text:
\section{Introduction}
Quantum thermodynamics \cite{Wigner1932, Esposito2009, Campisi2011, Binder2018, Deffner2019} is an emerging field studying the nonequilibrium statistical mechanics of the quantum dissipative systems \cite{Yu1994, Breuer2002, Hanggi2005, Caldeira2014}. Quantum work \cite{Kurchan, Tasaki, Talkner2007, Talkner2016, Ken2018}, quantum heat \cite{Saito2007, Aurell2017, Ken2018b, Hu2018, Hu2020a, Hu2020b}, and quantum entropy production \cite{Aurell2015, Weiderpass2020, Pucci2013, Esposito2010} are among the most basic concepts, which play important roles in the study of work extraction and heat transfer in quantum devices, such as the quantum heat engines and refrigerators \cite{Scovil1959, Geusic1959, Alicki1979, Kosloff1984, Geva1992, Geva1996, Bender2000, Scully2003, Kieu2006, Quan2007, Allahverdyan2010, Linden2010, Lutz2014, Dong2015, Beau2016, Karimi2016}.

A typical exactly solvable model used for addressing these problems is the quantum Brownian motion model proposed by Caldeira and Leggett \cite{Caldeira1983, Caldeira2014}. It consists of a system described by the Hamiltonian \begin{small}${\hat{H}_S}$\end{small} (often a harmonic oscillator \cite{Hu1992}), a heat bath of harmonic oscillators with the Hamiltonian \begin{small}${\hat{H}_B}$\end{small}, and the interaction Hamiltonian \begin{small}${\hat{H}_{SB}}$\end{small}. One can analytically integrate out the degrees of freedom of the heat bath, which brings important insights to the understanding of the thermodynamics of open quantum systems. In the studies about entropy production and heat transfer, previous efforts have been focused mainly on the entropy production in the heat bath. For example, in Ref. \cite{Aurell2015}, by adapting the Feynman-Vernon influence functional formalism, the change of the von Neumann entropy of the heat bath is computed. In Refs. \cite{Oono1998, Sasa2001, Qian2001, Qian2001b, Seifert2005, Imparato2006, Ge2006, Ge2009, Saha2009, Ge2010, Esposito2010a, Esposito2010b, Esposito2010c, R2012, Ge2018}, the expressions of some basic nonequilibrium thermodynamic quantities, such as the total entropy production rate \begin{small}${e_{\rm p}}$\end{small}, the heat dissipation rate \begin{small}${h_{\rm d}}$\end{small}, the housekeeping heat \begin{small}${Q_{\rm hk}}$\end{small} and the excess heat \begin{small}${Q_{\rm ex}}$\end{small}, are discussed for a classical isothermal process described by a master equation or a Fokker-Planck (Langevin) equation. More recently, the energy exchange, and thus the entropy exchange, between the system and the heat bath is calculated in Refs. \cite{Ken2018b, Hu2020a, Hu2020b}. However, the quantum-classical correspondence, and especially the quantum corrections to the entropy production in the quantum Brownian motion model, as well as the total entropy production, have not been explored so far (but see Ref. \cite{Weiderpass2020}).

In this article, we study the time evolution of the quantum entropy and its corrections to the classical entropy in an open quantum system. Specifically, we calculate the von Neumann entropy in the quantum Brownian motion model subject to a driving force, as well as the total entropy production. The von Neumann entropy of a quantum system described by the density matrix \begin{small}${\hat{\rho}}$\end{small} is given by \cite{von Neumann1927, von Neumann1955}
\begin{small}
\begin{equation}\label{eq:0000}
S_q=-\rm{Tr}\left[\hat{\rho}\ln\hat{\rho}\right].
\end{equation}
\end{small}Here, we have set the Boltzmann's constant to be ${1}$, and thus the entropy becomes dimensionless. It is difficult to calculate the von Neumann entropy through its definition Eq. (\ref{eq:0000}), because one has to diagonalize an infinite-dimensional matrix in order to compute the trace of a function. However, by reformulating the problem in the phase space the calculation can be significantly simplified \cite{Qiu2020}. By solving the quantum Langevin equation exactly, we obtain the analytical expression of the Wigner function at an arbitrary time \begin{small}${t}$\end{small} in terms of the initial Wigner function, which is valid in both the strong and the weak coupling regimes. Then by adapting the method in Ref. \cite{Qiu2020}, for initial states which have well-defined classical counterparts, we find that if we expand the von Neumann entropy in powers of \begin{small}${\hbar}$\end{small}, the zeroth-order term reproduces the classical Gibbs entropy. We can also obtain the quantum correction to the entropy of the system and its time evolution. Moreover, under the Markovian approximation, we obtain the expression of the quantum corrections to the total entropy production rate ${e_{\rm p}}$ and the heat dissipation rate ${h_{\rm d}}$. Our results bring important insights to the understanding of the entropy in an open quantum system.

This article is organized as follows. We begin in Sec.\uppercase\expandafter{\romannumeral2} with a description of the model and the quantum Langevin equation. In Sec. III we derive the general solution to the quantum Langevin equation of the driven quantum Brownian motion model. In Sec. IV we obtain the expression of the reduced Wigner function of a driven quantum Brownian motion model. In Sec. V, by using the method developed in Ref. \cite{Qiu2020}, we calculate the quantum corrections to the entropy of the system and its dynamical evolution. In Sec. VI, based one the results in Sec. V, we obtain the expression of the quantum corrections to the total entropy production rate ${e_{\rm p}}$ and the heat dissipation rate ${h_{\rm d}}$. Finally, in Sec. VII we make some remarks and summarize our results.

\section{The model and the derivation of the quantum Langevin equation}
We consider the quantum Brownian motion described by the Caldeira-Leggett model \cite{Caldeira2014, Caldeira1983, Hu1992}. The system that we consider is a harmonic oscillator subject to a time-dependent driving force \begin{small}${\hat{f}(t)}$\end{small} \cite{Xu2009}. The system is linearly coupled to a heat bath consisting of a set of harmonic oscillators. The Hamitonian of the composite system is given by \begin{small}${\hat{H}_{\rm{tot}}(t)=\hat{H}_S(t)+\hat{H}_B+\hat{H}_{SB}}$\end{small}, with
\begin{small}
\begin{subequations}
\begin{eqnarray}\label{eq:0001} &&\hat{H}_S(t)=\frac{\hat{\mathcal{p}}^2(t)}{2 m_0}+\frac{1}{2}m_0\omega_0^2\left(\hat{\mathcal{q}}(t)-\frac{\hat{f}(t)}{m_0\omega_0^2}\right)^2,\ \ \ \ \ \ \ \\
&&\hat{H}_B=\sum^{N}_{j=1}\left(\frac{\hat{p}^2_j(t)}{2 m_j}+\frac{1}{2}m_j\omega_j^2\hat{q}_j^2(t)\right),\\
&&\hat{H}_{SB}=-\hat{\mathcal{q}}(t)\sum_{j=1}^{N}C_j\hat{q}_j(t)+\sum_{j=1}^{N}\frac{C_j^2}{2m_j\omega_j^2}\hat{\mathcal{q}}^2(t),
\end{eqnarray}
\end{subequations}
\end{small}where \begin{small}${m_0}$\end{small}, \begin{small}${\omega_0}$\end{small}, \begin{small}${\hat{\mathcal{q}}}$\end{small}, \begin{small}${\hat{\mathcal{p}}}$\end{small} and \begin{small}${m_j}$\end{small}, \begin{small}${\omega_j}$\end{small}, \begin{small}${\hat{q}_j}$\end{small}, \begin{small}${\hat{p}_j}$\end{small} are the mass, angular frequencies, coordinates and momenta of the system and the \begin{small}${j}$\end{small}th harmonic oscillator of the heat bath, respectively, and \begin{small}${C_j (j=1,2,3,...)}$\end{small} are the coupling constants. Here, we have included the counterterm \begin{small}${\sum_j (C_j^2/2m_j\omega_j^2)\hat{\mathcal{q}}^2(t)}$\end{small} in the interaction Hamiltonian to cancel the negative frequency shift of the potential \cite{Caldeira1983a}.

The equation of motion of the time-dependent operator can be obtained by using the Heisenberg equation
\begin{small}
\begin{equation}\label{eq:0002}
i\hbar\dot{\hat{O}}=[\hat{O},\hat{H}],
\end{equation}
\end{small}which gives the time derivative (denoted by the superposed dot) of an
arbitrary operator \begin{small}${\hat{O}}$\end{small}. Then we have \cite{Ford1988}
\begin{small}
\begin{equation}\label{eq:0003}
\left\{
\begin{aligned}
&\dot{\hat{\mathcal{q}}}\!=\!\frac{1}{m_0} \hat{\mathcal{p}} \\
&\dot{\hat{\mathcal{p}}}\!=\!-m_0\omega_0^2\hat{\mathcal{q}}\!+\!\hat{f}(t)\!+\!\sum^N_{j=1}\!C_n\!\left(\hat{q}_j\!-\!\frac{C_j}{m_j\omega_j^2}\hat{\mathcal{q}}\right)
\end{aligned}\ \
\right.
\end{equation}
\end{small}for the system, and\begin{small}
\begin{equation}\label{eq:0004}
\left\{
\begin{aligned}
&\dot{\hat{q}}_j=\frac{1}{m_j} \hat{p}_j \\
&\dot{\hat{p}}_j=-m_j\omega_j^2\left(\hat{q}_j-\frac{C_j}{m_j\omega_j^2}\hat{\mathcal{q}}\right)
\end{aligned}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\right.
\end{equation}
\end{small}for the \begin{small}${j}$\end{small}th harmonic oscillator of the heat bath. Solving Eq. (\ref{eq:0004}) and substituting it into Eq. (\ref{eq:0003}), one can obtain the quantum Langevin equation of the driven Brownian particle \cite{Ford1988}
\begin{small}
\begin{equation}\label{eq:0005}
m_0 \ddot{\hat{\mathcal{q}}}(t)+\int^{t}_{0}dt' \mu(t-t')\dot{\hat{\mathcal{q}}}(t')+m_0\omega_0^2 \hat{\mathcal{q}}(t)+\mu(t)\hat{\mathcal{q}}(0)=\hat{F}(t)+\hat{f}(t),
\end{equation}
\end{small}where \begin{small}${\mu(t)}$\end{small} is the memory function and it is given by
\begin{small}
\begin{equation}\label{eq:0006}
\mu(t)=\sum_{j}\frac{C_j^2}{m_j\omega_j^2}\cos(\omega_j t),
\end{equation}
\end{small}and \begin{small}${\hat{F}(t)}$\end{small} is the fluctuating force operator and it can be expressed in terms of the initial bath variables
\begin{small}
\begin{equation}\label{eq:0007}
\hat{F}(t)=\sum_j C_j\left[\hat{q}_j(0)\cos(\omega_j t)+\hat{p}_j(0)\frac{\sin(\omega_j t)}{m_j \omega_j}\right].
\end{equation}
\end{small}It is straightforward to show that the correlation and the commutator can be expressed as \cite{Ford1988, Ford2001}
\begin{small}
\begin{equation}\label{eq:0008}
\frac{1}{2}\langle \hat{F}(t)\hat{F}(t')+\hat{F}(t') \hat{F}(t)\rangle=\frac{1}{\pi}\!\int^{\infty}_{0}\!d\omega\! \ \rm{Re}\{\tilde{\mu}(\omega\!+\!i0^+)\}\hbar\omega\coth{\!\left(\!\frac{\beta\hbar\omega}{2}\!\right)\!}\cos[\omega(t\!-\!t')],
\end{equation}
\begin{equation}\label{eq:0009}
\left[\hat{F}(t), \hat{F}(t')\right]=\frac{2\hbar}{i\pi}\!\int^{\infty}_{0}\!d\omega \ \rm{Re}\{\tilde{\mu}(\omega+i0^+)\}\omega\sin[\omega(t-t')],
\end{equation}
\end{small}where the bracket \begin{small}${\langle...\rangle}$\end{small} depicts the quantum expectation value, and \begin{small}${\tilde{\mu}(z)}$\end{small} is the Fourier transform of the memory function:
\begin{small}
\begin{equation}\label{eq:0010}
\tilde{\mu}(z)=\int^{\infty}_{0}dt\ \mu(t) e^{i z t}, \ \ \ \ \rm{Im}\ z>0.
\end{equation}\end{small}In the following, we will try to solve Eq. (\ref{eq:0005}) by using the Green function approach.

\section{General solution to the quantum Langevin equation (\ref{eq:0005}) }
For the stationary process, the system is held fixed at the origin in the distant past \cite{Ford1988}. From Eq. (\ref{eq:0005}), we obtain the quantum Langevin equation for the stationary process
\begin{small}
\begin{equation}\label{eq:0011}
m_0\ddot{\hat{\mathcal{q}}}^{(\!s\!)\!}(t)\!+\!\int^{t}_{-\infty}dt'\mu(t-t')\dot{\hat{\mathcal{q}}}^{(\!s\!)\!}(t')\!+\!m_0\omega_0^2\hat{\mathcal{q}}^{(\!s\!)\!}(t)\!=\!\hat{F}(t)\!+\!\hat{f}(t),
\end{equation}
\end{small}where \begin{small}${\hat{f}(t)}$\end{small} is switched on at \begin{small}${t=0}$\end{small}, and the solution can be written as \cite{Ford2001}
\begin{small}
\begin{equation}\label{eq:0012}
\hat{\mathcal{q}}^{(s)}(t)=\hat{\mathcal{q}}^{(F)}(t)+\hat{\mathcal{q}}^{(f)}(t),
\end{equation}
\end{small}where
\begin{small}
\begin{subequations}
\begin{eqnarray}\label{eq:0012a}
&&\hat{\mathcal{q}}^{(F)}(t)=\int^{t}_{-\infty}dt'\ G(t-t')\hat{F}(t'),\label{eq:0012a1}\\
&&\hat{\mathcal{q}}^{(f)}(t)=\int^{t}_{-\infty}dt'\ G(t-t')\hat{f}(t'),\label{eq:0012a2}
\end{eqnarray}
\end{subequations}
\end{small}and the Green function \begin{small}${G(t)}$\end{small} is given by
\begin{small}
\begin{equation}\label{eq:0013}
G(t)=\frac{1}{2\pi}\int^{\infty}_{-\infty} d\omega \ \alpha(\omega+i0^+)\ e^{-i\omega t},
\end{equation}
\end{small}with the response function
\begin{small}
\begin{equation}\label{eq:0014}
\alpha(z)=\frac{1}{-m_0 z^2-iz\tilde{\mu}(z)+m_0\omega_0^2}.
\end{equation}
\end{small}From Eq. (\ref{eq:0008}) and Eq. (\ref{eq:0012a}), we obtain the correlation
\begin{small}
\begin{equation}\label{eq:0015}
\frac{1}{2}\big\langle \hat{\mathcal{q}}^{(F)}(t) \hat{\mathcal{q}}^{(F)}(t')\!+\!\hat{\mathcal{q}}^{(F)}(t') \hat{\mathcal{q}}^{(F)}(t)\big\rangle=\frac{\hbar}{\pi}\!\int^{\infty}_{0}\!d\omega \ \rm{Im}\{\alpha(\omega+i0^+)\} \coth{\left(\frac{\beta\hbar\omega}{2}\right)}\cos[\omega(t-t')].
\end{equation}
\end{small}\par One can easily find that for negative times the Green function (\ref{eq:0013}) vanishes, and for positive times Eq. (\ref{eq:0013}) is a solution to the following equation,
\begin{small}
\begin{equation}\label{eq:0016}
m_0 \ddot{G}(t)+\int^{t}_{0}\!dt' \mu(t-t')\dot{G}(t')+m_0\omega_0^2 G(t)+\mu(t)G(0)=0,
\end{equation}
\end{small}with the initial conditions \begin{small}${G(0)=0}$\end{small} and \begin{small}${\dot{G}(0)=1/m_0}$\end{small}. Then we obtain the general solutions to the quantum Langevin equation (\ref{eq:0005}) \cite{Ford2001},
\begin{small}
\begin{subequations}\label{eq:0018}
\begin{eqnarray}
&&\hat{\mathcal{q}}(t)=m_0 \dot{G}(t) \hat{\mathcal{q}}(0)+G(t) \hat{\mathcal{p}}(0)+\hat{X}(t)+\hat{Y}(t),\label{eq:0018a} \\
&&\dot{\hat{\mathcal{q}}}(t)=m_0 \ddot{G}(t) \hat{\mathcal{q}}(0)+\dot{G}(t) \hat{\mathcal{p}}(0)+\dot{\hat{X}}(t)+\dot{\hat{Y}}(t),\label{eq:0018b}
\end{eqnarray}
\end{subequations}
\end{small}where \begin{small}${\hat{X}(t)}$\end{small} is the position operator relevant to the fluctuating force \begin{small}${\hat{F}(t)}$\end{small} \cite{Ford2001},
\begin{small}
\begin{equation}\label{eq:0019a}
\hat{X}(t)=\int^{t}_{0}dt'\ G(t-t')\hat{F}(t'),
\end{equation}
\end{small}and \begin{small}${\hat{Y}(t)}$\end{small} is the position operator relevant to the driven force \begin{small}${\hat{f}(t)}$\end{small},
\begin{small}
\begin{equation}\label{eq:0019b}
\hat{Y}(t)=\int^{t}_{0}dt'\ G(t-t')\hat{f}(t').
\end{equation}\end{small}In the following, based on Eqs. (\ref{eq:0018a}) and (\ref{eq:0018b}), we try to obtain the analytical expression of the reduced Wigner function of the system.

