Clifford Algebras, Quantum Blobs, 2-Time Theories — Part 1

0:00 completely gone and some of us were never there to begin with. The point that I'd do was just go through some points, summarising it as I see the work I did with what people have done. So I had a request that I'd go through more slowly from a different point of view, just to contrast the two views. I've got a presentation here which is not up to scratch, so please forgive me if it's a little bit better. Some of the mistakes you've had on yours, it's going to be even worse on this one. Okay? Well, I'll make you feel good at that. When is it all bad? No, no, no. I just noticed you were embarrassed, you know, putting your hand over things that shouldn't have been there remember what I was trying to do I was really doing my talk was really three messages I wanted to get first of all I was wanting to motivate why we should be interested in cliff and algebras from the point of view of the process but of course cliff and algebras are there in their own right so if you don't want to find the idea of principle There's the Clifford Algebra. The next thing was that, contrary to common belief, the Clifford Algebra is purely classical, and all this notation that goes with it is purely classical. And somewhere quantum mechanics comes in, the question is exactly where. And the where, for me, was when I added to this Clifford, the orthogonal Clifford, then I added in the symplectic Clifford and that's what Ray talked about a little bit and I introduced, in fact I went through the symplectic Clifford much too quickly, in fact I didn't go through it until Barbara asked me what was I supposed to be doing with it then I had to go through the other history but what I wanted to point out really was that there is this beautiful hierarchical structure in the Clifford Algebras and we all with the power spin matrices is really a cliff analogy where we're living here. And we know that Dirac is living here. And also, at the top, we have the conformal where all the twisters are.

2:30 But what I didn't realize, perhaps I'll just confess my own ignorance, was that you could actually do a whole of Schrodinger particle mechanics in this structure here over the real fields. You realize it's a mystery where you're in that one. That bracket shouldn't be the italic. Oh, if you're only worried about that. Yeah, no. OK, yeah. So the fact that all of these sit in front of each other, sit on top of each other, is great, because we can go down from the Dirac, which we had to introduce. I mean, Dirac sort of discovered it almost by accident, or rediscovered it almost by accident. Then you go to the non-relativistic with spin. You've got relativistic with spin, non-relativistic with spin, just non-relativistic, and it's nice if you can go right down through the Clifford algebras and see Schrodinger sitting at the bottom. But the other thing that I found rather interesting was that when I introduced this into a Clifford bundle, and I brought in the left and right connections, and used, in fact, I went smuggling in the Schrodinger equation in its dual, but But the equations I got was essentially just the equations that, in this algebra, Bohm started with the constellation of probability and essentially of quantum Jacobi equation. And then there's a natural sequence. Best way to look at it this way, you get a quantum potential for the quantum Jacobi equation. You've got a quantum potential here. You're saying when you go to the limit, it's exactly the quantum potential that turns up here. and it's exactly the whole potential that comes in here. So there's some essential feature there in this structure. Now then, for people who are physicists, let me just put this in. What I was using were the generators as written in books like Kumrol, and other who use these as the basis elements. To the physicists amongst us, this generator E1 here is playing the role of the square root of E1. Here are the power-e-spin matrices. Here are the gamma matrices. And inside this structure here, we have the twisters. Now, what I want to show you is the relation between these and the twister symbols that Roger was using.

