Symplectic Camel Properties & Gaussian Quantum States

0:00 So, Brian, have you decided who you're going to have to perform? Yeah. The slideshow is we not get to your position. Really? Well, for a good reason, economics, they seem to have access to this here. Uh-huh. Okay, okay, get you. Yeah. Thank you. Okay. We're waiting for you, right? We've still got one short. I think probably we should start there. It's a great pleasure to welcome Maurice Gossel from Karls Kroner in Sweden. You've written a very good book, most people know that, but for the mathematicians, a very good book on a very fine foreword and a very good, yeah, all right. What's it called, Morris, your book? Principles of Newtonian and quantum mechanics. The need for Planck's constant. And the need for Planck's constant. And today you're going to talk about quantum blobs versus quantum cells. Well, yes, this is an expression I coined. I don't know if it's... Well, my goodness. I should thank Charlene for allowing Morris to actually not be spending time in London. Good. Okay, Morris. Okay. I'd first like to express my gratitude for Professor Hardy for having to send me this invitation to give a talk at this famous college. college and uh okay i would like to entertain you today with some thoughts about quantum cells from a symplectic point of view uh you see uh well i think that everybody magicians or or uh physicists are familiar with uh things like the heisenberg uh uncertainty

2:30 the principle which in its very crudest form can be written like this here okay this is a very crude form of course we're going to do something better than in a while okay this is the most commonly known form of inequality but one should not forget that you have other inequalities also So, like this, for instance, as soon as j is different from k. This emphasizes the role played by pairs of conjugate variables, like x, j, p, j. Mathematically, it means that we have here a symplectic plane, and this is something very important. I'm going to try to explain to you why. Good. Okay, this can be rewritten, actually. be rewritten if you like it like 1 over lambda delta p j squared plus lambda delta x j squared larger or equal than h bar. h bar is a constant. It's h divided by 2 phi. Everybody knows, of course, this for all lambda positive. It's not perhaps immediately obvious that these both formulations are the same, but they are. It's easy to go, for instance, from this direction. You just write the delta j delta xj multiplied by 2 is anyway smaller or equal than 1 lambda delta pj squared plus lambda delta xj squared. This is obvious. Just 2ab smaller or equal to a squared plus b squared, okay? So, this implies this for all lambda is obvious. The other way around can be proved by elementary means. So, this shows that ellipses are playing some role here. have an ellipse here in the plane, in the pj, xj plane, and the ellipse which has exactly area pi h bar, that is, pi times h over 2 pi, that is, one-half of h, one-half of the quantum affection. Good. So, we've already seen here that if we talk about the answer to the principle, the ellipses up here, ellipses in a symplectic plane.

5:00 Now, okay, there's another thing which is well known in physics, a little bit less well known in mathematics because the result is in general false, mathematically unless one puts some very strict conditions on the Hamiltonian function. Suppose we have a Hamiltonian function H, of any type actually, and that we consider the hypersurface h equal to e, where hypersurface in space, phase space, I'm using of course these notations here, x, the n vector, p is the n covector here and I'm working in r to 2n here, r2n, xp, and, well, as we've also used the collective notation xp, z for xp. So this equation here, H, the Hamiltonian, is a function of x and p, that is, of z. This equation defines, under suitable assumptions on H, a hypersurface that is at a manifold of dimension 2n minus 1 in phase space. It turns out that if this hypersurface is bounded, is compact, a subset of, or rather, if it bounds a compact subset of a space R to N, here, Then, in quantum mechanics, one uses often the so-called Weyl formula, which says that the number of quantum states having energy, lambda, smaller or equal to E, is asymptotically, when E goes to infinity, be asymptotically equal to the volume of omega, where omega is the phase space region bounded by this energy shell divided by pi h-bar to the 2L. This is an asymptotic formula. It's hard to prove quite generally. You can do it using wild catheters in certain cases. Hermander has given a proof

7:30 well, not only him but has given a proof of this rather special circumstances and all, but this is not the point here the point is that this oh sorry, N, not 2N I think the point of Hermander was that he wanted to have an estimate for the error excuse me? Hermander's point was that he wanted to have an estimate for the error yes, yes, yes I'm talking here more for physicists I'm going to I'm not going to really really rigorous details. This is just to motivate what I'm going to say now. And, okay, in terms of dynamics or statistical mechanics, what I'll say is that, yes, okay, so the approximate number of states with energy less than E is the number of quantum cells you can pack into omega. This is actually the volume measured in units pi h bar here, okay? And, well, this if you open any book on thermodynamics or quantum statistics Kinshin does that. Well, they use an empirical tool which they call quantum cells. What are quantum cells? They're small hypercubes. Well, with size equal to pi h, pi h bar, and the volume of such a quantum cell is then pi h bar up to the node. It should be the square root here, of course. Okay, so this is, I think, what is commonly called in physics a quantum cell, okay? Of course, things like this here are very few symmetries, okay? Very few symmetries, of course. If you transform this by a linear transformation, the symplectic or not, you will distort that quantum cell, you will lose lots of properties. One thing, for instance, which makes this more or less consistent with the Heisenberg-Gamstack Unimprinciple, is that if you project a quantum cell with a basis parallel to a xj pj plane, then, indeed, you will obtain a square and a square with area pi h bar that is one half of h but as soon as you do something to this little cube as soon as you transform it by a linear transformation well the projection will no longer have this area so i mean it's not very trackable mathematical but you can bound the area you can know the area

