§ Normal operators: Decomposition into Hermitian operators

Given a normal operator AA, we can always decompose it A=B+iCA = B + iC where B=BB = B^{\dagger}, C=CC = C^\dagger, and [B,C]=0[B, C] = 0. This means that we can define 'complex measurements' using a normal operator, because a normal operator has full complex spectrum. Since we can always decompose such an operator AA into two hermitian operators B,CB, C that commute, we can diagonalize B,CB, C simultaneously and thereby measure B,CB, C simultaneously. So extending to "complex measurements" gives us no more power than staying at "real measurements"

§ Decomposing a normal operator

Assume we have a normal operator AA. Write the operator in its eigenbasis {ak}\{ |a_k \rangle \}. This will allow us to write A=kakakA = \sum_k |a_k \rangle \langle a_k|. with each ak=bk+icka_k = b_k + i c_k. Now write this as:
A=k(bk+ick)akakA=kbkakak+ickakakA=B+iC \begin{aligned} & A = \sum_k (b_k + i c_k)|a_k \rangle \langle a_k| \\ & A = \sum_k b_k |a_k \rangle \langle a_k| + i c_k |a_k \rangle \langle a_k| \\ & A = B + iC \\ \end{aligned}
B,CB, C are simultaneously diagonalizable in the eigenbasis {ak}\{ |a_k \rangle \} and hence [B,C]=0[B, C] = 0.