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Okay, I typed in "Spinors of the Pauli Spin Matrices" back in June and no one deleted it, I'm going to tempt fate and type in "Spinors of the Dirac Algebra" and see what happens. I also added the general solutions for spin in the (a,b,c) direction for spin-1/2 Pauli particles and also for Dirac particles and antiparticles.
As an aside, there are several versions of spinors based on operator theory that probably belong somewhere on wikipedia. The basic idea is to define spinors entirely through the projection operators themselves. The Cambridge geometry group calls this "density operator" theory: http://www.mrao.cam.ac.uk/~clifford/publications/abstracts/hd_density.html
These methods allow all the usual spinor calculations without having to specify a representation. It's far more elegant than the usual methods.
Carl Brannen Carl Brannen 09:15, 25 January 2007 (UTC)
From the discussion on the talk page it appears that at least some people want to further expand this article. Maybe my comments will go against the flow, but I feel very strongly that this article is already way too big and (a related issue) diluted. I think it can be improved a lot by removing much of inessential material, then adding some important points that are not adequately addressed (eg Dirac equation or spinor fields). If I had not known what the spinors were already, I would never even come close to finding my way to their definition through the infinite ramblings of this article! This may be one of those cases, where giving a definition and a couple of examples straight up would serve the reader much better than vague and convoluted preliminary comments about their "meaning". (Curiously, the stub of an article spinor bundle seems to be better in this regard.) There should be a paragraph close to the beginning to the effect that "Spinors are elements of the spinor representation of orthogonal group G, which is constructed from the Clifford algebra as follows...", then the example of SO(3) worked out with as little notation as possible. Normally, I would have done it myself, but given the size of the article and apparently enormous amount of effort invested in it, I wanted to discuss it first. Also, there are a couple of wrong or, at least, misleading statements right at the beginning of the article, which raise the issues of the consistency, hence not easily fixable.
It occurs to me that there are two ways to think about the quantity ψ M ψ~ -- two viewpoints which can clash, unless the article explains them. But at the moment it doesn't say a word.
The first way to think about ψ M ψ~ is to think of M as a bilinear operator, and ψ as "half a vector" (or half a multivector) -- so it and its twin ψ~ are just the passive objects being operated on.
Thus one can write
and then
because, if you like,
The other way is to think of ψ as the operator, and M as the passive object on the receiving end -- ie so that between them ψ and ψ~ define a transformation, which has the effect on M of mapping it to N:
because, if you like,
Now the first way is the point of view from which much of the article is written. (Indeed it seems to correspond to both of the approaches described in "Two basic approaches"!). It's also the way most of us probably first get conditioned to thinking about spinors via the wavefunction |ψ> in quantum mechanics.
The second way, with the spinor as the operator, is eg the way spinors are used to do rotations in computer algebra. It's the point of view now taken in the article by the sections on spinors in 2D and spinors in 3D. It's also possible to think of the wavefunction in this way, something that boosts an initial pure state |ej><ej| into a density matrix |ψ>|ej><ej|<ψ|, and any operator M into the corresponding matrix |ψ>M<ψ|.
Unless some sort of bridge is added, at the moment I fear there is a real dissonance between the different sections. Jheald 23:07, 26 March 2007 (UTC)
ψ is easier to see as an operator when it looks like a matrix. It is more difficult to think of ψ acting like an operator when it looks like a column.
When ψ looks like a column, the way it does in the last sections of the article, what is the hidden machinery in the formalism that allows it to act like an operator? Jheald 23:13, 26 March 2007 (UTC)
'My' lead section was completely redone, with physics buzzwords to excess. Charles Matthews 21:13, 18 April 2007 (UTC)
Thanks for the edits. The pain has gone. Charles Matthews 09:42, 20 April 2007 (UTC)
It's not just the lead section. There are whole paragraphs full of buzz words, such as 'polarization' used left and right without explaining anything.
Once upon a time, after Charles had gone over the article and cleaned it up, for a while, it was actually possible to find out what spinors *are*, and have a good view of the topic: first a short overview, then the details. Now a long 'motivation' section replaced the overview, the new overview is just a rambling. The old overview survives in the commented out form with a cryptic remark:
I am sorry if it sounds a bit harsh, since it looks like someone invested a lot of effort in editing the article, but while certain parts have been somewhat improved, the overall quality of article has gone way down. I definitely appreciate the time and effort, but it has to lead us in the right direction, I think.
Some specific comments:
This is (1) vague, (2) not particularly insightful (unless one happens to already possess suitable intuition), (3) mathematically meaningless, for the most part. For example, if spinors are geometrical objects, what can I do with them? Can I add them? Do they have length? How about angles? What does 'factorizing a vector' mean? How can a multivector be 'polarized'? And so on, and so forth. The next few paragraphs about things acting on each other are particularly unclear. In addition, some of the links, such as
are rather confusing (wrong link, in this case? I coudn't even be sure!).
