 So we are computing the displacement squared, we obtain this result plus the special value of x squared, now this is not very nice, okay, because as a matter of fact this is divergent and the reason essentially is that, well, because of the similarities of the problem, they can be arbitrary, distant, if you just move all the ions, there is this invariance and they, so if you go see the distance from the given points, then you can find something which is arbitrary, distant, okay, it's just the meaning of the quantity of your computer, it's not really physical, I would say, so in order to remove this kind of ambiguity, maybe we can just consider the, instead of the displacement, x squared, we remove the position of the central mass of the problem, okay, and then we compute this kind of displacement, because otherwise we have a way this kind of divergence, because of the arbitraryness in the position of the atoms, okay, it's just okay, this is a pathology, this quantity actually is pathological independently of this, but I didn't want to actually consider this kind of pathology, there is something that's more interesting, so let's just consider this, just remove the contribution from the position of the central mass here, let's consider this quantity, instead of x i squared, x i minus x squared, okay, we are removing that, let's do this, okay, so in this way we are removing this infinity that is kind of not physically meaningful, I would say, and we end up with this expression here, so this expression and we have seen before that this is just equal to 1 because of the commutator, so in the end you have that is equal to the sum from k over k from 1 to n minus 1 over 1 over 2 m capital N omega sine over, and so on, yeah, what is this? This is, I can tell you the asymptotic result in the limit of large n, if you consider many, I think this is asymptotic, many, many oscillators, so this behaves as log n divided by pi m omega, okay, this is just the result of the sum, if you write this sum and then you consider, you assume that n is large, then you find this expansion, okay, this is the leading term, so this term, yes, log n over pi m omega, okay, the coefficients are not important here, what you can see is that in the limit of large n, this approaches infinity, so as a matter of fact, even if we subtract the ground, the central mass position, then we find that this atom, this science can be arbitrary, arbitrary distant from this, like this size, okay, this is something that you don't find higher dimension, in particular if you solve the problem in 3D, then you find that they are close to the, to the lattice side, and this is why in 3D you can talk about solids, they are close, instead of here in 1D, but we'll see in 2D, I guess, I'm not sure about 2D, because this is a quantum problem, but without that in 1D, they can be arbitrary, far away on points, yes. This is somehow, where we are, we are at temperature equal to zero, that can be phase transition, temperature equal to zero, you can't have a finite temperature phase transition in 1D, and this is why one dimensional system at temperature equal to zero are similar to classical system in two dimensions at finite temperature, so you have this kind of correspondence, and in 2D you can have phase transition, the easy model, okay, the classical is more the element, no, not the quantum, okay, so, well, so this one just to tell you that it's something to diverge, but okay, maybe then we, instead of asking this kind of displacement, a single atom with respect to the lattice point, which was kind of an artificial point, we could actually wondering what is the distance between two atoms, what is the displacement squared between two atoms, so I mean something like that, something like this, yes, but this is useful, so I erase this, here it's not, so this other quantity is, so you see that now we don't have any more problem of the central mass, because it's the difference between two positions, basically, meaningful, and I mean the distance r, so two ions, so this is the first ion with a one plus r, okay, and, well, what's the meaning of this? Well, we are just saying if they are correlated, because if this is, if this is more, then this means that if this ions is to the right, if you want, of the lattice sides, it means that also this ions is to the right, so there is a correlation between them, if this is finite, okay, you can expect. The question is, is it finite in this one-dimensional model? Let's see, at least on large distances. So, how can we compute this quantity? The idea is the same as before, so we use this latter formalism, so the formants, and now this is more complicated than before, because we have this operator, actually it's the same as before, because it's the operator of squared, so again, we have to use this expression, yes. You can't read the indices, okay, should I change? It's right, five, okay, it's, I don't know, anyway, okay, I, maybe you want this arm, sorry, okay, okay, yeah. So, we use the same trick as before, so we have to compute the expectation value, now we have to write x1 minus x1 plus r. So, what you can see, x1 minus x1 plus r, so this term now simplifies, likely, and now we have to consider this expression with the l equal one minus the same expression with l equal r. Notice that this is independent of l, okay, so we can just write this minus, this for l equal one minus this for l equal r times the rest. So, this is equal to sum over k, you have e to the minus 2 pi, 2 pi i k over capital N minus e to the minus 2 pi i k1 plus r, sorry, everything divided by square root of 2m omega, sine of pi k over capital N times i a capital N minus k minus a times k. This times itself, because it's squared, this times the same expression on the value. Now, as before, when this operator acts on the left is equal to zero, because this data is annihilated by a tag, it's on the left, not on the right, this is why there is a tag. So, I can forget this and ignore this, and the same, you have an expression, you will have something a N minus k, instead of k, let's call k tilde minus a dagger k tilde here, or q, that's why we call it q, and then you will have to ignore this part. So, this annihilates the value there. Yes? At the denominator, I missed it, yes. Thanks. No, yes or no? I'm also here. I know there is an N here. Okay. So, the final expression, what is it? Again, here you have a, yes. Yeah, there is a sum over q. It's just the same expression. No, okay. You do what you've done before, and so you have here the vacuum, again. Now you have some, okay, let's write both, sum over k, sum over q, and here you have e to the minus 2 by ik over capital N minus e to the minus 2 by ik, ik 1 plus. Then here, we'll have the other with the q. So, there will be e to the minus 2 by iq capital N minus e to the minus 2 by iq 1 plus r divided by capital N divided by square root of 2m omega sine of iq over capital N N, and here you have the operator remain, which is minus i a dark q. I wrote everything, no? Now, again, we know this step. The only way not to move this on the right annihilates the vacuum is that n minus k should be equal to q, or the opposite q equal to n minus k. If q is equal to capital N minus k, this sine is exactly equal to this one, because sine is symmetric. Then, all the rest is equal. So, I can remove the square root here. This becomes n minus k. So, I change the sine, okay. This becomes n minus k. So, I change the sine k. This is n minus k. And, yeah, that is not q anymore. Now, again, a n minus k times a dark n minus k applied to the vacuum is equal to 1. You can simplify this to