So let me just briefly recall what we're doing this day. So remember we had this Poisson process on the real line. We cut it off between minus L over 2 and L over 2. This defines, so between two consecutive Poisson points, we have an interval that we call a piece, right? So delta k is xk xk plus 1. So I put twirls because at the edges, right? We have to take points minus L over 2, L over 2. And we considered the one particle random model, which was the direct sum of the Laplacian restricted to the delta k's with Dirichlet boundary conditions. And we considered the n particle model to be h, so omega. So this was L, and this was omega. To keep my notation of yesterday, n, U, L to be what? To be the sum.  And U was a non-negative function. Assume to decrease at a certain rate, I recall this this morning, at a certain rate at infinity. So roughly, it should decay at least as fast as x to the minus 4. So we had, I put two theorems on the board yesterday, actually one theorem. So I said that if I take so, remember just one definition, E, U, omega, n, U, L, that's the ground state energy of this operator. And psi omega, n, L, U, ground state is a ground state, as I don't know in general, where the ground state is unique. So it's any eigenvector associated to the eigenvalue E, U, L, omega, n. And I stated yesterday the fact that omega almost surely we have that if you take the energy per particle, sorry, particle number is larger than capital N, U, L, this converges in the thermodynamic limit rho, meaning that when n divided, so we are in dimension 1, so the volume is just the length of the interval, which is L. So this goes to rho, which is a positive number, and n goes to L goes to infinity. This is the meaning of thermodynamic limit rho. And so this converges to some function. And moreover, the important thing, let's say, the interesting result. So that's essentially because the model is ergodic. Just ergodicity is enough to prove such results. So it can be done in a much more general framework, it was done in actually Veniaminov's PhD thesis. But when it was in 3 or 4 years ago, he proved such results convergence for rather large classes of models of this type, not only this one. And so this had an expansion which was E0 rho. So this is for 0 potential plus remainder term. So let me just pi squared gamma star. I won't recall what gamma star. It's going to come up later on. Log rho to the cube times rho 1 plus little rho of 1. And this little rho of 1 goes to 0 when rho goes to 0. So that's one thing I didn't emphasize enough yesterday is that, of course, when to study these things, I'm not able to do it in fourth general. You see the model is already quite specific. But even with such specificity, I need some asymptotic parameter to say something. And the asymptotic parameter is going to be the density of particles, which is small. So I don't assume that the potential is small, but really I'm using something which is a reasonably physical parameter. In physics, typical densities are small. For example, not in the case of electrons, but if you take the case of bosons, the densities of bosons where people look at bosons and condensates are 10 to the minus 6. So they are reasonably small for such asymptotic theories to bear some meaning in physics. Oops. Sorry. And so I promised you some information on the ground state. So let me put a plural. Ground states as I don't have a result, a general result, whether it's ground states. Actually, I have a general result. So maybe it's the first thing I say, ground states. So there was first theorem. So remember, for you, we had some assumption of decay at infinity, but this assumption will presumably hold or can hold for some special use. And if u is actually analytic, then we can prove that for almost omega, almost surely, for any l and n psi. So this thing, actually I think yesterday I called it 5. The psi u omega n of l is simple. The ground state, if you know in addition on your potential that it is analytic, it's an analytic function, which may be satisfied this doesn't contradict the decay assumption that we have, then we know actually how to prove that this thing, it has a single ground state. Ground state is dimension 1. OK. So and in the sequel, when I will speak, I don't want to assume that you is analytic, but when I speak of the ground state, I speak of any ground state. This is going to hold uniformly for any state in the ground state, in the ground state subspace. So before actually telling you what the ground state for the interacting model is, I'm going to rewrite the ground state for the non-interactive, which in this case is almost surely simple, when u equals 0. You remember, we said we have to take the for the ground state, we have to take the n lowest eigenvalues, meaning the n vectors corresponding to the n lowest eigenvalues of this system. And we just take the anti-symmetric tensor product of these n states. This we can represent in the following way. In this graph, on this graph actually, I'm going to plot in the abscissa, it is the length of the pieces. I'll tell you what L rho is in a second. It's the length of the pieces. And in ordinate, it's going to be an integer. So 0 here, 1, 2, 3. So what is L rho? So remember, we have the Fermi energy. When we did our first computation for the free system, what did we see? We saw that in the ground state of a non-interacting system, all the levels up to the Fermi energy are filled. The particles are placed exactly in all the levels below the Fermi energy. Now, we have a first proposition. If you look at this model, take h omega L and take the counting function of the eigenvalues of h omega L less than e. Divide this by L. Then, as I said, this converges to what is called the density of states, n of e, when L goes to plus infinity. This is omega almost surely. And this thing is non-random. And the nice thing is that. I'm not sure I have the formula here. I think I have it somewhere. I just need to find it. Where is it? Oh, yeah, here it is. This thing, what is it? This is equal to exactly e to the minus L of e divided by 1 minus e to the minus L of e, where L of e is the square root of pi over e. So in this case, you can actually compute this explicitly. By the way, the proof, exercise 3. That's the exercise number 3. Compute this explicitly. And you get a nice closed formula, sorry? The capital AR, yes, is the length of the cube. Exactly. That's the length of the interval. It goes from minus L over 2 to L over 2. Capital L is the length of the interval. And what you do is you take the counting function, divide. And of course, this is correct. I didn't say for e, non-negative, right? For e negative, this is 0. I mean, the spectrum is positive. OK? So in fact, it's not a one-dimensional. This is one-dimensional. Here we have just a single line. That's a one-dimensional model, right? So it's one-dimensional. This is one-dimensional. This one is not, right? I mean, this has many particles, so each particle counts for one dimension in the limit of infinity, but this is just one-dimensional. OK? So you can compute this. And what do I call L row? Well, you remember, in the free system, we define the Fermi energy. It was given by N of E row is equal to row, right? N of E row is equal to row. This was the definition of the Fermi energy, E row. And now what I call L row is exactly L of E row, by definition. Well, L of E is this. What does this correspond to? This corresponds to this length is the length of the smallest piece having some energy below level E row. OK? What does it mean? So by what we said yesterday, it means that for any delta K such that delta K has lengths strictly larger than L row, the ground state, the free ground state, meaning no interaction, does not have any particle, or has zero particles, let's say put it before, or puts zero particles in delta K. Right? You put exactly zero particles in such a piece, OK? That's the definition of this Fermi length, OK? Up to errors, actually, right? Because we are not here. We did not assume that you satisfy N divided by L equals row, right? But up to the error that N divided by L converges to row, this is going to be correct, right? So here, actually, I should say, you shouldn't use exactly L row. What you should use is L row plus something which goes to zero as the difference between N over L and row. But I neglect this, OK? Here there's an approximation. I neglect this. And the other thing we know is that, well, psi 0 omega L N, sorry, you go way around, puts exactly K particles in pieces of length in two K, sorry, L row, K plus 1 L row. That's exactly the same thing. If you have a piece that is of length comprised between K L row and K plus 1 L row excluded, then how many levels does this piece have below energy E row? Well, it has exactly K levels, right? Because the eigenvalues for such a piece, remember, the eigenvalues are be given by pi J is an index, right? Integer divided by the length of the piece squared. And what you want, you want this to be less than E row, right? What does it mean? It means that this is equivalent to saying, looking at the definition of L row, L row, it's the same thing to say that delta K divided by J should be larger than E row. If you want the Jth particle, right, the Jth eigenvalue of the plation, to be inside the piece, you need this. So if you take a piece which is of size inside this interval, you're going to contain exactly K particle. So here is the picture I wanted to draw. You can sum this up in a nice density picture. In pieces between 0 and L row, your ground state, free ground state, puts 0 particles. In pieces of length between L row and 2 L row, your free ground state puts 1 particle. In pieces of length between 2 L row and 3 L row, 3 particles, and so on. You get a full description, OK? So this L row is just the distance between 2 particles? No, this L row is what? You have the following thing. You're in the free case, right? Nothing is interacting. Your electrons live all alone. What you want to do is you want to understand in your system, which is this direct sum, right? You want to understand what are all the levels between energy below, sorry, energy E row, OK? But what are the individual levels in the given piece? They