\Pi =\frac{\lambda}{4} (\tau_x^2 + \tau_y^2) \bigg\{ \frac{\tau - (1- \tau^2) \tanh^{-1}(\tau)}{\tau^3}\bigg\},$$ where $\tau = \sqrt{\tau_x^2 + \tau_y^2 + \tau_z^2}$. For a pure state ($\tau\to 1$), we get the particularly simple result $$\Pi =\frac{\lambda}{4} (\tau_x^2 + \tau_y^2) = \frac{\lambda}{4} \sin^2\theta.$$

We can also compare this with the von Neumann formulation in Eq. (\[Pi\_vN\]). For the case of a dephasing bath, given by Eq. (\[D\_dephasing\]), the target state $\rho^*$ will be any diagonal state in the $J_z$ basis. The entropy production given by Eq. (\[Pi\_vN\]) is then readily found to be $$\label{Pi_deph_vN}
\Pi_\text{vN} = \frac{\lambda}{2} \; (\tau_x^2 + \tau_y^2) \frac{\tanh^{-1} (\tau)}{\tau}.$$ A comparison of this result for the case where $\tau_x^2 + \tau_y^2 = \tau^2$ is shown in Fig. \[fig:dephasing\]. As it can be seen, both the Wehrl and the von Neumann entropy productions behave in a similar way. However, as the system approaches a pure state ($\tau\to1$), the von Neumann entropy production diverges, whereas the Wehrl entropy production rate remains finite.

![\[fig:dephasing\] The entropy production rate contribution of the dephasing bath for a spin-$1/2$ particle, as a function of $\tau$. In red we show the von Neumann entropy production rate, given by Eq. (\[Pi\_deph\_vN\]), and in black the corresponding Wehrl entropy 