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The answers in this question: What is spin as it relates to subatomic particles? do not address some particular questions regarding the concept of spin:

How are some useful ways to imagine a particle without dimensions - like an electron - to spin?
How are some useful ways to imagine a particle with spin 1/2 to make a 360° turn without returning to it's original position (the wave function transforms as: $\Psi \rightarrow -\Psi$).
When spin is not a classical property of elementary particles, is it a purely relativistic property, a purely quantum-mechanical property or a mixture of both?

Qmechanic
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Gerard
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    I've voted to close on this - as the question is stated I don't think there will be a single "right answer". Perhaps you might change it to "What are some useful ways of imagining a particle ..." instead of "How should". – j.c. Nov 03 '10 at 14:05
  • While answering this question I realized that it was not a real, not argumentative question... You should change the question as @j.c. suggested. – Cedric H. Nov 03 '10 at 14:12
  • Isn't this similar to, e.g. mathematical dimension, which is a generalization of the common concept? It's not fruitful to ask "How should I imagine the fourth (or the eleventh) dimension. Wikipedia states spin did originate from a classical interpretation: "Ralph Kronig, one of Landé's assistants, suggested in early 1925 that it was produced by the self-rotation of the electron. Pauli heard about the idea, he criticized it severely, noting that the electron's hypothetical surface would have to be moving faster than the speed of light..." – recipriversexclusion Oct 11 '11 at 21:22
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    "It is like a spinning ball. Only it's not a ball, and it is not spinning" – ziv Mar 13 '21 at 21:51

5 Answers5

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How should one imagine a particle without dimensions - like an electron - to spin?

You don't. If you want to imagine, then you think classically and it is just a particle spinning... Thinking like that doesn't give you any other insight of what spin really is (an intrinsic angular momentum, behaving like an [orbital] angular momentum).

How should one imagine a particle with spin 1/2 to make a 360° turn without returning to it's original position (the wave function transforms as: Ψ→−Ψ)

Just imagine it ... no big deal. Again, classically this is not possible, but quantum mechanically it is.

When spin is not a classical property of elementary particles, is it a purely relativistic property, a purely quantum-mechanical property or a mixture of both?

The spin of elementary particle is a pure quantum mechanical effect. Edit: See @j.c. comment. Relativity also plays a role.

Any other interpretation/calculation requires things like commutator, symmetry properties and group theory.

The parallel between "real spinning" and "spin" (which is just a name) comes from the fact that the spin operator needed to account for properties of elementary particles behaves (= has the same definition, based on commutators) like orbital angular momentum operator. This again comes from symmetry properties of ... nature.

The goal of quantum physics is to provide a way to calculate properties. If you want to calculate or go deeper in the problem, then you don't need this classical interpretation.

hft
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Cedric H.
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    It's arguable, but relativity does come into the notion of spin. Consider for instance the fact that photons have only two polarization states despite being "spin 1". I don't know of a way of understanding this without using the (relativistic, I'd argue) fact that the symmetry group of the universe is the Poincaré group. (And you could argue that the rest of the observed spins of particles ultimately arise from this symmetry group too). Of course quantum mechanics is essential for all this as well. – j.c. Nov 03 '10 at 14:16
  • @j.c. OK, I agree with that. – Cedric H. Nov 03 '10 at 14:17
  • +1 Trying to "imagine" quantum mechanics too deeply will likely lead you astray! Study the fundamentals, the principles, and the maths, and eventually you'll get a holistic picture. – Noldorin Nov 03 '10 at 14:26
  • This is a useful answer for me. Is it possible to elaborate on the connection between relativity and quantum theory here? – Gerard Nov 03 '10 at 14:46
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    @Gerard, it's something that's typically covered at the start of a quantum field theory course (meaning of course, that you should have a solid grasp of both before trying to understand the details of the connection). If I had to make it into a slogan though: relativity is one of the fundamental symmetries of the universe; quantum mechanics tells us how symmetries are reflected in properties of quantum states, in particular, an application of the mathematics of representation theory leads to the concept of spin. – j.c. Nov 03 '10 at 14:52
  • @j.c. I like your slogan ! – Cedric H. Nov 03 '10 at 14:53
  • @j.c. does your slogan imply that without the Poincaré group symmetry the 'spin' property could not be deduced in q.m.? – Gerard Nov 03 '10 at 16:35
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    Well, originally "spin" was deduced from experiment... but certainly the current theoretical framework for spin needs the Poincaré group. – j.c. Nov 04 '10 at 09:50
  • Actually it is not true that all symmetry groups in QFT are ultimately representations of the Poincare group (which arises from Special relativity) because the Fermion fields are transformed by representations of the double cover Spin(1,3) which is isomorphic to SL(2,C). This is precisely why the Spin is so unintuitive - if it would transform as representation of SO(1,3)+R(1,3) (the Poincare group), then a rotation around 360 degrees would result in no change because representations (which are homomorphisms from groups to vector spaces) map identities to identities. – exchange Jan 04 '18 at 20:18
  • But of course the underlying Lie Algebra of both, SO(1,3) and Spin(1,3), is so(1,3). However, it is harder to understand a Lie-Algebra intuitively and to say "dont imagine anything in QM because it is non-classical" is not satisfying for me and I suppose also not for the OP for otherwise he would not have asked such a question. – exchange Jan 04 '18 at 20:20
  • @exchange Thanks for the more precise explanation. Looking back at what I wrote 7+ years ago, I see that I was trying to keep things at a basic level and certainly left out some important details. Perhaps what I would add now is the adjective "projective" to my mentions of representation theory (to capture the spin representations that you mention). I might also add (for any other future readers) that the wikipedia page https://en.wikipedia.org/wiki/Representation_theory_of_the_Lorentz_group now seems to be a decent overview of the mathematics of spin in the setting of particle theory. – j.c. Jan 04 '18 at 20:39
  • @j.c. Alright, sounds much better then! Thanks for following up the comment on your old comment. – exchange Jan 07 '18 at 09:13
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About this:
What are some useful ways to imagine a spin 1/2 particle making a 360° turn without returning to its original position, i.e. Ψ→−Ψ ?