Some suggestions:
Arcfrk 01:22, 28 April 2007 (UTC)
To give some flavour of why definitions of spinors are often next to useless and at best highly confusing, consider the following from the current edit. I read: "a spinor must belong to a representation of the double cover of the rotation group SO(n,R)". So I think, aha, I know what a spinor is if I know what a representation is. Reading representation, I learn that a representation is a homomorphism from a group G to an automorphism group. So what that means is that a spinor is a group homomorphism?
No. That's doesn't sound right. Ah, the problem must be this phrase "belong to", what does "belonging to a representation" mean. No idea. No definition of *that* anywhere.
A similar problem occurs later in "In this view, a spinor is an element of the fundamental representation of the Clifford algebra Cℓn(C) over the complex numbers". What is an "element" of a representation?
When I was at college I was taught a representation was just as wikipedia defines it (a hom) but now I suspect that maybe we are talking about a module over the algebra and the spinor is a vector in that module. Is that right? Who can say?
I completely, 100% understand what a clifford algebra is -- and clifford algebra is pretty clear and lucid. I *think* I know what a representation is, and I could do character theory at college. I feel that I ought to understand or be able to deduce what a spinor is. But I'm afraid this page leaves me as confused as ever. Will someone help?
By the way, its no use giving an example. I'm not sure if my efforts to generalise an example will work. Eg, I am quite familiar with (say) the first few chapters of spinors and space-time by Penrose and Rindler. Great fun, but no definition of what a spinor is in general. I can manipulate different kinds of objects (called spinors) when calculating interaction cross-sections from Feynman diagrams in QED. I have seen examples. What *exactly* am I looking at? Anyone care to help? Francis Davey 21:34, 24 July 2007 (UTC)
Adam1729 10:51, 25 September 2007 (UTC)
This article lacks easily findable definition (which should be backed with motivation, informal explanation, etc...).
Somebody fix this, please! Thank you!-- 83.131.0.38 ( talk) 14:37, 26 November 2007 (UTC)
Probably very few readers of these pages have heard of geometric algebra. That fascinating and highly efficient mathematical tool originated 150 years ago in the works of Hamilton, Grassmann, Clifford. It was revived 40 years ago by David Hestenes, an american mathematician and physicist, who reinterpreted geometrically some abstract Clifford algebras, particularly those applied to real euclidian space and to Minkowski space. Among his most active followers are some astronomers and physicists of the Cambridge university (UK).
Geometric algebra (GA) has stunning simplifying effects on the manipulation and interpretation of the mathematics applied to quantum physics. By the way GA is more efficient and easier to work with than the hamiltonian quaternions, which are a subalgebra of GA.
It seems to me that everybody agrees with the fact that the mathematical object called spinor is not at all easy to understand and manipulate. But in geometric algebra the equivalent object is quite simple : in GA a spinor is an even multivector - nothing mysterious -, whose action when applied to a vector (v) by ψvψ~ means geometrically a rotation combined or not with a boost and a dilatation. In the same manner the effect of a rotation, passive or active, on a spinor is simply encoded by Rψ or ψR , where R is a rotor. Even a beginner in GA can quickly grasp these ideas ; there is no need to be an experienced algebraist.
Unfortunately todays only a few privileged students or lucky readers of internet stumbling on it when browsing, have the opportunity to study GA. Why ... ?? —Preceding unsigned comment added by Chessfan ( talk • contribs) 10:30, 24 August 2008 (UTC)
Sorry, coming back to this wiki discussion I see that I gave unadvertently a false idea of the possibilities offered by GA. Of course it is an easy task in GA to establish the restricting conditions we must impose on even multivectors to define the true equivalence of classical spinors. One must read Hestenes and/or the Cambridge Group : " They have ears but don't want to hear ... ". Jheald : thanks for your explanations, but looking for matrix calculations is indeed not in the spirit of GA. Chessfan ( talk) 11:44, 4 February 2010 (UTC)
A last word about something funny : One might say in 1912 Cartan anticipated Picasso ; he accidentally cut a Hamilton quaternion in pieces, put the first and last term together, and did the same with the second and third term, piled them up in a column matrix and labelled them (Pauli) spinor . It took more than 60 years to recognize the remnants ! And still the vast majority of Clifford specialists and of quantum physicists (for different reasons) refuse that simple truth. See Roger Boudet Annales de la Fondation Louis de Broglie n°26 special 2001 . Chessfan ( talk) 14:29, 4 February 2010 (UTC)
"[R]otation of a multivector is performed by the double-sided application of a rotor. The elements of a linear space which is closed under single-sided action of a representation of the rotor group are called spinors. In conventional developments a matrix representation for the Clifford algebra of spacetime is introduced and the space of column vectors on which these matrices act defnes the spin-space. But there is no need to adopt such a construction. For example the even subalgebra of the STA forms a vector space which is closed under single-sided application of the rotor group. The even subalgebra is also an eight-dimensional vector space, the same number of real dimensions as a Dirac spinor, and so it is not surprising that a one-to-one map between Dirac spinors and the even subalgebra can be constructed." (as they do, in their Appendix A).
