## § Normal operators: Decomposition into Hermitian operators

Given a normal operator $A$, we can always decompose it $A = B + iC$ where $B = B^{\dagger}$, $C = C^\dagger$, and $[B, C] = 0$. 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 $A$ into two hermitian operators $B, C$ that commute, we can diagonalize $B, C$ simultaneously and thereby measure $B, C$ 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 $A$. Write the operator in its eigenbasis $\{ |a_k \rangle \}$. This will allow us to write $A = \sum_k |a_k \rangle \langle a_k|$. with each $a_k = b_k + i c_k$. Now write this as:
\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}
$B, C$ are simultaneously diagonalizable in the eigenbasis $\{ |a_k \rangle \}$ and hence $[B, C] = 0$.