For warming up, let me start this post with some quotations. First, a snippet from an online article titled "Discovery of Electron Spin":
"The discovery note in Naturwissenschaften is dated Saturday, 17 October 1925. One day earlier, Ehrenfest had written to Lorentz to make an appointment for the coming Monday to discuss a "very witty idea" of two of his graduate students. When Lorentz pointed out that the idea of a spinning electron would be incompatible with classical electrodynamics, Uhlenbeck asked Ehrenfest not to submit the paper. Ehrenfest replied that he had already sent off their note, and he added: "You are both young enough to be able to afford a stupidity!"
Ehrenfest's encouraging response to his students’ ideas contrasted sharply with that of Wolfgang Pauli. As it turned out, Ralph Kronig, a young Columbia University PhD who had spent two years studying in Europe, had come up with the idea of electron spin several months before Uhlenbeck and Goudsmit. He had put it before Pauli for his reactions, who had ridiculed it, saying that "it is indeed very clever but of course has nothing to do with reality." Kronig did not publish his ideas on spin. No wonder that Uhlenbeck would later refer to the "luck and privilege to be students of Paul Ehrenfest."
A. Pais, in Physics Today (December 1989)
M.J. Klein, in Physics in the Making (North-Holland, Amsterdam, 1989)"
So, we know that the birth of the spin concept was not an easy one. Ideas that would revolutionize physics were initially dismissed, sometimes with sharp words, and only managed to take root under the shelter of intellectual bravery and a bit of recklessness. But here we are, nearly a hundred years later. Spin is no longer a child; it has matured into a cornerstone of quantum mechanics. How is it faring in its adulthood?
To answer this question, let us turn to another fascinating quotation, this time from the comprehensive Wikipedia article "Spinor." In the section titled "Attempts at intuitive understanding," we find the following:
"Nonetheless, the concept is generally considered notoriously difficult to understand, as illustrated by Michael Atiyah's statement that is recounted by Dirac's biographer Graham Farmelo:
No one fully understands spinors. Their algebra is formally understood but their general significance is mysterious. In some sense, they describe the "square root" of geometry and, just as understanding the square root of −1 took centuries, the same might be true of spinors."
Centuries? Do we really have to wait that long? Can we afford to? And what is it, exactly, that remains a mystery? Despite all the successes of quantum theory and the remarkable applications of spin physics—from magnetic resonance imaging to quantum computing—why is it that some of the brightest minds in physics and mathematics are still uneasy? Are they just inveterate malcontents, destined to grumble in the face of progress? Or is there a deeper puzzle lurking beneath the surface, one that defies our current frameworks?
I think the answer depends on the level of curiosity, which varies greatly among individuals. It has nothing to do with optimism or pessimism but rather with an insatiable desire to dig deeper. And so, with this in mind, let us make an inventory of what we have learned so far. Let’s take stock of the journey that has brought us here and explore the questions still waiting for answers.
Our starting point was a "vector space". Why vectors? Well, we
have to start with "something". Vectors is a reasonable starting point.
Not excessively simple and not excessively complicated. Why 3
dimensions? This is a good question. There must be a reason for three
dimensions. One possible way of dealing with this question would be:
"because organic life is possible only in 3D". But that would be an
answer that begs other questions. How could some intelligence, that
creates it all, know in advance what is possible and what not? By being
able to see into the future, even if only vaguely? Perhaps, but that
would lead us into speculations with no end in view. So, let us stay
with an empirical fact - we live in 3D space.
Then we added an Euclidean metric to V. Another empirical fact. Perhaps on a large scale the geometry is non-Euclidean, but infinitesimally is Euclidean enough. So, let us start with the Euclidean flat space and see where we can go along this path. That is what we are doing now.
Then we endowed V with "orientation". This is more iffy. Why this orientation and not the opposite? It hurts our love feelings towards some "perfect symmetry". It hurts badly. Yes, there is an empirical fact - we live in a universe with broken parity. But it hurts. So, we look for possible remedy. Perhaps our universe is a two-sided surface, a boundary separating two higher dimensions? On one side of this surface there is one parity, on the other side there is an opposite parity? Perhaps the surface is not necessarily of zero thickness? Perhaps, occasionally, the two sides can "communicate" somehow? That is for the future. For now let us accept the fact: we live in a 3D space with a preferred orientation.
