 I think today we will finish this part on quantum harmonic chain and then we will consider the quantumism model. This is a recap, but there is also something new about what we did yesterday. At least I changed a bit the notations, because yesterday with all the indices in the end it was a bit messy, so maybe it's better to use a vector notation. It's easier. OK, so we consider this a system of harmonicos letters. This system can model the movement of ions in some crystal, and in particular we just for the sake of simplicity, we imposed three-order boundary conditions, which means that the variable n plus one is equivalent to the variable at one. And OK, this is the Hamiltonian of the of the model. The i is the conjugate momentum of Xi, so they satisfy this combination relation, h bar is equal to one. Then, OK, let's introduce this kind of vector notations. We have n momenta and then coordinate position, so we can collect them in some vector. And I define this vector p hat, and this is the same for the x. Now, using this notation, it is kind of simple to show that the Hamiltonian is written this way. So you have a scalar product within these two vectors. In the scalar product, we have to multiply each element by itself squared, so it becomes p squared and so on. And then, instead, the other part is just given by the vector of x, the scalar product with the cx, where c is the matrix that I introduced yesterday. Why you find this matrix? Because if you expand the square here, you have x squared minus y xi xi plus one plus xi plus one squared. But then you have to consider that you get to consider all the contribution from both from i and i minus one. So you find the you want to see it or you are happy with the Hamiltonian and we can go. You want to see how this matrix arises or it's OK? I think it's OK. So this is our, or maybe there is a, I hope there is no wrong factor here. Let's check it, OK? Just to be sure, because now I'm not sure anymore about the factor. So if you write this term of the Hamiltonian, you have a sum over i of xi minus xi plus one squared. This equals sum over i of xi squared plus xi plus one squared minus twice xi xi plus one. So here this, this term is just the x. The second term is the same because we are just shifting the index. So we find twice x vector x minus sum over x of this. Sum over, over i, sorry, twice xi xi plus one. So our matrix, so if we, if you write this as x, scalar, some matrix applied to x, one xn. What we find, we find that the diagonal should be equal to one, two, two, because we are a scalar product. And then here there is a minus twice xi xi plus one, but in which can be written as minus xi. We can write this term as minus sum over i of xi xi plus one plus xi xi minus one. And so we have to minus one minus one. So partly this was correct. I guess it's the correct form. So this is the, I can erase this, no? It was just a check, because I wasn't sure. So this is the Hamiltonian, which is fucking this computation. It's a bit better. And then, okay, we, we try to find the excitation energies and the spectrum of the Hamiltonian. And we, the idea was to use the same formalism that we use in for the case of a single harmonic oscillator. So this ladder formalism. So the idea is to introduce some operators which are linear combination of the coordinates of the problem x and p. And then we impose that the commutator between the Hamiltonian and this operator, which can depend on some. This is just an index because, in principle, we can find more of these operators for the single harmonic oscillator. We are just one operators, but in this case, because we have a chain, we can expect that there are many of them. So we, the idea is to impose that the commutator between the Hamiltonian and this operator is proportional to the operator itself. Why do we do this? Because if you are able to find this operator and using both this relation and this other relation, which, which is essentially the conjugate of this one. So they are related. Then you find immediately that the ground state should be such that the is annihilated by all this operator, a case. Why? Because when you apply these operators to a 20 set of states, you find that the energy becomes lower. And it's lower by action k, which is by definition a positive. So this means that the ground state is determined by this condition. Analyze all the operators. But okay, we don't know only what is the ground state. We also can construct it. Yes, because of this. Okay, okay, okay. To be a letter for blackboard. Okay, so this is the ground state. What about the excited states? Because when you apply this later operator, a decade to an excited state to an idea state of the amuletona, what you find is that you find another state of the amuletona with energy, which is the same as before plus epsilon k. So in other words, we can construct all these excited state by apply this later operator to the ground state. So this is the idea, the general idea, but then, okay, we started doing the calculation yesterday and get some problems with the indices. Okay, but okay, and now we can completely this calculation, I guess. Okay, so we we have to impose the commutator between these and these are you finding innovations? These are just vectors. These are numbers. Okay, maybe if you prefer, I can change this the symbols here because I use X and P like the coordinates and maybe it's confusing. If you prefer, I change them. I do not because I didn't do that because I started with this notation, but are you fine if I change that? Okay, so let's call this you. Let's call this. You know, let's call this Y and let's call this Q, okay, I guess. Now, the reason why I'm choosing this name, you see, that is a linear combination and these two are different dimensions, P and X. So the idea is to write some letter here, which reminds me that they should have the dimension, for example, you have a length and you have the moment. Just to be sure is a check is a fellow check in the end of the calculation and I can I can see the dimension