operator. And so, we have our expression, which is a sum over k, from 1 to n minus 1, of this time this, of e to the minus 2 pi i k over capital N minus e to the minus 2 pi i k over capital N 1 plus r, everything multiplied by its conjugate. So, this is absolute value squared divided by 2 m omega sine pi k over capital N. And, okay, now I am below the line. Yeah, this is our result. How can you, okay, because this is user, you see, because this is the same expression with the plus sine instead of the minus sine, this is why I put the r sub r squared is the conjugate expression. Now, maybe you see immediately that this can be written as a sine, is there? Yes? Yeah? What? The meaning. Ah, the meaning, okay. So, now we are computing x1, the position of one, the displacement of one ions minus the displacement of the second of the other ions, which is actually the difference of the distance of the position squared. So, you can see that like you, you can see this, this displacement minus this displacement squared. If there is some correlation between the two, then this quantity should remain finite in particular, because, okay, if, if this is completely array to this one, if two ions are very far away, then this means that you can have one atoms in one, in one direction, some position, the other atoms in a completely different position, so it can be arbitrarily large. Instead, if there is some correlation, so if this is on the right, then also the other is in the right, then you can expect that this remain finite, this kind of quantity. This, this kind of quantity could remain finite, finite, yeah, this is the reason. So, the, what we are looking for is whether in the limit of large R, so large, large distance, far away ions, then there is some correlation between the position or not. So, if you have some structure in, or maybe they are just random, then, okay, this is the essentially, this here, I can tell you that in 3D, or maybe also in 2D, in this case, you have this kind of correlation, so you find something finite, there is some order, okay. In 1D, we are going to see that this doesn't happen. Okay, so this is the result, I, let's just erase here, also here, so this can be written as, so we are, this is equal to, as you see immediately, there is 1 over, actually it's a 2, over m omega capital N sum k from 1 to capital N minus 1 of sine squared of pi k r over capital N over sine of pi k over capital N. This is just another way to write this equation. Yeah, you see your face, and yeah, I wrote every interval of sine, you can do this, just a step. Then here, what we can do is, we can see the limit of large N in this sum, maybe you are not familiar with the Euler-McLean formula, you are not, I guess, so the way to write some in terms of integrals, I just tell you something that can be useful for your future, so yeah, I could just give you the result, maybe it could be useful for you. So if you have something like this, you have to compute some quantity like this, okay. And let's assume that this is actually a function of N over capital N, and this is a sufficiently smooth function. Then what you find is that in the limit N goes to infinity, you can replace the sum by an integral, and then you have that this is the integral over this variable x from 0 to 1 of x, you can prove this. This is, this is called, you can check if you are interested in the Euler. Actually okay, the formula is more complicated, I just show you a simple application of the formula, where A is a sufficiently smooth. So this means that in this case actually we can replace the sum because you have 1 over N sum of this by an integral, and what we have is that this approaches in the limit N goes to infinity, 2 over M omega, the integral from 0 to pi of vk over pi of sin squared of R yx, okay, k divided by sin k. And then you can prove that this is not at least the coefficient, it's not trivial to find, but I can tell you the result that if you consider the limit R large, so large distance between the two dv ions, R goes to infinity, R is very large, then you find that this behaves like 2 logarithm of R divided by pi M omega, just the result of the calculation if you want just, the coefficients are not really important over here, we care maybe about the behavior, that it behaves like a logarithm of the this distance between the reference points. So in particular what we find here is that this quantity, this displacement approaches infinity logarithmically when these ions are very far away from each other, so this means that they are not really so correlated, so this is not another proof that this is not solid, there is no order here in this system. Okay, what happens is that they I can tell you the classical result, if you study this model class, then you find that in 1d there is no order, order I mean you don't have the position order of the atoms, and in 3d there is position order, in 2d you don't have position order, but you have an orientational order, something very subtle in the classical case in 2d. In the quantum case, again in 1d, apparently we don't have any kind of order because of this divergence here, I don't know the behavior in 2d, as a matter of fact, but I think that there is some order like this. Okay, we use this model just as an example of a quantum antibody system, so not as an example of phase transition, so we have what we find anyway. Why? No, no, there is a potential. Okay, you mean external potential? No, no, no. Harmonic. If you change the potential view, you can find different... Ah, yeah, no, no, this is for this model, I'm not saying that in general you want this behavior. For the model we consider we have this not extremely interesting behavior. Okay, this was just to present to you the formalism of what is called the second quantization, so when you introduce the Spock space, now your states don't have a fixed number of particles, but you characterize the state using either bosons or fermions and you have to consider all the possibilities. So you consider, for example, you have n ions, but when you consider excitation you can have infinitely many bosons, because you have infinitely many excited states, so there is no constraints on the number of the particles. It's this clear, so it's not bounded by capital N. When you consider excited states you can apply a DAG how many times you want and you find always excited states. Like in the harmonic oscillator you have infinitely many excited states, so you can apply a DAG how many times you want. Not to stand you being only one oscillator. Okay, anyway, here we consider rather simple correlations because we were able to compute them just the one to really rewriting the expression and simplify a and a DAG, but you can imagine that, well, if we now compute slightly more complicated correlations, for example, this times something like this. You put R1 here, X1 minus X1 plus R2 squared. This starts being a bit complex now, because now you have two fermions from here, two bosons from here, two linear combination bosons, and two linear combination bosons here. So four linear combination bosons, so you have to use this trick of moving all the fermions from one side to the other. Yeah, it can be a mess, it can be complicated. And then maybe you could also be interested in other observer, so you say, oh, why don't put another one? And then you have six. So it can be, well, without doubt we are able to solve this problem. But the problem is that we are able to solve it in a decent time. Fortunately, there are some techniques, some theorems, mathematical theorems that allow us to compute these more complicated expectation