 Let's talk about the same thing for the topic. We want to clear up your last class before we come back. The first thing is this silly issue about signs. It's not going to be very illuminating since we've confused ourselves about the signs of the action in the last class. Let's get it back up. OK. So the first question you should ask is how do you know what is the correct sign? How do you know what is the correct sign of the action just starting with the first action, which, by the way, is the name. It's called the number of actions. So remember, we said that the action is equal to minus 1 by 4 pi, or 2 pi alpha prime times the determinant, well, square root of minus 10 del alpha x mu del alpha x mu. Is that the action? So just to convince you this is the right physical sign of the action, let's do the following. Let's take a configuration of this string that stretches infinitely long in the x-1 direction. So because string was a very strict configuration, it was stretching infinitely long in the x-1 direction. So the origin in all other spatial directions. And for this string, let's be completely physical. And let's use, again, such that x1 of sigma is equal to x1, and tau is equal to x0. And let's look at what this action becomes in this range. We're only using, say, great interpretation purposes here. So we also combine ourselves to an action around this equal to different variations. So we'll combine ourselves to an action that is quadratic in black directions. And let's see what this becomes. Useful exercise to make contact with something, you know. OK. So what do we get? Well, so first let's write out the stuff in the model here. So this s is equal to minus 1 by 0 to 5 alpha prime. And then that's integral. And then this fancy determinant, as we saw last time, could be written as minus x mu dot x mu dot times x mu prime. And the plus x mu dot, this is minus 1, this is plus 1, this is minus 2, this is minus 5. Now I want to take this action and expand it to quadratic order in black directions. So first question I ask is, is there any term that's not 0? That's 0. Is there a background reason for this action? And here it is. Because x mu dot dot x mu dot is not 0 because it gets contributions from x0. So that part is minus 1 because of the negative vector in the time direction. Space, the time direction of space. And x mu prime makes me prime is not 0, from the contribution in the x1 direction. So that part is plus 1. So we see that to 0 order, the action evaluates to square root n, any type of square root of 1. Now let's do the quadratic issues. This term is 0 in the background. And so this term is 0. OK, let me just take what comes from this term. Just to say that we get some addition. What part of that's very much? So very quickly, if we want to do what comes from this term, we'd have to take background n fluctuation here or background n fluctuation here. So background here and fluctuation here gives us minus x mu dot x mu dot. OK, now versus mu, this mu should become R. You see, because of these degrees of freedom, 200 case fixed. So they're not dynamically distributed. So these are the remaining d minus 2 that are dynamically distributed. So the part where mu runs over either 0 and coordinate or 1 v, coordinate where it gains space, only contributes to the background. There are no fluctuations in those notifications. Is this clear? OK, so suppose we're going to be writing three writing exactly that area, which is hurrying up. But suppose we're doing the rewriting exactly. Actually, this term won't even be contributing to what I wrote for this one. So suppose we're doing the rewriting exactly, what will this thing become? So this thing would become minus 1 plus x mu x i dot x i dot. This term would become 1 plus x i prime x i prime. And this term would become plus x i dot x i prime. This isn't exactly like the expression about making no assumptions. Now you see that this thing is quarter, it is quarter x. We really do what we need to. It's to quadratic what we forget about it. Now to quadratic all of what we get, we get 1 by 2 pi alpha prime. Then because we're doing quadratic, we're going to get half on the square root. And we get integral of, sorry, there was a minus sign here from this minus sign. So we get integral of 1 minus x i dot x i dot plus x i prime x i. We have written the square root of this in a dupe, which is equal to 1 by 4 pi alpha prime into the integral of 1, well, 2 minuses minus. And this is now minus x i dot x i dot plus x i prime x i. This is it. OK? So now we don't consider the reaction we get about it. Though it might be just for future purposes, let's remember that it's a negative constant. So this is 1 by 2 pi alpha prime minus plus 1 by 4 pi alpha prime into x i dot square minus x i prime x i. This is a physical action. Because this is your physical action because that's positive guidance. It gives you a physical action for fluctuations of the string. In a condensed manner, of course, if you write down an action for fluctuations of the string, you would write down this action with something not equal to attention. OK? So that determines the sign of what we start with. That there have been months. We saw that last time that this was equivalent, classically, to 1 by 4 pi alpha prime with a class of formulas, which there was a process point where we took the square root. We weren't sure which sign to take. 1 by 4 pi alpha prime square root minus g g alpha theta z alpha phi x mu z alpha phi x. OK? OK. Which sign should we be taking? Well, there are two arguments, both of which we did the same answer. We did them carefully. Though one of them was I was doing it confusedly in the aspect of getting the wrong answer. The first argument is whether we better have the kinetic term positive for the physical fluctuations. The