 So, since many of you asked me about references, I wrote something here. As a matter of fact, you don't need any of this for the exam. So, as I will give you some notes, but okay. Here in case you find something on the cover of the first lecture, so density measures and whatever in these two books, for example, that I partially followed, so Sakurai or Messia. And about the chain of harmonic oscillators or identical particle, bosons or fermions. Well, you can find some discussion in Messia, for example. General discussion on this formalism of second quantization, I think, is in this chapter as well. But you don't find the details of the calculation here. You don't hope to find something. That is not. And about the remaining lectures, there are some notes. Maybe I could give you some other references, maybe to some reviews. But so far, I prefer to do it later. Okay, so, yes? No, well, just, okay, h just depends on two variables. And I just brought it in this way because I would like to stress the similarities of the Hamiltonian. Then we will call it h without joining. No, because it's equal to... So the dog of this operator is equal to itself. So, otherwise, yes? So, indeed, okay, we were talking about the quantumism model. So this is the Hamiltonian of the model. And yesterday, I told you that... So we consider the case h larger than zero, no? And j larger than zero. Now, actually, I would like to point out that this is completely general. So this restriction is just a matter of convention. Indeed, you can always... You can find the unitary transformation that changes the sign of j and h. So we can just fix the sign. We show interest in the spectrum and study this particular region. Okay? And in particular, okay, you see, well, we saw this yesterday that the Hamiltonian commutes with this operator, which is the spin flip operator. Okay? And the symmetry in it is broken when h is sufficiently small. You have seen. But there are also other symmetries. So if you can see, instead of spin flip in this direction, you can see a spin flip with a x. Then you see that you can change the sign of the magnetic field in this way. And there is also another transformation, which consists of... Consider the product of a sigma x, sigma y, sigma y, and so on. And this changes the sign of j. Okay? So... So this means that we can focus on... Just on this particular case. Yeah. No. It's x, y, x, y, x, y, alternating executable. Okay. So yesterday, we introduced the Jordan v, no. Transformation. c dot l equal product of j smaller than n, sigma j z sigma l plus... So we have c l, j is smaller than l, sigma j z, sigma l minus. We also introduced other operators, pyonaphermias, the finest a to l minus 1, which is equal to c dot l plus c l. And this a to l, which is equal to i, c l minus c dot l. This operator satisfies the anti-commutative... anti-commutation relations. c dot l c n is equal to delta l n. c dot l c dot n is equal to c l c n, which is equal to zero. Then we are here instead of the anti-commutative a j and a k is equal to twice delta j k. And this is the Jordan-Binger transformation, and these are generally called Jordan-Binger fermions. If you use this transformation, we will discuss also the inverse. What happens to the Hamiltonian? I wrote the result of the mapping, but just consider one term just to see how it works, because yesterday we didn't do it, but maybe we should have. So let's assume that we consider the term sigma z of the Hamiltonian. It's the trivial, the simple one. So sigma z, you have seen that it is simply given by i times a to l a to l minus 1. You have seen that, no? If you write a in terms of the power matrices, then this corresponds to this as a sigma x, and this as a sigma y. And so if you use the sigma that i, sigma y, sigma x is equal to sigma z, then you realize that you find this expression. Let's write that in the inverse transformation. The transformation is a sigma, so a in terms of the power matrices, a to l minus 1 is called j is more than l, sigma j z times sigma l a to l is equal to j is more than sigma j z, sigma l y. So if you multiply the two, you have just an expression of sigma x and sigma y. So this is for sigma z, so we already know a part of the Hamiltonian. What about this other part? Now, if we consider sigma l x sigma l plus 1 x, we want to write in terms of the disperseance. So we have the same x, we have a string of sigma z up to l minus 1. So we have a string of sigma z up to l minus 1 times the operator a to l minus 1. And this is the first operator, sigma x. Then we have a sigma l plus 1 x, which is a string of sigma z up to l, l plus 1 minus 1. This sigma z is written in terms of the of the Majorana fermions in 12 and 12 minus 1. So this means that this, okay, now you can see. Okay, this clearly commutes with this operator, this product. The reason is that we have the sigma z up to l minus 1, but this is written in terms of operator is an n, l and l plus 1. So they commute. So you can put this string on the right and then you find that it's equal to a to l minus 1. Then when you consider the product of this times this, only one sigma z remain. So this is a sigma, sigma lz. Then you have this a to l minus 1 minus, okay, let's write one. What is this? This is 12, 12 plus 1. Now we rewrite the sigma z in terms of the Majorana fermions here. Okay, you don't need this, no? You already wrote this. You need the main lecture notes. Yeah, there was notes in capital letter. Okay, so this is equal to, let's continue here. This equal to a to l minus 1, i a to l, a to l minus 1. Then here you have a to l plus 1. What is this? Ah, 12 minus 1 is important. This is the subscript. And this is an i. We have this expression. Now we know that the Majorana fermions