by parts, we arrive at $$\label{Pi_dephasing}
\frac{{\,\mathrm{d}}S}{{\,\mathrm{d}}t} = \Pi =\frac{\lambda}{2} \bigg(\frac{2J + 1}{4\pi} \bigg) \int {\,\mathrm{d}}\Omega \; \frac{|\mathcal{J}_z(\mathcal{Q})|^2}{\mathcal{Q}},$$ which has the typical form of an entropy production [@Tome2010; @Spinney2012; @Landi2013b; @Santos2017b]: It is always non-negative and zero iff $\mathcal{J}_z(\mathcal{Q})=0$. This occurs only when $\mathcal{Q}$ is independent of the azimuthal angle $\phi$, which is the phase-space analog of requesting that $\rho$ is diagonal in the $J_z$ basis. Hence, this result establishes $\mathcal{J}_z(\mathcal{Q})$ as the current associated with the loss of coherence in the $J_z$ basis.

Equation (\[Pi\_dephasing\]) also shows that a dephasing bath has no associated entropy flux. This also appears in the context of the von Neumann entropy production and the Wigner entropy production for bosonic modes [@Santos2017b]. Moreover, it also agrees with the definition of dephasing as a unital map, for which the entropy can only increase [@265894] (whereas $\Pi \geq 0$, the sign of $\Phi$ is in general arbitrary and thus may lead to a reduction in the entropy. But when $\Phi = 0$, we ensure that the entropy can never decrease).

Spin-1/2 case
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We can find an explicit formula for the integral (\[Pi\_dephasing\]) in the case of spin-$1/2$ particles \[cf. Eqs. (\[rho\_spin12\]) and (\[Q\_spin12\])\]; viz, $$\label{Pi_deph_Wehrl}
