Anna asked " why is the spin in m-direction unitarily equivalent to sigma3 and what does it mean?"
This question was asked in a comment to the previous post, and here I will propose my answer. There is more than one way of answering possible, and I will chose just one, that relates to previous discussions.
Let us address a more general question: given any two unit vectors m and n in V, why is the spin in m-direction unitarily equivalent to the spin in n-direction?
Let us analyze the question first. What exactly do we mean by that? What is "spin in m-direction"?
The fact that we are asking about "unitary equivalence" indicates that
we have in mind operators, not state vectors. Which operators?
Introduction
Given
a unit vector m in V, we consider it as an element of the Clifford
algebra Cl(V). Then it acts, in particular, as an operator L(m) of left multiplication by m
on Cl(V) . But Cl(V) is 4-dimensional complex, and when we say
"unitarily equivalent", we mean in a 2-dimensional complex space. So, we
probably have in mind some irreducible representation of CL(V), perhaps
one provided by one of its non-trivial left ideals. But we can start
answering our question even before specifying the representation. And
this is what we will do now. Cl(V) is a Hilbert space, the scalar
product <v,w> is defined as the scalar part of v*w. We know that
L(u*) = L(u)*, where on the right-hand-side we have Hermitian conjugate
of L(u) with respect to the <v,w> scalar product. The spin
operator in m-direction is then L(m), the spin operator in n-direction is L(n). In matrix representation, if we select an oriented orthonormal basis e1,e2,e3 in V, and represent the basic vectors by the three Pauli matrices, L(n) would be represented by the matrix σ(n) = n1σ1+n2σ2+n3σ3.
But we do not have to use matrix representation yet. We can proceed on
an algebraic level, and descend to a particular representation only at
the very end. It will be more "geometrical" this way. Instead of using
almost mindless matrix multiplication we are going to use scalar
products and vector products, which have rather simple geometrical
meaning. Well, we will also use the exponential, but this will be just a
compact way of using sin and cos functions. It is a longer way, but it
gives some satisfaction.
In the past, using n2=1 (if n is a unit vector), we calculated, within Cl(V), exp(itn), with the result
exp(itn) = cos(t) + in sin(t).
But we do not have to calculate anything, we can just define u(t,n):
u(t,n) = cos(t) + in sin(t).
Then u(t,n) is in Cl(V), and we can verify that u(t,n) is unitary:
u(t,n)u(t,n)* = u(t,n)*u(t,n) = 1.
But the U(t,n)=L(u(t,n)) is a unitary operator acting on the Hilbert space Cl(V).
U(t,n)U(t,n)* = U(t,n)*U(t,n) = I (the identity operator on Cl(V)).
We will use this fact in what follows, remembering that it holds for any unit vector n and any real t.
Back to the original question
Let us return to the original question. We have two unit vectors m and n. There are two cases here. The first, generic case, is when m and n are not parallel. There are two exceptions from this generic case: m=n and m=-n. We will discuss first the generic case.
If m is not parallel to n, then the cross product m⨯n is non-zero. In fact we have
||m⨯n|| = |sin(θ)|,
where θ is the angle between these vectors. The vector k=m⨯n/|sin(θ)| is then in V, and of unit norm. Thus
u(t,k) = cos(t) + i k sin(t)
is unitary. We can use our formula for the Clifford product to calculate u(t,k) m u(t,k)*. This is a simple exercise with cross products. The result is:
Exercise 1. Calculate the result.
Exercise 2. Check that with t = θ/2 (or t = -θ/2, I am not sure which, since I did not yet do these calculations!) we get
u(t,k) m u(t,k)* = n.
So m and n are unitarily equivalent in Cl(V). But then, since L is a *-representation, we have that U(t,k) gives us unitary equivalence of L(m) and L(n) acting on Cl(V). This is in four complex dimensions. How to descend to two? Simple: choose two-dimensional left ideal. For instance choose e1,e2,e3, and choose the left ideal determined by p = (1+e3)/2, as we have done before. But any other choice will do as well. Since it is a left-ideal, it is invariant under the action of U(t,k) = L(u(t,k)). And U(t,k) being unitary in the whole Cl(V), is also unitary within any invariant subspace.
