So in this video, we're going to complete the derivation of the spectrum or dispersion relation of the quantum phonons. And well, this part as well as the previous part is a bit technical, so I will actually skip some of the calculations and we'll just show you sort of the method, illustrate the method. We just called, by the way, in the Bogoliubov transformation that is used to solve such a problem. And as a side remark, let me mention that well this material that I'm going to be covering is actually a pretty advanced, gradient material. So if you've got to this point, you should be proud of yourself. This is very complicated, very advanced material of quantum antibody physics you're actually studying. So in this slide, I would like to review the calculations we have done so far, sort of tell you where we have been and also where we're going with this. And as we know, we started with this classic oscillator chain which is characterized by the coordinates and momentum for the atoms in this one-dimensional crystal. And the first step in putting together a quantum problem was to quantize this, which means basically to replace the variables x and t with the operators, and so, we just put hats on top of x, p, and h. So the following step involved expressing this momenta and coordinates with the corresponding creation and emulation operators, which led to this complicated looking Hamiltonian. And here, this an and am dagger correspond to creating an oscillation in a particular site with a particular atom labeled by m. So the next step that we took in the previous segment was essentially to Fourier transform this creation and anti-annihilation operators and to express them in the q space in the wave vector space if you want. And the corresponding Hamiltonian is shown here, so it has, as you can see, two types of terms, a dagger and a and this aq a minus q. And so, now, we have to work with these operators, these new creation, and annihilation operators, and this Hamiltonian. So and now I'm coming to where are we going with this? So if we did not have, let's imagine for a second that we do not have this term. Let's say we have just this term. So if this were the case, it would have meant that we already solved the problem. The reason for this being that this linear combination, the inevitable altura is a linear combination of this A dagger a's. independent a dagger a's, they sort of correspond to a collection of independent harmonic oscillators, the spectrum of which we know perfectly well. So for each q, we have a, basically a simple harmonic oscillator such as in the previous lecture. And we know its spectrum and now we just collect these oscillators and write them as an integral, but all in all, we know the results. So this is essentially already a solution. And therefore, this A of q or more precisely actually do to the omission conjugate part here twice A of q, would have been the spectrum of our problem, the spectrum of this quantum force. But, of course, we do have this true. We do have this true. And we don't know what to do with this, right? So this, so the spectrum of this Hamiltonian is not at all obvious. So a way to solve the problem now is to actually to construct a new set of creation and annihilation process. So here, I have an operator B which is an annihilation operator, can you type? And there's also a B dagger of q, and so lets assume that we can construct this B and B dagger as a linear combination of a and a dagger in such a way that the new Hamiltonian won't have the bad terms. By bad terms, I mean the terms that we don't know how to deal with and instead we'll have only this standard combination of b dagger and b for which the spectrum is obvious. And so this essentially would mean to solve the problem. The ability to write the Hamiltonian in this form, in the form of a liner combination of some creation and annihilation operators of this sort, so would mean to solve the problem in this why's this a tilde would give us the dispersion relation of these bs. Which actually are the quantum phonons. So as we will see, so this b dagger hat and b ha, hat, then correspond to creation, and anagolation operators of quantum phonons. Essentially, the normal modes, the quantized normal modes, in this oscillator chain. But together is well, as you can see, is rather nontrivial for us, we have to do this quantization and then, well, which was simple, but then we have to do with this 3 Linear transformations to get to this point. And this last linear transformation is what is actually known as the Bugoliubov of the transfrmation. So here's a picture of a young Nikolai Bogoliubov who was a very influential and talentented mathermatical physicist. A ration mathermatical physicist. And he was an author of many very important works and also he came up with a number of very important and practical mathematical methods that are used up to these days. And one of these methods is this Bogoliubov transformation that we are going to be studying. And let me mention that this Bogoliubov transfomation is still used very actively in research papers. I myself have used it in a number of of my published works. And this is basically in some sense an everyday tool of theories working In quantum physics and then condensed matter physics as well as perhaps high energy physics. So this is an important and important method but well in the example that we're going to be considering to be sort of method is rather straight forward and I want to get you advertised, sort of a motivation for