Decoupled spin vectors A and B

[filip97]

Let we have $J_i \in{J_1,J_2,J_3}$ ,and $K_i \in{K_1,K_2,K_3}$, rotation and boost generators respectable .

$A_i=\cfrac{1}{2}(J_i+iK_i)$, and

$[A_i,A_j]=i\epsilon_{ijk}A_k$

$[K_i,K_j]=-i\epsilon_{ijk}J_k$

$[J_i,K_j]=-i\epsilon_{ijk}K_k$

How proof that $(m,n)A_i=J^{(m)}_i\otimes I_n$ ?

I was proof that $(m,n)(J_i+iK_i)=J^{(m)}_i\otimes I_n +I_m\otimes J^{(n)}_i \neq J^{(m)}_i\otimes I_n$

by definition that $(m,n)(J_i,K_i)=J^{(m)}_i\otimes I_n + I_m\otimes K^{(n)}_i$

I was read Weinberg QFT Foundations , page 253 .

 

[azntoon]

but i can't proof it .The result follows from the definition of the tensor product and the property that the generators of rotations and boosts commute with themselves. That is, the equation $$(m,n)(J_i+iK_i)=J^{(m)}_i\otimes I_n +I_m\otimes J^{(n)}_i$$holds because $$J_i\otimes I_n + I_m \otimes J_i = J_i\otimes I_n + I_m \otimes J_i$$ and $$K_i\otimes I_n + I_m \otimes K_i = K_i\otimes I_n + I_m \otimes K_i$$where $J_i$ and $K_i$ are the generators of rotations and boosts respectively.