values just by reducing them to the ones that we compute, the ones when you have only two bosons. This is called the Biggs theorem. You already know the theorem in high energy physics, but I state now the result in this context, in case you need, you can apply. Okay, so the theorem is the following. So, Biggs theorem, let's assume that you have operators of this form. A j equals sum over n of c j n a n plus d j n a dot n. So linear combinations of your creation and destruction of rados, or particle, just a linear combination of them. An arbitrary linear combination. And just I put this label j just to indicate a particular linear combination which depends on some index. Now, we have seen, for example, I give you an example. If I want to compute a j, a l, something like this, what do I find on the ground state? So on the vacuum here. What do I find? Well, I have just to write this expression. You have sum over n. Let's call the other one m of vacuum. And here you have a c j n a n plus d j n a dot n. So c l m a m plus d n. Again, as before, we only need to, okay. This term is equal to zero. It doesn't count. This term doesn't count because it is at the vacuum. So this term gives you a delta m n. So again, this is equal to sum over n of c c j n d l d l n minus. And minus nothing. Just this, okay. This is just the result. In the case of two operators, a j and l. But now, okay, we are more ambitious. I would say, oh, I would like to compute a generic. We want to compute a generic product of a operators. So it's kind of, yeah, very complicated. You are here. You can have a one, a two, and so on until a n. In principle, this is a lot of work that you have to do. But now, okay, there are some simple rules that you can immediately see. First of all, let's assume that I have an odd number of operators here. Then this will become equal to zero. Why? Because the only way to avoid that one operator moves and goes to annihilate the vacuum. Is that you must encounter another operator. The dog of itself must incant. But if you have an odd number of operators, there will be always one, at least one operator that is unable to find this is made. And so this means that this will be equal to zero. So the only way to make this different from zero is that here we have an even number of operators. Otherwise, this is equal to zero. You can see in the case of one single operator, you have this term. So one is annihilated on the left and the other one on the right and the other one on the left. And the same for three, five and so on. So we can focus on this kind of correlations with an even number of operators. And the theorem, the mixed theorem states that this is equal to, this is equal to the sum of the product K from one to N vacuum here. A, I, K, A, J, K, five, what is the sum? Some partitions, one to N, I, one, J, one, smaller than J, N. Now this is the theorem and then now I explain you what does it mean. Let's see what you do. So in practice, what do you, okay, I can think again. So what do you do in practice? So let's, let's apply it in some simple cases so you understand what was the idea. So let's assume there were four operators, a1, a2, a3, a4. Now the idea is that this is equal to the expectation value of a1, a2. You can see all the pair that you can form here or the pair, so you start with a1, a2. And it remains a3, a4, plus then you have the other possibility, a1, a3. We have a1, a3, and remains a2, a4. Plus the last one is a1, a4, and here you have a2, a3. What happens if you have six operators? Six operators, then you add again. A1, a2, and then you have here you have to consider the rest, which is a3, a4, a5, a6. So you can do it recursively if you want. Plus a1, a3, and the rest. A2, a4, a5, a6. Plus a1, a4, a2, a3, a5, a6. Plus a1, a5, a2, a3, a4, a6. Plus a1, a6, a2, a3, a4, a5. And for each one of these, you use the same decomposition of before. There are many terms, fine. But you see what is nice is that we are able to reduce this very complex. This is extremely complex because here each term is a sum. It's a linear combination of creation, destruction, operators. And there are six of them. And we are able to reduce everything in terms of the expectation values of just two operators. Wish we know. Wish we already computed. OK. How is this called? As a matter of fact, there is a name for the entire combination. A mathematical name for this kind of combination is called permanent. It's the permanent of a particular matrix. So it's a equivalent of determinant. So maybe you can recognize that this is similar to a determinant. If you interpret this as the element of a matrix, it's very similar to a determinant. But now we have all plus signs. And this is called permanent. Just to know if someone asks you what is a permanent. Something related to fixed theorem in Bosons. Permanent. OK. This is all fundamental. Just a curiosity if you... Is the determinant with the... Actually, it's not. More or less. So let's say it's not there. It's actually the permanent. The truth is that the square of this is the permanent of the matrix. OK. This is just... It's not important. Forget it. OK. That's not... Yes. OK. But this is not like a determinant. It's like a faffian. We'll see. Just forget this permanent. It's not really important. OK. So anyway, but there is a way in recursive way to... You just consider all the possible permutation. You have a way to access also this kind of correlations. I have to say that in this Bosonic case, you don't know permanent. But there is a reason why you don't know permanent. Because there are not many theorems on permanence. So this means that you... It's nice that we have this expression, but we are not able to do many calculations, I would say. If I give you... If I ask you, oh, let's consider product of 2n, 2n operator for generic n. I doubt that you are able to give me some expression. OK. Something interesting. Fortunately, this... If you consider fermionic analog of this powder, then situation changes. OK. Because in that case, we can actually compute also more completely correlation functions. Because instead of permanence, we have determinants. And there are many theorems for determinants. OK. Just a parenthesis. OK. So in these two lectures, we consider this... 3 now. OK. This harmonic chain, so a quantum mechanical problem. We start from that. And then we use a different formalism to... A different respect to usual formalism to solve the problem, introducing these ladder operators. And so we define this kind of fox space, which is constructed using some effective particles, bosons, with which you can indeed... You can construct the ground state and these are the states of the model. We... In this particular model, we found that this operator commutes one another, except for the commutation between A and A dagger, the same operator. So they are bosons. OK. OK. Now, maybe in the future, not next time, maybe also tomorrow, I don't know, we'll also see some dynamics in this problem, perhaps. But I think that now it's better if we... If I also present another model, another simple model, so only the static properties of this other model and see how to solve it and see some general properties of this other model. And this is a spin chain model. So instead of considering our harmonic oscillator, we consider again, in the first lecture, a spin system. It's the most expensive. So we start discussing the B-1-2 easy model for icing, if you prefer. I don't know. I think it's easy. It's better than I know. See that? One, two, easy. Model. Model. One dimension. One dimension of quantum. Easy model. Also known as