\section{Analytical expression of the reduced Wigner function}

In this section, we calculate the reduced Wigner function of the system. By tracing out the degrees of freedom of the heat bath, we get the time-dependent reduced Wigner function of the system \cite{Yu1996},
\begin{small}
\begin{equation}\label{eq:0020}
W(\mathcal{q},\mathcal{p};t)=\int d\mathbf{\boldsymbol{q}}\int d\mathbf{\boldsymbol{p}}\ \it{W}_{\rm{tot}}(\mathcal{q},\mathcal{p};\mathbf{\boldsymbol{q}},\mathbf{\boldsymbol{p}};t)
\end{equation}
\end{small}Here \begin{small}${\it{W}_{\rm{tot}}}$\end{small} is the Wigner function of the composite system, with \begin{small}${\mathbf{\boldsymbol{q}}=(q_1,q_2,...,q_N)}$\end{small} and \begin{small}${\mathbf{\boldsymbol{p}}=(p_1,p_2,...,p_N)}$\end{small} the coordinates and the momenta of the heat bath, respectively. Because the evolution of the composite system satisfies the Liouville-von Neumann equation, we have \cite{Ford2001, Yu1996}
\begin{small}
\begin{equation}\label{eq:0021}
W_{\rm{tot}}(\mathcal{q},\!\mathcal{p};\!\mathbf{\boldsymbol{q}},\!\mathbf{\boldsymbol{p}};t)\!=\!W_{\rm{tot}}(\mathcal{q}(0),\!\mathcal{p}(0);\!\mathbf{\boldsymbol{q}}(0),\!\mathbf{\boldsymbol{p}}(0);0),
\end{equation}
\end{small}where \begin{small}${\mathcal{q}(0)}$\end{small}, \begin{small}${\mathcal{p}(0)}$\end{small}, \begin{small}${\mathbf{\boldsymbol{q}}(0)}$\end{small}, \begin{small}${\mathbf{\boldsymbol{p}}(0)}$\end{small} are the initial values of the coordinates and the momenta, and \begin{small}${\mathcal{q}=\mathcal{q}(t)}$\end{small}, \begin{small}${\mathcal{p}=\mathcal{p}(t)}$\end{small}, \begin{small}${\mathbf{\boldsymbol{q}}=\mathbf{\boldsymbol{q}}(t)}$\end{small}, \begin{small}${\mathbf{\boldsymbol{p}}=\mathbf{\boldsymbol{p}}(t)}$\end{small} are the solutions of the equation of motion (\ref{eq:0003}) and (\ref{eq:0004}). We assume that the initial state of the composite system is in a factorized form, i.e., a direct product of the density operator of the system and that of the heat bath, and the heat bath is in a thermal equilibrium state at the inverse temperature \begin{small}${\beta}$\end{small}. Then we obtain \cite{Ford2001, Yu1996}
\begin{small}
\begin{equation}\label{eq:0022}
W_{\rm{tot}}(\mathcal{q}(0),\mathcal{p}(0);\mathbf{\boldsymbol{q}}(0),\mathbf{\boldsymbol{p}}(0);0)=W(\mathcal{q}(0),\mathcal{p}(0);0)\prod^{N}_{j=1}w_j(q_j(0),p_j(0)),
\end{equation}
\end{small}where \begin{small}${w_j(q_j(0),p_j(0))}$\end{small} is the Wigner function of the \begin{small}${j}$\end{small}th oscillator of the heat bath with the mass \begin{small}${m_j}$\end{small} and the frequency \begin{small}${\omega_j}$\end{small},
\begin{small}
\begin{equation}\label{eq:0023}
w_j(q_j(0),p_j(0))=2\tanh\frac{\beta\hbar\omega_j}{2}\exp\left[-\frac{p_j^2(0)+m^2_j\omega^2_j q_j^2(0)}{m_j\hbar\omega_j\coth(\beta\hbar\omega_j/2)}\right].
\end{equation}
\end{small}Substituting Eqs. (\ref{eq:0021})-(\ref{eq:0023}) into Eq. (\ref{eq:0020}), one obtains
\begin{small}
\begin{equation}\label{eq:0024}
W(\mathcal{q},\mathcal{p};t)=\int d\mathbf{\boldsymbol{q}}(t)\int d\mathbf{\boldsymbol{p}}(t) \it{W}(\mathcal{q}(\rm{0}),\mathcal{p}(\rm{0});0)\it\prod^{N}_{j=\rm{1}}w_j(q_j\rm{(0)},\it{p_j}\rm{(0)}).
\end{equation}
\end{small}Using the general solutions of the quantum Langevin equation (\ref{eq:0018}), we transform the integration variables from the final coordinates of the heat bath \begin{small}${(\mathbf{\boldsymbol{q}}(t), \mathbf{\boldsymbol{p}}(t))}$\end{small} to the initial coordinates of the heat bath \begin{small}${(\mathbf{\boldsymbol{q}}(0), \mathbf{\boldsymbol{p}}(0))}$\end{small}, while holding \begin{small}${\mathcal{q}(0)}$\end{small} and \begin{small}${\mathcal{p}(0)}$\end{small} fixed \cite{Ford2001},
\begin{small}
\begin{equation}\label{eq:0025}
d\mathbf{\boldsymbol{q}}(t)d\mathbf{\boldsymbol{p}}(t)=\frac{1}{m_0^2(\dot{G}^2-G \ddot{G})}d\mathbf{\boldsymbol{q}}(0)d\mathbf{\boldsymbol{p}}(0).
\end{equation}
\end{small}Substituting Eq. (\ref{eq:0025}) into Eq. (\ref{eq:0024}), we obtain\begin{small}
\begin{equation}\label{eq:0026}
W(\mathcal{q},\mathcal{p};t)=\frac{\langle\it{W}(\mathcal{q}(\rm{0}),\mathcal{p}(\rm{0});0)\rangle}{m_0^2(\dot{G}^2-G \ddot{G})},
\end{equation}
\end{small}where the bracket represents the average over the initial equilibrium distribution of the heat bath, and \begin{small}${\mathcal{q}(0)}$\end{small} and \begin{small}${\mathcal{p}(0)}$\end{small} in the integrand can be obtained by inverting Eqs. (\ref{eq:0018a}) and (\ref{eq:0018b}),
\begin{small}
\begin{subequations}\label{eq:0027}
\begin{eqnarray}
&&\mathcal{q}(0)=\frac{m_0\dot{G}(\mathcal{q}-X-Y)-G(\mathcal{p}-m_0\dot{X}-m_0\dot{Y})}{m_0^2(\dot{G}^2-G \ddot{G})}, \label{eq:0027a}\\
&&\mathcal{p}(0)=\frac{-m_0^2\ddot{G}(\mathcal{q}-X-Y)+m_0\dot{G}(\mathcal{p}-m_0\dot{X}-m_0\dot{Y})}{m_0^2(\dot{G}^2-G \ddot{G})}.\label{eq:0027b}
\end{eqnarray}
\end{subequations}
\end{small}We would like to emphasize that in obtaining Eqs. (\ref{eq:0027a}) and (\ref{eq:0027b}), we have performed the Weyl-Wigner transform \cite{Wigner1932, Hillery1984} over Eqs. (\ref{eq:0018a}) and (\ref{eq:0018b}). That is, the operators \begin{small}${\hat{\mathcal{p}}}$\end{small}, \begin{small}${\hat{\mathcal{q}}}$\end{small}, \begin{small}${\hat{\mathcal{p}}(0)}$\end{small}, \begin{small}${\hat{\mathcal{q}}(0)}$\end{small}, \begin{small}${\hat{X}}$\end{small}, \begin{small}${\hat{Y}}$\end{small} in Eqs. (\ref{eq:0018a}) and (\ref{eq:0018b}) have been replaced by variables \begin{small}${\mathcal{p}}$\end{small}, \begin{small}${\mathcal{q}}$\end{small}, \begin{small}${\mathcal{p}(0)}$\end{small}, \begin{small}${\mathcal{q}(0)}$\end{small}, \begin{small}${X}$\end{small}, \begin{small}${Y}$\end{small}. We can calculate the average in Eq. (\ref{eq:0026}) by taking the Fourier transform of the initial reduced Wigner function \cite{Ford2001}
\begin{small}
\begin{equation}\label{eq:0029}
W(\mathcal{q}(0),\mathcal{p}(0);0)\!=\!\frac{1}{(2\pi\hbar)^2}\!\int\!dQ\!\int\!dP\ \tilde{\it{W}}(Q,P;0)\ e^{\frac{i}{\hbar}(\!P \mathcal{q}\!+\!Q \mathcal{p}\!)}.
\end{equation}
\end{small}Inserting this into Eq. (\ref{eq:0026}), after some simplifications, one can obtain \cite{Ford2001}
\begin{small}
\begin{equation}\label{eq:0030}
W(\mathcal{q},\mathcal{p};t)\!=\!\frac{1}{(2\pi\hbar)^2}\!\int^{\infty}_{\!-\infty\!}\!dr\!\!\int^{\infty}_{\!-\infty\!}\!ds
\tilde{\it{W}}\!\!\left(m_0\dot{G}r+Gs, m^2_0\ddot{G}r+\dot{G}s;0\right)\ e^{\frac{i}{\hbar}(r\mathcal{p}+s \mathcal{q}-m_{\rm{0}}\dot{Y}r-Ys)-\frac{1}{2\hbar^2}(m_0^2\langle\dot{X}^2\rangle r^2+m_0\langle X\dot{X}+\dot{X}X\rangle rs+\langle X^2\rangle s^2)}.
\end{equation}
\end{small}Here, we have transformed the integration variables \begin{small}${Q}$\end{small} and \begin{small}${P}$\end{small} into \begin{small}${r}$\end{small} and \begin{small}${s}$\end{small} by
\begin{small}
\begin{equation}\label{eq:0031}
Q=m_0\dot{G}r+Gs,\ \ \ P=m^2_0\ddot{G}r+\dot{G}s,
\end{equation}
\end{small}and we, due to the Gaussian property of \begin{small}${X(t)}$\end{small}, have used \cite{Ford2001}
\begin{small}
\begin{equation}\label{eq:0032}
\langle e^{-\frac{i}{\hbar}(m_0\dot{X}r+Xs)}\rangle\!=\! e^{-\frac{1}{2\hbar^2}(m_0^2\langle\dot{X}^2\rangle r^2+m_0\langle X\dot{X}+\dot{X}X\rangle rs+\langle X^2\rangle s^2)}.
\end{equation}
\end{small}We would like to emphasize that Eq. (\ref{eq:0030}) is the one of the main results of our paper, i.e., the analytical expression of the time evolution of the reduced Wigner function for a quantum Brownian particle in a driven harmonic potential. In this evaluation, the Green function \begin{small}${G(t)}$\end{small} is given by Eq. (\ref{eq:0013}), \begin{small}${X(t)}$\end{small} is given by Eq. (\ref{eq:0019a}) and its correlations are evaluated using Eq. (\ref{eq:0008}), and \begin{small}${Y(t)}$\end{small} is given by Eq. (\ref{eq:0019b}).

