Symplectic topology, Wigner-Moyal distribution / Ray Brummelhuis talk

0:00 To you, to work with me, can you bypass here the ridiculousness by finding a necessary and sufficient condition for this to know? Nobody knows how to find a simple or relatively simple necessary and sufficient conditions for the positive. I suspect, in other words, that it's forbidden to be uncertain in the principle. You can find the form in the usual one, using the principle of the symplectic camera, which I tried to expose in a few moments, and most of you have already heard that before, so this is old stuff. Yeah, I did a remark here. If you put h-bar equal to zero in the condition of your slope, then you find this here. And this means that actually rho itself is positive. This means that if you can move slowly, not slowly, but continuously from h-bar to zero while keeping these conditions, then it will make sense of speed of the semi-classical limit. But as you can see, this is almost never true. You are almost never authorized to do it in reality, because for some values of h bar, then you will get a negative operator. And what is worse, what is worse, it's simple. Okay, so in the example, that's just Hudson's theorem. When you have the Wigner distribution of a pure state, well, then it's positive if and only if psi is impossible. Okay, so, uh, uh, no, no, no, no, no, because the, the, the necessary part of this is the chemistry of population and it's a sufficient part, again, which creates... No, no, no, no, no, no. Okay, so let's speak a little bit more about the unsympathetic principle. Okay, I defined the covariance matrix in the usual way as we do in classical statistical mechanics or statistics. This here, for instance, in the case n equal to 1, it would look like this here.

2:30 And it turns out that the uncertainty principle is equivalent to a condition on the covariance matrix sigma. And what is this condition? Well, it's just the condition that the covariance matrix plus one half of i h bar j is longer or equal to zero. Oh, j is the so-called standard symplectic. The matrix, it's a block matrix written like this here, the zero and I are n times n matrices, and J is then given by this here, J squared is equal to negative I, so J plays, in a sense, the role of the complex number of I, you have a complex structure associated with all this, and like that. This was actually proved, oh, I think about 20 years ago, by people working in quantum optics here. The proof is not very difficult, but it takes the use of linear algebra. You have to calculate a little bit, but you'll find that's strange enough. The whole complicated Schrodinger-Robinson uncertainty principle can be re-expressed to this very convenient form. And, using the fact that if s is a symplectic transformation, you have s, d, j, s equal to j, s, d plus the transpose. You immediately also see that this uncertainty principle exists invariant under canonical transformations, linear canonical transformations, which is not true, of course, of the usual Heisenberg uncertainty principle, and which is not visible directly, by the way, from the Roberts and Schrodinger inequalities. I mean, people have to calculate a lot to actually... But which is putting you right back into the Hamiltonian picture, of course. Of course, of course, of course, of course. You left the whole point. So, now, there's another way you can re-express this here. And this makes use of the notion of symplectic capacity, so perhaps I... Can you explain the definition of sigma overheads? The definition? The covariance measures, which I wrote down. The x and the p's are the position and momentum of the variables. You can do it for any operas or observables and so on, but I'm just talking here about x and b, position and momentum, vectors.

5:00 This is a matrix. Now, before I explain the other way we can express the uncertainty principle, I must perhaps recall a little bit what is usually called the principle of the symplectic panel. Suppose I have my big phase space, 2m freedoms and so on and so on and so on. This is a face-based ball. Now, anywhere. It doesn't have to be centered at a particular point. Now, I take a plane here of conjugate coordinates. Plane X, J, P, J. If you project this ball down to this plane, you get a circle, same radius R. This projection is just bi-arithmetic. It's brought up that I have a Hamiltonian flow, a totally arbitrary Hamiltonian flow. This Hamiltonian flow will move points in phase space. It will also move this ball, and this ball will be distorted and eventually get very strange shapes. And the higher the number of degrees of freedom, the more distorted it will become. Let us, let us project this distorted wall down, the XJ-PJ. Well, the principle of the symmetric canvas says that, okay, we will have a formidable distortion, but the area will be at least 5R squared. So, the area will not, the projection will not become smaller. Much finer than a property, or much more precise property than Newton's theorem, because what does Newton's theorem say?

7:30 It says that, okay, the volume of the distorted phase space ball will remain the same, but this is not typical of Hamiltonian flows, it's true of any flow which is divergence-free, okay? This property is only true for Hamiltonian flows. So in Roger's book, which was written before this property was discovered, it stands that yes, the fact that things can distort in such a way tends to show that Hamiltonian mechanics cannot be true of the real world. Actually, Hamiltonian flows are much more tame than one could imagine. Observe that this is, of course, a classical version of the answer to the principle. You have a projection, a certain area, and this area will never become smaller. What happens if you reverse time? Then what you can do is you take a ball inside this here. It's the area of this ball, or the projection of this ball, that will not decrease. Well, we can reformulate this property in various different ways. One of these ways is to say, okay, let's take a hole here, a projection here, a hole in the xj-pj plane, a hole with area pi r squared, okay? The radius of the ball. Then there is no way you can squeeze the ball through this hole using canonical transformations only. So what is the synthetic cannon? Well, this is a biblical reference. Of course, it's Mark 5. It's harder for a rich man to go to heaven than for a cannon, a synthetic cannon to go to heaven. This is why it's called this. How do you prove this? All known proofs are extremely difficult. And the usual way to prove this is to use Gromov's non-freezing theorem, which goes back to 1985. Good. So it's very useful. No wonder that it's not more known perhaps by physicists and mathematicians, by the way.

