Lecture

0:00 The ordinary quantum mechanics doesn't actually matter, and you can show that you've sort of figured that out. And these models are sufficiently complicated that isn't quite a lot of proof. And it turns out that the precise technical requirements of quantum theory, like true start and joints and so forth, very strongly restrict what you can actually do. This is something we really try to get to the bottom of. So this is one of the motivations for it. We could have also done that with any of the existing models. Now, the basic problem is that this, as I've said already, is a system of fields, and it has infinite numbers of degrees of freedom, and because of this, you can have any other problems. And the basic idea of these quantum problems is to introduce us to a system with finite numbers of degrees of freedom, and then quantize that. Now, I'll also explain how this happens, but I'll just try to show a diversion of all the quantum mechanics. And what quantum physics can be used in terms of quantum mechanics, rather than in terms of something that is not quite straight at the time. The whole thing of quantum mechanics, everybody knows, is the finite number of degrees of freedom. I emphasise the finite number of degrees of freedom. You're supposed to find, at least in one of those, you're supposed to find the parameters of certain parameters. That is, the finite number of degrees of freedom. And along with this, you have the sum of all the degrees of freedom. This is the mathematical side. From the physical side, the interpretation of this, of course, is that if you have some real set P of the Rn number, which takes one to the square, dx1 up to dxn, you call this Pd, which is the probability of finding a configuration of the system.

2:30 While the department is dealing with quantum mechanics, and this is going to be part of it, or this is something we might, this is something that Andrew is including, this interpretation, I emphasize, is what we will come back to at the end of the day. Thank you. And of course we know the standard representation of this, the training of representation, which I'll move on to right now, where xj goes to xj, and pj is the partial derivative, and of course you know also there's the sort of alignment, which shows you much less of a category, but it's essentially unique, that's what that kind of works. That's all there is to quantum mechanics. But the question is, what would happen if you were tempted to do quantum physics in the same way? In fact, we may do, but I think it's a great pity, because this is an aspect of quantum physics that I'm going to discuss a little bit more. Supposing you wanted to develop quantum physics, in particular, in a system like this, in a sort of scripted kind of form, kind of like a C. Well, you're supposed to, in a computation way... Of this form, these are now field operators, of course, one of those that are obvious in mathematics, and that's the only reason you're an operator, because that's what you're used to being asked to do. Now, at this stage, I'm going to be talking to you about computer-based textbooks, and what this is usually going to do is to instantly proceed into the proper space, and you can start with one of the first things I'm going to do for you at the start is building a business on it. Now, the score, obviously, is going to be mildly, not likely, mildly low, but it's going to be a comparatively low bid, so it's actually a good idea to do that. So that's what I'm talking about, I'm looking at it. The problem of finding representations of all these, which are kind of racist for the stomach. It must say, well, help us expand this view to an arbitrary set of modes. By the way, we choose some basis functions on our tree. And we expand something like this. Now, of course, the exact nature of Schroeder-Stein's expansion of the mathematical sense in which it converged, of course, depends on the fact that what we did at home was this gradient, and we typically used to use it in some sense, but it was a reconvergence in a sense of commutative space. It doesn't really matter, I think, at all. But he would do a similar thing, in a very general case, to use the expansion of the basis. This is really simply saying that our field is the system of infinite degrees of freedom and one can regard these as the most.

5:00 Now if this is a computer, what would that mean to sort of like a human language? And now if you quantize by saying these are the operators, what you get is about the same as you have over there. Except that i and j now go to one infinity. So in this point of view, you see quite clearly what you mean by a set of quantum fields here, it's the problem of infinite sets of uncertain relations between them. It's the idea that you can't clone a commutation of them by doing i and j go to one infinity. Now, of course, at this stage, you could say, let me find the representation, please. Now, the obvious analogue, thinking about what we did up here, is to take... Well, accusing people of what on one stage shows a removed consolation when I accuse them of being more than possible, it's got to be a lot of problems in terms of theory, they're not observables in quite the same sense that they are, they're more than possible. These are simply the normal moments of freedom. You see, in this picture, you're really thinking of the field itself as being observable. And beyond all that, quantum physics. It's not normal for all space. It's all about energy and quantum physics. And this is quite an appropriate language to use if you want to talk about quantum mathematics, because the notion of particles is intrinsically linked to quantum mechanics, and one doesn't want a fair right to include it. So we don't want to use it as a particle, actually. So I really want to talk about the fields here, the measure of 10. But the field, of course, is an interdimensional system, so it's only natural to expand it out in terms of what basis sets. But beyond that, that's all I want to say about it, really, about what the interpretation of this is. But let me start looking at you about representation of quantum mathematics. The obvious representation would be, as you might think, to take wave functions and infinite sets of variables, which you might call wave functions of lower infinity, and we can interpret this in a different way in the appropriate sense, the q's would act as unified multiplication, and the p's would act as...

