Is this operator Hermitian? Commutator of non-Hermitian operators In the derivation of a Master Equation, I am left with two additional terms:
$$ \sigma_j [\sigma^{\dagger}_k,\rho] - [\rho, \sigma_k]\sigma_j^{\dagger} \quad ,$$
where $\sigma_j = |g\rangle \langle e|_j$ and $\sigma_j^{\dagger}=|e\rangle\langle g|_j$ are spin lowering and raising operators on subsystem $j$ and $\rho$ is the density matrix. The second term is the Hermitian conjugate of the first and therefore, if the first term is Hermitian the result is zero.
Is there any way to tell if $$\sigma_j[\sigma^{\dagger}_k,\rho]$$ is Hermitian? Further to this, to make the question more general, can the  commutator of an arbitrary density matrix with non-hermitian raising/lowering operators be expressed in a different form?
 A: No, the operator is not hermitian in general; it suffices to give a concrete counter example. Consider $h\cong \mathbb C^2$, $H:=\otimes^N h$ for $N\geq 2$ and denote by $\{|e\rangle,|g\rangle\}$ some orthonormal basis on $h$. Define
$$\rho:=\rho_1 \otimes\rho_2\otimes\ldots\otimes \rho_N \tag{1},$$
for some density matrices $\rho_\ell$ on $h$ for $\ell=1,2,\ldots,N$
and
$$\sigma_j:=\underbrace{\mathbb I\otimes \mathbb I\otimes \ldots\otimes \overbrace{\sigma}^{j-\mathrm{th\, factor}}\otimes \mathbb I \otimes \ldots \otimes \mathbb I}_{N-\mathrm{factors}} \quad , \tag 2 $$
where $\sigma:=|g\rangle\langle e|$ and $\mathbb I$ denotes the identity operator on $h$.

To proceed, we compute (here for $1\leq j<k\leq N$):
$$ \sigma_j \,[\sigma^\dagger_k,\rho] = \rho_1\otimes \rho_2 \otimes \ldots \otimes \sigma\rho_j\otimes \ldots\otimes [\sigma^\dagger,\rho_k]\otimes \ldots \otimes \rho_N  \quad . \tag{3}$$
Choosing $\rho_j=\rho_k=|e\rangle\langle e|$ then yields
$$   \sigma_j \,[\sigma^\dagger_k,\rho] = \rho_1\otimes \rho_2 \otimes \ldots \otimes \sigma \otimes \ldots\otimes - \sigma^\dagger \otimes \ldots \otimes \rho_N  \tag {4} \quad ,$$
which is clearly not hermitian.