10:00 It's only people working in synthetic topology who know this problem. And this gave me the idea a few years ago to try to apply this to the uncertainty principle of quantum mechanics, but actually then also prove an uncertainty principle for classical systems. Okay, we're going to talk about this. So what did I actually do? Okay, I used what is called the notion of symplectic capacity. And what is a symplectic capacity? Sorry, I don't, sorry, I, near order, lost. In this case, I will use the whiteboard directly. Hold on, thank you, obviously. Okay. Could you go back over that last, that point in the sixth line of the last slide before? No, I'm just joking. Okay, so let me define a symplectic capacity. A subset of phase space, a simplistic capacity, associates to every such subset, omega, a number, c omega, well, a positive number or infinity, by the way. And this association, this function or whatever you want to call it, has the following properties. If omega is a subset of omega prime, then C of omega, the symplectic capacity of omega, is smaller or equal to that of omega prime. It's a monotone function. The second property is that if F is a canonical transformation, linear or nonlinear, then the symplectic capacity does not change, so it's a symplectic invariant. But even with the nonlinear, even with the nonlinear, yes, for all, then it is conformal, whether that's, this word has not the same meaning as the one we heard the other day here, it means just if you multiply omega by a scalar lambda, a dilation, then the centricity capacity is multiplied by lambda squared. So you see, C omega in a sense has a property which is reminiscent of area.

12:30 And then, this is the more subtle problem, the capacity of the ball Br is equal to pi r squared, that is, is equal to the area of the projection of the ball on any conjugate plane or any plane, and it's also equal to the synthetic capacity of any cylinder, Zjr. By a cylinder I mean that, or d, j, r, I mean that I take a plane of conjugate coordinates, and I take a cylinder just resting on, or going through, or what would it be, sorry, what is b? Ball, ball, baseball, baseball, baseball, which is r, in one eye. So, you see, a valvular set can have finite or has finite symmetric capacity. You also have infinite sets, like the cylinder, which has finite symmetric capacity as well. So it's not really area. Actually, if you want to relate the notion of symmetric capacity to something physical, it's a generalization of the notion of action. This is more general. This coincides actually with action when you have convex and bounded sets, but okay, I don't want to be too technical here. So, the natural question is, do symplectic capacities exist? The answer is yes, and their existence is equivalent to global non-squeezing theorem, actually. So, if you prove the existence of a single synthetic capacity, you have at the same time proven Gromov's non-squizy theorem, and vice versa. So, using Gromov's non-squizy theorem, it's very easy to define actually a synthetic capacity.

15:00 Let's call it CGR. GR for Gromov. You just write CGR omega equal to the supremum for all canonical transformations f of The number pi r squared where f takes b r inside omega. So what does this mean? It means that for some very small radius r, you can squeeze the ball vr inside omega, okay? Then you calculate the number of bitr squared. Then you take the supremum of all these bitr squared, and what you get is cgr of omega, and you can prove very easily, by the way, that this is indeed a symplectic capacity, using Rolling Stones, using theory. Okay! How do these things relate to Poincarean dynamics? They relate in the following way. Okay, very interesting question actually. Now, it turns out that if you take an ellipsoid, a phase-based ellipsoid, all symplectic capacities agree on phase-based ellipsoids. There are infinitely many symplectic capacities. In fact, this is the smallest. It's rather easy to prove that ellipsoids all agree. Now suppose I have a face-to-face ellipsoid. Good. Then you also have characteristic curves, I mean Hamiltonian trajectories, corresponding to a Hamiltonian having the boundary of this ellipsoid as energy level. It turns out that the capacity, the subjective capacity of this ellipsoid... Okay, let's call this gamma here. Here's the integral of Bx along gamma. So this is the smallest action you can have. This is actually also true for, more generally, for convex bodies, convex and double bodies here. So this is the easiest example to show that this is related to the action form. And this was actually already, well, okay, Narkovitch, which I mentioned and which has been the source of my inspiration, I'm going to do, has proven properties involved in the Poincare invariant and all this for ellipsoids in the context of the uncertainty principle, but at this time, non-squeezing theorem had not been proven, so, well, I have really a very good insight.

17:30 Good, so what has this to do with the uncertainty principle? Well, it turns out that this condition is equivalent to the following. I have a covariance matrix. Good. It satisfies this condition. You can easily prove that this implies that sigma is invertible, that actually sigma is positive definite, so it has an inverse. Consider the ellipsoid, called a W, because it's called a Wigner ellipsoid of the literature, one half of sigma rho hat minus one z dot z smaller equal to one. You understand this notation. If you want to use matrix notation, z, t, here, z, here. This is a quadratic form in the position and momentum coordinates. It turns out that this condition here now is equivalent to saying that the symplectic capacity of W is at least pi h bar, that is, one half of the quantum of action. So this is a very concise way of re-expressing the uncertainty principle of quantum mechanics, okay? And this led me then, as Bazino, to introduce a lot of quantum drops. From my point of view, you can use this to introduce a canonical invariant coarse-grained partition of phase space, and to apply that perhaps in statistical mechanics, that's something I would like to do. You know, in classical statistics or quantum statistical mechanics, you use quantum cells usually. They are small cubes. I forgot to mention that a cube has not many invariance properties. Even if you use a linear symplectic transformation, a cube is transformed into something that's not so nice and, well, anyway. If you take ellipsoids or if you make a coarse grain of space by sets, by just saying,