What remains is the exceptional case of m = -n. This exceptional case can be handled even simpler. Let k be any unit vector in the plane perpendicular to n. Let u = kn (Clifford algebra product). Then
unu* =(kn)n(nk) =knk = -n.
Moreover, u is unitary in Cl(V).
Exercise 3. Verify this last statement.
(left ideal) works as before.
L(u(t,k)). And U ->
ReplyDeleteL(u(t,k)). And U
are also unitary ->
is also unitary
case of m = -n ->
case of m = -n
Fixed. Thanks.
DeleteI am trying to do Exercise 1, but it does not want to come as I expected ....
DeleteAnd what did you expect?
DeleteSorry, I meant Exercise 2.
DeleteWhen talking about spin, it might be useful in general to outline the context in which we look at the concept.
ReplyDeleteFor example, in experiments testing the GDH sum rule, spin of a photon is clearly linked to classical property of the circular polarisation (of a photon as a quantum of energy of a e-m wave), while that of a nucleon is arrested in parallel or antiparallel state with respect to the photon orientation with NMR and cryogenic technics.
On the level of simple QCD, spin part of the u and d quarks' wave functions in calculation of proton and neutron magnetic moments makes important contribution that at the end gives results suprisingly, when the simplicity of the calculation is considered, close to the experimental ones.
On the level of SU(2) only, we introduced "spin" and called it isospin in various different fields of study, e.g. in nuclear physics we consider proton and neutron as same nucleon just with +/- 1/2 value of a projection of their isospins, and in radioactive decays we consider u and d quarks and electron and neutrino pair as the same particles just with different weak isospin projection +/- 1/2. In all these cases, dealing with isospins gives rather good agreement with experimental results, i.e. measurements and observations.
When looking into protons for finer structures, from the results of deep inelastic scattering of electrons on them, we deduced that 1/2 spin particles, three of them, might "reside" inside, but also that they are not completely alone, i.e. independent, indicating that those partons exchange for example gluons among each other.
So to talk about spin without a clear context attribution might lead us into misunderstanding from the start, even of ourselves by our very selves. FWIW.
In addition, it should be noted that contrary to widespread opinions and beliefs of the guys dealing with spin at the start, including Wigner, almost maximal parity violation in weak interactions was established by Wu in the experiment with cobalt atom decays. For that experiment, spin of the Co60 atoms that was "frozen" during their decays, was clearly treated as their magnetic moment.
DeleteSo, it would really help a lot to know the details in which context the concept of a spin is being considered, as the devil very often hides in those very details, as you are fond of pointing out.
https://en.m.wikipedia.org/wiki/Wu_experiment
We are interested in spin 1/2. Whatever are physical applications, it is always related to the unitary representations of SU(2) (or two-valued representations of SO(3)). I want to know where this two-valuedness is coming from? Why SU(2) instead of SO(3)? Do you have some answer to this question? What physics says about it?
DeleteI know a very simple explanation of two-valuedness. It stems from the fact that one can turn to an angle 'phi' in two different ways: turning left and turning right. Two opposite rotations give one and the same result.
DeleteGoing in opposite directions can lead to the same place only if the path is closed, and what we thought was a straight line is actually a circle.
"Do you have some answer to this question? What physics says about it?"
DeleteI was pretty much influenced early on during my studies by the claim that, "whoever said that he understood what the spin is, clearly showed by that very statement that he did not understand it at all."
Physics to my knowledge mostly followed that claim, that in fact originated from Pauli, that very same Pauli who ridiculed the idea of electron spin and strongly advocated that it has nothing to do with reality.
From Wikipedia:
https://en.wikipedia.org/wiki/Spin_(physics)#Models
Pauli's "classically non-describable two-valuedness"
"Wolfgang Pauli, a central figure in the history of quantum spin, initially rejected any idea that the "degree of freedom" he introduced to explain experimental observations was related to rotation. He called it "classically non-describable two-valuedness". Later, he allowed that it is related to angular momentum, but insisted on considering spin an abstract property. This approach allowed Pauli to develop a proof of his fundamental Pauli exclusion principle, a proof now called the spin-statistics theorem. In retrospect, this insistence and the style of his proof initiated the modern particle-physics era, where abstract quantum properties derived from symmetry properties dominate. Concrete interpretation became secondary and optional."
