Once upon a time, guided by a whisper of intuition—or perhaps a playful nudge from fate—we set out on a journey. At first, our quest seemed clear: to uncover the mysteries of the enigmatic spinors. We had a map (or so we thought) and a destination in mind. But as we wandered deeper into the unknown, the wide road faded into a meandering trail, and the trail became a wisp of a path. Before we knew it, we were in a forest—dense, shadowy, and alive with secrets.
The forest wasn’t on the map, but here we were. And while our grand quest felt like a distant memory, the forest itself had other lessons to teach. At first, we worried: how would we ever find our way? But then we noticed the sweetness in the air, the earthy scent of moss, and the rustling leaves whispering ancient songs. We saw plump berries, glistening with dew—some delicious, others mysterious. And then, as if out of a dream, a gentle roe deer emerged, its soft eyes urging us to follow. It led us to a crystal-clear lake, where the water was cool and refreshing, as though the forest itself offered us a blessing.
Forests, after all, are not just places to get lost; they are places to be found. They nourish the soul, if only we stop to look. So we paused, took a deep breath, and began to notice both the towering trees and the soft carpet of the forest floor. In this moment of stillness, we remembered our original quest. Yes, we were here to understand spinors, but perhaps the forest—the journey—was as important as the destination.
This particular forest is called Geometric Algebra A. It is a simple land, yet rich with wonder. To truly know it, we must not just walk its trails but see its beauty, smell its air, touch its textures, and listen to its tales. Some stories are soft whispers; others roar like waterfalls. This is one of those stories, told by the forest itself.
So, dear traveler, let us begin.
We are in geometric algebra A. It is simple, but it has a rich structure. We need to feel this structure by sight, by smell and by touch. We need to be able to hear the stories it says to us, sometimes silently, sometimes in a really loud voice. So this is on of these stories.
A is simple. In algebra saying this has a precise meaning: an algebra is simple if it has no non-trivial two-sided ideals. A two-sided ideal is a subalgebra that is at the same time left and right ideal. We did not consider two-sided ideals yet (and we will not in the future), but it is not difficult to show that A is indeed simple. But we did consider left ideals, and those of a particular form. To construct such an ideal we select a direction (unit vector) n in V, from this we construct p, with p=p*=pp:
p=(1+n)/2
Then we define, let us call it In:
In = {u: up=u}.
This is a left ideal.
Now, the defining equation up=u is equivalently written as un=u (convince yourself that this is indeed the case!). Since n2 =1 (n is a unit vector), and n*=n, it follows that n, considered as an operator acting on A from the right, has two possible eigenvalues +1 and -1. So the equation un=u means that In consists of eigenvectors of n belonging to the eigenvalue +1. This is our left ideal under consideration.
The first thing we notice is that p itself is an element of In. But that is not all In. In is a complex two-dimensional space. Thus there are two linearly independent (even mutually orthogonal) vectors in In.
Thus we proceed as follows: we choose an oriented orthonormal basis e1,e2,e3 in V in such a way that e3 coincides with n. Then e2 and e3 are perpendicular to n. Then we define a basis E1,E2 in In by choosing:
E1 = p
E2 = (e1 - ie2)/2,
Then magic happens: in this basis the left action of e1,e2,e3 on In is given exactly by the three Pauli matrices!
We came to the lake in a forest and it is time to fore out the thinking machine in our brains. We have arrived naturally at Pauli matrices, which is very rewarding. Except for the fact that there is nothing "natural" in this process! First we had to select a direction n in otherwise completely isotropic space V. This cannot be deterministic. No deterministic process can lead to breaking a perfect symmetry. It can be done only by a conscious choice. So consciousness is entering here (or it can be a random choice, but then consciousness is needed to define what precisely a "random choice" is). In practice the choice of a reference direction, and of an orthonormal basis is being made by a conscious "observer" (or by a machine programmed by a conscious "observer"). You can, of course, replace "observer" by "experimental physicist" or "an engineer", but that will not change the idea.
Then we decided to define E2 the way we did above. Another application of consciousness. Thus, temporarily, I am associating the right action of the algebra on itself, the action of p in up=u, with consciousness. It is not needed for further considerations, but it something that should be thought about: we have left and right actions of operators on our Hilbert space. In ordinary quantum theory only left actions are being considered, What can be the meaning of right actions, if any? But let us abandon philosophy and return to math.
We have the basis Eα (α=1,2) in In, we have the basis ei (i=1,2,3) in V and they are related by Pauli matrices σi by the following relation
ei Eα = Eβ (σi)βα. (*)
Notice that I write the right hand side by putting the basis vectors first, and coefficients after. That has the advantage that matrices transforming the components are transposed to those transforming the basis vectors. This way I do not have to transpose anything.
But what happens if we replace our basis Eα by some other orthonormal basis in In? Then the whole beauty and simplicity of Pauli matrices will be spoiled. And we like Pauli matrices so much! And here is the place to demonstrate the power of the desire. We want Pauli matrices, whatever the cost would be! So, we start thinking. When there is a desire, there must be a way! So we start looking at our equation (*) from a different point of view. The elements Eα and ei are related (or "correlated") by the Pauli matrices. If change Eα, perhaps ei also need to be changed so, that the correlation stays the same? We try our great idea of saving our love - the sigmas. The result is condensed in the following statement:
Proposition. There is one and only one way to have Eq. (*), with Pauli matrices in it, valid for all orthonormal bases in In. It goes as follows: if Eα is replaced by E'α related to Eα by a 2 by 2 unitary matrix A of determinant 1 (element of SU(2)):
E'α = Eβ Aβα,
then ei must be replaced by e'i, related to ei by a real orthogonal 3 by 3 matrix R(A) (an element of SO(3)):
e'i = ej R(A)ji,
where the relation between A and R is
AσiA* = σj R(A)ji.
Proof. Left as a straightforward, but needing use of indices, exercise.
And this way we have accomplished something that was left unexplained in October 16 post Part 3: Spin frames.
What we see now is that this correlation between spin frames Eα and orthonormal frames ei is not so "natural" at all. It requires certain arbitrary human-made choices. It has little to do with the "true state of affairs". Spin frames and orthonormal frames are two different realities. Yes, they can be "correlated", but this correlation is artificial. So, the question remains: what are spinors? Elements of a left ideal? But which one? And why this one, and not some other?