In the literature of mathematical physics there are elements called spinors, traditionally associated with the description of rotations in quantum mechanics. These are closely related to rotors. It is useful to understand this link, since some of the spinor literature is relevant to geometry.
Spinors are not introduced as geometric products of vectors, but as elements that preserve grade under a sandwiching product in a Clifford algebra. Consider the set of elements S that can transform a vector x into a vector by the operation S x S−1. (This is called the Clifford group.) When such elements are normalized to S S˜ = ±1 and of even grades, they are called spinors, making up a spin group (though some authors appear to permit odd spinors as well [51 (Porteous, I. R. Clifford Algebras and the Classical Groups. Cambridge: Cambridge University Press, 1995)]).
The special spin group is the subgroup of the spin group consisting of the elements for which S S˜ = +1. Its elements are most closely related to the rotors, but careful study shows (see e.g., [33 (Hestenes, D., and G. Sobczyk. Clifford Algebra to Geometric Calculus. Reidel, 1984.)], pg. 106) that there are some special spinors that are not rotors. They consist of the weighted sum of a rotor and its dual, but they are rare (they only occur in spaces of dimensionality 0 mod 4). So it is almost true that "special spinor" and "rotor" are equivalent terms. In summary:
All rotors are special spinors; almost all special spinors are rotors.
Our use of the word 'spinor' in reference to elements of Spin(p, q) is unusual. It is justified by our unusual general definition of 'spinor' , which is as follows: we say that an even multivector ψ in G(A(p, q)) is a spinor of A(p, q) if for each vector x in A(p, q), ψ x ψ† is also a vector. Versors satisfy this definition, so the question is, how much more general can ψ be? [Algebra omitted]... Thus a spinor is always an even versor unless ¼n = ¼(p+q) is an integer, in which case a spinor can always be expressed as the sum of two even versors.
At first sight our definition appears to be quite different from the conventional definition, but the two have been proved to be equivalent in the cases of physical interest (see [H2] (D Hestenes (1967), " Real spinor fields", J. Math. Phys. 8, 798-808), [H6] (D. Hestenes (1973) " Local observables in the Dirac theory", J. Math. Phys., 14 (7), May 1973, 893-905) and [H9] (D. Hestenes (1975), " Observables, operators and complex numbers in the Dirac theory", J. Math. Phys. 16, 556-572. ).
Our definition has the advantages of simplicity in its algebraic formulation and its geometrical interpretation. Thus (8.10) shows that a spinor determines an orthogonal transformation and a dilation (by a factor ρ)
R=scalar + relative vector a + dual of relative vector b + pseudoscalar pseudoscalar = 0 a . b = 0
L and D do not mention these conditions, but they are easy to verify. Chessfan ( talk) 10:40, 15 February 2010 (UTC)
Please note that a maximal isotropic subspace does not have a unique isotropic complement in dimension >2. For instance, in R^{2,2}, with quadratic form q(w,x,y,z) = w z + x y, the maximal isotropic subspace
has complementary isotropic subspaces of the form
for any real number c. Geometry guy 14:42, 7 September 2008 (UTC)
Please forgive my ignorance - I don't have Chevalley to hand - but can someone explain the following passage?
Thanks. Geometry guy 19:28, 9 September 2008 (UTC)
I notice that in the Comments field of the maths rating box above that Geometry guy suggests that the Clebsch-Gordan decomposition might be too detailed for this article. However the existing article Clebsch-Gordan decomposition appears not to be suitable as a move target as it was clearly written straight out of a quantum mechanics book (it is only applicable to SO(3), from the looks of it). Anyway, the Clebsch-Gordan decomposition is probably the most important single fact about spinors, since it encodes essentially all of the operations of interest on spinors (including triality, inner products, relationships with tensor representations of the orthogonal groups, and even the Dirac operator). While one could spend an entire book discussing each of these topics, it is all succinctly summarized in the Clebsch-Gordan decomposition. Weyl and Brauer spent a large part of their seminal paper on spinors in n dimensions discussing it. And I don't personally find it undue weight to have such a section here, but I would just as well happily move it elsewhere if there were a suitable move target. siℓℓy rabbit ( talk) 11:55, 10 September 2008 (UTC)
Sorry to be annoying, but is there any chance of putting a definition in the lead? All this deep discussion is very interesting, but in the first few paragraphs there should be something like "A spinor is a...." or "Spinors are members of ....". The best that the article can manage is to tell me that spinors are group homomorphisms (because that is what representations are defined to be in wikipedia). I am pretty certain spinors aren't defined as members of a group of homomorphisms, so how about a definition, one that doesn't begin "You can think of..." or some such weasel phrase? This is a real concern for me and I am sure many others reading an article like this. It may be that if you *know* what is meant you can interpret it. At the moment what is there is neither a definition a lay person can understand nor in internal terms a pedantically correct one. Francis Davey ( talk) 19:57, 10 September 2008 (UTC)
OK. I am still trying to work out in my mind what a spinor is (I don't think most of you have any idea how hard it is to approach this material - there are so many hidden assumptions, abuse of notation and logical leaps that its hard for someone even with a maths degree like myself to follow). In the examples we have the following:
The action of an even Clifford element γ ∈ Cℓ02,0 on vectors, regarded as 1-graded elements of Cℓ2,0, is determined by mapping a general vector u = a1σ1 + a2σ2 to the vector
- ,
where γ* is the conjugate of γ, and the product is Clifford multiplication. In this situation, a spinor [1] is an ordinary complex number. The action of γ on a spinor φ is given by ordinary complex multiplication:
- .
