 Thank you very much. Welcome back. I hope you all had a great weekend. I hope you didn't forget everything I told you about how this works. Nevertheless, let me remind you what the setup is and then today I want to come really to the theorem, right? I haven't stated the big theorem yet. The theorem that says that Ginsberg-Landau theory actually can be arrived from this PCS model. Okay, let me remind you also this PCS model. That's a minimization problem. You have somehow, I don't know, learned in high school how to minimize when you have numbers. Then in functional analysis, you learn to minimize over functions and now what you do here is we minimize over operators. Okay, that's perhaps a little bit unusual. You might not have seen this before, but conceptually there is nothing really complicated going on. Okay, so the operators that we're going to minimize over, they're denoted by capital gamma and this so-called free energy functional. Which depends on the temperature T. There, well, has three parts. The first part, so I apologize. I mean, as last time there will be a script H and a usual H. The usual H is the semi-classical parameter. The script H is the one body operator whose form I remind you in a second. Okay, so I should also remind you about the capital gamma. Capital gamma is, as I said, an operator. It's an operator on two copies of L2 of RD and we usually write these operators in block form. And so there's the little gamma is the 1, 1 entry. The alpha is the 1, 2 entry. And the other, I mean, this is because of this formalism. This is alpha bar, which is the same as alpha star, and there's a 1 minus gamma bar over there. Gamma bar denotes the operator conjugated by complex conjugation. All right, so the first part of the functional just couples the gamma, this is the 1, 1 entry, to this one body operator. Good. The third part in the functional involves an integral over an interaction potential, which we have scaled here. So this is x minus y over h. And now here's alpha of x and y squared dx dy. So remember alpha is an operator on L2 of RD and alpha of x, y denotes its integral kernel. This will be Hilbert-Schmidt operator and so there's no problem about the existence of an integral kernel. And I remind you also that these operators, the gamma and the alpha, they're periodic operators on L2 of RD. And so therefore here this integral is an integral over RD times the torus, TD, and this trace is interpreted in the sense per unit volume. Okay, so here these are the two terms, these are the particles, the ordinary particles, say electrons, and this involves the so-called Cooper pair function. And now here's a coupling between those two and that's given by the trace of this gadget here. Gamma log gamma and in front we write a T, which is the temperature, okay? And then let me remind you just to make the formulas appear a little bit nicer. I multiply by h to the power d, where d is the dimension. And now this object, I also have to remind you the script h, that is minus i h nabla plus h a squared plus h squared w minus mu. Okay, this a is a vector field and w is a real-valued function. And so I remind you that there are three parameters in the game, which are all coupled. The first one is this h here. That is the quotient between the microscopic scale, that's the scale on which the electron, so your particle live, and the scale on which the external fields, the a and the w, live. Okay, and we will take a limit where somehow the electrons they interact in very very small scales, but we watch the system on a scale of size one. Okay, that's the first h. The other h is the one that I wrote in front of the a and in front of the w. That means that the external fields that we apply are weak. And it turns out, and I tried to motivate this, that the right thing to do is to have them of size h squared to couple these things. And then the third thing that's in the game is the temperature. The temperature has got to be close to the critical temperature. Which critical temperature? Well, the one that we computed in the very first lecture in by the translation variant model. Okay, and yet again this difference, a difference between the temperature to the critical temperature, that also should be of size h squared. Okay, so all these three parameters that are a priori in the game are coupled in such a way that one one arrives at a nice limit. One can of course do the limit in all the other cases. They will just be easier or trivial. Okay, and then last time we also computed the energy of the normal state. Remember normal state means that we just think that alpha is equal to zero. So this term is absent. This is a diagonal matrix and then this is a little computation. Okay, and now let me make an assumption that will be important for all the following. And namely, I will assume, well, first of all, Tc is greater than zero. Tc is defined in the first lecture and I'll remind you in a second what this is. And remember there was this operator k Tc, okay, T was equal minus Laplacian minus mu divided by Tunch minus Laplacian minus mu divided by 2t. Okay, so that's a Fourier multiplier. It's a Fourier multiplier that depends monotonously on t and the Tc was just defined as that point of t where the eigenvalue crosses zero. And so we assume that this happens at some positive Tc, okay? For the first eigenvalue crosses zero. So I'm assuming that this kernel here is one-dimensional. And I'll denote the function in the kernel by a star and let's say we normalize the function to be one. I told you that this is not always satisfied, but it turns out that in most physically relevant cases it is satisfied. I mean, this is now really, you plug in numbers and you compare with temperatures that you have in experiments and then it turns out to be verified. There's no deeper theoretical meaning. Without this assumption, I think things would become much more complicated. Okay, so let me define now two things, namely the critical temperatures. And we've got to be a little bit more careful because in the translation variant model, we've seen there's one critical temperature below which you have superconductivity above you don't. Now, that doesn't have to be like this in the case for this more general model. So we have to be careful what we mean by critical temperatures. And really there are two critical temperatures. They're really two natural candidates. Let me write those down. So there's a Tc lower, which is the smaller of the two. And that's the one such that below you always have superconductivity. Okay, see the problem is you might have for small temperatures always superconductivity, then you're in a kind of transition regime where you have superconductivity, then not, then yes, then not. And then eventually you come to a regime where you