10:00 yes but as i'm going to show you can do even better now what is the mathematical object which has most symmetries? Well, it's just a ball. Ball B to N, R. One moment I take the radius equal to R. A ball has entered with many symmetries. Good. Of course. Very good. Balls are very nice things. What happens if I deform a ball by a linear transformation? Well, I get an ellipsoid, a phase-based ellipsoid. Good. Ellypsoids are interesting things, very interesting things, especially if I just, if I take ellypsoids, which I obtain from a ball using a symplectic linear transformation. This is what I'm going to show, and then I'm going to show you what kind of properties we get. Does anyone here know what a symplectic transformation is? Okay. Very quickly, define the objects I would use. Sigma will be the standard symplectic form. standard symplectic form on R2N for those who like differential geometry db wedge dx with an obvious means so called canonical to form observe that I write the p's before the x's here just because this is the differential of the human form exterior derivative of the human form do not like differential forms, we can do it like this here. I just use matrix notation. Okay, don't forget that, okay, z prime would be written as a column vector here, z also, So, and what is J? J is standard symplectic matrix here. Good. Okay, I'm working in one given basis. We don't care about any intrinsic statements here. So, this is the standard symplectic form. And the symplectic group then consists of all linear transformations which preserve sigma,

12:30 sigma s z s z prime equal to sigma z z prime this for all vectors z and z prime but if you translate this in matrix notation this is the same thing as saying that s d transpose d stands for the transpose of course s d j s is equal This is quite simple, everyone. Good. Okay. So why am I saying that it is interesting to transform a ball, an aniphase-based ball, by subtraction transformation? It is due to the following property, which is very closely related to the so-called Gromov's non-sweezing theorem, which gives a proof of it in the linear case. I'm going to show you why. See, let's take a ball here with radius R. If I project this ball, oh, yes, if I project this ball on any plane, I've got a circle, a circle with RFI R squared, okay? Okay, what does Gromov's theorem say, Gromov's non-squeezing theorem, alias the principle of the symplectic canon, Gromov's non-squeezing theorem, unlike in 85, but it says that if you take a symplectic transformation, linear or not, that is if you take what we call in mathematics, asymptotomorphism, and if you deform this ball using asymptotomorphism, thus obtaining some hard blob or something like that, when you project this deformed ball down on asymptotic plane, for instance, XJPJ plane, then the area of the projection will be at least by bar squared, but it will never decrease

15:00 the area of the projection. There's another equivalent statement of Bromo's theorem, which says that, okay, if you put a symplectic cylinder here on the plane, a symplectic cylinder means a cylinder based on a symplectic plane, you will not be able to squeeze that ball inside the cylinder if the cylinder has a radius which is smaller than that of the bottom. But you can do it if you base the cylinder on any other plane than a symplectic plane. All the known proofs of Gromov's theorem are extremely difficult. Gromov himself used what is called the theory of pseudo-holomorphic curves. Viterbo has given that proof using generating functions, not simpler, by the way. And alternatively, you can prove it by establishing the existence of what is called a symplectic capacity. Good. But all approaches are equally difficult anyway. Good. So here is something that is very deep, and you already see that this is reminiscent of Heisenberg's uncertainty principle. If you project down on a symplectic plane, you cannot go below the initial value of a projection. Actually, you can prove that there is also a classical hazard reconstruction principle to this I show in my book. But, okay, so this is Gromov's theorem, quite recently. This is, in fact, because it's possible. Yeah, of course it is, from a topological point of view. Yeah. But I'm going to do a little bit better now than this here. And actually, what I'm going to show you is going to give a proof of Gromov's So you have the linear case. I see the dream is to adapt the method and you have proof in the non-linear. Yes, please. You're actually proving the circle's a minimum size, aren't you? Sorry? You're proving the circle, you're going to prove the circle's a minimum size. You can have, at least it's saying the same thing, aren't you? Minimum, no, no. This is true for any R, any value of R. So pi r squared's a minimum value can be. Yes, yes. Okay, so let's now abandon for a while we're almost here Sorry, I just want to make sure I understand it correctly