physical fluctuations are the I fluctuations. So there are no signs from the spacetime metric, but the kinetic term does have signs of a word shift. That's how mine is from a word shift metric. OK? So if we want to make it positive, there's a better way to notice. Is this clear? When I was getting confused, I was the most diffidentness of this action last time. But is that stupid confusion? I think it's easy. OK. Let's take the same integration as before. And check that on the background. Just think it evaluates to something negative. OK? So the configuration where we can have the x0, we can tau and x5, we can go to sigma, some constant. x1, sigma. OK? Now let's see. You see, the x1 part is clearly positive, because both signs are positive. But the x0 part is also positive, because both signs are negative. Which is great for this case. In fact, value will also be, because you get 2, which can cancel the value. Is this clear? OK? So any point of view of force-to-correction, the last time I was remembering the minus sign of the world shift metric, but for getting the minus sign for x0 and spacetime. OK? So this is the right sign, but in the process, the derivation will be to the square of which it is. Any questions about? OK. So that was clearing up one for both. Now if we want to go back to our quantization, the sedation of the string, let me remind you of what we thought was right. We decided to do the quantization. We just had to do the quantization in this poly-outboard, the so-called poly-outboard, actually. So we decided to work with 1 by 4, 5. Alpha prime is minus, square is minus. G, integral of the front, d alpha beta, del alpha times mu. OK? And we fixed gauge. We gauge fixed this action. It was classically equal to the early action, only when both of your opposites, as well as wild transformations were treated as gauge. So we fixed gauge to set G alpha beta, equal to eta. The flat mid-touch came next to the lower sheet. There you go. But we realized that this gauge-fixing procedure did not completely fix gauge. OK? There were designing sigma plus and sigma minus, sigma plus is tau over sigma, sigma minus is equal to tau over sigma minus sigma. OK? There were gauge transformations that were not fixed, because of the form sigma plus is equal to f of sigma plus, till now, for instance. And sigma minus is equal to G of sigma minus. OK? Now this is actually a very small amount of gauge, a mid-touch, because it functions one value around 2. But if precisely the sort of gauge of rigidity that is relevant to classicalism, that is important to classicalism 6. And why is that? That's because we saw that solutions to the equations of motion, the classical solutions to the equations of motion, and all the x's being a function of sigma plus plus is function of sigma minus. OK? So nine-year appeared as if we had 26 functions of sigma plus, d functions of sigma plus, and d functions of sigma minus. Now, those d functions were each constrained to a real constrained equation, that of my gear of momentum. So that looked like it went to d minus 1. However, this reparameterization ingredients actually tells you that this d minus 2 functions of sigma plus and d minus 2 functions of sigma minus are parametrized equal to solutions. OK? That was an important crucial point. So the space of classical solutions was space of d minus 2 functions of sigma plus, d minus 2 functions of sigma minus. OK? Now, what do we do? Last time what we did was in order to fix this ambiguity, in order to fix this ambiguity, we have one of the functions which returns to d x minus. OK? And set all the non-zero mode parts of it to be y minus co-ordinary recognition. Because to be honestly in the sigma coordinate, you couldn't touch the zero mode parts. You could redefine that mode, because that would break. Pick out the k. So this was alpha prime, d minus tau, and then something I forgot to write last time. So there's a center of mass, just a constant. So we were able to use our gauge redundancy to set this into this one. OK? That made it manifest that once we in this gauge, it manifests that d minus 2 equals 3. That d minus 1, less than half a liter of this gauge and then this constant. Now, the question is, how do we take this classical phase space, this classical space of solutions, and what phase? OK? Can you find a reasonable procedure that will allow you to take this classical physics that we've solved completely, and to give it a new part of it? OK? So in order to do that, I'm going to give you a five-minute introduction, a five-minute introduction to a way of thinking of quantization that is possibly broader and more, but we make very little use of it. It's not possibly broader than the ways you have been introduced to before. Actually, we've looked at quantum mechanics one class. We use this method to quantize like that. They don't think it quantizes. OK, so this method, both have fancy names, it's called geometric quantization. We can use such a small part of it that we won't require much in the game of devinology or anything like that, but let's just see the method. You see, how do we normally do quantization? What we normally do is say, well, suppose we were just doing number one mechanics, we had x and b. We say that there are these variables, x and b, and they have a quantum practice such that this thing is 1. And then we raise this Poisson bracket to a commutator. And commutator, please, multiply Poisson by i and make them true as commutator relations. OK, so from the study of classical mechanics, as you know from the study of classical mechanics, this structure of a commutator, the structure of a Poisson bracket, or even if you didn't know any of this, I mean, the structure of the Poisson bracket is a structure that is geometrical, and it's not tied to a particular set