7:30 Now, that always can be done from this point of view. Quantum field theory can be regarded sort of as a problem to solve the Schrodinger equation, but now it's a Schrodinger equation, it isn't in many variables. So, in this point of view, the problem of quantum field theory to some extent is the problem of solving the partial differential equations into variables. This gives you some idea of what's right and what's wrong, because there's a lot of trivial stuff in the study. Now, there are, I might say, vast mathematical problems here. For Scott, it's no longer true that there's a unique representation of this in the computational relations. So you don't actually know whether this is the right thing to be doing. In fact, I can show you why it's not, because I'm going to adjust it to the slide. In a certain sense, the representation which you choose has to be geared to the countermeasure we're using, something which is totally different from all the quantum mechanics. In other words, in quantum physics, the kinematics and the dynamics get mixed up together, and you can't just put the representations of the countermeasure into a sort of routine. But apart from that, I'm just, I was really able to... Now, from this point of view, let's see at once where the quantum model idea comes in. One said, this is a very complicated system of equations that can't be solved in general, so the only thing to do is to make some sort of fluctuation equation. Now, if I wasn't terming the coupling classes on here, I would end up, actually, with the usual sort of term you feel to be, which is commonly my particle theory, which is known as Feynman diagrams. They wouldn't be Lorentz covariates any more, because I've used a non-equivalent composition, which is picked up into the choice of time, but in principle, in essence, they're the same thing, it's either or neither. Now, again, one of the strong reasons to believe that part of the ancient theory of coupling classes is the last thing you can do with general relativity. Not only that, of course, but a lot of people do contemplate it, certainly in the company. I mean, I've done that myself on many occasions, but nevertheless, there's a good reason to believe that it may not be the best thing to do to yourself. So you want to seek alternatives on the preservation theory. And one which sort of leaps to mind, if you look at this, is that rather than determine whether capitalism is a toke in the number of degrees that people attract to the context of it, In other words, we truncate the system. We truncate it up to, say, 1 degree 3, or 2, or 3, or 4. Each time we put together a more or more complicated one, which is a system of differential equations, but we might be able to solve them. Then we're supposed to discuss the limit. Now, needless to say, taking the limit would be far more trivial, but that wasn't even got to that stage yet. What problem... Has there ever been trying, let's say, class-based quantum mechanics? No, there hasn't been. We're talking about quantum mechanics.

10:00 I'm trying to get someone to do that, one of my students, and they're absolutely confused. It's more complicated than it looks. It's a funny thing. For some obscure reasons, I don't quite understand myself, it's easier to do quantized general relativity like this than it is to do quantized lambda phi. More than constructive, it feels, that it does do this, actually. In fact, it's a technique that's used a great deal to prove the existence of solutions of things. But it's an actual practical technique for doing computations, I don't see. But now, is it the real-time program to do that? Yeah, sure, as I said. Yeah, precisely. The modern, I mean, Jim Jeffery Mark Fogg, I think, was the owner of it. It's precise to do this. I mean, you can do that exactly. But not like this, which is the most sophisticated way. But, of course, you don't do that as an actual calculation technique. But I was really inviting that man inside. Thank you for your time, and have a good rest of your day. I mean, in the Schrodinger case, for example, there's several ways of finding the degrees of systems. There's several ways of finding the degrees of nature and all that. Well, I was going to come to that. I don't want to get stuck on this because, I mean, the years trying to do this somehow seem to be more prominent than they used to be. That's what I'm actually working on at the moment. But the model which I've been talking about, all fine, I can't be pleased with, because the problem is in the right. But it's kind of like the question Ryan was asking me, what the expectation of this is, is that you'd expect, by analog, that sort of... The following is true. The modulus of psi squared, dq onto dq, infinity, and I'll come back to that in a moment, that can be based on something like that. The P of B is a component of the number of measures of the field. You will find the various components of the normal modes lying in that specific cell. But this in itself tells the problem that if you make a measurement of a field, certain fields in the field measure. Now that's precisely the significance of these kind of measures. That's what this measure actually is. It's certainly not bad. You can prove quite easily that it's not a very good measure if it's in principal spaces. But there isn't a lot of adjustment measure. And in fact, you can construct quite accurately, quite clearly, a representation of non-computational measures on a space like our own, in which you do have such a measure.

12:30 But as I say, since I'm only talking about a finite number of degrees of freedom, that's not a problem which can be solved. Okay, so the idea then is to top off all but a very small number of degrees of freedom, just quantise those. And if you do this, both to get some idea into the peculiar technical nature of the quantum gravity system. And also, of course, to try and avoid any manipulation of the coupling constant. So you can use these models to talk about gravitational traps and so on, with some hope you might be doing something which is sensible, in which you could determine the coupling constant in order to study the problem. Now the actual technique which you can assume depends to some extent on the sort of version of general physics you are actually using. Since one of the aims of the exercise was... To try and see, as I said already, exactly what mathematical quantum mechanics were in mind when applying quantum mechanics in mathematics, we were very anxious to keep a very genuine canonical optimization scheme, because that's the one which is, in a sense, best understood in the interpretation of quantum mechanics. And so the idea, so natural precision quantum mechanics, if you start off, if you haven't said all this, you sort of forget this and go back to the beginning, then you write down Einstein's equations. With some methods of physics, we adjust a few degrees of freedom, so you choose some particular time frame, which has a few degrees of freedom in it, and similarly, you choose a few degrees of freedom for the rest of it. You then try and classically solve those equations, and this is what I would call the curve. You must solve it classically first to show that it actually is consistent. This is obviously important. I mean, in general, you can't take a typical type of metric, and only certain types of mathematics will excite that metric. For example, if your metric has some sort of continuity, then you usually expect that metric to have some sort of continuity, and you have to check the development systems. So you do that. You're then taking the system equations, treating absolutely, as far as you can, in the best possible terms. In other words, you're trying to reduce them to a genuinely complex set, and then you're trying to construct representations of the appropriate CTL. I'm going to talk about two models. The first one is more or less the simplest that you can imagine doing, which is giving you the first talk in the Robertson-Waterman method of the scalar field.