Once we have our starting point, we turn on the Clifford algebra machine. It rewards us with the complex geometric *-algebra A,
isomorphic to complex quaternions, it rewards us with three
involutions, with the group isomorphic to the group SL(2,C) of special
relativity, and with the group isomorphic to SU(2), usually employed in
the study of simple spinors. But there are no spinors yet.
In quantum physics spinors transform under the irreducible
representation of SU(2). We were not discussing group representations so
far, but we were discussing the representation of A.
Once we have a representation of A, we have also a representation of any
group contained inside the algebra. And, in quantum theory, to deal
with spinors we really need the algebra, not only the group.
To relate the Clifford algebra to quantum theory the method of
searching for minimal ideals has been invented by algebraists. Usually
just one left ideal is picked up and it is shown that this enables us to do
all standard tricks with spinors, write equations, add interactions,
etc.
We already know how to construct these ideals. We need to choose a Hermitian idempotent p. Each such nontrivial p is of the form p=(1+n)/2, where n is a unit vector in V, a direction in 3D space. Algebraists define then the ideal generated by p as Ap = {up: u∈A}. I have chosen another, equivalent, way, a way that looks as a solution of a (right-sided) eigenvalue problem:
Ap = {u∈A: up=u} = {u∈A: un=u}
Note: We can also look for the eigenspace belonging to the eigenvalue 0 of p. This would give us a complementary left ideal. It can also be obtained as the eigenvalue 1 subspace for p'=(1-n)/2, corresponding to the choice of the opposite direction.
We consider p as an operator acting on A from the right. Since p2=p, eigenvalues of p are 0 and 1, and we are looking for a subspace belonging to the eigenvalue 1. Similarly for n. Since n2=1, n has eigenvalues +1 or -1, and we are looking for a subspace In belonging to the eigenvalue +1.
What can be the meaning of selecting just one such ideal?
Perhaps it is with selecting a point in space to set up a reference
frame there, as Alain Cagnati suggested? We select a reference direction
in space to be able to quantify what we measure? This defines our
two-dimensional reference Hilbert space. Or it is like choosing a
certain perspective, so that we can map the 3D house on a 2D canvas, as
suggested by Anna?
So we select a reference direction, set up our axes e1,e2,e3 in V, and E1,E2 in In, and produce our 2D complex Hilbert space with a basis E1,E2. We get then Pauli matrices as representing left action of the basis vectors of V. Standard quantum theory of spin is reproduced within In.
Another choice of n would then give us
hopefully equivalent description. Except what "equivalent" is is not
completely clear. It needs to be clarified. We will come to this point
later on.
Given a projection p=(1+n)/2 we can act with it from the left or from the right (or both ways at once). Given two projections p=(1+n)/2, q=(1+m)/2, we can ask two questions at the same time: find all u in A, satisfying simultaneously both equations:
1. up = u
2. qu = u.
We can interpret 1. as setting a reference direction to be n, and interpret 2. as finding all spin states with spin direction m. This is intersection of one left ideal with one right ideal. It is always one-dimensional.
Exercise 1. Prove this last statement.
What if we do not want to select a reference direction n? We can ask 2. without asking 1. We get a 2-dimensional (complex) right ideal. Strange. We are dealing now with two-dimensional complex subspace of a 4-dimensional complex space A. 2D subspaces suggest using bi-vectors. Bi-vectors are elements of a Grassmann algebra or Clifford algebra. Which suggests using the Clifford algebra of the Clifford algebra. Why not. Since A is real 8-dimensional, this would be Cl(8) - the beloved Clifford algebra of many. But that is just dreaming.
In the next post I will use another toy, going from pure
algebraists to C* and von-Neumann "non-commutativists". They use so called
GNS construction for playing with reducible and irreducible
representations of *algebras. This will give us another perspective.