anticommute at different sides. So we have minus sign, let's put the i here. Then you have a to l minus 1 squared, a to l, a to l plus 1. But what is this? This is equal to 1 because of the anticommutation relation. So when you square the Majorana fermions, you find 1. So this is equal to minus i, a to l, a to l plus 1. Is it no or yes? Yes. Okay, yes. But there is a subtlety, you know? Because when you consider here the sum, you have also a case when you have l equal capital N. So what happens when l equal capital N? We impose periodic boundary condition. So this means that the sigma capital N plus 1 is equivalent to sigma 1. But we have to treat these cases separately from this. Because we have to see where this operator is mapped. This is sigma n, sigma n plus 1. So we know that for this simple operator for l smaller than capital N, we find this expression. We find that expression minus i, a to l, a to l plus 1. Now the question is what happens when you consider the term sigma nx, sigma 1x, which is part of the amethyst. Let's do the same. So sigma capital N, capital Nx is equal to the product of the string, the string of sigma z up to N minus 1. Sigma, this is a j. Sigma z. And you have a to n minus 1. And you have sigma 1, which is simply given by a1. So this is what we have. Now here we define an operator by z. Is it happening? So now we define this operator by z. So let's use this operator. This is a product up to N minus 1. So if we multiply on the left by that operator once, we find that this is equal to pi z times sigma capital N z times e, a to n minus 1, a1. Because when you can say pi z times sigma z, you remove one sigma z, the last one. So you find exactly this expression. So you have this expression. Now let's write sigma z in terms of the Majorna fermions. So this is equal to pi z. And then we have i, a to n, a to n minus 1. Then you have a to n minus 1, a1. a to n minus 1 times a to n minus 1 is equal to 1 as before because of the algebra. And so we end up with i by z, a to n, a1. So what's the difference between this term and this term? It's in this part, no? This pi z. Now pi z is commutes with the Hamiltonian. So what does it mean when you find an operator that commutes with the Hamiltonian? It depends. Yes, and it means that you can find the basis of the Hamiltonian, which consists of eigenstates of the operator, pi z. And on this basis, this operator acts like a number, and the number is just the eigenvalue. The eigenvalues of pi z are either 1 or minus 1 because it is a product of sigma z matrices. So this means that depending on which eigenstates we are considering, here we can never sign plus 1 or sign minus 1. And this is exactly why when we wrote the Hamiltonian yesterday, we had these two signs, plus 1 or minus 1, which depends, indeed, on the eigenvalue of this operator. You need to check the expression I wrote yesterday. I wrote something like that, 1 plus pi z over 2 times something plus 1 minus pi z over 2 times something. What's the meaning of this? This is a projector on the eigenstates with eigenvalue equal to 1 for this operator. Indeed, when you calculate this operator by minus 1, you find 0. So this term doesn't contribute. And the same for the other term, with minus 1. So I said, OK, if pi z is equal to 1, then the Hamiltonian is equivalent to this one. If it's z pi z is equal to 1, 2 minus 1, then the equivalency with the other Hamiltonian. So let's try it again. So we start from the Hamiltonian in this model. Then we are looking for the eigenstates of this Hamiltonian. We know that we can this operator communicate with the Hamiltonian. So we choose a quantum number, which is the eigenvalue of this operator. So we define something like this. So we write the eigenstates in this form, where you have this quantum number associated with pi z. And then there can be other quantum numbers that we don't know. And when you apply pi z to this eigenstates, you find plus minus the same. And now, depending on which eigenstates you consider, what you find is that this h applied to plus minus the equivalent is equal to h plus minus. Well, these are the matrices that we wrote yesterday, which are given by h plus minus is equal to the sum. No, no, this is not graphic h, but now are you ready? So this is 1 over 4, sum over l and n up to n and l, calligraphic l, n, a, n. So this is the meaning of the expression that I gave you yesterday. I wrote the Hamiltonian this way. So this means that I can find a basis of eigenstates in which I can choose a quantum number, the eigenvalue of pi z. So this operator has eigenvalues plus or minus 1. So the states have these properties. And now, depending on the sector where I'm working, so depending on the second value, this Hamiltonian is equivalent. It acts identically like this with a plus sign or minus sign. Just explain what the expression of this is. It's OK. And OK, I define this matrix yesterday. And now you can actually write everything because we computed all the elements, all the blocks, all the elementary constituent of this Hamiltonian. So you can reconstruct the matrix. This matrix is h plus minus if you want. And you realize that they have the form that I gave you yesterday. OK, fine. So now that we hopefully understand the Hamiltonian, what I wrote yesterday. So this is, you have this. And I remember it. So we don't need it. We don't need this. We don't need this. Now we don't need this. So let's see that it's not important. That's all right. H. OK, now this is a graduation. H is equal to 1 plus pi z over 2 h plus 1 minus pi z over 2 h minus. Where h plus minus equal sum over ln from 1 to 2 capital N of Al. That is 1 over 4 h ln. So this is the application of ln of ln plus minus. OK, the matrix. Now you