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  Spinor Intuition

+ 2 like - 0 dislike
249 views

I asked here before about finding examples of spinors in classical mechanics, but got little response, so I suppose I'll make a more general question.

I've done the rounds for awhile now and have read about all the algebra of spinor representations and how the Lorentz group $SO(1, 3)$ can be decomposed as $SU(2)\times SU(2)$, but I still feel that I don't know how to do reasoning with spinors as basic geometrical objects in the same way I can with vectors and tensors.

I'm limited to using all these wonderful group theory tools only in the cases that I know it's appropriate a priori. I don't get much satisfaction out of knowing I should use gamma-matrices here or this or that projection operator only because that's what's observed in the Standard Model. Is there literature I can read that reveals a deeper level of the underlying geometrical reasoning that is not solely restricted to specific contexts in field theory or quantum mechanics? I'm even welcome speculative ideas that attempt to do this at this point.

Edit: Wrong category

asked Oct 4 in Theoretical Physics by connornm777 (25 points) [ revision history ]
recategorized Oct 4 by connornm777

It is related, in my humble opinion, to the difference between Classical Mechanics (Physics) and Quantum Pechanics. The difference is deep, so deep that (nearly) nobody catches how from QM one obtains CM. Maybe I will elaborate my answer farther, tomorrow or later on.

Try https://users.aalto.fi/~ppuska/mirror/Lounesto/ for one alternative. There you will find links to a few of the many weird and wonderful ways to think about spinors.
Spinors are used classically in robotics because they can be used to encode the property that twisting a plate in your hand while keeping it flat requires two rotations to return to the initial position, as in https://www.youtube.com / watch?v=BknOhnxBL-c (I split up that URL because it didn't work as a single entity as a comment). There are many similar videos if you search for "spinor plate twisting" on YouTube. However I'd say that whether this particular geometry has anything whatsoever to do with why spinors appear in QFT is an entirely open question.

@VladimirKalitvianski

Please do, I'm interested to hear any perspective I can on this topic.

4 Answers

+ 3 like - 0 dislike

The first example of spinors in classical mechanics goes back to circa 1760 and the brilliant Leonhard Euler in the form of the Euler Rodrigues formula. This was long before William Rowan Hamilton came up with the quaternians in 1843.

He treats the two complex numbers of (what we call now) a spinor as four individual real numbers. He then comes up with a formula to extract a unique rotation matrix from any combination of the four numbers. Thus, any spinor maps to a unique rotation matrix.

The rotation matrix has three orthogonal vectors. One of three is associated in quantum field theory with s$^z$, the other two can be associated with s$^x$ and s$^y$.

So, a spinor can be associated with an orientation in space via the rotation matrix. I found a method to calculate the entire rotation matrix with a single 4x4 real value matrix multiplication. The derivation of the generators is here.

answered Oct 11 by Hans de Vries (80 points) [ revision history ]
edited Oct 11 by Hans de Vries

Your post gives the impression that Euler came up with the Euler-Rodrigues formula. Could you cite the corresponding work (from 1760 presumably)?

It seems to me that the formula is primarily named in honor of Euler, but not because he discovered it. According to Wikipedia, it was in 1840 that Rodrigues' worked out a formula for 3D rotations.

@GregGraviton The actual publication date is given as 1776 here on page 16

Nice! I didn't know about the Euler parameters, only about the Euler angles. Apparently, the original paper is:

  Euler, Leonhard, "Nova methodus motum corporum rigidorum degerminandi" (1776). Works by Eneström Number. 479. https://scholarlycommons.pacific.edu/euler-works/479

Not sure if the parameters are really contained in there, the notation is too old for me to grasp quickly.

+ 2 like - 0 dislike

The Wikipedia page on spinors mentions a quote by Sir Michael Atiyah:

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.@

So… if anyone understands spinors more fully, please post it here. :-)

But seeing them as 'square root' of Riemannian/Lorentzian geometry makes sense. Here is some evidence:

  • Let $q$ be an element of the spin group. It is an element of the Clifford algebra, which contains the gamma matrices $γ_\mu$. Since the gamma matrices are linearly independent, any vector $(v^\mu)_{\mu=1}^N$ uniquely corresponds to an expression $v^\mu γ_\mu$ (Einstein summation convention). Then, it turns out that the transformation $ν^\mu γ_\mu \mapsto q(ν^\mu γ_\mu)q^{-1}$ maps expressions of this form to themselves, that is we have $q(ν^\mu γ_\mu)q^{-1} = (R^\mu_\nu(q) v^\nu) γ_\mu$ for some matrix $R(q)$. This is a rotation matrix, so this is how the spin group acts on rotations. The key point worth noting here is that the left-hand side of this equation is quadratic in $q$, while the rotation $R(q)$ appears only linearly. So, in some sense, the elements of the spin group are square roots of the rotation matrices.
  • The space of two-tensors is obtained as, well, the tensor product of two spaces of one-tensors (vectors). Similarly, higher tensors are obtained by adding more factors to tensor products. However, a one-tensor itself can be obtained as the tensor product of two spinors (Commonly expressed in the equation $1/2 \otimes 1/2 = 0 \oplus 1$)! So, in some sense, the spinors are the 'square roots' of one-tensors, where 'square' now refers to the multiplication given by the tensor product. The spinors themselves cannot be decomposed further in terms of tensor products.