15:00 Now the motivation for doing this... Originally, it was that there was some work a couple of years ago by a partial tool on precisely the problem of quantizing the state of everything that was involved in the metric, which is not quite the same thing as quantizing everything that was involved in the metric, but what they did was the following, they determined Einstein's equations in the following sense, and this is a fairly obvious interpretation to make, that they said that we'll quantize the matter, but the metric itself, we are purely classical. And so in order to get the right-hand side-by-side equations, which doesn't get very classical, check the expectation value of the energy measurement tensor in some suitably chosen state, and then you solve the resulting combo system equations. And they did actually manage to show, by choosing the Robinson-Walker method, which is the form I'll remind you of. There's some three-space here, which is a fixed metric, or it could be a three-spare, but you'll have one three-space, but if that space is in the middle, it doesn't matter, I mean, it's some control. This is sort of the radius of the four-pronged components, and this is something that plays a big part in the choice of time and order. Now, they assume that the metric was of this form, and they put in a decline more than that. And they quantized this, but didn't quantize that. They managed to show that by choosing the suitable state, the system of equations could be made consistent. That's not true of the game, but this has a tiny bit of genetic, so obviously we're going to have to change our state of mind. They managed to make something of the show, so they did actually show that the system didn't collapse, but instead went down to a length, which was effectively the common wavelength of this particular mass here, which would be about 10 to the minus 13 centimeters or more. This was, of course, just one example. So we were thinking it would be nice to do the same sort of thing, but rather than putting these brackets around, which is a rather tricky thing, and we want to talk a little bit about this MLS term, free self-consistent theory, we're going to remove the brackets, which of course instantly makes it more complicated to do, quantize the metric, but then simplify things for ourselves again by reducing the number of degrees of things that we actually quantize. So in our approach it sometimes is sort of complementary to this, it's taking the same type of problem, but slightly different type of view. So this is one reason why we chose this particular one.

17:30 Right, so now we have to look at some more technical detail of what this means. So we are going to assume then that the metric we choose has just this very simple form. So our operators will be just this and this. This is fixed. There's only two operators here as well, fixed and divisive. And the question is, what does this tell us about the climate? Because we're going to quantify it as well, and obviously, in general, we're not going to look at how all possible moments are tied to it, because we've just stepped out certain moments. Now, if you look at the Einstein equations, there are lots of order metrics, and you'll find, purely classically, I don't mean to just talk classically, but as I say, you have to check the classically, the G0I equation... This is the user ID 0, and this in fact leads to one statement that the DI argument is not. In other words, that phi is simply a function of time, as you might expect. In other words, a field, a classical field, is not exactly like a field of ordinary sense. It is simply a sort of an average distribution of matter, which is what you would expect from an ordinary network. It's just that because you present it as a quantum field, you can quantumize it in a very well controlled way, as well as choose a term to do this, rather than yourself choose clouds as dust. These are what's been done previously. And the previous work on Robinson-Walker plus matter quantizing, in terms of using dust and things like that, the notion of quantizing the dust, it seems to be quite wrong. I mean, all the help that we need to do is to use a matter to be quantized. And this is what we do. So you end up then by, well... You get this at once, and then the question is whether the resulting system would be a very visual system, and the answer is yes it is, and if I show you this I'll actually write down the solution so you have some idea of what this classical system looks like. Supposing for example that we make a choice of time, remember that in general relativity we've got these three rows, n of t, r of t, and pi of t. But because of the general coordinates of events, they were not always going to reach a dynamic degree of degree, and in particular this time they had ribs, and the fact of fixing n in some explicit way corresponds basically to choosing twice the time, and it might be true about how you learn. So let's just choose to simply do n plus r, which is perfectly admissible for twice the time.