if they are right only for that. Okay, so we want to impose this constraint. What do we find? We find the following equation and times another way to change all the Q and vector plus Y and scalar product with P. This is equal to minus I, not P is Q, Q and divided by M plus one over four I, M omega squared C. Okay, now let's go one over four, omega squared C, not C, Y, M. It's how can you derive this? Actually, you know, it's kind of simple. You just have to take the derivative if you want in a with respect to the variable P and X. You can do it with the indices if you want, but this is essentially the just a quick trick to the end result. So we we have this equation and you have to solve it. This is just imposing. I am posting this using those notations. So how can we first of all, what does it mean here? You see that there's a linear combination X and P, but here and here. So the only way to satisfy this equation is that all the coefficients for X and P separately are equal on the left hand side on the right hand side. So this imply this question that epsilon and Q and should be equal to one over four I, M, omega squared C, Y and this is one question. Then we have the other question for the momentum. Epsilon n, Y n is equal to minus i Q and over M. And these are probably the questions that I wrote yesterday in the end. OK, so well, then OK, what do you want? We can now I don't really mind such. OK, maybe we should do this. OK, so I think I can erase everything. OK, so we have this question to solve, but actually we also would like to impose a normalization for this operator. Because you see this equation are invariant under where you can multiply this equation by a constant and you obtain the same. So we want to fix the normalization. And the way to fix the normalization is to impose that a dot a k, a dot k is equal to one, which is the same relation that we had for the harmonic oscillator. OK, so first of all, let's let's let's fix the normalization and see what what is this condition. This condition means that one should be equal to the commutator between. Hey, is the conjugate? Yes, yes, exactly. This is what we are going to do in a moment. And just OK, before solving this, there is another constraint given by the fact that this operator we can fix the normalization away. And it's convenient to fix the normalization. Such way that this is equal to one. OK, so my session is that this is so this is the conjugate of this. So this means that you have to take here the conjugate of this coefficient, the conjugate of the other coefficient. So you have the one should be equal to P and star vector scalar. Oh, there is no P. Sorry, this is called Q. Then a plus Y n star scalar P commutator with Q, n, Y, Y, not Y, X, and X plus Y n scalar P to be equal to one. Now the X variable commute P variables commute one another. So you have to take into account the commutator between X and P and P and X. We know that the commutator being X and P is proportional to a delta. So it gives rise to the scalar product of the coefficient. So we have one should be equal to Q. There is an eye. I because the commutator between X and P is equal to I. So we have Q star scalar product Y n minus Y n star scalar Q n should be equal to one. Why? Yeah, there is a minus because we have the commutator between P and X, which is minus I is not equal to Y. So this is our, this is normalization, normalization of the operator. You have this condition that we will impose and then we have to solve the other two questions. OK, let's not solve these equations. What can I do? OK, we can, for example, what I do, what I do here. So we could plug and choose. So we can plug, for example, view the first equation into the second one. So this means let's assume now that epsilon n different from zero. OK, if you assume epsilon n, epsilon k, epsilon n different from zero. So we can divide by epsilon n. So we find Q n equal to the expression divided by epsilon n. We plug Q n here. What do we find? Epsilon n, Y n, is equal to minus I over m epsilon n for I omega squared C Y n. OK, we'll find this, if I find this correct. So this can be written as C y n if you multiply by this coefficient. It is equal to 4 epsilon n squared over omega squared Y n. So we have to solve this equation. How do we solve this equation? As I told you yesterday, this is a circular matrix. And because of the symmetry, you can diagonalize this matrix in Korea, in full space, essentially. So this means that the eigenvectors are given by E to the I. So you have that C. OK, let's write this matrix. You can write this C l n in this form, 1 over n. We go in the free space, OK? 1 over n sum over j. It goes from 1 to n of E to the 2 pi i j l minus n over capital N. And here we put some coefficient, which is called C tilde j, if you want. So because of the properties of this matrix, we can actually rewrite the matrix in this form. And so we just have to determine which are the coefficients C j. Then C j, we know that in particular, OK, let's write C instead of C j C of 2 pi j over n. So what we have, if we find that the C tilde of 2 pi j over n is equal to 2. Yes? What's written in? In the x-point, OK, that is 2 pi i j, this is a j, multiplied by l minus n, and everything divided by capital N. What is this? This is a discrete free transform, because the size of the matrix is n, and you see this is periodic. And we're just going in the free space. And it depends on the difference of the indices, just because this matrix depends only on the difference of the indices. And there is this condition of periodicity. So we can find this C tilde. C tilde will be equal to 2, because when you replace C tilde by a constant, you find that the sum of phases. And the sum of phases is equal to is non-zero only if the phase is equal to 1, if the phase is 0. So you find the condition that corresponds to the first diagonal of the matrix, these two. Then you have to take into account the other diagonals. So you obtain minus 2 cosine of 2 pi j over n. Why is this the case? Because if you expand