17:30 So you're saying if you have a projection of a deformed sphere because of cycle symmetry, for some reason you end up with a minimum area in the projection? Well, that is, the area will, the point is that you will never be able to squeeze using a symplectic transformation. For instance, symplectic, yes. Notice that this applies to Hamiltonian flows, of course, because Hamiltonian flows consists of symphicomorphisms, you see, so if you want the dynamical point of view, you will be forearm, you will be forearm above, but the projection will never decrease below its original value. Now, this shows that in Hamiltonian mechanics you have much more rigidity, much less room for chaos actually, than people believe. If you look at Penrose, this excellent book, you know, The Emperor's Neonite and all this, he says that, ah, Hamiltonian mechanics cannot be true of our word because, actually, and then he takes a metaphor of the drop of milk in the tea which can spread out chaotic, but no, it cannot because of Gromo's theorem. On the other hand, he has an excuse, Gromo's theorem He's only known by mathematicians for 20 years, that's all. Well, Penrose is a mathematician. Even, I mean, only known by topologists. He was a professor here at Birkberg. Birkberg, Birkberg. He's here. No, but I was not saying anything. No, no, don't misunderstand, gentlemen. Okay, let's talk about something after. Good. No, but I think you're right in the fact that not many people know about it. No, no. Very few, very few, very few know. And the algebraists will know, most of the algebraists will never have heard about normal theorem. I mean, it's still confidential. Well, it's not a . Good. So, let's now take our ball again here, B to N, R. Okay. And let's deform it by using a symplectic group, syntactic linear transformation. I get an ellipsoid, okay, an ellipsoid, let's denote it by, well, let's denote it by S-E to the end of it. Well, an ellipsoid is only in a space, like a rugby ball, and of course if you cut a rugby ball,

20:00 any plane, any plane you will get an ellipse with a certain surface, a certain area. If you cut the rugby ball using another plane, you will get an ellipse or a circle with another area and so on. However, a very amazing property is that if you cut this by any symplectic plane, for instance, a plane of contract coordinates passing through the center, well, then you will always get an ellipse with area exactly 5 bar squared. Two requirements. S must be symplectic and I must cut it by a plane, a symplectic plane going through the center. I'm going to show you why there are several. Well, the most elementary proof would be for the mathematicians to say, ah, okay, if I pull back this section here to the ball, I will transform a symplectic plane into a symplectic plane, because symplectic planes are preserved by symplectic transformations, of course. This would cut the ball along that large circle, which thus has area by R squared. Now, the restriction of a symplectic transformation to a symplectic plane is still symplectic, hence area-preserving. That's why we have this . But I'm going to give that proof that it's more satisfactory for physicists than dynamical proof. First, before I give you the dynamical proof. I thought this proof was very satisfactory. Yes, yes. You're a mathematician. It's satisfactory for everybody. But it's also interesting to see the Hamiltonian point of view. But then you see the action integral. I was kidding. Okay. But I discovered that, well, it's only one month ago, actually. And, as you can see, you can immediately get from this a proof of Gromov's theorem in the linear case, of course, immediately pops out. So my secret dream would be to admit also a non-linear case, but there you have a difficulty, of course, because if you cut this by a plane, suppose this is a f of b to n, where f is a syntactylomorphism.

22:30 If you pull back, you'll get a syntactyl manifold, two-manifold, of course. so you can it's not as easy to evaluate the area, however I think that if you use Darbou's theorem because if you can motivate that you can shrink the ball that you can take any radius then by Darbou's theorem the cut, if the radius is small enough the cut will be morphed actually to a plane and then you will still still have Kushner areas there, but I think I could make the proof. Okay, that's another question. So, okay, Gromov's theorem immediately follows because if the section has area pi r squared, the area of the projection of the whole thing on a symplectic plane will have also at least area pi r squared. So, you get the proof of Gromov's theorem. Good. So now, a dynamical proof of this. Okay, here as the energy shell of the Hamiltonian. For instance, the Hamiltonian HZ equal to modulus of Z squared. Modulus, okay. By this, I mean the norm of the vector Z squared, the usual length of Z squared. Good. Fine. This is the energy of this here. For which energy? for the energy, yes, R squared. Now, let's look at this here. This section here. What? Like a mu, it is bounded by a closed orbit. A closed orbit of what? Well, just a closed orbit of the inner shell, which you see here, the ellipsoid, the boundary of the ellipsoid itself, which is the image of the boundary of the ball here good but this can be beautiful I'll turn once I can do this as a periodic orbit okay good what is the what is the action of this periodic orbit well the action is look I'm cutting here by let's say xjpj plane the proof also works for so let's call this gamma yeah so the action will just be pj xj because i'm lying in the xj pj plane

