Generalizations of Kochen & Specker's theorem

0:00 I met Rob Clifton when he was a graduate student at Cambridge. I was then on sabbatical in Cambridge. And this started a long friendship, and some collaboration between us at Cambridge and then in Jerusalem, he came a couple of times to Israel, and also at the University of Western Ontario later on. and it would always be a pleasure to work with him and we had long conversations about quantum mechanics and needless to say I learned a lot from him and we collaborated on a paper on a subject that I'm not going to talk about today but rather I'll talk about Koshin-Huspecha generalizations of Koshin-Huspecha my aim here is really to give particularly to the younger the audience, an algorithm for writing papers on quantum mechanics. I mean, we all expect that with the advance of technology, computers will write the papers for us. So this is a modest contribution. I can now reveal this, because I'm beyond promotion in my age. So I'll give it to you. And I'll just say that actually Rob discovered that algorithm credit that he only used it once. To my great shame, I used it twice. As you'll see, I'll mention it as we go along the way. The point is the following. If you really know Gleason's theorem, and you are not ashamed of reading the proof rather than trying to reinvent it in some form or another, as Bell was, you really see that Kosciuszpecker is actually a consequence. Kosciuszpecker, I mean not just the infinite Kosciuszpecker, the finite Kosciuszpecker is a consequence of Gleason's theorem. and here's the argument, it's a heuristic argument and what I'm going to do is turn it into a formal argument and show you the logical structure of Gleason's argument and then all kinds of finite outcomes of it

2:30 and here is the argument, it's a simple argument of what's called logical compactness and it says the following thing from Gleason's theorem in particular it follows that every state has to be a continuous function Suppose we're just looking at R3, just the real three-dimensional space. And Gleason's theorem is just assigning probabilities in a certain way to all directions in space or to all, if you want, rays in space. And it says that such an assignment has to be actually a bilinear form on R3. So not just a continuous function, even an analytic function, but certainly a continuous. So it means that when I say I can, that you can't color the unit sphere by zeros and ones because this is a non-continuous function. I mean, it's just jumping all over the place. Fine. So you can derive a contradiction from Leeson's theorem plus the statement. There is a state which gets only the value of zero and one. Now, a contradiction is something that can be proved. A proof is a finite set of statements. Every proof is finite. You write it. So what you do, you write the proof, the vectors that you are using in the proof and forget about all the others and you get caution perspective theorem this is the trick here I mean it's a kind of compactness argument you can actually formalize it in a certain way that I'll do it in a minute but once you do it you see that there must be a finite counterexample and you can actually forget about the rest you don't need to actually write down the counterexample instructive proof, you start to search for a finite set, and we know all this great race to find less and less vectors and so on and so forth, starting with 117. But the idea is really compactness here. So let me start by just looking at what the Leeson's theorem say, and make this thing into a tool for proving theorems, and then we go on from there. By the way, this paper is written, this paper is done with collaboration, in collaboration with a mathematical logician from the Hebrew University, Ehud Horchowski. So here's Gleason's theorem.

5:00 Gleason's theorem just say, oh, supposedly having a finite dimensional Hilbert space, of dimension three or more, we'll just concentrate on the finite dimension and actually on the three-dimension because everything that I'm going to say is generalizable easily. And even on the real three-dimension is generalizable easily to the other. And so a function that is defined on unit vectors in the Hilbert space of finite dimensions, a real O-complex is a state these two things up here, 1 and 2, that probability of, it just depends on the array, in other words, and it sums up to 1, the probability sums up to 1 on every orthogonal basis. As is well known, the theorem, Gleason's theorem, says that in In this case, there is a function, there is itself a joint or Hermitian operator on the space such that Px is x times Wx, the inner product. And this is the proof. The proof has three parts to it. One of it is the reduction to the three-dimensional real case, which we will concentrate on three-dimensional because all the reductions that have been done there can be done here. The other part is proving that, so it's just proving that if it's true for R3, it's true for every Hilbert space of, as I mentioned, greater or equal to 3, the Gleason's theorem. The other part is just saying that if the function is continuous, then Gleason's theorem is true. And the third part, which is just the proof that every function like that, that satisfy one or two, is a continuous function on the other states, as I mentioned, on the real dimension on it. Now, so I'm not going to prove it. I mean, I'm not going to run through the proof. I'm just going to say what are the assumptions that undergo, that actually enter into the proof. And the assumptions, the logical assumptions are the following. and we're just going to concentrate on R3. So we have S2, the unit sphere, and we have a function from S2 from the unit sphere in three-dimensional real space to the real, and it satisfies, and I'm writing the axioms.

7:30 I'm writing the axioms in a logical form, but I'll explain, you know, what they say. The first thing is that the function is not negative. It's a bounded function. The boundedness is really the important aspect here of the boundedness of the frame function. things that have weight 1. The other thing is this is just depending on the ray, so it's p of minus x is equals to p of x for all x. And G3 is just saying this, but let's write it as a schema, not as a universally quantified formula that says for all orthogonal triples x, y, and z, we have this and that. Rather than that, let's write it as a list of uncountably many, actually, a continuum of propositions, each for an orthogonal triple. So, formally, what we are doing, we are looking at a particular model of the reals, a standard model of the reals, and we formalize everything. we look at the theory of the model so we have we have their real numbers and then on this particular model from this particular model we construct R3 which is just the triples of real numbers and then we construct S2 and then we add a symbol to our theory which is a special symbol P which is going to be formal, it's a formal symbol, a formal function symbol. And we assume that P satisfies additional axioms to those of the real numbers. And so we have the additional axiom G1 and G2. And then you have, instead of one axiom, you have a continuum of axioms for each one of orthogonal triples you have. And this is enough. We don't need the universally quantified formula. We don't want it, actually. G4 is just a statement, never mind the logical formalization of that, that says that every set has an infimum every bounded set has an infimum every set that is bounded from below has an infimum we want the formalization of that that is first order so we don't want to quantify over sets this I say to people who are familiar with logic and so we just look at extensions of predicates in our theory, that's enough and we look at predicates in our theory at properties that are

