thermal conductivity and you look at the Friedman-Franc's ratio. And at low temperatures, you recover it with very high accuracy. And then you see a minimum, which is not unlike the minimum in copper. So this is a trivial evaluation with Friedman-Franc. But it tells you that scattering is inelastic. But in this case, the statement is non-trivial. The number or kind of order zero explanation of linear scaling would be that instead of phonons, we have some other kind of bosons. And these bosons have the typical energy solo that temperature is larger than this frequency, right? Then it doesn't matter what the nature of these bosons is, the number is going to be proportional to temperature. Okay, so for that there is some dark mode, we don't know what. And this dark mode has very low effective adiabatic energy. And then by a trivial argument, the linear anti resistivity is nothing more than it's the same as in copper is non-aluminum, but just push down the energy scale. Okay, the fact that you have a violation of Friedman-Franc's by trivial reasons in the same temperature range where you have linear anti resistivity tells you that linear anti resistivity is not coming from quasi-elastic scattering. So the simple-minded boson model doesn't work. I wish we had more data like that, but as far as I know, it was checked only twice. And this is another example. This is YBCO, a high-temperature superconductor. Actually, that's the same paper in which copper was measured, just to set an example. You go here in temperatures from TC, which is 90 Kelvin, or slightly above, to room temperature. Don't bother looking on the left, just look on the right. So this is the Lorentz ratio. The universal value in this unit is supposed to be around three, pi squared over three. And you see that the Lorentz ratio is significantly below the universal limit. So that's the same story as our before. You see suppression of Lorentz ratio in the same temperature range where resistivity is linear with temperature. Then you know that your scattering is inelastic. I forget about boson, but some with some characteristically small energy. Yeah, so this is optimally doped. Yeah, 90 Kelvin in YBCO is optimally doped. Yeah, I wish to have more data like this, overdoped, other compounds, but that's all we have because thermal conductivity is very difficult. And essentially, there are only a few groups in the world which can measure it. All right, so basically then I'm done. Well, I've already stepped into non-fermal liquids. Let me tell you a story which is in the category of a system. Question. Sure. Yeah, just to finish, to go back on what you just said. So if you have a quantum critical point, is it expected that even at the, exactly at the quantum critical point, you will recover at equal zero the Wiedemann-Frenzler? Or is there no theory of that? Experimentally what is shown, I mean. There is really no theory for that. But I wouldn't be really surprised that it does not evaluate Wiedemann-Frenzler because after all, what we're probing is elasticity of scattering. And no matter what, scattering is going to be elastic. Having said that, there is one example of the evaluation of Wiedemann-Frenzler. I don't even measure it, but I think theorists have no doubts that this is true. Imagine that instead of a bulk system, which we're talking about here, you have a ballistic quantum wire. Let's say with one channel. The conductance, the charge conductance of this wire is two square root of H, regardless of the interaction in other wire. Okay, now if you look at the thermal conductance, which would measure by producing the temperature gradient, then the thermal conductance will be also a universal number times some function of the interaction inside the wire. So now if you take the ratio, it cannot be universal. But this is a special case where scattering happens at the boundaries of the system rather than in the bulk. There is really no disorder in the system, so it's not special. All right, so now what I wanted to tell you is a story. So a system is a metal near a quantum critical point by formal definition of a non-fermal liquid in the system in which the imaginary part of the self-energy is larger than the real part of the self-energy. It is a non-fermal liquid. At least that's how I understand it. However, it maintains what is called coherent transport. That is, there is some long time of relaxation which is, let's say, longer than at the plank time. So I did not destroy coherence completely. So what I have in mind, two systems, there are more, but I will just show you two slides. One is that's something which I already showed you at the very beginning. This is nickel palladium. So we have almost a ferromagnet, which is our palladium and a little bit of nickel, which is magnetic and the system happily goes ferromagnetic at some critical concentration of nickel. The non-fermal liquidness comes about in two measurements. One is you measure the electron part of the specific heat. So delta CO over T, which is supposed to be a constant in the fermal liquid. And instead, it shows an upturn. And the upturn is most pronounced at the critical concentration of 2.6%. And you can fit it into a log. For reasons I will