25:00 there are no other contributions from the other coordinates good so this is the action this is equal to fire square of course cause because it's the area well then it's now if i pull this back by s minus 1, then I will get a curve, let's call it gamma prime here. the action along gamma prime will again be pi r squared. Now, it will be pi r squared. Why will it be pi r squared? Well, just because gamma prime is the image by S minus 1 of this of gamma and action is conserved by simpletomorphism linear or not action is a simplet action on closed curves is a simplething invariant so this must be by R squared by R squared as I said same thing again here so what was the role of the Hamiltonian? the whole role of the Hamiltonian was that I was to talk about the action of a periodic course. Actually, the Hamiltonian is a superfluous, but it gives a dynamic approach in the sense that, you see, every large, great circle of the sphere orbit here so when you left all these periodic orbits vary you get the whole fan of periodic orders here but you also get a whole fan of periodic orbits here so if I take in particular the cut here by the xj pj plane I get a periodic orbit which will pull back here disappear over here this is enclosed in an area of pi r squared and since action is conserved well this is equal to this here, you see, turns out that in this case, the action integral is exactly the area, because I only have of gamma, because I only have pj in d3 here. But of course, I mean, a much more concise proof is to say that the restriction of a symplectic mapping to a symplectic subspace

27:30 is still symplectic and is very preserving. By the way, the first proof yields much more, because I just cut my ellipsoid using a plane. I could cut my ellipsoid using any symplectic subspace. And what happens then? Then it happens that this cut by a subspace of the dimension, say, 2k, will be an ellipsoid again. And actually, this ellipsoid will be the image by, as a linear symplectic mapping of a ball in the space with smaller dimension I mean everything can be extended to higher dimension here I'm talking about areas because areas play a very important role in physics and also actually symplectic geometry because symplectic geometry is a sort of area geometry area you see you already see that on the level of the symplectic form itself while written for instance in differential forms is a sum of areas the notion of symplectic capacity which falls from grommos here and has the dimension also of an area what i wanted now to show is that this has much to do actually with the vignettes form of gaussian and the heisenberg answer affinity principles how much time do i have Okay. We have questions here. Please do not hesitate to interrupt me. One question. I know Roth's enormous work, mainly from his work on the isoperimetric inequality. Exactly. Was this what he was writing on at this time? Yes, it was. In his 1985 paper, actually his goal was to establish an isoperimetric inequality. and he introduced these methods of pseudo-harmonic curse, which I'm not very familiar with, or you have all the contributors also, and he's non-squeezing out of this. Actually, this theorem has had much more impact than the isoperimetric things. And actually, what immediately falls from Grommel's theorem is the existence of a synthetic capacity, which I... Have you defined here? Yes, yes, I haven't defined... You haven't defined it in front of us. I have a problem with a symplectic capacitor from the physicist's point of view. I don't see what is a symplectic capacitor.

30:00 Well, I can say a few words and so I suppose this is an informal problem. Okay. A symplectic capacitor is a matching which associates with every subset. Which is also to every subset of phase space, positive or non-negative number or infinity, good. So it has the following properties. First of all, there are several sets of more or less equivalent axioms for symplectic capacities. The first is that if F is a symplectomorphism, what is a symplectomorphism? Well, it's a nonlinear generalization of a methyl symplectic group. It's a diffeomorphism, because Jacobian is symplectic at every point. Then, this is of omega in omega prime, then C, C for capacity here, of omega or F omega is smaller or equal to the capacity of omega prime, that's a kind of monotonicity property here. And in particular, if omega is included in omega prime, then C of omega is smaller or equal to C omega prime. And also, also, C of f of omega is equal to C of omega. That is, this capacity is symplectic invariant. This is the first possible axiom. Axiom number two, axiom of conformality sometimes, is that C of lambda omega is equal to lambda squared C of omega for every lambda, or actually any lambda. So you see that the capacity has the behavior of an area. If you multiply the dimensions by lambda, you multiply the capacity by lambda squared. And third capacity, third property, sorry, C of B2NR is the capacity of a ball is pi R squared.

32:30 All this is trivial. And now comes the non-tribular thing. This is also the capacity of any cylinder, ZJR. What's this here? This is precisely a cylinder based on a plane of conjugated variables and with radius r. You see, as soon as n is larger than 1, the cylinder will be an unbounded set. And this unbounded set has the same capacity as the ball. And the existence of a symplectic capacity, of a single symplectic capacity, implies the problem of theorem because of the third property. But, it's extremely difficult to construct explicitly a symplecting capacity. From Glowman's theorem, immediate defaults that one can define one's symplecting capacity. What's this? Take a set omega in phase space. Good. and define c omega as being the supremum of all numbers pi r squared such that you can squeeze b2 n r the ball inside omega this is something called gromov's capacity gromov's area and so on And actually it's the smallest of all capacities existing, subletic capacities. But the other way around... Do you have any physical feelings for this? Yes, I think that it is a perfect candidate for being an adiabatic environment. A adiabatic environment? A much to make an adiabatic environment? And why this? Because there exists one, well not privilege, but one very interesting subletic capacity, that capacity which can be which can be defining terms of actions of periodic orbits you see if you take any hypersurface talking about a smooth hypersurface in phase space okay you can always view it as as the energy shell of Hamiltonian, right? And all Hamiltonians having the same energy shell have the same periodic orbits. This is a well-known fact in celestial mechanics, for instance. It's rather, yes, okay, I see that