transverses, transverses, field, ice, easy. OK. Also known as transverse field decision. I think that most of you know the classical easy model. I want to see everybody. So the classical easy model. For example, it's something like sub over i of s, i, s, i plus 1. Then you put... You want to put... Yes? Why not? You mean this? This is the classical. Classical is in model. So you have this term. Then you can have a magnetic field. OK. And well, we know how to solve this problem. OK. Maybe here instead of i, if you consider generic dimension, generally there is a sum over nearest neighbor spins. You can have it 1d to d over there. And then you have this magnetic field. We know the solution of this model in one dimension. It's easy. No? You know. No? Yes. And you can solve it using the transfer matrix formalism. It's a very simple calculation. Then we also know the solution of this model in 2d, more tricky solution in the classical case. We don't know the solution of this model in 3d. If you find a way to solve the model, if you want to tell me. OK. And what about the quantum version? Generally, when we discuss, when we compare classical models with the quantum models, what happens in d plus 1 in the classical case happens in dimension d in the quantum case. So for this model, we are able to solve easily the reason. We will solve the case, the one dimensional case. It will be is feasible. We are not able to solve the dimension of quantumism model as far as I know. So because we are these lectures, quantum lectures, so we have to confine ourselves in one dimension of quantumism model. So the model, I will consider the ferromagnetic model. What is the reason why I'm saying ferromagnetic? Because I put minus something. So j from 1 to capital N. Now, instead of having the spins, the classical spins, which are just variables that can assume values plus 1 or minus 1 or 0 or 1, OK? Now we have spins, spins that we can represent. We have seen the first lecture by Pauli Metrizes. So generally, the Hamiltonian is making this form. There is some interaction between near-neighbor spins in a given direction. For example, x. Let's call this x direction. Then we can put an external field. But in order to make this problem quantum, we shouldn't put a longitudinal field in the same direction. Because otherwise, it becomes just a classicalism model. Because then you have this. You can treat all this freedom-like classical spin because everything commutes here. If I put here sigma j x, this commutes with this. So we can solve this problem in the basis that diagonalizes sigma x. So this is actually a classicalism model in this case. So in order to see some quantum effect, we have to tilt the magnetic field. So instead of having the magnetic field in the x direction, we consider magnetic field, for example, in the z direction. So this becomes a quantum problem. This term doesn't commute with this one. So it's a non-trivial problem. Your choice. I prefer to put it here. So I can choose j as a dimension of energy. And then this will become a dimension without dimension. Just a simple choice to avoid. Because I don't want to write h with some dimension. So this is the model. So what do we know about this model? So without solving it, can we already understand something of this model? For example, what happens if h is very large? So you imagine that there is, OK, you have this. These are spins interacting between one another. And that there is some magnetic field in a different direction. And now this magnetic field is so strong. It's very, very strong. So what happens to the spins? Yeah. h is very large. It's very, very big. It's very large. What happens? All the spins tend to align along the magnetic field in the direction of the magnetic field. So this is why it's paramagnetic. It's paramagnetic in the case. So what happens is that you increase h. It tends to align in the z direction. And so what is the ground state of this model? The limit h goes to infinity. It's simply all the spins. Let's assume that h is larger than 0, OK? So it goes to infinity. And we consider the limit h goes to infinity. So we know that the ground state of the model in the limit h goes to infinity is nothing but all the spins in the z direction. Yeah, the sigma is just 2. It's a 2 by 2 matrix. So it can be either half or down. It's a spin of 1 half. It's the standard version of this model. Like the classical regime model, you have two possibilities. Otherwise, you call it a POTS model. So this is the ground state in the limit h goes to infinity. So we already know something. What about h goes to 0? If h is equal to 0, what do we know? We don't have this term. We have already some time. So what's the ground state? 1 half and 1 half. OK, 1 half, what? Y is the disorder. How do you infer that this is the disorder? That the ground state is the disorder. There is an interact. There is already this term. And you want to minimize this term. Because there is a minus sign, you want to maximize this. And the maximum of this actually is simpler than that. Maximum is just all spins are aligned in the x direction or in the opposite. Because here you can see the sign of the direction. So you have two possibilities. Either let's use this notation for saying that the spin in the x direction goes to the next here. This is z. You can have this or this. So my components is two. This is spin 1 half. Each spin. So this is spin 1 half. We are in the sigma set base. So this is spin in the z direction. We have two possibilities. So what happens when we have this kind of degeneracy now? What can happen? It's spontaneous symmetry breaking. And so what does it mean? What's the mechanism? How does it happen? So the idea is that we wrote the Hamiltonian in this form. But now maybe there are extremely small terms. Just the x. You know some really small that approaches 0. For example, it could be coupled with a magnetic field in the x direction. Why not? You cannot exclude this kind of terms. Maybe we are considering which epsilon is very, very small. We are unable to see it. Just a little bit. You say, OK, here there are other terms you are not considering. They are very, very small in the Hamiltonian. And the idea is that OK. But we would like to use this Hamilton to approximate this problem. We don't want to take into account all these terms. Nevertheless, they are important when you have this kind of degeneracy. Because there is a symmetry in this Hamiltonian. And which is the symmetry that you want to break? Because you introduce a symmetry breaking. So which is the symmetry of this Hamiltonian? Reflection with respect to what? You mean, OK, so you mean that you can rotate spins? So in particular, that's true around the z-axis, for example. You rotate the spin. And because this term goes in itself. And because it's a rotation about z. And this remains in itself. So it is a rotation of pi rotation. So the operator that generates the symmetry is the problem over L of sigma z. So if you apply this operator to the left and to the right of this Hamiltonian, this is invariant. But if you apply this operator here, it maps this state into this other state. So what we have is that the Hamiltonian is symmetric. And now we have a degenerate ground state. And then the problem becomes, OK, which ground state should I consider? Should I consider just the superposition and invariant superposition of the two ground states? Or maybe one is better than the other state? And what happens is that if you consider this fluctuation that I told you before, what happens is that there are two privileged ground states. So the ground state is not just a linear