20:00 And also, to put a measure, I'd certainly choose that the mass of the building can grow basically because of that linear element itself. I'll say it more like this today. It doesn't automatically take much of a moment. And if we do that, then we can... And if it goes wrong, then the point of balance is going to be... Yes. Excuse me. Yeah, that's right. In fact, we can do the positive and the maximally positive, right? It's totally completely different from how we indicate that in my country. So we just consider the masses... The classical equations of the mass not equal to zero resisted all the attempts to solve them, where the plane makes some problems, they're just non-linear, forming differential equations, and then they're just put back into a solution, so the plane, which shows that it's a distance, can actually run the equation. Well, the solution is simple. As you might expect, it's very similar to the Newton sort of thing. You can write it down. The square is the... Now, this is the solution for the positive curvature case. As you see, daily tilt's gravitation slaps. I say basically that you've got a sine-type solution for the radius, so it goes up from zero and increases in time and decreases, similarly to the five-tilt, and at the same time varies between minus 15 and plus 15. So the central point here is, and soon be a point to classify, is when r-tilt is zero, or also when the five-tilt is infinity, so the five-tilt becomes infinity, which is what we're going to talk about. Now, the fact that these solutions exist, and also that they have the right number of arbitrary constants in them, shows that the system of equations is consistent with the successful system of problems. So having said that, we now go back again to the starting point and write the thing we call a report. It's an essential part of the problem of problems. Isn't it the case that in a performance thesis they had trouble at knowing what are the variables? For example, if you chose the wrong variables, you even got the particle creation and the flat space, etc. Are you going to go into this? Tell us how you know why your variables are correct, etc. Well, the choice of canonical variables. Yes, I'll discuss that a little later on. Of course, no one knows the exact answer to that.

22:30 The question of particle creation, how does it arise from that? No, no, I understand that, but this is in plot. In other words, my question was, since you did run into this problem, how do you know you don't? Oh, no, everybody runs into that problem. One of the reasons to study these models anyway is always to be able to try and understand what happens And we have no good answers at the moment. At the moment we're just choosing symbols. Now the first thing is, what does this system look like in canonical form? That's the first thing. Now, I'll just tell you this. The problem is that it's called the living room. I'll tell you later. The word writing is down, but it's possibly true. The following is true. If I look at this diagram here. I've gone back to doing maths in time. And I'll leave the maths in writing until the last minute. Now this diagram here. We'll look at that. The statement is that if you vary this for Grangian, assuming that pi r, pi phi, pi and n are all independent variables, then you will recover completely the Einstein-Grangian scale of the system you were considering, in what is obviously first-order form. It says this is clearly contrary to that, and that is to that. At least it would be, should we say, if this were an absolute non-specific. But that's just a statement of fact, you have to check that. Varying these things leads to the G.I.J. equation. Varying this leads to the primordial equation, and varying n leads to the G0-0 equation. The G0-I equation is trivial in these theories, it just reads the primordial equation. Okay, so this is where the idea of different parts of the earth-side diagram comes from, and it's very important to have been written in this formula to come up with it. Well, they're just variables. The statement is that if you just take it to the fraction, it's just a variable. It's been described in I-5 and I-R. The basic variables here are just r and pi, right?

25:00 The resulting equations you get are first-order differential equations, but which are in fact exactly the same as the second-order differential equations, when the actual, strictly system-like assuming is capable of classifying the forces of the metric. But you have to take that one on face. These come from nearly every composition in the study, which you can work out as useful. But if you look at it as just the ordinary classical system, the finite number, what configuration stays the same? Well, the configuration space. This is really using that from the phase space point. If this were a canonical form, and in fact it's not, the reason it's not is that the very name gives you a constraint effect, as you can see at once. If it were a canonical form, then the face stretch would be four-dimensional, it would be this and this and this and this. In fact it's not, because you see you have these constraints, and there's some constraint on the face. That's why I'm going to count it straight away. So this is just the way it looks as if it ought to be a canonical form, and in fact it isn't. If this were a canonical form, this would clearly be the Hamiltonian. It's obviously the form pq dot minus h. However, this dimension is identically invariant as one of the equations of motion, so we end up with the well-known problem of the trigonometry, which is that some of the equations of motion are equally-cruely constrained in their particularity. I'm not telling you to cut back all this and incur the edge. If you incur the edge, it's equal to naught. Now, classically, that is true, and that is the G-zero-zero equation, but the question is how do you quantify it? It's at this point, really, that various people's approaches can be more or less practical. If you were John Wheeler, what you say is that we will pretend that these things are economic variables, we'll pretend they are, we'll give them a kind of a computational version, so this may now become an optimism, we will quantise in the sense that the allowed state vectors are simply those which satisfy the specific criteria that it does not. Having done that, you may want to try and get some sort of interpretation of it. And this is a complicated thing, and typically we'll notice that because of the accounts of pi r squared and pi phi squared, what you get is a second-order equation over here. So you get some sort of second-order time-organ time equation, which you then have to interpret. Now, a lot of work's been done on that approach, both in the fourth year and also in the second year. So the reason why we didn't want to use it is because it's not quantum chance, whether it is or isn't. It is not true quantum quantization, and once you've done this, you don't really know what to say anymore.