have the matrix. And I suggest you to rewrite it in a different way, which is this. You can write h plus minus. Now you express the indices in this form. 12 minus 1 plus j to N minus 1 plus j prime. It's equal to 1 over capital N sum over sum a set of numbers. That we will see in a moment. OK. Plus minus e to the i L minus N k plus minus. And then write this. This is a 2 by 2 matrix. K plus minus j, j prime. And then you have that j, j prime goes from 0 to 1. It can be either 0 or 1. And then you have L that can be equal to 1 and 2. And then which is equal to from 0 to N minus 1. OK. Now this seems to be a bit strange what I'm writing here. But I want to exploit the symmetry of the matrix of yesterday. OK. And you remember that it was a periodic, there was a symmetry of periodicity, there was a symmetry in variance if you want, on the row of the matrix. Or anti, it was either block-circulant or anti-circulant, block-anti-circulant matrix. Yeah, I just called h. OK. Let's use a different, different, because otherwise you can come and let's use some. OK. What is e? I forgot to write it. Humi is equal to, OK, is equal to, now, let's write just the matrix, 0, i h minus i e to the i minus i k. Minus i, OK, there's a title here. Minus i h plus e i e to the i k. So I would like to express the matrix of yesterday in this form. Why I can do that? OK. First of all, in the case of block-circulant matrix, then you can imagine, because of the symmetry, this kind of transition in variance, that you can express it in a simple way in the Fourier transform. So what is the condition to have this kind of symmetry of a circular matrix? The condition is that when you sum capital N to an index of the matrix, you obtain the same element. It should be invariant under this transformation. So in the case of k plus, so in the case plus, you should have that when you consider this expression and you shift one index by capital N, then you should find the same expression. There should be an invariance of this form. From L, L that goes to L plus capital N. Now, if you do this kind of transformation here, you find that this becomes equal to this space, becomes equal to e to the i capital N k plus times the same. So in order to satisfy this condition, so in order to be periodic, then you need to, that this expression should be equal to 1 e to the i n k plus. So this means that we have that the condition becomes this equal to 1, in one case. And in the other case, you have now the condition that when you sum capital N to the index, instead of having a plus sign, you must have minus the same expression, because it was anti-block-circular matrix. So the other condition becomes e to the i n k minus. This should be equal now to minus 1. So this means that you can always rewrite the matrix in this form if you choose this particular coefficient k, satisfying this condition. In practice, what does it mean? How can you satisfy this condition? e to the i n k plus equal to 1. This is satisfied by k plus equal to pi j over capital N, where j goes, for example, from 0 to capital N minus 1. If you substitute, if you plug this into this and you realize that you find all the possible solutions, these are the square root of k. I wrote the matrix yesterday, and you can compute it. And this matrix has the property that when you move along the, for example, along the first row, you have all different elements. But then when you go back to the second row, again you find that there is this kind of circular structure, like this periodic. So the idea is to solve this kind of problem like you would do for periodic function. So you go in free transform, in a free series. Now, if you have the circular structure, it corresponds to periodic function, so you know that the solution actually are e to the i to pi j over capital N, which is what we are obtaining here. And the condition is that the phase associated with the number of sides should be equal to 1. But then we have also another case when the matrix is unperiodic. So it takes a minus sign. So instead of this condition, which is the standard one, we have to impose the other condition. e to the i times minus equal minus 1. Which are the solutions of this? You have simply to shift. Instead of considering integers, now you have to consider half integers. Why is this a case? Because when you multiply by N, you find 2 pi j, which is irrelevant, because it's in a phase, and then you have 2 pi times 1 half, which is equal pi, and e to the i pi is equal to minus 1, which is our condition. So in other words, both matrices that we obtained yesterday can be written in the same, exactly in the same form. And the only difference is in these numbers, that they are quantized in different ways. So different quantization conditions. So what they are seeing is that if we restrict it to the sector when pi z is equal to 1, we will see that these are called the momenta, are quantized according to this case. If instead we consider a sector where pi z is equal to minus 1, the momenta are quantized in this way. So we are splitting the i and functions in two different sectors, and depending on which we consider, we have to apply one or the other. This is the complication of the easy model, with respect to, for example, the harmonic chain, because now we have, instead of just one sector, we have two sectors to consider, they are different. Okay, so in the end, I just give you a picture of this. Let's assume that we plot the eigenvalue of the, in some case, for example, h larger than 1, and let's plot the eigenvalue of the matrix, say h, this spectrum of the Hamiltonian. The spectrum of the Hamiltonian will be something like this, so if you have a random state, then you will have other states. Now, because we know