are given by this formula, right? So this is the formula I have over there, L row squared, OK? In a given piece, that's just a relation. You invert the relation that's written over there, OK? In a given piece, the eigenvalues are given by that, OK? Now you want to look at which pieces do contain actually particles. They, for a piece to contain a particle, you need what? You need to find the J, non-vanishing integer, such that pi J divided by the length of the piece squared be less than this, meaning that's such that this happens. So what does this tell you? Well, if J is non-vanishing, it means that this one needs to be larger than L row. Below L row, no particle, OK? If you want to contain two, you need this to be valid for one and two. So you need delta K to be larger than two L row. If you want to contain three and so on, OK? So what is this L row? This L row is a characteristic distance, right? Telling you what are the smallest intervals, the smallest pieces that can actually contain a single particle, or at least one particle. This is what L row is, right? So it's a physical characteristic of your system, depending, and it depends, of course, on the system, meaning on N, and on L row. We have an explicit formula for N in this case, so it can be easily computed, OK? Well, now. So it would be delta K smaller than L row. Oh, yeah, you are right, of course. That's correct. Definitely, thanks. If delta K is smaller than a row, it doesn't contain any particle. Definitely. OK, now let me, what you are going to do now, I'm going to do the same picture, but for the interacting case, before I state the theorem. Before stating the theorem, I'm going to show you a picture. OK, so theorem in quotes, because it's going to be only a single picture, right? And one thing, OK, I'm going to need a little bit more space. And then I write down the proper theorem, which is going to take up much more space than the single picture, but the picture sums it up nicely. Here we have this. OK, in blue, I repeat what I had in the non-interacting case. That's non-interacting case. And in green, it is going to be U non-vanishing, right? U non-vanishing, but of course, with our assumptions, OK? And don't assume that U is strictly positive everywhere. It's just to say, let me write it. So this is, let me write it this way, with our assumptions, OK? OK, so I need to, oops, where are my picture? Here there are a few numbers to remember, which of course I never know by heart. OK, so here we have one threshold. I put it in green. This is L rho minus, well, some constant that we already saw. It's the gamma star that came up in the formula above there, times rho. And here we have another thing, which is, oh yeah, one thing you should, OK, I should do this thing, but trying to keep the sizes of the objects. So this was green, sorry. This was green. And keeping the sizes of objects is this. And this is, well, L rho is large, right? Remember, rho is small, L rho is large. So of course, this is supposed to be, as you're going to see, order one. And so this is 2 L rho minus gamma star is a positive constant, right? So this thing is actually, this difference is going to be a sum of two positive things, right? And the picture is the following. In the free case, here you had two states, right? But in the interacting case, you have just a single state here. I superimpose it here. And here you have, again, two states. So that's the picture. This is what the ground state does. What it does, it does the following. And it's very easy to understand. Here in the ground state, you had two states together in these pieces, right? But of course, if there is no interaction, having two states in a piece doesn't cost you extra energy because the two states don't see each other. They don't interact, right? But as soon as you switch on the interaction, well, the interaction being non-negative, this starts costing you energy. So what happens is that it may be a good thing to actually take a state that was here and put it into some region that previously was forbidden because the sole energy of the Laplacian, the sole kinetic energy on the potential, is not large enough, right, to compensate for the loss of having two states here interacting. And this is exactly what you are doing, right? Is that the states that were here by two, they were actually moved down there, right? And they were moved down there because here, of course, there is a loss of energy, right? This is forbidden. This area here, this region here, is forbidden if you take non-interacting states because the energy is too large. It's larger than pi over LRO squared. But here, there is a gain compared to that. We're going to do the equation, right? We're going to write down how we gain things and derive this, it's not very difficult. And the one thing one should notice is that you see, the length, remember, rho is very small. So the length of this interval is much smaller than this one, right? But because of the exponential low of roughly, because you have a Poisson process, the distance separating two points, meaning the length of the pieces follow an exponential low. Because of the exponential low, actually the number of pieces having length in here is the same as the number of pieces having length here, right? Having length inside this interval. Even though this interval is much shorter. Simple to understand, what's the probability that the piece has length L, right? Is roughly e to the minus, or a length rather than L, it's e to the minus L, right? So here to have length in between here, it means that the probability is roughly e to the minus two LRO minus e to the minus the thing, okay? So this is roughly e to the minus two LRO times a constant, okay? But, so what I'm just trying to convey to you now is that we really did take all the particles that were by pairs in the free ground state here and translate them here into new pieces, okay? And, well, anyway, I'll come to that later. So let me now state, what do I need to state theorem? I need some space, that's for sure. Okay, now we've done this. Okay, now, yeah, one thing, the one thing I need to say is that, of course, here I represented, I said that I put two particles in here, but here the green and the blue line do not represent the same thing at all because in the case of the blue line, in the free case, these were two free particles inside this piece, whereas here in the case of the green line, these are two interacting particles, right? So these are actually really two states that you cannot separate from one another, okay? So, a little bit more space. Okay, how should I go about that? Well, I need to define something. So take a piece of length L, definition, okay? Let me call, well, I call it zeta two. No, I call it zeta, well, I don't need a zeta two. I call it zeta b sub L, b the ground state of, well, what you already saw, it's the operator that we had yesterday to define the constant gamma star. So it's anti-symmetric functions on the square zero L times zero L, right? And remember, we had, let me call phi L just to be the ground state of minus d two. So this is, of course, I didn't say, this is with Dirichlet boundary conditions at the boundary of the square, right? Put Dirichlet boundary conditions at the boundary of the square. So by symmetry, you can see that actually what you're looking at is minus Laplacian plus u on a triangle with Dirichlet boundary conditions on all the three boundaries of the triangle, right? This is because of the anti-symmetric condition tells you that the function has to vanish on the diagonal, okay? On with Dirichlet boundary conditions. Let me call this. And now I'm going to define, well, I'm going to define not the state but the one particle density matrix, right? Which is, we call it gamma optimal, which is not the one of the true ground state but which is the one that will enable, that we will compare to the true ground state. Part one particle density function. So definition. We call gamma one opt, where did I put the one up here? Gamma one opt, okay, what is it? It is exactly the sum for all the pieces of length between L row minus gamma star row and two L row minus log one gamma star, right? And here we put phi delta K. So when I replace the L here by your delta K, it means that I'm looking at the eigenfunction of minus Laplacian on the piece delta K, okay? Sorry, what I need to do is note this. I need to take just the gamma one for this state, right? I take the, and I take, so this is just the orthogonal projector on L2 delta K, right? Sorry, on phi delta K inside L2 delta K. That's the gamma one. So you take this one particle and you take the one particle density matrix itself, right? It's just a projector on exactly this vector. And then for the others with two particles, so this is for delta K of length, larger, two L row minus log, one minus gamma star. Let's say length less than, I cut it off at five halves of L row, you're going to see why. And here you take the gamma one, but now for the zeta of delta K. So that's a real two state, and you take its one particle density matrix, right? So you do the projection, you do this marginal, you take this marginal of this operator, right?  