of variables, not tied to a set of variables x and b. If you use any variables in phase space and you find a Poisson bracket for these variables. So let's see what I mean. So given a set of variables in phase space, let's say that the moment we're dealing with x and b, x and b is 2j, so that's more than 1x, more than 3x. Let's say that this thing is a omega ij. Where omega ij is delta ij is delta ij, and we also have. So if we had a system of nx's and nb's, we've got a matrix, this omega matrix, which is 2n cross 2n dimension, and it's an anti-symmetry matrix. Now to move to more general variables than x's and b's, it's better to think of this omega as a tensor. It's a tensor in both of those indices of alpha. Though in order to do the manipulations with it, I will find it more convenient to define the inverse of omega. This is a matrix, both of those indices are over. And to do my manipulations with that inverse of omega. An inverse of omega is a tensor in a more dimensional way. So what is the inverse of omega? Once we've got omega, we can find the inverse. So if omega, remember, x, i, p, j, so I'm writing x, i, and p, j to say that it's a Poisson product with x and b. It was equal to delta ij. It's easier to convince yourself that omega or the lower x, i, and p, j is equal to minus delta ij. Now this is the fact that 0, 1, minus 1, 0 is the inverse of minus cj. In order to make it square 1, we need to put it on this side. Make a power matrix. So instead of dealing with this object here, this object which is an anti-symmetry matrix and the on-phase space, I did look at this object here. Whenever you have an anti-symmetric object, an anti-symmetric tensor of any rank, but with lower indices, there is convenient notation to deal with this anti-symmetric object that we're now going to use. This notation makes it a form. It's just a notation. I'm going to encode these components, omega, x, i, p, j, into a geometric logic, which I'll call omega, which is omega, x, i, p, j, dx, i, d, and then the cylinder here, d, p, j. It's purely form. In order to understand what it means, it has the same formal content as the translation between the components of a metric and the expression for an infinite density. When you have a metric, it would characterize it by gij. It would characterize it by the focal length of the infinitesimal line element, which is dx, which is scale out. ds squared is equal to gij, dx, i, dx. Taking this equivalent of a metric, which is this infinite form, and multiplying it by some formal object. Basically, I have an interpretation for such infinitesimal lengths. This thing is a formal object, and the only property of this formal object that we will use, the only property of this formal object that we will use is that it's antisymmetric. Because we're doing this for an antisymmetric object we could avoid this is, we can call the fact that this thing is antisymmetric into a formal property of this d object. The formal property we will use is that dA, where gij is equal to minus gij, drawn in this course, in the middle of next semester, we will return to a study of this geometrical ideas in much more detail. Not for using quantization then, but in studying compact imagination, space dynamics. Let's quickly introduce this notation and these ideas anyway. At the moment, we don't want to go to systematically. All I need about this object is that it's... Okay. Now, it's a very nice way of thinking of the symplectic structure, the quantum bracket structure of classical theory. You see, as we have, as we discussed in the previous class, a good covalent way of thinking of phase space is to think of it as a class, okay, as the sake of all classical solutions of this system. It actually induces a quantum bracket structure, a symplectic form, on the space of solutions. If you label solutions by initial values of x and initial values of p, that symplectic structure is just delta p raised to x. It's not delta x raised to delta p, because remember, this bracket, once we lower it, you pick up an additional matrix. So, if you use x and y to label solutions, the symplectic structure on that labeling of solutions is just delta x raised to delta p. But the symplectic form is a geometric object and that means there's any covalence. So once you have a parameterization of phase space, once you have a parameterization of phase space, if you want to know what the symplectic structure or the Poisson bracket structure all in the space space is, what you have to do is to use canonical covalentization to say that in the x and p parameterization, it's delta x raised to delta p, but minus i. And then use the usual rules of tensor transformations to go to the symplectic form in any other way. This notation makes that particularly convenient because if you got a set if you got a set of solutions explicitly, you know what x i is as a function of parameters. You know what p i is for those solutions as a function of those parameters. In order to transform the symplectic form into form on these parameters all you have to do is to put in what dx is in terms of d of those parameters. d transforms like a differential. That's that induces the right tensor transformation properties in omega. So all you have to do is to substitute what delta x is in terms of delta of the parameters of your solution delta p is in terms of the parameters of your solution and then you have the Poisson bracket structure in terms of the parameterization of your face space namely solution space that is motion action. You can use that to invert this omega in terms of the parameters. On the objects that you would promote to operators that you like not necessarily in terms of x and p. Is this clear? Now actually this procedure often leads you into a symplectic structure that's not flat. It's not like in