Exercise 1. Do the calculations needed to prove the Proposition.
Exercise 2. For any x in A denote by Ax the set
Ax = {ax: a in A}
Show that Ax is a left ideal. Show that it is the smallest left ideal containing x. With In and p =(1+n)/2 show that In = Ap. Why Ap is not the same as An? What is An?
Exercise 3a. If u in A is invertible, it cannot be contained in any of the In's.
Exercise 4. Show that the * operation transforms every left ideal into a right ideal, and conversely.
Exercise 5. If Il1 and Il2 are two left ideals, is their intersection also a left ideal? If Il is a left ideal and Ir a right ideal, is their intersection a two-sided ideal?
P.S. 07-01-25 9:33
Рождество Христово:
В такой день существует обряд — поздравлять близких словами: Христос родился! Славим Его! Вечером в сочельник можно загадывать любые желания.
P.S. 07-01-25 18:47 Today I stumbled upon an unpublished paper submitted to a mathematical journal. The authors advocates the use of three-valued logic "True, False, Undecided". This should supposedly help us with the consequences of Goedel's theorem for physics. Perhaps this logic has something to do with Alain Cagnati's "trits" (-1,0,1) that he likes so much (see comments below). Suppose. So Nature leaves some questions undecided. Perhaps that is a room for consciousness and free will. If Nature leaves them undecided, it opens doors for us to "decide". Suppose. But how do we decide? What is the mechanism behind our decisions? Something to think about.
"by choosing:
ReplyDeleteE1 = p
E2 = e1 - ie2."
Halfy inconsistent.
In is given ->
ReplyDeletesubscript
decided yo ->
ReplyDeletedecided to
In σordinary ->
ReplyDeleten ordinary
in In, we have ->
ReplyDeletesubscript
All fixed. Thatnsk.
DeleteMaybe change "unitary matrix A" to "unitary matrix U", or change U to A in the expression of the exercise?
DeleteChanged U to A. I used before A in Part 3, so that is why I used A instead of U.
DeleteOK, but there's still an U in: UσiU* = σj R(A)ji. Is that also supposed to be an A?
DeleteArk, thank you for always keeping the course and showing the path-leading star for us. Alas, the way of cognition is like travelling 'through the looking glass': we have to run fast in order to stay at least at the same place.
ReplyDeleteFrankly speaking, I do not share your anxiety about unnatural character of spinors. Eventually, all our knowledge is subjective; yes, we choose a direction, violate the symmetry of space, and it is indeed an intrusion of consciousness, so these spinors are only OUR spinors, natural for us in our reference frame. Another observer can obtain his own spinors, with the same properties with respect to his frame. What is the trouble with this?
One more remark on the relations between matrices of different symmetries (unitary and orthogonal for example). I think of the Schur-Weyl duality, perhaps it can help to make this relation less artificial?
What is Shur-Weyl duality?
Delete"nother observer can obtain his own spinors, with the same properties with respect to his frame. What is the trouble with this?"
DeleteBut what is behind these different descriptions? We know that a given house can look different for different observers, if they look form different directions. But ther IS A HOUSE. So, what kind of a house is behjind our descriptions of a spinor?
@Ark "So, what kind of a house is behind our descriptions of a spinor?"
Delete"Not in the spinors is the truth, not among the symmetries
But in the rain, in the dew, in the noise
In the ocean that is,
And I don't understand it"
https://www.youtube.com/watch?v=TuRln6U8Zig
:)
OK. I know what Schur-Weyl duality is. Will look into it. Thank you!
DeleteI do not see how to apply Schur's-Weyl duality to our case. The algebra A has no natural tensor product structure. We have found F and its complement. This would allow us to build the tensor product, and to apply the duality theorem. But once we have F, we do not need the theorem anymore! We have the irreducible representation!
DeleteThe only option would be to start with something like F, and from this construct V. But I am trying the other way around. I want to start with the geometric algebra, not with an abstract complex vector space. Bad choice? Perhaps. But there is Kassandrov, and I am not forgetting his dream.
This comment has been removed by the author.
Delete"Not in the spinors is the truth, not among the symmetries
DeleteBut in the rain, in the dew, in the noise
In the ocean that is,
And I don't understand it"
So, we go need to switch to geophysics then. ;)
@Anonymous
Delete"Perché la gente (I want to know)
non dice niente (I want to know)
ai Mister Hyde, ai Dottor Jekyll,
i construttori di questi orrori
che senza un volto fanno le case
dove la carie germoglia già."
"Why do people I want to know
don't say anything I want to know
about mister Hyde, about doctor Jekyll
the creators of such horrors
who make the houses without a facade
where the cavities are already infected"
https://www.youtube.com/watch?v=JEuMIPVg4Lk
or
https://rutube.ru/video/c0105c278a7f976c63633e6ac4a8a3a3/
1:00:00
@Ark "Perché la gente (chcę wiedzieć)"
Delete"We must know, we will know" Hilbert as Hyde
Godel as Jekyll - impossible !!!!
Feynman as house builder -
"shut up and calculate"
https://youtu.be/N6KZi4egjRU?si=mnQpS1Lq7yAulubO
:)
Yes, we know about Feynman. But, sa my blog tries to demonstrate, it is possible to calculate and NOT shut up. Much better solution, and calculations, in fact, add fun to just thinking that we don't understand.
Delete@Ark "calculations, in fact, add fun to just thinking that we don't understand."
Deletefact,it is funny to presume that thinking and understanding can be related to calculations :)
@Saša "we go need to switch to geophysics then. ;)"
Deletewhy not - but this type of calculation would probably be too simple for Ark? :)
Probably not, as it concerns solving non-linear partial differential equations.
Deletehttps://en.m.wikipedia.org/wiki/Navier%E2%80%93Stokes_equations
@Saša "Probably not, as it concerns solving non-linear partial differential equations."
DeleteI'm not particularly familiar with probability :)))
However, I have seen that he uses programs such as Mathematica , so he can probably manage, especially since such nonlinear equations especially non-integrable systems are rather not solved analytically?