to cosine, you have e to the i, the first plus e to the minus i. The same. So now the condition becomes l minus n should be equal to plus 1 or minus 1, which is exactly what we have here. This is the minus sign. Up here, yes. So this is just another way to write the matrix C. See how this is equal to this? You can verify. That's the case. So we were interested in the eigenvalues of C, no? So let's forget it. OK, maybe it's better if I try to write this question somewhere here, normalization. OK, if you want to just get this matrix, you notice that it's just tilde. Yes? How do you do that thing? OK, if you want to do the anti-fuelet transform, but I just guess the solution because it's kind of intuitive. So let's imagine to expand this in all the Fourier coefficients. See, you have the generic coefficient which have this form e to the i to pi j over n times some integer. No, not l maybe, some integer m. Then you have to sum all the spaces. But again, in order to be different from 0, the space should cancel. So this means that l minus n plus j should be equal to 0. So this means that this mth Fourier coefficients contribute as the mth column of the matrix. You can just keep the first coefficient, the first e to the i phi. I have to see because we are just a tri-diagonal, more or less the diagonal matrix. But OK, you compute the anti-fuelet transform as a part. That's not so. OK, so we have a normalization there. So now that we wrote this matrix in this form, it's easier to diagonalize it because we have is that the eigenvectors are of the form vn equal to vk eigenvectors, coordinate j is equal to e is proportional to e to the 2 pi ijk over n. Let me be with the cross sign. So we just use the sunsets, which is equivalent to consider Fourier transform. And then we apply c ln, our matrix c, to the vector vk. And we suppose that this is an eigenstate. And what do we find? We find that we sum over the indices. We write all the indices. So we have sum over n of c ln times vk of n should be equal to lambda vk of n over l. That's what we have to solve. Now we use that representation there. So we have sum. This n is from 1 to capital N. And from 1 to capital N, we have 1 over capital N sum over j. It goes from 1 to n of e to the 2 pi ij over capital N l minus n. There is c tilde after that. So we have this contribution. e to the 2 pi i nk over capital N. Then we have c tilde of 2 pi j over capital C is equal to our eigenvalue of lambda k times vk l, which is e to the 2 pi i lk over capital N. That's it. We have to solve this. Now again, here there is the sum over n. If you consider the sum over j, there is also this contribution that we don't like. But we first can consider sum over n. When you sum over n, you have n only on the phases. So again, we can use that the sum of phases is equal to is non-zero only if the phase cancels. So this means that j should be equal to k. And you get the factor of capital N. So this becomes capital N and simplifies this. So you remain with sum from j that goes from 1 to n of e to the 2 pi i j l over capital N. Then you have c tilde. There is a delta k, sorry, j equal k. So here is a k. It's 2 pi k over capital N, which is equal to lambda k e to the 2 pi i lk over capital N. We guess well, you see? Because now we have that, indeed, these two terms. This is consistent with the c tilde to pi k over capital N equal lambda k. Yes, OK? So we have this equation, fine? Now we sum over n. When you sum over n, you realize that N appears only on the phases. So you have the sum of phases. And the only possibility to be non-zero is that the phase becomes equal to 0 to pi over there. And the only possibility is here is equal to 0. So you have l minus n should be equal to minus, sorry, you find j equal k because we are summing over n. So you have this condition, j equal k. So when you sum over j, actually, you just pick the value k. And you have just one contribution there. I'm using this. I'm using this. Sum j goes from 1 to n of e to the 2 pi ij k over capital N is equal to n delta k 0 for k equal 0 n minus 1. So in this way, we obtain the values of c tilde. No, sorry, no, c tilde, we had the values of lambda k. So we know that lambda k is just given by this expression, j equal k. So what we found is that lambda is the eigenvalue c. And so in this equation, we have this part here, which was the eigenvalue. So this part, for epsilon n squared over omega squared should be equal to this one, 2 times 1 minus cosine of 2 pi n over capital N, which we want is equal to 4 times sine squared of pi n over. OK, so this gives us the eigenvalues. We also have the eigenvectors. I guess the eigenvectors, which were this one, vk equal vk. Yes? OK, here you have. You see, we were looking for the eigenvalues. So this is the eigenvalue. And then you have to put the eigenvalue equal to c tilde of k of this special. Yeah, OK, well, it's very easy. Anyway, so we have 4 epsilon squared over omega squared is equal to twice 1 minus cosine of 2 pi n over capital N. And this is equal to, this is just a trigonometric identity, is equal to 4 sine squared of pi n over capital N. OK, so we have this. And we also know the eigenvalues of c, this is the eigenvectors of c, which are this one. These are the phases, these phases. So we also know that yn should be proportional to where I write this. OK, I know that this is important, maybe. I should erase something. I erase this one, this part. Can I erase this? So we have the eigenvectors, yn should be proportional to 2e to the 2 pi n ynj over capital N. So we have yn, we have the dispersion relation. Let's try this, please, the dispersion relation here. So we have epsilon n, epsilon n, which is positive, by construction, should be equal to omega times the absolute value of sine of pi n over capital N. Maybe now we understand why I chose the normalization 1 over 8 times omega squared, because just to simplify this for that reason. So this is our epsilon, and we know yn. But now, from this equation, we can get qn, because we need both, or from the other equation. No, from this equation, it's easier. So we have the qn, OK, let's put here some constant. This will be some constant, which could depend on N. That's true, depend on N. So we apply this dispersion there. And we find that the qn is equal to i m epsilon n times this. k is constant, kn is 2 pi i n j over capital N. We have this arbitrary constant, which is just because we have to impose the normalization. We didn't do it yet. So the normalization was the condition there. So we have that the, are you guys correct? Yes, we have that. There is written that this is the imaginary part. OK, let's write all this written there. 