Physics Asked by user286848 on June 9, 2021
Lorentz generators satisfy the Lie algebra
$$[J_i,J_j]=iepsilon_{ij}^kJ_k, ~~~~[J_i,K_j]=iepsilon_{ij}^kK_k, ~~~~[K_i,K_j]=-iepsilon_{ij}^kJ_k.$$
Now, define $$A_i=frac{J_i+iK_i}{2},~~~~B_i=frac{J_i-iK_i}{2},$$ and we can easily prove that
$$[A_i,A_j]=iepsilon_{ij}^kA_k, ,[B_i,B_j]=iepsilon_{ij}^kB_k, ,[A_i,B_j]=0.$$
We also see that the ${M^{munu}}$ Lie algebra is isomorphic to two $SU(2)$ Lie algebras and they have the Casimir invariants $A^2, B^2$.
Now, I do not understand what is the meaning of
$A^2$ and $B^2$ commuting the spinor representation provided by $sum^{munu}=frac{i}{4}[gamma^{mu},gamma^{nu}]$ evaluate the two casimir invariant.
Any help is highly appreciated.
This is a direct plugin, and I fear I am doing your homework for you, denying you the opportunity to figure it out. Rather than doing it, I'll demonstrate to you how more efficiently you can get the answer by the tensor product representation of Dirac matrices as 2×2 matrices tensored with each other, $$ gamma^0 = sigma^3 otimes I , qquad gamma^j = isigma^2 otimes sigma^j , $$ so, by sleepwalker's plugin, you find $$ -Sigma^{0j}=K_j= -frac{i}{2} sigma^1otimes sigma^j, qquad -frac{epsilon ^{jkl}}{2}Sigma^{kl}=J_j= -tfrac{1}{2} Iotimes sigma^j, ~~leadsto A_j=-P_A otimes frac{sigma^j}{2}, qquad B_j=-P_B otimes frac{sigma^j}{2}, $$ where, of course, the projectors to either Kronecker factor are $$ P_A= (I- sigma^1)/2, Longrightarrow P_A^2= P_A, P_B= (I+ sigma^1)/2, Longrightarrow P_B^2= P_B P_A P_B=P_B P_A=0. $$
It is then evident that $$ vec A cdot vec A = P_A otimes vec sigma cdot vec sigma /4 = tfrac{3}{4} P_Aotimes I, vec B cdot vec B = P_B otimes vec sigma cdot vec sigma /4 =tfrac{3}{4} P_Botimes I, $$ your desired 4×4 matrices.
Correct answer by Cosmas Zachos on June 9, 2021
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