tell you later, but the simplest version of the theory for the system tells you that specific heat CO over T has to diverge up logarithmically. At the same time, you measure other resistivity. And again, at the critical concentration, which is, okay, so the circles, right. So the circles is the same 2.6% the resistivity minus the residual value scales as T to 5 thirds. And as you step away, and you step away actually in this case, below this point and above this point, so you go ferromagnetic and you go paramagnetic and it doesn't scale as 5 thirds. Then there are good reasons why it has to be 5 thirds. Now, you don't see from the plot here how large is the vertical variation in the resistivity. And this is because the residual value was subtracted. And this was not published. But you can get the thesis, which others paper was a part of online. And you actually find that the variation in resistivity along the vertical axis is 10%. Not too much. Although palladium is a compensated cement. So in principle, it might have a variation of the resistivity due to electron interaction as high as possible. So there is another example, which I discussed a lot with Gil, and actually another lecture from today, or HIDA is also on this paper. This is kind of a poster child of ferromagnetic or quantum acryticality. It's a zirconium zinc II, which is what is called a weak ferromagnet. And the weak ferromagnet is a ferromagnet with low Tc. So at ambient pressure, Tc is around 30 Kelvin. And then you bring it down by applying pressure. So this is your quantum critical point. The resistivity of acryconium zinc II scales again as power of five thirds. So this is a fit to five thirds. But in this case, you see the whole data and you see that the vertical scale of the variation of the resistivity is large. Why I keep are coming back to the vertical scale because it is supposed to be a quantum acrytical point into a phase in which the other parameter is uniform. Which means that electrons interact with long wave lengths of fluctuations of the incident or the parameter. And by the second part of anti-unclub argument, unclubs are suppressed. Right? Then I have two choices. I can either say that a different surface is in the tropic. In this case, I will have a variation between two limits. And I wouldn't expect to have a strong a variation or I have a compensated semi-metal. And we started to talk about this material with Gil a year ago in our Santa Barbara. And I said, well, it would be so easy to understand. If this material were a compensated semi-metal. And Gil said, oh, but it is. So kind of unbeknown to us, this is a compensated semi-metal. Okay. So this is the good part of the story, which I think I will understand. The part which we don't understand is that it is somewhat too good to be true because the non-formal liquid scaling is seen not only at the quantum critical point, but also all over the phase diagram. So this is an exponent of resistivity. At ambient pressure, it's already 1.6, 5 thirds is roughly 1.6. So one has to say that the quantum critical R behavior is extremely strong. So that affects even the ordered part. And then there is another interesting puzzle. As soon as you go through the quantum phase transition at pressures which are slightly above 20 kilowatts, the exponent in resistivity drops. Not dramatically, but noticeably. And the feed is from 5 thirds in here to three halves in here. So three halves is an exponent which also shows up in some quantum critical theories. But I think we don't know yet how to relate one to the other. All right, so now let me tell you the justification of 5 thirds. Oh, sorry, before I forgot, and certainly he would not let me get away with this, this is actually even better than this because not only the charge, but also the thermal conductivity was measured. And if you convert the thermal conductivity and what is called thermal resistivity, and so what you need to do, you need to take the conductivity, sorry, thermal conductivity, one of these, multiply by temperature, this is called W alpha beta, this is thermal resistivity. Okay, so the purpose is that you want to get rid of the effector of tau, or of T, which you have in thermal conductivity in a comparison to the charged one. So you take it out, and then you convert tows, which you have in the conductivity, so we into one of those, which you have in other resistivities. And thermal resistivity scales linearly with temperature. Whereas in the thermal liquid, we'd expect it to scale quadratically. Okay, so what are the main features to explain? I'm not going to touch now on specific heat. I will try to tell you why resistivity scales as T to 5 thirds, while thermal conductivity scales as T. By the way, T scaling is actually what is called marginal thermal liquid. And a standard route, to this physics, is through self-energies. One calculates the self-energy, finds non-thermal liquid behavior, et cetera, et cetera. But because we're doing here everything in the Boltzmann language, I will bypass self-energies and diagrams, and I will go immediately to the Boltzmann equation. So this is my phase diagram. Thoromagnet, this is paramagnet, our quantum critical point, our control