10:00 definable in our theory and we say if the set of numbers or if the numbers that satisfy that property have a lower bound, they have an infimum. This is just the fourth. And from the usual axioms for the real, and from G1 to G4, you can prove that P must be a continuous function. And you can see that either from reading Gleason's theorem, original theorem, or looking at my formulation of this particular part of Leeson's theorem, which is actually just to show that you need actually a finite number of uses of the schema G3. So we actually use a finite number of trickles in the proof of that fact. You don't need more than a finite number of trickles. And this means immediately that, you know, you can get Kaushin and Specker in this way, just from Leeson's theorem. And now to complete Leeson's theorem, you need just one other axiom, which is called here G5. And it just says that if P is continuous, then it obtains its maximum at minimum. have it at the outset in our model because it's just a model of the numbers, not the functions of numbers, of the real numbers, not the functions of numbers. So it's just a statement that says that if the function is continuous, then there is a point at which it gets its maximum and minimum. With that, you can continue and prove that there is a bilinear form or a self-adjoint operator operating in R3 such that Px equals x times that. There's one subtle point here that we have to remember how this is done. This is done, for example, in Cook, Keene, and Moran, mathematical proceedings, that, you know, they gave a proof, which they call a constructive proof of Leeson's theorem, which is the most non-constructive proof I've never seen, because they use the axiom of choice there, which is a very non-constructive thing, but that part is very constructive, and then there is another not a constructive proof, I mean, a simple proof. They call it a simple proof, or elementary, elementary, sorry. Their paper is entitled Elementary Proof, and it's not very elementary.

12:30 And Rickman and Bridges has a constructive proof. This is a really constructive proof, and you see it's a point that I'll return later to it. It has, it's a long paper. It's a detailed proof, step by step, that you can actually prove Gleason's theorem only by using constructive mathematical principles. Okay, so here this is an outline of the axioms that you need in order to prove Cleason's theorem. These five axioms, you know, except for the axioms for the real, you need a function symbol that says this is going to be a state, a definition of the states, which are the three, what the state should satisfy, which are the first three axioms, and then two more axioms that are true for every real function. on the reels, that every set of reels that is bounded from below has an infimum, or every bounded set has an infimum or a superimum, and another action which says that if a function on a compact, if the function p is continuous, it has its maximum and minimum, which is true for every function on a compact set, so it's actually true here. Two on the reels, we just add it because we don't have it, and here are the consequences. what O you need, you have a compactness, a series of compactness results. So let's start with Kochen and Specker. Let's start to consider false results. F1, there is a state such that for all x, either P of x is 0 or P of x is 1. So there is a state which gets only the value of 0 and 1. So we know that F1 contradicts actually our first four axions, because the axioms already entail that f is continuous. And therefore there must be a proof, a formal proof of a contradiction from this set of assumptions. So actually go through the, this is what I said before, go through the proof of the contradiction and collect all the vectors that you use in the proof and forget about the others. Since you need only finitely many times to use the axiom schema G3, which says, you know, Px plus Py plus Pz equals 1, you know, just forget about the others and you get the Kaushin This is why Kaushin and Specker theorem is actually a consequence. It's a logical consequence, an immediate logical consequence. But you can have many more theorems like that. But before that, some history of that,

15:00 this is where actually Lifton discovered that principle because he had a proof. He did the same thing. There is a proof of a so-called Gleason theorem, is the one that you referred to in the last, the paper by Bell. The paper by Bell actually proves a Gleason's theorem just for functions that have zeros and ones, and it proves that there cannot be a function which gets the value of zero and one, and it's a state, and it's a state in Gleason's sense. And this is a very simple proof, so what Rob did was just to look at that proof, lift the vectors that appear by name in the proof, and forget about the others, and you get the This is an American Journal of Physics. I've done the same thing years before for a somewhat different group, but it's exactly the same thing. And just to mention on another paper by J.L. Bell, John Bell from the University of Western Ontario, the logician, who also actually got the compactness principle for this particular case, He used it in a somewhat different formulation as for gradients. Let's look at F2. F2 is the statement, the state P has exactly, now let's generalize it. The state P, instead of 2, the state P has exactly K into more than 2 values. So suppose that we are just, we want to look whether there could be states that have more than 2 values, but perhaps 17 values. Now we know that that's impossible because by Gleason's theorem a state should be continuous so if it has two values it has anything in between because continuous functions are continuous so therefore there must be a finite set on which you can't have 17 values. So there is a finite set like that, you know, I mean this is just a logical consequence, exactly the same thing. This particular result was used by Breuer, Breuer, it was used by Breuer to give another argument which was not mentioned before against the so-called nullification of the Caution-Espacket theorem, by showing that if you just take approximations over averages, you will still