In the way we define gamma one. And here, this is here, it is trivial because this is anyway only a single state, so you get itself back. But here it's non-trivial because this is a two state and it's a real two state because the two states are bound by this U, right? You cannot separate them. And of course, if you look at this and you look at the fact you put two states in here, you put one state here, the total number of states that you have inside this one particle density function is actually strictly smaller than the number you want to have, right? This doesn't count, it doesn't have enough states. If you look at the trace of this thing, this trace will not be large N. It's going to be N plus something small, right? So what you do is you add states taken here, for example, right? So here you have to add. So you take delta K's less than N. Well, rho minus gamma star rho, strictly. And such that the number of these K's be equal to, well, it's going to be actually smaller than N times rho, sorry, it's going to be, so the rho to the five halves, right? This is what you get, yeah, something. Okay, it's correct. Okay, because here the number of states that you have in here, that's a computation, we are going to do it later on, but the number of states that you are doing, that you have in here, right? This contains N minus large O, sorry, one minus, sorry, this contains, no, this is three halves, actually not two halves, five halves, three halves, because it's large N. This contains N one minus O of rho three half states, right? If you do the counting, you count the number of intervals, well, these intervals are random, right? But what happens is because of the independence inside the Poisson process, the number of intervals satisfying such an inequality converge, right, to the average number with a very good, I mean, rather fast, so it has a very good precision, okay? And you can compute this just by a standard, I'll do some computation like that a little bit later, by standard statistics for the Poisson model, okay? And what you see is that if you take these two together, all the intervals here, you put two, these intervals, way two, they contain two states, these interval way one, they contain a single state, then you sum, you get something which is this, where this is actually strictly positive, it's, there is an upper and a lower bound of this size here, right? And so what you do is you put the remaining particles below, but, and so, and here, of course, you need to put something and you should just take phi delta K. So the interesting thing is saying, well, what am I doing? I put any state here, right, yeah? I just put something random, right? That that's not already in there. I could put it again, I could put another state again, but then I would change the rank. What I wanted, I wanted this to be of total trace, but actually I could, because I wanted trace to be, I could put anything actually. So what's the idea? The idea is that, so what's the theorem? Sorry, you write down the definition of gamma one phi? Yeah, sure. In this case, because this is a single state, I'll put it over there. Phi one is just a single state, right? So phi one delta K is a function in L2 delta K. So gamma one of phi delta K is just, if you want, xy is just phi delta Kx phi delta Ky. I don't need to put bars, right? The things are all real, so I forget about the complex conjugates. OK, it's just this. This is just the kernel. Here, this is the operator that has this kernel. I identify operators and kernels, right? And if you take, remember, zeta two of delta K, this thing belongs to this space. And if you take gamma one of zeta delta K, this is equal in xy. It's also an operator in one particle. And it's equal to the integral of zeta delta Kxz zeta delta K yz, you don't really mind when I put the signs on it, dz over delta K, right? And I need to put, sorry, I have two particles, so what do I need to put? So I won't choose two, it's two in front, right? I want the whole thing to have trace two, right? That's the definition, OK? We had the definition in general on the board yesterday. Can I just rewrite that? So if you had complex eigenvectors, you need to put bars, but OK. Yes, sure. What is exactly the number of K smaller or equal to n? Sorry? There you have the sum over. Yeah, exactly. You take intervals that have length here, below this number, and you take just the number of them, right, to be less than this, meaning with a constant, well, actually, to be exactly the number that is missing here. Would you take fragment intervals? Yes, take any of them. Exactly, you can choose them as you want. Never mind, they will not change anything, OK? It gives you some better results if you take them as close as possible. Well, actually, maybe not anything. We can take them, I take them of length larger than 1 half of L rho, so that I have a minimal length, right? I don't want the length of the interval to be too small to blow up energy in an artificial way, right? So I can take them here because there are enough intervals. See, between L rho and 1 half of L rho, I have what? I have L times 1 over, so I have E. So what's the number of intervals? What's the number of Ks? Such that delta K belongs to 1 half of L rho and 3 quarters of L rho, right? This you can estimate. The number of Ks such that delta K is in here, this can be estimated. It's equal to L, right? That's just low of large numbers, right, for the Poisson intervals, times E to the minus 1 half of L rho minus E to the minus 3 quarters of L rho. So it is what? What is this? This is E to the minus 1 half of L rho, 1 minus E to the minus, oops, sorry, 1 quarter of L rho. But remember, just do a small computation. I have it still above there. Look at what's the definition. What is the definition of N of E, right? We computed it. But what do we have? We have that L rho. Remember, L rho is large. Rho is small. L rho is large because of the definition of, because of the formula for N of E. And so what it means, it means that N of E is roughly E to the minus L of E, right? Where E small, E rho is small. It's roughly this. So what do we see here? We see that we have roughly square root of N of E, right? And here we have 1 minus roughly 4th root of N of E rho. But what is N of E rho? By definition, it's rho. So this is L divided by square root of rho, 1 minus 4th root of rho, roughly, right? And what is this? In terms of the total number of particles, well, what is the total number of particles? It's L time rho. So this is N divided by square root of rho, right? And you see that you have many more such intervals than you actually have particles. So there are lots of space. You can actually put particles in without any cost. And you can do it in such a way, actually, that the particles should be a bit more precise. But the thing is, with my assumptions on you, particles even if they are very far away from each other, interact. But the interaction is decaying. And so what you can do is you can actually, because there are so many intervals, interspace them so that the interaction is also negligible. It's not completely obvious a priori, but it's really the case. But now what I did is I just defined this. I need a definition. And the reason why this is actually not important, the reason why this is not important, is because of the following. It's because I will never be able to give you an exact formula for who the ground state is, or who the complete one particle density of the ground state is. But I will be able to do it up to a certain precision. And that's exactly the cutoff. This size is exactly the cutoff. It's giving me the precision at which I want to work. So you first fix a certain precision, which is roughly in my formula that I have above there. Remember, in the formula of EU, you have this little o of 1, exactly the precision this is a cutoff that guarantees that the errors I make, if I put something in here, take the energy, it will not cost me more energy than the little o of 1 times the remainder term. That's the whole thing. So you're not allowed to be too greedy. Otherwise it's just untractable. And so the theorem is the following now. So fix this theorem. So there exists some constant such that for rho small omega almost surely, if I look at one over the number of particles and I take the true ground state, I take it's one particle density matrix, I take any true ground state. So this is the n is here, the l is here. I subtract the gamma up that I just constructed. And I look at the trace norm. So this is, I take the trace per particle. Then here I have to be a bit more precise. And I restrict myself. I'm using the pieces in the model. I restrict myself to the pieces, I tell you what this means, of length less than rho plus a constant. Then this is bounded by above c square root of rho times the max of, should have taken a board, a slightly larger board. I get another square root of rho divided by rho. And here I get my function. Remember I had a way to measure the decay. It was a function z. Remember the z, I just recorded the definition. And I assume that this goes to 0. This was my assumption. And so the thing, you have a bound, you say that the distance, the trace norm, so this has trace class. This is trace class and its trace is equal exactly to n. This is trace class and we built it so that the trace be also exactly equal to n. And what you are saying is that actually the trace is much smaller than this number times n. So that's on the pieces, I'm going to explain what this notation is, but that's on the pieces of length less than rho plus c. Now if you look at what's happening on larger pieces, you have another actually better estimate, which is important on larger pieces. You have that this is less than c times rho times the max of 1 over L rho, just one remark. So where I need to define, where what? Where you can look at h, right? So this is L2 of 0 minus LL minus L02 L02, sorry. This can be written as the sum for the L2 over the delta k's for the delta k's less than rho plus c plus direct sum L2 delta k's for the delta k's larger than rho plus c. You have the orthogonal sum, right? And what you are doing here, so you have an orthogonal projector, that's the orthogonal projector on these spaces, right? So you have two subspaces, orthogonal to two orthogonal subspaces in your little bit space. You have two orthogonal projectors corresponding to project on one orthogonal to the other one, OK? And the norm that trace less than rho plus c is just the trace norm for the operator. When you're restricted to this, OK? It's just this. So you're looking really at what's happening on the pieces of length rho plus c, less than rho plus c, and of length larger than rho plus c. And the first thing you can forget about this, imagine that you take now u. Imagine that you take u exponentially decaying. If u decays exponentially. So in particular, if u is compact support, imagine you have such an exponential decay, right? For x large, for x large. Then you see that if k is larger than c, this is e to the minus l rho. So it is going to behave, this thing is going to behave like square root of l rho, right? And you can get rid of those. And here again, exactly the square root of rho, right? Then you have, if you know this, then if