current operator for the density matrix, in the sense that Eq. (\[amp\_D\]) takes the form of a continuity equation. Moreover, one may verify that $f(e^{-\beta H}) = 0$, which allows us to interpret the stationary state as the one for which the current is itself zero. Moving to phase space, we have $$\mathcal{D}(\mathcal{Q}) = \frac{\gamma}{2}  \bigg\{ \mathcal{J}_{-}(f(\mathcal{Q})) - \mathcal{J}_{+}(f^*(\mathcal{Q})) \bigg\},$$ where $$\label{fQ_spin}
f(\mathcal{Q}) =  \frac{1}{2} \bigg[2 J \mathcal{Q} - \mathcal{J}_z(Q)\bigg] e^{i \phi} \sin \theta + \frac{1}{2}\bigg[ \cos\theta - (2\bar{n}+1)\bigg] \mathcal{J}_+(\mathcal{Q}),$$ \[see Eqs. (\[Jp\_orbital\]) and (\[Jm\_orbital\]) for the definition of the current operators $\mathcal{J}_i$\]. Alternatively, in terms of the TSS bosonic representation, the current $f$ becomes $$\label{fQ}f(Q) = \bigg[ \alpha^* \beta + (\bar{n}+1) \beta \partial_\alpha - \bar{n} \alpha^* \partial_{\beta^*}\bigg] Q    
$$

Identification of the entropy production rate
---------------------------------------------

To separate ${\,\mathrm{d}}S/{\,\mathrm{d}}t$ into the form stated in Eq. (\[dSdt\]), we recast all phase-space variables in terms of the relevant current in the problem, which in this case is $f(\mathcal{Q})$. One then notes that following standard thermodynamic arguments, the entropy production should be an even function of the relevant currents, whereas the entropy flux rate should be odd [@Callen1985].

It is more convenient to use Eq. (\[Q\]) in order to express $Q$ in 