# Why is angular momenum related to the spin?

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?

• Related question: physics.stackexchange.com/questions/582612/… Commented Jul 15, 2022 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. Commented Jul 15, 2022 at 17:33
• The transformation of vectors is not trivial. The transformation of scalars is trivial. Commented Jul 16, 2022 at 10:15
• I mean it rotates as much as the transformation matrix rotates it Commented Jul 16, 2022 at 10:16