5:00 So let's see the picture from this point of view. Okay. Now the way I started this particular presentation was really via the physics. When I was doing this I was thinking about neutrinos and then adding mass in to get the electron and then going on to the twister. So I had helicity built in right at the beginning, which is different from the way I presented it, whenever I presented it the other day. When I started from the mathematics, the structure of the clippin algebra, and then builds up from there. And the simplest thing about neutrino is that the only property it has is spin a half. No mass, no charge, nothing else. So then the question is, how do you describe the neutrino? well you've actually got this helicity operator which has eigenvalues plus one and minus one since the neutrino's moving at the speed of light it has a spin at half you either get the spin the spin parallel to the direction of movement or you get the spin anti-parallel to the direction of movement and what Roger did with his twisters was to use the photon case all zero-res mass cases always start with And then, one of the two equations put the energy in, because energy in this case is just plus or minus mod p, because m is zero, and for energy greater than zero, you get these two equations, one is an anti-neutrino equation, the other is an anti-neutrino equation. and then what you do is you say let's change notation because Roger uses a different he's been using a different notation from me and here is the P now unfortunately I've got my sigma on this side and it should really be on that side but that's the U dot U A and sigma and this is the introduction to the dots well the dots are at the moment can you go back to this side I want the dots in the... Oh, well, the dots. No, but even without the dots. Okay. They are not dots. Well, yes. No, here. This is the important thing. If you're worried about the dots, that U describes the antinutrient. That B describes the nutrient. Okay?

7:30 But now what I haven't said is because these are power-based spin matrices, they're two by two if I'm finding a representation, then this U is going to be a two-component object. And then what was R, H, L, U, L, Psi, R, R, U, A dot? Well, here? Oh, thank you. This is what I didn't spot as I was talking. I'm now going to change the notation. Okay? And I'm going to call what was called U here a right-handed spinner. Forget about my previous talk. This is right-handed spinner. and it's going to be denoted by u upstairs a dot and the left-handed neutrino is u downstairs it has the right transformation properties now you might say how did I know that I know that because I've been through the whole calculation and dug it out so that we can go from one picture to the other without any problems thank you for slowing me down Okay. So now then, what we want to know is what the left-handed and right-handed neutrino equations look like. And I attribute these to vial. They're called vial spinners and that's why I use them to vial. So all I'm doing here is taking the dotted and undotted notation and showing how it's related to what I've been doing there. I believe historically Dirac found the first, although he didn't publish it. Sorry, found what first? He wrote a vile neutrino equation. Did he? Well, he wrote a paper much later, which he referred to, I think it's called Pretty Mathematics or something. And he has an equation which I think must be the vile neutrino equation. I'm not sure. I wrote these things years ago. No, that's the conventional point of view, what do you say? Oh, I've got this right, haven't I? No, I'm not complaining about it. It's just that I think it's probably true that Hamilton knew it, actually. I wouldn't. The trouble with Hamilton is he didn't even know there was an electron. But that's maybe... You can write down the equation. Yes, I like it. Because... you know he had he had he had essentially the beginnings of quantum accounts with the iconol and all those relationships and but he just didn't know what the electron was so he couldn't push it through but he did have the square root of the uh of the the classic yes yes yes yes oh yes

10:00 in fact you're right the linearization of the quadratic form has been known for a long time I'm not going to quibble it, I'm not... No, no, no, no, I'm not going to quibble it, yeah. Okay, and then, again, just building up to the matrix representation, again, linking with what Roger did, then you find that you get this capital X, A, A dot, which stands for this object. I'm doing matrix representation there. My previous talks on the term not to put matrices in anywhere. But this is now putting matrices in. So the determinant of that x is just the Lorentz metric. And that's using this. When we get this distance equal to 0 down here, then that puts a condition on this. And I think this tells us that these things are moving. The particles are moving with the speed of light. So if they have light marks. It's not 1 over 2. The second line, this is not there, too. Well, I've got it up here for some reason. That's right, yes. It's 1.5, yeah. It's 1.5. All right. Yes, you're right. I'm sorry. No, you can feel even more comfortable, right? No, you're right. No, I'm sorry. Well, that's okay, Basil, because you've changed your notation halfway through a formula so often anyway. I know. I'm sorry. I'm sorry. I have a good excuse. I'm dyslexic. It's a nice word to use whenever I get into trouble. OK, so then the next step is to say, so what I've got there is the C01. And now I'm saying, let me go one stage further while I introduce a new generator. And these are the quaternions there. OK, so I'm going to represent these by power spin matrices. But now I want to go to the language I've been using lately. I'm going to create my spinners from left ideals and right ideals. And the way I do that here is because I'm not from Neumann algebra type 1. I look for my idempotents, and the Clifford algebras have plenty of idempotents. And that was the comment that Ray said, I couldn't find the idempotent in the Heisenberg algebra because it's nilpotent. If you've got a nilpotent algebra, you don't have any idempotents. okay so then the input here there are just