1 is equal to i, which multiplies qn star. So you have minus i m epsilon n k, this constant of star e to the 2 pi i n j over capital N. You have this, this is j, the component j of this. So you have to sum this over j times yn, which was this one. So you have kn squared e to the 2 pi i n j over N. So there is a minus here, thanks. Otherwise, it wasn't simplify. Then you have the other term, which is minus yn star. So we have this kn star e to the minus 2 pi i n j over capital N times qn, which is this one. So there is an n i here. This again, this simplifies the minus 2 pi i n j over capital N. Then you have m epsilon n and everything, I guess, hopefully. So this is equal to the sum over j gives n, that's right. Yes, so the sum over j gives n, because there is nothing depends on j in this expression. So you find i times minus i is equal to 1. The same for this, it stands cancel. And so you find twice m epsilon k, km squared. Maybe check, OK. Factors can be, maybe there is some time, but I guess it's correct. So this is our normalization. And this fixes this, km up to the phase, but phase is completely irrelevant, this problem. So we can choose this km real, no problem. So we have everything. I guess now we have really everything. So we can rewrite this operator a dot k. But there is a subtlety. These are our energies, the alien values of c, which are actually the expression energies, as we see, because they relate to the commutator between h and a. But now there is a zero eigenvalue here, because for n equals zero, this is equal to zero. While this normalization condition here seems to states that the one should be equal to zero. So this means that we cannot impose this condition for the zero mode here for n equal to zero. So we can impose the i, well, this was our convention to choose this normalization for it was for any n, but zero. So for zero, we have to treat it separately. OK, so what can I raise here? This again now. This again. Indeed, OK, we assume from the beginning that I'm trying most different conditions. Why did we do that? Why? So normalization is this, but we don't need this anymore. So let us rewrite this a dot k. So the a dot k is finally given by, I use now this pressure so that I avoid typos. So we equal to sum over j from 1 to capital N of e to the 2 pi i n of kj over capital N divided by square root of 2m omega sine of pi k over capital N times capital N. And here you have the momentum pj plus the same 2 pi i kj over capital N xxj divided by nothing. Raise here times the square root of raise n i square root of m omega sine of pi n over capital N divided by the square root of 2 capital M xj. x, yes, j, whether it's a k, yes, or n. OK, I think this is our final result for the a dot k. And we know we impose that a k a dot k is equal to 1 as in the single harmonics effect. And we have k different from 0. k different from 0. When we have k equal to 0, so this is k different from 0, let's now think a bit about the case k equal to 0. k equal to 0, everything was fine, but the normalization. And indeed what you find is that the k equal to 0, the solution was something just proportional to the total momentum of the particles. And this is clear because it's a pre-ordered system and the potential depends on the difference of the position. So this means that the Hamiltonian is invariant under rotation. Rotation, OK, translation, global translation. And so indeed k equal to 0 correspond to, what is it? k equal to 0, OK? We can choose instead of, it's just something proportional to the total momentum, which is equal sum over j from 1 to n of the particle, vj. So what I mean is that if you consider the commutator k, here we derive this asking for h a dot k should be equal to epsilon k, which is equal to omega sine of pi k over capital N times a dot k. Now we have that the momentum satisfy h capital P equal to 0. This is the difference. This concept for the moment, nothing more than that. OK, so we have all these operators. How many of them? How many operators we have? We have n minus 1 a dug, right? Then we have n capital N minus 1 a. And so the total is 2 capital N minus 2. And we have a momentum, total momentum, p. And operator is missing here because we need two capital N operators, like the original variables where capital N x is and capital N p is. So which is this variable missing, this operator missing, is the conjugate variable of the total momentum, which is the central mass position of the particle. So the operator that we are missing is, we can call, we can call x, which is somewhere j from 1 to capital N of the position divided by capital N. You can check that the commutator between this and this is i. So these are conjugate variables. And you can also check that this commutes with all the a. a, c, as well as p. So now we complete, I think we have all the operators. So we pass from the x and p to the new operator, this lateral operator, a dug and a. We have this subtlety because we lack an operator, which is the total momentum because of the symmetry of the Hamiltonian, total momentum, and the conjugate position, central mass position. So what do we do now? Well, we write the Hamiltonian in terms of this new operator. So we have to invert this relation. You can invert the relation, but I just write the result. It's not very useful to do really the calculation. It's easy. And what you find is the following. You find that the momentum j in the momentum