Note that the sentence beginning "In this situation, a spinor" plucks the notion of spinor out of thin air. The example has worked through the fact that you can embed vectors into the clifford algebra (in the natural way) and that the even graded elements act on those vectors by conjugation such that they each represent rotations. Fine. But then there is a leap where the "spinor" is mentioned without being introduced. When I first read it I thought that spinors were also the vector elements of the algebra but acted on by multiplication not conjugation, but reading the 3D example made me think that what is actually happening is that the spinors are the even graded elements, acted on by left multiplication. Each spinor corresponds to a vector in a natural way, but the rotations require 4pi rotation not 2. Is that a correct understanding? I.e. that (i) in this example the spinors are the even subalgebra and (ii) that there is a 1-1 correspondence between 2D vectors and spinors? If clarification could be given, the example could be tightened up a bit to explain that. Francis Davey ( talk) 17:43, 21 September 2008 (UTC)
I echo these comments, nearly 9 months later. I can't fix this article at this point, but I recognize that this section is broken. 420ftjesus ( talk) 14:23, 8 June 2009 (UTC)
POTENTIAL ERROR:
I think it is not correct to say that so(V,g) is embedded as a Lie subalgebra in Cℓ(V,g) because in particular Cℓ(V,g) is not a lie algebra. Rather, one should say that one can construct a representation of so(V,g) from one of Cℓ(V,g) called the spinor representation of so(V,g), by:
where \Gamma_a span a representation of Cℓ(V,g). These representations act on vector spaces of the same dimension.
Any associative algebra can be considered as a Lie algebra with the commutator as the Lie bracket. What you wrote down above is exactly such an embedding. ( RogierBrussee ( talk)) —Preceding undated comment added 21:06, 11 July 2009 (UTC).
There seems to be a reversion war occurring in the lead section. This is not good to see, and reverts withour even an edit summary are a bad idea. Everyone is agreed that spinors turn out to be needed, both in mathematics and physics; but the reason why they are needed can be explained from different angles (basically analytical, algebraic and topological). No one reason should be privileged in the lead: in particular the talk about the Lie algebra is only one way of looking at it, and the talk about tensoring up representations to find less than all representations is also only one way of looking at it. Everyone should bear in mind that this is an introductory section to an encyclopedia article, and not a place for pedagogical quarrels. Charles Matthews ( talk) 10:06, 11 July 2009 (UTC)
Introducing spinors from the point of representation theory of the lie algebra so(V, g) is by far the most natural way to introduce spinors. Once you see that non tensorial representations of the Lie algebra exist, you have to find a way to construct them, which is where Clifford algebra's come in naturally. Now you can argue that the Clifford algebra is perfectly natural, even without its relation to so(V,g) and that spinors therefore live independent of the Lie algebra and I would agree. However, searching for vector spaces on which the infinitessimal rotations (or Lorentz trafos) act, and which are therefore behave nicely under infinitessimal coordinate transformations seems as close to a pedagogical starting point as you can get. In addition the Lie algebra point of view forces you to deal with the fact that representations of the Lie algebra are representations of the universal covering of orthogonal group SO^+(V,g) rather than representations of the group itself, i.e. topology. As for the analytical side, for analysis you need to talk about spinor _fields_ i.e. spinor valued functions (or more generally sections of a spinor bundle) and Dirac operators. But to understand spinor valued functions you better understand what spinors are and to understand Dirac operators you better understand the representation of the Clifford algebra point of view.
Now you can disagree with all that but at least defining spinors as non tensorial representations of Lie algebra's that are subspaces of algebra representations of Clifford algebra's, seems like a definite improvement over a "definition" that reads "a complex vectorspace whose extra dimensions are needed to reveal the full structure of orthogonal groups". I realise that part of this edit war comes from different expectations between mathematicians and physicists, but could we at least get a definition of spinors back that is unambiguous or as unambiguous as possible please.
If people want to improve this article then in it would be much better to go through the low dimensional examples (dim 1 up to 4 say) in the concrete and abstract definitions of spinors that are given. The way the current low dimensional examples are done are not very nice in my opinion. ( RogierBrussee ( talk) 22:23, 11 July 2009 (UTC))
In addition to this page on spinors and the page on spin representations, there's also a small page on Clifford modules. These are closely related. But the different pages aren't linked very well. And I'm too lazy to do it. John Baez ( talk) 03:18, 7 October 2009 (UTC)
![]() | This is an archive of past discussions. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page. |
Archive 1 | Archive 2 | Archive 3 | Archive 4 | Archive 5 |
Okay, I typed in "Spinors of the Pauli Spin Matrices" back in June and no one deleted it, I'm going to tempt fate and type in "Spinors of the Dirac Algebra" and see what happens. I also added the general solutions for spin in the (a,b,c) direction for spin-1/2 Pauli particles and also for Dirac particles and antiparticles.