terms of new variables you have constant functions times d something d something. How do you do the quantization that is a complicated and delicate story of geometric quantization. Which we are not going to get in here but it's luckily in what we do at the moment we will always get flat strokes of flat spaces. So how do you do quantization will be great. Is this clear? So we are going to now implement this use these ideas to quantize our state and space solutions. The beauty of these ideas is that it uses when you can classically solve the system. You've done a lot of them. It uses the fact that you've correctly understood what the right space of classical solutions is. In order to promote the theory of quality. That's the situation we're in. We completely understand the classical theory. We should be able to relatively easily get a quality. Let's see how we do it. Is this clear? Let's move. 1 by 4 pi out of r. Integral equal to sigma. Now we let's say, even before fixing the gate there is G G alpha beta then I will find mu In order to get the symphlectic structure in space of solutions, what is the symphlectic structure? Just for every variable it's dx where is gpe? Somewhere over all the variables. The variables are continuous integrated. What are the variables in the game? There's G alpha, G alpha beta There are eight solutions to the symphlectic structure from G alpha. Somewhere. Zero. It's momentum going to get to G alpha. That deep part is just zero. Okay, what are the other solutions? Fine. So, you see, when we when we gauge fix we must be careful to use the symphlectic structure of the whole theory before sitting on the slice that we're interested in. But in this particular case we're fixing G something G itself has no symphlectic structure we can just do the quantization on that slice. It will not miss any it will not miss any quantization of the symphlectic structure. There you go. The next So, what is the symphlectic structure? Well, now we might as well switch to dots and primes. So, the action as far as quantization is concerned is the same thing as d to the sigma x Now, we're going to split up into x plus dot x plus dot and this probably has minus sign x minus sign dot plus x plus prime x minus prime and then plus x i dot squared minus x i minus 2. We've got signs right here. One would be Yeah, this is this right x plus is t plus x x minus t minus x the part that comes from d and d should appear with minus sign. Yeah, so that's the plus sign. So, this is our action So, what's the symphlectic form of the function of this action? So, the symphlectic form of the function of this action is 1 by 4 by 5 but now we need to complete the momentum conjugate to any variable ok that's fine 1 by 4 by 5 where integral now momentum conjugate to x minus ok, the momentum conjugate to x minus this is minus x plus dot ok, so we get minus x plus dot delta of this that class is d here ok, we are writing delta as this function ok wage delta of x minus ok, what's the momentum conjugate to x plus minus x minus dot so, plus so, minus delta of x minus dot wage delta of x plus ok, and the class sum over i ok 2 times delta of x i dot wage delta of is this clear? momentum conjugate to any variable factor of 2 is momentum conjugate to x i that's a 2 delta of the same symbol is d I'm using the symbol delta because it's a variation of function a function of a small variation of a number I will write it if you want it's just formula it's the basis just the basis behind which you write here this is exactly the same information as the statement that the Poisson bracket will be x minus and x plus dot is a delta function that's all that we are seeing standard denominator of this x plus it doesn't contribute to symplectic structure because symplectic structure is only derivative with momentum is derivative at the time we know what that right space of physical solutions is ok so what is the correct space of physical solutions? the correct space of physical solutions can be labeled distinct physical solutions ok, or what? we can write x i as equal to and I should write it the way I wrote it in the notes everyone writes it as an automated writing standard x i is equal to plus i x i plus x i minus where i plus is equal to alpha prime by 2 the arbitrary and arbitrary function of x plus so that respects video analysis so that's sum not n not equal to 0 alpha n if alpha minus i is plus and then I should also write 0 for the contradiction so that's x i is 0 plus alpha prime when alpha prime b i is done x minus i is equal to alpha prime by 2 sum n not equal to 0 alpha 8 tilde by x i n of x i I still have to write down some part of x plus and x minus we'll do that in a moment but yeah, what have I done? I've just taken an arbitrary function of x i and fully I've transformed it at the zero but that's my choice it's the choice of dimensions and you really find a new Fourier variable of a n as alpha n divided by x i n that these numbers alpha n alpha n tilde and x 0 r and b i plus 1, a couple of additional numbers that I've written down have to do with the x plus x minus 7 give you a full labeling of all classical solutions of course there are reality conditions alpha n will have to be a complex function not a complex function and real conditions so then we'll keep that in mind if you have any questions, comments please please, please, please stop me if it's not clear of anyone please the ideas are simple but they may be answered if anyone has questions like the algae we have the same question don't be embarrassed we've got to do that as you've seen before this does not exhaust all the basic solutions completely firstly, the constraint equations that we saw last class determines everything by the 0 mode of x plus once we fix this h because in terms of del plus of x plus in terms of