12:30 well I say there are just two of them there are a whole equivalence class of them but they are always of this nature plus whatever it is here minus whatever it is here so you get the unity and then you've now got and this is the interesting thing you've now got two left ideals and two distinct right ideals in this and then when you construct an element of your ideal on this ray. If you construct the element of the ideal, then you just have A times the E11, E21, and so on. So you've built up your left ideal, and I've translated these ideals. I've got the basis of the ideal written down here in terms of the parallel spin matrices, if you want to convert these into parallel spin matrices. And then I do this here. In terms of matrix representation, They're not yet powering spin matrices because the powering spin matrices, in this particular way of doing it, is you bring it here. I've got an eye outside here. I've got a complex representation in this particular case. And then your left ideals, the two left ideals you've got, essentially fill in the columns of the matrix. But of course, you can actually have an ideal with all elements in here. is, where the two elements are just down one column of the matrix, and your left ideal two fills in the other column. And now you can show that in terms of Hilbert's spaces, these two things are equivalent. And that's where you just use the one spin or something. The reason why you can use just one spin matrix side one upside two is because these two things are equivalent as far as the quantum mechanics goes. But they do rather interesting things. And then the right ideals, well, then they just come in as the rub elements in the matrix representation. And then if you multiply an element of the left ideal with an element of the right ideal, so that's a left-handed spinner and a right-handed spinner, and you just construct that object there, which is where I started. So what I'm doing here is tying up a notation. Hopefully you can see how it all fits together. Well, I've gone backwards. No, I'm not. I was using my other device,

15:00 and now I thought I'd go back to a fail-safe way, and I still can't press forward and backwards. Okay, so now let's go up to the Pauli algebra, which is a bigger algebra. The quaternions, I've got two generators and a third element. In the Pauli clipper, I get three generators, and then I generate an algebra of eight dimensions. Okay, so this is a much bigger algebra. primitive impotence. The reason for the primitivity is I'm looking for irreducible representations and that's the equivalent thing that came here. I then once again get an element, get the basis from my elements of my left ideal when I choose these two particular impotence. And remember we've got choice here. I've chosen a third component whereas I could choose either an X component, Y component, There's an equivalence class of ideals there. And I've got my right ideal very easily constructed there. And then we very quickly tie this in with the flag picture. Because if I just multiply an element of left ideal with an element of right ideal, and look at what I've got, I simply get a vector plus a by-vector. And then here's the vector, and the by-vector is the flag. But the way I construct it is I look at it as a stereographic projection with projection through the north pole there and I've got my flag is perpendicular to that. And then when I write it out in terms of the wave functions I get these results here. So this tells me what my wave functions are in the initial attention. Now, how do you get the Dirac spinner in this? Well, what you do is you take your left-handed neutrino and your right-handed neutrinia, and you couple them with the mass term. So you've got U, V, V, U, so they're coupled equations now. And then when you just manipulate these things, you can actually put it in the form that we're very clearly familiar with, with the gamma matrices. And then we find that the wave function for the Dirac theory is composed of two, two-dimensional weight functions, if you like to call them that. One is left-handed and the other is right-handed. So the Dirac theory is essentially coupling

