In QFT textbooks, representations for Lorentz group is constructed from $A$-spin & $B$-spin discussion. The Lie albegra of Lorentz group is $[J_i, J_j] = i\epsilon_{ijk} J_k,~[J_i, K_j] = i\epsilon_{ijk} K_k$, and $[K_i, K_j] = -i\epsilon_{ijk} J_k$. By defining $A_i \equiv (J_i + iK_i)/2$ and $B_i \equiv (J_i – iK_i)/2$, we obtain $[A_i, A_j] = i\epsilon_{ijk} A_k,~[B_i, B_j] = i\epsilon_{ijk} B_k$, and $[A_i, B_j] = 0$. The commutation relations of $A_i$ and $B_i$ are the same as $SU(2)$ Lie algebra.
Many textbooks assume $A_i$ and $B_i$ are Hermitian matrices, but, is there any non-Hermitian solution for $A_i$ and $B_i$? I think it is not so obvious whether $A_i$ and $B_i$ are Hermitian or not because of its definition $A_i = (J_i + i K_i)/2$ and $B_i = (J_i – i K_i)/2$. Trivial non-Hermitian solution is $A_i = P \hat A_i P^{-1}$ where $P$ is (non-unitary) invertible matrix and $\hat A_i$ is Hermitian matrices to satisfy $[\hat A_i, \hat A_j] = i \epsilon_{ijk} \hat A_k$. Is there any solution for $A_i$ and $B_i$ which is not classified to this category? Or can we show that any solution for $[A_i, A_j] = i\epsilon_{ijk} A_K$ falls into this?
(Srednicki's textbook says "The standard derivation assumes that the matrices are hermitian, but allowing nonhermitian matrices does not enlarge the set of solutions." in section 33 but there is no detailed information about that.)
Best Answer
This is an oblique reference to Weyl's theorem on reducibility, in particular to Weyl's unitarian trick:
Concretely, for any compact Lie group $G$ (such as $\mathrm{SU}(2)$ and a representation $\pi$ on a finite-dimensional Hilbert space $H$, you can pick any inner product $\langle -,-\rangle$ on it and define a new inner product by $$ \langle v,w\rangle_G := \int_G \langle \pi(g)v,\pi(g)w\rangle \mathrm{d}g$$ where $\mathrm{d}g$ is the Haar measure of $G$. Then $\pi$ is unitary with respect to $\langle -,-\rangle_G$ and consequently the induced representation $\mathrm{d}\pi$ of the Lie algebra $\mathfrak{g}$ is Hermitian.
So we can construct an equivalent Hermitian representation for every non-Hermitian representation of $\mathfrak{su}(2)$ and adding non-Hermitian options for finite-dimensional representations of $\mathfrak{g}$ doesn't add any solutions, just as claimed.