del minus x plus we can tell you the 0 so that's one of the remaining so that's x plus which is equal to something that's completely determined and it will turn out to investigate this formula so a new parameter that's important because it can parametrize the new solutions that's an initial value in the state of solutions zero x minus is more than h x minus has square root of prime p minus tau which we sometimes like to think of as sigma plus plus sigma minus by 2 and then there's little x minus so there are 3 new parameters there are 3 additional numbers that parametrize the new solutions in the x plus x minus sum then they make x0 plus x0 minus a few it's where we get from solving the constraint equations in terms of what it says x0 plus is the 0 mode it is undetermined by the constraint equations the constraint equations only tells you what derivatives are but the truth is all of this is n not good alpha n tilde by minus i n it will power minus i n sigma minus now we want to determine the symplectic structure on the state of solutions so all we have to do is to plug in these classical solutions into the selected formula we are going to start doing this in detail and we will do super but even before we start doing it we may be seeing something very nice because delta x minus so what is delta delta has a variation as you move on the state of solutions minus has only one parameter for 2 parameters in which solutions change x plus is much richer because this function is a function of all these alphas so delta x plus is something very complicated but basically because delta x minus is very similar delta x minus comes only from this part and these parts have no Fourier modes as Fourier mode equals 0 and it usually tells you that the only part of and then using the fact that if you have got a constant function and you Fourier integration from 0 to pi a constant function multiplying another function the only part of the function that is picked up is 0 more than that you see the only part that enters the symplectic form the only part of x plus that enters the symplectic form is the 0 more than x plus right so what do we have suppose we plug this space of solutions into the symplectic form let's look at this part of the symplectic form this delta x minus is what? it's delta p okay let's write down the contribution from from this stuff contribution of this stuff so if we get delta of x dot plus let's write it down from this stuff delta of x dot minus where the delta of x plus is integrated now this integral includes an integral over sigma what is delta of x dot minus okay delta of x dot minus is simply out of time delta of p minus because delta acts on some other parameters in various solutions okay and therefore waiting for delta of x plus while delta x plus is in general something very complicated when we also do the integral the only part of the x plus that survives is the zero mode it's Fourier transforms the two pieces in the same plot okay there's a part that comes from here and there's a part that comes from the time dependence the p plus, alpha prime p plus now it can't be very simple it will have some delta of x zero plus the time dependence is very complicated it's very complicated because it's a very complicated function of all of this stuff as we will see in a moment imagine doing the following solve for x plus we have delta of x plus is equal to constant times delta x i, delta x i, that's the constraint of it now look at the zero Fourier mode of that information that if you pin plus on the left-hand side is equal to everything that has harmonic oscillator n with harmonic oscillator minus n so it's binary as an alpha and that's very complicated quantization everything else is going to be very simple the part from the i is a very good answer but all the complication quantization comes from the fact that p plus is a very complicated function of everything else so if we actually have a plug that's a very complicated form what p plus is a function of everything else it's a very complicated form and it would look like that it would be very complicated but we don't have to because, let me say this more clearly as we will see that soon the variables here in the xi-sector appear in canonical conjugate ways you know give us a nice structure of this case ok but so what we left in it is these three different are these three different numbers characterizing solutions the electric structure and odd-dimensional states you can't put a non-degenerates in electric structure non-dimensional states you know that's from Poisson brackets Poisson brackets always come in pairs if you have a third guy it would have to be something in which the Poisson bracket, you know is zero now what I'm saying is that you take any symmetric anti-symmetric matrix and it's all there it's going to be degenerate if you want to promote you want to use some simplistic kind of structure and promote that you mark many consistency conditions must be created one of which is that the simplest of which is that you must be working in the even dimensions so when you got an odd-dimensional space when you got an odd-dimensional space space you can't use this method of quantization unmodified in this situation this is the second thing is what Dirac used to call well that's an example of what Dirac used to call first class constraints the second method that we've already seen in these lectures is just and the second method is the following let us forget about the fact that P plus is determined let's forget about the fact that P plus is determined by these constraints and let's promote that to a random delay so we say it's square root of our prime P plus time and then do the quantization so now we've got a new solution to be large-dimensional how we cure the fact of being large-dimensional space the