Furthermore, one can substitute the inverse of the Fourier transform (\ref{eq:0029}) into Eq. (\ref{eq:0030}) and rewrite the Wigner function in the form of a propagator acting on the initial reduced Wigner function \cite{Ford2001},
\begin{small}
\begin{equation}\label{eq:0033}
W(\mathcal{q},\mathcal{p};t)=\int^{\infty\!}_{-\!\infty\!}\!\frac{d\mathcal{q}'\!(0) d\mathcal{p}'\!(0)}{2\pi\hbar} P(\mathcal{q},\!\mathcal{p};\!\mathcal{q}'\!(0),\!\mathcal{p}'\!(0);\!t)W\!(\mathcal{q}'\!(0),\!\mathcal{p}'\!(0);\!0).
\end{equation}
\end{small}Here the propagator \begin{small}${P(\mathcal{q},\mathcal{p};\mathcal{q}'(0),\mathcal{p}'(0);t)}$\end{small} can be written as \cite{Ford2001}
\begin{small}
\begin{equation}\label{eq:0034}
P(\mathcal{q},\mathcal{p};\mathcal{q}'(0),\mathcal{p}'(0);t)=\frac{\hbar}{\sqrt{|{\bf A}(t)|}}e^{-\frac{1}{2}{\bf R}^{\rm T}(t){\bf A}^{-1}(t){\bf R}(t)},
\end{equation}
\end{small}where \begin{small}${|{\bf A}(t)|}$\end{small} denotes the determinant of \begin{small}${{\bf A}(t)}$\end{small}, and
\begin{small}
\begin{equation}\label{eq:0035}
{\bf A}(t)=\left(
  \begin{array}{cc}
    m_0^2\langle\dot{X}^2\rangle & \frac{m_0}{2}\langle X\dot{X}\!+\!\dot{X}X\rangle\\ \\
    \frac{m_0}{2}\langle X\dot{X}\!+\!\dot{X}X\rangle & \langle X^2\rangle\\
  \end{array}
\right),\ \ \ \ \ \ \ \ \ {\bf R}(t)=\left(
  \begin{array}{c}
    \mathcal{p}-\overline{\mathcal{p}(t)} \\ \\
    \mathcal{q}-\overline{\mathcal{q}(t)} \\
  \end{array}
\right).
\end{equation}
\end{small}Here the overline depicts the average over \begin{small}${X(t)}$\end{small} (\ref{eq:0019a}), and the operators \begin{small}${\overline{\mathcal{q}(t)}}$\end{small} and \begin{small}${\overline{\mathcal{p}(t)}}$\end{small} correspond to the solutions to Eqs. (\ref{eq:0018a}) and (\ref{eq:0018b}) with initial values \begin{small}${\mathcal{q}'(0)}$\end{small} and \begin{small}${\mathcal{p}'(0)}$\end{small},
\begin{small}
\begin{subequations}\label{eq:0037}
\begin{eqnarray}
&&\overline{\mathcal{q}(t)}=m_0\dot{G}(t)\ \mathcal{q}'(0)+G(t)\ \mathcal{p}'(0)+Y(t),\label{eq:0037a} \\
&&\overline{\mathcal{p}(t)}=m_0 \ddot{G}(t)\ \mathcal{q}'(0)+\dot{G}(t)\ \mathcal{p}'(0)+m_0\dot{Y}(t).\label{eq:0037b}
\end{eqnarray}
\end{subequations}
\end{small}

Now we calculate the propagator (\ref{eq:0034}) explicitly. In the case of an Ohmic heat bath, the memory function (\ref{eq:0006}) has the form
\begin{small}
\begin{equation}\label{eq:0038}
\mu(t)=2\gamma_0\delta(t),
\end{equation}
\end{small}where \begin{small}${\gamma_0}$\end{small} is the Newtonian friction constant. Substituting Eq. (\ref{eq:0038}) into Eq. (\ref{eq:0014}), one can obtain
\begin{small}
\begin{equation}\label{eq:0039}
\alpha(z)=\frac{1}{-m_0 z^2-i z \gamma_0+m_0\omega_0^2}.
\end{equation}
\end{small}Substituting Eq. (\ref{eq:0039}) into Eq. (\ref{eq:0013}), after the integration, one can obtain the expression of the Green function. It turns out that in the high damping regime (\begin{small}${\gamma_0/(m_0\omega_0)> 2}$\end{small}),
\begin{small}
\begin{eqnarray}\label{eq:0040}
G(t)=\frac{1}{2m_0\Omega_0}(e^{-\lambda_2 t}-e^{-\lambda_1 t}), \ \ \ \lambda_{1,2}=\frac{\gamma_0}{2m_0}\pm\Omega_0,\\ \nonumber
\end{eqnarray}
\end{small}and in the low damping regime (\begin{small}${\gamma_0/(m_0\omega_0)< 2}$\end{small}),
\begin{small}
\begin{eqnarray}\label{eq:0041}
G(t)=\frac{1}{2i m_0\Omega_0}(e^{-\lambda_2 t}-e^{-\lambda_1 t}), \ \ \ \lambda_{1,2}=\frac{\gamma_0}{2m_0}\pm i\Omega_0. \\ \nonumber
\end{eqnarray}
\end{small}where
\begin{small}
\begin{equation}\label{eq:0041a}
\Omega_0=\sqrt{\left|\frac{\gamma_0^2}{4m_0^2}-\omega_0^2\right|}.
\end{equation}
\end{small}It is straightforward to prove that in both the high and the low damping regimes,
\begin{small}
\begin{equation}\label{eq:0042}
G(-t')-G(t-t')=G(-t')-m_0 G(-t')\left[\dot{G}(t)+\frac{\xi}{2}G(t)\right]-m_0 G(t)\left[\dot{G}(-t')+\frac{\gamma_0}{2m_0}G(-t')\right].
\end{equation}
\end{small}From Eq. (\ref{eq:0012a1}) and Eq. (\ref{eq:0019a}), after performing the Weyl-Wigner transform, we have
\begin{small}
\begin{equation}\label{eq:0043}
X\!(t)\!=\!\mathcal{q}^{(\!F\!)}(t)\!-\!\mathcal{q}^{(\!F\!)}(0)\!+\!\int^{0}_{\!-\infty\!}\!dt' \left[G(-t')\!-\!G(t-t')\right]\!F(t').
\end{equation}
\end{small}Substituting Eq. (\ref{eq:0042}) into Eq. (\ref{eq:0043}), one obtains
\begin{small}
\begin{equation}\label{eq:0044}
X\!(t)\!=\!\mathcal{q}^{(\!F\!)}\!(t)-m_0\!\left[\!\dot{G}(t)\!+\!\frac{\gamma_0}{m_0} G(t)\!\right]\!\mathcal{q}^{(\!F\!)}\!(0)-m_0 G(t)\dot{\mathcal{q}}^{(\!F\!)}\!(t).
\end{equation}
\end{small}From the Weyl-Wigner transform of the correlations of \begin{small}${\mathcal{q}^{(\!F\!)}\!(t)}$\end{small}, i.e., Eq. (\ref{eq:0015}), one can obtain the elements of the matrix of \begin{small}${{\bf A}(t)}$\end{small} \cite{Ford2001} as follows,
\begin{small}
\begin{subequations}
\begin{eqnarray}
&&\langle X^2\rangle\!=\!\frac{1}{2}\left\{1+m_0^2\left[\dot{G}(t)+\frac{\gamma_0}{m_0} G(t)\right]^2\right\}s(0)-\frac{1}{2}m_0^2 G^2(t) \ddot{s}(0) -m_0\!\!\left[\dot{G}(t)+\frac{\gamma_0}{m_0} G(t)\right]\!\!s(t)+m_0 G(t) \dot{s}(t),
\label{eq:0045a}\\
&&\langle \dot{X}^2\rangle\!=\!-\frac{1}{2}\left[1+m_0^2\dot{G}^2(t)\right]\ddot{s}(0)+\frac{1}{2}m_0^2 \!\!\left[\ddot{G}(t)+\frac{\gamma_0}{m_0} \dot{G}(t)\right]^2\!\! s(0)-m_0\!\!\left[\ddot{G}(t)+\frac{\gamma_0}{m_0} \dot{G}(t)\right]\!\!\dot{s}(t)+m_0 \dot{G}(t) \ddot{s}(t),\label{eq:0045b} \\
&&\langle X\dot{X}\!\!+\!\!\dot{X}X\rangle\!=\!m_0^2\!\!\left[\!\dot{G}(t)\!\!+\!\!\frac{\gamma_0}{m_0}\! G(t)\!\right]\!\!\!\left[\!\ddot{G}(t)\!\!+\!\!\frac{\gamma_0}{m_0}\! \dot{G}(t)\!\right]\!\!s(0)\!-\!m_0^2 G(t) \dot{G}(t)\ddot{s}(0)\!-\!m_0\!\!\left[\!\ddot{G}(t)\!\!+\!\!\frac{\gamma_0}{m_0}\! \dot{G}(t)\!\right]\!\!s(t)\!-\!\gamma_0 G(t)\dot{s}(t)\!+\!m_0 G(t) \ddot{s}(t),\ \ \ \ \ \ \ \ \ \ \ \ \ \label{eq:0045c}
\end{eqnarray}
\end{subequations}
\end{small}where
\begin{small}
\begin{equation}\label{eq:0046}
s(t)\!=\!\frac{2\gamma_0}{\pi}\!\int^{\infty}_{0}\!d\omega \frac{\hbar\omega}{m_0^2(\omega_0^2-\omega^2)^2\!+\!\gamma_0^2\omega_0^2}\coth{\frac{\beta\hbar\omega}{2}}\cos{\omega t}.
\end{equation}
\end{small}Note that Eqs. (\ref{eq:0045a})-(\ref{eq:0045c}) are valid in both the high and the low damping regimes.

Before proceeding to the next step, let us make a self-consistency check about our results of Eqs. (\ref{eq:0045a})-(\ref{eq:0045c}). It is expected that in the high temperature limit, Eqs. (\ref{eq:0045a})-(\ref{eq:0045c}) will reproduce their classical counterparts (see Sec.10.2.1 in Ref. \cite{Risken1996}). In the high temperature limit, \begin{small}${\coth{(\beta\hbar\omega/2)}\to 2/(\beta\hbar\omega)}$\end{small}, then Eq. (\ref{eq:0046}) becomes
\begin{small}
\begin{equation}\label{eq:0046a}
s_{\rm cl}(t)\!=\!\frac{4\gamma_0}{\beta\pi}\!\int^{\infty}_{0}\!d\omega \frac{\cos{\omega t}}{m_0^2(\omega_0^2-\omega^2)^2\!+\!\gamma_0^2\omega_0^2}.
\end{equation}
\end{small}After taking the integration, one can obtain
\begin{small}
\begin{equation}\label{eq:0047}
s_{\rm cl}(t)\!=\!\left\{
\begin{aligned}
&\frac{1}{\beta m_0 \Omega_0 \omega_0^2}(\!-\lambda_2 e^{-\lambda_1 t}\!+\!\lambda_1 e^{-\lambda_2 t}),\ \ \ \ \ \frac{\gamma_0}{m_0\omega_0}\!\!>\!\!2, \\
&\frac{1}{i\beta m_0 \Omega_0 \omega_0^2}(\!-\lambda_2 e^{-\lambda_1 t}\!+\!\lambda_1 e^{-\lambda_2 t}),\ \ \ \ \frac{\gamma_0}{m_0\omega_0}\!\!<\!\!2.
\end{aligned}
\right.\\ \
\end{equation}
\end{small}Substituting Eqs. (\ref{eq:0040}), (\ref{eq:0041}), and (\ref{eq:0047}) into Eqs. (\ref{eq:0045a})-(\ref{eq:0045c}), after some simplification, one can obtain the classical limit of Eq. (\ref{eq:0035}),
\begin{small}
\begin{equation}\label{eq:0047a}
{\bf A}_{\rm cl}(t)=\left(
  \begin{array}{cc}
    m_0^2\langle\dot{X}^2\rangle_{\rm cl} & \frac{m_0}{2}\langle X\dot{X}\!+\!\dot{X}X\rangle_{\rm cl} \\ \\
    \frac{m_0}{2}\langle X\dot{X}\!+\!\dot{X}X\rangle_{\rm cl} & \langle X^2\rangle_{\rm cl} \\
  \end{array}
\right),
\end{equation}
\end{small}where
\begin{small}
\begin{subequations}\label{eq:0048}
\begin{eqnarray}
&&\langle X^2\rangle_{\rm cl}=\frac{\gamma_0 v^2_{th}}{m_0(\lambda_1-\lambda_2)^2}\!\!\left[\frac{\lambda_1+\lambda_2}{\lambda_1\lambda_2}+\frac{4}{\lambda_1+\lambda_2}(e^{-(\lambda_1+\lambda_2)t}\!-\!1)\!-\!\frac{1}{\lambda_1}e^{-2\lambda_1 t}\!-\!\frac{1}{\lambda_2}e^{-2\lambda_2 t}\right], \label{eq:0048a} \\
&&m_0^2\langle \dot{X}^2\rangle_{\rm cl}=\frac{m_0^2\gamma_0 v^2_{th}}{m_0(\lambda_1-\lambda_2)^2}\!\!\left[\lambda_1+\lambda_2+\frac{4\lambda_1\lambda_2}{\lambda_1+\lambda_2}(e^{-(\lambda_1+\lambda_2)t}\!-\!1)\!-\!\lambda_1 e^{-2\lambda_1 t}\!-\!\lambda_2 e^{-2\lambda_2 t}\right], \label{eq:0048b}\\
&&\frac{m_0}{2}\langle X\dot{X}\!+\!\dot{X}X\rangle_{\rm cl}=\frac{m_0 \gamma_0 v^2_{th}}{(\lambda_1\!-\!\lambda_2)^2}(e^{-\lambda_1 t}\!-\!e^{-\lambda_2 t})^2,\label{eq:0048c}
\end{eqnarray}
\end{subequations}
\end{small}and \begin{small}${v^2_{th}=1/(\beta m_0)}$\end{small}. One can see that Eqs. (\ref{eq:0048a})-(\ref{eq:0048c}) are exactly the same as Eq. (10.63) in Ref. \cite{Risken1996}, i.e., our results (\ref{eq:0045a})-(\ref{eq:0045c}) reproduce the results of the classical Brownian particle in the high temperature limit (please note that ${m_0}$ has been set to one and \begin{small}${f(t)=0}$\end{small} in Eq. (10.63) in Ref. \cite{Risken1996}). Substituting Eqs. (\ref{eq:0035}), (\ref{eq:0037a})-(\ref{eq:0037b}), (\ref{eq:0046a}) and (\ref{eq:0048a})-(\ref{eq:0048c}) into Eq. (\ref{eq:0034}), one obtains the classical propagator in the phase space, which is the same as Eq. (10.55) in Ref. \cite{Risken1996}:
\begin{small}
\begin{equation}\label{eq:0049}
P_{\!\rm cl}(\mathcal{q},\!\mathcal{p};\!\mathcal{q}'\!(0),\!\mathcal{p}'\!(0);\!t)\!=\!\frac{\hbar}{\!\sqrt{|\!{\bf A}_{\rm cl}\!(t)\!|}}\!\exp\!\left[\!-\frac{1}{2}[{\bf A}_{\rm cl}^{\!-\!1\!}(t)]_{\!pp}[\mathcal{p}\!-\!\overline{\mathcal{p}\!(t)}]^2\!-\![{\bf A}_{\rm cl}^{\!-\!1\!}(t)]_{\!pq}[\mathcal{p}\!-\!\overline{\mathcal{p}\!(t)}][\mathcal{q}\!-\!\overline{\mathcal{q}\!(t)}]\!-\!\frac{1}{2}[{\bf A}_{\rm cl}^{\!-\!1\!}(t)]_{\!qq}[\mathcal{q}\!-\!\overline{\mathcal{q}\!(t)}]^2\! \right].
\end{equation}
\end{small}Here, the elements of the inverse matrix of \begin{small}${{\bf A}_{\rm cl}(t)}$\end{small} are given by
\begin{small}
\begin{subequations}
\begin{eqnarray}
&&[{\bf A}_{\rm cl}^{\!-\!1\!}(t)]_{pp}=\langle X^2\rangle_{\rm cl}/|\!{\bf A}_{\rm cl}\!(t)\!|, \\
&&[{\bf A}_{\rm cl}^{\!-\!1\!}(t)]_{pq}=[{\bf A}_{\rm cl}^{\!-\!1\!}(t)]_{qp}\!=\!-m_0\langle X\dot{X}\!+\!\dot{X}X\rangle_{\rm cl}/(2|\!{\bf A}_{\rm cl}\!(t)\!|),\\
&&[{\bf A}_{\rm cl}^{\!-\!1\!}(t)]_{qq}=m_0^2\langle \dot{X}^2\rangle_{\rm cl}/|\!{\bf A}_{\rm cl}\!(t)\!|,
\end{eqnarray}
\end{subequations}
\end{small}and
\begin{small}
\begin{equation}\label{eq:0050d}
|{\bf A}_{\rm cl}\!(t)|\!=\!m_0^2\left[\langle X^2\rangle_{\rm cl} \! \langle\dot{X}^2\rangle_{\rm cl}\!-\!\langle X\dot{X}\!+\!\dot{X}X\rangle_{\rm cl}^2/4\right].
\end{equation}
\end{small}