The we ->
ReplyDeleteThen we
in teh ->
in the
that looks as a solution ->
that looks as one of the solutions
Thanks for "then". As for the "solution": it is a complete solution - the whole eigen-subspace. There are no other solutions.
DeleteOoh, I see what you mean. Eigenvalue 0 which gives the complementary left ideal. I will add something appropriate,
Deleteit shown ->
Delete?
Each such p is of the form ->
Each such nontrival p is of the form
in teh ->
in the
Fixed. Thanks.
DeleteTony thought of Cl(8) as a von Neumann algebra more precisely a von Neumann factor. He most certainly relied on not standard symmetry breaking for this. A complicated center would also have to come from the symmetry breaking. Following the effect on spinors for this sounds like a nightmare.
ReplyDeleteset up our axes e1,e2,e3, in V and in In ->
ReplyDelete?
Fixed. Thank you!
Deletee1,e2,e3, in V ->
Deletee1,e2,e3 in V
dimensioonal ->
dimensional
Done. Thx.
DeleteDear Ark. I would like to show you this web page (http://www.madore.org/~david/weblog/d.2015-02-03.2276.octonions.html). I'm sorry, in french ;=)) David Madore seems to me to be a very good mathematician who I have been following for several years. He speaks about split-octonions ! And for me, as you know, Cl(3, 0) is isomorphic to split-octonions (not complex-quaternions). At the end he speaks about "Isotypies et trialité"... And after "Bonus : un mot sur les octonions-déployés" maybe you will understand better than me ... There is a curious similarity with your words above. I'm not good enough to well understand...
ReplyDelete"for me, as you know, Cl(3, 0) is isomorphic to split-octonions"
DeleteHow can it be?
"There is a curious similarity with your words above. "
Can you specify? What is similar to what?
And yes, I met Madore. Very clever man! I have his book on non-commutative geometry.
I'm trying to be more explicit (but it's not easy for me ;=). I'm sure the KEY is there, at the end of the Madore's page ;=) You have two "idempotent" elements : e0+ and e0- because (e0+)(e0+) = (e0+) ... You can split your algebra in two parts +/- (I prefer LEFT/RIGHT) as a "superposition" of two algebras.
DeleteThe chirality is important.
For me, our Universe is not oriented ! He is like a Klein bottle, and we "live" at the hyper-surface 3D (in a sort of superposition of left and right part reflecting each other).
Other thing : Cl(3, 0) is isomorphic to Split-octonions because of the same neutral signature (+ + + + - - - -) ...
Finally, one more intuition, if you start with a finite field with 3 elements, E = {1, w, w^2}, w is the cubic root from ONE and is a complex number. (w^2 = w*) and (w + w^2 = -1) it could be interesting...
I know how my "theory" is imprecise, but we are very close. Believe me ! ;=)))
But split octonions form a non-associative algebra, while Cl(3,0) is associative. So they cannot be isomorphic as algebras.
DeleteТак то оно так, но есть ещё ассоциативная векторно-матричная алгебра Цорна http://www.madore.org/~david/weblog/d.2015-02-03.2276.octonions.html
DeleteКроме того, матричная алгебра Дирака имеет отношение к нейтральной метрике 8-мерного пространства представления этой алгебры.
Пардон, не та ссылка. Правильно - https://en.wikipedia.org/wiki/Split-octonion
Delete"they cannot be isomorphic as algebras"
DeleteIn this paper : https://quantumgravityresearch.org/portfolio/all-hurwitz-algebras-from-3d-geometric-algebras/
You have this Theorem 3 : "Let x, y ∈ G (p, q) with p + q = 3. Then (G (p, q) , •, N ) with • and N defined in (4.2) and (4.23) is an Hurwitz algebra, and thus isomorphic to either the octonions or the split-octonions."
With a new product "." (in bold caracter)
And a fourth involution = "full grade inversion" which changes sign of all non-zero grades.
From my point of view, it could be interesting to have a look on this ! ;=)
Thanks. I will look at this paper. Though at first glance it is not helping to understand spin. These authors have different goals.