20:00 okay, I want the sets to have a symplectic capacity of at least one after quantum reaction, you are much more free to do things. Because there are many signatures here. By the way, notice that, as I said, this condition here is invariant under symplectic linear transformations, but this condition, because of the properties of the definition of the symplectic capacity, is invariant under arbitrary symplectomorphisms. So, one can envisage that this is a more general form, a non-linear form of the answer of any principle. I don't know if it's useful or not, that's what I'm trying to see a little bit. As I told you, there are not many answers I'm going to give in this talk here, actually. Okay. So, let's go now to the Narcon-Witsch-Ligno spectrum. As I mentioned before, a given operator with trace one can be a positive semi-definitive for some values of Planck's concept and not for others, and in 86, Narkovitch introduced the notion of Wigner spectrum, which is called now as, sometimes, Narkovitch linear spectrum as well, but I don't have much of it, please tell me when I'm... How much time do I have? So, he made the following. He took a phase-space function rho, a candidate for being the symbol of a density matrix. Of course, this is natural here. Then, he defined the ringer spectrum of rho as the set of all known numbers eta, for which the KLM conditions are satisfied. We replaced h-bar, the conditions we saw a moment ago, by this parameter eta, and said, well, okay, if the corresponding matrix is positive definite, then we say that eta is in the Venus spectrum. This means that, from a practical point of view, this means that eta can be used as a new Brems constant. I mean, the operator you will get then will be a trace class operator. And the question is, yeah, well, how do we determine, give that function rho, how do we determine the Wiener spectrum?

22:30 Well, nobody has succeeded in doing it quite generality, but there are, there are partial results. First of all, what are the properties of the Wiener spectrum? Well, rho has this density matrix if and only if its Wiener spectrum contains H bar. Of course, this is just a definition of positivity here. If, ah yes, if the Wiener spectrum contains the number zero, that is, if you are allowed to put h-bar equal to zero, then it's at the same time a classical probability density, okay? We call that a quantum semi-classical state, a state which is both quantum and classical in a sense. Yeah, if eta is in the Wigner spectrum, then so is minus eta. Of course, in practice, we're taking a negative plane constant, so useful as it's a useful device, anyway, for making calculations. And it's bounded, bounded set. But if the Wigner spectrum is always containing an integral, that is, you are not allowed to let h-bar tend to fill in, for instance. In that sense, there are examples of bigger spectra which are really strange. They are discrete and consist of numbers converging, a sequence of numbers converging towards zero. The situation is very strange anyway. Ah yes, and the convolution. The bigger spectrum of the convolution of two rows contains the sum of the two sets, the sum of the... Well, are there states for which you get it really H-barred, go continuously to zero while keeping positivity? This is an important question because, okay, these are states which really make sense to talk about the limit when H-bar goes to zero, in this case, because you can read smoothly. Then, other question. What about this condition here? Oh, quantum and classical and with nothing between. These questions are unanswered, so... But Werner and Breuker, who joined the guys, have shown that you take a mixture of the three first states...

25:00 Ground state, the first excited state, and the second excited state, these are all going to be functions, okay? If you mix them in a proper way, you do not have this. So, in general, if it does not make sense to speak about the classical limit of a densiter operator, if you mix states of something as simple as a Hamlet oscillator, see? So, one result here, which I did not prove, actually, but which was proven by two Portuguese guys, recently, Diaz and Prato, and this is a nice result, it's approximately the only result which is well known, and it's about pure states. The linear spectrum is equal to non-negotiable and otherwise, that is, if psi is not a pure state, no, sorry, if psi is not a Gaussian, then the linear spectrum only consists of two points. This is, I find, very striking. I mean, it means that the only space for which... We are rigorously allowed to get banks constant n to zero by Lawson's. Otherwise, you can have trouble, or you have trouble, mathematically speaking, if you do so. So, I mean, it seems to me, I haven't studied yet the philosophy of all this, and I'm not a philosopher, but it seems to me that when you talk about the limit h-bar going to zero, it seems... Does it really have a meaning or not? But anyway, the first problem is very easy. Prove, because for Gaussian states, the uncertainty principle is both necessary and sufficient.

27:30 They are actually the only few states for which it is known that it is necessary and sufficient. And, all we have to do is to show that for every eta which is smaller than h-bar, Then, this will still remain positive, okay? But this is easy to do, because since h-bar anyway belongs within the spectrum, we will have this, which is the uncertainty principle. Now, you just notice that we have this for every bar between 0 and 1, you can write this. And, since sigma anyway is positive, This will be a positive matrix, and r times this will also be positive, so this is positive, and the sum of two positive semi-different matrices is also positive as semi-different, so of course, the truth of this case is triviality. What disguised Diaz and Plata then did to prove this is a rather subtle argument. They did the following. They introduced a family of Gaussians, phase-based Gaussians. Yes Ray? No, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no. It can contain more, but what is important is it contains minus h-bar, h-bar, and zero. This assumption leads to the fact that zero is in the Wigner spectrum of the convolution. Good. Why? Because we have these problems here. It follows, then, that g, eta, convoluted with rho is the Wigner distribution of some state. Of course, yes, since it contains h and minus h-bar. If you have KLM conditions, positivity conditions, you must have this, positivity of this here. Now we define the convolution of rho with gh-bar. This is called in the literature a

