Sum of commutator and Anticommutator Suppose $A$ and $B$ are Hermitian Operators. Then what will be the nature (purely real/purely imaginary/complex number of form $a + ib$, $a,b \in \mathbb{R}, a, b \neq 0$) of eigenvalues of  $[A, B] + \{B, A\}$?
Clearly $[A, B] + \{B, A\} = 2AB$. So, if $A$ and $B$ commute they have real eigenvalues and if they Anti-Commute, they will have purely imaginary eigenvalues. But what if they neither commute or anti-commute? Will they have eigenvalues of the form $a + ib$ where $a, b \in \mathbb{R}$ and $a, b \neq 0$?
 A: Suppose $A$ and $B$ are Hermitian operators: what is the nature of the eigenvalues of $[A,B]+\{B,A\}=2AB$?
First, "$A$ is Hermitian" from the practical point of view means that: $A=A^\dagger$,  it is unitarily diagonalizable and has real eigenvalues. The same for $B$.
Now, $(AB)^\dagger = B^\dagger A^\dagger =BA$: this tells us that, in general $AB\neq (AB)^\dagger$, meaning that $AB$ is not  necessarily Hermitian: its eigenvalues can be complex. Two particular cases:

*

*If $[A,B]=0$, then $(AB)^\dagger =BA=AB$: in this case $AB$ is Hermitian and has real eigenvalues.


*If $\{A,B\}=0$, then $(AB)^\dagger =BA=-AB$: in this case $AB$ is anti-Hermitian (or "skew-Hermitian") and has purely imaginary (and possibly zero) eigenvalues.
General case: assume that $v$ is an eigenvector of $AB$, namely $(AB) v = \lambda v$, where $\lambda$ is in general complex. However, we can show that the eigenvalues of $AB$ come in complex conjugate pairs or are real:
$$
0=[\det(AB-\lambda I)]^* = \det(B^\dagger A^\dagger-\lambda^* I) = \det(BA-\lambda^* I) = \det(AB-\lambda^* I)
$$
In the second equality we used the Hermitian property of $A$ and $B$. In the last equality we used the "Sylvester determinant theorem". We see that both $\lambda^*$ and $\lambda$ must be eigenvalues.
For example, if $A$ and $B$ are $3\times 3$ matrices, we will have a pair of complex eigenvalues $(\lambda^*$,$\lambda)$ and a third purely real eigenvalue.