That's true, but even numerically they are extremely difficult to solve to satisfactory level, so even Wolfram Mathematica won't be of much help with that. In fact, even modern HPC systems prove to be inadequate for the task, and that's also one of the reasons why meteorological models are rather unreliable, especially on bit longer time scales. FWIW.
Delete@Saša "they are extremely difficult to solve to satisfactory level"
DeleteNumerical calculations are easy - in the sense - we direct a query and see the results . However, what gives satisfaction to whom - is probably an individual matter :))))
In general, the point is that mathematical physics does not quite explain much - in terms of the non-linear real world - not even much can calculate and predict.
Probably that's why Ark deals with the linear world where you can have the impression and satisfaction that we can do something :))))
"Probably that's why Ark deals with the linear world where you can have the impression and satisfaction that we can do something."
DeleteNot really.
If you read his "Revisiting Wigner's mind-body problem" paper, you'll see quite the opposite argumentation there.
Maybe you should do it, before jumping to conclusions.
And funny thing, you yourself wrote in your previous comment:
"I'm not particularly familiar with probability",
and yet in this comment you assigned a likelihood why Ark's doing something. A bit contradictory, isn't it? :)
Certainly to have a real grasp of reality, we have to go for non-linear. But notice that there is no an autonomous word for non-nonlinearity. Non-linear is defined as something that is goes beyond linearity. So in order to have a firm basis for jumping into the non-linear jungle, we must first learn how to drive fast in a linear world. Even there we have already met a friendly mild non-linearity in conformal transformations. More serious non-linearity will come when I learn better Kassandrov's works. For now we have learned that spinors, if understood as elements of the whole algebra, and not only one of its left ideals, can be not only superposed linearly, but multiplied within the algebra. That is something exciting. Multiplication leads to powers, and these are examples of non-linearity.
DeleteArk, now i see why did you questioned about the meaning of the right ideals. They remind of the Kassandrov's half-derivatives! As far as i know, he introduced them using multiplication in the algebra of biquaternions (2x2 complex matrices), which led him to the generalization of the Cauchy-Riemann equations, which is essentially nonlinear.
Delete@Saša "A bit contradictory, isn't it? :)"
DeleteYes - you are right - as you can see not only with probability but also with logic I don't do very well :)))
@Ark "Non-linear is defined as something that is goes beyond linearity."
DeleteUntil now, I was under the impression that we have definitions of linearity and nonlinearity is everything else. That is, it is a whole ocean of possibilities and linearity is only a distinguished specific enclave. Beyond it there are systems qualitatively similar to linear systems - the so-called systems close to equilibrium, but also systems qualitatively different so-called systems far from equilibrium - to remind Prigogine .
The idea that mastering the court etiquette (linearity) is necessary to go into the wild jungles (strong non-linearity) somehow does not appeal to me :))))
It is all true. But do you have some particular non-linearity in mind that would help us to understand quantum mechanics instead of just "shut-up and calculate"?
Delete@Ark "help us to understand quantum mechanics"
DeleteThe problem is “understanding.” That is, the requirement to describe in a reductionist manner phenomena that cannot be so described.
For example : life,consciousness,chaos, etc.
Galileo blew everything really important from the field of science and we have what we have :))))
Understanding and reductionism are two completely different things. I understand what you wrote, you understand what I wrote. I am not reducing you, and, you are reducing me only in a mild way. You understand? See! Without reductionism.
Delete@Ark "See! Without reductionism."
DeleteOk! I understand.But how to convey it? How to turn intuition/subjectivity into objectivity of the message?
Then maybe, for the sake of example, convey to me your understanding of understanding :)))
It is easy. It is about degrees or levels. If you, for instance, do not know Chinese, and you see a text written in Chinese characters, you say "I do not understand at all what it is about". But when you see my comment, your mind is seeing some message, an attempt in communication, and you manege to reply in the same language and with the same words. It means "your level of understanding is higher than it was with the Chinese. We may have a wish, once in a while, to understand something better than we are understanding right now. This, of course, requires some effort form us, and we may not be ready for some additional effort at that very moment. But we may be readyat some later time, provided we understand what understanding is, and we know, from our own experience, that some psychological comfort may sometimes come form such an understanding something.
Deletethat is of interest for us, better.
Am I being understood now? Did I reply to your question?
@Ark "Am I being understood now? Did I reply to your question?"
DeleteYes! Yes! :)
Good. And please, forgive Galileo his possible sins. Dont be too harsh on him. I started one of my papers: with:
DeleteNote: We use the method chosen by Galileo in his great book "Dialogues Concerning Two New Sciences"[1]. Galileo is often referred to as the founder of modern physics. The most far-reaching of his achievements was his counsel's speech for mathematical rationalism against Aristotle's logico--verbal approach, and his insistence for combining mathematical analysis with experimentation."
@Ark "please, forgive Galileo his possible sins"
DeleteIf he or rather you on his behalf need it - I give it without hesitation. I wonder if I would also receive it for my - anti-scientific - heresies? :)))
https://www.youtube.com/watch?v=Pdl_zfzwXkU
I would give it to you, but there is a possibility of misunderstanding that needs to be avoided. If I would say my full "yes" to Okudzhava, someone could think that I am also approving evil. But I am not.
Delete@Ark "someone could think that I am also approving evil. But I am not."
DeleteA beautiful declaration - I wonder how about the implementation ?
Because with good intentions the road to hell (of quantum_theory ) is paved (too) :)))
This is a scientific blog. A little bit of diversion from science towards more general issues is tolerated as unavoidable in a fruitful discussion. But only a little bit.
Deleteneeding us\e of indices ->
ReplyDelete?
up=p -?
ReplyDeleteup=u
Yup. Not the first time! Thanks.
Deleteeigenvectors of n belonging ->
Deletebold n
@Saša, Bjab
DeleteFixed. Thanks.
From Tony: The later effort is that of Hestenes, Keller, et. al. [15]. They notice that full spinors are defined as 2^n-dimensional minimal ideals in 2^2n-dimensional Clifford algebras, so that full spinors can be defined as either left-ideals or right-ideals. In terms of 2^n x 2^n matrices, the full spinors can be chosen to be either 2^n x 1 column vectors or 1 x 2^n row vectors. To use the whole Clifford algebra, both the left-ideal column spinors and the right-ideal row spinors should be used. They then use the left-ideal column spinors to dene the spinor transformations of the vector spacetime, and the right-ideal row spinors to define the fermion spinor particles and antiparticles. I think that their general approach is correct, but that the 4-dimensional full spinors of D2 Spin(4) are not big enough for the 4-dimensional right ideal row spinors to account for all 8 types of fermion particles: electron; e-neutrino; red, blue, and green up quarks; and red, blue, and green down quarks.