30:00 Housini function. It's a smoothening of a linear distribution by a Gaussian, by convoluting with a Gaussian which has sufficiently large support in order to violate the uncertainty. Automatically, a Fusini distribution is always positive or equal to zero. This is a well-known, it's easy to prove by a direct calculation of Gaussian and integrals, I mean, these you find in many sources. We've got that. Also, using the semi-veil property of the Gaussians, we can write this here. But that is not the Gaussian, because rho isn't, we're just assuming that rho is not the Gaussian. And hence, there must exist a z0 such that f of z0 is equal to 0. But writing explicitly what f of z0 is, you get this. So what are we doing here, actually? Well, if this is equal to 0, and this is positive, sometimes it forces this convolution to be negative in a whole set of measures which is larger than 0. This is impossible. We know this assumption here. Well, it's a rather subtle, and actually it's very new to me because this guy sent me a pre-print one week ago that I was hesitant to write this number. It would be interesting. All the while I would have nothing to say. Thanks, this guy. Well, what can you do? Oh, yes, let's invent these stuff. One question, I mean, you mentioned whether it makes sense to take H going to zero in sort of... I mean, in some cases, H is not really H, but H is some sort of a rescaled version of H. Some what what? Some rescaled version. I mean, in the simplest case, if you look at the further equation, H is really H divided by the square root of 1. Of course, of course, of course, of course. What you do is you rescale. Of course, but this is another. Then you change the function itself by rescaling. Here I'm talking for a given, given law, see, with only H as a parameter. Of course, if you rescale, that's a lot of things. Of course, this is another question.

32:30 Here I do not change a given row. Okay, so what more do I have to do? Oh yeah, well, I will not have time to talk about this, and I talked about this last year anyway. No, the first part, at least. Sub-Gaussian pure states. I mentioned that last year here. Here's one more thing anyway. Suppose that I have a pure state satisfying, we call it like this, this is called a sub-Gaussian state for obvious reasons. Well then, the symplectic capacity of this ellipsoid is again larger than or equal to one half of the action. If this ellipsoid is the image of a ball with radius square root of h-bar, then you can conclude something very precise, namely that psi is a coherent state, or rather the image of a coherent state by a metaplectic operator. The proof of this, as explained last year, is based on Hardy's uncertainty principle. And, in this case, the biggest section of WSIMA is this interval. This is anyway implied by the result of the 14-piece die, as I said. In material calculation, you verify that rho, z, goes to the iridescent difference at the point C. And the gaussians seem to correspond to points in classical phase space, but that is certainly not the first one. I said, please, yeah, okay, the proof, the reference also was for 2007, and anyway, you have forgotten it, so I shouldn't have written that. Sorry, what's, how does it happen? How does that happen? This is something interesting. Well, in the simplest form it says that, suppose you have a function psi of x. Which is dominated by c times the exponential of minus 2 to a x squared, okay? And if its Fourier transform is dominated by, oh I should perhaps, no sorry, I should put an a here.

35:00 And if its Fourier transform is dominated by another, well then... You must have AB smaller or equal to 1. That is, just a pretty small statement of the fact that a function and its Fourier transform cannot be too sharply localized at the same time. Smaller or equal to 1? Sorry? Smaller or equal to 1? No, smaller than or equal to 2. Yeah, right. And, if AB, the product, is larger than 1, then psi is 0. Moreover, in the case of equality, And you have equality. Then psi is precisely equated as the, then you have an equality statement. When ab is smaller than one, then psi conveys all of our instructions. There are six ways, lots of people working with that from various standpoints now. You can generalize this actually to the n-dimensional case, and recently I have proved that actually One single condition of the beam of transform is sort of having a conjecture, one if not two, this paper that has, well, it's, anyway, sorry, but this is all the stuff, of course, but still, it's related. Oh yeah, and here's also a new result, but it's also very, very tentative. Generalization of the previous result. No, it's not, this is a generalization. I'm not very interested in what actually introduces itself to the Gaussian case. This is what I tried to do recently also. I replaced the mc squared by a convex function cube, strictly and uniformly convex. And I tried to look at putative densities satisfying inequality like this. But then I proved that if the operator is positive, then it is a prehabitant matrix.

37:30 Then, again, we have a set whose synthetic capacity must be at least one-half of the content of the action. Equivalently, and this is actually what Roger asked, equivalently, because of the subtropical synthetic capacity, this is equivalent to saying that the integral of the action form is at least one-half of H, for every Hamiltonian periodic order down the characteristic boundary. A little remark. If you have a convex set, not only an ellipsoid, it makes sense to speak about Hamiltonian ordens on this set. They are all the same regardless of which Hamiltonian you use to define the boundary as an energy hypersurface. This is a particular product of Hamiltonian systems. The fact that true normals to the surface at an even point are always proportional, so you can talk about Hamiltonian orbits as an intrinsic property of a convex bounded set, without reference to any true Hamiltonian. Okay, so this is an extension that's actually the proof. Reduces this inequality to a Gaussian, sub-Gaussian estimate. So in fact, when you have a condition like this, rho z is automatically also sub-Gaussian, so we have no real extension here. And there's nothing actually interesting we can say about the linear spectrum in this case here. Nothing, almost nothing is known about, all the negative results is known about the density matrix of mixed states. And it would be very interesting to find a procedure for finding the Venus spectrum of mathematics of a, of a, of a, of a, of a, of a, of a, of a, of a, of a, of a, of a, of a, of a, of a, of a, of a, of a, of a, of a, of a, of a, of a, of a, of a, of a, of a, of a, of a, of a, of a, of a, of a, The framework of the Grammar Squeezing or Non-Squeezing Theorem and the symplectic capacity relates to the behavior of coverings of the phase space and it's one of the topics you've talked about last year, the fact that the role of the metaplexic group as the