DeleteConcerning Exercise 1. 'Prove this statement: the intersection of one left ideal with one right ideal is always one-dimensional'.
ReplyDeleteIt seems to me that we have done already something very similar here: https://ark-jadczyk.blogspot.com/2024/12/spin-chronicles-part-28-left-and-right.html
Exercise 5, where we showed that the intersection of a left and right ideals is a 1-dimensional space of complex numbers C.
In matrix form, any element of a left ideal can be written as
{a, 0, b, 0}
and any element of a right ideal can be written as
{a, b, 0, 0}
It seems to be clear that their intersection is a 1d space...
Is this sufficient, or a more rigorous proof is needed?
True. This is a particular case. The main point in the exercise is to show that it holds for any two unit vectors m and n.
DeleteOf course we can always use the "continuity" argument to prove that if it holds for one particular choice of m and n, then it holds for any choice. But continuity arguments are more advanced, they need, in this case, group theory of topology. If you know how to use it - do it. Otherwise there is a pure algebraic way that requires calculations and solving systems of linear equations.
DeleteOne can also solve the problem "quantum-mechanically", without ANY calculations whatsoever.
Delete"One can also solve the problem "quantum-mechanically""
DeleteBe one.
"Given two projections p=(1+n)/2, q=(1+m)/2 ..."
DeleteWhat does this have to do with physics experiments?
"Ask, and it will be given to you."
DeleteSo we have m and n. We can always choose our axis so that n coincides with the third space axis. In the ideal In, which now we may call I3, we choose a basis so that Re(E2) coincides with e2. So, we are in the standard quantum mechanical environment. We have m, which is just spin m1 sigma1 + m2 sigma2 + m3 sigma3 in m-direction. It is unitarily equivalent to just sigma3, which eigenvalues and eigenspaces we know. Or we can solve the eigenvalue problem for m directly, as it is usually done in qm.
If something in the above reasoning requires expansion - do not hesitate to ask.
Delete"What does this have to do with physics experiments?"
DeleteIn physics experiments we set the laboratory orientation, set axis, prepare our complex Hilbert space for mapping the state vectors. This is p. Then we "prepare" the spin state. This is q. The intersection is one-dimensional. If we normalize the state vector, it is uniquely determined up to a complex phase. We somehow suppose, taking into account other quantum-mechanical mysteries, that this phase, even if unobservable, has some deeper meaning, is unavoidable for this or another reason that we do not understand yet.
This is one possible interpretation of the mathematical structure. But there may be another one, as the structure may have several different uses.
DeleteWe will see a different use of left and right multiplications after I fully understand Kassandrov's work.
DeleteI don't fully get it. Below I write what I can think of it. Please correct me if I am wrong.
Delete"In physics experiments we set the laboratory orientation"
- means we establish coordinate system with base vectors e1, e2, e3 in the room.
"set axis"
- means orient Stern-Gerlach magnets in direction n
"prepare our complex Hilbert space for mapping the state vectors."
- I don't get it. We prepared the experimental equipment by former doings.
" Then we "prepare" the spin state."
- means we send particles to our experimental equipment from some device (may be from another Stern-Gerlach magnets set it direction m
Very good question. It will take me a while to formulate my thoughts on this subject. They are not yet completely clear to me.
DeletePerhaps this remark for now: experimenter must have some rough idea what is it that he is measuring. If there is a "theory", he wants to compare his results with the predictions of some theory. Otherwise he will see just some meaningless data and wonder: what is the meaning of all that? Theory of spin involves selecting some Hilbert space. He may decide to take just pairs of complex numbers and be happy with that. Or he may ask a quite reasonable question: what is the relation of these complex numbers to reality? Why?
DeleteWhich way would you choose?
Theory is not everything that should satisfy us. I know how it was with me regarding the special theory of relativity. The special theory of relativity is, just like quantum theory, a mathematical theory. We have its axioms, and if we shut up and just do calculations, the results will be in wonderful agreement with experiments. This did not satisfy me. Only the realization that the mathematical axioms of the theory of relativity result from the existence of a medium (aether) together with its properties (including those resulting in shrinking the distance between atoms of bodies moving at high speeds) satisfied me.