do not have a superconductivity anymore. So therefore we have to be more careful use these upper and lower critical temperatures. So that's the largest temperature such that below, right, so below, so I have a ft prime, below that I always have superconductivity for all t prime less than t. And then we also have an upper critical temperature. You can imagine what that will be. That's the one above which you never have superconductivity. Okay, so it's a smallest t and actually in a very strong sense. So whenever you take a state which is not the normal state, then we require that the free energy is strictly greater than the energy of the normal state and for all t prime greater than t. Okay, so that's just, I mean, if you don't know that you have a unique point where things change, then you have to define a smaller one and a larger one. But we'll see in a second that that distinction is actually not really needed for practical purposes. And this is the first of two important theorems here. So, well, so these temperatures of course depend on these parameters h. And what this first theorem says is, makes mathematically rigorous, is this fact that you really want to change the temperature by order h squared in order to see something. So this theorem says, let's look at tc, say the lower one. And tc is the one that's defined in the translation variant model. Then, well, not only does the tc lower converge to the tc, but we can also say what the deviation is, namely the deviation is proportional to h squared and given by a constant that we know. And moreover, the same is true for the h upper and the deviation is the same. Now, okay, so where dc is equal to 1 over lambda 2 in spec minus i nabla plus 2a star lambda 0 minus i nabla plus 2a plus lambda 1w. Okay, so this thing here, let's look at this thing here first. This thing is an operator in l2 of the torus with periodic boundary conditions. And it depends on certain numbers. There's a capital lambda 0, that's a matrix. It's a d by d matrix is positive definite. There's a lambda 1, which is a real number. And there's a lambda 2 that we've already seen in the first talk. It's exactly the one from the first talk where I wrote down some very complicated expression. I will not write down the expressions for lambda 0 and lambda 1, but it's exactly the same thing. So you look at this guy here, the alpha, the a star, and you take the square of this, or the square of the Fourier transform, you multiply it by some complicated function of xi and you do the integral. And that defines these numbers. How do we interpret this result? Well, this result says that, so you have a microscopic theory. The microscopic theory depends on mu and v. Out of that, the only thing you need to remember from the microscopic theory is this thing here. This operator, you have to compute its kernel and then you get this function a star. Once you have that, you can forget everything about what the electrons do. Then you can compute these numbers lambda 0, lambda 1, and lambda 2. And then, once you have done these three coefficients, so I mean this really matrix, but then you have these coefficients and now you do something macroscopic, right? This thing lives on the size of your system. And on that size of the system, you have to compute an eigenvalue. The landers do not depend on the external fields A and W, but here they enter. Okay? So this says that to leading order, the critical temperature coincide with the translation variant temperatures. But then the external fields shift these temperatures by an amount h squared and this amount h squared can be computed from a macroscopic model. Okay? So that theorem really says that if we want to understand something, I mean get a non-trivial limit, we better take some temperatures which are an order h squared away from Tc. So then if A is W or 0, then Tc is 0. Exactly, absolutely, right. And in general, you see somehow the magnetic field, usually by the time magnetic inequality raises the eigenvalue, say W is equal to 0, then this eigenvalue is really strictly positive. Okay? So which means then, remember there's a minus Tc, that the critical temperature is actually lowered due to a magnetic field. Okay? Lower to order h squared. Is lambda 2 also given by some formula? Yes, exactly, right, right, exactly, absolutely. I mean, yeah, I mean I could write them on the board, but I don't think there is much, yeah. Good. You see, I mean what this says is somehow we're not able to prove that the critical temperature is unique, but we know that it's almost unique. Somehow if there is a range where superconductivity disappears and then reappears and then disappears for good, say, then this range is a little o of h squared. Okay? So it's, and it would be something that could not be explained by Ginzburg-Lander theory. So we cannot rule this out. There's no general monotonicity argument, like in the translation variant case, but at least this is somehow such a small range that it's not important really. Okay. So that was the theorem of the critical temperatures. And now the other theorem, that's the theorem about Ginzburg-Lander theory. So I call this derivation of GL theory. So I'm still working under this assumption, right? See, the assumption is simply needed in order to have these landers well defined. If there would be several elements in the kernel, there would be several choices about these parameters. So here's this theorem. So now we take some arbitrary D and we look at temperatures which differ exactly by the amount h squared from Tc. And we, the proportionality constant, we denote by D. Okay? That's what this theorem, this theorem told us that this is the interesting range. This is where things might happen. So let's do this. Let's take the limit where temperature is coupled to the scale h. And then as h goes to zero, the free energy or the minimum of the free energy. Well, to leading order, of course, it's given by the normal state, right? Because we're close to the critical temperature where the superconductivity disappears. And then there's a correction. And yet again, it's an h to the 4 correction. And it's given by some number that can be determined by a variational problem. I'm writing here little o of h to the power 4. We have remained estimates. There would be some power. So I think for the upper bound is h to the power 5. And for the lower bound, it comes from some not so optimal things. But let's, let me just keep it like this. Where the ed is equal to the infimum of the Ginsberg-Lander functional. Okay. And let me write down what the Ginsberg-Lander functional is. So it's a Ginsberg-Lander theory on the torus. As I told you, we're not able to deal with boundary terms. And there is the gradient term. Okay. It's okay. Thanks. So it's like this term