17:30 get that the sum on every orthogonal triple should sum up to something, and you can't get more than finitely many results, and by using this type of argument, you can say, okay, I have a different cause, I respect it. So this is not, if anyone, you know, entertained the idea of having more than, more than bi-valued logic, say three-valued logic or four-valued logic and get around quantum mechanics by adding values to the, and, and still maintaining the design, you know, you can't, you can't do it because you, you get the same question, only on much larger graphs. And I'll show you how to actually construct these graphs perhaps a bit later if we have time to make a sequence of constructions like that. So this is something about how to use this result, this particular. Some further results of that time. I mean, these are just finite sets. what you could say, let's go over this one. This is not very interesting. Oh, here. Let's go to this result. This is a result that I actually constructively proved in this paper that I mentioned before. And this is what I call the uncertainty principle, the logical uncertainty because all that is tied up nicely with quantum logic because it gives you a finite propositional structure on which the probability functions that you are looking at behave in a way that is quantum mechanical. So one thing that you can ask yourself is, can you have in quantum mechanics two non-orthogonal rays, both of which obtain extreme values by a state? And the answer is, of course you can, are obtained the value zero. Because you see, if, say, a ray that is orthogonal to both of them is getting the value one, then both of them are getting probability zero. But if one of them is one, the other should be strictly between zero and one. Cannot be, this is just because of Leeson's theorem. And this is just a statement of the uncertainty principle. The uncertainty principle says,

20:00 give me two rays or two subspaces, or let's talk just one-dimensional subspaces, that are non-orthogonal to one another. them is getting the value 1, the other should get a value or probability which is strictly, I'm talking about probability, probability which is strictly between 0 and 1, and vice versa. If the other one is getting 1, the first should get a value which is strictly between probability 0 and 1. And the compactness version, the compact version of that is that there exists, you know, because, you know, this statement is false, it's a contradiction to Gleason's theorem, the statement that there is a state such that, oh, there are two directions, such that their inner product is not zero and not one, so they are not identical and nor are they, you know, nor are they, nor are they orthogonal to one another, and B, meaning our state, is giving them Either both of them are the value one, or one of them the value one and the second zero. And we know that this statement is false because it contradicts Gleason's theorem. Therefore, there must exist a finite set such that, give me any two directions, x and y, which are not orthogonal, I can add to them a finite set such that every state on that finite set should satisfy that if px and py have extreme value, the only possibility for f, p, x, and p, y to have extreme values is for both of them to be 0. So this is a generalization of caution and speckle in the following sense. It tells you that you have a stronger result here because it tells you something about probability functions on finite set of rays, not just truth functions. Because I mean, the other values of p could be anything between 0 and 1. I mean, on the rest of the collection of directions which are in this finite set, p can get any number between 0 and 1, you know, 1 over 17, 1 over square root of 2 or whatever, and still, you can't force px and py to be, to have extreme values other than both of them 0, which is a kind of, I call it the logical, the logical, you know, the quantum logical uncertainty principle, because you can force a finite set of propositions, already a finite set of propositions can force uncertainty on the, on the, and the most dramatic result is

22:30 actually that you can actually force, you can actually force on a finite set the values of, of, of P to be very, as close as you want to the quantum mechanical value. So suppose that we take the extreme value, then the extreme case were of a pure state, and suppose that we know that for some Z0, for we have it here, for some z0, the value of p is 1. So p of z0 is 1. So suppose that we know in this direction, we are talking about r3, right? So in this direction, the probability for this direction is 1. Then by Gleason's theorem, you get the Born rule, which says that the probability in any other direction is just given by an inner product between z0 and the other direction, squared. This is just Born rule which follows, you know, Leeson's theorem immediately. And then suppose that you want to force on a finite set. You want to force this rule on a finite set. You can't strictly force this rule on a finite set, but what you can do, get any close to it as you want, you say, okay, give me another direction, x, and I'll say the distance between px, p of x, and this Born rule is greater than epsilon. And this contradicts Gleason's theorem. Therefore, there should be a finite set, such that already on this finite set, give me any state which is just restricted to this finite set. It's not defined on the other vectors. It will, if it satisfies this, it must be less than epsilon. The difference between px and z0 must be less than x. So this is the consequence. For all epsilon, give me any triple epsilon number, and then I can find you a set of directions, a finite set of directions such that give me any p which is defined on this finite set of directions and satisfy the Gleason's rules on this finite set of directions, the difference between this is going to be less than epsilon. And this is, of course, true for x. It's true for any finite number of x's. So you can actually interpolate quantum mechanics on a finite set which just dance as you want on the sphere and force the values to be those of quantum mechanics, not others. Well, this is important for me. Firstly, it's important because of the approach that,

25:00 the Bayesian approach to quantum mechanics, which you can show that quantum mechanics can follow from the assumption probability is non-contextual, it's the same no matter in which context you measure the, the, the, if you assume the probability is non-contextual, you can look, you can look at finite sets of, finite sets of gambles, of, of actually each measurement, a finite set of measurements, incompatible measurements, and present them as a gamble to someone. And if that one is following this rule of non-contextuality for this finite set, quantum quantum mechanics already is forced upon that person. And not just quantum mechanics, you know, quantum mechanics to any degree that you want depends on how large the finite set is. So you actually don't need, in order to force quantum mechanics to any degree of accuracy, you don't need the entire set of directions. You can do that for mixed states as well, only then you don't, you have to specify the value of p, not just in one direction, but in five directions is a 3 by 3 matrix, which is self-adjoined and has trace 1, so it's completely identified in five sets of directions. So if you specify the five directions, then you can get anywhere, as close as you want to. So these are all results. Now, another interesting result is just the one very old result of mine, which is in the spirit of the last of the last from 1985 and you can ask what is the inverse question how large a set of the S square how large a set of directions can we find on which we avoid the caution and speck of contradiction such that on which high valued states exist how large can such a set be and the answer is You don't have to confine yourself just with the denumerable set of directions which have rational coordinates. You can have a much, much bigger set, and how big it is depends on the axioms of set theory. But you can have a set omega of directions on which there is a bivalent, on which there are just zeros and one.