27:30 You can't use any of that, because you've completed a few days of working on it. So what we prefer to do is, in fact, to proceed along the strict conundrum itself, which is to actually solve this constraint equation explicitly for some of the variables, substitute them back into the Lagrangian, and get a true conundrum for the very most complex of variables. Therefore, and I think the easiest way to actually show you how this works is to write down some of the conundrums and then we'll see what actually happens. Then we'll see the nature of the quantum theory problem as well. First of all, this theory is... Still, general problems can vary, particularly since doing it under the choice of two dimensions of time might appear to be a little bit discerning. And that's all tied up with the fact that this is an Australian equation. So in order to get to the truth behind the problem, you have to make an explicit choice of time before you contact it. Now, the first possibility is to check the tables out. That's the main actual choice of time. One knows what happens to that. It turns out, when you make a choice at a time, you have to solve the constraint equation for that variable which is conjugate for this particular choice. In other words, if you choose t equals r, and you solve the constraint equation for the conjugate variable r, which is higher, explicitly solve it, put it back into the algorithm, and if you do, you will get a system which is actually economic at all. And that's it. So, if we solve this... And we end up, which you can now see is a canonical form, now really, fairly hairy, in the form pi pi pi dot plus or minus sine, which needs to be known, the k here is the curvature, it's plus one, it's a sphere, it's naught, it's a flat space, and minus one, it's a sphere. That's what you're actually going to get, and that Lagrangian certainly now is in constant form, this is just the form p q dot, there's only one variable left now, minus one, which is going to be the Hamiltonian. So this, now, this whole lot, once you've decided on the sine, of course, plus or minus, really is going to be counterintuitive.

30:00 And there are two things to observe about this which are really interesting in these models, which do require some intuitive care. The first one is that H is the form of the square root of something. It's not like the Schrodinger equation, normally, sort of half the only square root of something. It's the square root of the model complicated expression. And secondly, that H is time-dependent. Now, strictly speaking, we should be able to have an angle time-dependent from autonomy, it's just that unfortunately they've had to crop up very much in these standard technical terms in quite this form, and it's impossible to go astray, and we've got to be careful. And both of these problems have to be handled, because at one time there was a spelling attack, taking care of the fact that this is a square root, and you have this time-dependent, and then you get an answer. Now the question, first of all, at least would be problem-dependent actually, the first question is how to handle the square root? Now what tends to be done, or what has been done, and sometimes we can't do it properly, and that's the question we're trying to solve. The great temptation of what we observe as h as a square root is to multiply on both sides by h, so that h squared times sine is minus the integral of sine of identity squared. Now, if that worked through, I think this would make my pretty much simple, because you can have a really good script, and this thing looks now just like a sort of an ordinary type of quantum mechanics work, more or less, E squared, the red part, the time determinants, and you've got a second group on the right-hand side. And that's usually what's being done. Unfortunately, it's quite wrong. And the reason it's quite wrong is that this is only true if the Hamiltonian system is independent at the time. That's pretty obvious when you think about it. If you really emulate, you've got an H dot. And an H dot depends on the sphere root, and that's the continuous one there. So the fact that you can't do this is quite wrong. You could, of course, if you want, maximize this as what is correct. And that's a charitable way of explaining what quite people believe. But it clearly is wrong under all circumstances. So we don't do that. What we choose to do is we're actually defining the square root using the spectral theorem. This is the first instance where the genuine requirements of the operator theory come in. Now, spectral theorems will tell you that, provided this is a genuine positive, the self-adjective operator theory actually reflects this theorem, and you can take the square root of this theory. So, that's what we do. But that does mean that we have to prove that this is genuinely positive and genuinely self-adjective. You can't just do the solution and then do the left and then do the right part of it. So we're going to solve this thing by using the spectral field. Okay, well that's the first thing. Now, the second point, the time dependence,

32:30 the solution to this problem of time dependence isn't quite well known, you can just leave it out. You know, so the analogically, normally one says that psi of x E to the i h times t minus t zero times psi of x and zero. That's the integrated form of the Schrodinger equation. Now, that is not true of the nature sign. What is correct is the time of the part of the minus sign there. E to the minus i over h bar, the interval between t and t, the nature of s to the s times psi of x. And that is actually always correct. If we integrate the Hamiltonian between T1 and T2, and also take the time order quite well, then the whole thing doesn't access the Hamiltonian equation because it's the correct integrated form of the Hamiltonian equation. So that's what you actually have to use because it's possibly more complicated to actually use Hamiltonian to solve the Hamiltonian equation. Now, there are various ways in which the time... Well, basically because the Hamiltonian equation... Age events in general may not commute with ages at the time, and if you can just work it out, just work out what differential equations satisfies an archetype this time around, you're almost likely to be able to prove that this satisfies the Schrodinger equation, especially using the question of the Hamiltonian commutator. Now, if the Hamiltonians have commuted with each other at different times, if h of t, h of t, not, then h of t is not a commuter. Then, actually, we've got two nice things to cover. First of all, we've got the time-ordering of the symbol, because it's down on the matrix. But secondly, and perhaps even more useful, is that the fact that these things commute means there must exist some set of simultaneous eigenstates at all, including general theory. This is really what I've been using. Otherwise, it means that there must exist some... this equation actually makes sense. In other words, there's some sort of eigenfunctions which is fixed, which has the property that in fact some of your ancient states, you always get an eigenvalue, which is equal to, depending on the time. And this comes about here because they commute from each other.