combination, a generic linear combination of these two. It is not. It is one of the two. The symmetry is broken. And you have either the spins are aligned the x direction or in the opposite. OK. Now I want to, OK, this may be my view. Why can it be? OK, because there is some reason that it's not stable under these fluctuations. But there is also another physical reason. Because, OK, let's assume that the state is in a superposition of the two. So we have a state with some coefficient of this other state divided by root of 1 plus alpha squared, for example. Let's assume we have this our ground state because we want to consider a superposition. And now let's compute correlations. For example, we could compute the correlation, let's call this psi. And we could compute psi sigma 1z, sigma 1 plus rz psi. OK. So this is the correlation between two spins at distance r. And then I remove from this term the value of the spin at position 1 psi position r. Why am I doing this? Sorry? Yeah, you know, but because the question was why I cannot be in the state. And I'm trying to give you some physical reason why you shouldn't expect it. So let's assume that we are in a state like this. Let's compute this kind of correlations. And let's see the meaning of what we find. Because we compute this correlation, so the state is straight. So it's easy to compute this correlation. And in particular what we find is, OK, this stands for this. Both spins in the x or the opposite direction. So here you have, let's write everything. Write everything. Yes, let's write everything. So we have this state plus alpha star this value for root 1 plus alpha squared. No, this is the magic of this. Then you have here sigma 1z, sigma 1 plus rz. And here you have this plus alpha this, the value by square root 1 plus alpha squared. This is the first. Then you have the other two and we compute them. So now when you consider this term, for example, then you have the operator. And then you consider the other term, the opposite direction. Now, here it means that all the spins are in the given direction. And this one, all the spins are in the other direction. Sigma z is able to reverse to flip one spin, but only one. So this sigma z is flipping the first spin. This sigma z is flipping the 1 plus r spin. There are all the other spins which are in the opposite direction. So this product between two is equal to zero. So this means I'm going to consider this term, the contribution from this and this is equal to zero. The same from this and this. So the only terms that contribute are this with this and this with this. Oh, sorry. Yeah, but it's x. It's x which flips the spins. Yeah. As a matter of fact, I have to correct myself because I wrote the opposite. So let's consider this term and this term again. Now we have sigma x. This is a negative state of sigma x. Sorry. It's a negative state, but you see that these are two different tiger states. So when you consider the overlap between the two is equal to zero. It's a negative state. The same for this, but now you have to consider the other contribution that the contribution. What is this? In this case, you have the spin, which is one given direction. So you have one times one. So you have this is equal to one over square root of one plus alpha squared alpha. Square and then you have the other contribution from this other term, which is the same as this. But there is a one plus alpha squared. I would say that is it. Yes, I would say so. Why minus because there are two of them. So you're minus one times minus one and she will be equal to one. I think this is correct. I guess if I did I do something wrong because they are there is okay. It's equal to one. Fine. Okay. It's a number. No, let's compute this. What is this? Now we have the same as before square root one plus half square sigma one X times the same. Again, the same. This gives a contribution one. Now this gives you a contribution minus minus one. Yeah. So you have here one minus alpha squared divided by one plus alpha squared. Is it right or am I wrong? Yes, it can be correct. I think so. I think it's correct. Now you have to multiply this. You must compute the square of this because we have two of these terms. So in the end, you find that this is two are equal to one minus alpha squared divided by one plus alpha squared squared. And this other term is there. It was equal to one. So now you have this term minus this term. Well, the difference depends on alpha. And this is something different from zero for any distance. Now, what is this correlation? We are considered so the correlation between one spin in a given position. The other spin even far away from it. And you'll see that even if they in the limit August infinity, this doesn't take up. So in other words, it's like that there is no locality in this deal. Because you have that what's what's happening here depends on what's happening on the moon. So this is why this is not a physical, a physical state. This kind of state break cluster decomposition. So you cannot, you cannot just describe your state looking on the local properties because everything depends on what's happening very far away from here. This is the reason why you don't. I think the physical reason why these are not physical states. So this doesn't matter because okay, in this simple case, you have a that is not an entangled state. But you have a question where you are. You can have a time. For example, if age is more, you have a time. Okay, it's enough about the. But okay, but you see that. So we are describing two very different situation for age goes to infinity. We said that this being just a line in a given direction instead now for age equal to zero. We see that there is some symmetry, which is broken. And so we don't have just a paramagnetic phase, but we have a firm magnetic phase because the spins are broken, not in the direction of the magnetic field. An arbitrary small infinitesimal magnetic field in the x direction. So this is why it's for magnetic. So it is a magnetism that remains even if you remove the magnetic field. So the system is for magnetic in the limit age goes to zero. So this paramagnetic age goes to infinity. Now, because we are describing different phases, you should expect the existence of some phase transition in the meat. Now, what is different with respect to what you know that the classically generally you you have a similar situation, but the parameter that you vary is the temperature. So you generate your class from a disorder phase at a temperature and you decrease the temperature and then you you end up in order phase like in a two dimensional model. Yeah, you don't have temperature. We are at the temperature because we are describing the ground state of the model. The parameter becomes, in this case, just the Hamiltonian parameter, the magnetic field for them. So you change some parameter and you can completely change the properties of the of the ground state and also of the excitations. This is essentially the correspondence between classical phase transition and quantum phase transition. Okay, so it's equal to. Okay, I give you five minutes. But I require I request some explanation of your answer. Okay, so you choose you you take the what you think is correct. Then you explain why. And maybe if you don't explain why I have the score of that question, so you should do it. And then maybe there could be also open questions. So we don't have answers. Most of them would be just the multiple choice question. You should be able to answer this question using your you can use our notes. No problem with that. So you use books, papers, no electronic