ReplyDeleteback to me: As I previously mentioned, my understanding of ideals is less than ideal so I can't really say I like or dislike the Hestenes general approach but Spin(4) and Cl(4) are too small I think. That said the Hodge duals in the even smaller A aka Cl(V) aka Cl(3) stays important I think even in larger Clifford algebras where there are different Hodge dual relationships.
Hestenes and Tony are examples of ambitious real geniuses. My goal is much less ambitious: I simply want to understand what spinor is? And what quantum theory is about. And I don't.
DeleteDear Ark, you will know soon ;=) You, and your group, are close to THE solution... You know how I'm not good for calculus, but I can maybe help you. To start with Real-Numbers is not the good way to create Complex-Numbers (Wildberger). You have to start with finite-field {0, 1} and the second one is "maybe" ? {1, w, w^2} with w the cubic-root from one. So "maybe" the 3x3 orthogonal real matrix could be diag{1, w, w^2} !? (I know it is not Real, you prefer this one ((0 0 1),(1 0 0), (0 1 0)) who cube to ONE (Identity) and is just one of the six permutations of ONE) It's not so difficult to make emerge Complex-numbers from 2x2 binary matrix ... I read all the comments and my intuition is there. You know how the Eisenstein numbers are in my head ;=)) As the golden ratio...
Deletehttps://quantumgravityresearch.org/portfolio/quantum-walk-on-a-spin-network-and-the-golden-ratio-as-the-fundamental-constant-of-nature/
Delete"It's not so difficult to make emerge Complex-numbers from 2x2 binary matrix "
DeleteHow?
OK. I see, you can do it. But what is the advantage of playing such games? "It from bit?" That would be a dead "It".
DeleteIn the annex of the paper above, you can see how to make emerge Complex-numbers, Quaternions, Cl(3, 0) 8D, etc from "bits" (binary matrix). Yes ! You can see emerge you favorites matrix Pauli and Dirac... All you need for physic. I'm pretty sure the Nature make the complexity from simple things. Has the Nature built the fibonacci sequence by add 0 + 1, 1 + 1, 2 + 1, and so on. The golden ratio is just a rationnal number from this sequence. To construct the World, God use just few symetries (or permutations of the 3D vector space). I remember that Grothendieck's dream was to construct "topology" from a "combinatorics" way. The paper above use {-1, 0, 1} as trits. A three elements field. And I remember how much Wildberger insists on reconstructing the "true" Complexes from an algebraic method (without going through real numbers, the poorly defined infinity, etc.). At the end of my little paper about Dihedron Algebra, we can find epsilon = ((0, 0) , (1 ,0)) 2x2 matrix. Epsilon is the new "infinitesimal" very usefull for calculate derivation (rigth and left). You can have epsilon' = ((0, 1) , (0 ,0)) who square to zero. It is a lot of things that makes me believe we must construct from "bits" {0, 1} without any irrational numbers... as Wildberger recommends. And as Constructivist mathematicians advocate. I tell you this, not to prove but "maybe" give you a new point of view ;=)))
DeleteYes, I have looked at the annex. But I do not see how this will help me to understand quantum theory? Do you have any idea in this direction?
DeleteElementary Cellular Automata (ECA) has a Cl(8) symmetry so it is easy to think of Clifford algebra as bits. ECA starts with 3 bits Cl(3)-like and ends with 256 8-bit rules Cl(8)-like.
DeleteStill want to know what a spinor is because via ECA it can seem like a vector since odd grades are for spinor fermions but grade one is for vectors. I think Tony was OK with this via triality and symmetry breaking? I also think EEQT is as ambitious and ultimately the same thing as what Tony was trying to do.
This comment has been removed by the author.
DeleteThere is a "funny" analogy, once again, with the Toffoli gate, also known as the CCNOT gate, an Universal logic quantum gate for quantum computers, with 3 qbits in and 3 out. The link between "logic", "quantum calculation", and Cl(3, 0) 8d is strengthened. The "topology", "geometry", "real-numbers", and space-time-continuum could emerge from this binary logic. In my head, I see all this like a superposition of all the permutations of the three vectors (x y z). A sort of permanent "trembling movement" or in german a ”Zitterbewegung”. Hestenes said a lot about this, but always with "Space Time Algebra", never with the simpler version "Algebra for Physical Space"...
DeleteThat is true. But I still do not know what spin is, what spinor is, and what "quantum measurement" is. And I want to know.
Delete>That is true. But I still do not know what spin is, what spinor is, and what "quantum measurement" is. And I want to know.
DeleteА можно ли об этом узнать изучая лишь алгебраические структуры? Думаю этого не достаточно, для понимания необходимы метафизические представления этих алгебраических структур. Кстати, ещё раз спасибо за рисунок, и если интересно во что это вылилось, то посмотрите раздел 2.3 (Колебания метафизического маятника). Это к вопросу о пользе моделей.
Забыл дать общедоступную ссылку
Deletehttps://www.researchgate.net/publication/322369062_Matematiceskie_zametki_o_prirode_vesej
или
https://www.researchgate.net/publication/329252706_MATHEMATICAL_NOTES_ON_THE_NATURE_OF_THINGS
Thank you Igor. But I am looking for a simple picture. Algebra gives us the group that we need: SU(2). But it does not give us a natural irreducible representation. To get such a representation we need to select a left ideal, one out of infinitely many possible, since there are infinitely many ways of choosing a direction in space. I am not happy with this picture. In your paper you are addressing different questions, more complicated and using more advanced tools.
DeleteПрежде чем выбирать направление в пространстве стоит выбрать само пространство. Затем необходимо заполнить пространство материей, которая и укажет нам нужное направление.
Delete"необходимо заполнить пространство материей"
DeleteHow? Where are you getting this matter from? And what exactly is "matter"?