35:00 you're surprised. I'm going to show you. Yes, I was also surprised when I saw this for the first time it's the proof goes back to Morse actually see I take let's take a region Omega here with smooth boundary the Omega C infinite manifold and so on okay assume now let's assume that the Omega is this represented by an equation H equal to E, okay, energy shell, give it, all right, and suppose that it can also be represented by another equation, K equal to E or E equal to E prime, this doesn't matter, okay, well, then both Hamiltonians have the same trajectories on the omega, and the same trajectories or periodic orbits or semi-orbits with different velocities you see? oh yes no no no just because the Hamiltonian vector fields are the tangent vectors you see what's the normal vector to this it is lambda lambda h calculated point z sorry? No, no, no, no, any dimension. Look, look here. The normal vector at a point, let's say at point Z, is the gradient, okay? This is a normal vector. Let's multiply this by J. Ah, well, then I get a vector, which is a tangent here. This is the Hamiltonian vector field. So, this vector is XHZ. if I do the same thing here I would take X, K, Z but they are both proportional so after a reparametrization they are exactly the same I was surprised also because it's not so obvious perhaps immediately so it doesn't matter about which Hamiltonian I talk in other words so the Hovett-Sinner capacity says that yes, okay they construct it actually for convex and compact

37:30 They assume that this is a convex and complex set. And they say that we find a capacity for such a set by saying, by... Thank you. Bye, Robert. Congratulations. Thank you. By saying that their capacity is the action of the minimum periodic orbit lying on the boundary of America. They show that this defines a capacity. Yeah. Well, the orbit, which has the smallest action integral along this boundary, if you take a convex and compact set, you know that there will always be periodic orbits. Periodic orbits is extremely sophisticated and difficult anyway, and, well, I want to touch that too much still lots of open problems, but it's important in the semi-classical mechanics or things like this. Well, so what was your question about it? What shocks you here? Nothing shocks me. I was just trying to get a handle from the physicist's point of view of what the capacity is. The universe is a hologram. Why am I defining it? I think it's called capacity because of analogs with potential theoretic capacity. Certainly, certainly, certainly. I don't know why it's defined, really. Potential theoretic capacity is going all the way back to capacity of capacitors, so when you do electrostatic stuff, but I can't see the analog here beyond the very formal one. So there's the minimum theoretic orbit on the boundary here, though. All is for convex and compact sets. In the general case, they give a definition which is much more sophisticated. It's in their book, by the way, I think, Hamiltonian Dynamics, Hamiltonian Dynamics, published by Birkheuser-Ferrag in 1992, something like that. It's a very good book, actually. Okay, it could have been a book, but it's a monograph, actually, with lots of good ideas and deep things. there. Okay. So, you've opened a little window. Yeah. Yeah, I'm not sure because it was

40:00 static capacity. I never really understood that spirit capacity in the physics sense. So, I guess this is something mathematicians, you know, sometimes mathematicians have this, they see something, and they take hold of it, and they run away with it. Well, you know, mathematicians often misuse terminology. Sure. Who cares? Like, geometric quantization. Well, I guess, potentially, that now in the meditation center, I don't have a hold of it. Of course, of course, of course. So, do you not want to hear about my talking about something? No, no, no, no, no. This was just feeding, trying to get something that was puzzling me when I read your paper the other day. No, but all these things are new, I mean, since my work. Today's talk is something I have submitted, well, part of it in Physic Letters A, and in the Journal of Physics A, Math in general. That's the paper you sent me. Yes, yes, yes, yes, yes. If you need a file, I can't tell. I can read the papers you sent to Morris. Okay, okay. You might not understand them, I'm not doing anything. No, no, this is okay. This is okay. So what do I call a quantum blob? Well, it's something I denote by a blackboard cube here, and which is simply the image by any phase space balls centered at Z0 and radius, radius equals the square root of h-bar here, S in S-V-N. Good. And in view of that property I explained, I proved here, the section, the section of Q by any symplectic plane for instance a plane of conjugate variables a symplectic plane through the center through S of Z zero is an ellipse with area equal to pi H bar one half of h. That's pi r squared here, of course, yeah. Good. So what can one do with

42:30 this stuff? Lots of things. If you have questions, or if you are dissatisfied, please, um, I've never got any way with it, but this is pushing it forward. let me know see there's a very close I was trying to work these blobs out in 1968 because in 68 I hope this is not a misuse of the English language you see blob is something a cell is something small when a blobs can have enormous spatial extensions of course because you can extend very much one dimension and squeeze and all the other as well I was doing it actually and this is very consistent actually with with the uncertainty principle I'm going to show you very quickly if I have time that that is a projection between what you states, Gaussian states, and quantum blobs. The deum of a quantum blob is the same thing as the deum of a Gaussian state. And I need to recognize this here. So let me, and by the way, another interest of the quantum blobs, but they are new from me since I invented this, and only one month ago, Wiener function, on the quantum block you get something positive. So they are in a sense really quantum cells, the smallest objects with which you can coarse green phase space. Did I say something stupid? No, no, no, it's just that I've got a couple of papers which we could marry. Okay, let's do it. Let's do it. But let me first show you the thing. Okay. Um, um, um, um, um, um, um, um, um, um. Where are my... Okay. Yeah, yeah, that's exactly right. Yeah.