that, depending on the eigenvalue of pi z, we can be in one or the other sector. So this means that in one case, the second values will be the same of the eigenvalues of the Hamiltonian, h plus, but it can be also the eigenvalues of the other Hamiltonian, h minus. So you will have a subset, a subset of the eigenvectors, and it will be the same as for the Hamiltonian, h plus, and the other nearest will be the same for the Hamiltonian, h minus. So this is Hamiltonian, h minus, for example. No, as a matter of fact, this is h plus, and this is h minus, and this is the energies of h. This is the meaning of the equivalence. Is it clear? Okay, sorry. You can have this, because I draw some particular... I choose the color in a particular way, but it's just a matter of that. It can be any value of h, and you can always select... For each eigenstate, you can determine whether it comes from one sector or the other. The important thing here is that there are two possible sectors, and you can identify which one it is. And in each sector, the Hamiltonian can be written in the simple form that we wrote here, just the quadratic form of fermions, like in the harmonic chain. Okay, so... You understand this? I can erase something. Like in the harmonic oscillator, we like to solve the problem. Commutator to h, and h plus minus, say, and a linear combination of the Majorana fermions, and some linear combination, which we had called b-dag k, is equal to epsilon k, b-dag k. So we want to... Because the h plus minus are quadratic, and I said it's by simple combination relations, we can expect that if we compute the commutator with this quadratic operator and something which is linear in the Majorana fermion, we will find something which is linear, like in the harmonic oscillator in the Majorana fermions. So the idea is to choose just an ansatz for b-dag k, so I only know that there will be a linear combination of all the fermions. Let's call this b a l a n. And then I try to find a solution or some solution to this equation. Why? Because let's assume that I, as before that I found the solution for this equation, that this means that given an eigenstate of the Hamiltonian, again, we know that when I... when I apply b-dag k to this eigenstate, so let's assume that you have... OK, let's put this plus minus here and somewhere plus minus... h plus minus, so the quadratic Hamiltonian. So if we are able to find this operator, then we know that when we apply h, if h plus applied to psi e is equal to e psi e, so it's an eigenstate, then we know that when we apply h plus to b-dag k psi e, what we find using the computational... this computational action, that this is equal to e plus epsilon k of psi of b-dag k psi e. So like in the harmonic oscillator, what we added is that for each eigenstate of the Hamiltonian, we can construct another eigenstate by applying this operator and it will have this kind of disenergy. OK? And so this means that also we can define the ground state of this Hamiltonian h plus as the vacuum of the operator bk, the adjoint of bk, because we also know that if this holds, it also means that h plus minus... this depends on plus minus, so I should choose different... I can do that. minus epsilon k bk. So the ground state is defined as the state such that the bk applied to psi ground state will be equal to zero, exactly as in the harmonic chain. So, well, we have a different... OK, this holds for h plus and h minus. Now you could say, we are interested in the Hamiltonian of Bayesian model. So we have actually to see which are the eigenstates of h plus minus which are also eigenstates of h, because there are these two sectors. So depending on the eigenvalue of pi z, we can be in one or in the other sector. OK? So what I mean, let's assume... let's assume for a moment that the ground state of the Easing Hamiltonian, let's call this C is in ground state. Let's consider the state and let's assume that this state is a negative state of h minus. OK? Because we have the possibility either it's h minus or h plus. We don't know now, but let's assume that it's h minus. Yes. Well, OK, yes, there are... If you want to keep in mind that here there is a dependence on plus or minus, this is not needed as a matter of fact because as we saw here, they have different quantization condition. So the k are different for the two. This is why I can actually write in this way for both cases. Only the k's are different. So if you want, you can also put this label plus minus, but it's not really... just keep in mind that you will find different... So... Yeah. This is what will happen, yes. So let's assume that this psi ground state in the notation of before is just given by minus. And then we have some other quantum numbers. So this means that when you apply h minus to psi ground state, we find e ground state psi as ground state. So far we are not. Just assume that we are in this situation. If you're... You could also consider the other k's. No problem. So let's assume that we are in this situation. So the ground state is in this sector. Now let's consider the first excited state of this Hamiltonian. So the first excited state, we know that the first excited state of h minus is just b dagger. You have to apply b to the Hamiltonian, to the state. So we could guess that the excited state would be something b dagger of k applied to psi as in ground state. This is a negative state of h minus. We know that when we apply h minus to this, we find e ground state plus epsilon k psi that is b dagger k psi ground state. This we know for sure. The problem is that, okay, is this an excited state of h or not? This is the question. And in order to see whether it's an excited state, we have to apply pi z. So we have to check whether