parameter. And as is usually assumed in this game, I will take the susceptibility of the divergent error fluctuations, which in this case are thermo-magnetism, in the Orenstein-Zernicki term. Qc is nothing more than the inverse accrylation lengths. Now the formula starts to look as if they're written by Andrey Achebukov, we're getting into the same domain. So that would be the susceptibility of magnetic degrees of freedom in the system, but they're coupled to electrons, and electrons provide damping. Damping is a term which is proportional to the ratio of the frequency to the wave number. This is this lambda of damping, and there is some coupling constant gamma. At the quantum critical point, psi minus one is equal to infinity, OQc is equal to zero. So there is a very important line in this phase diagram, which is not a phase transition line, but it is a crossover. The meaning of this crossover is that if I am at low enough energies or below this line, I can forget about dynamical term in the conductivity. Low temperature, low energy scale, I'm going to low energy or energy scales here. This term is irrelevant. Well, and now I have system of fermions which interact via some potential, which actually is precisely screen cooling potential in 3D. It is a ephemeric liquid, right? So in here, I will have a ephemeric liquid phase where dynamics of the acrylator of fluctuations is not important. When I go to high enough energy scale, I can forget about final deviation from the quantum critical point. Everything is quantum critical. I can forget about Qc here, put it to zero. And then now you will see why this model is called z equal to three acryticality because you have certain scaling between the wave number, which is the inverse at inverse distance and the frequency which inverse time. And because you have Qc versus frequency scaling, this is called z equal to three acryticality, okay? And I'm going to do a D3 case. D equal to three, z equal to three, okay? This line is obtained by our comparing the dynamic term with the static one. So when omega gamma divided by the characteristic wave number which is the inverse acrylation length is comparable to Q estimated at Qc, this is where your acrossover happens, okay? Now, suppose that I'm here. I am in a Fermi liquid. And as long as I'm in a Fermi liquid, I will use the Boltzmann equation. So the Boltzmann equation is going to be of the same variety which we looked at before. The scattering probability, so this is the scattering amplitude is the anti-symmetrize sum of two diagrams. Here my effective interaction potential is a function of Q. Here it is a function of K minus P plus Q. The scattering probability, which enters the Boltzmann equation, this big W, is mod squared of this, okay? Because I have long range interaction, the exchange diagram which contains the interaction potential at roughly 2KF is small. I can forget about it. And so my interaction is just the square of, my probability is just the square of the effective interaction, which up to a coupling constant between electrons and spin fluctuations is just the square of the spin susceptibility. And because I am in a Fermi liquid domain, I can take this susceptibility to be finite. So this is my interaction. Well, also because I want to have a model in which I can talk about charge conductivity and thermal conductivity simultaneously in a meaningful way, I will break the Galilean invariance by saying that the system is semi-metal compensated one, but that's what it actually is, right? So I will have in mind a very simple model of a compensated semi-metal with Fermi energy cutting in between. The masses may be different, but the number density is the same. Then I will have a system of Boltzmann equation, band number one, band number two. If I want to find the electric conductivity, I will apply electric field and I will have usual term on the left-hand side for one band and there will be two collision integrals. One is within one band and the other is between band one and two. Likewise, band number two will have interband plus interband. Notice that I use the same sign of charge, why? It is a compensated semi-metal. I have one whole band, one electron band, and yet I'm using the same charge on the left-hand side of the Boltzmann equation. Sorry? So holes in electrons here differ by the sign of the curvature of the mass. The charge is a physical charge. When we form a drudy model, while we don't know anything about band structure, we flip the sign of the charge. But if we have a more or less realistic band structure, we don't need to worry about the sign of the charge. It will be taken care of automatically by specifying other dispersion. Now, this is really fun to solve the system and it's even fun to solve a similar equation for one band case and just to kind of map out the complexity of the problem, which was first formulated by Abracosov and Arhalotnikov in 1959 when they parameterized at the collision integral and then was solved by Bucher and Sykes in 1968. The problem is produced to an effective Schrodinger equation, second order, with a potential, which is now famous supersymmetric potential, which is one over Cauch's squared. So the solutions of this