40:00 I guess the very vague question I'm asking is can I get a little bit more of a geometric feel for what is behind all these rather lovely analytic predictions. You're not the only one in the room I think of whom that's true. As I told you, all the proofs of robot theory are very difficult. Here is roughly the idea behind it. Suppose I take, again, a face-based ball, radius r, and I deform this by a linear canonical transformation. Fine. I get an ellipsoid, okay? Good. So this is the deformation of a ball by a symplectic transformation S. If you cut this every sort by any plane of symplectic coordinate, you'll get an ellipse, of course, which has exactly the same area as the section of the surface. It's like having an egg. It doesn't matter how you cut it, as long as you're using... Symplectic cuts of area will always be the same. Why is it so? Well, it's interesting because if you cut this by symplectic plane, you can come back to the circle using S minus 1, of course, which will give, of course, a big circle here, but the restriction of S or S minus 1 to a symplectic plane is symplectic itself, hence area preserving. This is why this here and this will have the same area. In the non-linear case, what Gromov did, he said, yeah, okay, more or less, I'm really, this is really big talk right now, but yeah, okay, if I transform a cut like this here, this is not necessarily, you know, it's already something, you know, up here. If I come back, I will get something which is not flat.

42:30 But I can study the properties of this binary by integrating and all this, and then the procedure is very complicated, it uses the motion of pseudo-homomorphic curve and all this, and it's really very difficult, but the result is the same. Here's one way of proving monotheism. Another way, and this anyone can try to do as an exercise, is to try to invent a symplectic capacity, because as soon as you have a symplectic capacity, you also have monotheism. So the more I cannot say, I'm going actually to Tel Aviv in November, I just need to meet Paul DeRouge, which is one of the specialists there, and we're going to discuss about these things and possibly other applications, because he has written a paper with Elias Berg and Kim, where they use my ellipsoids to prove something in contact geometry, which is a generalization of simple geometry, and in fact, I would like to understand what they have in mind with that here. There are some feedback on them, so don't ask me too much. I know how to use this, but it's very easy. Sometimes people talk about the classical world as a, you know, sort of like a singular limit. That's how I feel, that I can moderate this reading. I was going to say, do you have any... What I have? Not all. This synthetic camel property seems to indicate that you have, anyway, a classical or a quantum world emerging from the classical world. My first book, whose best part is Basel's foreword, I mean, he explains this very well. Much better than I could, so I will send you the foreword of Basel. Yes, absolutely. I have the impression that the classical world is more quantum, in fact, that one believes, because you have properties like the principle. I mean, it has not been believed that screening of weight back is a very quantum phenomenon, because that is not true. Lil John already wrote in the late 70s. The paper in physics report where you study classical spreading, I mean, classical spreading, sorry, take a bunch of projectiles and go on the screen, take a Gaussian distribution, it will spread exactly in the same way as, in the same format.

45:00 Spreading, spreading, spreading, spreading, spreading is not particularly a quantum phenomenon, no. Optics, I mean, Hamiltonian optics, one has similar effects. Hamiltonian optics, you have the... You could sort of let h bar go to zero and you'd sort of seem to get past the world, but there's obviously other things where no matter how small h is, you're never actually the same as h equals zero, so some things are kind of singular in that sense. Yes, yes, wait, this is obvious, in my abstract I actually mentioned that I would like to talk about something else also, but I don't have time, but it turns out that you can replace We're adjusting the matrix by an operator of this type here. x bar sub i h bar over 2 e p, p minus i h bar 2 e x. Okay, these here operators are called box shifts. Physical literature. And you can totally, and this is a very recent word of mine, you can totally motivate this just by playing with the symbols. These are wide operators but in four n dimensions, you can say, you see. And it turns out that rho tilde applied to a function psi of z, because we have no variables here, is just... What is called the Moyal product. So this is actually information quantification. To be very, I mean, very tangible here, very heuristic, if you let H-bar tend to zero here, of course you get the classical Hamiltonian. And the Moyal product also then degenerates into… A normal product. What is deformation quantization? Well, you just deform the ordinary, plus some brackets and so on and so on. Another open question, totally open question, is how do density matrices or Wittmann distributions behave under chemionic quantum transformations, which are not linear? Many physicists believe that Dirac's proof only works in the linear case, otherwise you must have h-bar corrections and lots of things like that.

47:30 So this is a big open problem, which is of course related to the notion of the linear spectrum, because the linear spectrum is invariant under linear subjective transformations, so I suspect it's also the case for nonlinear. I don't know. I don't know. I don't know. I know nothing. To answer Tim's question about his limit, and there are these other semi-classical theorems which, these trace formulas, which, what they do is when you look at them, you look at how they give an energy, and you look at an interval of about... Thank you for your attention and see you in the next lecture. Okay, but they are, they are treated, this is an interesting, has been an interesting topic. This would be very interesting, and a new comedy on this topic anyway, I mean, it seems really nice. But, but I, I'm sort of desperately trying to remember what the, what the relation was between Degner distribution and Moyel function, etc. But, but Bessel had these things about... Well, I've got them on some slides somewhere. Yeah. So are these, does positivity play any role in the things which you are trying to do? Because you have sort of these two equations. The vegan solution is just the vial symbol of an operator whose kernel is a tensor product, that's all. And the Moyer product is? Ah, okay, what's the Moyer product of A and B? Well, take two bio-operators, A and big B. The symbols are A and B, and this is E. Compose them. You get an operator C. So this is the symbol of this, this is all. But if people are interested, I've got some slides which I was going to show this afternoon. Perhaps this product defines it, the way we physicists like to see it.