we are in the same sector as before. If you apply pi z to the state, what do you find? By definition here, pi z applied to the state is equal to minus itself. We were in this sector. Now if you apply pi z to this state, so we are as writing everything clear. Assumption, pi z applied to the ground state is equal minus psi ground state. We are assuming this. Then I saw this means that psi ground state means that the h minus of psi ground state is equal to e ground state psi ground state is implied. Now we consider the state b dug k psi ground state. Let's check what is the again value of this pi z applied to the state. Okay? Pi z is the product of sigma z. And we have that the bk is a linear combination of a of Majorna fermions. So, pi z was defined as the product of 1 to n of sigma z and if you write it in terms of the Majorna fermions, this is equal to minus i to the n product over all the Majorna fermions that you have. This is an m. This is pi z. Now when you consider pi z the commutator between pi z and a generic Majorna fermions you can easily check that it's anti-commute with the Majorna fermions. Why is the case? Because we know that all the Majorna fermions commute with the different Majorna fermions. Sorry, anti-commute. So it should be equal to this let's multiply this expression by pi z to the right, for example. We have pi z a j pi z should be equal to this expression here. Okay, now it's a written like product l goes from 1 to 2n of a l a j times the product from l goes from 2n to 1 over a l. You can write this product in this way reordering the operators, if you want. Okay, this is just the detail but just to tell you that if you do this then you start moving the Majorna fermions and realize that you pick just the minus sign. Okay, yeah. So what I'm claiming here is that when you write pi z applied to b-dug what you find is that minus b-dug k pi z times a gram of state so we find b-dug k psi gram of state. So in other words I'm saying that when you consider a single excitation when you add the fermion to the gram of state of the model you change sector. So instead of being in the same sector you go in the other sector and so this means that this is not really an excitation of the easy model. So this is an excitation of the Hamiltonian H minus but doesn't belong to the spectrum of H. Yes? Yes, okay. You have seen the spectrum is divided in two sectors. Then we in sector the Hamiltonian is equivalent either to one Hamiltonian H minus or H plus. Now let's assume that the gram of state is one of the two sectors, in particular the sector with the pi z equal minus one. Now, okay, we we can consider the excitation of H minus for example, because we know that some excitation of H minus will be excitation of the easy model. Okay, the simplest excitation that we have is just a single particle excitation you put a fermion. You add a fermion there. And so the question is, okay, is this an excitation of the easy model? You try. In order to be an excitation of the easy model it should be in the correct sector which is the sector minus one. Then the idea is we should check it. This is this sector. And now I applied pi z to the to the new eigenstate and let's see whether I find minus one. Unfortunately I don't find minus one. I find plus one. So this means that the excitation is in the wrong sector. So why this is an excitation of H minus is not an excitation of H. How long can I obtain an excitation of H? I simply can apply two. I can put two fermions instead of just one. Because if I put two fermions then I remain in the same sector. So this means that among all the excited states of H minus only the excited states with an even number of fermions are also excited states of the easy model. Only a part of the spectrum of H minus is also a spectrum of H. And the same for H plus which gives the remaining eigenvectors. Just to the calculation if you want. Let me think a simple way to see. Yeah there is this problem. We know that pi z is equal product from 1 to 1 to sigma Lz. That's right in this way. Now we have pi z multiplied by a Majorana fermions A. The Majorana fermions A as you have in your notes is equal to the string of sigma z times something which can be sigma x or sigma y. Right? It's written in this way. It depends on which operator it is. So we have here that the operator A I want to compute something like this with some L. And what I'm saying is in terms of spins in terms of spins will be for example A to L minus 1. This will be 12 plus 1. So this will be given by the product from 1 to L or sigma z times sigma x for example in this case L and then here pi z here. This string of pi z commutes with sigma z. It commutes with themselves at every position. So you can move this here pi z. So this is equal to product that j goes from 1 to L of sigma z. And then you have here to compute pi z sigma L x or y or whatever. You have one of the two. Pi z. The only term here sigma x is the one corresponding to the side L. So in the end you have on one side you have one sigma z in the middle you have a sigma x or y and on the right we have sigma z. So here you will end up with I hope you can see here with sigma z L, sigma L x or y sigma L z. These two probably may be some type of mute. It doesn't work. You can find this and this is exactly equal to minus sigma L x, y. This is related to the algebra of the polymerases. It's just a check you can do. It's not a problem it's just that the first single particle excitation of H minus is not an excitation of the easy model because if it were an excitation then the the eigenvalue of Pi z would have been equal to minus one. This is the problem. We have said H H plus minus something equal to E, some energy plus minus and something. We have this. We