We now take the thermal equilibrium initial state as an example to calculate the time evolution of the Wigner function of the system (Eqs. (\ref{eq:0033}) and (\ref{eq:0034})). Please note that the initial state can also be any state other than the thermal equilibrium state. But for simplicity, we use the thermal equilibrium state as an example to demonstrate the effectiveness of our method. We assume that the system is initially prepared in a thermal equilibrium state at the inverse temperature \begin{small}${\beta'}$\end{small}, which is different from the temperature of the heat bath \begin{small}${\beta}$\end{small}, and the Wigner function of the initial density matrix can be written as
\begin{small}
\begin{equation}\label{eq:0051}
W({\mathcal{p}'}\!(0),{ \mathcal{q}'}\!(0);0)\!=\!2\tanh\!{\frac{\beta'\!\hbar\omega_0}{2}}\exp\!\!\left[\!-\frac{1}{2}\!\left(\mathcal{p}'\!(0)\ \ \mathcal{q}'\!(0)\right)\!{\bf B}\!\left(\!
  \begin{array}{c}
    \mathcal{p}'\!(0) \\ \\
    \mathcal{q}'\!(0)
  \end{array}
\!\right)\!\right]\!,\ \ \
{\bf B}\!=\!\!\left(\!
  \begin{array}{cc}
    \!2\left(\!m_0\hbar\omega_0\coth{\frac{\beta'\!\hbar\omega_0}{2}}\!\right)^{\!-1\!}\! & 0 \\
    0 & \!2\left(\!\frac{\hbar}{m_0\omega_0}\coth{\frac{\beta'\!\hbar\omega_0}{2}}\!\right)^{\!-1\!}\!\\
  \end{array}
\!\right)\!.
\end{equation}
\end{small}Substituting Eqs. (\ref{eq:0034}) and (\ref{eq:0051}) into Eq. (\ref{eq:0033}), one can obtain
\begin{small}
\begin{eqnarray}\label{eq:0053}
W(\mathcal{p},\mathcal{q};t)&=&\frac{2\tanh{\frac{\beta'\hbar\omega}{2}}}{\sqrt{|{\bf A}(t)||{\bf B}+{\bf \Lambda}^{\rm T}(t){\bf A}^{-1}(t){\bf \Lambda}(t)|}}\exp\left[\frac{1}{2}\left(\mathcal{p}-m_0\dot{Y}(t)\ \ \ \mathcal{q}-Y(t) \right)\right.\nonumber \\
&&\left. \cdot\left\{{\bf A}^{-1}(t){\bf \Lambda}(t)[{\bf B}+{\bf \Lambda}^{\rm T}(t){\bf A}^{-1}(t){\bf \Lambda}(t)]^{-1}{\bf \Lambda}^{\rm T}(t){\bf A}^{-1}(t)-{\bf A}^{-1}(t)\right\}\cdot\left(
  \begin{array}{c}
    \mathcal{p}-m_0\dot{Y}(t) \\
    \mathcal{q}-Y(t)
  \end{array}
\right)\right],
\end{eqnarray}\end{small}where
\begin{small}
\begin{equation}\label{eq:0054}
{\bf \Lambda}(t)=\left(
  \begin{array}{cc}
    m_0\dot{G}(t) & m_0^2\ddot{G}(t) \\ \\
    G(t) & m_0\dot{G}(t) \\
  \end{array}
\right).
\end{equation}
\end{small}From this result, one can easily find that the variances of the Wigner function is independent of the external force \begin{small}${f(t)}$\end{small}.

Furthermore, we would like to show how the equilibrium solution arises in the long time limit, i.e., the relaxation process from \begin{small}${T'=1/\beta'}$\end{small} to \begin{small}${T=1/\beta}$\end{small}:

First we recall that, so long as the angular frequency of the system \begin{small}${\omega_0}$\end{small} is nonzero, the Green function will vanish as \begin{small}${t\to\infty}$\end{small} \cite{Ford2001}, thus \begin{small}${{\bf \Lambda}(t)={\bf 0}}$\end{small} when \begin{small}${t\to\infty}$\end{small}.

Next, from Eqs. (\ref{eq:0045a})-(\ref{eq:0045c}), we have
\begin{small}
\begin{equation}\label{eq:0055}
\langle X^2\rangle=\frac{1}{2}s(0),\ \ \ \langle \dot{X}^2\rangle=-\frac{1}{2}\ddot{s}(0),\ \ \ \langle X\dot{X}\!+\!\dot{X}X\rangle=0.
\end{equation}
\end{small}

For simplicity, we consider the weak coupling limit. In the weak coupling limit,
\begin{small}
\begin{equation}\label{eq:0056}
\lim_{\gamma_0\to0}\frac{2\gamma_0}{\pi} \frac{\hbar\omega}{m_0^2(\omega_0^2-\omega^2)^2\!+\!\gamma_0^2\omega_0^2}=\frac{\hbar\omega}{m_0\omega_0^2}\delta(\omega-\omega_0).
\end{equation}
\end{small}Then we have
\begin{small}
\begin{equation}\label{eq:0057}
{\bf A}(t)=\left(
  \begin{array}{cc}
    \frac{m_0\hbar\omega_0}{2}\coth{\frac{\beta\hbar\omega_0}{2}} & 0 \\ \\
    0 & \frac{\hbar}{2m_0\omega_0}\coth{\frac{\beta\hbar\omega_0}{2}} \\
  \end{array}
\right).
\end{equation}
\end{small}Substituting Eqs. (\ref{eq:0054}), (\ref{eq:0055}), and (\ref{eq:0057}) into Eq. (\ref{eq:0053}), one can obtain the asymptotic expression of the Wigner function in the long time limit
\begin{small}
\begin{equation}\label{eq:0058}
W(\mathcal{q},\mathcal{p};t)=2\tanh{\frac{\beta\hbar\omega_0}{2}}\exp\!\!\left[-\frac{[\mathcal{p}-m_0\dot{Y}(t)]^2}{m_0\hbar\omega_0\coth{\frac{\beta\hbar\omega_0}{2}}}-\frac{[\mathcal{q}-Y(t)]^2}{\frac{\hbar}{m_0\omega_0}\coth{\frac{\beta\hbar\omega_0}{2}}}\right].
\end{equation}
\end{small}This is the familiar form of the Wigner function of a dragged harmonic oscillator, which is independent of the initial temperature \begin{small}${\beta'}$\end{small}. Please note that when the driving force vanishes, i.e., \begin{small}${\hat{f}(t)=0}$\end{small}, our result (\ref{eq:0058}) reproduces the result in Ref. \cite{Ford2001}.

\section{Quantum corrections to the entropy of the system}
Based on the above results, we now calculate the quantum corrections to the entropy of a dragged harmonic oscillator which is undergoing quantum Brownian motion. First, we assume that the initial state of the system has a well-defined classical counterpart \cite{Qiu2020}, i.e., when we expand the initial Wigner function in powers of \begin{small}${\hbar}$\end{small}, there are no terms in negative powers of \begin{small}${\hbar}$\end{small},
\begin{small}
\begin{equation}\label{eq:0059}
W(\mathcal{q}'\!(0),\mathcal{p}'\!(0);0)=W_{\rm cl}+(i\hbar) W^{(1)}+(i\hbar)^2 W^{(2)}+o(\hbar^2),
\end{equation}
\end{small}where \begin{small}${W_{\rm cl}(\mathcal{q}'\!(0),\mathcal{p}'\!(0);0)}$\end{small} is the corresponding classical probability distribution in the phase space. Next, we expand the propagator (\ref{eq:0034}) in powers of \begin{small}${\hbar}$\end{small}. Because
\begin{small}
\begin{equation}\label{eq:0060}
\hbar\coth{\frac{\beta\hbar\omega}{2}}=\frac{2}{\beta\omega}-(i\hbar)^2\frac{\beta\omega}{6}+o(\hbar^2),
\end{equation}
\end{small}we have
\begin{small}
\begin{equation}\label{eq:0061}
s(t)=s_{\rm cl}(t)+(i\hbar)^2 s^{(2)}(t)+o(\hbar^2),
\end{equation}
\end{small}where \begin{small}${s^{(2)}(t)}$\end{small} is given by
\begin{small}
\begin{equation}\label{eq:0062}
s^{(2)}(t)\!=\!-\frac{\beta\gamma_0}{3\pi}\!\int^{\infty}_{0}\!d\omega \frac{\omega^2\cos{\omega t}}{m_0^2(\omega_0^2-\omega^2)^2\!+\!\gamma_0^2\omega_0^2}.
\end{equation}
\end{small}After taking the integration, one can obtain
\begin{small}
\begin{equation}\label{eq:0062a}
s^{(2)}(t)=\left\{
\begin{aligned}
&\frac{\beta}{12 m_0 \Omega_0}(\lambda_a e^{-\lambda_1 t}\!-\!\lambda_2 e^{-\lambda_2 t}),\ \ \ \frac{\gamma_0}{m_0\omega_0}\!>\!2, \\
&\frac{-i\beta}{12 m_0 \Omega_0}(\lambda_1 e^{-\lambda_1 t}\!-\!\lambda_2 e^{-\lambda_2 t}),\ \ \ \frac{\gamma_0}{m_0\omega_0}\!<\!2.
\end{aligned}
\right.\\ \
\end{equation}
\end{small}From Eqs. (\ref{eq:0045a})-(\ref{eq:0045c}), we know that
\begin{small}
\begin{eqnarray}
\langle X^2\rangle\!\!&=&\!\!\langle X^2\rangle_{\rm cl}+(i\hbar)^2\langle X^2\rangle^{(2)}+o(\hbar^2),\label{eq:0063a}\\
\langle \dot{X}^2\rangle\!\!&=&\!\!\langle \dot{X}^2\rangle_{\rm cl}+(i\hbar)^2\langle \dot{X}^2\rangle^{(2)}+o(\hbar^2),\label{eq:0063b}\\
\langle X\dot{X}\!+\!\dot{X}X\rangle\!\!&=&\!\!\langle X\dot{X}\!+\!\dot{X}X\rangle_{\rm cl}\!+\!(i\hbar)^2\langle X\dot{X}\!+\!\dot{X}X\rangle^{\!(2)\!}\!+\!o(\hbar^2),\label{eq:0063c}
\end{eqnarray}
\end{small}where the expressions of \begin{small}${\langle X^2\rangle^{(2)}}$\end{small}, \begin{small}${\langle \dot{X}^2\rangle^{(2)}}$\end{small} and \begin{small}${\langle X\dot{X}\!+\!\dot{X}X\rangle^{\!(2)\!}}$\end{small} can be obtained by substituting Eq. (\ref{eq:0062}) into Eqs. (\ref{eq:0045a})-(\ref{eq:0045c}). Then the matrix \begin{small}${{\bf A}(t)}$\end{small} can be expanded in powers of \begin{small}${\hbar}$\end{small} as follows,
\begin{small}
\begin{equation}\label{eq:0064}
{\bf A}(t)={\bf A}_{\rm cl}(t)+(i\hbar)^2 {\bf A}^{(2)}(t)+o(\hbar^2),
\end{equation}
\end{small}where \begin{small}${{\bf A}^{(2)}(t)}$\end{small} is given by
\begin{small}
\begin{equation}\label{eq:0065}
{\bf A}^{(2)}(t)=\left(
  \begin{array}{cc}
    m_0^2\langle\dot{X}^2\rangle^{\!(2)\!} & \frac{m_0}{2}\langle X\dot{X}\!+\!\dot{X}X\rangle^{\!(2)\!}\\ \\
    \frac{m_0}{2}\langle X\dot{X}\!+\!\dot{X}X\rangle^{\!(2)\!} & \langle X^2\rangle^{\!(2)\!} \\
  \end{array}
\right).
\end{equation}
\end{small}Then we can similarly expand \begin{small}${|{\bf A}(t)|}$\end{small} and \begin{small}${{\bf A}^{-1}(t)}$\end{small} in powers of \begin{small}${\hbar}$\end{small},
\begin{small}
\begin{subequations}
\begin{eqnarray}
|{\bf A}(t)|&=&|{\bf A}_{\rm cl}(t)|+(i\hbar)^2|{\bf A}_{\rm cl}(t)|\ {\rm Tr}[{\bf A}_{\rm cl}^{-1}(t){\bf A}^{(2)}(t)]+o(\hbar^2),\label{eq:0066a}\\
{\bf A}^{-1}(t)&=&{\bf A}_{\rm cl}^{-1}(t)-(i\hbar)^2{\bf A}_{\rm cl}^{-1}(t){\bf A}^{(2)}(t){\bf A}_{\rm cl}^{-1}(t)+o(\hbar^2).\label{eq:0066b}
\end{eqnarray}
\end{subequations}
\end{small}