27:30 this with zero and one, satisfy the caution and speckle, satisfy the, you know, go against caution and speckle on this big set, and this big set has the property that it's intersection with, it's intersection with, with any major circle. So you have, here is the unit, here's the unit sphere, which looks more like that. Like here's the unit sphere, this is a major circle and you have a set and you ask yourself how much of the set is the section of any major circle? How big is it? And so it's all the circle except for countably many points. And this is true for every major circle. And you can find the set as big and avoid the Cossian-Specker theorem on that. This is from 1998. So Cossian-Specker theorem can be avoided on a huge set. The continuum of positives. If you want this to be true. But Gernel thought the continuing hypothesis was false. Well, so if you want something else, you can take, if you want this to have measure zero, you can take a weaker axiom, which is called Martin's axiom. And this will have just measure zero, but perhaps will not be. You can get all kinds of results. So you can avoid the Kosh-Nuspecker contradiction on a very, very large set, not just the countably many of the rational, very large set, which is, uh, which is, uh, you know, it's quite terrible in, in many respects, but, uh, so now the question is whether we're going to go to some constructions or not. Uh, all these results are actually non-constructive, because I just, I didn't, I didn't show you where these sets are. I just told you they exist, and they must exist because of this compactness argument, right? They must exist because of this compactness argument, but I didn't show you what they are. I can show you some of them can be actually constructed. But in all cases, what I'm going to show you in the end, I hope I'll get there, is that there is an actual algorithm to construct these sets. You can have an algorithm for constructing these sets, and that's my promise, right? To give you an algorithm such that you can write papers in front of the camera. I'll show you that there is this algorithm. Well, it would be difficult to look for it, but I'll tell you why it exists.

30:00 So let me perhaps go to that before I go to the construction. The reason why there is this algorithm is interesting. You know, everybody really thinks, everybody, when thinking about mathematical logic, thinking negatively about Gettel's theorem, because it's really the greatest achievement in mathematical logic in the 20th century. For sure, perhaps, you know, since, I've read somewhere, since Aristotle or something like that. And it's really a great, great result. But there is another great result in logic, which is a positive result. And since we, the 20th century, in particular the late 20th century, had the tendency to stress negative results, it's less known, less well-known. And that's the result of Tarski. if Gödel proved that there is no general algorithm to decide whether a statement or proposition in number theory is true or false it's one of the consequences of Gödel's theorem and there is a theorem by Tarski that says that there is an algorithm to decide whether a statement in geometry is true or false there is an algorithmic decision for geometry and it's not just geometry as we know it of straight lines and circles is that the algebraic geometry, the first order algebraic geometry of every algebraic curve, of all the algebraic curves. So it's a very, very powerful result. And he actually gave an algorithm. And using this algorithm, I'll just say it, it's hard to explain here exactly how, using that algorithm, you can actually show the following, using that type of consideration. so this is it you can show the following statement that if Gleason's theorem is true so I mean, let's compare this with the actual proof of Gleason's theorem constructively, there is a constructive proof constructive proof of Gleason's theorem that takes about 30 pages very meticulous, you know, step by step construction 24,000 layman's unbelievable but there is a simple proof of the following statement if Gleason's theorem is true it must be constructed A very, a very short, a very short, and, and even, even a statement of what it means to, to be constructive in this sense. And, and this actually relates to a very, a very, a very interesting, a very interesting exchange that was, was, you know, exchanged mostly in the Journal of Philosophical Logic.

32:30 around, you know, in the 90s. On whether Gleason's theorem has a constructive proof or not, Geoffrey Hellman thought it didn't, and this was just to show that, for all the constructive mathematicians, that physics cannot be done in ways that constructive mathematics preach, preaches, and therefore constructive mathematics has its limitations, let's say. But that's not true, and as I said, there is a particular, actual step-by-step proof of Gleason's theorem. by constructive means, but you can do it, you can do it, you can do it, as I said, with this formal theory, just with mathematical logic, with mathematical logic. First of all, let's think about Gleason's theorem in a different formulation. Let's relax the idea that, let's relax two of the, let's relax two of the assumptions about P, and say that P is just bounded, not not necessarily positive, but it's bounded. And the other assumption, let's relax the other assumption, that says that px plus 2, py plus bz equals 1. I would rather say that it equals some constant that is independent of the direction. There is some constant such that, which is the weight of the function. And then it's easy to show that Euler's theorem is actually is actually Gleason's Theorem, which is A here. For this particular function, Gleason's Theorem for this particular case, which is called brain function by Gleason, which is just statement A here, is equivalent to the statement that if you give me a P, which has the following property, it's zero on the standard base. it's 0 on the standard base it's also 0 on standard basis which is just, you know let's call it 1, 0 and minus 1 ok, this is the standard basis and it's also 0 on 1 over square root of 2 of 0 plus 1 and so on on the 1 square root of 2 plus minus 1 and the third one