17:30 left-handed and right-handed spinners together, and I'll use the, I think this is the standard representation, I'm using the, that's Dirac's representation. That's one of the problems when you're using representations, you've always got to decide which representation you're going to use because there's a number of different representations. And this becomes more acute in the Dirac theory. Okay, and then finally tying it up with the pictures I put on the board last time, the left and right are also related to the future and past light currents, so you can get a It's a complete geometrical picture of those spinners. Now, I now want to move up to the conformal group. And the definition of the conformal group consists of the following transformations, a translation, a Lorentz transformation, a scale transformation, and an inversion. and that's my conformal group and then in the standard representations I've got the generators here for these groups and then I've got the Neartable Structure and yes I've got this rather messy Neartable Structure there, okay so what have I got I've got the Clipid Conformal I've now got six generators in this And I'm now writing this in terms of betas, because I think someone who influenced me in this uses betas instead of gammas. So you've got signals, gammas, and betas as you go higher up. And then you've got the, here's the metric for you. In this case, the metric, I'm sure I've got something wrong here. Yeah, I've changed my notation again. I've got my, this is my time, which I'm now calling for. or occasionally I go back to zero and call it time, or sorry for the confusion, the minus. No, no, no, sorry, forget the feature, you put it right. That's the time, and here's the spatial parts, and then I've got two more components, four and five, which I've got to be able to deal with. Now, if I introduce my mapping onto a manifold, from the algebra onto a manifold,

20:00 then is this object as being the invariant under these transformations, and this is a vector instance space. And we'll see that this is a projective representation, which is why Roger had to use those projective things. And the matrix representation for this, I'm just putting it down here again, that's the matrix representation, the relation between the quasars, the signals, and the cameras. Now, this six-dimensional sphere I've got here, this one here, it's part of the program which was known as the Hypersaurical Geometry. Is it a sphere? Sorry? Is it a sphere? Is it a sphere? A sphere, yes. It's a sphere in six dimensions. Well, it's the hyperloid or whatever, whatever the equivalent, where you've got two negative signs in your metric. If they were all positive, then it would be a sphere. But I think it originally came from, was it Klein or somebody who was looking at hypospheric geometry? That's ancient history, which I can't remember. Anyway, let's have a look. how do we relate spheres in space-time or in space with this object here well if you write down a sphere you can write an equation in following form if it's centre is at a and its radius is on up expand it and then introduce projected corners where these excises are the excises on the previous slide and I'm dividing by a where a should be the sum and fifth component, and the B in here is the difference between those two. So what I'm using is the projective coordinates with respect to A, shove it in here, clear up, and then what I get is that this equation here becomes this equation here. And this is why the spherical side comes in, because I'm describing this sphere. So it's a sphere in this space, but it's a sphere where the origin is at some other point and now in terms of this notation by infinitesimal generators, I might have to use them later on just very simply express them in that form

22:30 the genes are sorry? the genes oh they're the metric the one which is the abstract I put up here, which was plus, plus, minus, minus, plus, whatever order I put in the image. It's just that generalized method, six metric. OK, and then we've introduced these projective coordinates. Do I really need this? This is just to show, I want to have to show it anyway, that it goes, this group turns light rays into light rays. Oh, I don't want to waste time on that. Okay, now I want to pick out the generators of this conformal group. Here are the translations, here is the scale transformation, and here are the inversions. So I've now changed the notations that I'm using in these excites rather than the original X and Ys I've got. And I'm doing that because I want to find the spinners in my conformal geometry. So I start with my sphere, I'm still calling it a sphere, my sphere in the sixth space. And the spinners are always defined by linearising the quadratic form. and then this upside now is the spinner in the conformal group and then what I do is put it into this equation here with the matrix representation this is all matrix representation I'm doing there and now I get these equations and you'll notice now this object is an 8 dimensional object and it consists of two left-handed spinners and two right-handed spinners and we'll see those are the spinners which relate to the twisters in a minute and then the origin is when x0, x1 is equal to 0 and that corresponds to this relationship here so the origin comes about when psi4 minus psi5 is equal to 0 so that's the origin in the projected coordinates now then, at the origin, if we go back again we should be able to see, suppose we put