Substituting Eqs. (\ref{eq:0066a}) and (\ref{eq:0066b}) into Eq. (\ref{eq:0034}), we obtain the expression of the propagator in powers of \begin{small}${\hbar}$\end{small} as follows,
\begin{small}
\begin{equation}\label{eq:0067}
P(\mathcal{q},\mathcal{p};\mathcal{q}'\!(0),\mathcal{p}'\!(0);t)=P_{\rm cl}(\mathcal{q},\mathcal{p};\mathcal{q}'\!(0),\mathcal{p}'\!(0);t)+(i\hbar)^2 P^{(2)}(\mathcal{q},\mathcal{p};\mathcal{q}'\!(0),\mathcal{p}'\!(0);t)+o(\hbar^2),
\end{equation}
\end{small}where \begin{small}${P_{\rm cl}(\mathcal{q},\mathcal{p};\mathcal{q}'\!(0),\mathcal{p}'\!(0);t)}$\end{small} is given by Eq. (\ref{eq:0049}), and
\begin{small}
\begin{equation}\label{eq:0068}
P^{(2)}(\mathcal{q},\mathcal{p};\mathcal{q}'\!(0),\mathcal{p}'\!(0);t)\!=\!\frac{1}{2}P_{\rm cl}(\mathcal{q},\mathcal{p};\mathcal{q}'\!(0),\mathcal{p}'\!(0);t)\left({\bf R}^{\rm T}(t){\bf A}_{\rm cl}^{-1}(t){\bf A}^{(2)}(t){\bf A}_{\rm cl}^{-1}(t){\bf R}(t)-{\rm Tr}[{\bf A}_{\rm cl\!}^{\!-\!1\!}(t){\bf A}^{\!(2)\!}(t)]\right).
\end{equation}
\end{small}Substituting Eqs. (\ref{eq:0059}) and (\ref{eq:0067}) into Eqs. (\ref{eq:0033}), we obtain the time evolution of the Wigner function of the system in powers of \begin{small}${\hbar}$\end{small},
\begin{small}
\begin{equation}\label{eq:0069}
W(\mathcal{q},\mathcal{p};t)=W_{\rm cl}(\mathcal{q},\mathcal{p};t)+(i\hbar)W^{(1)}(\mathcal{q},\mathcal{p};t)+(i\hbar)^2 W^{(2)}(\mathcal{q},\mathcal{p};t)+o(\hbar^2),
\end{equation}
\end{small}where
\begin{small}
\begin{subequations}
\begin{eqnarray}
&&\!\!\!W_{\rm cl}(\mathcal{q},\mathcal{p};t)\!=\!\int^{\infty}_{-\infty\!}\!\frac{d\mathcal{q}'\!(0) d\mathcal{p}'\!(0)}{2\pi\hbar} P_{\rm cl}(\mathcal{q},\mathcal{p};\mathcal{q}'\!(0),\mathcal{p}'\!(0);t)W_{\rm cl}(\mathcal{q}'\!(0),\mathcal{p}'\!(0);0),\label{eq:0070a} \\
&&\!\!\!W^{(1)}(\mathcal{q},\mathcal{p};t)\!=\!\int^{\infty}_{-\infty\!}\!\frac{d\mathcal{q}'\!(0) d\mathcal{p}'\!(0)}{2\pi\hbar} P_{\rm cl}(\mathcal{q},\mathcal{p};\mathcal{q}'\!(0),\mathcal{p}'\!(0);t)W^{(1)}(\mathcal{q}'\!(0),\mathcal{p}'\!(0);0),\label{eq:0070b}\\
&&\!\!\!W^{\!(2)\!}(\mathcal{q},\mathcal{p};t)\!=\!\int^{\infty\!}_{\!-\!\infty\!}\!\frac{d\mathcal{q}'\!(0) d\mathcal{p}'\!(0)}{2\pi\hbar}\! \left[\!P_{\rm cl}(\mathcal{q},\!\mathcal{p};\!\mathcal{q}'\!(0),\!\mathcal{p}'\!(0);t)W^{\!(2)\!}(\mathcal{q}'\!(0),\mathcal{p}'\!(0);0)\!+\!P^{(2)\!}(\mathcal{q},\!\mathcal{p};\!\mathcal{q}'\!(0),\!\mathcal{p}'\!(0);\!t)W_{\rm cl}(\mathcal{q}'\!(0),\mathcal{p}'\!(0);0)\!\right]\!.\ \ \ \ \ \ \ \ \ \ \label{eq:0070c}
\end{eqnarray}
\end{subequations}
\end{small}One can see that \begin{small}${W_{\rm cl}(\mathcal{q},\mathcal{p};t)}$\end{small} is the corresponding classical probability distribution at time \begin{small}${t}$\end{small} in the phase space, while \begin{small}${W^{(1)}(\mathcal{q},\mathcal{p};t)}$\end{small} and \begin{small}${W^{(2)}(\mathcal{q},\mathcal{p};t)}$\end{small} are the first- and the second-order quantum corrections at time \begin{small}${t}$\end{small} to the classical probability distribution, respectively. It is worth mentioning that this result is a dynamical extension to results in Ref. \cite{Wigner1932}.

Finally, using the methods developed in Ref. \cite{Qiu2020}, we obtain the quantum corrections to the classical Gibbs entropy of a dragged harmonic oscillator which is undergoing quantum Brownian motion,
\begin{small}
\begin{equation}\label{eq:0071}
S_q(t)=S_{\rm cl}(t)+(i\hbar)S^{(1)}(t)+(i\hbar)^2 S^{(2)}(t)+o(\hbar^2),
\end{equation}
\end{small}
where
\begin{small}
\begin{subequations}
\begin{eqnarray}
S_{\rm cl}(t)\!&=&\!-\int\frac{d\mathcal{q}d\mathcal{p}}{2\pi\hbar}\ W_{\rm cl} \ln W_{\rm cl}.\label{eq:0072a}\\
S^{(1)}(t)\!&=&\!-\int\frac{d\mathcal{q}d\mathcal{p}}{2\pi\hbar}\ \left[W^{(1)} \ln W_{\rm cl}+W^{(1)}\right],\label{eq:0072b}\\
S^{(2)}(t)\!&=&\!-\!\int\!\frac{d\mathcal{q}d\mathcal{p}}{2\pi\hbar}\ \left[W^{(2)} \ln W_{\rm cl}\!+\!W^{(2)}
\!+\!\frac{(W^{(1)})^2}{2W_{\rm cl}}-\frac{W_{\rm cl}(\stackrel{\leftarrow}{\partial_{\mathcal{p}}}\stackrel{\rightarrow}{\partial_{\mathcal{q}}}
-\stackrel{\leftarrow}{\partial_{\mathcal{q}}}\stackrel{\rightarrow}{\partial_{\mathcal{p}}})^2W_{\rm cl}}{16W_{\rm cl}}+\frac{G(W_{\rm cl})}{12W_{\rm cl}^2}\right].\label{eq:0072c}
\end{eqnarray}
\end{subequations}
\end{small}Here,
\begin{small}
\begin{equation}\label{eq:0073}
G(W_{\rm cl})=(\partial^2_{\mathcal{p}} W_{\rm cl})(\partial_{\mathcal{q}} W_{\rm cl})^2+(\partial^2_{\mathcal{q}} W_{\rm cl})(\partial_{\mathcal{p}} W_{\rm cl})^2-2(\partial_{\mathcal{q}} W_{\rm cl})(\partial_{\mathcal{p}} W_{\rm cl})(\partial_{\mathcal{q}\mathcal{p}} W_{\rm cl}).
\end{equation}
\end{small}One can find that \begin{small}${S_{\rm cl}(t)}$\end{small} is exactly the corresponding classical Gibbs entropy of the system, while \begin{small}${S^{(1)}(t)}$\end{small} and \begin{small}${S^{(2)}(t)}$\end{small} are the first- and the second-order quantum corrections to the entropy, respectively.

As a demonstration, we take the thermal equilibrium initial state as an example to show our results (\ref{eq:0071})-(\ref{eq:0073}). We assume that the system is prepared initially in the thermal equilibrium state at the inverse temperature \begin{small}${\beta'}$\end{small}. The initial Wigner function is given by Eq. (\ref{eq:0051}), which can be expanded in the form of Eq. (\ref{eq:0059}) and
\begin{small}
\begin{subequations}\label{eq:0074}
\begin{eqnarray}
&&W_{\rm cl}(\mathcal{q}'(0),\mathcal{p}'(0);0)=\beta'\hbar\omega_0e^{-\beta'\epsilon(\mathcal{q}'(0),\mathcal{p}'(0))},\label{eq:0074a}\\
&&W^{(1)}(\mathcal{q}'(0),\mathcal{p}'(0);0)=0,\label{eq:0074b}\\
&&W^{\!(2)\!}(\mathcal{q}'\!(0),\!\mathcal{p}'\!(0);\!0)\!=\!W_{\rm\! cl}\frac{{\beta'}^2\!\omega_0^2}{12}\![1\!-\!\beta'\!\epsilon(\mathcal{q}'\!(0),\!\mathcal{p}'\!(0))],\label{eq:0074c}
\end{eqnarray}
\end{subequations}
\end{small}where
\begin{small}
\begin{equation}\label{eq:0075}
\epsilon(\mathcal{q}'\!(0),\mathcal{p}'\!(0))=\frac{{\mathcal{p}'}^2\!(0)}{2m_0}+\frac{1}{2}m_0\omega_0^2{\mathcal{q}'}^2\!(0)
\end{equation}
\end{small}is the Hamiltonian of a single harmonic oscillator. Substituting Eqs. (\ref{eq:0059}), (\ref{eq:0067}), (\ref{eq:0068}), and (\ref{eq:0074a})-(\ref{eq:0074c}) into Eq. (\ref{eq:0069}), one can obtain the expansion of the reduced Wigner function at an arbitrary time \begin{small}${t}$\end{small}. Finally, we obtain the the quantum corrections to the entropy by substituting Eqs. (\ref{eq:0070a})-(\ref{eq:0070c}) into Eqs. (\ref{eq:0072a})-(\ref{eq:0072c}). One can find that for the thermodynamic equilibrium initial state, all terms odd in \begin{small}${\hbar}$\end{small} are exactly zero due to \begin{small}${W^{(1)}(\mathcal{q}'\!(0),\mathcal{p}'\!(0);0)=0}$\end{small}. The evolution of the classical Wigner function \begin{small}${W_{\rm cl}(\mathcal{q},\mathcal{p};t)}$\end{small} is given by Eq. (\ref{eq:0070a}), and \begin{small}${S_{\rm cl}(t)}$\end{small} reproduces the classical Gibbs entropy for a dragged Brownian harmonic oscillator. And the lowest order quantum correction to the entropy of the system is given by Eq. (\ref{eq:0072c}).

\section{Quantum corrections to the total entropy production rate and the heat dissipation rate}
Based on the results (\ref{eq:0072a})-(\ref{eq:0072c}) in Sec. V, we can get more information and physical insights about the entropy production in a nonequilibrium quantum stochastic process. In Refs. \cite{Ge2009, Ge2010, Esposito2010b, Esposito2010c}, the expressions of the classical total entropy production rate \begin{small}${e^{({\rm cl})}_{\rm p}}$\end{small} and the classical heat dissipation rate \begin{small}${h^{({\rm cl})}_{\rm d}}$\end{small} are defined in a nonequilibrium stochastic process. Then the second law is reformulated in the nonequilibrium form, and the total entropy production rate can be further split into two nonnegative parts,
\begin{small}
\begin{equation}\label{eq:0076}
e^{({\rm cl})}_{\rm p}=\beta Q^{({\rm cl})}_{\rm hk}+(-\beta f^{({\rm cl})}_{\rm d}).
\end{equation}
\end{small}Here, \begin{small}${Q^{({\rm cl})}_{\rm hk}}$\end{small} is the classical housekeeping heat, which represents the irreversible work done by the surrounding to the system that is kept away from reaching equilibrium, and \begin{small}${f^{({\rm cl})}_{\rm d}}$\end{small} is the classical free energy dissipation rate associated with spontaneous relaxation \cite{Ge2010}. Nevertheless, a quantum version of the total entropy production rate and the heat dissipation rate has largely been unexplored so far, except for the open quantum systems described by the Lindblad master equation \cite{Spohn1978a, Spohn1978b}. Based on the results in Sec. V, we now derive the expression of the quantum corrections to the total entropy production rate and the heat dissipation rate.

We consider an Ohmic heat bath with the memory function (\ref{eq:0038}). Under the Markovian approximation, the time evolution of the reduced Wigner function of the system satisfies the Caldeira-Leggett master equation \cite{Caldeira1983, Hu1992, Yu1996, Ford2001}, which has the same form as the Kramers equation \cite{Risken1996}
\begin{small}
\begin{equation}\label{eq:0077}
\partial_t W(\mathcal{q},\mathcal{p};t)=-\frac{\mathcal{p}}{m_0}\partial_\mathcal{q}\!W(\mathcal{q},\mathcal{p};t)\!+\!m_0\omega_0^2\left(\!\mathcal{q}(t)\!-\!\frac{f(t)}{m_0\omega_0^2}\!\right)\partial_\mathcal{p}\!W(\mathcal{q},\mathcal{p};t)+\frac{\gamma_0}{m_0}\partial_\mathcal{p}\!\left[\mathcal{p}W(\mathcal{q},\mathcal{p};t)\right]+\frac{\gamma_0}{\beta}\partial^2_{\mathcal{p}}W(\mathcal{q},\mathcal{p};t).
\end{equation}
\end{small}In the overdamped limit (\begin{small}${\gamma_0\gg 1}$\end{small}), the solution to Eq. (\ref{eq:0077}) can be written as \cite{Risken1996}
\begin{small}
\begin{equation}\label{eq:0078} W(\mathcal{q},\mathcal{p};t)=w(\mathcal{q};t)\mathcal{W}^{(eq)}(\mathcal{p}),
\end{equation}
\end{small}Here, \begin{small}${\mathcal{W}^{(eq)}(\mathcal{p})}$\end{small} is the  Wigner function in the momentum space which is always in the thermal equilibrium state,
\begin{small}
\begin{equation}\label{eq:0079}
\mathcal{W}^{(eq)}(\mathcal{p})\!=\!\sqrt{\frac{\beta}{2\pi m_0}}\exp{\!\left(\!-\beta\frac{\mathcal{p}^2}{2 m_0}\!\right)},
\end{equation}
\end{small}and \begin{small}${w(\mathcal{q};t)}$\end{small} is the Wigner function in the coordinate space which satisfies the Smoluchowski equation \cite{Ge2009, Esposito2010c},
\begin{small}
\begin{equation}\label{eq:0080}
\partial_t w(\mathcal{q};t)=-\partial_{\mathcal{q}}j(\mathcal{q};t),
\end{equation}
\end{small}where\begin{small}
\begin{equation}\label{eq:0081}
j(\mathcal{q};t)=b(\mathcal{q};t)w(\mathcal{q};t)-\frac{1}{\beta\gamma_0}\partial_{\mathcal{q}}w(\mathcal{q};t)
\end{equation}
\end{small}is the probability flux and\begin{small}
\begin{equation}\label{eq:0082}
b(\mathcal{q};t)=\frac{1}{\gamma_0}\left[-m_0\omega_0^2\mathcal{q}+f(t)\right]
\end{equation}
\end{small}is the drift coefficient.