35:00 1 over square root of 2 of 1, of minus 1 plus 0. If it's 0 on this triple, and it's 0 on this triple, then I'll have a turn of it. Every frame function which is 0 on those is identically 0. This is just Leeson's theorem. If you can prove that every frame function, meaning function that satisfies g, and it's bounded, every bounded frame function which is 0 on these 6 vectors is identically 0. follows from Gleason's Theorem, but it's also the other way around. So if we look at this formulation of Gleason's Theorem, which is just identical to the previous one, now let's not panic when you see the following, then the following is true, I'll read the corollary, the corollary here of the theorem, that there is a construct, an algorithm for the following thing. You can generate an algorithm with the following property. Give me any direction x0, I'll find a finite set, gamma, which includes this x0 and these six vectors, and perhaps some more, of course, and which has the following property. On this particular set, on this particular set, which has a fixed size, there is an n which is fixed once and for all. It depends on the dimension, 3 in this case, but it's independent of x0 or the vectors that we chose. It's just a particular number, n. This set has the following property. First of all, it's algorithmically generated. It could be algorithmically generated effectively, meaning there is an algorithm to generate. Second of all, it has the property that if you have a function which satisfies, on this set of directions, which satisfies our assumption about the Gliese assumption, that on all triples it sums to zero. And the second one that is zero on all these ones, then there is, you know, there is an x such that at that particular x0, which we chose at the outset, it's less than half the value of the absolute value of one other x. Now, this gives you immediately a constructive proof of Gleason's theorem. Why?

37:30 Because what you want to show in Gleason's theorem is that the function that is zero on all these is identically zero. So, you say, okay, you start with x0, and you say, well, on x0, it must be half than some value on that particular set now you take that one which is half and you construct another set like that and then it's a quarter of another one px0 is a quarter of another one and then you construct another set like that using the same algorithm with the same number of direction and then it will be 1 over 8 of the other one and by iterating this construction you show that px0 must be 0 why it has to be 0? because it's less than half of px For every x0, for give me any, any, give me any x0, for any direction x0, I can find a finite set with the same number of directions such that px0 and such that every state which satisfies this on this set, zero on this, which is zero on these six directions, will be, will be, will be, will have this property that in this particular finite set, for px0, there is one x such that dx0 is less than half of the value there. And this will be true also for this x in another set. And this will be true for another set, and so on and so forth. So you get 1 half, 1 quarter, 1 eighth, 1, which makes this go to zero. Now what makes the proof constructive is the fact that you can actually generate, I mean, what's written here is the generation, the constructive generation. It says that what will be generated by it is the proof of, if you can prove this thing that is written here, whatever that is, the proof of that will give you the results. And the proof is a constructive proof because of Tarski's theorem. You look at the theory of what is called real closed fields, or in the complex case of what is called the theory of the moment. Then you get the result. The only trouble with this algorithm, which makes it a bit troublesome, is that it runs in worst case exponential time, so it may take a long time to generate these sets, which still gives some credit to the graduate students who will do it, but there is an algorithm

40:00 to do it, and you even have a running time for this algorithm. and there is such a set there is such a set which is a kind of an iterative process by which just by finite means you get any approximation that you want and it's a uniform approximation in the sense that the n here, the number of directions in the set, is independent of what you started with so this is the constructive aspect of Leeson's theorem. How much time do I have? Have I applied to your time? So, finite is the important word here, you're right. And use it constructively. Right, so let's do real construction here. Now we're going to get our hands real dirty because, oh, you know, I mean, there was another pair of words are very important in the discussion so far there exists I just said there exists and I didn't show you anything so it's time to show you something so for some of these things you can actually go and construct them and the easiest way to do it and I'll do it for say for this logical or finite uncertainty principle the logical uncertainty principle what I call it and I'll do it here in a much simpler way than I did the first time around by using the weaker version of Gleason's theorem, which was proved by Piron years ago. And I'll just show you the exercise of how to, you know, just take the vectors that appear in the proof from Piron and use it to prove such theorems about finite graphs. And forget about the others and show you why the infinite case is just like the finite case. Or what is implied by the infinite and finite case. So here's the uncertainty principle. If you have any two non-autogonal rays in Hilbert space H or finite dimension greater than 3. By the way, all the results that I mentioned are true for any Hilbert space or finite dimension greater than 3. Hilbert space of the reals or complex. Everything that I mentioned is just true for all these results. So this is a statement that says that, you know, two non-orthogonal rays cannot get extreme values, zero and one, unless both of them are zero.

42:30 I mean, if one of them is one, the other is strictly between zero and one. So how do you prove something like that? Okay, so a couple of constructions. First of all, this is the basic concern. This is the basic graph for everything, you know. I mean, this fact that I just mentioned here can be used to prove Gaussian and Specker and everything else. I mean, this is, it's also a very basic part of the proof of reason. Suppose that I know that P, here you have Z is our north pole, okay? We're just looking at the northern hemisphere. Z is the north pole. And suppose that Q is some other vector different from Z. And we look, and Q prime is the vector orthogonal to both Q and Z, okay? and we look at the great circle that goes from Q prime to Q here, this is the great circle and here is the following statement which is almost trivial to prove which is the following, suppose that you have a state which gives Z the value 1 the value of that P on Q is greater than the value where R is any other number is any other direction on that great circle but with, with, to the south of Q, all right? Any value, any vector to the south of Q is going to get a value less than Q on this particular circle. And the reason for that is the following, because you can, you know, if you look at Q, what we have is Z, Q, Q prime, and R. So, let's, let's Q double prime be the one that completes Q and Q prime and let L be one orthogonal to R and to Q double prime, okay? So this is the orthogonality graph for them. This is the orthogonality graph for them. This is the orthogonality graph for them. It's just taken from there. Now suppose that PZ is equal to 1. Then once you know that PZ equals 1, you know that on this triple you have to have 1. On this triple you have to have 1. But you know that q prime gets the value zero because pz is one, okay? So, and, and, and, so you have pq prime plus pq plus pq double prime equals pq double prime