In the following paragraph of that Wikipedia article, we read:
Circulation of classical fields
"The first classical model for spin proposed a small rigid particle rotating about an axis, as ordinary use of the word may suggest. Angular momentum can be computed from a classical field as well. By applying Frederik Belinfante's approach to calculating the angular momentum of a field, Hans C. Ohanian showed that "spin is essentially a wave property ... generated by a circulating flow of charge in the wave field of the electron". This same concept of spin can be applied to gravity waves in water: "spin is generated by subwavelength circular motion of water particles".
Unlike classical wavefield circulation, which allows continuous values of angular momentum, quantum wavefields allow only discrete values. Consequently, energy transfer to or from spin states always occurs in fixed quantum steps. Only a few steps are allowed: for many qualitative purposes, the complexity of the spin quantum wavefields can be ignored and the system properties can be discussed in terms of "integer" or "half-integer" spin models as discussed in quantum numbers below."
Spin as such is evidently related to electromagnetic properties, or more precisely, to magnetic moment of a particle, so it could be envisaged that its states are quantized, because electromagnetic energy is quantized, i.e. we have photons with energy E=hf or E=hbarω.
DeleteIn that sense, we also see that the quantum harmonic oscillator, has discrete energy levels, given by, En=ℏω(n+1/2), that is that the minimum or its lowest energy level, ground state or zero-point energy, is hbarω/2 above the minimum of the potential well, with the variance in both position and momentum of that state being exactly sigmax sigmap = hbar/2, just like the "minimum" value of a particle spin, which is as you noted in the comment to previous post Part 33 one of two ways where Planck's constant entered into fundamental physics. Coincidence? Probably not.
https://en.wikipedia.org/wiki/Quantum_harmonic_oscillator
To that, I'd just like to add that Planck's constant entered into our reality with realization that electromagnetic energy is quantized, i.e. as a quantum or a photon with energy E=ℏω. I don't know if that qualifies as fundamental or not, but it seems to me that it all started there and that at the end it all returns there as well.
Another issue with the analogy with classical spinning, apart from being quantized, is that the electron would need to spin faster than speed of light, as stressed in the SA article:
https://www.scientificamerican.com/article/what-exactly-is-the-spin/
"Spin is a bizarre physical quantity. It is analogous to the spin of a planet in that it gives a particle angular momentum and a tiny magnetic field called a magnetic moment. Based on the known sizes of subatomic particles, however, the surfaces of charged particles would have to be moving faster than the speed of light in order to produce the measured magnetic moments. Furthermore, spin is quantized, meaning that only certain discrete spins are allowed. This situation creates all sorts of complications that make spin one of the more challenging aspects of quantum mechanics."
If we dispense of that limitation (speed of light), would the analogy with the spinning work in the light of the quantized energy of the light?
Spin of a photon certainly is related to the "spinning" of the electric and magnetic field of the electromagnetic wave, i.e. to the circular polarization of a photon.
Integer spin can be understood geometrically. But what is half-integer spin? Where it comes from? Why SU(2) and not geometrically clear SO(3)? This is the question that I have no answer for. There is a mystery here.
DeleteAs for E=ℏω - that is another mystery, since we have no idea why this is so? It seems to work, but no one understands why it works.
It is my strong feeling that two-valuedness is connected with the dimension. We live in a 3d world, but light naturally exists in the 4d space (Maxwell's equations look perfectly in 4d). The step between spaces with adjacent dimensions can cause two valuedness, an example is my 'simple explanation' with a circle above. Imagine a 1d straight line, and let one moves right by a distance 'phi' and another moves in the opposite direction by quite another distance 2'pi' - 'phi', and both arrive at the same place. This is possible only if 1d line appeared to curl up into a circle, which lives in 2d. Plus one dimension gives rise to two-valuedness.
DeleteDoes that sound weird?
It seems right, but I think two-valuedness requires a knot, not just a circle.
Delete"Integer spin can be understood geometrically."