Boltzmann equation are the eigenfunctions of this effective Schrodinger equations which are associated with gender polynomials. Okay, why one over Cauch's squared? Well, this is the formula which you all know. If I take a Fermi function and I take derivative of the Fermi function with respect to energy axis, is my energy, you get one over four Cauch's. Cauch's squared, so once you have the Fermi functions, they played the role of the effective R potential. Now, it wouldn't be, I wouldn't even there to show you how the solution works, but there's a very simple case which one can understand without going into associated gender polynomials and that is the case of long range interaction. So what long range means is that energy is being relaxed quickly while it takes some time to relax at the moment. So suppose that this is my Fermi surface. I take an electron in here and there are two steps in this process. First, it goes down to the Fermi surface in a time which is energy of relaxation. That's fast, right? And then it starts to diffuse over the Fermi surface each time changing the direction of its motion only slightly, that's long, right? Okay, so in general, I can claim that if I want to find the distribution function in each of the bands, F12, okay, I will start with an equilibrium value in here and then I will write a R correction as something which is proportional in this case to the electric field because it is electric field which takes me out of equilibrium but because it is a vector and they need a scalar, I will multiply it by a velocity. I can take out a factor of charge if I want. Well, then I would like to multiply it by the Fermi function, the derivative of in the same band and this again totally arbitrary times some function of the energy of the electron. And this supersymmetric solution of the Boltzmann equation is an equation on this function but because in this particular case which is sub case of a general problem, energy is being relaxed quickly, this function does not depend on energy. It's a constant, okay? So now I plug it back, I have an algebraic system for two numbers, G1 and G2, so a little bit. So the resistivity is T squared because I said that I'm still in a Fermi liquid and there would be certain integral of the scattering probability W which is a function of momentum transfer with a weight equal to RQ squared and they will have some pre-factor here which will contain masses, et cetera, et cetera. Okay, now if I want to solve the thermal conductivity problem, I will apply temperature gradient, right? And I will say that my distribution function which enters the left-hand side of the Boltzmann equation is an equilibrium one. That's a Fermi function epsilon minus mu over T but T is now a function of R. So V, the del dot T term in the left-hand side, the F term in the left-hand side of the Boltzmann equation will be replaced by V del T, the derivative of the Fermi function, factor of epsilon minus mu over T squared. And again, I will parameterize my solution. It's something which starts with an equilibrium one. It is proportional in this case to other gradient of temperature dotted into other velocity. We'll take out the distribution function and there is some other two functions. Well, let's call them, I don't know, H one and two, which I gain in general a function of energy but because of the argument about slow momentum of relaxation, fast energy relaxation, this is another set of two constants. Plug it back into the Boltzmann equation, solve two way to assist. Okay, in this case, I will find that the thermal resistivity, which is temperature over kappa, is again T squared with some other number, C bar, multiplied by an integral of the scattering probability without this extra factor of active squared. All right, so this factor is there for a good reason. Let me remind you that if you are solving even a simple problem, motion of electrons in a fixed quenched disorder, there are two times which one can think of. The other one, the first time, it's also sometimes called a quantum time or single particle time. This is a time which takes into account all collisions, no matter how effective they are in momentum of relaxation. So in classical terms, it is proportional to the total cross-section. So this sigma is a cross-section rather than the conductivity. But the conductivity is going to be determined by transport time, which discriminates against small angle scattering events and it does so while this filtering factor, which is one minus cos theta, so if the scattering angle is small, the corresponding factor is also small. So for small angles, for small angle scattering, this factor is theta squared over two. Well, but this is related to a change in at the momentum. If you are changing the momentum from K to K, this is Q. Q is twice K times sine of the scattering angle over two. For a small angle, this is K over sine K times theta. So this angle is proportional to the change in at the momentum squared. Okay, so that's the origin of this factor here. Whereas when we talk about thermal conductivity, any scattering would do. We don't need to go back to the other side of a different surface, so I don't have this filtering factor. Okay, so