As an aside, there are several versions of spinors based on operator theory that probably belong somewhere on wikipedia. The basic idea is to define spinors entirely through the projection operators themselves. The Cambridge geometry group calls this "density operator" theory: http://www.mrao.cam.ac.uk/~clifford/publications/abstracts/hd_density.html
These methods allow all the usual spinor calculations without having to specify a representation. It's far more elegant than the usual methods.
Carl Brannen Carl Brannen 09:15, 25 January 2007 (UTC)
From the discussion on the talk page it appears that at least some people want to further expand this article. Maybe my comments will go against the flow, but I feel very strongly that this article is already way too big and (a related issue) diluted. I think it can be improved a lot by removing much of inessential material, then adding some important points that are not adequately addressed (eg Dirac equation or spinor fields). If I had not known what the spinors were already, I would never even come close to finding my way to their definition through the infinite ramblings of this article! This may be one of those cases, where giving a definition and a couple of examples straight up would serve the reader much better than vague and convoluted preliminary comments about their "meaning". (Curiously, the stub of an article spinor bundle seems to be better in this regard.) There should be a paragraph close to the beginning to the effect that "Spinors are elements of the spinor representation of orthogonal group G, which is constructed from the Clifford algebra as follows...", then the example of SO(3) worked out with as little notation as possible. Normally, I would have done it myself, but given the size of the article and apparently enormous amount of effort invested in it, I wanted to discuss it first. Also, there are a couple of wrong or, at least, misleading statements right at the beginning of the article, which raise the issues of the consistency, hence not easily fixable.
It occurs to me that there are two ways to think about the quantity ψ M ψ~ -- two viewpoints which can clash, unless the article explains them. But at the moment it doesn't say a word.
The first way to think about ψ M ψ~ is to think of M as a bilinear operator, and ψ as "half a vector" (or half a multivector) -- so it and its twin ψ~ are just the passive objects being operated on.
Thus one can write
and then
because, if you like,
The other way is to think of ψ as the operator, and M as the passive object on the receiving end -- ie so that between them ψ and ψ~ define a transformation, which has the effect on M of mapping it to N:
because, if you like,
Now the first way is the point of view from which much of the article is written. (Indeed it seems to correspond to both of the approaches described in "Two basic approaches"!). It's also the way most of us probably first get conditioned to thinking about spinors via the wavefunction |ψ> in quantum mechanics.
The second way, with the spinor as the operator, is eg the way spinors are used to do rotations in computer algebra. It's the point of view now taken in the article by the sections on spinors in 2D and spinors in 3D. It's also possible to think of the wavefunction in this way, something that boosts an initial pure state |ej><ej| into a density matrix |ψ>|ej><ej|<ψ|, and any operator M into the corresponding matrix |ψ>M<ψ|.
Unless some sort of bridge is added, at the moment I fear there is a real dissonance between the different sections. Jheald 23:07, 26 March 2007 (UTC)
ψ is easier to see as an operator when it looks like a matrix. It is more difficult to think of ψ acting like an operator when it looks like a column.
When ψ looks like a column, the way it does in the last sections of the article, what is the hidden machinery in the formalism that allows it to act like an operator? Jheald 23:13, 26 March 2007 (UTC)
'My' lead section was completely redone, with physics buzzwords to excess. Charles Matthews 21:13, 18 April 2007 (UTC)
Thanks for the edits. The pain has gone. Charles Matthews 09:42, 20 April 2007 (UTC)
It's not just the lead section. There are whole paragraphs full of buzz words, such as 'polarization' used left and right without explaining anything.
Once upon a time, after Charles had gone over the article and cleaned it up, for a while, it was actually possible to find out what spinors *are*, and have a good view of the topic: first a short overview, then the details. Now a long 'motivation' section replaced the overview, the new overview is just a rambling. The old overview survives in the commented out form with a cryptic remark:
I am sorry if it sounds a bit harsh, since it looks like someone invested a lot of effort in editing the article, but while certain parts have been somewhat improved, the overall quality of article has gone way down. I definitely appreciate the time and effort, but it has to lead us in the right direction, I think.
Some specific comments:
This is (1) vague, (2) not particularly insightful (unless one happens to already possess suitable intuition), (3) mathematically meaningless, for the most part. For example, if spinors are geometrical objects, what can I do with them? Can I add them? Do they have length? How about angles? What does 'factorizing a vector' mean? How can a multivector be 'polarized'? And so on, and so forth. The next few paragraphs about things acting on each other are particularly unclear. In addition, some of the links, such as
are rather confusing (wrong link, in this case? I coudn't even be sure!).