DeleteHow exactly, that is what would be geometrical interpretation of for example spin 2 hypothetical graviton particle? And what would be geometrical interpretation of spin 0 pion triplet of so called pseudoscalar particles?
Spin 1 is a vector (or a co-vector), spin 2 is a symmetric tensor. Vectors and tensors are simple geometrical constructs. Spinors are a mystery.
DeleteOh well, it seems it's high time to invest some time into properly understanding covectors and tensors in general in the geometrical sense, which remained hanging all the way since met them for the first time in GTR course. Then maybe I might have something useful to say about spinors from geometrical point of view, which at first glance appear like fractal objects considering their geometrical non-integer dimensionality.
DeleteA wild attempt at picturing what a spinor might represent in geometrical sense.
DeleteA vector is a line with its beginning and end in our ordinary 3d space; a spinor then might be "half" of that, i.e. a part of a line that has its beginning in our real 3d space, but its end "out" of it, where we only have an information in which "direction" that end might be found, either going towards us or away from us in real 3d space. In that sense it would appear to us just like a point at first glance, but not exactly dimensionless as an actual point when taking a closer look at it.
So, you got an idea. Now: how it relates to SU(2)?
DeleteWell, in a sense that's either or, either towards us or away from us, no something in between. But, to have it in proper math framework, a bit more thinking and polishing of that idea is needed. Will get back after it, if the idea survives the sanity tests.
Delete@Ark, 'It seems right, but I think two-valuedness requires a knot, not just a circle.'
DeleteSurely, a circle is only a simple example minded to illustrate how two-valuedness appears in the interplay between 1d and 2d spaces. To understand spinors, one has to grasp the same effect in the interrelation between 3d and 4d spaces.
I don't know about knots, but think that the figure of eight very likely plays the central role there.
Ark, i would not ask you why "spin in m-direction unitarily equivalent to sigma3" if understood that by 'sigma3' you meant vector n={0,0,1}. If i realized that, it would be clear that you conceived of one more approach to the correspondence between rotation of vectors in 3d space and unitary transformation of the related spinors. You have already shown us in details how this relation is constructed.
ReplyDeleteWell, this post made things much clearer for me, even though I didn't know they were unclear probably because I didn't understand them. So, thank you very much for asking.
Deletek=m⨯n/|sin(θ)| ->
ReplyDeletek=m⨯n/sin(θ)
Of course |sin(θ)|=sin(θ)
DeleteArk and Saša, if you have a couple of minutes, i would like to ask you to look at our discussion on Ex. 1 at the end of Part 34. With the help of Bjab and Incognito reader we had arrived at some result, and I'm curious what you think about it.
ReplyDeleteThat Anonymous was probably Bjab. :)
DeleteWell, the term "one dimensional" in, "This is intersection of one left ideal with one right ideal. It is always one-dimensional.", is a bit 'stretchy' for my taste, as we saw in Part 32, that a left ideal corresponding to p=(1+n)/2 is spaned by E1=(1+n) and E2=u2=iu2×n, and a corresponding right ideal would be E1'=(1+n) and E2'=-u2=-iu2×n, i.e. that would be a right ideal for m=n, which should be perfectly valid choice. In that case, the intersection of those left and right ideals is given by at least (1+n), so if we call that "one-dimensional", then OK.
Similarly if we choose m=-n, as showed in the comment to Part 33, then the right ideal for that choice becomes a set spanned by E1'=(1-n) and E2'=u2=iu2×n, which for the intersection with left ideal for (1+n), gives at least u2=iu2×n, i.e. a (sub)space or a whole set Tn^C, so if we consider that to be "one-dimensional", then again OK.
In general, we need to solve up=u=qu, which means:
(u0,u)(1/2,n/2)=(u0,u)=(1/2,m/2)(u0,u),
and gives the basis E1=(1+n) and E2=u2=iu2×n for the left ideal, and the basis E1'=(1+m) and E2'=u2=-iu2×m for the right ideal, so the intersection of those two sets can be found by solving,
aE1 + bE2 = a'E1' + b'E2',
or if we use Pauli's sigma matrices, as a linear combination of sigma's for the left ideal on the left-hand side and a linear combination of sigma's on the right hand side for the right ideal. As sigma matrices are linearly independent, the conclusion would be that equality holds only in the case when their coefficients are equal, i.e. the intersection of the left and right ideals in general would necessarily be just one possible linear combination of 4 Pauli's sigma matrices (up to the factor of proportionality), where the fourth sigma0 or sigma4 is 2×2 identity matrix.
