@Brunelli2016; @Frank2013; @Muller-Lennert2013; @Audenaert2013; @Abe2003; @Esposito2010a; @Pucci2013; @Li2016c; @Solano-Carrillo2015; @0295-5075-94-3-30001; @Oliveira2016a]. Here we shall be interested in systems described by a master equation of the form $$\label{M}
\frac{{\,\mathrm{d}}\rho}{{\,\mathrm{d}}t} = - i [H,\rho] + D(\rho),$$ where $\rho$ is the system’s density matrix, $H$ is the Hamiltonian, and $D(\rho)$ is the dissipator describing the effects of the bath. In Refs. [@ALICKI1976249; @Alicki1979], Alicki suggested a relation for the entropy production in terms of the dynamical semigroup $\{\Lambda_t|t\geqslant 0\}$ generated by Eq. (\[M\]) and its corresponding invariant state $\Lambda_t\rho^*=\rho^*$. The relation is given by $$\label{Pi_vN} 
\Pi_\text{vN} = - \frac{{\,\mathrm{d}}}{{\,\mathrm{d}}t} S_\text{vN}(\Lambda_t\rho(0)||\rho^*),$$ where $S_\text{vN}(\rho||\sigma) = \operatorname{tr}(\rho \ln \rho - \rho \ln \sigma)$ is the von Neumann relative entropy. Clearly, with this definition, $\Pi_\text{vN} \geq 0$, with the equality holding only for $\rho = \rho^*$. In the case of a thermal bath, the invariant state becomes the Gibbs state, $\rho^* = \rho_\text{eq} = e^{-\beta H}/Z$. Using Eqs. (\[dSdt\]) and (\[Pi\_vN\]), one may then show that in this case, the entropy flux rate $\Phi_{\text{vN}}$ becomes $$\label{Phi_vN}
\Phi_{\text{vN}} = \frac{\Phi_E}{T} = - \frac{1}{T}\operatorname{tr}\bigg\{H D(\rho)\bigg\},$$ which is the familiar Clausius relation between entropy and heat, therefore providing a more physical basis for Eq. (\[Pi\_vN\]).

However, Eqs. (\[Pi\_vN\]) and (\[Phi\_vN\]) 