DeleteI am not saying that theory is everything. I never said so. What I am saying is that an experimenter always should have some theory. Otherwise his actions are meaningless and nothing follows from his actions. A theory may be a simple one: "look, whenever i let the apple fall, it falls, always down". So he has a theory that there is "down", that there is "fall", that there is "apple". A primitive theory. He may go further, and imagine that there are things like "speed", "acceleration". Then he decides that there may be something like "gravity". Then he will model this "gravity" writing down some formulas. He will invent "force" and "vectors". This will enable him to model not just one experiment but a whole bunch of experiments.
DeleteWith spin it is probably similar. Perhaps at some point in the future we will be able to understand spin better, as some kind of microrotations in the aether. But what would the model of this aether that allows us to model a whole bunch of new phenomena? We have no idea and we should be open. Right now I do not know of any model of spin beyond the one quantum-mechanical, which I consider satisfactory, but leaving out a lot of questions.
"an experimenter always should have some theory. Otherwise his actions are meaningless and nothing follows from his actions."
DeleteWhat about the actual collection of result data (which can be analyzed by another entity that can figure out the compression system (theory)? It cannot be said that nothing results from such actions.
You are right. It is not a good theory. Something more creative is needed.
DeleteFrom https://johncarlosbaez.wordpress.com/2023/04/26/bosons-fermions-and-clifford-algebras/
ReplyDeleteHow do we make the Clifford algebras into superalgebras? We just decree that the square roots of -1 we throw in are fermionic. In Cliff₂ this means that i and j are fermionic and k is bosonic. That may seem weird, but that’s because we’re getting the quaternions from studying 2-dimensional space, which is also a bit weird. (In 3d space it turns out that the quaternions are the bosonic part of Cliff₃, and this is closer to Hamilton’s original thoughts.)
back to me: so does that mean for the superalgebra idea that the even grade quaternions of Cl(3,0) are bosonic and the odd grade is for fermions? I think this superalgebra version of a star algebra is more the one for the theory of Tony's I like. Tony kind of had multiple theories going on including ones that fit the von Neumann algebra idea where SO(16) was for M16(R) for the 8-fold real Bott periodicity while SU(2) was for M2(C) for the 2-fold complex Bott periodicity.
Fermions, Bosons - that is going into many-body systems (spin-statistics). . I want to understand a single spin. I do not understand it and it bothers me. Other people are being bothered by more complicated things. Perhaps they understand spin. But if so, why didn't write it for others to understand?
DeleteА чем плох такой вариант - спин образован вращением в 4D?
DeleteHow?
DeleteКак вращение Клиффордова тора - по широте и по долготе тора.
DeleteAnd what it has to do with spinors in 3D?
DeleteЭто уже другой вопрос. Но и на него можно ответить. 3D в 4D соткано из нитей в виде спиралей, образованных линиями тока векторных полей чисто мнимых кватернионов и единичного векторного поля.
Delete"Это уже другой вопрос."
DeleteBut it is the most important question. All the rest is of secondary importance.
"3D в 4D соткано из нитей в виде спиралей"
How?
Координатную сеть плетём из линий тока трёх векторных полей. X из e+i, Y из e+j, Z из e+k.
DeleteAnd what these three vector fields have do with spin and rotations in 3D? If you know answer to this question, why not to write it in extenso, in a clear way, so that everybody can understand and check for possible errors in your reasoning? Why?
DeleteСейчас я занят другими делами. Но в качестве упражнения каждый может вэять плоскость и задать там координатную линию в виде логарифмической спирали, полученной с помощью линейного векторного поля алгебры комплексных чисел. Расстояние на спирли нужно мерять логарифмом от длины радиус-вектора на плоскости, но у этой линии-спирали всё равно есть скрытый параметр - угловое положение спирали. Вот вам и росток для квантовой механики.