have identified the ground state we assume that the ground state corresponds to the sector minus one. Now we say I want to I want to consider the I want to focus on the sector with Pi z equal minus one. I just choose that particular sector. I know that in the particular sector the Hamiltonian is completely equivalent of H minus in that sector. So now I can compute the spectrum of H minus all the eigenfunctions and eigenvalues but now I must be careful because this equivalence holds only if Pi z applied to the eigenstate is equal to minus that eigenstate because I'm assuming that I'm in that particular sector then I check it so I said okay let's consider a single particle excitation in H minus and I realized that no, I'm in the wrong sector. The eigenvalue of Pi z is not equal to is not equal to minus one but is equal to plus one. So there is a problem. I should reject the eigenstate. Yes, yes. We are constructing the eigenstate and we say okay we will see that it's easy to find the eigenstate of H plus H minus and now we have to identify which one are the which are the correct one the eigenstate of this model because we have these two possibilities and so in this case we immediately see that if the ground state is in the sector with minus one then we see that the single particle excitation state is not it should be in the other sector. So we have to diagonalize the matrix H plus to get that excitation not H minus. From H minus we can extract all the excitation with an even number of fermions because when you have an even number of B here this operative commutes with them because it takes a minus sign every time so minus one to the to the 20 is equal to one. This is not complicated it's just a problem of understanding one another so I won't probably a graphical representation is better so we have all the eigenstate of H some eigenstate of H are eigenstates of okay this is H this is H plus and this is H plus some of these eigenstates are the same as H minus other are the same of H plus but it doesn't mean that these are all the all the eigenstate of H minus you can have other eigenstate for example this one this one this one the same here you can have this eigenstate this eigenstate H plus when I compute the spectrum of H minus or H plus I compute all these eigenstates but then I should understand whether it is the correct one is there is a corresponding eigenstate of this in model I have to ignore all these eigenstates because they are in the wrong sector and this is what we did we checked if the wrong state is here the first excited state is not there the single particle so we have to excited two particles two particles in order to be to be to remain the same set we see clear now we cannot reach from the wrong state from which from what we infer this for each eigenstate each eigenstate of H has a definite definite value of of pi z so either one or the other here I'm alternating them this is not true in general so they can also be you can have more states condense that you have to neglect it's just a drawing don't take it seriously so how can you we know that the the single particle excitation of H minus is not in the spectrum of the easy so this means that we have to diagonalize H plus again we diagonalize H plus and then we have to check what is the eigenvalue of pi z and now should be equal to plus one and then we select all the second state now you can do this kind of calculation what do you find now I'm telling you the result of this the result is that when you are interested in the ground state let's consider just the ground state of the model for any finite value of N capital N independently of H the ground state is always H minus pi z equal to minus one independent of H the ground state is always there what about the excited states of the model this depends on the value of H so if we now consider our phase diagram this is H this is H equals one zero so we have that the ground state here is is in the sector minus one and then what we find is that the first excited state there is a gap here general and the first excited state is I think I think is in the sector plus generally and there is a gap now if you consider the case A the case H is more than one you find that the ground state is still the sector minus but B there is a very small gap between the ground state the first excited state which is now in the other sector and this gap scales exponentially is exponentially small in the system in the system size so these two eigenvalues become degenerate these two these two eigenstate becomes degenerate in the in the thermodynamic limit in particular we know that this gap is equal to zero when H equals to zero because the two states are really degenerate we check yesterday that we have the spin either are in the X direction or in the opposite direction so in the case the gap is zero in the in general case for H larger than zero you find that they are closed in the final system but when you take the thermodynamic limit then it becomes degenerate and then here there is a gap this is the situation the qualitative the qualitative properties of the spectrum not the low energy spectrum not the ground state of the model we shouldn't be so surprised about this because we know that here there should be a phase so in the thermodynamic limit there is a symmetry which is broken so this means there should be some degeneracy in the ground state notice that this degeneracy exists only in the thermodynamic limit and if the symmetry is broken only in the thermodynamic limit if you diagonalize the model for final chain you use your mathematical for example to obtain the energy you find that there is just one ground state but this doesn't mean that that is the that is the correct ground state in the