this is all good. Now, I wanna crawl to a non-fermal liquid case. Crawl means that I solve the problem here and because my scattering probability is parameterized by at the correlation lengths, I will take a limit, not a mathematical sense, but in physical sense of the correlation lengths going to infinity. Okay, let me first do it in the thermal liquid case when I need to substitute my scattering probability. My scattering probability, which I just erased, that's unfortunate. My scattering probability is some constant Q squared plus Qc squared all squared. Okay, now I just need to power count at this integral. I have three factors of the wave number divided by four wave numbers, so the resistivity is proportional to T squared divided by one factor of the critical wave number. If I do the same for the thermal resistivity, I don't have a factor of a Q squared, so instead of Qc, I will have Qc cubed, okay? That's all I can regularly say being within the thermal liquid. But now I will claim that really without any good reason that I will reach at a quantum critical point from here by recalling that I have z equal to three a criticality in which the wave number scales at the third power on the energy scale, one third power of the energy scale. And I'm going to apply it to the critical wave number and I'm going to replace this frequency by temperature. So when I'm getting very close to this point, I will claim that there will be some dynamically generated accrylation lengths. As long as I'm at finite temperature, there will be some thermal disorder in the system and it is this finite temperature which will determine at a scale of a decay of our fluctuations. Well, now if my Qc scales, as t to minus one third, the resistivity scales as t squared divided by t to one thirds, and that's t to five thirds. And thermal resistivity is t squared divided by t to one third, but it's all cubed. It's t squared over t, five thirds scaling of charge resistivity, linear scaling of thermal conductivity. And we need it, of course, in a hand waving way as far as a crossover to the non-former liquid point is a concern, but at least we obeyed all conservation law. We didn't do what is often done in the field when you calculate the self-energy. And you say, well, let me multiply the self-energy by a typical Q squared, that will be my transport time, that will go into the resistivity, that will go into thermal resistivity. We actually solved the Boltzmann equation. If you don't like my way to going through at the quantum critical point, I will make a statement, okay? I cannot prove it in the time which I have. Actually, this problem allows the Boltzmann equation, even in its non-former liquid phase. There's something which is called Prandtl-Caddonov argument, may have heard of it. Okay, it actually was cooked up for phonons, but this pin R fluctuation sign, one sense similar to phonons, in that they're slow. Okay, for slow bosons, there's something which is called Mieckdel theorem, which tells you that you can really resound the perturbation theory and neglecting vertex R corrections. Once you can forget about vertex R corrections, you can actually derive at the Boltzmann equation in the Kildesh R-Technik, and it will have the same form as we're using in the thermally. How I'm doing on time? When did we start? The minutes, okay. So this is DC. I wanted also to talk about optical conductivity. In the context in which I was going to talk about it, is if we go back to the Drude formula, just for a second, but this time is going to be the Drude formula at finite frequency. Well, let's start with our MA equal to fourth equation. My field now oscillates in time, so will my R velocity, and if I solve this, I will find that the conductivity, which is now a function of the frequency, can be written as something which is called plasma frequency squared divided by four pi, and then there will be a denominator, one over tau minus i times the frequency, so that when we go back to the static case, omega is zero, this is the Drude conductivity. Now, I'm going to be looking at the limit when omega is larger than whatever tau we want to bring in. So this is called collisionless, or dissipationless up-regime, right? But I still have a small but finite real part of the conductivity because I'm going to expand in one over tau, one over omega tau. If I do this and I keep the main term, sigma prime is going to my mega p squared over four pi, and then there is a mega squared tau. Now, suppose that we have a Fermi liquid, in the Fermi liquid, up loosely speaking, I'm now speaking very loosely, one over tau can be associated with the self-energy to some extent, and we know how the self-energy scales with frequency and temperature, it does so in this way, so there is almost total symmetry between a frequency and temperature, except for some of you who went through this R calculation, the self-energy as a term, which is not simply R temperature, but temperature times pi. You may recall that pi t is the first fermionic motor body frequency. This is other decay rate of single fermions, so it is natural that R temperature enters as the fermionic motor body frequency, whereas the effective time for the optical conductivity