here, right? So there's a minus i nabla plus 2a psi squared plus a lambda 1w psi squared minus lambda 2d times psi squared plus lambda 3 psi 4. And these landers are the same as over there. And in particular, the lambda 2 and the lambda 3, they're exactly those given in the first lecture. In the first lecture, remember I told you I can derive the Mexican head shape of this thing. And so the new version of this term, when you break translation variants by these external fields, then you get these additional terms. So the lambda 0 and the lambda 1 are the new things here. I want to emphasize one thing, which is already up, appears already up there. When you look at the one particle operator in the definition of f t, this script age operator, right? You have a A, the external vector potential, is coupled with a one relative to the gradient. Over here, there's a two. That's a very important thing. So the physics explanation of this is somehow that this really describes pairs. It's a Cooper pair wave function. Okay. And so if you remember what you should write in front of the vector potential is the charge of the particle. And so if the charge of this thing couples to gamma, which means electrons, if they have charged one, then the pairs have charged two. So that explains the factor two. I know whether this really explains. I mean, it's rather the mathematics. You do the mathematics and you find this and then you have to interpret the two. And then you come up with that. This really describes pairs. So I don't know, I mean, you know, which is first. But it's certainly an important fact. And it's perhaps a little bit surprising because at first sight, it looks a little bit like it would somehow break gauge invariance. But it does not. I mean, it has to be. And it's really one of the fundamental things in the interpretation of this theory. Let me compare these two theorems. It's at least on a formal level. You could say that this theorem up there follows from this theorem. Why is that? That's because, let's talk about the minimal energy, the minimal Ginsberg-Landau energy. Now, when the quadratic part of the functional is positive, that's exactly when D is smaller than this Dc. Then certainly the nonlinear functional is positive because I'm adding something positive. And the three remember the positive number. Okay? Now, conversely, if I can make the linear part negative, well, then I multiply the psi by a tiny constant, the psi quarter to the power 4 becomes even tinier, so it doesn't matter and doesn't destroy the negativity. So this Dc is, so to speak, the critical temperature in Ginsberg-Landau theory. Okay? That's exactly where, somehow, where in Ginsberg-Landau theory, also you have a transition. Namely between a psi which is identically zero and a psi which is non-zero. The transition happens at this Dc, which is up there. And somehow, what this theorem up there says is that the change in the PCS critical temperature is given by this Ginsberg-Landau critical temperature. Okay? And so therefore often, I mean, this is usually interpreted as a temperature in Ginsberg-Landau theory, but really you should think of it as kind of this renormalized temperature where you somehow subtract the real thing and then pass to a limit. And that's what it is. Okay? That's how it comes about. Good. Are there any questions at this point about the statement of the theorem? Okay? Good. Then let me describe... Well, okay. No. Let me... Perhaps I should continue this thought. So, I mean, why are these two things related? So if D is such that this functional is positive, or I mean, this would mean that this Ed is equal to zero, you can always... This Ed is always less or equal than zero. So if this thing is equal to zero, right? Then it says that from this theorem we conclude that there is no superconductivity to leading order in H to the power 4. That, however, is a weaker statement than the one that I make up there because there I say there is really no superconductivity at all, to no order in H. Really, some quantity is really positive. Okay? There is no somehow smallness statement. Therefore, I mean, while they look similar, one really... The proofs are kind of orthogonal or similar. And similarly, this theorem somehow, in any case, only gives local information close to Tc. And for that theorem, one has to control completely different temperatures as well, and one has to find different arguments. So they're related, but they're really logically independent. So now let me tell you what we really do in this theorem, what we really prove. So I told you when one wants to prove such energy asymptotics, usually one does an upper bound and a lower bound. For the upper bound, one guesses a certain gamma and tries to make this thing as small as possible, namely as small as this. And then for the lower bound, one has to show that for any gamma, this thing is actually a lower bound. And we do this in a... We prove something more general, namely for the upper bound, we do the following. So for every psi in H1, but when I write H1 on the torus, I mean, it's a periodic function, right? Everything here is periodic. I'm simplifying things here a little bit because there are some high-frequency problems, but let me ignore those. So given a psi against Boglanda psi, we show there exists a gamma admissible for BCS, such that the BCS energy of this gamma, well, it's equal to the energy of the normal state, but then plus H to the power 4 E d of psi. Plus, and again I write little o of H to the 4, and by this I mean that this little o only depends on the M, okay? So for any psi, I can build up a BCS state which achieves such this energy inequality. And then of course, what do I want to do? I would want to take psi, the minimizing against Boglanda theory. But I want to say that you don't need to take the minimizer. This really holds for everyone. And then we have an inequality, namely the two leading order, alpha minus. So alpha is the 1, 2 entry of the gamma, and I'll tell you in a second what the alpha against Boglanda of psi is. And 2, 2, this is equal to little o of H squared, okay? So somehow this, I think I'm missing, this is 4. This thing is order H squared. No, no, sorry, no, it's because it's a little low. This thing alpha will be order H, just like in the translation variant case. Alpha will be order H. This is the leading term and that this statement says exactly it's the leading term. This thing is smaller than the remainder. And now the second statement that's a lower bound for every gamma, for which we know that its PCS energy is less than the PCS energy of the normal state plus m times H to the power 4. There exists psi such that, well, the norm of psi is equal to, well, again, such that I have this identity. And yeah, I mean, I'm writing down