using this method so lets lets look again at this Hamiltonian that we have for the Original the creation and annihilation operators a sub q dagger and a sub q. So and again, as I said, so our sort of strategy would be to correct in some sense, the definition of the creation and anogolation operators, so that we can get rid of this unwanted true. And, if you look at, at this Hamiltonian, you will see that this Hamiltonian involves essentially, so if we didn't have this term again, so we would have had just a linear combination of independent q operators. But this term, the appearance of this term couples the q mode with minus q mode. So whatever transformation we come up with, it should probably involve a linear combination of at least a q and a minus q. So let me just preset a form, sort of knowing the result of course is easy but it will save us some time. So let me just present a form Well if this generic Bogoliubov transform that accomplishes what I want. Namely that it introduces new operators in this case these are these operators b sub q and b daggers of q which are linear combination of a and a dagger or vice versa a and a dagger are linear combinations of b and b-dagger is here. And so here I will assume that this assumption will be verified, sort of a posteriori. And that, let me assume that this, these coefficients can be chosen to be real numbers well, functions better to say. And that there even functions of q. So, u of q is equal to u of minus q. So in this case actually this second definition here is redundant because the second definition of the creation operator can be obtained from the Definition of the annihilation operator, simply by applying the permission conjugation. But in any case, so, let us assume that we can construct such operator. So the first thing we need to do, and this is, I'm basically describing the strategy of constructing Bogoliubov transforms Is to make sure that this b and b hat ID creation and anti-annihilation operators. So, I can call them such, but actually, this should have a precise mathematical meaning. And the precise mathematical meaning would be that this guy. So, let's see if I have b of sum q and b of b, let's say dagger. So these guys should be the commutator of this guy should be zero if q is not equal to b and it should be a delta function other wise. So infinite otherwise in some sense. So this what I'm looking for. So my a and a dagger, so the reason I call the creation and annihilation operators was that they satisfied this relation. Now I want the new operators which are actually a linear combination of those to satisfy the same commutation relation so and this would involve a calculation which would result in a constraint on these coefficients u and v. So if I calculate this commutation relation you already know. So this is a the commutator of a sub q and a dagger sub b. So i simply plug in these definition to here and this guy to here with replacing q with p and I just calculate the commutators term by term. So what I can write, I can just write so we'll have one two three four different terms. So the first term lets say will be the commutator of these two guys so such is here and using this constraint I will require that the commutator of these two guys will be two pi delta function of q minus p Now these two commutators which result in this term should give me zero because these both are action and emulation operators. So likewise these two guys resulting in this term should give me zero as well because these are commutators of the creation operator. And so finally I have a term which couples b dagger of minus q and b of minus b and well since the commutator of b b dagger is equal to well this delta function the commutator of b dagger and b is basically minus the commuator of b and b dagger so it should give me this function. And let me remind you that, the delta function volt here, is actually an even function, so it doesn't really matter that I have here minus signs, etcetera. So, and if I put everything together, so these guys basically go away, right. So these guys go away, and I have these two terms, which essentially all involve the same Delta functions and also since there are delta functions here, Q is actually pint to piece so the delta function is zero if Q is not equal to P and so there are only terms, here I can simply write it as let's say UQ U of Q squared and this guy will V of two squared Now with a minus sign. So if I put everything together from this expression so this is what I get, okay. And now I have to required that the well the right hand side reproduce what is supposed to be in the left hand side. That is this two pi delta function of q minus p. So which means that this coefficient here. So this u squared minus v squared must be identically equal to one and this introduces a constraint, an important constraint on this Bogoliubov transformation. So basically what we are doing here again so let me sort of go back to where we are going with this, so we want to put together this new operators. And so we want to find explicitly the expressions for these u sub q and v sub q, that first of all wouldn't re-, will, would ensure that these b and b dagger nd that annihilation and creation operators of something. And second another constraint on this this guise would be that the resulting Hamiltonian now expressed in terms of b and b dagger would be simple enough so that we know its solution. And this basically the essence of the Bogoliubov transformation. In this case this Bogoliubov