50:00 Can somebody clarify for me what one is supposed to expect about the classical universe? I mean, there are far more quantum states than there are things that look like classical states. So we don't expect to see just putting H bar over 0 to get a classical... Or do we? I mean, I don't expect to see it. We don't expect to see a classical state. We might have things that are superpositions, I mean, Schrodinger cat states. Well, they are certainly very high. You can take H bar. You can take H bar. If you make H bar zero, it's still impossible. I mean, they don't disappear just because H bar is not zero. No, but somehow you have these things where you should collect lots of these, lots of Schrodinger and Katz which are sort of very close together, you might think. Well, that's why I'm saying for some purposes it must be, you know, no matter how close you are to zero, you're not sort of at zero, so it is a separate area. Yeah, they're just far more quantum states than classical ones. I just don't see, I mean, are we supposed to expect that we only see states that are not classical states when we integrate them? I think these results are rather amazing. I don't know about you. Maybe you've seen them before. It's a philosophical question. Mackey, for instance, his view of quantum mechanics is very fine, and I'm close to his point of view. No, of course, but we have to answer the question, why won't the steam rise in these things? I just don't see quite what one is supposed to expect. I mean, you take states that are very large and thin, just look at the angular momentum, and the vast majority of those states are nothing like classical states. So, what is the expectation? And we don't get the classical world by just taking h bar going to zero. But there may be certain statistics, and I agree with what you're saying, but maybe there are certain things, this is sort of my question to you, are there some things which do look like the classical world as h bar? All of these are pure states he's talking about here. He doesn't know what to do when you've got mixed states.

52:30 Again, I'm not the resident of this seminar. But you're investigating a mathematical deformation, which is something you can do. I'm just a mathematician here. I'm sure that I didn't mention any physical claims. I mean, no physical claims, I just mentioned mathematical properties. Mathematicians say, well, I can't expect to answer because I'm just a mathematician, physicists say, well, I'm just a mathematician, physicists say, well, I'm just a mathematician, physicists say, well, I'm just a mathematician, physicists say, well, I'm just a mathematician, physicists say, well, I'm just a mathematician, physicists say, well, I'm just a mathematician, physicists say, well, I'm just a mathematician, physicists say, well, I'm just a mathematician, physicists say, well, I'm just a mathematician, physicists say, well, I'm just a mathematician, physicists say, well, I'm just a mathematician, it's one of the reasons we haven't been getting very far, isn't it? But actually, you mentioned very briefly, you did mention the connection or very broad kind of convergence of this, the motivation for this program with that and Mackey theory, and I was thinking a little bit about, you know, there's issues about Mackey convergence. Now you think of analytic functions in the Mackey setting in terms of the behavior of triangular matrices. I mean, is there any connection there with what Mackey was trying to do? I don't know. Yeah, probably, because in one of his books, the complete book, like the Unitary Representations and so on, he vaguely mentioned also the limits when he was close to zero, but I don't find that he... I don't understand exactly what he wanted to do. Are you happy, Maurice? I think we ought to stop here. And have just a little break and then... Okay, absolutely. Okay. So life is a very natural kind of catapult of constructions and cases. Two things together are one and a common thing. So what we're going to do first is we're going to have a real sense of time. Thank you for your attention.

55:00 Because he was criticizing the book that David and I wrote by saying we assumed the Schrödinger equation. And I was wondering why we were to be blamed for our book by assuming the Schrödinger equation, but I thought everybody assumed it. And then I wrote to him and said, well what did you mean? He said, well I was taught quantum mechanics from Dirac. So I went back to look at Dirac, and he actually assumes the Schrodinger equation. He's got a hand-waving argument which makes it plausible. And then when I wrote to him, I said, well, look, it's in the symplectic structure, because you can get a one-parameter group in the metaplectic group, which is just the Schrodinger equation. And he wrote back and said, I don't know what you mean by the symplectic group. Well, I don't know if it was that confusing. That's physics. I mean, maybe I should have used canonical transformations. I was amazed because, you know, he's a smart guy. Very strange. Then I began to wonder whether most physicists have never looked at, have not taught the symplectic group. I mean, there is that confusion, because it doesn't mean two different things. It means actually several different things, only two of them are usually mentioned. I mean, there's a signature, that's all. Yeah, yeah. You don't say, I haven't heard of it. You just confuse them. It's a very negative statement. I don't know, there are all sorts of misunderstandings that can happen at different stages, and it's often difficult to track down what it is. Also, I was very upset that the problem with our work was being able to show the equation. I can see that's a problem, actually.