Some suggestions:
Arcfrk 01:22, 28 April 2007 (UTC)
To give some flavour of why definitions of spinors are often next to useless and at best highly confusing, consider the following from the current edit. I read: "a spinor must belong to a representation of the double cover of the rotation group SO(n,R)". So I think, aha, I know what a spinor is if I know what a representation is. Reading representation, I learn that a representation is a homomorphism from a group G to an automorphism group. So what that means is that a spinor is a group homomorphism?
No. That's doesn't sound right. Ah, the problem must be this phrase "belong to", what does "belonging to a representation" mean. No idea. No definition of *that* anywhere.
A similar problem occurs later in "In this view, a spinor is an element of the fundamental representation of the Clifford algebra Cℓn(C) over the complex numbers". What is an "element" of a representation?
When I was at college I was taught a representation was just as wikipedia defines it (a hom) but now I suspect that maybe we are talking about a module over the algebra and the spinor is a vector in that module. Is that right? Who can say?
I completely, 100% understand what a clifford algebra is -- and clifford algebra is pretty clear and lucid. I *think* I know what a representation is, and I could do character theory at college. I feel that I ought to understand or be able to deduce what a spinor is. But I'm afraid this page leaves me as confused as ever. Will someone help?
By the way, its no use giving an example. I'm not sure if my efforts to generalise an example will work. Eg, I am quite familiar with (say) the first few chapters of spinors and space-time by Penrose and Rindler. Great fun, but no definition of what a spinor is in general. I can manipulate different kinds of objects (called spinors) when calculating interaction cross-sections from Feynman diagrams in QED. I have seen examples. What *exactly* am I looking at? Anyone care to help? Francis Davey 21:34, 24 July 2007 (UTC)
Adam1729 10:51, 25 September 2007 (UTC)
This article lacks easily findable definition (which should be backed with motivation, informal explanation, etc...).
Somebody fix this, please! Thank you!-- 83.131.0.38 ( talk) 14:37, 26 November 2007 (UTC)
Probably very few readers of these pages have heard of geometric algebra. That fascinating and highly efficient mathematical tool originated 150 years ago in the works of Hamilton, Grassmann, Clifford. It was revived 40 years ago by David Hestenes, an american mathematician and physicist, who reinterpreted geometrically some abstract Clifford algebras, particularly those applied to real euclidian space and to Minkowski space. Among his most active followers are some astronomers and physicists of the Cambridge university (UK).
Geometric algebra (GA) has stunning simplifying effects on the manipulation and interpretation of the mathematics applied to quantum physics. By the way GA is more efficient and easier to work with than the hamiltonian quaternions, which are a subalgebra of GA.
It seems to me that everybody agrees with the fact that the mathematical object called spinor is not at all easy to understand and manipulate. But in geometric algebra the equivalent object is quite simple : in GA a spinor is an even multivector - nothing mysterious -, whose action when applied to a vector (v) by ψvψ~ means geometrically a rotation combined or not with a boost and a dilatation. In the same manner the effect of a rotation, passive or active, on a spinor is simply encoded by Rψ or ψR , where R is a rotor. Even a beginner in GA can quickly grasp these ideas ; there is no need to be an experienced algebraist.
Unfortunately todays only a few privileged students or lucky readers of internet stumbling on it when browsing, have the opportunity to study GA. Why ... ?? —Preceding unsigned comment added by Chessfan ( talk • contribs) 10:30, 24 August 2008 (UTC)
Sorry, coming back to this wiki discussion I see that I gave unadvertently a false idea of the possibilities offered by GA. Of course it is an easy task in GA to establish the restricting conditions we must impose on even multivectors to define the true equivalence of classical spinors. One must read Hestenes and/or the Cambridge Group : " They have ears but don't want to hear ... ". Jheald : thanks for your explanations, but looking for matrix calculations is indeed not in the spirit of GA. Chessfan ( talk) 11:44, 4 February 2010 (UTC)
A last word about something funny : One might say in 1912 Cartan anticipated Picasso ; he accidentally cut a Hamilton quaternion in pieces, put the first and last term together, and did the same with the second and third term, piled them up in a column matrix and labelled them (Pauli) spinor . It took more than 60 years to recognize the remnants ! And still the vast majority of Clifford specialists and of quantum physicists (for different reasons) refuse that simple truth. See Roger Boudet Annales de la Fondation Louis de Broglie n°26 special 2001 . Chessfan ( talk) 14:29, 4 February 2010 (UTC)
"[R]otation of a multivector is performed by the double-sided application of a rotor. The elements of a linear space which is closed under single-sided action of a representation of the rotor group are called spinors. In conventional developments a matrix representation for the Clifford algebra of spacetime is introduced and the space of column vectors on which these matrices act defnes the spin-space. But there is no need to adopt such a construction. For example the even subalgebra of the STA forms a vector space which is closed under single-sided application of the rotor group. The even subalgebra is also an eight-dimensional vector space, the same number of real dimensions as a Dirac spinor, and so it is not surprising that a one-to-one map between Dirac spinors and the even subalgebra can be constructed." (as they do, in their Appendix A).