transformation is relatively simple, and actually most cases you will see in the literature, even the research literature, it is relatively simple, but in general, it may be more complicated. So it doesn't necessarily have to include just two, two, two, two terms, it may include more terms. But in all these cases, so the strategy will be the same. So basically to ensure that the operators you introduce are indeed creation annihilation operators and second then they simply our Hamiltonian. So now we can sort keep this constraint in mind or we can resolve it once and for all by just using new barometrization if you want of this u and v. So and this can be achieved by using the following mathematical identities. So if we have, if we use, the if we introduce u of q being a hyperbolic cosine. So let me remind you that the hyperbolic cosine x is equal to e to the power x plus e to the power minus x divided by two and the hyperbolic sign of x is equal to e super x, minus e super minus x, divided by 2. So, this is a real and sometimes real version of the sign and cosine in, in which case I would have had And the metric[UNKNOWN] but the main case of this guy's so satisfy the following identity which is sort of an analogue of the identity that you most of you know that sin squared plus cosine squared is equal to one. So here for the hyperbolic functions I have the hyperbolic cosine squared minus hyperbolic sin squared is equal to one and this is exactly the identity that I want For my ??? if I use this ??? we should resolve this identity once and for all. Which basically reduces the problem with 2 parameter, to a problem with a single parameter. In this case this is [UNKNOWN]. So my sort of expression for the creation and angulation operators. A and a dagger now becomes this ok? So I have a of q is equal to cosine hyperbolic of Lambda b q plus sin hyperbolic of Lambda b of minus q dagger and the corresponding a dagger. Alright so now what needs to be done is that we ahve to just do a brute force analysis. We have just go ahead and plug in this expression this expression sent to the Hamiltonian, and I will leave it for you for those of you who are interested. For those of you for instance who are interested in becoming theorists, or who are theorists, or who are mathematicians or are just curious about the technical part so you can go ahead and do the simple, relatively straight, at least I'm not going to call it simple, its a straightforward exercise of just plugging in lets say this and this guy. Here we're going to put this linear combination, instead of this guy we're going to put this linear combonation, etc and so if you do so well you will see some unpleasant algebra but this is an exercise I'm sure you can do it. So here I'm showing the result of the calculation, if you did everything correctly this is what you should get and the result here is a Hamiltonian which now is expressed in terms of these new operators. B of q and b dagger of q. And as you can see it has the form which is pretty much the same as we had before just with some new coefficients which I call now i a tilde and b tilde and here is the expressions for these coefficients involving hyperbolic cosines of two Lamba and hyperbolic sins of two Lambda and these are the regional coefficients. And the Hamiltonian for A's. And so, you may ask me what, you know, why did we bother to you know, introduce new operators, if we got exactly the same thing? So, well, the answer to this, and actually this brings me to the second step in this Bogoliubov transformation is that well lambda here is a free parameter, so, better to say a free fun, function. We can choose this function Anyway we see fit. In particular we can choose it such that this unwanted term, the bad term that we don't want to see is identically equal to zero. So we can esssentially can set by hand this b tilde of q equal to zero which leads to an equation for lambdance. So this equation as you can see from here if we just move this guy to the, to the right-hand side and divide by cosine, so sine hyperbolic over cosine hyperbolic is tangent hyperbolic. So the tangent hyperbolic of 2 lambda of Q, is equal to minus B of Q versus over A of Q. And so this essentially gives us a specific result for this Lambda of q which ensures that our Hamiltonian in terms of this new creation, annihilation operators looks simple. So now let me clean it up a little bit and sort of summarize things again. So this is the result that we got and also let me remind you that this a of q and b of q, the regional coefficients we already know. So we derived them in the previous part of this lecture and so let me remind you of their explicit forms. So here is their explicit form. So this explicitly b of q and this is explicitly a of q and well I wrote it sort of explicitly this way but as you can see lot of things here cancel out including this H omega and there loads of other simplifications involved. And if you play around with this, well, with this The coefficients if you do a little bit of algaebra you will see that well first of all Lambda of q actually determines as you can see from here. So this guy is zero as we wanted it to be but now this guy is determined by Lambda and we do know Lambda. So we can simply plug it Y'know in here and if you, once everything is said and done you actually has an expilict expression for this a tilde of q. Which becomes in the end of the day surprisingly simple. So again this calculation is not obvious from here. It requires a little bit of work and