35:00 Oh, no, of course. Oh, no, no. I mean, no, again, you have to be absolutely, I mean, you have to be real, properly defined, separate from the operators. So, I mean, these may not, of course, be, these may not generalize much. Of course, but they, in general, this would be, in fact, in our case, it's always a one-second. Nevertheless, I mean, if you understand this in a general sense, then this is true. Okay, now, if that happens, then things simplify considerably, because then, if you ask, how does one of these things evolve in time? You can first of all drop the t, and say you know what this thing does at once, and you end up with a state of psi e of x and t, which is minus pi over h bar, and this gives you 2 more than 2, and p of s, g of s, is pi of x and t naught. And that's very much like the usual time evolution of an energy eigenstate, except you've got the usual, but at least it should be that soluble. So if you happen to have this in science, you can now proceed as normal. If you haven't, there's too little of that on computers. Okay, so those are the two comments I'm making about the form of time. Now, let's come back to how I've actually done this. If you choose t equals r, which I've already... I haven't told you you get to the square root of this. Now, first of all, you've got to make sure that your canonical variables are self-adjacent to each other. That's the same as 5. Well, you can always find a representation of phi and pi-phi, which are sub-quadrant operators. It's just that you can trade it for them. The reason it makes sense to choose that is that classically, phi and pi-phi do range between plus and minus infinity. And this is the problem which is going to occur later on, is that if you have a choice of canonical variables, and you want to quantize them by associating them with sub-quadrant operators, it's very helpful. In this case, we're okay. So technically speaking, then, we can choose the subject of the course. The usual quadratic representation will work, and everything's okay. And now we have got the term termination.

37:30 Now, in order to see the spectral film, it has to be a positive self-adjuvant operator. Now, you can see, well, I can't see instantly, but it should be fairly clear that this negative term in general is going to match right. If k is positive, this is going to give you a negative contribution to the spectrum. In fact, if we work out what the spectrum is going to be for our operator, it's only positive if k is less than zero. Otherwise it's negative. So this is the first example, then a real concrete example, of the quantum mechanics restriction, in some sense, of what we do. This says that for negative curvature spaces, you cannot chronically quantize the system using t equals r as the time. It just doesn't work. You get a non-positive continuum. But k negative, you can. k is maybe less than or equal to the order of k, but it's just an equation. So the first thing one learns in it is that, I'm sorry, k is greater than zero is the sum from s3, and then k is equal to zero is r3, and k less than zero is the sum from Witten space. And this operator is only positive if k is negative or zero. In other words, it doesn't work if you take a free spin. So the roles can walk out. Compact, close-based situations like the free sphere. You can't chronically quantise using the choice variables, it just doesn't work. However, if you choose the other case, it does. Now, I want to give you some idea of what the, of how it then goes. Again, this way, just to be sure that you can... You choose the minus r, not quite, you might have some problems here then. There's an obvious choice between these two types, what's perfect and what's not perfect. What would you now swear? Well, of course, there's no limit to the question. Yes, every choice of type makes a difference. Z equals R squared. What's really important is not to... We made a choice at the time, but what are the choice at the time, what are the variables you have to choose for it?

40:00 I said we made a choice at the time, we had to eliminate, fond it at the time of termination, it can't be compared with that. And then what's left is what turns out to be the actual canonical version. And that's really what matters, the choice of canonical versions. I don't know what particular aspect of it is. By and large, if you choose a term that's after the end, it doesn't do much for you. It seems to be a good three-and-a-half square from time to time. I'll come to as many of them as I can, because I'm very old. Okay, well, suppose you choose n as a knot, and you choose k as an integral. The virtue of choosing n as a knot is that you now do know the situation that the Hamiltonians commute at different times, and therefore it works in all their states, and you can get the time of evolution. The only thing that you need to note is that the equation is related to t squared e to t squared. So if you look at something like this, you can especially calculate what these continuous spectra look like, that's what you actually get, and the time evolution, as I just said, has got by integrating this between two numbers of two, and that indicates I've got a dot. Anyway, in any case, you integrate it between the two n's, and that's given you a time evolution. Okay, it doesn't tell you very much, but at least it tells you how to do it. If you're concerned about this... In fact, the choice of time limits you. You can, as I say, choose to use time, which is an obvious thing to choose. If you don't know how to write that on the board, you do choose to use time. It's exactly the opposite of what you find out now. Only the positive pair of two pairs does not make sense. The negative pair of two pairs is not what we should find. Okay, well, there is, as I say, obviously no limit to what we can choose. Let me just give you one more example. An interesting natural one. And that's to choose the time expressed in terms of the matter field. And this is rather similar to the initial calculation of the price of the wind. So you can consider the freedom of the universe and the count of clocks. And, well, you've got to play by the sum of this. The clock is the clock of time. It's the measure of time. Because you know the measure of time most of the time. And he considered this to be the count of clocks. And he used the time based on them. Now, how long should be very similar to that?