is going to be two pi times t, and two pi t is the first non-zero bosonic frequency, well, we're talking about current up relaxation, so it better knows about bosonic degrees of freedom. And if I send that temperature is equal to zero, which means that frequencies high are compared to everything, then in the Fermi-Liquid regime, you will have a pre-factor, which is a constant, and then there will be a cancellation of two frequency dependent terms, one is omega squared, which is coming from the formula, and the other one is coming from one or two, so it's a constant. So high frequencies are compared to temperature, the real part of the conductivity of a Fermi-Liquid at finite frequencies is a constant. Whereas many systems show, let's say, cuprates, show that the conductivity is actually a decreasing function of other frequency, and it depends on the group on who you talk to, but this is always omega to minus alpha, and alpha is less or equal than one, but let's say of larger vehicle than 0.5. So there's something interesting going on. There are models which allow you to find, to come up with a non-fermal liquid scaling of the conductivity, and one of them being the same model which I talked before, a ferromagnetic or a quantum critical point. However, that's not the other subject which I wanted to discuss. The subject is somewhat more subtle. When we think about temperature dependence of the conductivity or frequency dependence of the conductivity almost by default, we tend to assign, yes, I know, I wouldn't say that it will apply. Although actually if you talk to Patrick Lee, he believes that there is an animatic quantum critical point in the recuperate. I can come up with a model for some non-fermal liquid which will have a fractional power there. I can also come up with a model for the recuperate where the exponent is going to be one. We always tend to associate the temperature and the frequency dependences of the conductivity with other scattering rate. But it is also effective mass in the root of the formula. Effective mass is abnormalized by the interactions. And it is natural to expect then when you move to a quantum critical point, the effective mass acquires strong temperature dependence. So in what I showed you about a nickel palladium adopted with nickel, silver T, which is proportional to the effective mass over the bare mass was scaling as log of one over T. So this is this indication that the effective mass diverges towards the quantum critical point. And then if I want to continue, I need to combine the temperature dependences which are coming from the effective mass and temperature dependence of the conductivity. And then my wonderful quote-unquote an explanation of the experiment would fail, right? So the question is which mass? What do we stick into the root of the formula when we're near at the quantum critical point? And as far as I can tell, there is no unique answer. That is, it depends on the system. But one answer one can get a rigorously is to go back to the quantum critical point we discussed here. Now I'm going to be, I'm going to be doing it at finite frequency. Of course, I'm not going to solve anything. I will just formulate other paradox and I will tell you the answer. By the same argument or roughly the same argument as I have here, I can tell you that one over tau is a function of omega. And I'm going to specify now to 2D because this agreement between two ways of looking at the same formula is more pronounced. It is omega to the power of two thirds. Okay, so you can do your math. You will have one over omega squared, one over omega two thirds. The conductivity will then scale as omega two thirds. And this is actually at the result which is known for many, many years. At least theoretically, unfortunately, no one has done optics on this kind of system. So if you think that we need to do optics or zirconium thing too, okay? But then if you try to derive it in a very naive way starting from a single bubble for other conductivity, we immediately find that every frequency which enters into the formula is being divided, is being divided by what is called the fermion optimization factor or quasi-particle residue which is up related to the mass is that z in the systems is just the inverse mass, right? If I power count this with factors of z, factor of z will get in here. Now because my mass is divergent, and it's also divergent as a function of the frequency, mass is a function of the frequency blows up as omega to one thirds. If I keep this in, it will cancel the frequency dependence whatsoever. You will have conductivity as in a fermi liquid which is independent of the frequency. So what's the right answer? The right answer that one has to be very careful. There are vortex or corrections of this type which are related to the defective mass or to the quasi-particle residue by the word identity. And each green's function in this diagram brings you factor of z because that's what it is proportional to but each vortex our correction gives you factor of one o z. So there is no z in this formula. That means that actually bare mass enters the conductivity, right? Ask me about spin density wave. And with this, I will be done. Thank you.