constraints that would have determined P plus and impose it as an operational equation fictitious and slightly enlarged transversal equations like we did one time do the quantization and impose the one non-trivial equation that would have determined P plus as a function of everything else as an operational equation on a little space there are other modes of next class but those are completely solved better to say they just don't enter the symplectic structure as we want it the space of solutions doesn't care about all those other modes the only way in which those other modes could affect the next phase is by contributing to the symplectic structure on this equation but we see that there's no more path that can go on, there's zero more methods to deal with with this extra P plus and then imposing it's roughly as fishy as the kind of things we did in quantum mechanics in the first place you increase the number of variables and take constraints and impose them as operating in the text of a good quantization is it a does it give you a consistent quantum theory as all the problems now we will do a more systematic quantization a path integral quantization which lesson begins we will do a more systematic quantization in the next few lectures with the A-look and formula theory but at the moment let's let's hear the answer about such five things is this here as I said last class is very important as a research physicist to be able to do something rough and you know not completely you know something that more necessarily satisfies mathematics but yet the answer fast because that way you see whether something interesting is happening if something interesting comes out there needs to be a lot of dynamics interesting in my point it's very important to do rough investigation before being very far now let's completely cut to one decision so what we're going to do is to take this class of solutions with this inflecting form this class of solutions with this inflecting form and do the quantization imposing that one or rather two things there's one in the x-less sector, one in the x-line sector remaining constraints so let's first do the quantization and then volume of constraints okay so let's first do the x-i sector the x-i sector has been planted like this now we have to plant that into the supply okay so what do we do okay let's first deal with the zero zero was of x-i simple what we want to do is to plug the zero expansion into one by two alpha okay one by two pi alpha but now you do the integral over sigma that gives you a factor of two pi so one by four pi alpha then you do the integral over sigma and that gives you a factor of two pi so that's one by two alpha times integral of there's a two here so that gives that and then we have integral of we have integral of delta of alpha times pi which delta of delta of x-i there's this here we've got a quadratic expression which only gives you something non-zero if a mod n meets mod minus n in sigma so the zero force can only click on themselves so I'm isolating out the contribution the zero is this here okay I've just taken this expansion plug it in here and see what I get in zero when I do an x dot I don't get anything from x zero okay and I'm doing it at tau equal to zero so I have tau equal to zero this thing is zero unless I take a time limit did I have a square root of I had a square root of the parenthesis expansion should have been without the square root sorry about that should be basically chosen to cancel this alpha prime there's enough square root okay so this is simply delta pi where delta is zero i this is technically x commutation relationship for quantum mechanics so the quantization of the zero force sector is very simple then here we've got a space square and technical functions of x zero i and momentum is operated this is the we get landed off the zero force the same problem as ordinary quantum x plus x minus x the x plus x minus x is all zero so if I wanted to move back to the coordinates x zero and x one then the quantization of the zero force is exact for each of these things is exactly like any of the exercise so I don't need to say anything new about the quantization of these components it's exactly the same business is that clear? so we quantize all the zero forces what we've landed off with is functions of 20 of d vectors with differential operators we implement it what we're trying to do is rather the subtracted structure on the zero moments then we have to do the subtracted structure on all the lines we're doing that x zero and p enter the subtracted structure only in this one the quantization of that sector is clear it's decoupled the zero moments don't talk to the other things so we've got one here first place that is just functions of d vectors square integral functions of d vectors is this clear? power x so let's move to that because we've got x zeroes and p's for all the variables we haven't imposed a constraint rule that will lock things down but before imposing the constraint it's all you have to do to a second method of quantizing the particle remember we had functions of all d variables and then there was a constraint that gave you a martian edge this has been exactly this one I did in the quantization of zero moments is that it's the same now let's return to the normal let's say simply let's write it in the same procedure so let's plug into this object and see what we get now we've got a 2 and a 4 pi so that becomes a 2 pi then let me do any technical 2 pi when I'm going to take okay so we'll have 1 by alpha prime in our expansion there's an explicit factor of square root alpha prime behind each of the oscillators everything is quadratic so the alpha prime will cancel we've also got a factor of 2 further than x because as you may remember we've absorbed these two opposite so