57:30 There's lots of criticism that you can make, but none of that level, yes. No, he actually put it, they don't say where it comes from, they put it out of the air, and then he says, oh no, it comes out of Schrodinger's head. Well, he did come out of Schrodinger's head, because he didn't prove it. You can't prove it with a theory which isn't fun again. Well, except, of course, I think that's what Morris thinks he's trying to do at one level. Well, he thinks he can actually get the Schrodinger equation directly out of this kind of Hamiltonian flows on phase space and via the metaphylactic grid, doesn't he? As I understand it, if you go to his book, but I think, I agree. It's very, very interesting. It's very interesting, indeed, mathematically, to see what one can do at the level of all this rich topological structure, which is carried by the phase space and the way that it connects up, obviously, with this very, again, topological way of thinking. But I agree, it's talk of the absolute. No rock-solid problem that there are far more quantum states than classical ones, and it's just, you know, you can't change that unless, well, unless of course there is some kind of subtle conceptual issue about how you very define the notion of state as such in this setting. I don't want to go down that road. Yes, I'd rather crack and go into the funny farm than go down that road, I'm afraid. No, no, no. I know, it's astonishing how many of them there are. But indeed, as we were saying yesterday, at least a couple of them are my very close personal friends, but I can't, still can't get my head around what has led them down that particular path. Well, it's a good path to go if you want to make a reduction out of serving. Yes, I suppose. Not to realise that you've served. Exactly, not to realise that's what you've done there is worrying, especially when you're... Well, it absolutely fulfills, I think it was at Russell's dictum, I think it's Russell, that there is no position so absurd that a philosopher has not been found to defend it. I have to say, the whole Everett interpretation is due to a classic illustration of that saying. But I do find this whole symplectic topology, well, personally, to me, it just appears to be a very, very rich and beautiful subject mathematically, and even if it turns out, dare I say, a little bit like what people say about twistor theory, even if it turns out not to provide the correct…

1:00:00 When we route it to the physics, it will still stand as an absolutely beautiful piece of mathematics, because it does seem to give such a lovely geometric handle on the structures in the phase space and the way particularly that the dynamics... and it obviously relates to the way one should understand dynamics at some, perhaps, purely abstract mathematical level. I think it varies on anticipations as well, so if you want it to be the real world, there are other ways you can see that the classical world doesn't behave like that. But it is very interesting to see the shadows of the uncertainty relations and of other quantum effects in the classical world. We recognise more and more how they were already there, and we see them particularly in optics. The focal, the equation of the foci of... Well, optics I suppose is... It's much closer to quantum mechanics, I mean, from my point of view, it's because the math displacement is very trivial. Well, it's just about to say, isn't it? It's also, of course, mathless. It deals with literature, mathless. Yes, it is. Really, what you're seeing in the classical world is almost the same. It's much more, yes, yes. And of course, since Hamiltonian dynamics really is the framework for optics, or at least, well, Hamiltonian approaches is very much the natural framework for actual optics. It's natural that people should say, well, maybe we should get back and look at how far it... One can think of quantum mechanics as a kind of special case of the Hamiltonian approach, but I agree that you do then have the problem that there are far more quantum states than classical ones, so trying to get everything back inside the Hamiltonian framework, even with all this additional... You know, rich mathematical machinery, symplectic topology, I think it's probably not going to give the answer, unless there's something to do with our understanding the very notion of state and transformation that's going on in the background. And that's just helpless hand-waving. By the way, I did ask Vanessa, and unfortunately the dates don't work for... The dates don't work unfortunately for each of them. Yeah, I did ask her this morning at breakfast. They don't work because of the, but I was thinking if it's okay with you, I could just get back to Pavlov and mention, chat it to you and see whether he couldn't sort of maybe next year shift his date, his conference dates so as to fit with you and the school holidays.

1:02:30 It was me who nagged them into doing that. I said, look, they put in a kind of philosophy of math, Degner's problem, unreasonable effectiveness, and also section on symmetry principles in geometry, algebra, and physics, I think with the intention of, because they wouldn't have got Mark Laschews raised to come to talk to them if he had just been a Finsler Fandango. No, no, no, I think I'm with a bit of help from Mark. Lou, I am getting them to steer more and more in the direction of broadening it to other things, but I'll ask him if there's, ask Vanessa, in fact, when the right dates would be for the midterm. Well, she knows that it's not going to work for those dates for the first week in November. She'd know about next term, but she might not know next year. It probably doesn't sit there very much, because of course it probably moves rather than giving this glitch-like... Yes, a bit like... a bit like cohomology, you know. Poison here, incidentally. Well, of course, the canon is subject to those glitches anyway, because of the... That's why we have leap years. That's why we have leap years. Are you noise ready? Oh, yes, sorry, my brain wants to come on. There's a time problem with the machine. Oh, of course, okay. Yes, this is a big problem for one hour. Okay, gentlemen, we're starting. Okay. Because the, um, there's a, sorry, back to the gentleman involved. Keep the same fuel, which is multiplication by 10, I will still have the same amount of computation relations. You can rewrite this a little bit in, in, in saying that you have the... Ordinary operator P on a slightly different domain of functions for which a value in 1, you should take the value in pi, All of this is equal to 2 pi gamma times the value of minus pi. So for some reason, minus pi pi got replaced by 0, 2 pi got replaced by 0, 1. We'll just have a mapping on the slide.

1:05:00 But in any case, these are the things where you do solid state physics and you look at these block conditions. And then you also know that if you look at these block conditions, you can get L2 of R as sort of a diagonal to all of these subspaces. In some sense you get these other canonical, if you take the spaces of these other canonical pairs and you sum them all up, you get the answers from the mathematics. I'm not sure because I haven't time to read this paper. I haven't time to read this paper. It may be. Now, there are many more, in fact, there are many more canonical pairs. There's a, Gerard Pond has written, has proved a theorem which is on one hand quite surprising and on the other hand is quite surprising that nobody seems to have thought of it before. And this is presumably because nobody actually looked because of the result of Pauli, which says that you couldn't do this. So let's, instead of talking about P and Q, let's talk of operators X and Y. You can look at it quite a little bit more abstractly. And suppose that you have a self-adjoined operator on a Hilbert space, with eigenvalues e and u, which should have been e, and they should, to make this construction work, they should have constant multiplicity, and I'm thinking of multiplicity one over here, and you put the technical condition that the eigenvalues go to infinity sufficiently quickly, in the sense that the sum of one of the eigenvalues squared converges, An operator which is self-adjoined and quantum canonically conjugates to its original operator in the sense that the commutator is equal to i on some dense subspace. So x is self-adjoined, y is self-adjoined, and its relation is satisfied on a dense subspace. And moreover, under this condition, the operator Y is bound to the vector.