k is larger than c, you have that zl rho over c square root. When rho is small, it's going to be less than square root of rho. And you plug this into these estimates, right? This is going to be less than square root of rho. You plug this into these estimates, and you see that you get, well, actually, you can even, because I take strictly larger, I can get 1 plus delta, right? It depends on the ratio k over c, right? And what you get here is that these two things become the same now, and they just become this, rho divided by rho, right? You get that 1 over n. Sorry, the trace norm, gamma 1 psi u omega nl minus gamma is dropped in a trace class. Norm is less than c l divided by rho, rho, sorry, divided by rho, okay? In this case, you get this. So the two coales, for example, if u is compact support, you get this, right? So you see, there are still a lot of particles you do not control, okay? This is just an approximation. But there's two remarks I want to make for this approximation. No, okay, this, maybe I can erase this here. I'm going to erase this here. Yes, that's the most stable. Okay, let me have this one. I want, good. This one goes down, okay? So this gives me a description of the ground state. Of the ground state of the interacting system. But of course, this description is through an approximation, right? I have an approximate ground state. The question, a reasonable question, is the following. Do I actually do anything? Did I do better, for example, as if I would have taken the free ground state? I mean, the free ground state by the very formula for the energy that I actually just erased is also an approximation, right? Because the error term is rho divided by log rho cubed. And so it's reasonable to ask if you, well, you did all this work, right? But did you do anything better? Okay, is gamma one opt any better than the gamma one of psi zero, omega nl. And of course, you guessed the answer by the fact that I asked the question, right? It is indeed, it is. If you compare, if you take the same trace class norm per particle between gamma one opt and gamma one psi zero omega n, right? This thing is of size what? It is of size. You can compute it exactly. This is actually equal to n, sorry, n times what did I say, n times rho times gamma star to one plus little of one, right? And you see that, where is it, it's up there? In the case, sorry, there's no n here, thanks, right? In the case when you have U exponentially decaying, this is much larger than rho divided by L rho, right? Because L rho is large, this is of size rho. So you really have measured something more precisely, right? You have constructed a better approximation than the trivial one, which would be the free ground state, okay? And of course, in the picture that I just slid down, this, these states are exactly the ones that make up this difference, right? Because these are anyway the ones that they have in here, right, the remaining ones that I put arbitrarily, they play there in the little of one here. But the ones who really play a role is these difference. Okay, and you can do the same thing for the gamma two. So for the two particle density, so theorem, you have an analogous result, so the same thing, here I assume, no, same assumptions. There is this C positive such that for rho small, omega almost surely, well sorry, I forgot one thing. I said this, right, this of course, if you want this to be true for omega almost surely, for almost every omega, it's not going to be the case, right, if you take a fixed n. So what you have here is it's the lim soup in the thermodynamic limit, right? It's this lim soup, right, that's going to be small. Because for fixed n, you will have a set of positive measure, but small, for which this thing is not going to be correct, okay? So here this is the lim soup in the thermodynamic limit, okay? And you have the same kind of things, so lim soup in the thermodynamic limit rho of one over, but now how many states we are talking about two particle densities? Well, you have two types of states. So here you have the gamma two of psi u omega nl minus, and here what is this? How should you construct the two particle? Well, it's going to be constructed exactly from this one. It is one over two identity minus exchange. I'm going to explain what these two operators are. Gamma one opt, tensor, gamma one opt, okay, and this is in trace class norm, so this is trace class, and again, you have to, well, let's assume for anything. Assume, just to make it simple, let me put as an assumption, assume H u tilde satisfied, H u tilde satisfied, and what is H u tilde? Just going to shorten the thing, this is H u tilde, where is it? It's up there. Oops, I have to slide this down again. This is H u tilde, right? Assume that the thing is exponentially decaying with a sufficiently fast exponentially decay. I'll tell you in a second why I do this. Well, then you get that this is less than c times rho divided by rho. If you don't assume H u tilde, if you just assume the initial H u, meaning polynomial decay, sufficiently fast polynomial decay, then you need to cut it off into two pieces again, looking at the smaller pieces and what's happening at the larger pieces. And so what is identity? We all know what it is, right? And exchange, the exchange operator, is just the following thing. It's just if we take F tensor G, it's just G tensor F, right? For F and G in your Hilbert space, right? So that's, in this case, the Hilbert space is just the exterior tensor product, right? Of L2 of the interval by itself, right? This is living on the whole interval, right? So the Hilbert space I'm considering here is exterior tensor product of L2 by itself. This would be the exchange operator so that the operator is anti-symmetry. This one plays exactly, this is why this one, this operator as a whole makes the whole thing anti-symmetry, right? You can do the computation, but the exchange operator itself does not preserve anti-symmetry, but if we take identity minus exchange, this does. So this preserve anti-symmetry. Okay, well, how does one prove this? So of course I won't give complete proofs, right? It's a bit long, but I hope to be able to give the main ideas. So the first idea, the first very important step is decomposition by occupation, meaning that we can locate particles inside pieces in the following way. So as I said, we have that L2 of minus L over 2 L over 2 is decomposed as the direct sum of the L2 delta K. Sum of the K, this is a random number, but roughly the number is size L, right? Because I have a personal process of intensity one, let's say that L plus some lower order terms was a good probability. And of course this means that if I look at, remember, I was looking at the exterior tensor product of this and times, well, if you plug into this, into the exterior product, you get that this is a big orthogonal sum of the following. You choose how many particles you put in each delta K. So you do the following thing. You take Q equals, so let me call this from, I order the delta K, this is a random number, right? But for any given omega, almost surely, for almost any omega, I have this random number. It's the number of the pieces. And I can do the following. I take the M tuple, Q1 QM, which is in N to the M. And of course I have a total number of particles. So the sum of these QMs should be equal to N, right? And what I do is I put exactly Q1 pieces in the interval delta one, Q2 in delta two, QM in delta M. So it's going to be the exterior tensor product, right? Actually because these are, I can take the following, I could take this, yeah, I'm writing it in this way, but I'm going to see that it doesn't change anything. And here I take from K equal one to QJ, L2 of delta, delta what? I call it this K, actually. And I take this, Q, K here, delta K, okay? And this I'll call H sub Q. That's just because of this decomposition, if I put this into the anti-symmetric tensor product, I just expand. And if I expand, I get this. But of course here I could replace this exterior product by a cross product, would be the same thing, right? Just a tensor product, standard. Because the delta K is at this joint, okay? That's because of the jointness of the delta Ks. Okay, and now the nice thing is the following result. Okay, so I call this HQ. This is a closed Hilbert space, right? These are closed Hilbert spaces. You have this direct often known sum. And the nice thing is to notice that proposition, well, for any, for any Q in Q, I'm sorry, in N to the M, such that the Q is Q1 QM. Q1 plus QM equals N. We have that. If I take my operator, Hu omega L plus I inverse, right, so this is a bounded operator. This is self-adjoint. This is the bounded operator. I can apply it to HQ, well, I stay in HQ. What does it mean? It means that the HQs are invariant through these operators. This comes from the fact that the pieces at this joint, right? What does it mean? This immediately tells you, comes from the fact that it's true for the Laplacian in an obvious way. In an obvious way, it's true for Laplacian. And it's true for the potential. Why? Because the potential is multiplication. So on two disjoint intervals, right? They get respected if you multiply it by some function. Right, that's all. That's the only thing there is to it. But this means the following now. This means that my operator, H omega L U, can actually be decomposed. It's a direct sum of H omega L U, restricted to HQ. It's a direct sum over the occupations, right? It's actually over the applications of its restriction to a fixed occupation number. Right? So yes? H omega L U is H omega N U. Oh, yeah, exactly. Oops. Okay, you have this. So it means that if you want to look at the ground state of this one, this is a direct sum, you just need to study the ground state of those, okay? And in those, you have fixed the fact that you have exactly K1 or Q1 particles in interval delta one, Q2 in interval delta two and so on, right? So they can. Well, they cannot in the free case, but once you put the interaction, they can. But what this does say, right? It just says, it doesn't say you, when you think of hopping, you think of dynamics. Okay, but here there is no dynamics. Okay, this is not dynamical. It just tells you that the