45:00 plus pq plus 3L. And then you just drop one of them, you know, pq double prime, it feels both here and here. One of them is zero, pq prime is zero, and therefore that's what you have, I mean. PQ equals PR plus PL which is greater or equal than PR. Now this is the basic fact that derives everything else because once you have this you can show that if PZ equals 1 then PQ and if Q is to the north of R not necessarily satisfying this relation not necessarily for an R that is to the south of Q this circle but some any other r that is to the south of q if pz equals 1 then pq is greater than p greater or equals than pr and how you do it you just repeat this construction as much as you want so by repeating this construction you do it let's see how we how we complete it this is all the first of all is perhaps it's better to see it on the projective plane you know we take within the plane perpendicular to z. At the plane, sorry, tangent to z, perpendicular to the vector that goes to z, but, you know, tangent at the point z to s2. And we project every point. You see, we take from the center of the circle, given any point r, we just project it on the plane here, so we get here what's called a projected plane. I think most of you are familiar with that. Where the point on the equator goes to infinity. This is the infinity point of projected geometry associated here. So what we've seen before, the figure that we've seen before, is just the following. This is Z, as before. This is where Q has been projected. And R, and the great circle from Q to Q prime, which is the point at infinity here, this great circle is just turning into a line on this projected plane. Into a line. And R is just on that line. So the relation is depicted in this way. You have a line from Z to Q. You have an orthogonal line, which is orthogonal to this particular line on the plane, on the projected plane. And R is here. R is here. This is in the previous picture. But you take any other R, this is Q and R,

47:30 you can find points in between. Suppose that Q is to the north of R, you can find points in between such that you can repeat the construction sufficiently many times, such that if pz equals 1, then pq is greater than pq1. And if that, if pz equals 1, then pq1 is greater than pq2, and so on and so forth. So you have this graph, as before, between this and this, and then you have another graph which connects this to this, and then another graph that connects this to this, and another graph that connects this to this. And you just do it sufficiently many times, finitely many times, you don't need any more, that if Q, if Pz equals 1 and Q is to the north of R that PQ is greater or equal than which means that now you know that every frame function which satisfy Pz equals 1 is monotonically decreasing as you go from the north to the south which means if you just go to the infinite case again, we're not going to use it that every monotonic function is continuous at all points except countably many. So you're always there just by this fact alone. This was very easy. You just have to take care of the other points where it's not continuous. Once you have that, you're almost there probably. What you need, you return to these two vectors that we had before on which the theorem goes that these are A and B, you know, the points A and B. These are the points A and B. And we want to show that every, we want to construct a finite set and graph, an orthogonology graph, a finite set such that if P is a function, is a state on this finite graph, I mean, just to satisfy the relations of this finite set, then P A and P B cannot get extreme values. So if P A is 1, P B is strictly between 0 and 1. P B is 1, then P A is strictly between 0 and is really rather very much like what you do in the caution and speckle thing. You go and complete this set until you get to 90 degrees, and each time you make sure that the jump, say, from B to Z1, will be such that the angle will be a little bit smaller than this original angle. And the next jump will have a little smaller angle, and the next jump will have a little smaller angle, and so on,

50:00 until you get to 90 degrees. You can always do that. The number of jumps depends, of course, on the angle between original A and B. And now what you do, you're saying, okay, you force a finite graph. You put a finite graph which forces the forcing. If PB equals 1, then PA is greater than PC. You look at B as the North Pole, as the North Pole, and you have a finite graph that because B is to the south, relative to B, A is to the south of C1, it has a lower latitude because the angle here is Then if Pb equals 1, then Pc1 is greater than Pa. We can't see anything when you keep your hand on it. Oh, oh, now we can show you all for me. Does anyone have a laser stick? Okay, is this not okay? So suppose that we consider B to be the north pole. Then since the angle between A and B is greater than the angle between B and C1, it means that A is to the south of C1 relative to B because it has a greater angle you need to walk to go further from A to B it takes you a longer time to go from A to B than from C1 to B because the angle is small and therefore you can have a finite graph that's the one that we constructed before that says that on which every function that satisfies Pb Pb equals 1 also satisfies PC1 And similarly, if you take C1 to be the North Pole, the same is true with respect to B versus C2. B is more to the south than C2 relative to C1 at the North Pole, because the angle is bigger. That's how we describe it. So this is how you go. You say, suppose that both PA and PBs are 1. But then PC1 has to be 1, because PC1 is greater than PA, because P3 was 1. And therefore, also PC2 has to be 1. Therefore, PC3 has to be 1, and so on and so forth, until you get to 90 degrees and you get a contribution. And this is all on a finite graph. This is all done on a finite graph. So you already know that you can't have both PA and PB, both 1, by a finite construction. And this is just taking or lifting P1's proof. Now, in order to complete it, you do the same thing for A and B prime,