In other words, the intersection of a left ideal and a right ideal is one dimensional in the sense that is spanned by just one particular linear combination of Pauli's sigma matrices.
Or, at least, it seems so to me.
Correction:
Delete"Similarly if we choose m=-n, as showed in the comment to Part 33, then the right ideal for that choice becomes a set spanned by E1'=(1-n) and E2'=u2=iu2×n, which for the intersection with left ideal for (1+n), gives at least u2=iu2×n, i.e. a (sub)space or a whole set Tn^C, so if we consider that to be "one-dimensional", then again OK."
Not "a (sub)space or a whole set Tn^C", but one particular combination in Tn^C, i.e. one specific complex vector u2 determined upon choosing n and Re(u2) or Im(u2).
@Saša
DeleteIntersection of two linear subspaces is always a linear subspace, not a "specific complex vector", so I am lost here.
@Anna The particular case you have considered is the case when the left ideal consists of all matrices with second column zero, while the right ideal consists of all matrices with second raw zero. The intersection then consists of all matrices that have, in your notation, a2=b1=b2=0, only a1 can be non-zero. You have calculated it in a roundabout way. But the conclusion is just that.
This comment has been removed by the author.
DeleteThis comment has been removed by the author.
Delete@Saša, thank you for your detailed explanation! I only missed the moment when you take the second basis element in the form E2=u2=iu2×n. Before, we guessed the form of E2 and then showed that E1 and E2 actually form the basis. Could you say a few words on this please?
DeleteDear colleagues, i beg your pardon, every my comment has been published twice, i am tired to delete one copy of it. Perhaps, this is a magic sign to us as we are coming close to the mystery of two-valuedness and doubling ... :)
Delete@Ark, 'You have calculated it in a roundabout way. But the conclusion is just that'.
DeleteYes, i hoped that the result was meaninful, although it was not evident for me at all how to get it in general case of arbitrary m. Thank you for confirming that it is 'just that'!
"Intersection of two linear subspaces is always a linear subspace, not a "specific complex vector", so I am lost here."
DeleteMy clumsy and imprecise wording at 3AM in the night, sorry for that. What was meant with that was that the basis for the intersection set was formed by a specific complex vector in Tn, i.e. E2 once it was chosen as a second basis element for left and right ideals.
"Before, we guessed the form of E2 and then showed that E1 and E2 actually form the basis."
DeleteWell, I think we didn't just guess the form, but showed in Exercise 2 in Part 30 that u2=iu2×n is the solution for up=u within Tn, and later in Exercise 5 that -u2 is the solution for pu=u, which for (1-n) becomes E2=u2 for that right ideal. Then the intersection of that left ideal, for (1+n), and that right ideal, for (1-n), has the basis in the form of common basis element of both sets, i.e. E2=u2, which is a specific complex vector once we have chosen particular combination for n and Re(u2) or Im(u2).
Remember: if you solve some equation, you are looking for the solution manifold, not one particular solution. So, if your unknows are a,b and the solution is b=a, that means that the manifold of solutions is one-dimensional. It is now that "we consider it 1-dimensional". It IS 1-dimensional.
Delete"It is not", not "It is now". Sorry.
DeleteUnderstood it first time.
DeleteThank you for the explanation and reminder, it helps a lot to rectify things in my mind and improves my understanding quite a bit.
Thank you! I should evidently return to Exercises of Part 30, still have something to do there.
DeleteRe. Ex.1
ReplyDelete(thousand typos but finally:)
u(t,k) m u(t,k)* =
= (0, cos(2t) m + sin(2t) (1/sin(θ)) (cos(θ) m - n) )
which for θ = -2t equals n
We can also choose another unitary operator:
Deleteu(t,-k)
to get the same result.
u(t,-k) m u(t,-k)* =
= (0, cos(2t) m + sin(2t) (1/sin(θ)) (n - cos(θ) m) )
which, this time, for θ = 2t equals n
One can also consider t = θ/2 + π
DeleteOf course it would be more elegant if I have defined
Deletek=n⨯m/sin(θ)
instead of m⨯n.
