![image](plot5.pdf){width="30.00000%"}![image](plot6.pdf){width="30.00000%"}![image](plot7.pdf){width="30.00000%"}

A thermal quench
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Now we consider a thermal quench. We assume the bath temperature is $T$, but the system begins in a different temperature $T_0$. The exact solution of the Lindblad master equation will continue to be a thermal Gibbs state, but with a time-dependent temperature $\beta(t)$. It is more convenient to work with $\tau_z(t)=-\tanh(\omega\beta(t)/2)$. Since $\tau_z(t)=\langle \sigma_z \rangle_t$, it follows that this quantity satisfies the differential equation $$\frac{d\tau_z(t)}{dt} = \frac{\gamma}{\bar{\tau}_z}(\tau_z(t)-\bar{\tau}_z),$$ whose solution is $$\label{quench_sol}
\tau_z(t) = \bar{\tau}_z + e^{-\gamma t/|\bar{\tau}_z|}(\tau_z(0)-\bar{\tau}_z).$$

Thus, we may readily apply Eqs. (\[Phi\_damp\_W\]) and (\[Pi\_damp\_spin12\]) to compute the entropy productions and fluxes for both the von Neumann and the Wehrl entropies. Examples of these curves are shown in Fig. (\[fig:thermal\]).

![\[fig:thermal\] The Wehrl entropy production rate \[Eq. (\[Pi\_damp\_spin12\])\], the entropy flux rate \[Eq. (\[Phi\_damp\_W\])\] and the total rate of change of the entropy, \[Eq. (\[dSdt\])\], for a thermal quench, computed using Eq. (\[quench\_sol\]) for (a) $T/\omega = 1$, $T_0/\omega = 2$ and (b) the inverse. ](plot8.pdf)

Spin-$1/2$ in an oscillating magnetic field
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Here we consider again the problem defined by the Hamiltonian \[Eq. (\[H\_example\])\], but now subject to the amplitude damping dissipator. We assume that the system starts initially in the eigenstate 