In order to derive the quantum corrections to the total entropy production rate and the heat dissipation rate, we expand \begin{small}${e_{\rm p}}$\end{small}, \begin{small}${h_{\rm d}}$\end{small}, and \begin{small}${j(\mathcal{q};t)}$\end{small} in powers of \begin{small}${\hbar}$\end{small} as
\begin{small}
\begin{eqnarray}
e_{\rm p}&=&e^{({\rm cl})}_{\rm p}+(i\hbar)e^{(1)}_{\rm p}+(i\hbar)^2 e^{(2)}_{\rm p}+o(\hbar^2),\label{eq:0083a}\\
h_{\rm d}&=&h^{({\rm cl})}_{\rm d}+(i\hbar)h^{(1)}_{\rm d}+(i\hbar)^2 h^{(2)}_{\rm d}+o(\hbar^2),\label{eq:0083b}\\
j&=&j_{\rm cl}+(i\hbar)j^{(1)}+(i\hbar)^2 j^{(2)}+o(\hbar^2),\label{eq:0083c}
\end{eqnarray}
\end{small}and Eqs. (\ref{eq:0080}-\ref{eq:0081}) become
\begin{small}
\begin{subequations}
\begin{equation}\label{eq:0084}
\left\{
\begin{aligned}
&\partial_t w_{\rm cl}(\mathcal{q};t)=-\partial_{\mathcal{q}}j_{\rm cl}(\mathcal{q};t), \\
&j_{\rm cl}(\mathcal{q};t)=b(\mathcal{q};t)w_{\rm cl}(\mathcal{q};t)-\frac{1}{\beta\gamma_0}\partial_{\mathcal{q}}w_{\rm cl}(\mathcal{q};t),
\end{aligned}
\right.\ \ \ \ \ \
\end{equation}
\begin{equation}\label{eq:0085}
\left\{
\begin{aligned}
&\partial_t w^{(1)}(\mathcal{q};t)=-\partial_{\mathcal{q}}j^{(1)}(\mathcal{q};t), \\
&j^{(1)}(\mathcal{q};t)=b(\mathcal{q};t)w^{(1)}(\mathcal{q};t)-\frac{1}{\beta\gamma_0}\partial_{\mathcal{q}}w^{(1)}(\mathcal{q};t),
\end{aligned}
\right.
\end{equation}
\begin{equation}\label{eq:0086}
\left\{
\begin{aligned}
&\partial_t w^{(2)}(\mathcal{q};t)=-\partial_{\mathcal{q}}j^{(2)}(\mathcal{q};t), \\
&j^{(2)}(\mathcal{q};t)=b(\mathcal{q};t)w^{(2)}(\mathcal{q};t)-\frac{1}{\beta\gamma_0}\partial_{\mathcal{q}}w^{(2)}(\mathcal{q};t).
\end{aligned}
\right.
\end{equation}
\end{subequations}
\end{small}According to Refs. \cite{Qian2001, Qian2001b, Seifert2005}, the quantum heat dissipation rate can be written as
\begin{small}
\begin{equation}\label{eq:0086a}
h_{\rm d}=\beta\gamma_0 \int\frac{d\mathcal{q}d\mathcal{p}}{2\pi\hbar}b(\mathcal{q};t)J(\mathcal{q},\mathcal{p};t),
\end{equation}
\end{small}where \begin{small}${J(\mathcal{q},\mathcal{p};t)}$ ${=}$ ${j(\mathcal{q};t)\mathcal{W}^{(eq)}}(\mathcal{p})$\end{small}. By substituting Eq. (\ref{eq:0083c}) into Eq. (\ref{eq:0086a}), one can obtain
\begin{small}
\begin{subequations}
\begin{eqnarray}
&&h^{({\rm cl})}_{\rm d}=\beta\gamma_0 \int\frac{d\mathcal{q}d\mathcal{p}}{2\pi\hbar}b(\mathcal{q};t)J_{\rm cl}(\mathcal{q},\mathcal{p};t),\label{eq:0089}\\
&&h^{(1)}_{\rm d}=\beta\gamma_0 \int\frac{d\mathcal{q}d\mathcal{p}}{2\pi\hbar}b(\mathcal{q};t)J^{(1)}(\mathcal{q},\mathcal{p};t),\label{eq:0090}\\
&&h^{(2)}_{\rm d}=\beta\gamma_0 \int\frac{d\mathcal{q}d\mathcal{p}}{2\pi\hbar}b(\mathcal{q};t)J^{(2)}(\mathcal{q},\mathcal{p};t),\label{eq:0091}
\end{eqnarray}
\end{subequations}
\end{small}where \begin{small}${J_{\rm cl}(\mathcal{q},\mathcal{p};t)}$ ${=}$ ${j_{\rm cl}(\mathcal{q};t)\mathcal{W}^{(eq)}(\mathcal{p})}$\end{small}, \begin{small}${J^{(1)}(\mathcal{q},\mathcal{p};t)}$ ${=}$ ${j^{(1)}(\mathcal{q};t)}$ ${\mathcal{W}^{(eq)}(\mathcal{p})}$\end{small} and \begin{small}${J^{(2)}(\mathcal{q},\mathcal{p};t)}$ ${=}$ ${j^{(2)}(\mathcal{q};t)}$ ${\mathcal{W}^{(eq)}(\mathcal{p})}$\end{small}. The quantum total entropy production rate is given by
\begin{small}
\begin{equation}\label{eq:0087}
e_{\rm p}\equiv\frac{d S(t)}{dt}-h_{\rm d}.
\end{equation}
\end{small}By taking the derivative of \begin{small}${S_{\rm cl}(t)}$\end{small}, \begin{small}${S^{(1)}(t)}$\end{small} and \begin{small}${S^{(2)}(t)}$\end{small} in Eqs. (\ref{eq:0072a})-(\ref{eq:0072c}) and using the dynamical equations (\ref{eq:0084})-(\ref{eq:0086}), one can obtain
\begin{small}
\begin{subequations}
\begin{eqnarray}
e^{({\rm cl})}_{\rm p}&=&\beta\gamma_0 \int\frac{d\mathcal{q}d\mathcal{p}}{2\pi\hbar}\frac{J^2_{\rm cl}(\mathcal{q},\mathcal{p};t)}{W_{\rm cl}(\mathcal{q},\mathcal{p};t)},\label{eq:0088}\\
e^{(1)}_{\rm p}&=&2\beta\gamma_0\!\! \int\!\frac{d\mathcal{q}d\mathcal{p}}{2\pi\hbar}\frac{J^{(1)}(\mathcal{q},\mathcal{p};t)J_{\rm cl}(\mathcal{q},\mathcal{p};t)}{W_{\rm cl}(\mathcal{q},\mathcal{p};t)}-\beta\gamma_0\!\! \int\!\frac{d\mathcal{q}d\mathcal{p}}{2\pi\hbar}\frac{J^2_{\rm cl}(\mathcal{q},\mathcal{p};t)}{W^2_{\rm cl}(\mathcal{q},\mathcal{p};t)}W^{(1)}(\mathcal{q},\mathcal{p};t),\label{eq:0092}\\
e^{(2)}_{\rm p}&=&2\beta\gamma_0\!\! \int\!\frac{d\mathcal{q}d\mathcal{p}}{2\pi\hbar}\frac{J^{(2)}(\mathcal{q},\mathcal{p};t)J_{\rm cl}(\mathcal{q},\mathcal{p};t)}{W_{\rm cl}(\mathcal{q},\mathcal{p};t)}\!-\!\beta\gamma_0\!\! \int\!\frac{d\mathcal{q}d\mathcal{p}}{2\pi\hbar}\frac{J^2_{\rm cl}(\mathcal{q},\mathcal{p};t)}{W^2_{\rm cl}(\mathcal{q},\mathcal{p};t)}W^{(2)}(\mathcal{q},\mathcal{p};t)\nonumber\\
&&+\beta\gamma_0\!\int\!\frac{d\mathcal{q}d\mathcal{p}}{2\pi\hbar}\left[\frac{J^{(1)}(\mathcal{q},\mathcal{p};t)W_{\rm cl}(\mathcal{q},\mathcal{p};t)-W^{(1)}(\mathcal{q},\mathcal{p};t)J_{\rm cl}(\mathcal{q},\mathcal{p};t)}{W_{\rm cl}(\mathcal{q},\mathcal{p};t)}\right]^2\frac{1}{W_{\rm cl}(\mathcal{q},\mathcal{p};t)}\nonumber\\
&&+\int\!\frac{d\mathcal{q}d\mathcal{p}}{2\pi\hbar}\frac{\partial}{\partial t}\!\!\left[\frac{W_{\rm cl}(\mathcal{q},\mathcal{p};t)(\stackrel{\leftarrow}{\partial_{\mathcal{p}}}\stackrel{\rightarrow}{\partial_{\mathcal{q}}}
-\stackrel{\leftarrow}{\partial_{\mathcal{q}}}\stackrel{\rightarrow}{\partial_{\mathcal{p}}})^2W_{\rm cl}(\mathcal{q},\mathcal{p};t)}{16W_{\rm cl}(\mathcal{q},\mathcal{p};t)}\right]-\!
\int\!\frac{d\mathcal{q}d\mathcal{p}}{2\pi\hbar}\frac{\partial}{\partial t}\!\!\left[\frac{G(W_{\rm cl}(\mathcal{q},\mathcal{p};t))}{12W_{\rm cl}^2(\mathcal{q},\mathcal{p};t)}\right].\label{eq:0093}
\end{eqnarray}
\end{subequations}
\end{small}Here, \begin{small}${h^{({\rm cl})}_{\rm d}}$\end{small}
is the heat dissipation rate and \begin{small}${e^{({\rm cl})}_{\rm p}}$\end{small} is the nonnegative classical entropy production rate. One can see that Eq. (\ref{eq:0089}) and Eq. (\ref{eq:0088}) are exactly the same as the results in Ref. \cite{Ge2009, Esposito2010c}, thus our results reproduce their classical counterparts in the classical limit (\begin{small}${\hbar\to 0}$\end{small}).

By rewriting Eq. (\ref{eq:0000}) in the phase space formulation, one can obtain the exact expression of the von Neumann entropy \cite{Zachos2007},
\begin{small}
\begin{equation}\label{eq:0094a}
S(t)=-\int\frac{d\mathcal{q}d\mathcal{p}}{2\pi\hbar}\ W(\mathcal{q},\mathcal{p};t) \ln_\star W(\mathcal{q},\mathcal{p};t),
\end{equation}
\end{small}where
\begin{small}
\begin{equation}\label{eq:0094b}
\ln_\star W(\mathcal{q},\mathcal{p};t)\equiv-\sum^{\infty}_{n=1}\frac{[1-W(\mathcal{q},\mathcal{p};t)]^n_\star}{n},
\end{equation}
\end{small}and the Moyal product \cite{Hillery1984}
\begin{small}
\begin{equation}\label{eq:0094c}
\star\equiv\exp{\left[-\frac{i\hbar}{2}(\stackrel{\leftarrow}{\partial_{\mathcal{p}}}\stackrel{\rightarrow}{\partial_{\mathcal{q}}}
-\stackrel{\leftarrow}{\partial_{\mathcal{q}}}\stackrel{\rightarrow}{\partial_{\mathcal{p}}})\right]}.
\end{equation}
\end{small}If we expand Eq. (\ref{eq:0094c}) in powers of \begin{small}${\hbar}$\end{small} and take the lowest-order approximation of the Moyal product, the von Neumann entropy is approximated as \cite{Santos2017}
\begin{small}
\begin{equation}\label{eq:0094}
\mathcal{S}(t)=-\int\frac{d\mathcal{q}d\mathcal{p}}{2\pi\hbar}\ W(\mathcal{q},\mathcal{p};t) \ln W(\mathcal{q},\mathcal{p};t),
\end{equation}
\end{small}and then the approximate expression of the quantum entropy production rate \begin{small}${\mathcal{e}_{\rm p}}$\end{small} and the quantum heat dissipation rate \begin{small}${\mathcal{h}_{\rm d}}$\end{small} can be written as
\begin{small}
\begin{equation}\label{eq:0095}
\mathcal{e}_{\rm p}=\beta\gamma_0 \int\frac{d\mathcal{q}d\mathcal{p}}{2\pi\hbar}\frac{J^2(\mathcal{q},\mathcal{p};t)}{W(\mathcal{q},\mathcal{p};t)},\ \ \ \ \ \ \ \
\end{equation}
\begin{equation}\label{eq:0096}
\mathcal{h}_{\rm d}=\beta\gamma_0 \int\frac{d\mathcal{q}d\mathcal{p}}{2\pi\hbar}b(\mathcal{q};t)J(\mathcal{q},\mathcal{p};t),
\end{equation}
\end{small}where \begin{small}${J(\mathcal{q},\mathcal{p};t)}$ ${=}$ ${j(\mathcal{q};t)}$ ${\mathcal{W}^{(eq)}(\mathcal{p})}$\end{small}. Moreover, one can split \begin{small}${\mathcal{e}_{\rm p}}$\end{small} into two parts \cite{Ge2009, Esposito2010c},
\begin{small}
\begin{equation}\label{eq:0097}
\mathcal{e}_{\rm p}=\beta\mathcal{Q}_{\rm hk}+(-\beta\mathcal{f}_{\rm d}),
\end{equation}
\end{small}where \begin{small}${\mathcal{Q}_{\rm hk}}$\end{small} is the quantum housekeeping heat which is given by
\begin{small}
\begin{equation}\label{eq:0097a}
\mathcal{Q}_{\rm hk}\!=\!\gamma_0\!\int\!\frac{d\mathcal{q}d\mathcal{p}}{2\pi\hbar}\!\left[b(\mathcal{q};t)\!-\!\frac{1}{\beta\gamma_0}\frac{\partial\!\log{W_{\!\rm st}(\mathcal{q},\mathcal{p};t)}}{\partial\mathcal{q}}\right]\!W(\mathcal{q},\mathcal{p};t),
\end{equation}
\end{small}and \begin{small}${\mathcal{f}_{\rm d}}$\end{small} is the quantum free energy dissipation rate which is given by
\begin{small}
\begin{equation}\label{eq:0097b}
\mathcal{f}_{\rm d}\!=\!\frac{1}{\beta^2\gamma_0}\!\int\!\!\frac{d\mathcal{q}d\mathcal{p}}{2\pi\hbar}\!\!\left[\frac{\partial}{\partial\mathcal{q}}\!\log\frac{W_{\rm st}(\mathcal{q},\mathcal{p};t)}{W(\mathcal{q},\mathcal{p};t)}\right]^2\!\!W(\mathcal{q},\mathcal{p};t).
\end{equation}
\end{small}Here \begin{small}${W_{\rm st}(\mathcal{q},\mathcal{p};t)}$\end{small} is the reduced Wigner function of the steady state of the system, i.e., the long time limit (\begin{small}${t\to\infty}$\end{small}) of \begin{small}${W(\mathcal{q},\mathcal{p};t)}$\end{small}. Note that Eqs. (\ref{eq:0097})-(\ref{eq:0097b}) are not the exact results. They are valid only under the approximation (\ref{eq:0094}). Moreover, by expanding \begin{small}${\mathcal{e}_{\rm p}}$\end{small} in powers of \begin{small}${\hbar}$\end{small}, we obtain
\begin{small}
\begin{equation}\label{eq:0098}
\mathcal{e}_{\rm p}=\mathcal{e}^{({\rm cl})}_{\rm p}+(i\hbar)\mathcal{e}^{(1)}_{\rm p}+(i\hbar)^2 \mathcal{e}^{(2)}_{\rm p}+o(\hbar^2),
\end{equation}
\end{small}where
\begin{small}
\begin{subequations}
\begin{eqnarray}
&&\!\!\!\!\!\!\!\!\mathcal{e}^{({\rm cl})}_{\rm p}\!=\!\beta\gamma_0 \int\frac{d\mathcal{q}d\mathcal{p}}{2\pi\hbar}\frac{J^2_{\rm cl}(\mathcal{q},\mathcal{p};t)}{W_{\rm cl}(\mathcal{q},\mathcal{p};t)},\label{eq:0099}\\
&&\!\!\!\!\!\!\!\!\mathcal{e}^{(1)}_{\rm p}\!=\!2\beta\gamma_0\!\! \int\!\frac{d\mathcal{q}d\mathcal{p}}{2\pi\hbar}\frac{J^{(1)}(\mathcal{q},\mathcal{p};t)J_{\rm cl}(\mathcal{q},\mathcal{p};t)}{W_{\rm cl}(\mathcal{q},\mathcal{p};t)}\!-\!\beta\gamma_0\!\! \int\!\frac{d\mathcal{q}d\mathcal{p}}{2\pi\hbar}\frac{J^2_{\rm cl}(\mathcal{q},\mathcal{p};t)}{W^2_{\rm cl}(\mathcal{q},\mathcal{p};t)}W^{(1)}(\mathcal{q},\mathcal{p};t),\label{eq:0100}\\
&&\!\!\!\!\!\!\!\!\mathcal{e}^{(2)}_{\rm p}\!=\!2\beta\gamma_0\!\! \int\!\frac{d\mathcal{q}d\mathcal{p}}{2\pi\hbar}\frac{J^{(2)}(\mathcal{q},\mathcal{p};t)J_{\rm cl}(\mathcal{q},\mathcal{p};t)}{W_{\rm cl}(\mathcal{q},\mathcal{p};t)}\!-\!\beta\gamma_0\!\! \int\!\frac{d\mathcal{q}d\mathcal{p}}{2\pi\hbar}\frac{J^2_{\rm cl}(\mathcal{q},\mathcal{p};t)}{W^2_{\rm cl}(\mathcal{q},\mathcal{p};t)}W^{(2)}(\mathcal{q},\mathcal{p};t)\!+\!\beta\gamma_0\!\!\int\!\!\frac{d\mathcal{q}d\mathcal{p}}{2\pi\hbar}\!\!\left[\!\frac{J^{(1)}W_{\rm cl}\!-\!W^{(1)}J_{\rm cl}}{W_{\rm cl}}\!\right]^{\!2}\!\!\frac{1}{W_{\rm cl}}.\nonumber\\ \label{eq:0101}
\end{eqnarray}
\end{subequations}
\end{small}One can see that the lowest and the first order quantum correction of \begin{small}${\mathcal{e}_{\rm p}}$\end{small} (Eq. (\ref{eq:0099}) and Eq. (\ref{eq:0100})) are the same as Eq. (\ref{eq:0088}) and Eq. (\ref{eq:0092}), respectively. The second order quantum correction of \begin{small}${\mathcal{e}_{\rm p}}$\end{small} (\ref{eq:0101}) reproduces the first three terms while misses the last two terms of Eq. (\ref{eq:0093}). The last two terms just arise from the Moyal product when we rewrite Eq. (\ref{eq:0000}) in the phase space formulation \cite{Qiu2020}.