Why this works? I don't know. But keep in mind that these are not general statements about vector spaces, but about vector spaces with a scalar product, i.e. a "Riemannian" geometry. This structure is important; you have to know what a rotation is before you can talk about spinors.

answered Oct 4 by Greg Graviton (755 points) [ revision history ]

I remember very well the lecture delivered by Atiyah with the remark that "no one fully understands spinors." You can find it (French courtesy from IHÉS) in Youtube here.

I may sometime write what I think about these matters from my Cartan-biased viewpoint. I still think that Cartan's book on "The theory of spinors" is one of the best introductions!

Now, I would like to know what do you think about the intuitive geometric notion of a twistor.

+ 1 like - 0 dislike

Ok, let me say a word about spinors. They first were introduced in Physics to describe something quantum, discrete. Note, geometry has also been introduced in Physics, in Classical Physics, to be exact. The Classical Physics deals with inclusive pictures: any image contains many "pixels" (points) taken together. The number of points must be large enough to create a clear continuous image, but not too many in order not to burn the human eye. And the number of points must not be too small in order not to produce a dim and uncertain image. These numbers are governed with inequalities. But after inventing mathematics (geometry), those inequalities were forgotten. In Mathematics the Sun shines as one likes ;-). Continuity is taken as granted, but it is an unjustified extrapolation.

When we take too few points, we get into the realm of Quantum Mechanics. The truth is that we "obtain" information from elementary (indivisible) points. You may find many double slit experiment cartoons showing how the interference image is progressively built from collecting points.

Now vectors. Vectors are classical objects. They may have any length $L$. As you may now guess, the "continuous" character of vector length is illusory - it consists, in Physics, of many elementary pieces taken together. Trying to consider shorter and shorter vectors in Physics encounters the same effect: there is the smallest length (I am here speaking of angular momentum vector). Thus spinors were introduced.

It is not really correct to say that $1/2\otimes 1/2=0\oplus 1$ means obtaining a vector from spinors. Because such a product is of a certain length $L=\sqrt{1\cdot(1+1)}=\sqrt{2}$ in $\hbar$ units and it has three discrete projections on the $z$-axis. It is a spin-1 state, not a vector in a classical sense! A vector in a classical sense - a quantity of arbitrary length and projections - can be obtained as a product of many-many spinors (I omit how one gets a vector rather than a tensor of a higher rank from such a product, see QM textbooks for that), and, maybe, from a superposition of such spinor products.

About rotations: In Physics rotation transformations describe recalculation rules of the results obtained in one reference frame into another RF. They do not describe rotations of the object itself. There is no rotation velocity involved in such transformations: the group $SO(3)$ contains three angles and nothing else.

So I think your questions stem from "too mathematical" perception of objects introduced in Physics and forgetting the inequalities limiting validity of mathematical abstractions for physical notions.

answered Oct 6 by Vladimir Kalitvianski (97 points) [ revision history ]
edited Oct 7 by Vladimir Kalitvianski
+ 0 like - 0 dislike

The classical limit of spinors in quantum mechanics are grassmann numbers. 

In quantum mechanics, spinor fields satisfy anti-commutation relations

$$\left\{\bar{\psi}(x),\psi(y)\right\}=i\hbar\delta(x-y)$$

Taking the classical limit, $\hbar\rightarrow\infty$, they become anti-commuting fields. These fields take values in grassmann algebra. Mathematicians usually say that the quantization of Grassmann algebra is Clifford algebra. 

More rigorously, one needs to check the classical Poisson bracket, but then it can be proven that classical spinors and their canonical conjugate variables are Grassmann numbers.

answered Oct 6 by New Student (155 points) [ revision history ]

Just to troll you: why not to take a limit $i\to 0$ or $\delta\to 0$? The occupation numbers for fermions are $0$ and $1$. It is quite different from bosons where a small dimensional quantity $\propto \hbar$ can be neglected next to a huge another dimensional number.

@Vladimir Kalitvianski why do you want to troll me? If you stop trolling me, I will send you One Bitcoin.

Forget it. I just do not like limits expressed like $\hbar\to 0$. I work in units where $\hbar=1$.

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