Лично я представляю материю как векторное поле скоростей её частичек, а динамику каждой частички материи я представляю как ту, которая минимизирует длину её траектории. Самое главное это подбор пространства, в котором движутся эти частички. Тогда геометрическая алгебра должна проявится естественным образом.
Delete"я представляю как ту, которая минимизирует длину её траектории."
DeleteBut there are infinitely many such solutions. How you select just one?
Вы правы, даже глобально минимальный поток эволюционирует к абсолютно минимальному состоянию, а ещё имеют право на жизнь и просто локально минимальные потоки частиц. Но для алгебраического описания материи достаточно ограничиться линейными векторными полями.
Delete"статочно ограничиться линейными векторными полями."
DeleteStill there are infinitely many such vector fields. How do you select just one of them?
Я рассматриваю простейшие линейные векторные поля в пространстве наделённом одновременно евклидовой и псевдоевклидовой метрикой, следовательно выбор ограничен изотропным конусом.
DeleteBut there are infinitely many such vector fields. How do you select just one of this infinitely many possibilities? How?
DeleteЕсли вы имеете в виду то, что существует бесконечно много линейных векторных полей, удовлетворяющих алгебре, то это как раз хорошо, это соответствует тому, что существует бесконечно много решений системы уравнений квантовой механики, удовлетворяющих алгебре Дирака. А если вы о выборе направлений изотропного конуса, то и это нормально - такова структура выбранного пространства.
DeleteBut I am asking: how do you select one of these vector fields. I am not asking about what people do in quantum mechanics. How do YOU select just one vector field? Tell me how you do it? How do you choose to plot just one particular and not some other?
DeleteМои базисные векторные поля задаются выбором базиса пространства. А выбор базиса пространства произволен, но ограничен метрикой. В свою очередь, метрика порождается динамикой векторного поля замкнутого многообразия.
Delete"completely homogeneous space V"
ReplyDeleteDoes this mean isotropic?
Fixed. Thanks.
Delete@Ark, 'I do not see how to apply Schur's-Weyl duality to our case'.
ReplyDeleteIn fact, I meant something more general when mentioned the Schur's-Weyl duality. As i see it, it is a relation between two types of symmetries, where one defines the other. In the traditional form, it is duality between unitary operators and permutations. Isn't it remarkable that applying a group operator we obtain just another element of the same group, which can also be expressed as a permutation of elements? The S-W duality has modern generalizations, i could send you by mail my short note on this subject, if you find it worth looking at.
This comment has been removed by the author.
ReplyDeleteMessed up with 'bolding', it should be OK now.
Delete"First we had to select a direction n in otherwise completely isotropic space V. ... It can be done only by a conscious choice.
...
Then we decided to define E2 the way we did above. Another application of consciousness. Thus, temporarily, I am associating the right action of the algebra on itself, the action of p in up=u, with consciousness. ... In ordinary quantum theory only left actions are being considered, What can be the meaning of right actions, if any?"
There's an additional thing that might be considered, apropos the right action and right ideals, in relation to the left ones that are standardly worked with.
Selection of direction seems to involve in fact a double choice, first the "line of action", n, and then also the "real direction", i.e. +n or -n, as p'=(1-n)/2 also qualifies for the solution of p'p'=p'=p'*, that is for a general form of Hermitian idempotent. Is that an independent solution in relation to p=(1+n)/2? Well, maybe and maybe not, because we can always "choose" in the start a direction pointing in the opposite "side" of the "line of action" and say that's our +n.
When determining the left ideal for this solution, we again get two independent basis elements, u1'=z(1,-n) and the corresponding u2' = -i u2'×n. Interesting thing to note there is that the second part, i.e. E2 for left ideal in case of p'=(1-n)/2, turns out to be like the E2 for the right ideal in case of p=(1+n)/2, i.e. u2=i n×u2 (and vice versa). So, maybe instead of associating the right ideal with consciousness per se, we could perhaps start with associating it with antiparticles first?
Moreso, as these two solutions, p and p', commute one with another and their product is zero: (1,n)(1,-n) = (1 - n.n,n - n - in×n) = (0,0), which then for the product of their left and right ideals (or vice versa, as they also commute one with another, according to my calculations) gives (u1+u2)(u1'+u2') = (u2^2,0), where u2=u2' as noted above, can we maybe associate this type of the product with the particle-antiparticle annihilation? If so, does that suggest that light (and maybe information) could be perhaps described as a scalar part of our elements in A, i.e. scalar parts of (bi)quaternions as four dimensional geometrical objects?
According to the definition of an ideal, element (u2^2,0) as a product of a particular left ideal and a particular right ideal should be element of both of those sets, making light particle or photon its own antiparticle, just "going" in the "opposite" direction, i.e. with opposite polarization, that is with opposite spin.
Would that bring us closer to understanding consciousness per se? Perhaps, as we see that light (and maybe information) would not really be there in the corporeal world in the real sense as part of our ordinary 3d space, but comes into play only when we go to higher dimensional environment, at least 4d for starters, either in the scope of geometric algebra or by simply switching to quaternion case scenario. Additionally intriguing, and a bit puzzling, is that for our choice of E2, that is of u2 = e1 - ie2, u2^2 = 1-1 = 0, while only when taken as Hilbert space product of these two complex vectors within Tn, it would return the expected value of |u2|^2. Would that be suggestive of difference between mass and energy as complex scalars in our 4d descriptions? I don't know, and maybe all this is just another instance of my mind playing games with me, no matter how attractive these ideas and interpretations seem.
So, a word or two from our wise teacher and guide in these adventures and quest would be most welcome and highly appreciated.
@Saša, a very interesting observation about u2^2 behaviour and especially its interpretation involving physical terms of mass and energy. It somehow correlated with my feeling that mass is a natural attribute of objects in 3d space, while energy is the same in 4d space.
DeleteBesides, I wanted to tell Ark that the view of a house is different for every observer since it is a projection to the 2d plane normal to the line of sight. In order to see the house as a whole, one needs to raise to the 3d space. This analogy can be evidently extended to the 4d dimension. It is so trivial that i refrained from saying it before, but yours 'at least 4d for starters' resonated distinctly, so i could not keep silent any more :)
And - yes, I also wait impatiently a word from our wise teacher on the subject.
Added Exercise 2.
