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Why do no particles have a 1/3 spin? Why are all particles' spin either a half-integer or integer? How would a particle with such a spin behave, as a fermion, boson, or neither?

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marked as duplicate by Qmechanic Jan 16 '16 at 0:13

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    $\begingroup$ See physics.stackexchange.com/q/1/50583. A spin of 1/3 doesn'T exist because there simply is no representation of $\mathrm{SU}(2)$ corresponding to it. $\endgroup$ – ACuriousMind Jan 15 '16 at 22:49
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    $\begingroup$ @ACuriousMind - Not just that. You also need to consider representations of $\text{ISO}(2)$ (which is the little group for massless particles). I think the more general answer is that the Lorentz group is doubly connected (not triply connected) and so only allows for integer or half integer spins. $\endgroup$ – Prahar Jan 15 '16 at 22:58
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    $\begingroup$ Related: physics.stackexchange.com/q/29655/2451 and links therein. $\endgroup$ – Qmechanic Jan 15 '16 at 23:35
  • $\begingroup$ @Qmechanic there is a difference between my question and the "duplicate" I also ask how a particle with a 1/3 spin would act, which the other question does not. $\endgroup$ – tox123 Jan 16 '16 at 14:52
  • $\begingroup$ Btw, in 2+1D, there are anyons, cf. e.g. this Phys.SE post. $\endgroup$ – Qmechanic Jan 16 '16 at 15:19
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This is not my answer, it's one of the answers you can find here

Is there a reason why the spin of particles is integer or half integer instead of, say, even and odd?

I just wrote here (and re-posted) the work of @Siva which I found a very good answer. However, follow the link to read more interesting useful answers

The "spin" tells us how the wavefunction changes when we rotate space (or spacetime). Just because I double all charges by convention, the behaviour of the wavefunction will not be any different. What will happen is that the "doubling" or charges will lead to the "halving" of your definition of angles such that the physical results (which depends on angle multiplied by spin) remain the same.

Wrt. the observation on "odd" and "even" functions -- that is not an accident and it works quite like you think it does.

The crux of the matter is that a "full rotation" corresponds to $2 \pi$ so the phase picked up by a spin $\frac{1}{2}$ wavefunction is $e^{i \pi} = -1$.

Recall that (even in classical mechanics) "angular momentum" is the generator of rotations. So if I start using different units, eg: $\tau \equiv 2 \pi$ to represent a half rotation (instead of $\pi$) then the values of charge will halve to maintain the value of $e^{i q \theta}$


If you understand some representation theory, here goes:

  1. Representations of $SO(3)$ have integer charges. Since we're referring to the group of rotations, we call that charge as "angular momentum" or "spin". The representations correspond to scalars (spin 0), vectors (spin 1) and tensors (in general of spin 2 or higher).

  2. $SU(2)$ is a "double cover" of $SO(3)$ so representations of $SU(2)$ can have the "charges" as $SO(3)$ representations. Thus we also get half-integer spin. The new representations correspond to spinors.

  3. When we consider quantum relativistic physics (aka QFT), all physical fields/particles must form kosher reps of the Lorentz algebra, which happens to be $so(3,1) \sim so(4) = su(2) \oplus su(2)$. So (up to the "unitary trick") reps of the Lorentz group can be written as a tensor product of reps of the left and right-handed $SU(2)$ algebras. Based on #2 above, these continue to have integer or half-integer spin.

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  • $\begingroup$ I don't understand the last half there, but I think I get the point. So spin is not an action of the particle but describes the symmetry group of the particle's wave function? $\endgroup$ – tox123 Jan 15 '16 at 23:08
  • $\begingroup$ @Qmechanic I'm sorry! I thought that one could re-post old answers to questions which may be identical to a new ones.. $\endgroup$ – Les Adieux Jan 15 '16 at 23:39
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    $\begingroup$ No - don't re-post. Instead, post a link. If necessary, add the essence of the content of the link. But never, ever - here, or anywhere else - make it look like something is your work, when it is not. Attribution is key. Without it, you are committing a grave "academic sin". May I suggest that you edit your answer accordingly? $\endgroup$ – Floris Jan 15 '16 at 23:41
  • $\begingroup$ @Floris I'm sorry, I edit the answer immediately! $\endgroup$ – Les Adieux Jan 15 '16 at 23:43
  • $\begingroup$ @KimPeekII - alright, I'm holding off with the downvote button. We're all here to learn; perhaps this was not what you came here to learn today, but life is full of surprises. $\endgroup$ – Floris Jan 15 '16 at 23:44

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