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What I know about spin ½ particles is that they are represented by spinors, and thus, you need to apply a 720° rotation in order for the spinor to return to its original value. Spin 1 particles are vectors, and their transformation is trivial.

Also, spin ½ have $\frac{\hbar}{2}$ or $-\frac{\hbar}{2}$ angular momentum, while spin 1 have $\hbar$ or $0$ or $-\hbar$ angular momentum.

But why are these related? Why should a particle that transforms like a vector have integer angular momentum, while spinors have half-integer angular momentum? Is it just an observation?

Habouz
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  • Related question: https://physics.stackexchange.com/questions/582612/rotation-of-a-spinor?rq=1 – Andy Chen Jul 15 '22 at 17:15
  • A short answer: if we rotate our 3 dimensional coordinate, we expect a spin-${n \over 2}$ particle to transform according to the rotation. The former transformations are the Lie group $SO(3)$, while the later are the ${n+1}$ dimensional irreducible representation of the Lie group $SU(2)$. If ${n+1}$ is even, the corresponding irreducible representation of $SU(2)$ will be the double cover of $SO(3)$ (for example, $SO(3) \cong SU(2)/ {\pm I}$). The property of double cover makes us require $4\pi$ rotation to impose the isomorphism between $SO(3)$ and the corresponding rotations of spin. – Andy Chen Jul 15 '22 at 17:33
  • The transformation of vectors is not trivial. The transformation of scalars is trivial. – KlausK Jul 16 '22 at 10:15
  • I mean it rotates as much as the transformation matrix rotates it – Habouz Jul 16 '22 at 10:16
  • @AndyChen To post a short answer, please post an answer, not a comment. – rob Jul 16 '22 at 17:49

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A very simplified, hand-wavey explanation is to say that in the same way that linear momentum is related to one over the wavelength of the wavefunction as you move each linear direction, so the angular momentum (spin) is related to one over the 'wavelength' (really we need a new word, like 'waveangle' for this idea) of waves as you rotate in each angular direction. (Have a look at some pictures of the spherical harmonics to see what this looks like. We fit a number of sine-waves 'around' a sphere.) If you spin around in a full circle, you have to be able to fit a whole number of waves into that circle to avoid any discontinuity, so spin is quantised.

With a vector, spinning 360 degrees has to get you back where you started, so the number of angular 'wavelengths' has to be a whole number. With a spinor, spinning 360 reverses sign, and you have to turn 720 degrees to get back where you started. Fitting an odd whole number of angular 'wavelengths' into 720 degrees means fitting a half-integer number into 360 degrees.

Nullius in Verba
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