1:07:30 Just a side remark, any such canonical conjugate operator is not unique, because I can add to Y anything which commutes with X. For example, not to the Y of X. I can add an arbitrary element of the common to the Y of X. So, why do I say that this should have been, this is not, this should have been, it's surprising that it's... The reason is that the why you put down is quite simple. If you work on an eigenbasis of x, you just write down this matrix. You put zeros on the diagonal and you put basically one of the differences of the eigenvalues on the half-diagon. And here you sort of use that. This is the case of when the eigenvectors of multiplicity 1, as if eigenvectors would not have the same multiplicity, we couldn't really write down this. That's a bit of a prescription. So my claim is this is the operator y. So you compute the commutator of x and y. And it's sort of easy to see what happens. And what I will get is something like EI times I divided by AJ minus EI minus I divided by AJ minus EI times EJ. I is not equal to J, so you just get equal to minus. If, okay, but then you think, okay, but this is totally wrong, right? Because if I look at the commutator, I always get 1 on all the elements except on the diagonal where I get 0. So it doesn't look like the identity at all. Aha, but the trick now is to apply this to vectors in a certain subspace. You apply it to a vector and you cunningly choose a vector in such a way that all its components sum to zero.

1:10:00 Now if all its components sum to zero, if I apply it to a vector, I just get all of its components which are different from the j component. But if all of its components sum to zero, this is exactly minus the j component. And I'm exactly where I want to be. It's a very clever trick. So this is Galapagos. And then, now the rest is sort of functional analytical routine in the sense that you have to prove that this thing is well defined. It's a self-adjoint operator, well it's bounded and it's sort of symmetric. And the other thing is to prove that this thing is actually defined as a dense subspace. But this is sort of, you know... Anybody who has a functional analysis will be able to do this, if he sits down for some time, to do this. So the key idea is to write this down, but the key idea is to restrict this to a thoroughly defined substance. My watch sometimes stops and it's... Okay. Now, so if you think of X as an Hamiltonian, then Y will be commonly conjugate to the Hamiltonian, and you might think of this as being a time operator. If I put H bars at the proper places, then all these numbers over here will have the dimension of a time. My question is, what does this mean, something physically? And what does it mean, it means? So, the way to go, probably, is to look at some examples. I mean, you know that if you have an operator, then the physical interpretation is supposed to somehow come from the spectrum of you. So, I mean, I don't know, this is sort of, if you, the matrix elements are sort of made up of the frequencies of, you know, one of the frequencies of transitions between the two, between different energy levels. Well, I think that would be nice, that would fit in with the general overall structure of it. I'm not complaining, but I'm not sure what it means.

1:12:30 I'm not sure what it means, but it seems the last sort of thing you're going to do in this game. Yeah, no, no, I'm not complaining at all. Why do you think I'm complaining? Could you keep looking at me? Because I'm nervous for you. You're treading criticism and you're profound. So you... So if you think of X as M or T, you think of T as a diamond. And now the question is, can we do any explicit examples? And what is the physical meaning of T, if M? Now, as concerns the latter, there is a slightly disturbing example, which is if these energy levels even go to infinity too quickly in some sense. So if you think of a particle in a box, right? A particle in a box with various conditions. The eigenvalues are k squared. So I take my operator T and I want to check whether it's something more than a bounded operator. So I check whether it's a Hilbertian operator. So I take the sum of all matrix elements squared. And you do a little computation. So this is j squared minus k squared. This is j plus k squared, j minus k squared. And this plus should be a minus. So one over j plus k squared, one over j minus k squared. And you sum up all the not k and you think a little bit about it and you see it's finite. So it means it's a Hilbert-Schmidt operator, which means it's a compact operator, which means it's a compact salvage operator, which means it has a set of eigenvalues condensing, accumulating at zero. Which is not exactly what one would expect if you think of the speculations people have made about chronos. This is sort of a minimal length of time and here it seems this operator just seems to point to a maximum length of time. So the question is of course is what does this operator actually mean? Let's go to a less frightening example, the point of the harmonic oscillator. So we take the harmonic oscillator on L2 of R, omega, and put here with the ambition of making everything apparently true all the time and then at some point in the next line we put it in group R.

1:15:00 You look at the harmonic operator, for example, L2 of R, or you look at it in the Batman picture on L2 intersectional functions on C with its weight e to the power minus c squared, maybe e to the power minus c squared divided by two. For example, in this picture you know that the eigenfunctions are emission. Polynomials and mission functions are quite complicated. In this they are very simple, z to the power of k divided by the square root of a. Okay, so you, okay, let's try, the first idea is let's take this time operator associated to this oscillator. So you get 1 over k minus j off the diagonal and 0 on the diagonal. And you try to brute force some of this. And then it doesn't really work. Perhaps I should ask this question. Because you can probably do this if you know something about this entire function. I don't know whether anybody can tell me anything about this entire function. I'm not really very familiar with special functions, but I couldn't find any. I even went through the trouble of going to Mathematica, and Mathematica gave me the answer.