Indeed the absolute value in |sin(θ)| is not needed, if iwe precise that θ is between 0 and π.
Thank you very much for your help.
@Bjab, I also obtained this result
Deleteu(t,k) m u(t,k)* =
= (0, cos(2t) m + sin(2t) (1/sin(θ)) (cos(θ) m - n))
and for a long time was searching where i lost one minus sign.
Thought that it must be (n - cos(θ) m). Thanks a lot for publishing your result!
Exercise 3. Verify that u=kn is unitary in Cl(V).
ReplyDeleteIt was shown in the main text above that
unu* = (kn)n(nk) = knk = -n
To verify that, i had to remember that Clifford algebra is associative. After that, it seems not hard to show that
uu* = (kn)(kn)* = kn nk = kk = 1
Or the task implied anything more tricky?
That's it. Nothing tricky except that it is tricky simple!
DeletePerhaps I should say it explicitly: given m and n, we take a 2D plane containing these vectors. We take a vector k perpendicular to this plane. We need to make a rotation around k. This is exp(itk). We must remember that spinors rotate half the speed of vectors. The big mystery is: why is it so? Maybe it is not so big, since we have in Cl(V)! But something is still missing: why in physics do we need a left ideal? Perhaps we do not really "need" it? David Hestenes was asking the same question. But Hestenes quickly moved to space-time algebra, while the mystery of spinors is already in space, without "relativistic time".
DeleteRelevant paper by David Hestenes:
Deletehttps://www.researchgate.net/publication/265240377_Clifford_Algebra_and_the_Interpretation_of_Quantum_Mechanics
"We must remember that spinors rotate half the speed of vectors. The big mystery is: why is it so? Maybe it is not so big, since we have in Cl(V)!"
DeleteWell, in Part 23 we got that the action g x tau(g), where g=exp(ite3), resulted in rotation of basis e1 and e2 for 2t, indicating that vectors rotate two times faster than the initially conceived "transformation" parameter t value. So the question might be posed the other way around, why do vectors rotate two times faster than planned value for t at the start? In that sense spinors behave as expected and vectors do not.
Will read Hestenes' paper with interest, thanks for the link.
> We must remember that spinors rotate half the speed of vectors. The big mystery is: why is it so? Maybe it is not so big, since we have in Cl(V)!
DeleteЕсли отождествить противоположные точки окружности exp(i\phi)=exp(i\phi + \pi), то при полном обороте исходной окружности мы сделаем два оборота на окружности проективной прямой. Не тут ли собака зарыта?
But we have a 3D space. Where exactly is this circle?
DeleteВ 3D мы имеем точку, которая в 4D уже вовсе и не точка, а скрученная восьмёркой окружность. С другой стороны, проективную прямую геометрически можно наблюдать в обмотке классической сферы без полярных шапок (рисунок которой вы редактировали). Там это (1,1)-торический узел в виде окружности соединяющей полярные круги, которая идёт на поверхности сферы и под ней.
Delete"В 3D мы имеем точку, которая в 4D уже вовсе и не точка, а скрученная восьмёркой окружность."
DeleteHow?
Этой невидимой в 3D окружностью может быть траектория векторного поля мнимого кватерниона.
Delete"скрученная восьмёркой окружность."
DeleteWhere is figure 8?
Like Igor, i also try to represent figure 8 in the 4d space. Physicists used to work in 4d Minkowski space but still cannot visualize the picture, and this bothers many of us, naturally.
DeleteIt was a bit unexpectedly when Ark said that "the mystery of spinors is already in space, without "relativistic time"". Probably, i misunderstand something, but i think that the mystery of spinors must engage the 4th dimension inevitably. And the key may be somewhere in Lobachevky space, where the e-m vector is a complete analogy of a mechanical screw (see A.P. Kotelnikov paper 1927).
