Product of Representations of Lorentz Group

[filip97]

How to prove that direct product of two rep of Lorentz group $(m,n)⊗(a,b)=(m⊗a,n⊗b)$ ?

Let $J\in {{J_1,J_2,J_3}}$
Then we have :
$[(m,n)⊗(a,b)](J)=(m,n)(J)I_{(a,b)}+I_{(m,n)}⊗(a,b)(J)=$
$=I_m⊗J_n⊗I_a⊗I_b+J_m⊗I_n⊗I_a⊗I_b+I_m⊗I_n⊗J_a⊗I_b+I_m⊗I_n⊗I_a⊗J_b$
and
$(m⊗a,n⊗b)(J)=I_{(m⊗a)}⊗J_{(n⊗b)}+J_{(m⊗a)}⊗I_{(n⊗b)}=$
$=I_m⊗I_a⊗(I_n⊗J_b+J_n⊗I_b)+(I_m⊗J_a+J_m⊗I_a)I_n⊗I_b=$
$=I_m⊗I_a⊗I_n⊗J_b+I_m⊗I_a⊗J_n⊗I_b+J_m⊗I_a⊗I_n⊗I_b+I_m⊗J_a⊗I_n⊗I_b$, and we have that

$[(m,n)⊗(a,b)](J)\neq [(m,n)⊗(a,b)](J)$

Where $I_a$ is unit matrix $(2a+1) (2a+1)$, matrix, identical is for $J_b$

How this is work ?

 

[azntoon]

We can prove that the direct product of two representations of the Lorentz group is equal to the product of the individual representations by using the fact that the direct product of two matrices is equal to the product of the individual matrices. Therefore, if we have two representations of the Lorentz group, say $(m,n)$ and $(a,b)$, then the direct product of the two is equal to $(m⊗a,n⊗b)$. To prove this, we can use the definition of the direct product of two matrices, which states that given two matrices $A_{m*n}$ and $B_{p*q}$, the direct product of the two is given by the matrix $C_{(m*p)*(n*q)}$, where each entry of the matrix is given by the product of the corresponding entries of $A$ and $B$. Thus, for our two representations of the Lorentz group, we have $(m,n)⊗(a,b)=(m⊗a,n⊗b)$ This follows directly from the definition of the direct product of matrices.