of the particle j of the eigenj is equal to some from n because from 1 to capital N minus 1 of e to the minus 2 pi i. OK, let's call this k. 2 pi k over capital N square root of m omega sine of pi k over capital N divided by 2 capital N. And here you have a k plus a k, no, no, a k, a n minus k, plus everything, so all this, plus b total moment divided by n for the xj. You find some k from 1 to capital N minus 1 e to the minus 2 pi i k n over capital N. And you have 1 over square root of 2 m capital N omega sine of pi k over capital N times i a k times minus a, you're not going to be wrong, a n minus k minus a dagger k plus x. These are the inverse of those relations. And again, I already wrote them again because it should be clear. So the algebra of the operator a k a dagger q is equal to delta kq a k aq is equal to 0. Yes? Because again, we have to exclude the case k equal 0, which is like a k equal N, capital N, because it's always in the phase. It's 2 pi k. So it's a matter of convention. I could have written this as, in this way, I have to remove the case k equal 0, so you have n minus 1. It's original. We have this. Yeah, yeah, 1 and it is. So if we start from, we have to remove it at 0 capital N. You're checking this. It's completely equivalent. So we have a question, because there is no L in any other place. I don't know why I wrote L, because here is written L. So here, OK, so let's do it in this way. L. So there shouldn't be j. Indeed, there is no j there. That's a typo. OK, so we have this ration, but then we have also that xp is equal to i. Then we have all the other, which is x a dagger k is equal to pa p a dagger k, which is equal to 0. So in other words, every commutator is equal to 0, but the commutator between capital X and capital P and the commutator between a k and a dagger k. These are our operators. And let's rewrite the Hamiltonian I'll do for you. So you can check. And you should, because I hope it's correct, but no. So the Hamiltonian is written in this form. Total momentum squared divided 2m capital N. This shouldn't be surprising, because it's the kinetic energy of the central mass of the particles. So this should be something that you should expect. Then there is plus sum over k goes from 1 to capital N minus 1 of omega psi k over capital N a dagger k a k. Plus, there is a constant, if you use this for a real important constant, cotangent of pi over 2n 2n. What is this? Oh, this is part of the other one. So this is something you should expect. This is the same as in the harmonic oscillator. Now, but now we have more modes than the other case. So we have a slightly more complicated system. And there is a constant, like 1 alpha omega in the single harmonic oscillator here. We have 1 alpha omega cotangent of pi over 2 capital N. You have to leave it there. No, this is just the original variables. Excel was the displacement of the ions from the left side. And P are the conjugate moment of the velocity of the particles, the moment of the ions. What's the position of one phonon? Because it's a collective, OK. You already called the original phonon. I didn't say anything about phonon. OK. Anyway, OK, let me tell everybody this. So first of all, we should understand this Hamiltonian. So we found this Hamiltonian. And it has the same form as in the single harmonic oscillator. But now we have the sum of all the modes, which are independent because they commute between one another. So really, we were able to reduce the Hamiltonian to independent oscillators. And it was coupled for each other. So this is what we did. And so what is the ground state of this Hamiltonian now? We know, no? The ground state is the state such that it's annihilated by all this a. So it's the vacuum for all these a's. And it has also another condition. We should have zero momentum, zero total momentum, because otherwise we have a positive contribution from here. So the ground state of the model, which I write the ground state, is such that it can be written as the vacuum of all the a's. And then we have, let's write, zero. Well, we're going to put zero, OK. Vacuum, come on, zero. Well, this is momentum zero, the eigenvalue of the total momentum. This is our ground state. I give you four minutes, because some of you ask me, really strict four minutes. Then I start again. So don't go. Here in problem for the Hamiltonian, we have a sum. Yes. It's from k people who want and minus 1. Yes. We have omega side, but the second term? No, this is just a constant. This is a constant. So it's outside the sum. Oh, OK. And because of that, what is this omega cot in here? Cot engine, yes, of pi over 2 capital N. And what is the symbol that is besides the t? 2. No. Oh, it's a g, a cot. OK. This is cot engine. Cot. It was cot engine, yes. Pi over 2 capital N. Sorry? How did you get this cot engine? Well, what you obtain is the sum of this, and you can solve this. Just a serious sum, you can re-sum it and you obtain cot engine. If you want, you can write it as a sum, so yeah. Because you are using it. Then it's already, can you get? No, there is a reference for this case. What is it? Reference. Reference for you. I don't know. Now I don't know where you can find this exactly. I know what you're looking for. There are lechernos, OK? Maybe I will give you. But without doubt, you can find references for the singular monocledo. I should check which book treated the capital monocledo in this way. Because generally, OK, these problems are considered for this matter. And they are solved in maybe a slightly different way. So you just write the wave function and you try to solve the equation like we are doing. So I should think about the book, for instance. I don't know now. Sir, I didn't understand. I had this problem, too, of monocledo. And I found that it is unsolved. I thought about it. I got a note from my advisor that all these things are published. So you discovered today that they can be solved? No. No. Thank you. Use it. Use it. Use it. OK, so we will start again. I guess I see more than time. OK. So shall we