DeleteYou are regularly talking like Nostradamus or this lady
Deletehttps://ru.wikipedia.org/wiki/Ванга
@Ark, i'am trying to approach a pure algebraic solution of Ex.1.
ReplyDeleteSo, we have two ideals, which are 2d spaces (surfaces), determined by vectors n and m through the conditions
p=(1+n)/2, q=(1+m)/2, up=u and qv=v, where u, v are in A.
Let n = {0,0,1}. And m = {m1, m2, m3} - arbitrary values.
The defining equations for both surfaces are linear equations with 8 coefficients u0, u1, u2, u3 and v0, v1, v2, v3 and three unknown variables m1, m2, m3.
We know that two equations of plane
a1 x + b1 y + c1 z + d1 = 0
a2 x + b2 y + c2 z + d2 = 0
define a line of their intersection.
There are also 8 coefficients and 3 variables. Seemingly, we can get in this way a one-to-one correspondence between coordinates of vector m and the line of the intersection of the ideals.
Is this direction of thought a no-go one?
But m are assumed to be known (given in advance), the 8 coefficients are unknown.
Deleteindeed... thank you. i will revise the concept.
Delete@Ark, 'We have m, which is just spin m1 sigma1 + m2 sigma2 + m3 sigma3 in m-direction. It is unitarily equivalent to just sigma3, which eigenvalues and eigenspaces we know.'
ReplyDeleteMay i ask why is the spin in m-direction unitarily equivalent to sigma3 and what does it mean?
If you did not intend this as an Exercise #2 for us...
Very good question. I am changing my original plan and devote today's new post to this question. We should not go forward until all questions are answered. GNS can wait.
DeleteStill trying to do Ex.1
ReplyDeleteIt goes easier in the matrix form. Using Pauli matrices, we can write the product of element 'a' from a left ideal and 1/2(1+n) as follows:
{a1, 0, a2, 0} {1, 0, 0, 0} = {a1, 0, a2, 0}
The same element 'a' must be simultaneously the right ideal for 1/2(1+m), that is:
1/2{1+m3, m1-i m2, m1+i m2, 1-m3}{a1, 0, a2, 0} = {a1, 0, a2, 0}
This leads to a system of two linear homogeneous equations
with the determinant D = m1^2 + m2^2 + m3^2 - 1
which is always zero since m is a unit vector.
Is it a trivial result?
I do not fully understand your reasoning. Will return to it later on.
Delete@anna
DeleteSo we get:
a2 = (m1 + i m2)/(1 + m3) a1
(BTW Could you please use at least semicolon to distinguish vector from matrix - like
{a1, 0; a2, 0})
@Anna
DeleteSorry.
@Bjab, yes, but at the same time a2 = (1 - m3)/(m1 - i m2) a1,
Deletewhich is possible only for zero a1 and a2. That is why i suppose that i've got a trivial result, instead of proportionality of a1 and a2, which was the goal.
@Anna:
Delete"but at the same time a2 = (1 - m3)/(m1 - i m2) a1,
which is possible only for zero a1 and a2."
Why do you think so?
It is a bit inconvenient for me to communicate with an anonymous partner, could you please introduce youself?
DeleteA homogeneous system of equations has nontrivial solution iff its discriminant is nonzero. In my case, the discriminant = |m|^2 - 1 and it is definitely zero (m is a unit vector with |m|^2 = 1), so the trivial result seems unavoidable.
But the trivial result is not the only one.
DeleteSo equations
a2 = (1 - m3)/(m1 - i m2) a1
and
a2 = (m1 + i m2)/(1 + m3) a1
(which are mostly the same equations)
are fulfilled not only by a1=0 and a2=0.
@Anna
DeleteA homogeneous system of equations has nontrivial solution iff its discriminant is nonzero. ->
A homogeneous system of equations has nontrivial solution iff its discriminant is zero.
Oh, Bjab, you are right! I see at last that these two equations are really the same. Then, we've got that a1 is proportional to a2 and, hence, the space of intersection of the left and right ideals is 1-dim, which might be relevant to our goal.
Delete