42:30 What is the analogue of his choices of time? Well, the nearest thing that people would choose to see actually equal to the number of words at a time is actually the value of the matter. And it turns out that if you do this, you find that Hamiltonian becomes A squared is equal to r squared is equal to pi r squared, plus rho of kr is equal to four, which is squared m squared. This is an interesting one, for a number of reasons. That's Hamiltonian. The account that the conjugate of rho equals r... To quantize this you have to do it with care, because classically this verb will rise to be norton, and this should make you a little bit weary because the usual spectrum of arguments and the kind of quotations is normally a customized affinity, and you might find yourself in the midst of problems trying to do that and do this, and in fact if you try and choose delta norton affinity in the operator space of space, you can define r-alt, but pi-alt, which you presumably want to call d-bar here, just isn't so much right, is it? When I say it's not something I've heard of before, I mean there's no such thing as extensions before, so there's no way to define it as something I've heard of before. So whatever else you can or can't do, you can't do a mutual function computation with it. Now the various ways you might think it gets announced, one is simply to drop r, because they can't be called variable. For example, if you call r e to the e to the omega, then r goes between, like I was describing, when r goes between lots and infinity, this thing goes between plus and minus. And that gives you something which you might regard as . Again, that seems cheap, and we should face up to one or two spots, so you've made a choice in time. Well, it turns out the way to pursue is the following, that this here, although it doesn't make any sense written like that, because that's prior to the self-hydration operator, if you just formally look at the expression r squared, d2 by dr squared, it's possible to weigh the self-hydration operator out of it. Here, one is making a definite choice of operator orders. But you can show what this thing really is to pop these kinds of algebraic systems out of one place, and that in the course of our age zero, that really does exist. And what we've been trying to do is to actually define pi r. You can first of all show that this exists as a positive spectrum.

45:00 Take the square root of 1 pi over 1 over r, which exists, and result in actually defining pi r, and that's pi's spectrum. You get this problem, you say, in ordinary quantum mechanics. There's nothing to the dust. If you go through all the quantum mechanics in the world, you do actually have this, but it is there, and you do have a certain amount of it. So this is one way of seeing it. The only interesting thing is that if you look at this thing, first of all, this is negative. That means if you look at the square root of the number of the mass in here, or at least, it looks to be minor. So if we drop the mass... Everything makes sense, but writing K is positive. So again, this is the choice of dynamic learning models. But in actual fact, if you can't see it like this, it is possible to redefine the variables in the case where the mass is there. So the whole thing looks like a time-dependent simple harmonic system. And then it is actually possible to use the time, to use this for very simple values of time. Basically, the universal harmonic oscillator, the spectrum. It's discrete rather than continuous. This thing goes right down to zero. But we have the discrete spectrum with a little bit of leeway to ignore the first value, and it's just possible that a very small range of time, in the order of the time associated with it, can get to this part of the term. So, to use this kind of term, I don't know whether that's a good idea or not, I really can't imagine. It seems to be a good idea. We should use this course of time. But anyway, it is just possible. Regardless, the thing that goes through, you can work out the diagonal functions or what have you. Now, one advantage of this approach is that you don't have to ask questions about gravitational collapse, because here there's a molecule world involved. The away functions are side by side. And because you've got a decent color space, which is one of the big advantages of taking this approach upon revelation, you actually do have a probabilistic interpretation. Or at least, you have the possibility. It may not make any sense physically, but at least it's possible mathematically. You don't even have a positive-definitive method at all, so you sort of have a problem in terms of all these things. Here you do. So you could look up, for example, P of sine, which would be north of sine, which is the probability of finding the system in a configuration within the sine of the sine wave. And you can ask yourself, well, what is this option? And how do you do that? Now, there's a lot of contentious remarks about this in the literature of the second day of the class.

47:30 It has become customary to say that if a wave function manages a singular form, this means that there is no singularity required in the observation of facts. Now that doesn't seem to me to be really sensible. If it operates on a discrete spectrum, it would. But if a wave has a continuous spectrum, it's right down to the point. There's not very much you can say. All you really need to be discussing is the y which b-epsilon goes to zero, that's epsilon goes to zero. Now, the question is, is what do you mean by that? It goes to zero as epsilon, is that what that is? It goes to epsilon as a part, is that what that is? And this is some sort of physical input, which is something we have to know. The most you can say is that if time tends to diverge you towards zero very rapidly, as you would expect, but there's nothing sort of higher than that. As a matter of interest, in that particular case, the energy-hanging functions have the property of being all senior to each other. That's not surprising, because it's a continuous spectrum of non-stereotypical functions, and they aren't ultimately tuned for non-stereotypical functions, not by a factor of infinity, but by a factor of zero. So all these weight functions are not one and the same. So, this theory is infinite. What did you say, Wayne? Yeah, the energy ion function is it. And also you can't call white packets a problem. The energy ion function concerns have a problem. They all can't see the light. That would have been, say, some time ago, an indication that this theory could have changed the facts. Indicator by the way, that doesn't seem to make much sense. Someone could do a butt-prop analysis of what is, what seems, what are the actions that you want to do. It's a little bit doubtful, but they work on this a little bit in the sense that you want to do something. OK, so that's that case. Just to give you some examples of some of these non-energizing factors, which is interesting. Let me show you how a very simple way to calculate this is. Whether you call it a gravitational factor or whether it's an RDS function. If you choose the flat-space case for simplicity, K is not, which is about a dozen quarters of the case of the problem of similarity as another observation that you might expect to be true, which is a curvature that doesn't affect the season behavior.