Delete@Saša Anti-particles may fit. But then is there any deep relation between consciousness and anti-particles?
DeleteThere might be, but I'm leaning more towards the connection with light (and information), because antiparticles as we know them in our environment, are not very much different than particles, that is they are the same as particles just with some of their quantum numbers of opposite values. Except the mass, by all known observations the mass of antiparticles is the same as mass of their particle counterparts.
DeleteSo, if we ascribe a consciousness to a particle, no matter how rudimentary it might be, the same could be ascribed to corresponding antiparticle, no real reason why we should treat it any differently. In that sense, when "joined" together, particle-antiparticle 'pair' as such might have some deeper relation(ship) to consciousness, and then we might perhaps talk about consciousness being "antiparticle" to matter in general, because what is called antiparticles in today's physics is frankly still just matter, as concerns their physically measurable masses.
In which way would you relate antiparticles to consciousness, in contrast to particles, or even without any relationship between particles and consciousness?
Antiparticles "are not very much different than particles,"
Deleteand yet they annihilate each other.
Exactly.
DeleteBut they do not "annihilate" each other to "nothingness", they together "annihilate" to light or create light by joining their counterparts, and they also come into being by light or from light, if that light is energetic enough to sustain the pair, that is to sustain both of them together in the same place at the same time.
Why would you assign deep connection with consciousness only to antiparticles, and not also to 'ordinary' particles?
I am not assigning consciousness to antiparticles. I am thinking about the possible relation of consciousness to the fundamental duality.
DeleteI am not saying you assigned consciousness to antiparticles, that's closer to what I would do, but that you implicitly suggested "deep relation between consciousness and anti-particles", and so I'm asking why just "deep relation" with antiparticles and not also with particles? What does make antiparticles more special than particles, as both are just matter from the point of view of their material masses?
DeleteI was not precise enough. I am not trying to relate consciousness to antiparticles per se, but to their existence. In other words: would their be no antimatter, there would be no light and no consciousness. Why? I don't know.
DeleteBut that is not exactly true, that there would be no light and consciousness without antimatter. Well, except if we consider any light as a potential origin for neutrino-(anti)neutrino pair, that is, if neutrino has a minimal non zero mass which seems to be so.
DeleteWhat I'm trying to relate here is that going with the same reasoning we can say that there would be no light if there would be no matter, as matter plays equally important role as antimatter in this picture. If one is lacking, the other wouldn't be able to create light, i.e. to annihilate itself by itself.
It seems there's something more to light and consciousness than just existence of material or physical antimatter and matter.
But as this consciousness of mine needs some material fuel in a form of lunch, will come back to this subject after it. :)
We are living in the material part of the universe. Anti-matter is rare in this part. Is it really necessary? For what?
DeleteYou might be surprised about the abundance of antimatter in our world; Earth's atmosphere is full of antiparticles from its interactions with cosmic rays, especially cosmic ray photons.
DeleteBy all the science today there should have been equal amount of matter and antimatter in the universe, the currently observed difference is still not properly understood and described.
So the question might be modified to "is material, in the sense of both matter and antimatter, universe or existence really necessary, and if so what for?"
One possible answer might be, for learning purposes and for freedom of expression, so that a consciousness, as a part of the overall total consciousness of all there is, might express itself and learn.
Evidently opinions differ here. See e.g.
Deletehttps://www.scientificamerican.com/article/why-is-there-more-matter-than-antimatter/
Why do you think so?
DeleteIn the article you posted link to, it's clearly stated:
"While we still cannot completely solve the mystery of the universe’s matter-antimatter asymmetry", so who's opinions differ from who's, and in which part?
What I am pointing at is that maybe instead of focusing on matter-antimatter duality, as both are in fact a facet of material existence, perhaps we might look for consciousness relation to matter-energy duality?
DeleteA rather rough example in the form of question, is the material content of my body, in the sense of its matter-antimatter concentrations, any different now while my consciousness supposedly inhabits the body compared with when I die and that consciousness leaves the body?
Why do you focus exclusively on a relation between consciousness and antimatter and not also on relation between matter and consciousness? What makes antimatter special in that respect?
The thing is, I see consciousness in every material existent thing there is, no matter how small and insignificant it might seem from the point of view of normal human or even animal one. Every electron and positron as well, every living cell and a planet or a star or a galaxy has a consciousness of some kind corresponding to the being of that existent thing. For things I am not aware they exist, I can't say, but would guess they have a consciousness too. In that respect I make no difference between matter and antimatter, and trying to understand your point of view, that is why do you see them differently and relate consciousness to existence of one and not the other, or both of them equally? Can you explain that?
Delete"abundance of antimatter"?
DeleteRather inadequate estimate I would say. The article states " the universe we see today is made entirely out of matter", and you seem to disregard that fact by stating that instead there is an "abundance of antimatter". Using the term "abundance" seems to me suspicious in this context. The article states that matter-antimatter asymmetry "is one of the greatest mysteries of modern physics". And I agree.
Agree, on both accounts.
DeleteThe term "abundance" was an exeggeration, a bit over the top expression that in every high energy cosmic ray, especially photon induced atmospheric shower of secondary particles there are millions and even billions of antiparticles, mostly positrons. Apologies for that. Thought that it didn't affect the main theme of the discussion, which is the question posed at the beginning, what makes antimatter or its existence any different from matter in relation to consciousness?
Perhaps antimatter opens windows to other dimensions? Minimal quantities are needed, like with an enzyme? But it is just my speculation.
DeleteIn fact Ark, I wrote this:
Delete"You might be surprised about the abundance of antimatter in our world; Earth's atmosphere is full of antiparticles from its interactions with cosmic rays, especially cosmic ray photons."
explicitly stating the context of the antimatter presence in the Earth's atmosphere, and presented that more like a fun fact than as a contrast or as "instead" as you later compared to one of the first paragraphs of that article.
On the other hand, in that article they didn't say that actually we don't know for certain that the universe is made only out of matter, as our measurements reached only as far as the end of the heliosphere with Voyager and similar probes. Outside of the heliosphere we can only infer from today accepted and tested in the lab theoretical considerations.
But, again, that's deflecting from the main question I'm still trying to understand your answer to, why consciousness plus antimatter and not also consciousness plus matter?