start? When you say vacuum, it's the vacuum of the particles that we define. So we have another degree of freedom. So it's not, from my point of view, it's not so surprising. What's the physical meaning? So you know for particles having moment of freedom, why don't you think it means first of all for the vacuum, we can have p equals to 0 with the vacuum. But this means simply that it's localized. The central mass is completely localized. And you have p equals 0. Let's see. There shouldn't be problems. I hope. I think the center of mass is moving with some less So if we can move to some other place, I will turn on the momentum. Let's talk a little bit. If there is something here, like I said sitting down. What I mean is that these are completely independent terms. They commute. So you can create this part just like a standard particle without a potential, so a free particle. It's just a return of a free particle. And so just free because it's a p squared over 2m. So free particles at the ground state is that the momentum is 0, but it's completely localized. It's not an unnormalizable solution. OK, so let's start again. This is our ground state, which is written as the vacuum of this lateral operator a k. And it has 0 total momentum. OK. Now, again, how do we construct this set of states? We apply this operator, the creation operator. We understand why I call it a creation operator. And so we have a generic set of states can be written as in this form n1nj. Let's consider just a set of states with a 0 total momentum of the single simplicity. So this will be even by trust me about the normalization because it's something that you could guess easily from the result from the harmonic oscillator, single harmonic oscillator. So you have this a dagger 1. No, what is this? Yes, 1 to be n1 a dagger. This is 1 is k1, 1, k1, yes, n1. OK. No, OK. Sorry, sorry, sorry. Let's write it inside the form n capital n minus 1. So we have a1n1, a capital n minus 1 to be n capital n minus 1. 1, yes, n minus 1 factorial, vacuum, 0. This is a sector of our space. It's a sector just because we chose capital P, the total momentum to be 0. Otherwise, we have other eigenstates of the problem but just by changing the total momentum, increasing the total momentum. Just focus on this 0 momentum sector with this. How is this called this? OK, now, so far this formalism was just useful for some of the problem. So we define this kind of creation operator. And we were able to diagonalize the Newtonian, as we know that we also know the energy, clearly. Because the energy of this, the energy of the state, I will call it n1, n capital n minus 1, is just given by the sum over the sum. Now, let's write it well. So we have the sum of n from j equals to 1 from n minus 1. And j epsilon, the energy of k, which is the energy associated with this, which is already the energy, is omega sine of j pi over capital N. And I guess that's z i is this the energy. This here, yeah. Maybe that's right. OK, because it's consistent with the previous equation. I think that's it. OK. Yeah? Make it? 0, yeah? Why do you speak to the state? This is just another quantum number. It's the quantum number corresponding to the total momentum. And we are in the sector with 0 total momentum. You have also, again, you have a state with different p. Yeah? So now, we have created one vacuum, n1, 1, 2, n, n minus 1, so all of this is one more particular state. This is not a particle. This is 0. This is the momentum. It corresponds to the total momentum. It's a different, I would put a sprite in this way, just to. I see. Why is it that it's pointing the vacuum after we have created the particle? Yeah, but OK. If you want to characterize the eigenstates, you have to consider all the quantum numbers. So we have these quantum numbers, which are the number of particles with a given momentum. And then you have the total momentum, which is equal to 0 in this case. Yeah? The total momentum is 0. The chain, OK, it was kind of ring so that all the ions are just rotating. So you can always find the reference system where momentum is 0. But the sort of problem you have like all the oscillators for were coupled, and then we are. We decoupled the oscillators. Yeah. I can't find the explanation for that. The explanation is that, OK, originally when you consider the harmonic oscillator, you are describing the oscillation between two ions, just two ions. But then you realize that this is not a normal mode. This is not a, this doesn't have a fixed energy. This kind of oscillation is the system. Instead, there are other oscillations that have a fixed excitation energy and are collective excitation, where all these ions are moving either together or in the opposite phase or whatever. And this describes this kind of excitation. Yeah, yeah. OK, so now that is a nice interpretation of this formula. And the idea is to, indeed, reinterpret this operator here, this lateral operator, as a way to create particles, create a particle. So the idea is to say, OK, when I apply this a-dug, so this is, indeed, this I call the vacuum, so absence of particles. And then I'm saying, now, when I apply this a-dug, OK, I'm creating this particle with a given momentum, k. Just an interpretation. And using this interpretation, you realize that here, the state of the system do not have a fixed number of particles. Because here, you can construct, OK, we have the vacuum, zero particles. Then if you just apply one a-dug, you find one particle. So our system, generally, in quantum mechanics, you say, OK, I have a fixed number of particles. You've got the Hamiltonian, and that's it. So this is your states I've given, fixed number of particles. Now, if you use this interpretation here, we have to change, in a sense, our space. Because now, we have to take into account that you can have an arbitrary number of particles. In this space, it's called the Fox Space, OK? When you have this situation. So in other words, we interpret this operator as, operator that creates this kind of particle. So in particular, for