50:00 If you choose K as not, you can form a wave packet with an idea like delta. I'm choosing the coordinates where y is equal to my value of simplicity. Now, you can start off with an idea like a wave packet with a delta function, and ask how does it evolve in time, because that would be interesting. So you ask for this quantum, which is the weight function times T, which is times T0, which is the delta function. And what you actually get... That's what we're actually getting. The limit of the sine of this is 0, plus the constant steps of distribution, because the left-hand side of the equation says it might be 1. But the interesting point about it is if you look at this, it has a peak at a certain point, and the peak is, in fact, essentially where this minus is equal to 0. So holding the peaks up, we're able to draw a curve over our goal, which is basically 10 minus 10 to the 0, which in fact is a classical motion. Except that there's a plus or minus sign in time. What actually happens is this thing starts off with a certain peak, like this. It starts off with two weight peaks. One goes that-a-way, one goes that-a-way. The weight pack is instantly tailed down to zero. This is why I say the risk of a gravitational collapse is rather difficult to decide. It starts off at a zero peak, but the peak instantly, the weight, tails right down to the sine wave. And these two weight peaks... You can see why they're there too, because you've got an eigen space of positions in the momentum you're actually beating down circles, and particularly equivalent to the one on the left, it's a respect that's fed up with certain anatomical problems. Now in finite time, this picks that singularity. Presumably, the matrix system does exhibit gravitation on a plan, although it's not clear. Well, I don't know if it's not what I've been saying the first time or not, but I know it is. I wanted to talk a little bit more about what happens in what. What happens between the same thing and the other part of it. And the reason is that science has picked out a spectrum. The whole thing turns out to be totally inside out. You asked a question about the epsilon. That's a function of time.

52:30 Well, amongst other things, of course, but... This is the, if at a given time you have a consistent surface plane, you may create a position that's given properly to you from the value of your measure. Of course, this in general is a function of time. Yes, I do too, really, it's the epsilon of time. Suppose you take some finite time. Do you always have values, psi, or arbitrary values? Well, you do here. As I say, the bit is turning, like going right down to zero. I mean, the function always goes right down to zero. It doesn't disappear in any neighborhood. The original delta function now actually repeats around and is separated by a finite amount of zero. But if the brownness at any given decimal point of time, it develops a claim that it went down to zero. Does that mean, suppose we try to measure r of t now, there would be a chance that we could get a value of 10 to the minus 13 centimeters? Oh, 10 to the minus 13 centimeters, yes. Could you put in here 10 to the minus 13 centimeters yesterday? There would be a chance. Well, I don't think so. The sentence suggests it's more than that. So let me just take ten minutes if that's okay. Cool. Now this one is actually a lot more fun. This one is fun if you like plane and operations. It doesn't seem very much fun to me. But this one is much more interesting because a lot of other things you might expect. The first thing is that the ground view for a, we're still going to take the obstacle geometry, but we're now going to introduce a mathematical view which is not at all in scalable, but is rather the derived view, and I apologize for the falseness of this. That's the ground view, which is general quantum invariant, and additionally is invariant for the local problem, and that's what you'll see in the explanations. These are the fear points. So they're squared directly from that tensor. This is the usual Dirac thing. This is the Dirac Hall Spinner. And this right here is the covariance Spinner connection.

55:00 And that was just there. Ah, well, we'll come to that. At the moment, I'm going to say this is the Dirac Lagrangian in the present sequence. Um, now... And that goes back to what Roger said, what happens to the Stokes symmetry. What you've got to do to show that this lot makes sense, if I want to draw some more metric, I've got to show the sum of the solutions again for a couple of equations, just like what I thought of in the case. Now you might indeed expect something like a three is going to happen, because this has some sort of spin degrees between them. It's not entirely obvious how that's going to mesh in with the complete Stokes symmetry. In fact, there is, but you can show that there is no solution. So the coupled derogatory equations, which is positive culture or negative culture, which means the flat freedom of the equation, which is the perfectly sensible set of stories, in which you simply have psi over psi over t, just as you thought you had phi over psi over t, to know you've got sort of an average spin of t, which is the math. And of course there's a very big difference between an average spin figure and an average boson figure, and an average boson figure sometimes we could describe as meta, because boson figures are not physical, but an average spin figure is not real. It's not quite real, it's not physical. There is a solution in which this palace is not real, and you can even write down that solution. I'll write it down. It's that r is equal to 1-0, a plus t to the 2-3rds, that's the usual factor to recognise when platforms are all symmetrical. Because 1 over a plus b, i.m.c over r times d, e to the plus i.m.c over r times p, that's finished, but these are two different things, these are the actual other parts of the theorem. Because 1 over 1, a plus 2, apart from that, looks like more than the plane height we've got. Now in fact, this is what we get from the G.I.J. equation, this is what we get from the theorem. If you were just looking at the theorem equation in the given platform.