# Non-Hermitian solution for $SU(2)$ Lie algebra?

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.)

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.
• Thank you for the useful information! At the time when $[A_i, A_j] = i\epsilon_{ijk} A_k$ is obtained, I think it is not so obvious this A really generate SU(2), for example, it is not obvious that $\exp(2 \pi A)$ is identity. So what we know is only commutation relation $[A_i, A_j] = i\epsilon_{ijk} A_k$. Even in this situation, can we apply this Weyl's theorem? May 30, 2022 at 16:29
• @rsato Every representation of $\mathfrak{su}(2)$ induces a representation of $\mathrm{SU}(2)$ and vice versa, this is part of the general correspondence between Lie algebras and Lie groups. May 30, 2022 at 16:35
• Does this mean that $\exp(4 \pi i A_3)$ should be identity matrix if $[A_i, A_j] = i \epsilon_{ijk} A_k$ is satisfied? May 31, 2022 at 2:43