25:00 this equal to zero in here what we'll find there is the following that these two objects are zero whereas these two objects are not zero, so now at the origin light rays where light cone is now described in terms of one left-handed one and a right-handed one, which is where I started when I put them together in the direct case, but now I've got them in the eight-dimensional representation. And then elsewhere, when I don't put x0 equal I've got a relation between a left-handed 2 and a right-handed 1, and a right-handed 2 and a left-handed 1. So I can now rewrite my conformal spinner as left 1, left 2, r1, r2, right 1, right 2, and then they split into 2. with one you can't see these things we have the two direct spinners but I can also write them in this form L1 R2 L2 R1 and these are the twisters I haven't shown you yet but that's the right because when Roger first put this in front of me we had these objects with just two objects in and I was asking him what's the difference between this object with just two elements in But in fact, they're contained in the 8th dimensional representation, but they're just picking out different pairs of them. And what I'm going to show you is that this actually translates the light cone from here to here. So you've got a left-handed relicone here, a left-handed one here, and you're relating them by means of this x. Now, how do you do that? Well, in the clippant algebra, you have a translation operator. So if you write the translation operator in this form, then your p is just equal to the betas. And now if you apply that to this, this is the full twister.

27:30 I call it a bind twister because it's got both of Roger's twisters, semi-twisters, you use semi-twisters in this language. Yeah, you're using ambi-twisters in our language. Oh, okay, I'm using the eight-dimensional stack. Yes, yes. Well, the ambi-twisters, if they satisfy a relation, they'd be on the, never mind, they'd lose one dimension on the scale of product. It'll appear somewhere. Yes, don't worry about it. What I'm trying to do here is just to show the relation between how it appears from the Clifford Algebra and how you deal with it, just to show the structural relation between the two sets of arguments. I find this much more enlightening but because you created it from what you were doing, you found yours more enlightening. It's just that I recreated it in terms of this language and I felt much more comfortable with that. I think I just want to catch, I mean these aren't clippant algebra elements, are they? I'm getting matrix representations of the elements of an F ideal. It's the ideal, okay. These are ideal, yes. I picked out an ideal. They're the vectors. They're the vectors in that. I'm sorry, it's very obvious, but one or two people wanted to see the relationship between the two. It's always obvious when you know the answer to these things. So what you do is you take this bi-twister, operate with this, like the matrix representation on it, and you find you get this. In other words, this displacement operator displaces the two Dirac spinners so that you leave this one. If you start with this one at the origin, it will remain at the origin, and you shift the other one, and that's the picture that I had here. That shows you how I get this picture from okay and then so what I've got here is I can now pick out a pairing left one and the translated left one right one and the translated yeah the translated right one which is which of course is right two in the general thing and so I've actually got a pair of twisters here and these

30:00 That's an incredibly fine industry. I'm sorry, yes. It's an off the line. I'm trying to look at them here to see what they are. As I said, I did this a long time ago, so please forgive me. Now the reason why you can split this in half is because you can go to a representation where the beta matrices actually have zeros down the diagonal, so you can actually pull them apart in this language. the spinner splitting into the two semi-twisters, which is your twisters, and here's your set up that you were using with my Psi L1, Psi R2, and Psi L2, Psi R1, and this is where you get the dotted and undotted indices, and then it just translates straight away into your, But here's, I think this is the relation you were looking for, in terms of the pies. And then, in this representation, you find that your, this is what I should have said, that your algebra actually fractures into two independent parts, and that's why you couldn't work with the semi-spheres and the twisters. okay and then that's spin representation and that's fine and then to tie it up with the momentum and so on roger makes the ghost of the peace space and puts the structure in the peace space but you that's what his his own purpose is now i hope that helps mike yes it did a great deal I'd always wanted to try and get clear about how your route into the twister through the clippet algebra connected with Rogers and I've got much clearer now. We're talking about exactly the same structure and we're just coming at it in different ways, that's all. The reason why I felt it might be useful here is because I'm getting quantum mechanics out of the clippet algebra in a way that's different from most people. What I want to do is, I'm going to lose a sleep at the moment, but what I want to do... I want to show him how he can get some of his results. I'm sorry, I don't want to disturb you.