the same, really the exactly same things. Alpha minus alpha GL psi equal to little o of H squared and psi itself is order 1. Okay, now I should tell you really what the alpha against Boglanda is and I tell you what its integral kernel is. So it's of order H. There's a two because I'll write it as a mean of two things. And the first thing is psi of x 1 over H to the d A star. Remember A star is this guy in the kernel x minus y over H plus 1 over H to the d A star x minus y over H psi of y. Okay, so this thing I want to think of right A star is a fixed function. So this means if I look at this function at x minus y over H, this means that x and y are very close together. So instead of writing it psi of x plus psi of y over 2, I can write psi of x plus y over 2 1 over H to the d A star x minus y over H. And this thing you might recognize as the y quantization of the operator with symbol psi of x A star hat of x psi. Okay, so that depends on what you like. There are three things. This thing here, this is I think the important part, this form. So what this says is somehow that you have, you should look at the minimizer, actually at the all of gamma, separately in the relative coordinate and in the center of mass coordinate. In the relative coordinate, it lives on the scale H and it's given by exactly the same thing as in the translation variant case, right? So the formula from the first lecture was simply the one that psi is equal to zero. That was just this operator. Now in what happens when you turn on the external fields, well the alpha requires a modulation in the center of mass variable. This is a modulation which is of order one, both in size and in scale. And its shape is determined by the Ginzburg-Landau minimization problem. The proper way to think about this and also about what's going to follow is somewhere in terms of while quantization, okay? That somehow, so the idea is you have a function on face space. It's a function of x and psi. And now you want to make out of this an operator. And there are several ways of doing this. And the one that's most natural is actually the while quantization, which gives you, I mean if this is symmetric, it gives you a self-adjoint operator, okay? Now that being said, we can do and we can explain the whole theorem in terms of the while calculus, but we cannot prove it in terms of this. That's because of regularity issues that I try to avoid at this point. So therefore we do something like a poor man's quantization and that's the first line here. The first line just says, well take this. This is somehow, right? This is somehow an operator of the momentum variable. Multiply by function of x and then you do it because you want to have it self-adjoint. You do it the other way around, like this, okay? And so we have to work with this because later on in the proof there will be some some cutoffs in the size and somehow we can control this in this direct way much better. And I mean in the while calculus there's just such a high loss of derivatives. You need much more derivatives than you actually have for the final result that we cannot make this work. But I want to really present these three things to you, okay? And so what this statement here says is roughly speaking that there's a bijection between states which satisfies such an energy condition with states in H1 with a bounded H1 norm, okay? And so there's a bijection given for a psi. For every psi I can give you a gamma and conversely for every gamma I can extract a psi that does this job, okay? So and therefore it's a more general statement than just a statement about the ground state energy. I'm saying this because it was already in one of the first lectures it was somehow asked whether we can do similar things about equations, right? And so for equations you would probably like to prove something similar. Now probably H2 would be a better space and instead of some energy asymptotics you would have an almost solution of the, say, psi is a solution of the Ginsberg-Landau equation, you get an almost solution for the BCS equation and conversely, okay? But that is in my opinion the right way of stating such a theorem improving something like this. And I would also like to say that this is also, I hope, gives some perspective on how a time-dependent theorem should look like. So somehow you have an equation on the gamma and using such a theorem, perhaps this equation works in an energy setting, you want to give that somehow a meaning in the psi sense. And it should somehow commute with this bijection. I mean this thing, right? This gives you a map between gammas and psi's and if the gamma's involved then the image is the psi should also evolve, okay? And especially in the time-dependent situation you don't want to be at the ground state. So you need some theorem like that, which is somehow higher up, okay? So you can formulate this as a gamma convergence? Exactly, yes. Except that in gamma convergence I think, right, so first of all they live in different spaces, but more importantly somehow we give really explicit maps. I mean, right, I mean you can summarize it in terms of gamma convergence, but the thing is really we tell you somehow once you have this, apply this operator and this is what you get. But exactly, that's somehow the philosophy behind it. Okay, good. So let me tell you a little bit of what goes into the proof. And I will more or less talk about the first part of the proof, which is the simpler one, right? I mean this is the one where I'm given the psi and I just have to construct the gamma. Because I mean I have the psi is a small object and I have to make a big object. A really complicated thing is gamma is a big operator and I have to somehow compress it into a single psi. But I will tell you afterwards why when you have understood the first part it's not impossible to understand the second part either. So how to prove one? So we will look at states of the following form. Now this will be terrible notationally because the delta here is not the Laplacian, okay? It's just like in the first lecture every physicist calls this function delta, okay? And so I want to stick to that notation. But I hope it's clear somehow when the delta is this off-diagonal entry and when it's a Laplacian. And we again, we try to do it in terms of a Fermi Dirac distribution. You might remember we've done something similar in the translation variant case. So I put here an H, this one particle operator from there. I put a minus H bar over there. And then I put an operator delta up here. And how will this operator look? Delta look like delta of x comma y. This will be two times v of x minus y times this alpha gl which I just defined psi of x comma y. Okay, so that's an explicit map that goes from psi's to operator's gamma, right? Because I