The Dirac scissors is an example of such objects:

Dirac scissors

This picture is from the book by Penrose and Rindler "Spinors and space-time." I suggest reading it.

pushkine
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Kostya
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There's a paper$^{1}$ by Battey-Pratt and Racey with an intuitive model of spin 1/2. I'm not sure if it related to reality, but is an interesting read and attempt at an intuitive understanding.

--

$^{1}$ E.P.Battey-Pratt and T.J.Racey, Geometric model for fundamental particles, International Journal of Theoretical Physics 19 (1980) 437-475.

Qmechanic
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DarenW
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It's possible to do correct quantum mechanics without believing that particles get altered by 360 degree rotations. Use the "density matrix" form instead of "wave function".
http://en.wikipedia.org/wiki/Density_matrix

To convert a quantum wave state $\psi(x)$ or $|a\rangle$ to a density matrix, multiply the ket by the bra as in:
$\psi(x) \to \rho(x,x') = \psi(x)\psi^*(x')$
$|a\rangle \to |a\rangle \langle a|$

Since the bras and kets take complex phases, i.e.
$e^{+i\alpha}|a\rangle \equiv e^{-i\alpha}\langle a|$
the complex phases cancel. This is more general than the -1 you get by a 360 degree rotation, but a factor of -1 is also a phase and so it's also canceled.

In short while the state vectors or wave functions take a -1 on 360 degree rotation, the (pure) density matrices are left unchanged.

Carl Brannen
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  • However, relative phases between different pasts of a system will produce physical results. If you rotate one particle through 360$^\circ$, then you can interfere it against a non-rotated particle to get a shifted pattern. The overall density matrix will reflect this: all the (inter-particle) coherences will change sign. – Emilio Pisanty Jun 06 '12 at 20:15
  • To have an interference requires that the rotated and non rotated particles begin in a coherent state. Then you can use the density matrix form for them, but it has to be the two particle density matrix. So what you're doing is rotating one particle while not rotating the other. The result is indeed observable, one way of describing it is as a quantum phase, a topological phase, or Berry-Pancharatnam phase. But for a single particle, or a pair of particles, or any number of particles, rotating them all (i.e. rotating the system) by 360 degrees is not detectable. – Carl Brannen Jun 07 '12 at 09:40
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I find it useful to think about different spaces having different "sizes", such that one complete rotation through space "A" requires only 360 degrees of turning, but one complete rotation through space "B" requires 720 degrees of turning, making space B in some sense "larger" with respect to complete rotations.

Spin 1/2 particles live within the larger space B, where 720 degrees is a full rotation. As objects within that space, the spin 1/2 particles have essentially four "sides" to them, one side per 180 degrees. As observers, we live in space A, and every (classical) object around us in space A reveals all of it's sides in a single 360 degree rotation. But when we try to rotate spin 1/2 particles through their four sides, it takes two full rotations of our space to see one complete rotation of their space.

The trick is that a "full rotation" must be a fundamentally different concept than an "amount of rotation". Full rotations depend on the space, amounts of rotation are invariant across different spaces.

If this explanation makes sense, it is immediately obvious why we shouldn't think of particles as being "dimensionless". In fact, certain aspects of the particle are more dimension-full than the space we are used to thinking about.

Greg L
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