32:30 I've got something here which I'd like you to have a look at. What I've got here is that underlying the Clifton algebra are a pair of dual Grasman algebras. So I've got, I'm dealing here with the, I'm doing here with the Dirac algebra problem. Is that what I'm looking at? I think it probably, well think about it as the Dirac algebra. and I've got a pair of Gras-Renari and they're dual in the sense that the anti-commutator between an annihilation operator and a creation operator I'm using words at the moment but you'll find they actually do create and annihilate in your story and they create and annihilate which I think is something rather interesting and that's G.I.J. so this is like a pair of annihilation and creation operators electro-annihilation-creation operators, but I can build my gammas out of sums and differences of those annihilation-creation operators. But what I've done when I do that is I double up my Clifford algebra. So I don't use the complex Minkowski space. I use two real Minkowski the spaces. And the doubling up comes because I've got this pair of graphs lying underneath the structure. And so I construct essentially a phase space. And it's in that phase space Roger's got it as well, but what I haven't done here is to add a symplectic structure. I'm just saying there are two phase spaces here and I can add the symplectic structure later, just as Roger did with the twisters but I'm not doing it and this is why I don't know whether this is important or not I'm not using the complex numbers to do it but I have the complex structure in the cleft algebra anyway. What's your second you say about two Wincastry's basics what's the second one? The second one has the minus sign in the metric I've got So I've got one lot of generators with the plus sign, so I've got enough to have space-time there when I wrap down onto a placed manifold.

35:00 And I've got the minus ones and they have isomorphic structure but with the metric sign change. and then these two commute with each other so I've really got how do I put this I would have four generators there and I would have another four generators there so I've got an eight dimensional space and I'd say one as being the position and one as being the momentum because I just want to distinguish them I think they're momentum and position. The position doesn't have any... No, but it's the sort of thing which I'd like to move towards to say I've got two algebras present. It's just the doubling of the algebra. Now that it's with I thought Lou would probably be this lecture for him. Here is this idea of yours with a distinction, inside, outside, and I'm using your notation and I'm using your product where it's A times C, B times D. And then what you do is you introduce an operator which changes the content of these objects here. You have a G times this which changes these two things round. You have a P times this, which puts a minus sign in front of there. You've got a Q in here, which puts a minus sign in front of the A. And these are isomorphic G is equal to sigma X, P is equal to sigma J, and I, P, Q, sigma Y. OK? Now then, I have my Clifford algebras built out of a pair of Grossman algebras. So the question I ask myself is, what do these elements do to loose structure? Well, it's easy to find out. Use them in terms of the gammas and so on. And what I find is that A multiplied into AB actually destroys the inside, destroys the outside, pulls the inside, pulls the outside inside.

37:30 Yeah. Of course I should have said, destroys the inside, destroys the inside, and then puts the inside, the outside inside. So I get that result. I then take the creation operator which says, destroy the outside and put the inside outside. So these Grassman elements actually have an activity on the structure that you deal with, which is interesting. Now then, these algebraic spinners, these elements of the left ideal, are in the algebra and therefore they are going to be active. So what happens if I operate ideal on lose a b. But what you find is that that can be written as a a dagger plus a dagger, operated on there, and what it does, it destroys the outside and removes the distinction. And then I use the other left eye. Remember in the quantum physics, we say these two are equivalence classes of ideals but in the algebra they're not equivalent, they're doing something different in Lue structure so they're essentially removing the distinction I just thought you might be amused of that, Lue, yeah? now of course if you now now I'm going to relate that to would be happy with, and that is suppose I take my annihilation operator and I'm operating on A0, I choose a special element. Well what that does is to annihilate this, which means that the whole thing is, is this a zero and you all go, no? It's just annihilating, spin down if you like, it's annihilating, it's a vacuum stone, I'm sorry I'm going to call So this is a vacuum state. If you translate from your language into physics language, it's a vacuum state. So what does the creation operator do? The creation operator destroys the outside. So you've got essentially two vacuum states in your structure. And I'm putting vacuum in quotes.