example, the state, one and all zero here correspond to having just one of this particle with a given momentum, k1. And which kind of particles are these? Because we know two different species of particles. We know that there can be bosons. There can be fermions in general. So what are we constructing here? You can easily verify that we are constructing bosons. Why? Because of this algebra. So we have this commutator commute with one another. So if you interchange two particles here, you find the same wave punch. The same space, one equal to one. So these particles are bosons. And this is why sometimes you could hear that the way to diagonalize the harmonics later is to map it to a free chain of bosons. These are just bosons, free bosons. And we have a name for these particular bosons in this harmonic chain. They're called phonons. And maybe you already know about phonons from other curses, maybe you derive it in a different way. And they are exactly this model of oscillations. So the concept of particle is different from what you imagine usually. So it's not something that has a given position that you imagine of this moving, a particle. Here it can be also a collective motion. It's an effective description. So a single boson is something which includes all the degrees of freedom of our system. But still we can call the boson and we can study its properties. OK. So that's nice. But yeah, we like to diagonalize a metronome. But we would like also to investigate at least some properties of the state. We did all this work, at least the way we could compute something of the ground state. And yeah. They are all the same. Yes, indeed. Indeed, all the operator a tag commutes with one another because of this. So it's completely similar. And what is nice of this interpretation, because of this formalism, is that when you include, you consider bosons in identical particles in quantum mechanics, you have to symmetrize the wave function. So it's something kind of complicated, complicated map. Here instead, you obtain vectors which are already symmetrized. You don't have to do anything. You already obtain the final result. The symbol is already inside the properties, inside the operator. Now that we obtain the spectrum, we obtain the ground state, it could be nice to see some properties of this harmonic chain, just to do some physics at some point. And yeah. Well, you see from here. In this particular case, there are degeneracies. Because if you plot a sign of this as a function of k, this is nk pi over capital N. And this is the energy. What you find is this one. There is the degeneracy here. Yes. Yes. In general, you don't do that. Now as a matter of fact, every time that you consider chains, you have this kind of degeneracy. You have a similar degeneracy. It's always present. No. OK. We restricted ourselves to this sector. But you should imagine that you can actually consider. You can put every moment in p. So you will ask, OK, I'm interested in the ground state with a given moment, different from 0. Not the real ground state, but you choose another sector. And then you just change p. And everything is the same. So p is just a phase. When you write in real space the eigen function, the fact that you choose a different p means that you have a phase in front of everything which is proportional to the capital X. I, you choose some p, some particular p. And you will have a contribution like this. Yes. Times something. So the only difference between this and this with a p is that in the real, in coordinate space, you have this kind of phase multiply. It's completely the same. What you should expect is, because the interaction depended only on the difference of the position. It's not this momentum. It's a different kind of momentum. It's not the momentum of the ions here. It's the momentum of this effective particle, which has a completely different meaning. It must be clear about this. So we started with this variable. All the phonons, which are other particles, we correspond to this oscillation. So it has a different meaning. It's not a motion of the ions. It's something more complicated. OK. But now it's time to do this kind of simple calculation. And in particular, we could be curious, for example, to see in the ground state of this model, where are the ions? For example, so what is the average displacement of the ions with respect to the lattice? And in particular, clearly, if we compute the average displacement we find 0 because of symmetry, all the problem. So if you want to see the absolute value of displacement, we compute the squared of the difference of the x. We want to have some knowledge about how far the ions are from this artificial lattice sites. So in particular, we compute, OK, this can be useful. This we know by heart. I don't know if this is a boundary condition. No, no, we did. We did when we computed this. And when we computed the eigenfunctions and the eigenvalues of C. Because if you considered the matrix C as those particular eigenfunctions e to the phases, only if you have periodic boundary conditions. As I was telling to your colleague, if you consider open boundary condition, for example, just j, you have sine instead of the phases. As a matter of fact, you have a very similar result that you still can write in this way. The only thing that changes the argument when you change the boundary conditions. The reason why is this the case. Because if you imagine to take the thermodynamic limit here. OK, we will do. But now I just very quickly I tell you this. Then you send n to infinity to have infinite oscillator. And then this gives you the dispersion relation of this particle. But now if you are taking the thermodynamic limit, the boundary condition don't play any role in the bulk. So this is why you should expect the functional form here should remain the same. At the most, you should imagine that some change in the quantization of the moment already in there. The moment of the system, whether the state number n is fixed, that's why we don't consider this last index. OK, the