Consciousness is evidently using matter. We have abundance of data to support this statement. Then we may ask another question: would consciousness exist in a hypothetically perhaps possible universe with matter alone, no antimatter at all? One possible answer: it is pointless to ask such quations. Another possible answer: why not? And my working hypothesis is: not!
Delete"Perhaps antimatter opens windows to other dimensions? Minimal quantities are needed, like with an enzyme? But it is just my speculation."
DeleteIntriguing and interesting speculation. I like it, and it basically comes back to energy; antimatter in contact with matter produces high amounts of energy, or to be more precise high energy. That's why we need only a little bit of it.
Before LHC there was LEP, which worked exactly on that principle, matter-antimatter collisions, while LHC replaced that for mainly proton-proton collisions, both aiming for higher energies.
So I would look more into relation between consciousness and energy, light and information, than just purely antimatter. More energy, larger the phase space of possible configurations that might come out. Although, when energy is too high, basic configurations are not attenable any more.
"And my working hypothesis is: not!"
DeleteThat's quite a daring and provocative hypothesis.
As you maybe know, I graduated on theoretical thesis of BAU and leptogeneses when that was popular thing to study some 20+ years ago, but then moved into experimental waters, so would be very interested to hear and try in practice, if possible, any ideas how to test that intriguing hypothesis. If you have any that is.
There are two ways the Planck constant enters fundamental physics:
Delete1. in [Q,P] = i hbar
2. minimal spin is hbar/2
Which one is THE fundamental? I chose option 2. Then 1 should somehow be a consequence of 2. So I want to understand spin. Completely. All the rest must wait.
Understood.
DeleteAnd now I also see why do you put so much emphasis on understanding the spin. Will do all in my power to help you as much as I possibly can on that quest and journey.
Thank you for sharing.
I have a problem.
ReplyDeleteWhich one of two formulas is corect?
uv = (p0q0 + p·q, p0q + q0p + i p⨯q)
or
uv = (p0q0 + p·q, p0q + q0p - i p⨯q)
*correction
Deletecorect ->
correct
As long as e1e2=e1e2e3e3=ie3 and e1 x e2 = e3, the first formula i "correcter" than the second one.,
Delete@Anna
ReplyDelete"The S-W duality has modern generalizations, i could send you by mail my short note on this subject, if you find it worth looking at."
Please, do it!
Exercise 1 is more or less done. With one uncertain assumption that we can raise and lower indices using the most simple metrics delta_ij, i.e. vectors e_i and e^i are transformed similarly. It this admissible?
ReplyDeleteOne more 'thin place' is that the inverse matrix A^-1 is equal to the Hermitian congugated A*, which is true if A is unitary.
Yes, I implicitly assumed that we use delta_ij. Very good!
ReplyDeleteBut then we need exercise 3: Show that if two orthonormal bases in a Hilbert space are connected by a matrix A, then A is automatically unitary.
"by choosing:
ReplyDeleteE1 = p
E2 = (e1 - ie2)/2"
Still inconsistant.
E1 = (e0 + e3)/2 = (1 + e3)/2
E2 = (e1 - ie2)/2
would be better.
Or
E1 = p = (1 + n)/2
E2 = f(n)?
"E1 = p
DeleteE2 = (e1 - ie2)/2"
Still inconsistent."
Why inconsistent?
"E2 = f(n)?"
Which f? There is a free choice involved.
"Why inconsistent?"
DeleteIs E2 = (e1 - ie2)/2 the couple for E1= p = (1 + n)/2 for n not equal e3 ?
"Which f?"
Some.
But I wrote it explicitly:
Delete"Thus we proceed as follows: we choose an oriented orthonormal basis e1,e2,e3 in V in such a way that e3 coincides with n. "
In general we can chose E1 = p(n) and Re(E2) any vector in V perpendicular to n. Then E3 is automatic. Right?
"But I wrote it explicitly..."
DeleteI'm a bit picky, optically p is more associated with (1+n)/2 (with any n) than with (1+e3)/2
"Then E3 is automatic."
?
You mean E2?
Delete""Then E3 is automatic. ?""
DeleteWe automatically choose
E3=E1xE2.
Which completes E1,E2 to a unique oriented orthonormal basis. I have forgotten to stress the normalizations.
E3=E1xE2. Is it vector or Clifford multiplication?
DeleteIf we take, as we did, E1 = e0 + e3 and E2 = e1 - ie2, then Clifford product E1 E2 = 0. How can it complete E1, E2 to a basis then?
I messed up again. Should be E3=nxE2
Delete"Then we define a basis E1,E2 in In by choosing"
ReplyDeleteIf we chose n then for simplicity we no longer have a choice about E1 and E2.
"Then we decided to define E2 the way we did above."
We no longer have a choice as to E2.
For the real part of E2 we can choose any vector in the plane e1,e2.
DeleteThe process of doing exercises is like fighting a hydra - one head is cut, three more appear :) I think Ex. 3 could be made rather straightforwardly, using definitions. Tomorrow i will gather my thoughts together and do it accurately.
ReplyDeleteFor now, the comparison of two cases up=u, where p=1/2(1+n)
and un=u. The task was to convince oneself that these conditions of being in the ideal are equivalent. First condition leads to
u = {u0+u3, u1+iu2, u2-iu1, u0+u3}
the second to
u = {u0, u1, -iu1, u0},
which are seen to be the same in essence.
In the first case, the structure is suggestive, one can guess the basis consisting of E1=e0+e3 and E2=e1-ie2.
When turned to Ex.2, i decided that the ideal-defining equation is
axx = ax, or xx = x
Whence we get x0 = 1/2, as expected, and the condition for scalar product: (x x)=1/4. That is
(x1)^2 + (x2)^2 + (x3)^2 = 1/4
Since components can be imaginary, a whole set of solutions is possible. For example, x1 = i x2 and x3 = +-1/2. Notice that (x1)^2 + (x2)^2 = 0 and they are isotropic vectors on the x3=+-1/2 planes.
What could this mean? Probably, a link between spinors as isotropic vectors and spinors as ideals has appeared? Or it just seemed to me.
The solution is much simpler. It is so simple that it is difficult to see it....
DeleteRe Ex. 2
DeleteThe set Ax is an ideal because it's the left action of the whole A and whichever element of A acts "again", it will just be like another left action of an element in A. FWIW.