ground state for this one is going to be achieved by one or more than one. Ground states for that one. When you say the spaces are invariant by the resolvent, it means it's invariant also by the propagator. Yes, that's correct. The propagator cannot change the number of particles in intervals. That's correct. The propagator cannot change that. That's correct. That's correct. And again, hopping. Okay, in this way. Yeah, I was thinking hopping when you switch on the interaction. Okay, but I agree with you. Sure, there is no hopping. Indeed, that's correct. You can just look at the propagator on each of these spaces and do the direct sum and you get the propagator in the whole space. Definitely. That's correct. Yeah, in this sense, I was thinking of hopping when you switch on the, there is a change of configuration when you switch it on. So, actually what you can show is that for a given Q, the ground state on this one is going to be done to generate. It's simple. This one is always simple, but what may happen is that you have many Qs that are giving you the same ground state. This is something that may happen. But one thing that is true, lemma, is that ground state of H omega n, UL, is simple. Of course, not all of them are going to give you the ground state of the whole thing. And so the idea now is to understand, can we find, can we find a characterization of the Q that is giving you of the occupation number that is giving you the lowest of these ground states? Now we're going to reason on the occupation numbers. So, therefore, I'm going to do the small computation I had in mind. Construction of psi-opt. So this is a state which is corresponding to the gamma one-opt, if you want to take gamma one-opt to be the gamma one of the psi-opt. So what I said is, what we want to do is the following. So we have the intervals of lengths between arrow two arrow and then two arrow three arrow, right? And we know that we have one state in the free case, we're going to start from the free case. So, first, the first thing I should say is the following. Maybe as a preamble, I should say the following. The first rule, and this is due to the fact that you have fermions, as fermions or fermions implies, few pieces can contain only a few states. What does it mean? Be it for the interacting or non-interacting system? What does it mean? Of course, I can construct a state, I can pile everybody into the same piece. It's just that this is something which is going to cost you a lot of energy, right? The basic idea is the following. Imagine, so energy, what is the cost in energy of K states in a piece of length L? Well, it is just the lowest state, right? The lowest K state in a piece of length L is the one that is filling up the K first levels inside this piece. And so this is the sum of J equals one to K of pi J divided by L squared. So it is of order K to the cube, right? Times divided by L squared, right? So this order. So remember, we have, at our hand, if we start with the ground state for the free system, a lot of empty pieces. We have more empty pieces. Remember the computation I did before? In here, the size, so this is what? One half of L row. Within here, we have more empty pieces, many more empty pieces than we have actually particles to place, okay? And what's the cost of an energy? What's the energy cost to put one particle in such a piece? Well, it's just filled the ground state, right? So it's going to be at most two over L row, two pi, sorry, over L row squared, okay? So imagine that I have a P in which I have already K states. And I want to choose, should I put a K plus first state in there or should I put it into some of these empty pieces that I have here, right? I have so many pieces down there that I cannot fill them all. So there will be empty pieces, as we already saw. Well, what is the balance equation for that? It's just to say, if I put one more state here, how much do I pay? I pay K plus one, right? If the thing is of size L squared, that's what it's going to cost, okay? There's a pi in front of the pi, okay? And if I put it in here, it's going to be two divided by L row squared, right? And so we're going to put it in there as soon as this is that, right? As soon as this number is larger than this, we're not going to put it inside this L length L piece. We're going to put it down there into something which is less costly. But what is this? This tells you that K plus one is larger than two L divided by L row. So what you see is that you are not going to put more than two L over L row, right? States inside the piece of size L. The maximum number of states that you put inside the piece of size L is going to be two L over L row, okay? And so if you have only a few pieces, you can only put a few number of states in there, right? Because you have this maximum number for each piece. You have a few pieces. It tells you that you only have a few states, okay? So in particular, what does this tell you? This tells you that, look at, what is the number of pieces of length larger than five halves of L row? Well, you can compute what this is. This is less than N times row to the three half with a very good probability, right? And so if you do this summing, you cut it into multiples of L row. You do the summing. You see that these pieces contain at most, what? Something like row to the three halves times N particles, right? You can show using this, this gross estimation plus the counting of this that to show that these pieces will not contain more than this many particles, okay? This comes from the exponential decay. It's the first term that's more important one, right? And once you know that they don't contain many particles, of course, they don't contribute for a lot of energy, right? They don't contribute for a lot of energy and this size of energy, when you divide this by N, this is row to the three halves, which is much smaller than the remainder term you actually expect to achieve, okay? So what I want to aim at is to say that anyway, all the particles that live in pieces above this you just don't care about because they contribute too little energy because they can contain all these pieces, can contain only too few particles in the ground state, in any state actually near the ground state as well. So we want to understand what's happening in here. This is the region of interest, right? The pieces, the particles contain inside these pieces, okay? What you're saying is that the long pieces, they can, each long piece can contain many particles, but they are very few, so please. Exactly, this is what I'm saying and that's the exponential low, right? If you take another process, right, that divides the intervals in a completely different way such that long pieces have a much larger probability of appearance, something completely different would come out, okay? But if you take the usual random model, right? Usual meaning if you take a model, imagine you take, yeah, one thing. I started here, I'll come to this tomorrow, but the underlying model I have in mind here is Poisson, right? But Poisson has some universal features. If I take, for example, imagine you take an Anderson model, right, what is an Anderson model? It would be a potential of the following kind, omega gamma, and you put at every integer some non-negative potential, right? Let's say V is compactly supported and I would like to look at minus the pressure input of this thing, okay? Same Poisson model. No, it's not Poisson. This time what you do is you do Anderson. So imagine that I do the following thing. I take a collection, I take a single side potential, right? And I put a copy of it multiplied by some random variable at every side of the lattice, okay? And now again, you're interested in, and actually I talk about this tomorrow, but I'm a bit ahead of myself, but imagine just to speak about the quote and quote universality feature of the Poisson case. And you're interested in the following thing. You're interested in what are the low-lying eigenvalues of this thing, right? So you have this random potential, random bumps, and so here it's smaller, here it's smaller, and so on, right? So this is supposed to be the lattice, okay? If you want to have a low-lying eigenvalue, the only way is that, well, the bumps need to be small, right? Because otherwise the potential is going to push you up. You need large stretches of bumps for the people among you that know the work of Snitman, one motion inside the Poissonian traps or random traps, right? This was exactly studied. What you need is large clearings if you want to have low eigenvalues. But you see the probability of such large clearings because of the independence of the omega gamma is also exponential. It's of the same nature, right? So actually this exponential thing does not really come from the Poisson thing. It comes from local independence. The fact that events locally at two different sides are independent. If you take faraway events that are independent from one another, this is what gives you this exponential decay, right? So Poisson is one instance for that, but it's not the only one, okay? Every time you have local things that are independent, if you take things of a large stretch and you want to impose conditions on this large stretch because of this local independence, you'll have this exponentially decaying feature, right? Just because product of independent events is multiplicative. So if you want to impose many local conditions, if they're independent, it means that you have some exponential decay, okay? So it's not really a feature of Poisson. It's something much more general than that, which is due to the local or the dependency is that the dependence is only local. So let me come back to the main topic. So what I want to do now is the following thing. I can do it here. I don't have much time left, but let's do for today. Let me explain where the gamma star comes from. So remember, remember if we take minus laplacian plus ux minus y on L2 0L tensor itself, or L, little L, we know that the ground state E0 of L behaves like five pi over L squared plus gamma, yeah, I think it was gamma L cubed, one plus little of one. This is what I had before on the board, okay? We had this. Now, the thing is when the interval L, you