devices because otherwise you start chatting. So I don't know. Otherwise I wouldn't allow you that. So if you want to cheat, try not to show it. Okay, so not be smart. I'm telling you this because okay, we are researching. We use all material that we can find. So in your case, the material is the notes, the books and your neighbor. If you're able to use the intuition, it's okay. The phone or not because I know that you start chatting. No, no. How can I check? Only for that. The structure. Do you want exam now? Okay, there would be several questions. I will ask something which is related to, maybe all the lectures somehow in a way or in the other, starting from the first lecture. There will be some questions that you can answer if you follow also the first lecture. And if you simple, very simple questions sometimes, so you will be surprised of how simple can be the question. But sometimes we don't need to answer complicated questions to see whether you understand or not. But okay, there will be also some problem, maybe, that is less simple. Okay, just because otherwise you get bored. So I give you also some nice problem to solve. This is the idea. Yeah, you want the exam as well. So just read your notes. And you should be able to answer the question. Okay, without solving this problem, we already understand something about this problem. So we are saying that we are considering H, the model for H that goes from zero to infinity. And without ending, we already know that the temperature equals to zero. Here, the model is paramagnetic, paramagnetic, para, para, para. Okay, and here we have a paramagnetic model at zero. And our intuition, physical intuition, just thinking of a classical easy model is that there should be some transition, some place transition, some point which I will call one. And fortunately, we can solve this model. So we have access to the ground state of the model for any value of H. So we are not restricted to H equals zero H equals to infinity, which are kind of trivial levels. But we can compute the entire spectrum, the collection, correlation functions, dynamics in this model. It's an amazing model because how can you tune? Why do you want to tune? Now we are doing static. So you just fix the magnetic field and then we... Then it's not dynamics. It's H. What? No, what? Are you solving with H2? Ah! Okay. There will be a more interesting representation here. We're doing classical physics and quantum physics. Okay, well. Okay. So how do we solve this model? Now we already solved the harmonic chain. And the idea was to introduce this kind of ladder operator, creation operator. Yeah, we would like to do something similar. And the reason is that when you can't do something similar, you have to use a very complicated formalism and if it works. So we prefer that it's possible to solve the problem in that way. And this is a model that can be solved in a very similar model to the quantum harmonic chain. What's the idea? Okay. In the quantum harmonic chain we mapped the degrees of freedom x and p in two bosons. So we would like to do something similar here. So we would like to find a mapping in two fermions. As we'll see in a moment. How do we do this? So, the idea is the following. So let's now consider generic spin one-half state. Spin one-half means you can have spins in any direction, okay? And in particular, let's consider a particular state which is all spin down. Okay. All spin down. And then maybe we already have an intuition that we could interpret this state with all spin down as a vacuum. Then we could say, well, every time that we flip a spin, it's like to put, to include a particle in the state. This is just a rough idea, no? So let's assume that we do something like this, okay? Let's guess. Let's start guessing. So we say, okay, let's interpret this state of all spins down as the vacuum of some particles. And then, okay, so using this kind of rough idea, we'll say, and remembering the harmonic oscillator, we'll say, well, I would like to define the particle, the one that flips the spin. How do you flip spins? Okay. Yes, you are right. Yes. Sigma X or? Sigma Y. No. No. If you apply sigma X, no, then you end up with something which seems to be a mission. So we want something, you want to create, you want to consider an operator that creates a particle, and you want the conjugate will destroy the particle. So this is, yeah. Sigma X does the job here in the state, but apparently there is something not perfect. We could do the same thing with sigma Y, no? Sigma Y also. So for example, we could do this, for example, what happens if you apply this operator? So sigma X plus sigma Y is equal to this state. It's representing this way. When you apply it to spin down, what happens if I spin up? So it does what we want. Now let's assume that we consider the conjugate of this now, because we want to destroy, we want to flip the spin again. So we consider this, which is now 0, 0, 1, 0. Now let's apply it to spin up to see what happens. To spin up, it is equal to down, and it's okay. And we also see that when you apply this to spin down, which is our vacuum, we find 0. So actually, this has nice properties, very similar to the properties of the ladder operator in the Bosnian case. So this operator, so rise the spin when it is down. Then instead of the conjugate, it flips again the spin from up to down, and it also destroys the state when the spin is down. So it's nice. Now we'll see that there is some problem. Why? Because we would like to define our ladder operator. For example, in this way, let's call this D, for example. And then this index is just the position in the chain. Then we could consider two of these operators, which are two different particles in our P, the picture before. So we could have, for example, D-dug L and D-dug N, and X plus i sigma N y over 2. So what happens if you consider commutator between these two? These are defined in different sites. They don't commute. They commute, like in the Bosnian case. You can be ethical, because we are finding the same as before. So we find commutation, different values of the indices. We find that they seem to do the right job for the vacuum. And now, OK, let's check the commutation relation of this. The commutation relation of this form, D-L, D-dug L. If you consider commutator, what happens? So what we find is that if you apply D-L and D-dug L, what you find that this is equal to 1 minus D-dug L D-L. You can check this. You see that this is kind of strange, because now if you consider the commutator of these two, D-L D-dug L. Now you find 1 minus twice D-dug L D-L. So before, we had just 1. And now we have also this term. This term, which counts the number of particles at position L. So it's something more complicated than what we found. Now, what am I writing? Sorry, what am I writing? No, it's right. Yes, I'm right, right. So we find something, a commutation relation, which is kind of strange. But now you realize that if instead of considering the commutator between the two, the two operator, you consider the anticommutator. What happens? You find it is equal to 1. Anticommutator should make you think that maybe instead of trying to map the problem to bosons, you should try to map it into fermions. The problem here is that we found the mixed algebra. Because when we consider now the same site, we have that the anticommutator indeed between the L and D-L is equal to 1, as you would expect for fermions. But now when you consider different L, then we find that the commutator is equal to 0. So these are strange particles. And they are called hard core bosons. Not because they