Saša So far so good.
DeleteAdded Exercise 3.
DeleteThis comment has been removed by the author.
ReplyDeleteYes, good enough!
DeleteA part of my text disappeared somewhere in the road when being posted. I deleted it and reposted below.
DeleteEx.3. 'Show that if two orthonormal bases in a Hilbert space are connected by a matrix A, then A is automatically unitary'.
By definition, orthonormal basis = delta _ij
Transformation A should preserve orthnormality:
= delta_ij =
In Hylbert space, scalar product of operators has the property:
(Ax | Ay) = (x | A*y) =>
= < e_i | AA* e_j> = , whence it follows that
AA* = 1. And this is the definition of unitarity operator A.
There was a small cheat when replacing matrix by operator, but unitary matrix should correspond to unitary operator and vice versa, though i am not sure that it is accurately done above.
Sow that ->
ReplyDeleteShow that
Ooh! Thanks.
DeleteAdded Exercises 4 and 5.
DeleteRe. Ex. 2
ReplyDelete"Why Ap is not the same as An?"
n is in An, n is not in Ap.
But isn't Ap in An?
Delete"But isn't Ap in An?"
DeleteDoes it contradict the "n is in An, n is not in Ap"?
No. But why p is An, but n is not in Ap?
Delete"p is An" ->
Delete?
Is in An. Sorry.
DeleteI don't know if p is in An.
DeleteYou should know. n is invertible.
Delete"You should know. n is invertible."
DeletePlease explain.
un is in In for any u. But nn=1.
DeleteSo 1 is in In.
But then u=u 1 is in In for any u.
"un is in In for any u. But nn=1.
DeleteSo 1 is in In.
But then u=u 1 is in In for any u."
Don't understand. Did you mean An instead of In?
Yes!
DeleteA comment to Exercise 2. May i take n in the form (0,0,1)? Calculation is a bit too cumbersome in general form.
ReplyDeleteIf we can take in Ex.2 n=(0,0,1), then An = {a3, ia2, -ia1, a0}
ReplyDeleteand p = (1/2, 0, 0, 1/2) belongs to An,
with a0 = a3 = 1/2 and a1 = a2 = 0.
While n is not in Ap. Checking:
Ap = 1/2 {a0+a3, a1+ia2, a2-ia1, a0+a3}
n = {0,0,0,1} and we have an inconsistent conditions
a0+a3 = 0 and a0+a3 = 1.
Therefore, n is not in Ap.
Metaphysical meaning of this result is not clear for me yet.
"then An = {a3, ia2, -ia1, a0}"
ReplyDeletePlease explain.
i meant to take an element a = {a0, a1, a2, a3} and multiply it by n = {0, 0, 0, 1} using the sacramental formula for Clifford multiplication: a n = a0 n0 + (an), a0 n + n0 a + i [axn].
DeleteAfter substitution of n = {0, 0, 0, 1} the product a n becomes simply {a3, ia2, -ia1, a0}
But i am not sure that such simplification is valid if one wishes to prove the statement in general form.
And this covers the whole of A, right?
DeleteIn general: suppose I is a left ideal. Suppose it contains an invertible element. Then I=A.
DeleteWhy? let u be invertible in I. Then u^-1 exists. But u^-1 u =1. Since u is in I, which is a left ideal, also u^-1 u is in I. That means 1 is in I. But then any element v of A is in I since it can be written as 1multiplied from the left by v.
Such a nice and unexpected result! I will wrap my mind around it. Thanks a lot for this refined explanation.
DeleteSome considerations concerning Ex.3a
ReplyDeleteBy definition, ideal is In = {u, up = u}
If u is invertible, then there exists nonzero u^-1 and one can multiply both sides of the defining equation by u^-1.
Then, one gets u^-1 u p = 1 and the projector p = 1 is trivial.
Therefore, an invertible u should not belong to any nontrivial ideal.
Looks too easy to be true... but to begin with )
That is a correct reasoning.
DeleteExercise 4. Show that the * operation transforms every left ideal into a right ideal, and conversely.
ReplyDeleteIf i remember right, we showed somewhere in this Blog that *operation on a product permutes the factors (ab)* = b*a*.
If this is correct, then the solution suggets itself.
Given the equation for left ideal: up = u, one obtains for u*:
u* = (up)* = p*u*, and this is the equation for the right ideal.
One thing is confusing: now the projector is not p, but p*.
But p*=p.
DeleteBut for p we used p=p*=pp, i.e. p is real, so it seems OK.
DeleteHope that resolved the confusion.
Oh, really. Thank you, Saša, thank you, Ark. I missed that from my sight.
DeleteExercise 5. If Il1 and Il2 are two left ideals, is their intersection also a left ideal?
ReplyDeleteLet us write for the first left ideal Il1={u: u p1 = u}
and for the second left ideal Il2={u: u p2 = u}
Their intersection is a set defined by {u: u p1 p2 = u} or {u: u p2 p1 = u}
and it looks like also a left ideal with p = p1p2 or p = p2p1.
We have even two left ideals, if p1 and p2 do not commute.
There is a subtle point here. We know that if p is idempotent then the set {u: u p1 = u} is a left ideal. But we do not know if every left ideal is of this form. We did not study this question. So the answer to the exercise should use a general definition of a left ideal.
DeleteYes, i presumed that p is an idempotent. If it is not, then the situation is even more interesting. I am taking a timeout then.
DeleteRe Ex. 5
DeleteAnswers to the questions would be "yes" and "no".
For a left ideal S, left action of any u in A on any s in S gives result in S, therefore by definition, left action of any u in A on elements of intersection of two left ideals has to be in both of those two ideals, i.e. in their intersection, so intersection of two left ideals is again a left ideal.
Similar reasoning works for intersection of left and right ideals; the results of left and right actions of any u in A on elements of intersection of left and right ideals have to be in those two sets respectively, but as those two actions do not generally give identical results, i.e. us and su for s in both left and right ideal are not necessarily the same, then in general they will not be in their intersection, so the intersection of left and right ideals is in general neither left nor right ideal, thus also not a two-sided one.
This is a very satisfactory and logical reasoning. I wish Anna would read it and accept.
Delete