see because the L is going to be a row, okay? A row is playing the role of this L, but it's when L goes to infinity. And so now, what do we have? Well, imagine now we, so we have this, and now I started with my ground state non-interacting in here. If I switch on the interaction, I'm going to have to pay this price because as soon as you switch on this interaction, right, there is an additional cost due to the fact that the two things are repelled from one another by you, okay? So what I want to do now is I want to compare, I want to compare the cost of, well, I can avoid this interaction by taking one of the particles, the one with larger energy in here, and put it down there, okay? Okay, so energy gain by moving one particle from, what, from delta K which is length in two arrow, five half of arrow, to some delta K which is of length less than arrow, okay? So imagine that my delta K, initial one, so this is initial, this is final, right? That's my destination. So imagine that delta IK is two arrow plus A. Right, that's the length of this thing. So if this is two arrow plus A, how much do I gain? Well, when I have two particles inside this thing, right? The asymptotic that's, oh yeah, this doesn't slide any more up, okay. We have to slide this one down. The asymptotic that I have here tells me that before I move it, I have the interaction, so it's going to be five pi squared, two arrow plus A, five pi, sorry, this squared plus gamma of U, I forget about the little of one, and here I forget about the A, you're going to see arrow to the cube, right? Which is also, this I can rewrite as, so that's the energy if I have the two together in here, so this is five pi. I'm going to write this as, sorry, the five is not in here, the five is outside, it's four plus one, it's one square plus two square, so it is, the five is outside. So this is five times this, so this is pi divided by two, how should I write it? Pi divided by two arrow plus A squared plus pi divided by arrow plus A over two squared plus gamma U eight L row cube, right? That's the energy I have in the beginning, so this is also, I can write as, as what? I have pi over L row squared, right? Minus A over two pi squared over L row cube, right? Plus pi over two L row squared, minus pi over, pi squared, sorry, over eight L row cube, and here I have A, right? Plus gamma of U eight L row cube. Just do a first order expansion, right? Taylor expansion, near A equals zero, right? Or A much smaller than L row. And of course to do this thing, I need to add, right? So that's the initial situation. Yeah, that's the initial situation. And then what's the final situation? So that's initial. And final is, well, I have pi divided by L row minus B. So this is delta final, k is L row minus B squared. I put one particle here, plus I put another particle alone in here, which is pi divided by two L row plus A squared, right? The two particles are far away from one another and each contributes for some energy, okay? And so this is, well, I can keep, you see I shouldn't have, I separated it correctly, but I shouldn't have split this. So this is pi over L row plus A squared plus pi over L row squared minus, plus, sorry, B pi squared over L row cube, okay? And I want to understand when is this one, the initial one, right? More energetic than the final one. So I want E initial, right, to be larger than E final. This is going to tell me when I have, when I can do this move, okay? So what does it mean? You see that if I do E initial minus E final, this term goes away, it's here and it's here, okay? This term here goes away and what do I get? I get B pi squared divided by L row cube. Minus A over two, was it, sorry, here there is, sorry, it's not this one. This one I don't, it's this one. Minus A over eight L row cube, right? Plus gamma of U divided by A L row cube. This should be positive, right? Sorry, E final, no, this is, sorry, there is a B here and there was a minus, I have plus here and minus here. Yeah, I got it right, right? No, I have initial, sorry, so this is, you need to get signs correct. So I have initial, so this is plus. I have final, so this is, yeah, it was correct actually. It was correct, it's E initial minus E final. Okay, this needs to be positive, which tells me that I have A plus B should be less than gamma of U, right? That's one thing. But the second thing I need to satisfy is that, well I want to do this for all the particles, right? Between what, two L row and two L row plus A, that I want to transfer to L row, L row minus B. So if I want to do this for all these particles, I need that the number of three pieces in here be the same as the number of particles I have here, right? Each of these, so what is the number of pieces where I have two particles in here? It's E to the minus two L row minus E to the minus two L row minus A times L. So this is going to be using what we said before to be N times row one minus E to the minus A, okay? And I want this to be equal to N to be N and here I have one, sorry, E to the B minus one. B is positive. I want these two things to be equal. What do I see? I see first that this is, I don't know what, but this is small. So B has to be small and it has to be of all the row. Because this is less than one, right? So we see that B is of all the row. Row is small, okay? But B being of all the row, this tells us that A has to be of size gamma at most. Be equal, you need the first one to be smaller than the second one. You need, yeah, okay, yeah, you're right. You just need enough space, okay? So you can do it in this way, but it's going to give you the same result, right? You just need that B, it's going to be smaller than row. So B is very small, row is small, right? This is size B. This thing is B. Yeah, no, it's given the way around. Why? You need more small interval. Oh yeah, you're right. Yes, exactly, this is number of small, no, oh yeah, you're right, you're right, you're right. This is correct. Okay, but so you can do this if you choose this of that size, right? And so this means that you can go up to A equals gamma of U, right? You can go up, you neglect this, right? Because this is going to be a number of intervals which is again, B of order row squared because of the additional factor row. And well, what does it tells you? It tells you exactly what we had before is that. Sorry, I missed, oh, sorry, there was an eight. Oh, sorry, this is not gamma, I forgot. I did my computation wrong because here it's not gamma, it is going to be, I multiply by, what do I have? Let me think. Do I have the, this is A over, this one comes from this. So it's not, sorry. Tick, tick, tick. This one is taken care of, it's this one. So it's not, this one doesn't play a role, it's this one, so it's A over two. I did the mistake of expanding both sides, right? The only thing, this one I should keep. So this one I should keep because I'm going to kill it directly using this. And this way I should expand. So when you do the expansion, you just get this, right? And actually, do I really get this? No, sorry because of course, I forgot the factor two, so there is two missing here, okay? That's better. And so here what I get is this. And what I get is I have A pi square is less than gamma U over eight, right? Which says me that A is gamma U over eight pi squared. I forgot the factors, right? This one has a factor one. This one is neglectable, multiplied by L cubed, right? And so what I have is this. It's not gamma of U, it's gamma multiplied by this factor. And so what I do is I go, I do the following. So what I do is I move from two L rho minus, plus, sorry, two L rho, two L rho plus gamma U over eight pi squared, right? And these I go down. And what do I need? I need to be this exactly that. So this is L rho and L rho minus and rho times one minus, and here I have E to the minus gamma U over eight pi squared, right? And here I depleted for all of these, I move one particle down there, okay? And this gives me the best possible configuration up to this error level. Okay, and I think I'm gonna stop here, right? Okay, thanks very much for your attention. Here are the other questions. Yes, please. You have an interaction between particles in different boxes, so is it negligible? No, it's actually, well, what happens is that they are indeed negligible, and this is what the assumption on Z, remember the decay of interactions here. This decay is less than Z to the minus four, right? This guarantees that these interactions are negligible. Imagine that you had, what can happen is that you're gonna have two neighboring boxes, right? But that's an, actually I talk about this later on, but imagine you have two neighboring boxes where these things are roughly of size error, right? This happens, the probability to have two boxes like that is about the same as the probability to have one box of size two error, right? But what makes the difference between these two things is that you see here you have one eigenstate in the free case, living here. The other one here, living here, okay? But when they interact, because of this decay, of the assumption of decay, the interaction energy is of order at least error to the minus four. Whereas in this case, the interaction energy between this state and the next one, right? Is going to be error to the cube, okay? And this is what's going to make the difference, okay? This is why these do not need to be taken into account actually in the further step of the proof, okay? So indeed, X to the minus four is roughly a threshold, meaning that if you go above X to the minus four, right? If the decay is slower than that, then what is going to give you the main contribution, the main correction to the free ground state? So the free ground state is still going to be correct picture up to a smaller number of particles, right? A number of particles, which is small compared to N in the limit rho goes to zero. It's going to be of size N, but N times the small constant, the smallest of the constant depending on rho. What is doing to contribute now if you assume that Z does not go to zero or if you assume that U does not decay at infinity as fast as X to the minus four is exactly being the interaction between pieces, not of this type, but remote pieces of