In the literature of mathematical physics there are elements called spinors, traditionally associated with the description of rotations in quantum mechanics. These are closely related to rotors. It is useful to understand this link, since some of the spinor literature is relevant to geometry.
Spinors are not introduced as geometric products of vectors, but as elements that preserve grade under a sandwiching product in a Clifford algebra. Consider the set of elements S that can transform a vector x into a vector by the operation S x S−1. (This is called the Clifford group.) When such elements are normalized to S S˜ = ±1 and of even grades, they are called spinors, making up a spin group (though some authors appear to permit odd spinors as well [51 (Porteous, I. R. Clifford Algebras and the Classical Groups. Cambridge: Cambridge University Press, 1995)]).
The special spin group is the subgroup of the spin group consisting of the elements for which S S˜ = +1. Its elements are most closely related to the rotors, but careful study shows (see e.g., [33 (Hestenes, D., and G. Sobczyk. Clifford Algebra to Geometric Calculus. Reidel, 1984.)], pg. 106) that there are some special spinors that are not rotors. They consist of the weighted sum of a rotor and its dual, but they are rare (they only occur in spaces of dimensionality 0 mod 4). So it is almost true that "special spinor" and "rotor" are equivalent terms. In summary:
All rotors are special spinors; almost all special spinors are rotors.
Our use of the word 'spinor' in reference to elements of Spin(p, q) is unusual. It is justified by our unusual general definition of 'spinor' , which is as follows: we say that an even multivector ψ in G(A(p, q)) is a spinor of A(p, q) if for each vector x in A(p, q), ψ x ψ† is also a vector. Versors satisfy this definition, so the question is, how much more general can ψ be? [Algebra omitted]... Thus a spinor is always an even versor unless ¼n = ¼(p+q) is an integer, in which case a spinor can always be expressed as the sum of two even versors.
At first sight our definition appears to be quite different from the conventional definition, but the two have been proved to be equivalent in the cases of physical interest (see [H2] (D Hestenes (1967), " Real spinor fields", J. Math. Phys. 8, 798-808), [H6] (D. Hestenes (1973) " Local observables in the Dirac theory", J. Math. Phys., 14 (7), May 1973, 893-905) and [H9] (D. Hestenes (1975), " Observables, operators and complex numbers in the Dirac theory", J. Math. Phys. 16, 556-572. ).
Our definition has the advantages of simplicity in its algebraic formulation and its geometrical interpretation. Thus (8.10) shows that a spinor determines an orthogonal transformation and a dilation (by a factor ρ)
R=scalar + relative vector a + dual of relative vector b + pseudoscalar pseudoscalar = 0 a . b = 0
L and D do not mention these conditions, but they are easy to verify. Chessfan ( talk) 10:40, 15 February 2010 (UTC)
Please note that a maximal isotropic subspace does not have a unique isotropic complement in dimension >2. For instance, in R^{2,2}, with quadratic form q(w,x,y,z) = w z + x y, the maximal isotropic subspace
has complementary isotropic subspaces of the form
for any real number c. Geometry guy 14:42, 7 September 2008 (UTC)
Please forgive my ignorance - I don't have Chevalley to hand - but can someone explain the following passage?
Thanks. Geometry guy 19:28, 9 September 2008 (UTC)
I notice that in the Comments field of the maths rating box above that Geometry guy suggests that the Clebsch-Gordan decomposition might be too detailed for this article. However the existing article Clebsch-Gordan decomposition appears not to be suitable as a move target as it was clearly written straight out of a quantum mechanics book (it is only applicable to SO(3), from the looks of it). Anyway, the Clebsch-Gordan decomposition is probably the most important single fact about spinors, since it encodes essentially all of the operations of interest on spinors (including triality, inner products, relationships with tensor representations of the orthogonal groups, and even the Dirac operator). While one could spend an entire book discussing each of these topics, it is all succinctly summarized in the Clebsch-Gordan decomposition. Weyl and Brauer spent a large part of their seminal paper on spinors in n dimensions discussing it. And I don't personally find it undue weight to have such a section here, but I would just as well happily move it elsewhere if there were a suitable move target. siℓℓy rabbit ( talk) 11:55, 10 September 2008 (UTC)
Sorry to be annoying, but is there any chance of putting a definition in the lead? All this deep discussion is very interesting, but in the first few paragraphs there should be something like "A spinor is a...." or "Spinors are members of ....". The best that the article can manage is to tell me that spinors are group homomorphisms (because that is what representations are defined to be in wikipedia). I am pretty certain spinors aren't defined as members of a group of homomorphisms, so how about a definition, one that doesn't begin "You can think of..." or some such weasel phrase? This is a real concern for me and I am sure many others reading an article like this. It may be that if you *know* what is meant you can interpret it. At the moment what is there is neither a definition a lay person can understand nor in internal terms a pedantically correct one. Francis Davey ( talk) 19:57, 10 September 2008 (UTC)
OK. I am still trying to work out in my mind what a spinor is (I don't think most of you have any idea how hard it is to approach this material - there are so many hidden assumptions, abuse of notation and logical leaps that its hard for someone even with a maths degree like myself to follow). In the examples we have the following:
The action of an even Clifford element γ ∈ Cℓ02,0 on vectors, regarded as 1-graded elements of Cℓ2,0, is determined by mapping a general vector u = a1σ1 + a2σ2 to the vector
- ,
where γ* is the conjugate of γ, and the product is Clifford multiplication. In this situation, a spinor [1] is an ordinary complex number. The action of γ on a spinor φ is given by ordinary complex multiplication:
- .