40:00 And this, of course, is what my dear friend David Finkelstein calls the planet. so you've got these two structures and they cross you start with the vacuum state and you build it up with all these daggers you start with the plenum and you destroy it and you keep going down the two streams just go past each other without stopping so you've got a copy of both the plenum and the vacuum of this and then if you introduce this projector onto the vacuum this is just on this vacuum is zero, and V operating this way on the vacuum is zero, and so this thing is behaving as if it's something like that, which is what I was using in my talk in my first talk. So this is, as it were, more information. I'm not really talking out of my backside or where you probably think I am. It actually is all consistent. Are these related at all to the Bob or you Bob truck? Not yet, no. Actually, tell me when you've had enough, because I'm just, I'm just, I'm hot. Okay, because that's the end of the spin. I was going to do, do we want to stop here or do you want to, I just want to do a little bit. I want to do quantum blobs just for Maurice's sake. Are you happy? Shall I do quantum blobs? Yeah. Sure. I'm rounding up and directing comments that people who have been in the audience will never see again. And I encourage you, Maurice. This is really deformation algorithms. So I'm now going to do something entirely different. And I'm very fascinated with your blobs, as you know. I mean, because I had the idea of blobs as well. Or structures, I call them. In fact, my very first painting was about structures in face space. It's very nice to see you actually supporting the probes.

42:30 OK, so how do I... Right, in my talks that I've used before I'm very interested in the density matrix. Because I've got an element of the left ideal times the element of the right ideal and that's my density element in the Clifford Algebra, but now I'm doing it in ordinary quantum mechanics. What I'm going to do is talk about time, if I may, introduce a time operator which I hope might be related to... I'm not sure. Yeah, but if you want to shut up, just tell me to shut up. There should be enough time for you to talk about time. OK, but the idea here with this density matrix is that these wave functions have two different points in space. And in a minute, I'm going to say, well, why have just one time, why not have two different time values as well? So I've got a blob in the sense that not only is it ambiguous as to where it is, it's also ambiguous as to what time it is taking place. So you've got a notion of a moment which is ambiguous. quantum mechanics is about ambiguity. This is where I've got it from now. Because when we were analysing Niels Bohr's work, it seemed that he was saying that nature is intrinsically ambiguous and that's why you've got the problems you've got. And we're trying to pin it down all the time and it's not working. so then let's now go to coordinates which are mean so we've got a cell and I'll just do P and X and X and P at the moment leave E and T at the minute so I've got a cell in my phase space and it's got a mean position and a difference so it gives me a kind of a measure this eta gives me a kind of measure of the width of the cell and now then Moritz had, which is the Vigner distribution. Okay. And so the Vigner distribution is really about cells in phase space, not about probability distributions, as we normally say. Because this is just another representation of this.

45:00 You're just representing the density matrix in a way which is using these coordinates. Okay. It's not really probability distribution. What Moyer wanted to do was to see if he could reconstruct quantum mechanics out of probability distributions in function space. And then we've got negative values, and you're worried about the negative values. The reason why you get the negative values is because you're just using the density matrix anyway. There's nothing classical here. In this way of looking at it. No, then I come in with there's also an uncertainty in time Right, now I'm going slightly to one side, remember I talked about this very briefly last time this is the K calculus using Lue's construct gives you the range transformation I want to and say, all right, instead of just having a light signal with a clock, or a radar set with a clock, let me now take notice of the fact that the light may be a surgery function. And I'm doing radar. So I've got a radar gun that is a real radar gun now, And not an apology for one that Bondi, well in fact it wasn't Bondi, it was done in 1935, Page did this in 1935. Okay, so now what you do is you say, alright, let me have the signal goes out, I'm going to be more like...