size error because there are many such pieces. And now the interaction between these pieces is not going to be smaller than that. So this will contribute the main term. There's no proof of that, right? This is just conjectural. There's no, we didn't analyze this, but this should be correct. And then there is a next threshold, right? If you decay the interaction, if you take Kullo interaction, one of regs, then actually there is going to be a complete change in what the ground state looks like. It doesn't look like any more if you do some back of the envelope calculations, you can convince yourself that it doesn't look like any more like the free ground state because you're really in this case want to place the particles as far away from each other as possible, meaning that the Laplacian is going to be negligible, right? But that's still, there are no theorems as far as I know on this, right? So you have phase transitions depending on the speed of decay of the interaction. Yes, sure. In fact, you deal with unparagraphed systems. Sorry? In fact, you deal with unparagraphed systems with interaction. With unparagraphed systems. Unparagraphed systems, yeah. But just now when you mentioned the two-particle systems, basically what does that mean? Oh, what it means, it means that, so this unparagraphed state, right? It's a state on the n tensor product, right? So it's a state in a very big Hilbert space. So one way to look at, n goes to infinity. So one way to look at this state is actually to look at the partial trace so as to keep only two particles, right? You have this function. This is what I explained yesterday. So you have a true state of my system, right? Is something which lives in this big Hilbert space, okay? Now, but if I let n go to infinity, it's going to have more and more variables. So there is no easy way to take a limit of this, right? And you want to have a limiting statement. So what you do is you look at the projector, right? So phi of x, psi of y, so that's the kernel of the projector on this state, okay? And x is equal to x1, xn. You have n particles. Y is equal to y1, yn, right? And what you do is you take a trace. You integrate dz1 or dx1 or x2 dxn, taking y1, yj equal xj for j between two and n or between k and n. So you project down. You integrate a number of particles, right? It's just taking a partial trace. You do a partial trace of this thing. So this one now will only depend on k variables, the ones that you haven't integrated, k minus one actually, on k variables in this case, right? So this is what is called the reduced density matrix or k particle density matrix, right? So this is what this thing is over there. So this is for one particle and on the blackboard. Below that, it's for two particles. Yeah, I have a question. When you look at the two-point function. Yes. Somehow you don't use the two-point function of the pair of particles. Yeah. You could use the two-point function for this pair. Exactly, I agree with you. But the thing is that the number of these pairs of particles is below the error estimate. This is why you don't use them. If you take the pairs, right, it's the precision of the estimation is not good enough for you to be able to measure this effect. That's the main problem. But you can do. So that's something that's not written either, right? All of this is written, but you can, you see we fix the precision, which is a power of rho in the energy. But what you could do is you could fix an arbitrary power of rho and do the same analysis, right? And this can be done actually. If you fix an arbitrary power of rho, if you take, if you want to measure things up to groups of k particles, k being fixed, you can actually do the same analysis and do the same kind of computations, right? To understand what the states look like. What I don't know how to do, right? Meaning the statement would be fix a k. There exists some rho small enough, but depending on k for which you have such statements up to an error, rho to the k. What I don't know how to do is do the same thing with a rho for region and rho, which is independent of k, which presumably should be feasible. But this I don't know how to do. And that's a real challenge actually, because this, of course, by increasing the power of rho independent, in a way independent of the choice of rho, would enable you to measure things up to errors that actually are little o of n when n goes to infinity. Not playing with rho. And that's the real challenge. Getting precise enough information for this. I'm not sure this is feasible. At least I don't know how to do it. Thanks very much.


Trainer Road. I did not want to make this video, but this has probably been one of the most requested topics since I did a review on Zwift training plans over two years ago. And at this point, I've lost count of the number of people who have contacted me saying that they've over-trained or burnt out using a Trainer Road training plan. Part of the reason I've been so hesitant is because I actually like and listen to the Trainer Road podcast. There's a lot we agree on and if we don't, it's always good to hear different perspectives. And every once in a while, I find myself learning something new. Coach Chad often references research, which I obviously appreciate, but when it comes to the actual training plans that they sell, it's as if they forgot that research on this topic even exists. Trainer Road is not alone here. There are plenty of sweet spot coaches that do the exact same thing. But given that Trainer Road reaches the most number of people with their platform, I'm going to be addressing them specifically in this video. Wow, you're really going to do this, huh? Hold on, I need to get some popcorn first. This video is sure to get a lot of hate from TR fans. I'll probably lose some subscribers and there'll be a lot of angry people down in the comments. But as always, I'm simply here to report on what the science has to say. And I've done my best to curb my own personal bias and look objectively at this question, just like I try to do with every video I make. And as always, when I make a controversial video, I'm not here to bash Trainer Road or steal business from them. I'm simply pointing out a rather large inconsistency with what the research says optimal training should look like and what they prescribe. All right, enough of that. What exactly is the problem with Trainer Road training plans? You need a Trainer Road subscription to see their plans. And I want to respect that and not reveal any information that you have to pay for to thousands of people on the internet for free. So therefore, any screenshots that I show you in this video are ones that you can find on the Trainer Road website for free. That being said, I did take an extensive look at the plans that you get with a Trainer Road subscription to make this video. Let's start with the infamous sweet spot based plan. As the name suggests, this plan has you doing a lot of sweet spot work operating under the assumption that this intensity is high enough to give you a lot of physiological adaptation, but low enough that you can do it many days throughout the week. The low volume plan has three rides a week and every one of these rides is an interval session. The mid volume plan has five rides a week and of these five rides, four of them are intensity days, leaving just one day where you're riding at endurance zone. Finally, we have the high volume plan and it's not pretty. This plan has you riding six days a week and for five of those six days, you're performing intervals. The six day is not a recovery ride, but a zone to endurance day. So you're essentially going six days in a row without recovery. And for five of those six days, you're doing some sort of intensity and wait for it. You do this for five weeks straight before you get a recovery week. Well, two things. One, I did three weeks of the high volume plan to kick off this Cape Epic endeavor. And honestly, I emerged from that. I recognize I wasn't going to be able to make five without completely breaking down. So I did three, but I came out of that dang, you're bulletproof. I'm sorry, you're going to have to give me a second. I'm starting to feel over trained just thinking about that. You know, I don't say this often, but you're absolutely right. This plan is horrendous. I mean, one recovery day a week, that is just way too much recovery. If you watch a lot of my videos, you were probably cringing just hearing me describe that high volume training plan or you were waiting for me to have an aneurysm or something. If not, then let's jump into some science so we can get a better idea of why this plan is a bad idea. But before we do that, the research that I'll be referencing and most research in general on this topic uses a three zone training model. Zone one is zero to roughly 75 to 80 percent of your FTP or the point at which lactate levels first start to rise. Zone two is between this point and your FTP, which is where lactate levels spike and zone three is any intensity above FTP. To avoid confusion for the rest of this video, when I refer to training zones, I'm referring to the three zone model. With that, let's see what the science has to say about what optimal training should look like. This study on training intensity distribution and performance adaptations in train cyclists had subjects follow both a polarized and threshold distribution for six weeks. When the subjects followed polarized training, they spent 80 percent of their time in zone one, zero time in zone two and 20 percent of the time in zone three. During threshold training, they spent 57, 43 and zero percent of the time in these respective zones. So the threshold condition in this study more closely resembles what trainer Rhodes sweet spot based plan looks like in that you're spending a lot of time in that in between zone two or sweet spot. So what did this study find? Endurance performance, peak power output, lactate threshold and high intensity exercise capacity saw greater improvements following a polarized distribution. I'm going to get into some more research to really hammer this point home in a minute, but first let's use this study to address one of the most common criticisms of polarized training and a criticism that I've seen the guys at train road make, which is that with polarized training where you spend a ton of time in this endurance zone one, you need a lot of training time in order to make it work. Rather, it was this is what the fast professional riders do. And it's really a product of they spend so much time training that obviously they're going to have to do most of their training in a really easy territory if the productive, the other type of productivity is going to come from the really high intensity training. In this study that I just referenced on train cyclists, how many hours per week do you suppose the polarized group had to ride in order to see fitness gains? 