half times delta of x i dot where it should be delta of x where we've done the integral over 6 this is a bad condition let me start it we only get contributions when moving what is delta alpha it's just plugging in here so when we plug in here what we get we get delta when we get dot this factor controls exactly the cancel what we get by differentiating so the dot part is delta alpha n into the power minus i n sigma plus delta alpha delta alpha n into delta into the power minus i n sigma minus wage and then there's a factor of delta x not delta x, no, mister okay wage let me be more precise let me do it add 1 by 4 pi and I'm not getting the integral so this is sigma of e to the power of n sigma wage delta alpha n divided by minus i n is by minus i n sigma plus delta alpha n into the power minus i n is by minus i n sigma minus i n sigma okay, we're doing this summation over n here summation over n here summation over n here summation over n here n and that's it now we're going to see about sigma present sigma minus basically sigma that is a minus sign yes, so sigma plus and sigma minus are sigma and minus okay, so let's make a reflection okay, sigma plus is sigma and sigma minus is minus sigma now that we take it with that derivative okay, sigma plus this is plus and this is minus okay, so this is sigma plus is sigma i n sigma and minus plus i okay, sorry for the unsystematic nature of this statement if anything is unclear, please ask the summation is both over n and n but the summation because of the integral we get a kick only when the formula both add up to zero both alpha in terms of alpha and alpha tilde and alpha tilde I think that the terms between alpha and alpha tilde add up to zero okay, so it only takes when n is equal to n delta alpha n which delta alpha tilde n plus delta alpha tilde alpha tilde n which delta alpha n both divided by both divided by minus i quantization is done on a constant time it doesn't matter which time we're just choosing it you can check that the symplectic form is invariant under time it doesn't matter because if the AND is an entry, this is just it you might think, well is the other set of terms also going to give us zero because of AND is an entry but it's not we get non zero clicks between alpha and n and alpha minus n n plus n has to be equal we get 1 by 4 pi then delta alpha n which delta alpha minus n now the 4 pi and 1 by 2 because of the 2 pi that we get so we'll take it over so 1 by 2 delta alpha n and divide it by minus i of n but n was minus n so plus i this is summed over n n and we get the same thing with the other n n is summed over all non zero values but when you take minus the contribution from let's say look at the contribution from n equals 20 n equals minus 20 these 2 contributions add instead of subtracting because the denominator interchanges sign and you get a sign change then you flip the positions again the key point was clicking between n and minus n instead of n and therefore this thing can be written as n is equal to 1 to infinity of delta alpha n which delta alpha minus n divided by i n plus delta alpha till the n which delta alpha minus n divided by i n ok sum of all the sectors now also totally trivial you see we got a bunch of terms each of which is called rapidly decoupled decoupled from everything else alpha n is also alpha minus n I should also have been putting in i exactly I'll do it now any one scalar all the x i's I should have been putting that I just put it in here what is the definition of the symplectic form well we know that if we had delta something which delta something that is the commutator the second to do the first has commutator k that's where omega n is with lower indices so omega with lower indices has an additional factor of i n in the denominator therefore omega with upper indices has an additional factor of i n in the numerator that this with this has a commutation relation of x and p times i therefore we conclude that alpha minus n i commutator alpha n i put the i or alpha n i commutator alpha minus n at reverse position state in a minus side is equal to in fact that I made this more general of course there is only commutator between alpha and the statement write this more like write this more generally as delta i mean delta i j that completes our quantization we've got we've got a Hilbert space with these alphas that have to stop this is commutator density this is basically harmonic oscillator creation operator creation operator the remaining part of our Hilbert space is a bunch of harmonic oscillators with frequency that goes from 1, 2, 3, 4 up to infinity 1, 2, 3, 4 up to infinity and there are d minus 2 copies of these actually d minus 2 times 2 copies of these because although I've mentioned that everything that we do for the alphas also goes to the alphas oscillators times 2 sequences of harmonic oscillators with frequencies 1 to infinity so these Hilbert spaces they result in each other they result in the Hilbert space of functions square and technical functions that is the whole Hilbert space that we get from harmonic oscillators it's a scare now I want to go on with the analysis of constraint but I don't want to do that until I get some feedback from you I have this to clear up let me do the analysis of the constraint then we come back so I use slightly fancy techniques to do the quantization here the techniques for fancy and the reason why I use these slightly fancy techniques is because it's much faster than standard case and say that for this way of proceeding it's much clearer but however what I have done in context is exactly the same as the copy start from the action we wrote down right at the end of last class there is an action for 0 moles all coordinates and oscillators for d-2 coordinates and do the quantization of the system anyway you