take my psi, I form this operator, a star is fixed once and for all. I define this operator, then I do this functional calculus and I compute this gamma. And the statement of the first thing is, well, that we have both this energy expansion. And secondly, that the one two entry of this operator somehow recovers actually the thing that I've put in there. Okay, that's already kind of an inverse. And I also want to emphasize that this is exactly the same thing as what we did in the second lecture in the translation variant case, right? If psi is equal to zero, that's simply two v a star. That's what we did back then. And again, my motivation for choosing this operator is I could look at the Euler equation satisfied by minimizer. And I would find that exactly solves such an equation except that here I don't have the alpha gl, but I have the true alpha. Okay, so this motivates this. I mean, and now I, yeah, good. And so I use this and I plug this into my functional up there. And I do a computation. This is an identity. There is nothing to be afraid of. Ft of gamma delta minus the energy of the normal state. Not by the way that this notation is consistent, right? If the delta is equal to zero, then this is exactly the normal state. Okay, so that's given by minus t over two. There was a h to the d which I multiplied through. Trace of logarithm of one plus e to the minus one over th delta minus the logarithm of one plus e to the minus one over th naught. Now there's a minus term, another minus term, h to the d integral r to the d times tau to the d v of x minus y over h the alpha gl psi of x comma y squared dx dy. And then there's a third term plus h to the d, which looks similar. You see that somebody has completed a square here in this formula v of x minus y and now this is alpha of x comma y minus, I'm sorry alpha delta alpha gl psi x comma y squared dx dy. Okay, let's stare at these terms for a while. I would like that the last term is a remainder term. This means that I want to show that if I take this alpha gl here, plug it into this operator, compute its one two entry, then I get back the alpha gl psi up to low order terms. Okay, so this means I have to compute somehow asymptotically as h goes to zero such an entry. Okay, there are the one two entry of this. And well, this can be done. It's a little bit messy. I mean, you really, it's not a trace that you compute but an operator, but that's fine also somehow since this is only form bounded, you need really an h one control, not an l two control, as I said. But let's ignore this remainder term because at the end of the day, this is a remainder. Let's talk about the second term. If in the second term, we would have, if you believe me that we could plug this in, this is the real while quantization, well, what would we do? We would change variables into an x minus y integral and an x plus y integral. The x minus y integral, well, let's say first the x plus y integral, that this does not depend on it. So it just gives you the integral of psi squared. And then I do the minus y integral, and this gives us the v times a star squared, which is the term that we've seen before. Okay, and everything happens with a power of h squared, which just comes from the fact that this thing is of what h squared. Unfortunately, there's a little bit of a price to be paid because we took the wrong quantization. Right, I mean, when you do different quantizations to leading order, they coincide, but then you get some more differences in the commutators. So there is actually, because we do it, there is here one term that should not be there, but which is canceled by another term which should not be in there. Okay, if you would do it according to what the books say about the wild calculus, then this would not be there. But let me ignore that issue. So for us, this term is really just psi, integral psi squared. So let me write this down. So third line is a remainder. Second line is something like minus h squared psi squared times integral v of x, a star x squared dx. And then plus terms we don't want to talk about, which leaves us with the first line. Okay, and now I want to tell you a little bit more about the first line and what goes on there. And I want to define a function f of z as the logarithm of 1 plus e to the z. And this is a nice and analytic function in a strip around the real axis. And so therefore I can write, just by functional calculus, a logarithm of 1 plus e to the minus, what am I doing? It's minus here, sorry, minus h delta over t minus logarithm 1 plus e to the minus h naught over t as equal to, well obviously it's just this h delta over t minus f of h naught over t and then in terms of resolvents, this becomes 1 over 2 pi i integral f of z over t resolvent of the first one z minus h delta minus 1 over z minus h naught d z. Okay, and now what's this contour here? The contour is somehow, this is the real axis. You see this function has a pole at 2 pi i, right? So I just take here a strip of width pi, so it's pi at i pi it has a pole. So therefore I take a strip of width pi t over 2, right? Because there's still a t there. And now I integrate in that way from plus infinity parallel to the axis and I go back. So the formula is perhaps not as easy as it is because this function grows linearly. And so there is actually a contribution from infinity, but because I look at the difference the contributions cancel. But apart from that it's the usual thing. Okay, now we're going to expand. So we have the h delta, this is the h naught plus a delta bar zero zero. And the important thing is if you remember what h zero is, that's this h operator, the minus h bar operator. This is a diagonal operator and this thing is an off diagonal operator. So therefore it's very easy when you expand such a thing you can go to higher order than I mean with less effort to higher order than if you would not have such a structure and you'll find that this is z minus h delta minus 1 over z minus h naught is equal to, so it's 1 over z minus h naught. Let me do it, let me do it right away for the 1 1 entry. Okay, so for the 1 1 entry I have an h here, a delta 1 over z plus h bar, a delta bar 1 over z minus h, now plus 1 over z minus h, delta 1 over z plus h bar, delta bar, 1 over z minus h, delta 1 over z plus h bar, delta bar, 1 over z minus h, plus some things which are small. So what I want to emphasize this is really comes from the algebraic structure of the setup is that there are no terms linear or cubic in delta. Okay, that's really what what what we will see later on in the in the in the Ginsberg-Lander functional. Remember also that the delta is proportional to alpha, the alpha is proportional to psi, so in other words this is a psi squared term and this is a psi to the power four term. Okay and well