Furthermore, as a demonstration, we take the thermal equilibrium initial state and the linear external force as an example to show our results (\ref{eq:0088})-(\ref{eq:0093}). We assume that the system is prepared initially in the thermal equilibrium state at the inverse temperature \begin{small}${\beta'}$\end{small}. The initial Wigner function is given by Eq. (\ref{eq:0051}). The linear external force can be written as \begin{small}
\begin{equation}\label{eq:0102}
f(t)=m_0\omega_0^2\cdot vt,
\end{equation}
\end{small}where \begin{small}${v}$\end{small} is the velocity of the driving. Under the Markovian approximation and in the overdamped limit, the solution to the Wigner function of the system can be obtained from Eq. (\ref{eq:0033}),
\begin{small}
\begin{equation}\label{eq:0103}
w(\mathcal{q};t)\!=\!\frac{2\pi\hbar}{\sqrt{\frac{2\pi}{\beta m_0\omega_0^2}\!\left[1\!+\!\left(\!\frac{\beta\hbar\omega_0}{2}\!\coth{\frac{\beta'\hbar\omega_0}{2}}\!-\!1\!\right)\exp{\!\left(\!-2\frac{m_0\omega_0^2}{\gamma_0}t\right)}\right]}}\exp\!{\left\{\!-\frac{\left[\mathcal{q}-vt+\frac{v\gamma_0}{m_0\omega_0^2}\left(1-\exp{\left(\!-\frac{m_0\omega_0^2}{\gamma_0}t\right)}\right)\right]^2}{\frac{2}{\beta m_0\omega_0^2}\!\!\left[1\!+\!\left(\!\frac{\beta\hbar\omega_0}{2}\!\coth{\frac{\beta'\hbar\omega_0}{2}}\!-\!1\!\right)\exp{\!\left(\!-2\frac{m_0\omega_0^2}{\gamma_0}t\right)}\right]}\right\}}.
\end{equation}
\end{small}By expanding Eq. (\ref{eq:0079}) and Eq. (\ref{eq:0103}) in powers of \begin{small}${\hbar}$\end{small}, one can obtain
\begin{small}
\begin{subequations}
\begin{eqnarray}
&&\!\!\!\!\!\!\!\!\!\!W_{\rm cl}(\mathcal{q},\mathcal{p};t)\!=\!\frac{2\pi\hbar\ \mathcal{W}^{(eq)}(\mathcal{p})}{\sqrt{\frac{2\pi}{\beta m_0\omega_0^2}\!\left[1\!+\!\left(\!\frac{\beta}{\beta'}\!-\!1\!\right)\exp{\!\left(\!-2\frac{m_0\omega_0^2}{\gamma_0}t\right)}\right]}}\exp\!{\left\{\!-\frac{\left[\mathcal{q}-vt+\frac{v\gamma_0}{m_0\omega_0^2}\left(1-\exp{\left(\!-\frac{m_0\omega_0^2}{\gamma_0}t\right)}\right)\right]^2}{\frac{2}{\beta m_0\omega_0^2}\!\!\left[1\!+\!\left(\!\frac{\beta}{\beta'}\!-\!1\!\right)\exp{\!\left(\!-2\frac{m_0\omega_0^2}{\gamma_0}t\right)}\right]}\right\}},\label{eq:0103a}\\
&&\!\!\!\!\!\!\!\!\!\!W^{(1)}(\mathcal{q},\mathcal{p};t)=0,\ \ \ \ \ \ \ \ \ \ \ \ \ \label{eq:0103b}\\
&&\!\!\!\!\!\!\!\!\!\!W^{(2)}(\mathcal{q},\mathcal{p};t)=0.\label{eq:0103c}
\end{eqnarray}
\end{subequations}
\end{small}Note that all the quantum correction terms of \begin{small}${W(\mathcal{q},\mathcal{p};t)}$\end{small} are exactly zero because all the correction terms of \begin{small}${\mathcal{W}^{(eq)}(\mathcal{p})}$\end{small} are zero. By substituting Eqs. (\ref{eq:0103a})-(\ref{eq:0103c}) into Eqs. (\ref{eq:0088})-(\ref{eq:0093}), we obtain the quantum corrections to the total entropy production rate and the heat dissipation rate. One can see that the even order correction terms of \begin{small}${e_{\rm p}}$\end{small} are nonzero due to the last two terms in Eq. (\ref{eq:0093}). Therefore, the Moyal product has played an important role in the calculation of the quantum corrections to the entropy production rate.

From the above analysis, one can see that under the Markovian approximation, the quantum corrections to the total entropy production rate \begin{small}${e_{\rm p}(t)}$\end{small} and the heat dissipation rate \begin{small}${h_{\rm d}(t)}$\end{small} arise from two different physical origins. One is the difference between the definition of the von Neumann entropy and Gibbs entropy (the Moyal product), the other is the difference between the initial Wigner function and the initial classical probability distribution.

\section{Dicussion and Summary}
Before concluding this paper, we would like to give the following remarks.

(I) In calculating the von Neumann entropy, we trace out the degrees of freedom of the heat bath, and ignore completely the entanglement between the system and the heat bath. It can be seen that when neglecting the entanglement, the von Neumann entropy of the system reproduces its classical counterpart in the classical limit (\ref{eq:0072a}). However, it is unclear to us if it is proper to neglect the entanglement in the study of quantum information related problems, e.g., the Landauer's principle \cite{Horhammer2008, Horhammer2005}. How the entanglement between the system and the heat bath will influence the Landauer's principle in an open quantum system is still an open question.

(II) Exactly solvable models can bring important insights. The dynamics of the quantum Brownian motion model under a time-dependent Hamiltonian is of great importance in the study of nonequilibrium quantum thermodynamics, e.g., finite time quantum heat engines, quantum Landauer's principle, and quantum fluctuation theorems. But it is usually extremely difficult to solve exactly due to the huge number of degrees of freedom of the heat bath. Luckily, for this specific model, we obtain the analytical results of the time evolution of the Wigner function and the von Neumann entropy. The exact solutions of the quantum corrections to the entropy will be helpful for analyzing the interplay between quantum mechanics and thermodynamics at extremely low temperature.

(III) We also notice that in Refs. \cite{Agarwal1971a, Agarwal1971b, Agarwal1973}, the author presented a method to calculate the von Neumann entropy of quantum states whose Wigner function is in a Gaussian form. So this method can also be applied to calculate the entropy of Eq. (\ref{eq:0053}). However, their method is not applicable when the Wigner function is non-Gaussian. Nevertheless, the method for calculating the von Neumann entropy presented in Eqs. (\ref{eq:0071})-(\ref{eq:0073}) is valid for whatever states as long as they have well-defined classical counterparts.

(IV) Our results about the quantum correction to the total entropy production rate \begin{small}${e_{\rm p}(t)}$\end{small} and the heat dissipation rate \begin{small}${h_{\rm d}(t)}$\end{small} are valid under the Markovian approximation. In Ref. \cite{R2012}, the authors extend the definition of the classical nonadiabatic entropy production given in Ref. \cite{Esposito2010a} to an arbitrary non-Markovian systems. We plan to extend our results (\ref{eq:0088})-(\ref{eq:0093}) to arbitrary non-Markovian stochastic process in the future.

In summary, in this paper, we study the time evolution of the von Neumann entropy of a quantum Brownian particle under a driving force, as well as the total entropy production. By solving the quantum Langevin equation, we obtain the analytical expression of the Wigner function at an arbitrary time \begin{small}${t}$\end{small}. As an example, we obtain the evolution of the Wigner function explicitly when the system is initially prepared in a thermal equilibrium state, and it reproduces the classical probability distribution in the high-temperature and the weak-coupling limit. Based on the above results and the results of the \begin{small}${\hbar}$\end{small} expansion of the von Neumann entropy in the phase space, we prove that the zeroth-order term reproduces the Gibbs entropy, and we obtain the explicit expression of the time evolution of the quantum corrections to the Gibbs entropy. Moreover, under the Markovian approximation, we obtain the expression of the quantum corrections to the total entropy production rate \begin{small}${e_{\rm p}(t)}$\end{small} and the heat dissipation rate \begin{small}${h_{\rm d}(t)}$\end{small}.

In the classical stochastic thermodynamics, fluctuating work (heat) is defined along individual stochastic trajectory in the phase space \cite{Sekimoto2010}. Nevertheless, it is elusive to define a trajectory-dependent work (heat) in open quantum systems, because there is no well-defined trajectory in the Hilbert space due to the Heisenberg uncertainty principle. We plan to extend our current investigation to these problems and we believe that further studies along this line will advance our understanding about the relationship between the quantum and the classical work and heat and may bring important insights to some fundamental problems in quantum thermodynamics.

\section{Acknowledgment}
H. T. Quan acknowledges support from the National Science Foundation of China under grants 11775001, 11534002, and 11825001.

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text:
Q: Do electromagnetic beams interference to each other when they meet? Let's asume a practical case, we have four 5ghz directional wifi antenas and there is a connection between two of them. If I place the other two and make a connection that the beam goes and meets (or crashes?) to the other beam (from practical let's asume in a center point, distance from of 1km).
Will that crash have any noticeable impact on Tx/Rx ratings?  Of course beams are perpendicular.

A: The crash would not have any effect on the Tx/Rx rating.
The two beams will interfere only at their intersection, but will have no effect
on each other before or after the intersection.
This is due to the principle of superposition of the EM waves in a linear medium which, for this case, could be re-stated as follows:  the EM field at the receiver will be a sum of the effects of the the two beams if they acted alone.
Since the interfering beam would not have any effect at the receiver of the first beam if it was acting alone, it should not have any effect when both beams are present.
Of course, if this was not the case, wireless broadcasting or communication would not work.

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Well-cared for hands, feet, and nails are an essential part of achieving the complete you. Bukés nail technicians have undergone extensive training in sanitation and mastering the art of beautiful nails. Our service offering includes a variety of carefully selected hand and foot treatments with the latest products and color collections.
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Earthbag walls are placed on 6 to 8 inches lifts/courses/rows line masonry walls.
Earthbag walls are built in running bonds, the same as masonry walls are.
Low-rise masonry construction does not require high strength masonry units, the same as earthbags construction. In addition, earthbag walls are always thicker than masonry walls, so the stresses will be lower.
Most designers who design Rammed Earth use the masonry design equation. Since earthbag is Flexible Form Rammed Earth (FFRE), masonry equations can be used for earthbag wall designs.
Plaster/stucco is usually used on each side of earthbag walls.
This solves the issue of lack of shear strength of earthbag walls parallel to the wall.
Also, it helps with bending moments due to loads perpendicular to the wall's surface. In our practice at Precision Structural Engineering, we use stucco on both sides/surfaces of the earthbag walls.
The Building Code requires a minimum amount of reinforcement to be used for earthbag construction.
A vertical bar every 4 feet on center that is required by the Building Code for masonry construction for Seismic Design Category D, E and F could be used for earthbag wall construction. The bar could be added within the stucco thickness, or through the earthbag wall itself.
The steel mesh in the plaster/stucco on each side of the wall.
The steel bars in the bond beam at the top of the earthbag wall.
The steel bars in the top of the foundation/ bottom of the earthbag wall.
The barbed wire or the joint reinforcement used between the courses/rows of earthbags can be included in the required amount of steel.
The earthbag walls can, and must be anchored to the roof or floor at their tops.
BUILDING CODES GENERAL PART 4 2006 NEW MEXICO EARTHEN BUILDING MATERIALS CODE, sanco-bg.com/NM_Earthen_Building_Code, Attached, is a good source of information.
The earthbag wall could be treated as a Masonry wall.
An acceptable way to resist the horizontal force between courses such as barbed wire or masonry joint reinforcement, pending test results, or simply lime or cement mortar between the courses.
Reinforcement could be added in the center of the earth bag itself at 4 feet on center or it could be place in the stucco at the outside surfaces of the wall.
Concrete bond beam at the top of the wall with at least two No. 4 bars with lap of 30 inches and corner bars are essential.
Roof must be attached to the top of the wall to resist the seismic forces.
Please look for details that show the above requirement coming soon.
The International Building Code and the International residential Code, or any other building Code, do not give me the power to give generic prescriptive structural design that could be applicable to any building. So, Please see the disclaimer below.