45:00 Remember my 2.5. It's essentially the same, you know, much cruder way of... Much cruder way of doing, but this is inappropriate. what did you do with my Gaussian okay okay yes let's take a let's take a function psi of x because I put the letter psi because it has a connotation of being that wave function in physics what we're is exponential of minus 1 over 2 h-bar, and then x transpose large x plus i y, x. Okay. I take a Gaussian like this one. It's normalized, and we normalize the integral of the modulus of this square is equal to 1. It's normalized Gaussian here. X is a positive definite matrix and Y is just symmetric. X and Y are real matrices. Good. Very well. Okay. And you really do mean that X plus IY is XT and X, they're the same axes? no no this is unfortunate I should have taken a and b but all right okay fine well as long as I know I know what's when you type that you see it on the blackboard okay okay okay fine now I take the I think the Wigner transform of this. Wegener transform, just in case you don't know what this thing is, okay, is equal to this here integral of the exponential minus five. This is what Schoep is doing. Well, you have many people who are doing this because the Wegener transform is, of course, very closely related to the Vibe calculus and actually it allows one to associate your function in X base a functional phase base good Peter Holland

47:30 hates that because he says it's a totally a page of 145 I this in book He was one of my students, so I should know what he's talking about. Okay, okay, okay. I'm confused by the little x's and y's, they are what the integral is over. We've got this convolution type object here. Yes, exactly. It's a mixture between a convolution and a Fourier transform. Small x's and small y's. No, this is the general definition of a vagrant transform. The x is fixed and p is fixed, so the integral will depend on x and p on, yes, on x and p. I'm integrating in y. I'm only integrating in y here, right? See, if I take t equal to 0, I get a kind of convolution. Sorry? I have two X's. Sorry? Physically, you can think of the X as a sort of a mean pointer, the Y's as the difference surrounding the block. So you integrate overall. No, it's just a conversation. Sure, but I'm trying to tell you what physics behind the thing is. I just need physics. Okay, yeah, well. Wigner, sometimes it was called the Wigner-Moyal transform. Actually, the point is that it associates to a function in x, a function of facepa. The point is with this, this is always real. This is trivial, okay? I think that physicists, by the way, put the bar on the first term. It doesn't matter. This is always real, easy to see, of course, because if you put a bar on all this, the bar comes here. It amounts to changing y to minus y, but then you have a change of sign here. It's a real function. And the interest, actually, of theta function comes from the fact that, you see, if you integrate W psi z in the NP, well, then you get the modulus of psi squared. Of course, you must assume that W psi is L1, L2 and L1. So psi could be any way from G. Yes, yes, yes. This is independent of the form I have here. It's surfaces that psi is in L1 into section L2. And if you integrate this in X, then you will get the square of the modulus of the Fourier transform psi.

50:00 So that's why physicists call this a quasi-probability distribution, okay? But it's not a real, a true probability distribution in general because it can take negative values. Actually, that's a theorem of Hudson, 20, 25 years old. This would say that this is positive if and only if Psy is a Gaussian. It's of this type. You can also have lower terms here. This doesn't matter. It's positive for Gaussian. Sorry? It's positive. I didn't know this. Yes, yes. Yes, it's positive. If and only if psi Segocian. Otherwise, it can take negative values. Aren't you going to need some extra condition like for all dimensions n? Yeah, this is true for all. It is true for all n, provided that this here, that you have this condition here. Okay, you have to be able to perfect. Well, actually, what you should think of, take side in a short space and then you're happy. Yes, I would be happier then. So if I think of Psi being a Schwarz-Fitz... No, you can't prove it, actually. The proof is rather well known. It has been done by Folland and other people that suffices to assume that Psi is in L1 intersection or two. But it's not going to make a big difference to just do it in Schwarz-Fitz. No, no, of course, of course, of course, that's what I'm doing. Did you say that, perhaps I misheard you, if I take my psi, and the theorem is that w psi is a non-negative function, if it only psi is a Gaussian? And it also, actually, constant times exponential of ax squared plus bx, something like that, and it will always be positive. And if you find that W psi is positive, then time must be of this time, yeah. The reason I'm asking is because of the Bochner theorem. Ah, you're talking about both my integrals, yeah. Well, if I think of this just as the Fourier transform of the psi x plus half y psi bar x minus half y, W psi being positive is equivalent to the function in the integrand that your Fourier transform is being positive. It's very surprising to me that should imply that Psi is against you.