thermodynamic limit and in particular I can tell you that yesterday we talked about which is the correct ground state to consider so we said that you cannot pick any linear combination of the states in x plus and x minus but there are particular combinations which is just the spins given in x direction or opposite direction here you cannot consider any linear combination of these two the correct linear combination is actually the superposition like this is plus plus or minus minus divided by square root of two these are the physical ground state the physical ground state is not in the particular sector it mixes the two sectors okay and this is a very important complication when you compute correlation functions because everything is simple when you remain in the same sector but when you have to consider the presence of two sectors it makes the calculation more complicated so is it clear the picture or less this is just a description of the results that you will find computing the expectation value of pi z okay but we were talking about how to solve this question I give you 5 minutes she maybe not I said something wrong thanks because yesterday we we compute the ground state in the limit h cost infinity and we had that all the spins are aligned along z now if you apply pi z to this state you find plus pi z per plus the state with those spin up the sigma z applied to up is equal to up so I was wrong when I told you that the ground state is in the sector minus is in the sector plus sorry my mistake your problem is not a problem okay when you apply pi z to all spin up you find all spin up with the plus sign so this means that the state with all spin up is in the sector with pi z is equal to 1 we already know that this state is the ground state when h cost infinity so this means that I was wrong saying that the ground state is in the sector h minus the h was equal to h minus in the ground state this is the opposite I said something wrong here you have a plus I repeat I was wrong when I said that the ground state is always in the sector represented by h minus is in the sector h minus h plus okay so okay so so we still have to solve this problem like in the harmonic chain so what should we compute we should compute the commutator between h plus minus which is written like 1 over 4 some ln of al h calligraphic h ln am and some linear combination probably tired and I if you want I give you I tell you the idea you know you have to commute these operators then you find when you anticommute a n a j you will find delta jn and so on it's very simple I'm not doing it because I guess you are tired but the idea is that if you write this you see a matrix notation let's introduce a matrix notation because everything is easier so if we introduce this vector of a or the major affirmants a 1 8 1 okay again then this can be seen as a matrix so you can write this as 1 over 4 then you have some vector a scalar product this matrix h plus minus vector a is just this term and you can interpret this as b k if you do this what do you find when you consider commutator so I tell you that if you have the commutator of a generic form of this kind a is a matrix and here you have a vector a scalar a you find a very nice result that this is equal to this is equal to mv the scalar product a so you just have to apply the matrix to the vector which is very similar to what we found it's actually the same what we found in the in the harmonic oscillator and this you can prove at ease a and then you will find when you commute this with this you fix the index the colony index of m and then and so on so why this is what this is so nice because we wanted to solve this problem and now this problem is equivalent to again to compute the eigenvalue and the vectors of this matrix of the matrix h plus minus in other words that's this equation here is equivalent to to find h to solve the problem h plus minus applied to the vector which is called b bk so this problem is equal to epsilon k bk I'm saying that here when you write h plus minus in terms of the major you can just you you can reinterpret this as a scalar product between a vector of a and another vector which is h plus and here I'm saying that polygenetic polygenetic anti-symmetric matrix you have this relation notice indeed that our h plus minus are anti-symmetric and why you can always choose anti-symmetric matrices because the major and a fermions anti-commute so the symmetric part of this matrices don't play any role so you are always free to choose to put anti-symmetric matrices here and indeed you have this theorem holds for anti-symmetric matrices for any anti-symmetric matrix you have that the commutator between this quadratic form and a linear combination of the all the major and a fermions can be written just as the matrix applied to the vector corresponding to a linear combination scalar the vector of it so the other linear combination you can really maybe you can show you can show something you can see the eigenvalue and the vector of these matrices which are this one this is not so difficult because like we guessed the solution for the harmonic oscillator we can do the same here and the idea is that to again consider phases now as a solution but now we must take into account that the matrix is block so this means that we have to treat we will just we will seek for a solution of the form of the following form for the vector bk is equal to bk 2l minus 1 plus j is equal to l that is correct this times some vector b dimensional vector ok so I am saying that I am divided I split the indices in pairs 2 by 2 here you see this label the pair of the initial matrix and then these other indexes for the elements inside the two the pair and then what I am saying is that if I