52:30 where B prime is the vector orthogonal to B on the AB plan, the plan that contains A and B. So you do the same construction for AB prime, and you do the same construction for B and A prime, where A prime is not this picture, but it's the one orthogonal to A And when you do all these things, you can show that if P A equals 1, then P B has to be strictly between 0 and 1. And if P B equals 1, then P A has to be strictly between 0 and 1. And this is just one construction like that. From that, you can also construct graphs on which you can't have many values. As I mentioned before, this is a bit weaker. Just by using this fact and writing two lines, you'll get the following thing. Any state that is a pure state, given any state, given any z, I can find the finite set such that if p, such that if p is a state of that finite set, and it satisfies p z equals 1, then p has at least 17,000 values. How do you do that? You do that because when you go to this, when you use this construction that we did just a moment ago, then using this construction, you can force not only that PQ is greater or equal than PR, you can actually prove that, you know, by adding a graph here that connects Z and L and some other vectors, a lot of other vectors, you can actually show that PQ has to be strictly greater than PR. so you can actually force a strict inequality since you can force strict inequality you can have as many values as you want so the function, any probability function should have as many values as you want and so on and so forth there are many, many such constructions that could be added explicit constructions and all of them the point of all of them I think is becoming only clearer in one particular context or another in which they are being used, but more generally in the context where you want to actually derive quantum mechanics or the rules of quantum

55:00 mechanics from just considering finitely many measurements or finitely many possible measurements. Instead, that your set of measurements is just finitely finite and you have finitely many possible measurements, and you could have quantum mechanics, or you could have the values of the probability as close as you want to quantum mechanics, forced only by these, you know, observables, and you don't need all the rich infinity of observable quantum mechanics to do that. And the reason why this is important, I think, is basically for both, Well, for two things, for quantum logic, to which I cannot go now, and for the Bayesian approach. One is the probability that is in a paper, that is in many papers that people have written, people here have written, but it's particularly the one approach that I've developed and will come up in studies in the history of philosophy and quantum physics. I think the near issue was the one after that. Thank you very much. Early in your talk, you mentioned the logical uncertainty principle. What is that, exactly? Well, that's what I mentioned. I just said what it is. It's meaning the following thing. It's the question of the uncertainty principle. Look at the uncertainty principle as a principle. A qualitative, not a quantitative principle, which says the following thing. as for a matter of fact the quantum state is this you don't have any certain knowledge about any other quantum state except the ones that are orthogonal to it. Okay? Which means that if you know that the probability of this is 1 then the probability of anything that is not orthogonal to it is strictly between 0 and 1. Okay? So what this principle just says is the following give me any pair of directions that are not orthogonal I can force this relation just by considering finitely many directions. If you just look at finitely many one-dimensional subspaces, there is a set of finitely many subspaces such that if you consider all the probability functions on them, you already have that fact. That if the probability assigns this one,

57:30 it assigns the other one something strictly between 0 and 1. Itamar, I thought that part of what makes it logical is that you, not only the finitarianess of it, but the fact that you exploit only the orthogonality. Right. This is just true by the definition. The only thing that counts here for forcing all these relations is just the orthogonality relation. So this is how it relates to this Bayesian view because each one, each experiment that you can make is just a selection of a basis in Hilbert's space. Suppose that it's a maximal measurement, then it's a It's a basis, you know, it uniquely depicts a basis in a hilltop. And again, there's the following. I mean, suppose that there is a casino, and somebody comes to the casino. I mean, and the following thing is operating. First of all, somebody says, one of the following measurements is going to be performed. And then, that's the bookie. And the book, he says, and it gives you a finite set of measurements that can be performed. Suppose that all of them are maximal measurements. I mean, finitely many measurements. Some of them are maximal, some of them are not. Not necessarily. Okay? So I'm going to perform one of these measurements. Now you have to bet, he tells the other person, on the outcomes of each one of these measurements. This is the second part. The third part. Now I'm going to choose the measurement that I'm going to perform and return the money that you put on all the measurements that I'm not performing. Okay, I just give you back the money. And the fourth part is, now you win or lose according to the outcomes that you betted on in this particular measurement that I chose. Now once you assume that the probability function is non-contextual, it does not depend, the probability assigns the same probability the probability function assigned the same probability to this direction, regardless of whether it's in this triple or in that triple, you're in the game, because you have this relation that is forced by Gleason's theorem. But, as I said, as I showed here, there are finite gambles like that that already force something as close as you want to quantum mechanics. So you can actually, without going into the problem having infinite gambles and so on and so forth, you can

1:00:00 actually have a kind of Bayesian, if you want, the finite type of analysis of probability of something that actually can be done in a casino. I mean, you can actually do in Las Vegas something like that, you know what I mean? Tell them, you're either measuring this triple or that triple or that triple or that triple, OK? And if you have sufficiently many triples and the person is rational, and I have a different argument why it's rational to have non-contextuality of But that's a more subtle point, or difficult. There are some difficulties of that. But suppose that you have non-contextual probabilities, then any rational person should have the rules of quantum mechanics. This is the point, you know, on finite remaining, or very nearly the best of quantum mechanics. So after all this beautiful mathematics and history, I want to give you a completely unimportant but no to the history it's the construction by two graduate students at Boston University named Debo de Obaldia and Whittle and they adapted the construction that's in Belen Fonte's book but they made it more rigorous and they made it simpler they use great circles as you did in the argument drawn from Perron they only needed three great circles and they produced finally one point giving the polar coordinates that was assigned because of the inference, both the values 0 and 1. What's so important about that? The important thing about it is they gave me a draft of their term paper. It looked like this just with red and blue pins. It's in a grapefruit. The only edible term paper ever created. Well, a bit sour, I mean. Very good. I want your comment on something I recall from my own efforts a long time ago to study Gleason's proof and try to simplify it. Because I remember that you could get quite some distance with Piron's construction. Right. But I recall that I ran into trouble on the following point. Everything is easy, more or less easy, once you start with the assumption that there is some point that takes the value 1.