the remainder is hopefully small and so this should give the the quadratic terms in the Ginsberg-Lander functional this the the power four terms. Now the next thing we do, I mean I guess you can see this already, this h is defined up there and I want to think of this as minus h squared Laplacian minus mu and now there is a plus with an h in front minus i nabla h a plus a minus i h nabla plus h squared a squared plus w. One thing if that if you haven't seen the these type of computations before the important thing is an h in front of a differential operator is not small you always want every differential operator wants to keep an h but then if there's some left over h this is truly small so this really is a term of order h this is a term of order h squared okay and so now we call this thing h naught and now if in the last term we replace the the h naught term every h by an h naught and if we ignore that we have commutators right I mean this delta is a function of of this is a function of x this is a function of xi if we ignore this and we put all the all the psi's on one side and all the I mean all the x is on one side and all the xi's on the other side then this term here gives us the the the psi to the power four term it's really the same computation as before okay so I mean no I should not say it's the same well you can do the the computation and translation variant case such that you see that at the end of the day you compute the same integral okay so the delta the term with four deltas gives psi to the power four with the the the lambda three here all commutators can be ignored so let's talk about the other term so now we want to expand 1 over z minus h as 1 over z minus h naught plus 1 over z minus h naught now we have all this stuff right so there is a I don't know how much I want to write so there is an h times this term which goes in is proportional to a plus an h squared times a squared plus w squared h naught and then there is yet this term I have to take into account once again z minus h naught h this a stuff 1 over z minus h naught h a 1 over z minus h naught plus higher order terms okay now once again let me perhaps tell you a little bit more of what I meant up there so the first term the one where everything is now expanded is this term here where every h is an h naught so that's a trace of 1 over z minus h naught delta 1 over z plus h naught I don't need to write a bar because the the h naught is a real operator z minus h naught and now I what do I have I remember that delta is really where is it 2 there's an h over 2 there's an operator psi of x plus an operator t of minus i h nabler plus the other way around okay what I said was and these are also operators 2 of minus i h nabler so to leading order this thing is just 1 over z minus h naught squared 1 over z plus h naught and now there is a t minus i h nabler squared okay from here and then there is of course I mean there's an h squared in front just because it's a delta squared and then there's a psi of x squared so that is this thing here gives you exactly the this gives you exactly a star 1 over k t k t c 1 over k t a star if you think right there's so there's a temperature difference involved if we ignore that to leading order this is a minus h squared a k t k t a this goes together with that term which I called the second line these are the h squared terms the ones that we want to cancel they go away because a star is in the kernel so there's psi squared is just right sorry I forgot to write psi squared here this is of course psi squared times that okay now this term and that term can't switch other they equal to zero because I'm in the kernel of this operator that's exactly like in the translation variant case now we have to go on now we now so far everything we've seen was like in the translation variant case just written in a more complicated way now when I wrote this equality I made a mistake right because I just pulled the the psi somehow through this derivative this derivative when you when you pull it through this gives you a commutator which is of order h now by some even oddness you can actually see that the term of order h is equal to zero and what really contributes is a term of order h squared okay and this comes now when every psi has a gradient okay so the I don't know how much in how much detail I should explain this plus so there's a h to the power four and now I mean at the end of the day what you get is exactly this lambda naught one half minus i nabla plus two what not just just this yes it's really it really comes just from from from looking at at the difference of these two you can write it out say in in Fourier series if you really want to do it by hand because I mean I know one can do this in a very nice and elegant way but our point is we will not do this computation but we have lots of cutoffs in there and we have to do everything very very carefully and somehow remove the the cutoffs in the in the right way or some some of some terms blow up actually and and therefore I wanted to really show you the basic really way of doing it and then you adapt this and and and generalize this let me just continue so now this gave gave us this term there and now you see for instance there's a w term which is quadratic which I guess you can see from here there's a h square w square term right so that's again a very easy term because you have those and similarly the a terms you have to to put together and the fact that the that the two in front of the a appears it is something like a miracle if you look it from this point of view you I mean because you you separate all the terms and then you put them together and then it works it's a little bit more obvious when you really look at this formula where you differentiate and you see somehow when you that this is because this one over two right so I mean this is done in order to have a Jacobian one when you when you change variables and now this two somehow is eventually responsible for the two in front of the gradient when you expand such a function um so the yes one thing like the formal calculation using y yes quantization you will not get these two terms that cancel exactly that's correct exactly exactly and I mean it's yes yes exactly I mean that was also how we how we do did originally I mean right before you do all the computation you check whether there's a chance that it works and then you plug everything into the vial calculus and you you just compute according to this and then you get a nice expansion for all the terms and then it turned out when you do this um then then actually there were some extra terms which is not surprising because we these different quantizations differ by by lower order terms and then but one can actually find them together and and put them together I