Bears have become an endangered species. According to the American Association of Individual Investors (AAII), the percentage of individual investors with bearish outlooks over the next six months has been steadily declining since the 2008 financial crisis. It's hard not to be optimistic when the world's most powerful central bank is doing everything in its power to prop up financial assets.
Of course, the AAII survey polls retail investors, or what some refer to as "dumb money." Interestingly, the smart money of the world has been just as complacent. According to the National Association of Active Investment Managers' Exposure Index, active money managers are still historically overweight in equities.
Will more bears come out of hiding now with a rate hike looming? I wouldn't bet on it. The Fed may have abandoned its "patient" stance on lifting rates, but Fed Chair Janet Yellen made it clear that the Fed is seeking a pickup in wage growth and general inflation before raising rates.
Yellen's tacit commitment to dovish monetary policy should support equities, however these same policies only reinforce the importance of owning precious metals over the long haul.


text:
Another Texas Mother Arrested For Leaving Her Children to Die in a Hot Car
Parker Co. Sheriff's Department
A Texas mother is in custody after her 2-year-old and 16-month-old children died when she left them in a hot car to "teach them a lesson".
24-year-old Cynthia Marie Randolph was arrested Friday in connection with the deaths of her two children last month. According to CNN, Randolph found her children playing in her car around noon on May 26th. When her 2-year-old daughter refused to get out of the car, Randolph reportedly locked them in to teach them a lesson, saying she believed her oldest child would be able unlock the car to get she and her brother out when they were ready to come in.
Officers were notified at 4pm that afternoon after the children were discovered, but Randolph provided them with a fake story. Randolph told investigators that the children had been playing on an enclosed patio before she noticed them missing. She said she then spent nearly 40 minutes looking for the children, eventually finding them locked in her car, which she broke the window to in an attempt to save them. However, after several interviews with the Parker County Sheriff's department, holes and variations in Randolph's story quickly led investigators to not believe her account.
After an investigation by the Texas Rangers, it was found that after Randolph locked her children in the car, she went inside and smoked some marijuana, then took a nap for a few hours. When she discovered the children were still in the car, the temperatures outside had reached the mid-90s. Randolph later admitted to breaking the car window in an attempt to make her story seem plausible.
This marks the second time in just a few weeks that a young Texas mother has been arrested for the deaths of her own children, from having left them in a hot car. Earlier this month, a 19-year-old mother in Kerville,TX was arrested after her two toddler children died from being left in the car for 15 hours while she was spending the night at a friend's house.
Randolph is facing two felony charges for causing serious bodily injury to a child.
NEXT: Texas Mother Charged after Leaving Children in Car for 15 Hours
Source: Another Texas Mother Arrested For Leaving Her Children to Die in a Hot Car

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I missed this Rembert Browne deep read of this photo of Nicki Minaj at a bar mitzvah when it came out in April, but it is truly a timeless work that deserves to be memorialized one more time here.
Even after reading Rembert Browne's piece, I'm filled with questions. What is going on here?


text:
Irish Heartbeat is the eighteenth studio album by Northern Irish singer-songwriter Van Morrison and is a collaboration with the traditional Irish musical group the Chieftains, released in 1988. It was recorded at Windmill Lane Studios in Dublin, Ireland, and reached number 18 in the UK album charts.

Recording
The album was recorded on dates from September to December 1987 and in January 1988. The Chieftains and Van Morrison had met years before at the Edinburgh rock festival.  They joined up in Belfast during Morrison's No Guru tour and afterwards, Morrison and Paddy Moloney discussed recording an album together during a walk.  They each had a list of songs and reached a consensus to cover two of Morrison's previously released tracks (the title track was one) and the rest from traditional Irish songs.

Recalled Moloney:I think at that time Van was searching for his Irish roots. It was this man of blues, of rock 'n' roll, jazz and more importantly soul, coming home to his Irishness with The Chieftains and the music we'd been playing for so many years. Musically we were going to meet each other half way.

In October 1987 they performed together at Balmoral Studio in Belfast. The concert was broadcast on St. Patrick's Day in 1988.

Composition
The album consists of eight traditional Irish songs, plus re-workings of the Morrison songs "Celtic Ray" (which first appeared on 1982's Beautiful Vision) and the title track "Irish Heartbeat" (which first appeared on 1983's Inarticulate Speech of the Heart). "Carrickfergus" is described as "a melancholic air worthy of Otis Redding" by critic Denis Campbell. "On Raglan Road"  was adapted from a poem by Patrick Kavanagh and is the story of "a man ensnared by a beautiful revenant whom he had mistaken for 'a creature made of clay'."
In 1994, Billy Connolly recorded a live cover of the song "Irish Heartbeat" during his World Tour of Scotland. The performance was used as the closing theme to the BBC series.

Critical reception

Irish Heartbeat received positive reviews from most critics, one of whom called it "some of the most haunting, rousing, downright friendly music of the year". Rolling Stone magazine's David Browne said it has "splendor and intense beauty",<ref>[https://www.rollingstone.com/music/albumreviews/irish-heartbeat-19880811 Rolling Stone, Aug. 11, 1988]</ref> while John Wilde from Melody Maker hailed it as "a bloody considerable marvel", having "awakened [Morrison's] roisterous spirit". In The Village Voice, Robert Christgau was more critical and believed that Morrison, suffering a creative block, was "misguided" in his attempt to reconnect with his traditional Irish music roots.

In the annual Pazz & Jop critics poll, Irish Heartbeat was voted the 29th best album of 1988. The NME named it the second greatest album of the year. In The Rolling Stone Album Guide (2004), Rob Sheffield said it showed Morrison in a livelier, more enthusiastic state than on his previous records during the 1980s.

Track listing
All songs traditional, arranged by Van Morrison and Paddy Moloney, except as indicated.

Side one
"Star of the County Down" – 2:41
"Irish Heartbeat" (Morrison) – 3:52
"Tá Mo Chleamhnas Déanta (My Match It Is Made)" – 3:31
"Raglan Road" (lyrics by Patrick Kavanagh) – 4:43
"She Moved Through the Fair" – 4:44

Side two
"I'll Tell Me Ma" – 2:29
"Carrickfergus" – 4:23
"Celtic Ray" (Morrison) – 3:47
"My Lagan Love" – 5:19
"Marie's Wedding" – 3:17

Charts

Personnel
Van Morrison – lead vocals, guitar, drums
Paddy Moloney – uilleann pipes, tin whistle
Martin Fay – fiddle, bones
Derek Bell – harp, keyboards, tiompán
Kevin Conneff – bodhran, co-lead vocals on "Star of the County Down", "Tá Mo Chleamhnas Déanta" and "I'll Tell Me Ma"
Matt Molloy – flute
Seán Keane – fiddle
Ciarán Ó Braonáin – bass
Mary Black – backing vocals on "Marie's Wedding" and "Tá Mo Chleamhnas Déanta"
Maura O'Connell – backing vocals on "Marie's Wedding"
June Boyce – backing vocals on "Celtic Ray", "Irish Heartbeat" and "Marie's Wedding"

Notes

References
Glatt, John (1997). The Chieftains: The Authorized Biography, Da Capo Press,
Heylin, Clinton (2003). Can You Feel the Silence? Van Morrison: A New Biography, Chicago Review Press,
Hinton, Brian (2000) Celtic Crossroads: The Art of Van Morrison'', Sanctuary,

External links

The Chieftains albums
Van Morrison albums
Collaborative albums
1988 albums
Mercury Records albums
Albums produced by Van Morrison

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text:
Question: What Major Composer Was Influenced By Mozart?
1 What composers did Mozart influence?
2 What major composer was greatly influenced by Mozart's work?
3 Who was Mozart's biggest influence?
4 How has Mozart influenced music?
5 What Really Killed Mozart?
6 Who was a better composer Mozart or Beethoven?
7 Who killed Mozart?
8 Who is the greatest composer of all time?
9 Who did Mozart marry?
10 Who was Haydn influenced by?
11 What age did Mozart die?
12 What Made Mozart music special?
13 How did Mozart influence society?
14 What does the K mean in Mozart music?
What composers did Mozart influence?
His work influenced many composers that followed — most notably Beethoven. Along with his friend Joseph Haydn, Mozart conceived and perfected the grand forms of symphony, opera, string ensemble, and concerto that marked the classical period.
What major composer was greatly influenced by Mozart's work?
Wolfgang Amadeus Mozart (1756–1791) had a powerful influence on the works of Ludwig van Beethoven (1770–1827). Beethoven held Mozart in high regard; some of his music recalls Mozart's, he composed several variations on Mozart's themes and he modeled a number of his compositions on those of the older composer.
Who was Mozart's biggest influence?
The largest influence on Mozart had to have been his father, Leopold. His father not only gave up his own musical ambitions to tutor him but he also exposed him to a great many things to open up his mind (KBAQ). Leopold concentrated most of his life on making Mozart a child prodigy.
How has Mozart influenced music?
He almost single-handedly developed and popularized the classical piano concerto. He wrote a great deal of religious music, including large-scale masses, as well as dances, divertimenti, serenades, and other forms of light entertainment. The central traits of the classical style are all present in Mozart's music.
You might be interested: Readers ask: Mozart In The Jungle Why Cancelled?
What Really Killed Mozart?
Beethoven's disability: He was blind Mozart went deaf though.
Who was a better composer Mozart or Beethoven?
With 16 of the 300 most popular works having come from his pen, Mozart remains a strong contender but ranks second after Ludwig van Beethoven, overtaking Amadeus with 19 of his works in the Top 300 and three in the Top 10.
Who killed Mozart?
But today Antonio Salieri is best remembered for something he probably didn't do. He's remembered for poisoning Mozart.
Who is the greatest composer of all time?
Ludwig van Beethoven (1770–1827) The German composer and pianist Ludwig van Beethoven is widely regarded as the greatest composer who ever lived.
Who did Mozart marry?
Today is Mozart's wedding anniversary. It was on Sunday, August 4, 1782, that Wolfgang Amadeus and Constanze Weber were married in St Stephen's Cathedral in Vienna. He was 26, she just 20. The story of how they fell in love and eventually became man and wife would fit well in one of his operas.
Who was Haydn influenced by?
Haydn was very much influenced by developments made by Bach's son C. P. E. Bach who, moving away from his father's teachings, brought drama and originality to his keyboard works and helped to establish early Sonatas and Symphonies in 3 movements.
What age did Mozart die?
At 12:55 a.m., 225 years ago, Wolfgang Amadeus Mozart drew his last breath. Later, he was unceremoniously buried in a common grave — as was the custom of his era — in the St. Marx cemetery, just outside the Vienna city limits. Mozart was only 35.
You might be interested: Often asked: How Did Wolfgang Mozart Die?
What Made Mozart music special?
He composed masterfully in every musical format. Operas, choral works, concertos, symphonies, chamber music, solo songs, sonatas … Mozart was one of the few composers in history to compose masterworks in every conceivable musical genre.
How did Mozart influence society?
Well, Mozart played a role in the creation of today's society. Mozart also sparked ideas for other composers and philosophers, whose ideas may have shaped governments or leaders. Modern-day society has also been impacted by this amazing composer and musician, by creating new styles of music.
What does the K mean in Mozart music?
Köchel (K) numbers are assigned sequentially according to the date of composition. For example, Mozart's opera The Magic Flute is given the Köchel number 620, and is (approximately) the 620th piece of music Mozart composed. Compositions completed at the same time are listed K69, K69a, and so on.
Quick Answer: What Is A Freelance Compser Mozart?
Quick Answer: How To Trill In Mozart?


There are many people that as soon as the Thanksgiving leftovers are gone they're in full celebration mode for the next holiday. Others, even long after the lights are on the tree, are still struggling to really "feel the season." A good way to help you get in the mood for the holiday season is to go out into the community and take advantage of the different holiday events going on. Do that and before you know it you'll find yourself humming Christmas carols to yourself! Celebrating the Season in Dance: The holiday performances by the Litchfield Dance Arts Academy students. Dec. 12, 3 and 7 p.m., Litchfield Dance Arts Academy, 97 Otis Dr. $15/adults The Sounds of Christmas: An evening of worship, praise and fellowship by the music ministry. Dec. 13, 5 p.m., Mount Zion Missionary Baptist Church. Visitors welcome. Band concert: Christmas performance by the Pawleys Island Concert Band under the baton of Carson Turner. Dec. 15, 7 p.m., Pawleys Island Community Church. Free. Holiday memories: Model trains, period Christmas trees and other items that evoke the spirit of Christmas past. Through the end of the year, Brookgreen Gardens. Free with regular admission. Nights of a Thousand Candles: Illuminated sculpture gardens and seasonal entertainment. Dec. 10-12 and 17-20, 3-10 p.m., Brookgreen Gardens. $18/adults, $10/ages 4-12 A Fashionable Christmas: Candlelight tours of historic home decorated for the holidays by local designers. Dec. 10 and 17, 5-7 p.m., Kaminski House Museum. $5/adults, $3/youth, free/10 and under. Daily tours at 11 a.m., 1 and 3 p.m., $12/adults, $10/seniors, free/children 5 and under Santa Paws: Photos of pets with Santa in a benefit for animal rescue. Saturdays through Dec. 19 from 11 a.m.-2 p.m., All 4 Paws. Call 843-237-7297 for appointment. All of us at Litchfield Real Estate hope you and yours have a very merry Christmas and a happy, healthy and safe New Year!


I used to think ordering wine at a restaurant could define me, until I was caught underage drinking at 19. I ate cornichons at the Chateau Marmont one summer and wondered if this was a beginning or an erasure of my upbringing. Who was I to be anyone but myself—how dare I? Go to law school, drive a Nissan, knock down a spider's web that covered the porch of our rental. I wanted so badly to be someone else; I thought I was for three long years in California.
And now I am back home and the memory of who I was remains foggy in memory. But that's what happens when you live in fiction. I hardly eat out anymore; instead, I make small snacks and graze throughout the day. My mother likes to make casseroles. I ate three nectarines because I was lazy one morning. I wasn't myself then and now I'm too much of myself now, I get exhausted. So I don't bake, try to keep the oven off. I make small meals and take nothing for granted.
I made one such meal this week—local pickles, local olives, homemade crackers, and fresh and delicate goat cheese. A meal I would have spent $30 on when I was trying to create memories in California and a meal I will gladly allow to reify my Appalachian existence today.
Makes four large flatbread crackers that can be broken up into portions. Adapted from here.
Bake these two first. You will most likely have to use two racks.
Assembling your platter: This is the fun part. Grab your most favorite goat cheese (mine is, of course Vermont Creamery's) and top with a generous helping of honey or comb honey—the more flavor the better. Sprinkle a little fresh black pepper over this. Next, enjoy your cheese and crackers with various small bites. I did spicy habanero pickles, an assortment of kalamata and green garlic olives, and fresh red grapes. I would also recommend a robust or smoky cheddar, an apple butter or jam, or make it interesting with cubed watermelon and a drizzle of balsamic and blackberry juice reduction.