52:30 Oh, you know, there are hundreds of proofs of that. This really has been really well known since 1965. I can hear you lots of references. Actually, well, if Psi is a Gaussian, you have to do some calculations. It's not very difficult, and you end up with something which is positive. It's a little bit more difficult to prove that. If it's positive, then you must say it also. Actually, the first to prove this was Hudson, probably, and then you have Janssen, and then you have and so on and so on and so on. Well, the usual modern, so to say, reference is in Gerald Follin's Harmonic Analysis in Base Base. But I think that we are talking about Hernández, even take a look at Hernández's book, I mean he, I don't know whether, if he calls this the Viva Transform, he calls it something else that he produces in her, so, I mean this is really very, very well known, it has been well known for a long time. And actually, you were talking about Bohmian integrals, it's right, Mr. Fallon does the right to place, he sees the integrals from one of your Bohmian integrals. No, he was meaning a theorem of Bohmian, which characterizes functions which have a positive political form. There's a classical theorem dating back to the 1930s, which effectively says... Yes, this is a classical theorem also, so you find it's not a small marginal result. I mean, you find proofs of it everywhere. It's just... So the statement of the theorem is, if this W psi is positive for all z, then psi is Gaussian. But the other way around is very easy to see because you just gather for a chance to answer. This is not the point of question. Good. Can I make a trivial comment? My question, which is a question of facts. Do you know that quantum potential is in my own paper? No. Already there. I mean, 49 then. Three years before Bohm. Oh. Bohm didn't fight. I'm joking. Okay, let's...

55:00 It's not written as such, but it's there. It's implicitly there. I have my newspaper, so I have it. You haven't read my paper, have you, Maurice? Sorry? I show it in my paper. because they're in hidden form good okay so now let's suppose that psi is a gaussian like this if you do the calculations what you find is that you get a constant I don't, rather complicated, I don't write it. Then determinant of x minus 1 half and then x minus 1 over h bar z t g z. So this is the vaginal transform of this Gaussan. And what is G here? G is a matrix. X plus Y, X minus 1, Y, Y, X minus 1, X minus 1, Y, X minus 1. But this is what you get after performing about Triegel, but you're rather boring calculations. Now, two things. First of all, this is easy to check. G is a positive definite matrix. positive definite, easy to check, because if you calculate ZDGZ, you get XT large X, X plus YX plus B transpose X minus 1 YX plus B. this is obviously positive because capital X here is itself a positive definite matrix okay so you have a sum of two squares one of them being really positive definite another property which is more unexpected in the sense is the G is symplectic actually you can immediately check that G T J G is a

57:30 So this is a simpleton matrix. So when your Wigner transform a Gaussian, well, you get a Gaussian, of course, but you have something to look at. Now, let me perhaps state, if I have time here, that I have a theorem made here. Okay, so let me, yes, okay, perhaps, the mapping, let me write it here, the mapping, Q dot which to Gaussian Psi associates, oh, you noticed that my Gaussian was centered at the origin, but this is supposed to be relevant here. Gaussian Psi associates the quantum block Q equal to S of B to N square root of H with S defined equal to SST minus 1 ejection. Okay, the proof is almost trivial if you know some ways of factor, factoring, mappings, and so on here. See, S, which I have here, okay? I can define, since G itself is sublective, I have no problem in defining the matrix S, of course, using power decomposition theory.

1:00:00 Of course, you might argue that S is not uniquely defined. So here's a little remark, of course. the unitary group UM can be viewed as a subgroup of a separate group and how that? if to a unitary matrix A plus IB you associate a real matrix A minus B Well, then you obtain a bijection of U-N onto a subgroup of S-B-N, which is then identified with the uniget group itself. Good. No, the S above is only defined up to an orthogonal transformation, well, unitary, sorry, unitary, to an element un. Why that? Well, you see, this is because un is actually intersection of 2n. You're talking about a group into invariables. s is one good inverse square root of g, so to say, here, then with u in um will also be one that this does not matter at all for our mapping here why because s u applied to this ball is the same thing as s applied to this ball the elements of um are rotations symplectic rotations so if you rotate a ball you just won't change anything so the mapping this way is well defined now how to prove uh yes Okay, how to prove the existence of this, of this S here. Well, you can do it very explicitly, actually, here. Okay, I believe perhaps the formula here for G. Oh, yes, okay, like this.

1:02:30 This, you can, I'm going to try to do this, okay. Okay, yes, I do not have very much time left there. Oh, you just take the time you want. Oh, well, it's late and it will start here. Well, you're tired, I know. Oh, okay. Okay, good. All right. So, wind up when you're ready, Boris. Uh, yes, okay, right here. Hmm, yes, what was I doing here? Ah, yes. Well, yes. The proof of the subjectivity and the injectivity actually relies on the following fact. If I take an arbitrary symplectic matrix S, there are different ways to factor it. One way, which is used from here, central here, is to know this, that every symplectic matrix, Every subjective matrix S can be written as a product A0, 0, C0, A0, minus 1, and then a unitary matrix of this star. How do you do this? Okay. So for every S in SPN, there exists A0, C0, and A and B, so we have this. Of course, we in addition have this condition here, A0, A0, T. not really, of course, and A zero C zero symmetric, that is C zero A zero T. Why do we have this kind of decomposition? It's rather elementary. It can be viewed as a kind of pre-equals of a decomposition, but it's much easier to see it another way here. You see, take Lp equal to zero R and P. This is a Lagrangian plane or the vertical polarization vertical Lagrangian plane. Good. What is, what is the stabilizer of LP? Well, stabilizer of little groups, I think, also isotope subgroup or so on. The stabilizer of LP consists of all matrices of the type A0, C0, 0, no, C0, sorry, A0, A0, minus 1.