should say that even even if you do it with vial calculus it's quite a lot of work because if you look in the books what the books tell you somehow well they exist on an infinite asymptotic expansion in H and then they give you the first term and perhaps they tell you that the the I mean the zeroes order term then they tell you perhaps that the first term is equal to zero if the sub principle symbol is equal to zero and then they don't tell you anything else perhaps in some research paper you can find the h squared term but really to go to this order h to the power four I mean to fifth order doing all the computations is actually quite quite heavy and it especially because we have the magnetic field so we also have two terms that do not contribute to the trace do contribute to the state we have to keep those there are books where you can find the the whole expansion and even the the exact symbol in terms of an integral okay okay okay that's that's good I mean we we use mostly this um by didier robert this autour de la la approximation semi classic and okay okay well anyway so we we sweated somewhat here but um let me end another so there are two difficulties I mean um which I perhaps explained this somehow when you when you do these commutators then then you get derivatives this is how this comes about but to all control the higher order terms you somehow would need more than that and the other difficulty is also that this operator is not bounded from below but it has this this structure this is somehow a spectrum from zero to infinity and this from minus infinity to zero somewhat and what's so that is something and then this that this function is gross linearly means that because once you have this in analytic integral representation you you put in absolute values so somehow you have to to beat a linear growth here so the difficulty is not the closeness to the the real axis but somehow the growth of this function in one direction and again this can be done some by an a talk method using the the special structure that we that we have there that was what I wanted to say about one how to prove one let me say something about two well which is actually the the main part the really difficult part of the paper how to prove two that uses somehow these semi classical analysis ideas together with calculus of variation ideas so you somehow want to use say a gap inequality or what some cohesivity away from some ground state in order to get some regularity for for for functions above okay that you use this in order to say that for your minimizer alpha alpha the function has to be of such a form in a certain sense that everything which is projected away would have a higher energy and therefore has to have a very small coefficient okay once you have this you can do can run such a semi classical argument to a certain limited order now you plug this back into the non-linear equation you do some non-linear bootstrap to gain a regularity and then you can run this thing almost fully almost fully is again you need to to cut off some some high frequencies and and do something and there is of course you can in there you cannot get anything for free so you have to find some cancellation somewhere which allows you to put in a cutoff I know that this is all very way I just wanted to tell you even for the lower bound you need to understand this computation you need to understand this computation where the limitations are and and how to deal with those so I mean let me just to summarize here use use non-linear analysis non-linear analysis to show that gamma is almost of the form in one and then you can do the one computation okay my time is almost up I wanted to conclude these lectures by stating some challenges for future work as you see I mean this is a rather recent development there are not so many works and I think there are there are many interesting questions both difficult and and more straightforward and I think it would be nice if as I already mentioned if I could have motivated you to look into some of these the one that certainly closest to this workshop is the the to derive the time-dependent dependent ginsburg-lander equations I said so there are works by gorkov eliasberg so gorkov is the one who somehow proved in quotation marks this theorem he used a completely different strategy of what we do I mean I don't think I don't see whether there's any way of making rigorous what he did so what we do is perhaps closer to what the gen did although the gen did it on an equation level anyway so perhaps looking at our proof comparing it with gorkov's proof in the stationary case would tell someone how to look at the gorkov eliasberg proof this is like gorkov's stationary proof in the same framework and then there's a paper by a german physicist who's called schmidt who was actually before gorkov eliasberg and there would be nice I mean just think about it physically what you want to say at least in one regime is that you if you're above the critical temperature so your system does not want to be superconductivity superconducting however you start with a state which has some superconductivity built in then you should lose this over time okay so that's I mean a very very naive thing and this would mean that somehow the alpha propagates into the gamma in some sense okay so that is a little bit and there was this question that you asked him or this a dispersive or a dissipative equation so above the critical temperature I would expect the dissipative equation which makes the psi disappear but there are certainly other regimes and there is also a dispersive regime yes excuse me which means psi disappear or alpha disappear they're the same I mean if you look at this formula the a star is a universal thing and then psi is right I mean we always think somewhere in this framework so this would also be kind of a semi classical theorem where where somehow there is basically if psi is not there alpha is not there yes exactly right psi see psi just describes the variation of alpha on the center of mass scale and there's always a universal a star and the a star cannot move for energetic reasons it during time doing time evolution so that's one the other thing that I mentioned of course that's um very difficult is somehow derive bc s from many body quantum mechanics well from yeah from quantum mechanics from real quantum mechanics let me write from first principles there is some work by uh trubowitz and canura in this direction but I think it's fair to say that this is not yet uh satisfactory understood and I don't really know what to say about this let me come back here closer to this framework one thing if you remember how you might have seen the ginsburg-lander functional in the literature people usually also minimize over the a right so the physical