How to derive or understand the quantum detailed balance condition for Markov open system?

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In the "On The detailed balance conditions for non-Hamiltonian systems", I learned that for a Markov open quantum system to satisfying the master equation with the Liouvillian superoperators, the detailed balance condition will be

Definition 2: The open quantum Markovian system ($dim(\mathcal{H}) < \infty$) obeys the detailed balance principle if the generator $L$ in Heisenberg picture is a normal operator in Hilbert space $\mathcal{B}_{\rho_0}(\mathcal{H})$ (see Definition 1).

Definition 1: $\mathcal{B}_{\rho_0}(\mathcal{H})$ denotes the Hilbert space of all linear operators on the finite-dimensional Hilbert space $\mathcal{H}$ with the scalar product defined by the formula $$\langle A, B\rangle = Tr(A^\dagger B \rho_0), A,B \in \mathcal{B}_{\rho_0}(\mathcal{H})$$ where $\rho_0$ is a fixed state (density matrix) and $ \rho_0 > 0$.

The $L$ is the adjoint operator, defined with respect to definition 1, of the Liouvillian superoperator $\mathcal{L}$, such that $$ \frac{d \rho}{d t} = \mathcal{L} \rho \\ \frac{d A}{d t} = L A, A\in \mathcal{B}_{\rho_0}(\mathcal{H}). $$

The author started from the classical detailed balance condition $p_{ij}\pi_j = p_{ji}\pi_i$ and finally get to definition 2.

For me, I will write the quantum analogy of detailed balance as $$ \langle A,L(B) \rangle = \langle B, L(A)\rangle = \langle L(A), B\rangle. $$ Then, $L$ is hermitian and normal. However, normal operator is not necessarily hermitian. My question is that how can we get to definition 2 starting from the classical version of detailed balance?