12 hours per week, 15, 20. Nope, the polarized group average a little over six hours per week. What's even more shocking is that the threshold group actually spent more time training per week than the polarized group at an average of seven and a half hours per week and had a higher training load. And yet results were still better for the polarized group. As always, though, this is just one study. Where does the balance of evidence tilt on this issue of polarized versus sweet spot training? This study on recreational runners came to the same conclusion. Runners who ran the most polarized saw the greatest improvement. Even in Ironman triathletes who spent a significant amount of time racing at zone two between the two thresholds show slower race times when they spend more time training at this intensity. This 2019 meta analysis, looking at many studies on polarized versus threshold training, confirmed these findings. Looking objectively at the science, you simply cannot conclude that traditional sweet spot training is the best approach. Key word there being best. I can already hear people in the comments saying, I follow trainer Rhodes plans and I improve my FTP by 40 watts.  Great. I'm not arguing with you there. Clearly you can gain fitness using this approach. The research I just referenced showed this. The question here is not, can you get fit or following a sweet spot approach? Because the answer to that is undoubtedly yes. The question is, is it the best approach? Or is there another strategy out there that would lead to better results? When you take a look at the science, the answer to that becomes pretty clear. No, no, it is not in this review on current evidence for polarized training. They found an effect size of 0.85 to 2.8, which is a wide range, but is greater than the range for traditional threshold training, which was between negative 0.42 and 2.16, indicating that in some athletes, it had a large positive effect. And in some cases, actually decrease the performance of the athletes. That's right. The sort of between threshold training that trainer Rhodes promotes actually decreased performance in some athletes. And I've heard about this firsthand when athletes have emailed me saying that they tried trainer road and quickly felt burnt out. But why exactly does following this approach often lead to overtraining, burnout and staleness? This study on the autonomic recovery after exercise gives a pretty good potential explanation, finding that exercise below the first ventilatory threshold or that endurance zone one causes minimal disturbance in A&S balance. However, training at zone two or zone three induces a significant, but essentially identical delay of HRV recovery. This threshold between zone one and zone two demarcates a clear threshold for A&S recovery. Basically, if you ride over endurance zone, you'll start to build up autonomic stress. If you do this day after day or every single time you ride, then you may start to feel burnt out, overtrained, lack motivation or just generally fatigued. You may be able to get away with it for a week or a month, but over time it'll start to wear on you. What's ironic is that this is how most cyclists naturally train. They just want to ride kind of hard to feel like they got in a workout that day, but never really easy, which in turn means they can never ride really hard, which is where the real gains are made. This is essentially what a trainer road plan looks like. Lots of focus on intensity while completely ignoring longer and lower intensity rides, which is a problem. This review on high intensity versus high volume training stated that both high intensity and low intensity training are important components of the training program. When training does not have an appropriate blend of both high intensity and high volume training inserted into the program, performance ability can stagnate. What does this mean in terms of your training? You want no more than two or occasionally three high intensity or threshold intensity sessions per week. More than this does not induce further improvements and tend to induce symptoms of overreaching or over-training. That's right, two or maybe three intensity days in the week and the rest should be easy endurance riding. We've already seen that this is far from the case with trainer roads sweet spot base plan, but if we take a look at their other training plans, we see the exact same problem. For example, the short power build plan has three days of intensity in the low volume plan, five days of intensity in the mid volume plan, and five days of intensity in the high volume plan. And this pattern is fairly consistent across most of their training plans, three days of intensity in the low volume plans, four or five days of intensity in the mid volume plans, and five days of intensity in the high volume plans, which is still absolutely blowing my mind that any coach in their right mind would prescribe something like this, let alone a fairly credible company like TrainerRoad. This high intensity workout frequency is really the heart of the problem here, and I know I've spent a large portion of this video taking a dump on sweet spot training, but training in that zone isn't even the real issue if you're doing it sparingly, which TrainerRoad doesn't do. Training mostly in zone one, but then spending some time in zone two instead of avoiding it completely like you would if you were following strict polarized is called pyramidal training, and actually has some research to back up its effectiveness. For example, this study on Ironman triathletes which found no difference in race times between the polarized and pyramidal group, or this systematic review which found that polarized and pyramidal training was more effective than threshold training. This is not really an issue of polarized versus sweet spot, because it appears that if you use sweet spot training in the right proportion then you can see great results from it, potentially as good as following strict polarized training. The issue is that TrainerRoad simply overprescribes intensity, and I don't see how anyone looking objectively at the science could come to any other conclusion. Now to be fair, they have a traditional base plan which is mostly endurance zone riding, but in the description for that plan they state, it requires a large time commitment to give you significant gains. Unless you have at least 10 hours per week to train, we do not recommend the long low-intensity traditional approach. And then we have the description for the sweet spot base plan, which states the sweet spot block is the most effective form of base training for the majority of cyclists. It's what we recommend. I don't know what layperson would read these two descriptions and think, hmm, let me go with the inferior traditional base plan. You know, the one that they say they don't recommend. And despite offering this alternative plan for base training, they don't offer a polarized or pyramidal plan on their platform despite a large body of evidence supporting these approaches. Let me address one last potential criticism before I go, which is anecdotal evidence. This argument runs along the lines of, there's this guy at the Tuesday night group ride who always rips our legs off and he just rides hard every single day. Or there's this pro that I know who rides sweet spot all day every day and just look at how fast they are. Yeah. And what about Rick's teammate's three-legged dog's friend, Josh? I mean, there was some random dude on the group ride who told me that that guy got three KOMs in one day by using beer as his recovery drink. I mean, call me crazy, but that's all the evidence I need right there. These are all examples of anecdotal evidence. And in all of these cases, we don't know how much faster these riders could have been had they followed an optimal approach. As we know, though, there's a huge genetic component to adaptation to training, meaning that no matter how good our training is, there will always probably be someone faster than you, even if they don't train optimally. But again, clearly it is possible to make large fitness gains while spending a lot of time training in tempo and sweet spot zone. Probably even more so for pros who are naturally gifted with the ability to recover very quickly between hard workouts. But is it the best way to train? While we're talking about N of 1 anecdotes, I personally have used both training methods. I used to be the guy that would train at tempo or sweet spot all day every day. And then I did my research and I transitioned to a polarized slash pyramidal model. And I saw my best power numbers across pretty much all durations when I did this. In 2020, I got my FTP up to 370 watts, which is a solid 20 watts higher than I had ever gotten it using a tempo sweet spot approach. Does this mean much? No, I'm just one person, and there are a ton of potential confounding factors here. What I am saying is that very fast riders and even pros make training mistakes, but how much faster could they be if they trained perfectly? Maybe a lot, maybe not much at all, but we don't really know, so using this as some sort of checkmate is really a weak argument, especially compared to all the evidence we have for what a proper training intensity distribution should look like. Thanks for watching. If you enjoyed this video, be sure to give it a like, subscribe for weekly science-based cycling videos just like this one, and share this video with your cycling friends. I'll see you in the next one.