We should keep in mind that Lobachevky space is the absolute of Minkowski space, and 'absolute' is another guise of spinor.
"We should keep in mind that Lobachevky space is the absolute of Minkowski space, and 'absolute' is another guise of spinor."
DeleteThis is something that I probably should understand, but I do not understand at all. Zero understanding. Sure I will have to read the Kotelnikov paper carefully. So I adjust my plan accordingly. Today I will write about GNS construction, next post about Connes-Rovelli idea of time. Then Kotenikov ideas. OK?
@Anna
DeleteAnd the 4th dimension may have nothing to do with time dimension.
@Ark 'And the 4th dimension may have nothing to do with time dimension'.
DeleteYes, in theory we may concieve of any number of dimensions, 248 for example, like Garrett Lisi. But i would prefer first to put things in order within the observable 3+1 dimensions, including time. The more so that most of mainstream physicists are happy with the state of affairs in (3+1)-space and consider that a slight lack of visualization is not a serious problem.
Don't you think that "mass", for example, is also an observable dimension? What about "charge"? Or, "strength of gravity"? Or "action" as in Rumer? And why do you think that "time" is a "dimension"? Perhaps it is a derived and rather complicated "quantity"?
Delete'Today I will write about GNS construction, next post about Connes-Rovelli idea of time. Then Kotenikov ideas. OK?'
DeleteNo, i dare say, it is too fast for me. The promised pleasant adventure in the wonderland turns into a wild race through the jungle.
'Don't you think that "mass", for example, is also an observable dimension?...'
DeleteYes, but then we would have so much degrees of freedom to consider and just no hope to come to a reasonable final theory.
Furthermore, such dimensions like mass or charges are highly different from the three spatial ones. The time dimension looks much more like space dimension than the mass dimension does.
By the way, Vadim Varlamov touched upon the problem of space and time in his interview. I like his idea of extension and its first derivative 'dimension' as fundamentals of reality.
Do not worry, I am not going to "race". I will make it a pleasant visit to an aquapark.
Delete"Vadim Varlamov touched upon the problem of space and time in his interview"
DeleteWhich interview?
> Where is figure 8?
DeleteЕё надо ещё сложить пополам. Кроме того, под траекторией кватерниона я имел в виду траекторию, образованную воздействием на кватернион унитарного оператора, то есть поворотом в 4D.
Да, и поворот в 4D на самом деле это не обычный поворот в плоскости, а двойной поворот, сразу в двух плоскостях.
Delete"Which interview?"
DeleteFor example, here:
https://vadimvarlamov.blogspot.com/2024/12/about-space-and-time.html
Igor, Igor! Again you are talking like Nostradamus. Are you really trying to become famous for the same reason as he became famous?
Delete"I will make it a pleasant visit to an aquapark."
DeleteReally looking forward to it 🙂
@Ark, 'given m and n, we take a 2D plane containing these vectors. We take a vector k perpendicular to this plane. We need to make a rotation around k...'
ReplyDeleteWhy do we need a rotation around k? What does it mean? We still associate vectors n and m with the direction of observation and spin direction, respectively, right? Then, this rotation around k perpendicular to the plane (m,n) should be associated with something invariant in our arrangement?
This rotation is used just to prove the mathematical fact that m and n are unitarily equivalent. It is not supposed to represent anything taking place in reality. This is how we prove in geometry that any two unit vectors are related by an orthogonal rotation. So it is a constructive proof of the fact that SO(3) acts on the unit sphere transitively.
DeleteAs for physics I am not any longer sure about the interpretation of left and right actions. Bjab rightly criticized my initial attempt. So teh interpretation is hanging in the air. For a while we are simply collecting mathematical facts. There is a possibility that after collecting enough of such facts, there will be not much room left for alternative interpretation. If there is one - it should emerge from facts.
'So it is a constructive proof of the fact that SO(3) acts on the unit sphere transitively'.
DeleteOk, that is just what i wanted to learn. Thank you.
At the same time we showed that this transitive action of SO(3) on the unit sphere corresponds to SU(2) action on the unitary 2d complex space, right?
Right. Except that this complex 2d space is something still mysterious. Something like square root of -1. Impossible, yet useful. Why?
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