like this kind of music, but because they are bosons that behave like fermions, when they are close together. So they would like to move bosons, so they commute different sites, but then when they are on the same site, they are hard like fermions. But we don't know the... At least I don't know how to... I don't have much knowledge of these hard core bosons, so I don't want to use these kind of strange particles. So let's try to map this problem to some familiar particles. So either bosons or fermions, something with a very precise algebra. And we can do it. And in this case, you can do it into fermions. Why we should expect fermions are not bosons? In the case of bosons, as we've seen before, we have infinitely many excited states, because you can always create a new boson in the state. Indeed, in the Armonic oscillator, you have infinitely many states. Here, we are considered a chain, a spin chain. So the number of excited states is finite. The maximum number is 2 to the n. So you cannot put an arbitrary number of particles in this theory. So this is why you should think, oh, maybe bosons are not the right particles. So we should consider fermions, because we know that there is the exclusion, the Pauli exclusion principle. And so you cannot put more spins in the same state. And so you have a finite number of states. So they could, in principle, describe our basis, at least. How can we fix this algebra, this problem of algebra? Actually, it's kind of simple. And it was discovered by Jordan and Wigner, probably first with Jordan or Jordan. So, and the idea is to consider this operator, c dagger L, which is equal to the product for j smaller than n. So j from 1 to L minus 1 of sigma j z, everything times sigma j x plus i sigma z L, x plus i sigma L y divided by 2. So like before, but now we have this type of string of sigma z. The string of sigma z in this state doesn't do anything, because it just, it's a number. It just counts the number of spin down. So we are just, OK, we are modifying this operator in such a way that here it doesn't do anything from a number. But it changes the algebra of this operator. Why the algebra is different now? I write also C L. So it's the same string here. Sigma j z sigma L x minus i sigma L y. Now let us consider L different from n. Sorry, wait a second. OK, because before, we are just this, OK? And this operator, they probably make commute when they are different sites. But when you consider the operator with the dagger at the same site, then it has the identical mutation, which is not, so it's not a boson. It's not like a fermi. And then we observe, oh, but here if I consider the dimension of the inverse space, I realize that instead of having an infinite number of states as in the harmonic oscillator, the other number of states is finite. And so this means maybe I shouldn't look for bosons. But it's actually more, it's clever to look for fermions in this case. So this is why I would like to change the algebra, preserving the anti-computation relation at the same site. And this is a way to fix the commutation at different sites. And if we consider now the commutator between, no, the commutator. If you consider c dot L and c n, for example, for L different from n, what you can check is that if you consider the anti-computator, I don't know if you, I hope you know. So the anti-computator AB is defined as AB plus BA. So it's like the commutator with a plus sign. And so you find that this is equal to delta Ln. And then you find that c dot L, c dot n is equal to delta, no, it's equal to 0. You can check it. And probably you will do after this lecture, the tutorial. So why this is nice? Because now we can actually use the same interpretation as before. The only difference is that now instead of having mosons, we have fermions. So if we call this the ground state, okay, this is not the ground state. Just a state, a vacuum of the fermions that we define. Then given this vacuum, we can generate all the bases using this kind of creation operators. So in other words, each state corresponds to a given state in the fox space of these fermions. You can interpret it as a state in the fox space. For example, this is the ground state. Now let's consider, for example, the state with the spin L, just one spin, position N. What does it mean here? I have to count, I have to multiply the string of sigma z up to L minus 1. So here I find minus 1 to the L minus 1, because I have spin down. And then I have to flip the spin there. So I have here down, up to L minus 1, then have an up, and then all the other down. This corresponds to a fermion, a position L. Okay, this is not an ordinary fermion. It's a matter of fact, because maybe you learned that fermions generally have an half-integer spin. So in particular, fermions can have, the simplest fermions have a spin one half. So you can find them in two states. So given these are spinless fermions, because in this notation you have that you can have just one fermion per state. So in L, you cannot have two fermions. How can you see this? If you apply two creation operators at the same site, you find that it's equal to zero. Why? Because this, the square of this, is equal to zero. So it means that you can't have two fermions in the same state. Yeah? So this is the same position. Yeah, yeah. We are talking about position, but it's not rigorous, because here there is an operator which is a non-local operator. So we, it's a position, but we should be always careful. So we are changing the space, so now it's just a simple way to present it. But then what you realize is that when you generate these states, always you have this kind of non-trivial phase, okay? Because of the non-locality of the transformation. Okay, this is not relevant. Just that we are just trying to reinterpret all the bases of our space in terms of some fox space of fermions that we defined. So-called, I wrote this. Jordan, this one. You see this is completely general, independent of the model. So we, I introduced the model, but now this is really general. And for any given spin chain model with spin one-half, we can define these operators. And they satisfy the correct algebra. They can interpret the fermions, and the vacuum of the fermions is just this state. We will spin down. Completely general. Okay, yes. That was a question. I applied this operator to the vacuum. Now you have this string of sigma z. And you have that here, you have all the spins points in the down direction. So you have minus one to the, one minus one per spin up to L minus one. So minus one to the L minus one. Okay. So now maybe you will understand the idea. The idea is to map this Hamiltonian in terms of this operator. Okay, just that. And see what happens. So let's start the transformation. Okay. Do you know this notation? Sigma plus. Sigma plus is equivalent to say sigma x plus i sigma y over two, which is this matrix. Analyze one defined sigma minus. Sigma x minus i sigma y over two, which is this matrix. It's represented by this matrix. Know that this is not good to change notation many times, but sometimes it's usable. And it's, it's bad. Because then I remember that every time it's a plus. Okay. And I know it's not good to change notation, but I would like to define now. Well, no, it should be up there. It should be up. Not sigma plus. No, it's defined this way. It's going on. In the literature, you always you find sigma plus with a plus. Why not? Okay. This is not an order. I don't want. No, no, no. It's not important. Come on. Because you, sigma is a, sigma plus is not defined in other way. So you have sigma x. Why is that sigma plus? And sigma