setup is you apply a magnetic field and then the system has a response magnetic field which actually typically wants to cancel the the magnetic field that you put on okay so somehow there there's an external field and then there's one that's generated by the system so this would mean that in our first equation up there we would have to minimize also over a but then somehow add a penalization term which somehow says that this a does not want to to go away too far from the given a or something okay we have not been able to to derive this ginsburg-lander functional with this um what's called self-generated magnetic field this would be very interesting because this would explain the mice in effect which is a phenomenon that that is typically associated with superconductivity it's essentially based on the fact that semi-classics is very hard with an a with a first-order perturbation and that you really need to use that where did i write it i mean that some terms um right that this somehow vanishes when you integrate over xi roughly speaking so you need to use such a cancellation property which is rather hard so that's so um self-consistent magnetic field and the last thing that i also mentioned before in the course was that we ignore a boundary whereas um the boundary actually plays an important role in ginsburg-lander theory there are these normal boundary conditions or even they're called duchenne boundary conditions in this context because duchenne really somehow explained how important they are in particular if you couple two materials and we don't understand how this works um but it would be certainly a very interesting problem so i think that's all i have to say i thank you for your attention and please ask questions if you have any you said that you excluded the boundary from the very beginning we comment on that what what what is the difficulties in this approach okay um good i think i think the difficulties are not so much in the semi-classical approach i mean we know that semi-classics is very difficult with the boundary and that might be i mean even worst problem but we don't really know how to start so i mean with the boundary it's if you see somehow we got only an h squared term and h to the power four term boundary would give you terms in between okay now what what is however the the the bigger problem is that we don't know um if the the gamma satisfies um directly boundary conditions okay why is it enough that the psi satisfies Neumann boundary conditions okay why do in other words somehow you why do you pay this localization error only once why if you paid it somehow for the gamma so the electrons have to pay the localization error they have to disappear when they go to the boundary why do the cooper pairs not have to pay the price but some are in in in physical terms why it does it somehow not matter for the psi even so the psi perhaps might go down at the boundary somehow on an h dependent scale but there is a must be a cancellation that that this which means that you go down so you have a big change in psi is not um does not appear dysfunctional i i guess i cannot say much more it's okay um right exactly so i mean what i've written down there is not completely true i've oversimplified the situation right and i ignored regularity issues with the h2 but i guess yes i guess that's true right exactly right so can you comment on this problem specific i mean it's see whenever i i wrote here um where were we for instance i mean the the next terms so to say i i need to control those and morally i need one derivative more because there's somehow one more commutator here to control so if i want to derive an h1 theory i would need an h2 assumption in order for for the next term to be finite and i think that would be something that one would have to understand much better somehow to work directly with these terms just stay at the level of of derivatives one one has and it the only way as i perhaps said i mean we prove h2 bounds but then there is coming from the equation there's a structure there's a certain term that is zero and this gives you the space to really remove the cutoff okay you need a kind of more specific regularity very very specific yes right exactly i mean the reason perhaps i should say there's the reason why one can expect anything better in in general at all is somehow because we work with the laplacian the main term is the laplacian and this just if you take three derivatives of xi then this is gone or even if you two then you don't pay any commutators anymore so therefore this is much better than general theorems about commutators would would allow you to to to conclude but this is somehow still not enough this gives us some way but still there's at the end there's still a little bit where we need much more about the structure of the problem okay thank you we have a question i had is about well it's a time evolution so question if you okay if you put not it but okay and schroninger or even perturbed with a dissipative term near the near the grand state is there somebody at the scribe the smp in large time or things like that is there any work serious work on the qualitative work on that i should perhaps mention so a feature that is crucial here is i guess that we work at positive density okay this means that that there's somehow a background if you you see what i mean now there is a simpler version of this problem which is in the vacuum okay which and their heintzel and schlein so this is this is now in the bec regime which has the other simplification see this mu for us the mu is positive so therefore this is inside the continuous spectrum so if there is some if we expect some um dissipative terms to occur then they would come from a fermi golden rule that there's some something goes in there however in the simpler regime where you are at at zero density um mu and mu is negative so therefore you you're away from the continuous spectrum there there is a work of heintzel and schlein schlein where they derive from this uh setup so again there's an alpha and there's a gamma and in this setup they derive the the cross ptersky equation cross ptersky equation from and i say similar but it's i mean it's significantly simpler from similar bcs model and this is our so cross ptersky this is a dispersive equation right i mean it's cubic cubic nls dispersive um defocusing and so this this per sif equation and here the difficulty would be somehow to see that for positive mu you really get these dissipative effects i mean that that would be the really new feature there which makes superconductivity disappear that's not not present there but still it might be helpful if people want to look at this how is the mahal one works with such gammas and alphas on a time dependent level if not as far as the speaker