1:02:30 Right. Right. It's also true, given Gleason's theorem, that for every pure frame function, some point must attain the value 1. However, you want to prove Gleason's theorem, so you're not allowed to use it. Of course. And you would like to have a proof also of the fact that every, an independent proof of the fact that every pure frame function must attain the value 1. Right. And I was never able to get to that. Well, at least at that. Well, I mean, the point is the following. You first, as I said, I mentioned that actually on the axiom that you need for the proof. You first prove that every frame function is continuous. And this is the difficult part. But it turns out to be a finite part, which means the following thing. What he actually does, although this is not what he calls by name, It's the following thing. You give me, he says the following thing, I mean like in every continuity group. Give me an epsilon, right? I'll give you a delta. So give me an epsilon greater than zero, and I will produce you a delta greater than zero, okay? such that if you give me any two directions, z1 and z2, with difference less than delta, well, what he says, I'll show to you that p of f1 minus p of f2 will be less than epsilon. But what he actually does in order to show that, he constructs a finite set, a huge finite set, which is a combination of many lemmas that you can put together. Every P on that set have the property that it must have Tz1 minus Tz2 less than epsilon. So it's a finite proof. Now, the number of points in this finite set depends on the distance. You know what I mean? If the epsilon is very small, so the delta should be very small. So it actually derives a modulus of continuity for the function. And it's an ingenious sort of nested proof that it is very complicated. And Piron just avoided all that by assuming that you have a maximum point, which he does assume.

1:05:00 And then everything follows much easier, in a much easier fashion. But if you don't assume it, you have to assume, you have to actually prove directly that it's continuous. But it's a complicated thing. But it's a finite, it's already, it's a finite construction actually. And it's an interesting fact that it's finite. uses a quantified formula for every triple x, y, and z. He never uses the generalized. He just uses these geometrical facts. Jack, did you signal? Yeah, I just wanted to ask, suggest really a point of information. At some point, near the beginning of your talk, you referred to a result of filling out the whole Hilbert space with ones and zeros, you know, how big, oh, oh, oh, oh, yeah, yeah, I just wanted to know what's the relationship between that result and another result of Rob's actually, that if you start off by assigning ones and zeros to propositions associated with a particular basis, and then ask how about extending that assignment to the rest of the Hilbert space, subject to certain constraints, you get a unique result. I mean, there's only one way of doing that, subject to symmetry constraint and other constraints. I just wondered if there was any connection between that? You mean the result, my result, that you can have a bivalence almost everywhere? This is a very wild construction, so it's not constrained at all. No, it's not constrained at all. I can say something about that. That's a very old result. That's relates to my PhD. You were a supervisor. Yeah, it's one of the things that is the outcome of this thing, where I proved many more things such as it's not only that you have this almost everywhere contradiction to Cauchon and Specker, but also the distribution on every circle, not just major circle, little circle, is just the distribution that is given on quantum mechanics. So you have cos sine of square theta half of plus on every circle and sine of square if this is plus one and the inverse if this is a minus one and what you do for this is really you know, I mean, what set theory gives you, for example the continuum hypothesis gives you, is that you can take suppose that you just look at major circles you can take all the major circles

1:07:30 put them in what is called the well order there is an order, would you say this is the first, this is the second, this is the third and so on and so forth get, when you get, when you get to omega, you get, you know, I mean, to the first, in transfer finite order now, you get beyond that and go on and go on. And, and the continuum hypothesis says, just a second, I didn't finish, and the continuum hypothesis says that every initial segment is countable. This is the continuum hypothesis. There's a well-ordering of the set of all major circles such that every initial segment of the order is countable. this is the continuum. And now you just go, so there is C1, C alpha, C alpha plus one, then you have some limit ordinals and so on. But every initial segment, every one you get it, you have less or equal. And then you go by induction. On the first circle, you just take the vector orthogonal, so you decide that this is going to be plus one. You divide into four quarters, to this you give zero. And to all this circle, you know, sorry, to all this circle, you give zero, OK? And then by induction, you go. Now suppose that you got to a certain circle, which is C alpha. Now, the intersection with all the previous circles that you have already considered, because the intersection of two circles is just two points, two major circles, just two points, then the intersection of everything so far is just countable. So what you have to do is on the new circle to define things in a way that will conform with Cauchon and Specker. And then you go on, I mean. And since this is a well-defined induction set theory, and since the continuum hypothesis is consistent, by the way, perhaps Gettle didn't like it, didn't think it's true, but he proved that it's consistent, which is a big problem for him. Then you have this construction, which means that this construction is completely wild. I mean, you don't know anything. I mean, the set as a whole on S2 is just non-measurable. And this is a much more dense set than the one considered by Mary and Adrian and Rob. This is almost everything. I mean, it's just that you have to pull out the things that can lead to a contradiction.

1:10:00 And in every stage, the things that you pull out is the most countable. Charlie, it's the waiting station. Yeah. I should share your prediction. It's really very nice mathematics. I'm not so sure about the physical significance, but I thought it would be good, not just for your talk, but for many of the talks today, to read a statement of Bill, which is certainly not unrelated. It says, concerning hidden variable theorems, a final moral concerns terminology. Why did such serious people take so seriously axioms which now seem so arbitrary? I suspect that they were misled by the pernicious misuse of the word measurement. Thank you. Thank you.