due to the format of the lectures And lack of time I'm not presenting it but those of you who are interested can of course check that this is indeed the result. So this is by construction but a of q has this form. So if we plot it so this a of q has this form which is actually a very familiar form so this is a of q which we have seen before in the classical problem in the context of acoustic forms and this is exactly what they are but now they are quantum formula. So sort of to summarize the result and also to emphasize a very important point so this is what we actually got. We have found explicitly a series of linear transformations Of the regional Hamiltonian, of the regional operators involved with the Hamiltionian. So I should have the Hamiltonian can be sort of was casted in this form. Which is the so called diagonal form of the Hamiltonian. Sometimes if you happen to read research papers, the reasearchers call about diagonalization of a Hamiltonian. So today in lies the Hamiltonian, and, and in sort of in the problem, in a problem which involves mainly quantum particles, essentially means, the ability to write it, as a linear combination of independent oscillators or some other particles, the sort of behavior of which we know very well. And so, this is exactly what we accomplished here. And so here we have also this Hermitian conjugate, so if we couple it, it will be simply the same term, so we can write it integral over q, well twice A tilde of q b dagger q b q. So, and well, I should also mention one thing that I sort of swept under the rug and this thing is that I haven't paid attention to constants that appear in the Hamiltonian so, there could be constant here. If I had done it, so the correct result that I would have found is well, this is correct, but there also is a constant term here, which corresponds to zero point motion. And this term happens to be exactly one half, so if you remember the solution to the simple harmonic oscillator, we have essentially h omega and plus one half and so similarly it appears here. So in this two a tilde of q is exactly the energy modes of the oscillator. So we can call it so this guy is equal to h m n of q and this would be the spectrum of the quantum phonons, and this is as presented here. So the spectrum of the quantum phonons here is twice omega. Omega here, is the frequency of a single oscillation in some sense. In the original classical picture it would have been the frequency, the natural frequency of this spring connecting the atoms in some sense. And so, the remarkable thing, that I want to emphasize in the very end, and this will be the end of the, today's lecture is that, it turns out that this spectrum of quantum phonons, is actually identical to the spectrum of classical phonons. There is absolutely no difference whatsoever. So if you go back to the problem that we solved earlier today, in today's lecture, so namely the classical problem, so it was slightly more complicated in the sense that it involved two types of atoms, but if we have just one type of atom so this is the result which we would of got. So an important conclusion of this sort of discovery that quantum and classical phonons turn out to be exactly the same is the following so we can say that we could have quantized our problem at any stage. So in the solution I present it we did so at the very beginning. So we essentially converted the original coordinates and momento of the atoms into quantum operators, x hat sub n and p hat sub n and then we went ahead with this quantum calculation with the newer transform. But instead of doing that sort of there is a second version of the solution. One could have solved the problem, the classical problem first and which means finding the trajectories of particles and the corresponding waves. And so one who could have quantized this normal mode. So basically so the normal modes can be converted then into this Essentially operators, h omega q, sum b dagger q, b q plus one half. So we could have done that. And the results, basically we have proven that the result are, are exactly the same. And this is very important. It shows us something about the quantum to classical correspondence, and this quantum to classical correspondence sort of persists Also in more complicated problem. Let's say problems in three dimensions with more degrees of freedom. With different types of atoms. And this is true as long as we have non-interacting theory, which in the language of this creation and annihilation operators essentially means a theory which is bi-linear in this b's and b daggers. So as long as we have this bi-linear forms we're fine with this quantum two plus equal correspondents but once we have y'know more complicated nonlinear terms, so this situation becomes much more complicated. So the last thing I'm going to mention is that the appearance of such terms, which we're not going to discuss at all but you know if we had this quadratic terms, b dagger b dagger b b for instance. So they would have meant that the photons interact with one another. So in some sense it would have responded to scattering of waves off of each other, waves running through the crystal. And this is a very complicated challenge to understand the behavior of such an interacting system. And there is still on going theoretical challenges in describing it. So it's definitely beyond the scope of this course, but this theory is well understood and I hope some of you got an idea about phonons and collective modes and classical and quantum physics.