# Does dimension of irreducible representations of the double cover $SU(2)$ of the 3D rotation group define spin of particle?

In quantum field theory, does dimension of irreducible representations of the double cover $$SU(2)$$ of the 3D rotation group conclusively define spin? In other words, Is spin 1 particle only thing that vector field in 4D spacetime can generate? Does spinor field always generate spin 1/2 particle?

## 1 Answer

1. In relativistic theories, an irreducible representation with spins $$(j_L,j_R)\in\frac{1}{2}\mathbb{N}_0\times\frac{1}{2}\mathbb{N}_0$$ for the [double cover $$\tilde{G}_1=SL(2,\mathbb{C})\times SL(2,\mathbb{C})$$ of the] complexified proper Lorentz group $$G_1=SO(1,3;\mathbb{C})$$ decomposes into a direct sum $$(j_L,j_R)~=~\bigoplus_{j=|j_L-j_R|}^{j_L+j_R} j$$ of irreducible representations with spins $$j\in\frac{1}{2}\mathbb{N}_0$$ for the [double cover $$\tilde{G}_2=SL(2,\mathbb{C})$$ of the] restricted Lorentz group $$G_2=SO^+(1,3;\mathbb{R})$$.

2. In turn, an irreducible representation with spin $$j\in\frac{1}{2}\mathbb{N}_0$$ for the [double cover $$\tilde{G}_2=SL(2,\mathbb{C})$$ of the] restricted Lorentz group $$G_2=SO^+(1,3;\mathbb{R})$$ is also an irreducible representation with the same spin $$j\in\frac{1}{2}\mathbb{N}_0$$ for the [double cover $$\tilde{G}_3=SU(2)$$ of the] 3D rotation group $$G_3=SO(3,\mathbb{R})$$.

3. Examples. A Lorentz vector is a $$(1/2,1/2)$$ representation. Left & right Weyl spinors are $$(1/2,0)$$ & $$(0,1/2)$$ representations, cf. e.g. this related Phys.SE post.

4. The particles corresponding to a field are usually classified in terms of the little group, cf. e.g. this & this related Phys.SE posts and links therein.