consider this index just the index of the pair then I can use the symmetry of the matrix which is periodic and so I know that the solution is just going to be a phase the usual phase which is the Fourier series but ok I must take into account that there is this the degrees of freedom of the pair the fact that now this is not a scalar so here I should expect that I have to multiply by some vector that I have to determine which diagonalize the 2 by 2 part of the matrix I am just saying I have this matrix I see this matrix as a block matrix so these are 2 by 2 we have seen that the blocks have the periodic periodicity so this means that I can seek for solution where for each block you have a scalar phase and so on but now here we have to find which is the correct the correct vector to dimensional vector you are seeking a solution of this form because this solution has this form as a matter of fact you can convince yourself or you can just read somewhere how to compute the eigenvalues and get back to the block circle matrix so what do you think if we plug this expression in the question for the eigenvalues what do we find h plus minus 1 over n sum over k plus or minus of e to the i l minus n k q plus minus this 2 by 2 matrix equal e of q plus minus index j j prime and then I have to apply this to this particular vector so I have here i i n k plus minus and I have this more vector v j so this is the first part this is this okay and this should be in the exponent here is written this and with this one but now I just changed the name instead of using k I am using q because then I have k on the other side okay 1 over n sum over q plus minus of this j j prime and then I am applying to this vector and we have to sum over n because it is a multiplication matrix time vector so sum over n sum over n when we sum over n now we have the sum over n of e to the i just consider sum over n we have sum over n of e to the minus i n q plus minus plus minus sorry q plus minus you have this no right now you remember q plus minus are given by 2 pi no 2 pi j over capital N so as a matter of fact you will find that independently of plus and minus the sum is just equal to the delta of q plus minus times n I repeat so e to the i when we define k plus minus they were defined by the condition e to the i q plus minus is equal to q plus is equal to 2 pi as this form 2 pi j over capital N instead of q minus as the form 2 pi j plus 1 half over capital N these were the two possibilities the two sectors, the two quantization now independently of the quantization because there is a minus sign here you simplify the 1 half so in the case you have the when you sum all the phases sum over n, especially you find the chronicle delta so the 2 q plus is equal to k plus and q minus is equal to so for this reason we first sum over N this expression and what do we find when we sum over N so we find we are computing h plus minus b the element and this is equal to 1 over N simplifies with this N and q plus minus is equal to k plus minus pi L k plus minus times e k plus minus applied to the vector b k plus minus and here there is an index and this is 2L minus 1 plus plus j prime let's see and here we have j prime j then we have the vector v j and there is a sum over j what I mean I am just writing this expression so this which I which is nothing but this thing here now is equal to we sum over N so this gives this constraint it has the q plus minus is equal to k plus minus I am using this here and so I end up with the iL this matrix multiplied by the vector vj b you are confused because there is j prime there and there is a sum over j prime here this is what I find and I want to impose that this is equal to epsilon k vk this should be equal to epsilon k and vk was e to the iL k plus minus and a plus minus and there was v vj this is what we are imposing what we are trying to solve now you see that this term simplifies so this means that our answers was correct otherwise you would have added some dependence on L here so this means that in the end what we have to solve is a negative model problem for this 2 by 2 matrix here e e of k times v should be equal which is v of k v of k should be equal to epsilon of k times v of k I know that this is late and it is confusing but what I mean is that we started with a very complex problem because a native array system if you count the degrees of freedom we have 2 to the capital N degrees of freedom which is a lot then somehow using the algebra of the Majorana fermions we were able to reduce this problem to a problem of finding the eigenvalues of a matrix which is 2 capital N times 2 capital N matrix so we reduce the the complexity of the problem exponentially from 2 to the n to n now here using the properties of our matrix if the factor is periodic or anti-periodic we were able to reduce the problem to find the eigenvalues of a 2 by 2 matrix this is the magic of the solution of the easy model that we started with a extremely complex problem and then you finally manage to reduce everything to a 2 by 2 problem you can you have the matrix to compute the eigenvalues of this do it not now but the eigenvalues of the eigenvector then next time we will tomorrow we will finish this and then we will start talking about dynamics don't be scared about all this part you don't need to be able to do all the steps so the diagonalization is just to know how to do it and just to understand the meaning of the fermions how to construct the space and that is important I hope that that you can understand that the structure then the details are just details technical details maybe you need more time you have to do all the steps and all the passages in a few weeks you will realize that this is simple I can understand now