1:05:00 Oh, this is immediate to check. The stabilizer of LP in the symplectic groove. Okay, that's a show. Good. So, on the other hand, the symplectic group acts transitively on the set of all agronomic planes of the agronomic mass management. So, this means that for any matrix S, you can write SLP equal to, let me see now this, I don't say anything stupid. this will be equal to the yes, okay, sorry, let me take that once again the symplectic group acts transitively on the set of four Lagrangian planes so does UN, the unitary group, also acts transitively on the Lagrangian classmate, so if now I take if I consider the image of LP by S There will exist some U in UN such that SLP is equal to UNP. This implies, of course, that S is equal to U modulo, an element R, where R is in S DLP. Okay? So, in other words, S is just a product of a unit of a matrix by a matrix like this. Now, yeah, this is in the other way. Specific. get it there's a specific factorization yes yes yes yes of course of course yes yes yes of course but then that's equivalent to what you want so you're done all right that's very clever well well now that this is done look i can find the s i need from this here by just putting that this is to SSD minus 1. It gives me equations. I can just extract A0, C0, A and B from this. I don't know the A and B. I don't really need. And what do I get as formulas? I'm sorry, I'm going more and more messy here. This is Oz's fault. What I find that he

1:07:30 get is that x is equal to a zero minus two and y equal to minus minus a zero minus one c zero t well this is just your technical stuff and you extract this now the point is that we have got here i just finish quickly now because now enough is enough we have a bijection between quantum Gaussians, centric Gaussians also, and there's a very obvious relation with the uncertainty principle, but perhaps I will tell you that next time when you enlighten me again. Okay, thank you very much. Thanks, Morris. Or should I say any questions? I think Morris has had enough, actually, because he's been examining all day. And then I drag him in here to... No, it's okay. That's okay. Thank you again very much. Thank you all for your questions also. I can say I was way too tired, couldn't follow me. Yeah, me too. Well, I was having a follow-up bit of it because of having read his book, but that's the only way. And, you know, I couldn't have too tired. Yeah, so there's a similar shit up in there. That's a very nice surprise. You didn't get it? No, really? And there's one of those things. What do you think? Any positive. That's the thing. Oh, no! In other words, the major things are the same thing. By being positive. That's exactly the same thing. Every semester

1:10:00 And what's the definition of a triangular? Triangular, it's just one with zero on the triangle, it's lower triangle, it's the other way round. Geomatically, what it says is that the unit of the first J basis factors is contained Yeah, I can see the connection there, but it's stumped. Rose and I have just got to discuss this. I hope so very much indeed. Thanks a lot. You know, from pure mathematical, it's not functional analysis, it's really, you know, it's just a lot from the back of it, but it's not the truth, I mean, a lot of people I know, you know, in connection with the physics, and we'll obviously with this, and so I think, yeah, that's a lot of stuff, and that's where we're going to go, and that's where we're going to get you, but it's just something that came to us, because these do seem to be coming up the same. Thank you. Are you going off to eat tomorrow? Yes, yes, yes. Even me. Would you be my guest? I'd be more than welcome if you'd like. I'm in. You already have.

1:12:30 No, you know I've got my wife. But in that case. That was excellent. Thank you. With luck, I'll see you in Sweden for each time. I'm not certain. I have to try to get something organised to cover a meeting in Canterbury, which is taking place at the same time. If I can get something to cover that meeting for me, then yes, I would very much like to get to Sweden. But I'll know in a few days, and I'll certainly let you know in a few days. I would very much like to. Thanks a lot. Listen, I'm sorry. Yeah, I know we did. You're absolutely not. So why don't we look at me? Well, I'm going to Cambridge tomorrow for two days. I'm going to be back in London. I'm going back in London on the 2nd and I'm going down to Bristol to stay with a friend of mine a guy called John Mayberry and then I'm not sure if I'm going to Sweden I might stay around here and go directly out to Sweden on the 5th rather than going back to France because there wouldn't be much point in my going all the way back to France in which case just depending on how busy you are we could maybe get together on the 4th or the 5th and what I have to do is talk to the guy well there are two employers involved in this one of whom is an accountant well just give me a bow to that what I'd like to do if you're willing is to pitch your brains to get some advice Yeah, go, go, go, go, go, go, go.