minus. This is the notation. Okay. But I want to introduce other, other operators just because it's convenient. Not because I want to volume with other notation, but let's introduce this operator, which you call a, like before. I'm sorry for that. Like in the, in the, in the harmonic chain. And we define this operating this way. A 12 minus one, which is equal to C, plus C L and A 12, which is I, C L minus C dot L. Now, maybe you recognize something very similar to what we found in the harmonic oscillator. When we do this kind of combination, the harmonic oscillator, we, we had something like X and P. It was the inverse relations. Okay. Now, what happens when we consider this kind of, of transformation? So what algebra satisfy this A? You can check that now if you consider the anticommutator of A L and A N, this is twice delta L N. So again, the anticommute, like fermions, there is only difference. They are a mission. Okay. Maybe you, sometimes you were about this, about minor and a fermions. These are minor and a fermions. So they are, they are their mission. They, they are, they own antiparticles. So the, the, the, the adjoint of the, of the fermions itself. And they satisfy the same algebra of the fermions. Okay. These are, it's not important. It's just a question of terminology. So they are called minor and a fermions. Okay. They are useful because it's easy. They, they allow to, to rewrite the Hamiltonian and also the correlation functions in easier way. Just for that. Okay. So let, let's use this operator on that note. And let's write, for example, in terms of the spins, what are these operators? Well, you can easily see that the A to L minus one is nothing but the string of sigma, of sigma Z. L minus one, j z, times sigma L. And then you have A to L, which is product j from one to L minus one, sigma j z, times sigma L one. So we don't have plus and minus and it will be your idea. Okay. Just the definition. They satisfy this algebra. Okay. I think now it's time to, to apply this transformation to the Hamiltonian and see what we find. Ah, by the way, what is the inverse transformation of this? Okay. If you multiply this operator or this other pair, this is the same by the same string of sigma Z, then it simplifies the string because sigma Z squared is equal to one. So what you find is that sigma L X is equal to the, the string there, times A to L minus one. Okay. I did just half of the job because now I should express sigma Z in terms of the Majoran fermions. That's it. And what is sigma Z? Sigma Z is written like I, sigma Y times sigma X. Sigma Z, I, sigma Y, sigma X. You can check this. So this means that if I multiply by I, multiply this by this, I actually find sigma Z. So I have a sigma L Z is equal to I, A to L, A to L minus one. And so I can put this here. And then if you want, I have an expression for sigma Z in terms of the Majoran fermions, which is something like, something like minus one to T L minus one, Y minus one. Something like minus I to be L minus one. And then I have a product from J from one to to L minus two of AJ times A to L minus one. Okay. As a matter of fact, this is not really important. This is even, it's easier to work with this expression generally just because we don't want to work with spaces. So this hybrid notation. Anyway, so we have, this is sigma X and sigma Y is equal to the, again, the string times now A to L. Okay. So we plug now this expression to the Hamiltonian of this model and see what happens. I just write the result. You can do the, the substitution. We have to play a bit with the, with the algebra of the Pauli matrices when you move them, but it's a, it's five minutes. So you can do that. And what you find is this H is equal to one plus PZ over two, IZ over two times one over four sum over L and N up to two N of A, L, H, L, N, A, N plus, plus one minus Pi Z over two, one over four sum L, N from one to two N A, L, H minus L, N, A, N where Pi Z is the product from J one to two N of sigma J Z like something. Can I, can I erase here? You have the definitions. So what is, what is H? I introduce this kind of calligraphic H and H plus minus is a matrix and is even by zero H minus H zero zero zero one zero that is two IJ two IJ zero zero zero minus one zero zero zero H minus H zero zero H minus H zero zero H minus H zero zero zero one zero zero zero one zero zero minus one zero zero zero zero zero zero zero zero zero zero zero okay okay zero zero zero zero zero minus one zero zero 0, minus 1, 0, 0, 0, 0, 1, 0, and everything here is 0, and everything here is 0, here everything is 0, ok, ok, now, ok, some, some current, so yeah, ok, here all the block elements are equal on the diagonal, and are equal to 0h minus h0, and now if you consider the other diagonals, block diagonals, 2 by 2 blocks, you have that, oh, you see also these elements are equal, oh, now I have another color, sorry, they are equal, each other, and are equal to 0, 0, 1, 0, and also this element is equal, and then finally you have the last one, this one, which are equal, so it's exactly the same as in the, in the harmonic oscillator, the structure is similar, it's not exactly, but you see that it has kind of a circular structure as before, so you find the main diagram, the other diagram and these elements are equal, this one, like in the harmonic chain, and here this element is equal to this one, like in the harmonic chain, but now we said of having just entries of the matrix, we have blocks, 2 by 2 blocks, this is called indeed, indeed, maybe you can guess the name, the previous matrix was a circular matrix, this is called the block in circular, ah, sorry you're right, I forgot this, the difference is that, depending on the, what's minus, plus minus, okay, yeah, sorry, there is indeed, there is a difference with respect to the harmonic chain, but here you have plus minus, and here you have minus plus, okay, there is indeed a small difference, yes, h plus minus equal to ij times, everything else, okay, I always show you the matrix and then I write explicitly within this, but just because, so you understand, maybe the structure of the matrix, yeah, and okay it's late, but what I want to tell you is that, like in circular matrices, circular matrices are diagonalized by the Fourier modes, so you can expect the same here, because you have the same structure, just that instead of diagonalizing the entire matrix, you are just, you can block diagonalize the matrix, so if you now write here, you go in Fourier transform, now you could expect that you still have to diagonalize 2 by 2 matrices, so you can get rid of this big matrix by using the periodicity, like before, but now you will have also some stretch, some 2 by 2 stretch, which remains, okay, this is just to tell you, then another ratio of the sign, yeah, there is indeed a difference now, because in the harmonic oscillator we had a, indeed a block-circuit or arm tie, okay, we have this kind of a circular structure that these elements was exactly equal to this one, now we are the case, the case of minus, where you find the structure with a minus sign, but then you have also another matrix, where you have the wrong sign, the wrong sign is called anti-circular matrix, it means that when it's like when you put, instead of periodic boundary conditions, in a problem you put anti-periodic, so the elements n plus 1 is equal to minus the first element, all the difference, so these are two different kinds of matrices, which are simple matrices because it can be diagonalized easily, okay, and then I'll tell you how to do this, next time we'll finish the calculation, because I think it's time, yeah.