Note that the sentence beginning "In this situation, a spinor" plucks the notion of spinor out of thin air. The example has worked through the fact that you can embed vectors into the clifford algebra (in the natural way) and that the even graded elements act on those vectors by conjugation such that they each represent rotations. Fine. But then there is a leap where the "spinor" is mentioned without being introduced. When I first read it I thought that spinors were also the vector elements of the algebra but acted on by multiplication not conjugation, but reading the 3D example made me think that what is actually happening is that the spinors are the even graded elements, acted on by left multiplication. Each spinor corresponds to a vector in a natural way, but the rotations require 4pi rotation not 2. Is that a correct understanding? I.e. that (i) in this example the spinors are the even subalgebra and (ii) that there is a 1-1 correspondence between 2D vectors and spinors? If clarification could be given, the example could be tightened up a bit to explain that. Francis Davey ( talk) 17:43, 21 September 2008 (UTC)
I echo these comments, nearly 9 months later. I can't fix this article at this point, but I recognize that this section is broken. 420ftjesus ( talk) 14:23, 8 June 2009 (UTC)
POTENTIAL ERROR:
I think it is not correct to say that so(V,g) is embedded as a Lie subalgebra in Cℓ(V,g) because in particular Cℓ(V,g) is not a lie algebra. Rather, one should say that one can construct a representation of so(V,g) from one of Cℓ(V,g) called the spinor representation of so(V,g), by:
where \Gamma_a span a representation of Cℓ(V,g). These representations act on vector spaces of the same dimension.
Any associative algebra can be considered as a Lie algebra with the commutator as the Lie bracket. What you wrote down above is exactly such an embedding. ( RogierBrussee ( talk)) —Preceding undated comment added 21:06, 11 July 2009 (UTC).
There seems to be a reversion war occurring in the lead section. This is not good to see, and reverts withour even an edit summary are a bad idea. Everyone is agreed that spinors turn out to be needed, both in mathematics and physics; but the reason why they are needed can be explained from different angles (basically analytical, algebraic and topological). No one reason should be privileged in the lead: in particular the talk about the Lie algebra is only one way of looking at it, and the talk about tensoring up representations to find less than all representations is also only one way of looking at it. Everyone should bear in mind that this is an introductory section to an encyclopedia article, and not a place for pedagogical quarrels. Charles Matthews ( talk) 10:06, 11 July 2009 (UTC)
Introducing spinors from the point of representation theory of the lie algebra so(V, g) is by far the most natural way to introduce spinors. Once you see that non tensorial representations of the Lie algebra exist, you have to find a way to construct them, which is where Clifford algebra's come in naturally. Now you can argue that the Clifford algebra is perfectly natural, even without its relation to so(V,g) and that spinors therefore live independent of the Lie algebra and I would agree. However, searching for vector spaces on which the infinitessimal rotations (or Lorentz trafos) act, and which are therefore behave nicely under infinitessimal coordinate transformations seems as close to a pedagogical starting point as you can get. In addition the Lie algebra point of view forces you to deal with the fact that representations of the Lie algebra are representations of the universal covering of orthogonal group SO^+(V,g) rather than representations of the group itself, i.e. topology. As for the analytical side, for analysis you need to talk about spinor _fields_ i.e. spinor valued functions (or more generally sections of a spinor bundle) and Dirac operators. But to understand spinor valued functions you better understand what spinors are and to understand Dirac operators you better understand the representation of the Clifford algebra point of view.
Now you can disagree with all that but at least defining spinors as non tensorial representations of Lie algebra's that are subspaces of algebra representations of Clifford algebra's, seems like a definite improvement over a "definition" that reads "a complex vectorspace whose extra dimensions are needed to reveal the full structure of orthogonal groups". I realise that part of this edit war comes from different expectations between mathematicians and physicists, but could we at least get a definition of spinors back that is unambiguous or as unambiguous as possible please.
If people want to improve this article then in it would be much better to go through the low dimensional examples (dim 1 up to 4 say) in the concrete and abstract definitions of spinors that are given. The way the current low dimensional examples are done are not very nice in my opinion. ( RogierBrussee ( talk) 22:23, 11 July 2009 (UTC))
In addition to this page on spinors and the page on spin representations, there's also a small page on Clifford modules. These are closely related. But the different pages aren't linked very well. And I'm too lazy to do it. John Baez ( talk) 03:18, 7 October 2009 (UTC)