last index. OK, the last index, because we have defined this operator only for index k different from 0, or capital n. So they are completely undefined. It's not something I remove. They are not defined. And the remaining operator, there was two operators remaining, corresponding to this 0, which are this one. So I just change the notation. This I call a. And this becomes x and b. And the reason is that a, you see they are not their mission. You have a and a dot. Instead, b is their mission. So this is why I change the completely undefined. And also because we know the total moment, we use b to indicate the total moment, which is an x to indicate the position. OK, so the idea is to compute this displacement, which is, let's use the notation. So from now on, I assume the total moment equal to 0. So I'm not writing 0. I'm always assuming total moment equal to 0. So the first question is to compute this displacement. What is the expectation value of this? Because x was the displacement of xi squared. So first of all, well, our system is invariant under cyclic permutation of the ions. So this means that this is independent of i. So we already know that it should be equal to x1, for example, and fix the particular side. Then how do we compute this? This is now when this kind of formalism becomes very, very useful. Because now we can express x in terms of the A operators. And we know what happens when you apply this A to operators to the vacuum. So we just have to write this expression and then use this computational relation and be able to obtain the result. So let's do it. We are here vacuum. Now we have to write this x1 squared. That's why I pulled this. OK. So we have, I want to understand, because m is i a n minus a minus a dot k. Then we have plus or plus x. And we have the same here, OK? This is what we want to compute. Now, OK, we have several terms here. Let's immediately see which terms go to 0, either go to 0, the last term, x1, 0, 1. So here we have, OK? Anyway, first of all, we are sometimes like this, OK, multiply by itself. And this we have to compute. Then we have a term, which is this, multiplied by x, here. But now what happens is that this commutes with x, because the two operator commutes with x. So this means that you can move this operator to the right, if you want, of x or not. Now, when you move this operator a to the right, then it annihilates the vacuum. So it is equal to 0. On the other end, this operator is an a dot, and annihilates this one. So you have that there is zero contribution from this times x, OK? Then we have, I should see, now, it should simplify in some way, the contribution from x, for x squared here, OK? So let's compute it, because I did it, and I didn't have that. Let's compute this first term, so this first times the second. And we already know that we can neglect this, because this annihilates the vacuum. And in the same way, we can remove this operator from here, because it annihilates the vacuum on the other side. So we have that is equal to, this is equal to 2, OK? The other is the vacuum. Then we have some k from 1 to n minus 1 e to the minus 2. Let's consider first this term times this, so 8 pi k over capital N, 1 over square root of 2m n omega sine k pi over capital N times i a n minus k. Then we have here sum over, I call this q, 1n minus 1 e to the minus 2 pi i q over capital N, 1 over square root of 2 omega sine of q pi over capital N times i minus i a dot q. Right to the vacuum, and then we have also the other term. OK, so now it's true that this doesn't annihilate the vacuum here, but if we are able to move this operator here, then we annihilate the vacuum. So it becomes 0. So the only possibility for this term to be non-zero is if this n minus k is equal to q, because only when k equal to q, the commutator is different from 0, OK? So this means that this is equal to pi. Now we have the condition q equal n minus k here. So we have a sum of k from 1 to capital N minus 1. And then we have e to the minus 2 pi i k over capital N, 1 over square root 2m n omega sine of k pi over capital N. Then we have this term with the q equal capital N minus q, which simplifies this term. And we have this term when, what is this, pi? When q is equal to capital N minus q, which is equal to itself, because the sine is in variant under reflection about pi over 2 sine. So this is the same term. It means that we can remove the square root here. And then we remain with this a n minus k, a dot n minus k vacuum. Yes? Yes. We know that the vacuum is defined as a state such that when you remove, there are no bosons, that's 0. It's equal to 0. OK, so now you consider all the terms here. You immediately see that this term here gives zero contribution. Why? Because you can apply it to the left. And so this annihilates that part. Analogously, if you consider the other term, now you have in this other term, you have this contribution a that annihilates this vacuum. So you can remove those immediately. Then, OK, you remain with these terms. And now you realize, oh, I should be careful, because now I could move this to the right and annihilate the vacuum. And the only possibility not to do that is that n minus k is equal to q. So this means that I can just consider that term in the last expression. And this is what I obtain. Now, if I move this operator to the right, what do I find? 1, sorry, I find 1, because the commutator is equal to 1. And then the remaining term is just the a applied to the vacuum. What I mean is that this is, OK, I cannot write here. Maybe I'm sorry. I can write here. So this is equal to the commutator between a n minus k a dagger n minus k plus a dagger n minus k a n minus k, just by definition of the commutator. This term is equal to 1. And this term is zero contribution, because it's applied to the vacuum, OK? So in the end, you find that this is equal to 1. And we have an expression plus the rest. What about x squared? We will come back this afternoon. We will continue. And yeah, I think it's enough. Yeah, yeah, we will talk about this. OK.