The representation $(\frac{1}{2},\frac{1}{2})$ of the Lorentz group correspond to a four- vector or a spin-one object. Right? Does it imply that any four-vector is identical to a spin-one object or any scalar is identical to a spin-0 object? This can't be correct, right? Because although $A^\mu$ is a four vector and a spin-one object at the same time (which is photon), there is no concept of spin associated with $p^\mu$ or $J^\mu$. I'm confused by terminologies of representation.
Edit- How can I show that $A^\mu$ represent a spin-1 object?
Answer
The problem here is with the identification of the $(A,B)$ values of a representation with spin. $A$ and $B$ do not correspond to spin (they are not even Hermitian!), they just happen to obey $SU(2)$ Lie algebras, and as such they add up in the same way that spins do. When we say that $A_\mu,J_\mu,p_\mu,...$ are all in the $(\frac{1}{2},\frac{1}{2}) $ representation of the Lorentz group we mean that they transform as a four-vector, that's all. People may get lazy and say they are spin 1 objects, but what they really mean is $(A,B)$ spin 1 objects.
No comments:
Post a Comment