want quantization of 1 plus 1 direction how are you going to get this answer so anyone on the room is in both cases it's not a constraint last time I was so secretly enlarged the space a little bit without seeing what was going on this is much clearer because we see what was going on but anyway so what I want to say is that the quantization we've done so far is exactly the same as what we've done if you start with the action we wrote down right at the end of last class 0 moles for all coordinates, oscillators for d-2 and do the quantization anyway parameterize configuration space in some set of variables turn these into harmonic oscillators and you'll find that it's exactly the same but it means I'm comfortable with the way we without all these data on this and the symplectic forms and I don't know all this stuff just take that action we started with the end of last class this is a very useful exercise use your favorite method of quantization and find the answer you'll get this answer please do that I'll give you that one too it's not necessary to be jazzy except for the logic where it's much clearer ok, now this is the first one now about the the thing we want to do now is to so the constraint was yeah so the constraint was the equation then plus x mu then plus x mu and the similar one with minus so that's the even of the plus what we get from this zero-mode part you see the zero-mode part of the constraint you don't remember what x mu was x mu would equal to x zero mu that carrier thing is the constraint equation plus alpha prime p mu into sigma plus plus sigma minus right here ok, plus also plus so like this the oscillator part is only accurate for the i's where it's alpha prime by 2 alpha prime by 2 and then alpha n minus i n e to the power minus i ok, where this part is correct only if you put it where it belongs you understand what I mean please the first two terms were x zero x let's evaluate this constraint let's first see what the zero-mode is so what we get from the zero-mode so from the zero-mode we get the only contribution to the zero-mode is the plus hits sigma plus so we just get alpha prime p mu p mu by tau was sigma plus sigma minus by 2 we get two factors of that interested in that part of this constraint equation is a function equation for sigma but as we've seen before if the part of the constraint is zero-mode or in sigma so we are only interested in the zero-frequency Fourier component we are interested in the integral of this constraint over sigma because everything else relates oscillators are sigma plus to the coordinates of interest and oscillators are sigma plus and just in all that quantization it's true but we don't care the only term that's constraint equation that's of interest to us that's not relating something outside what we are interested in is the zero-mode of that constraint the integral of that constraint over sigma and that's what I'm evaluating if you take the part that is no Fourier more Fourier frequency what do we get from this other stuff okay from the other stuff I'm getting thousands of alpha prime which is okay let's oh it's alpha prime squared 2 half is alpha prime now expansion of alpha prime is not squared now from the zero-mode what do we get we get alpha prime by 2 we've got something squared and then we get sigma you see now when we take del plus of this these i n's vanish so we just get sigma n i alpha minus sum alpha is n equal to alpha n is not equal to 0 zero Fourier frequency part of that constraint x plus x yes all that is letting this mu run over all the matrices because the zero-mode is this so let's just say this is summed only over i i equals 1 to 1 to mu minus 2 but this is summed already the x minus yes doesn't have any any this yes but x plus and oscillator both which get in which group give a contribution well x plus oscillator both are determined by the constraint equations it comes back to this yes so they're not they're not labeled as phase space the space of solutions is completely determined by what the oscillator mode is it's not a complete characterization of the space of solutions now the only way in which the fact that x plus does oscillate that is the only way in which that effect of monetization is in contributing to some technique form in terms of the idea a part of the zero-mode is determined what x plus is doing is determined by what the oscillator is doing so there are no new degrees of freedom in x plus it's just a question of whether what acts of mapping around affects the symplectic form for the labels of our solutions and we pass through this argument to demonstrate it doesn't well it does in a certain way from the speed class business but we take back and forth by imposing the constraint it's the only way good, please keep passing this it's very important it's not dear to me but why x plus it's very important it's very important it's our function it will be from alpha plus to alpha minus alpha where alpha plus is given by finding a function of alpha and i is okay so now how do we but that function we enter no we enter okay not a zero-mode of the constraint an integration over sigma so anything with non-zero frequency drops right it would be non-zero frequency because sigma x minus is only 0 okay that's because if one of them is x plus another is x2 understand what this thing is see we've got the summation over all n okay but the summation over all n if you go to n positive if you go to n positive and n negative now when n negative is to the right we've got the creation operators to the right remember we saw that alpha n with n positive was like destruction okay when n positive is to the right we've got the destruction operators to the right now it's always a good idea to write things non-zero so let's do that to zero we put alpha prime square p mu p mu by 4 plus