text:
\section{Open quantum systems}
\subsection{Alternative expression of the Lindblad master equation}
Here we show that the Lindblad master equation can be written as
\begin{equation}
\Dr{\rho}=\mathcal{K}_\rho(-\ln\rho+\ln\rho^{\rm eq}),
\end{equation}
where $\mathcal{K}_\rho:\nu\mapsto i\beta^{-1}[\nu,\rho]+\mathcal{O}_\rho(\nu)$.
The operator $\mathcal{O}_\rho$ is defined by
\begin{equation}
\mathcal{O}_\rho(\nu):=\sum_{\mu,\omega}e^{-\beta\omega/2}\alpha_\mu(\omega)[L_{\mu}(\omega),[\rho]_{\beta\omega}([L_{\mu}^\dagger(\omega),\nu])].
\end{equation}
For any density operator $\rho=\sum_np_n\Mat{n}$, where $\sum_np_n=1$ and $\{\Kt{n}\}_n$ are orthonormal eigenvectors, we can express the tilted operator as
\begin{equation}
[\rho]_\theta(X)=e^{-\theta/2}\int_0^1e^{s\theta}\rho^sX\rho^{1-s}ds=\sum_{n,m}\Phi(e^{\theta/2}p_n,e^{-\theta/2}p_m)\Br{n}X\Kt{m}\Kt{n}\Br{m}.\label{eq:tilted.operator}
\end{equation}
Here, $\Phi(x,y)$ is the logarithmic mean of two positive numbers $x$ and $y$, given by $\Phi(x,y)=(x-y)/[\ln(x)-\ln(y)]$ for $x\neq y$ and $\Phi(x,x)=x$.
As $\rho^{\rm eq}=e^{-\beta H}/Z_\beta$ and $[\ln\rho,\rho]=0$, we have $i\beta^{-1}[-\ln\rho+\ln\rho^{\rm eq},\rho]=-i[H,\rho]$.
Thus, we need only show that
\begin{equation}
\mathcal{O}_\rho(-\ln\rho+\ln\rho^{\rm eq})=\sum_{\mu,\omega}\alpha_\mu(\omega)\bras{2L_\mu(\omega)\rho L_\mu^\dagger(\omega)-\brab{L_\mu^\dagger(\omega)L_\mu(\omega),\rho}}.\label{eq:operator.O}
\end{equation}
To this end, we first show that $[\rho]_{\theta}([X,\ln\rho]-\theta X)=e^{-\theta/2}X\rho-e^{\theta/2}\rho X$ for an arbitrary operator $X\in\mathfrak{L}(\mathcal{H})$ and $\theta\in\mathbb{R}$.
This can be achieved through the following transformation:
\begin{subequations}
\begin{align}
[\rho]_{\theta}([X,\ln\rho]-\theta X)&=e^{-\theta/2}\int_0^1e^{\theta s}e^{s\ln\rho}(X\ln\rho-\ln\rho X-\theta X)e^{(1-s)\ln\rho}ds\\
&=-e^{-\theta/2}\int_0^1\bras{e^{\theta s}e^{s\ln\rho}(\ln\rho+\theta)Xe^{(1-s)\ln\rho}+e^{\theta s}e^{s\ln\rho}X(-\ln\rho)e^{(1-s)\ln\rho}}ds\\
&=-e^{-\theta/2}\int_0^1\frac{d}{ds}\bras{e^{(\ln\rho+\theta)s}Xe^{(1-s)\ln\rho}}ds\\
&=e^{-\theta/2}(Xe^{\ln\rho}-e^{\ln\rho+\theta}X)\\
&=e^{-\theta/2}X\rho-e^{\theta/2}\rho X.
\end{align}
\end{subequations}
Next, applying the relation $[L_{\mu}^\dagger(\omega),H]=-\omega L_{\mu}^\dagger(\omega)$, one immediately obtains
\begin{subequations}
\begin{align}
[\rho]_{\beta\omega}([L_{\mu}^\dagger(\omega),-\ln\rho+\ln\rho^{\rm eq}])&=[\rho]_{\beta\omega}([L_{\mu}^\dagger(\omega),-\ln\rho-\beta H])\\
&=-[\rho]_{\beta\omega}([L_{\mu}^\dagger(\omega),\ln\rho]+\beta[L_{\mu}^\dagger(\omega),H])\\
&=-[\rho]_{\beta\omega}([L_{\mu}^\dagger(\omega),\ln\rho]-\beta\omega L_{\mu}^\dagger(\omega))\\
&=e^{\beta\omega/2}\rho L_{\mu}^\dagger(\omega)-e^{-\beta\omega/2}L_{\mu}^\dagger(\omega)\rho.
\end{align}
\end{subequations}
Consequently, as $L_{\mu}^\dagger(\omega)=L_{\mu}(-\omega)$ and $\alpha_\mu(\omega)=e^{\beta\omega}\alpha_\mu(-\omega)$, one can verify Eq.~\eqref{eq:operator.O} as follows:
\begin{subequations}
\begin{align}
&\mathcal{O}_\rho(-\ln\rho+\ln\rho^{\rm eq})\\
&=\sum_{\mu,\omega}e^{-\beta\omega/2}\alpha_\mu(\omega)[L_{\mu}(\omega),[\rho]_{\beta\omega}([L_{\mu}^\dagger(\omega),-\ln\rho+\ln\rho^{\rm eq}])]\\
&=\sum_{\mu,\omega}e^{-\beta\omega/2}\alpha_\mu(\omega)[L_{\mu}(\omega),e^{\beta\omega/2}\rho L_{\mu}^\dagger(\omega)-e^{-\beta\omega/2}L_{\mu}^\dagger(\omega)\rho]\\
&=\sum_{\mu,\omega}\alpha_\mu(\omega)\bras{-e^{-\beta\omega}L_{\mu}(\omega)L_{\mu}^\dagger(\omega)\rho+L_{\mu}(\omega)\rho L_{\mu}^\dagger(\omega)+e^{-\beta\omega}L_{\mu}^\dagger(\omega)\rho L_{\mu}(\omega)-\rho L_{\mu}^\dagger(\omega)L_{\mu}(\omega)}\\
&=\sum_{\mu,\omega}\brab{\alpha_\mu(\omega)\bras{L_{\mu}(\omega)\rho L_{\mu}^\dagger(\omega)-\rho L_{\mu}^\dagger(\omega)L_{\mu}(\omega)}+\alpha_\mu(-\omega)\bras{L_{\mu}(-\omega)\rho L_{\mu}^\dagger(-\omega)-L_{\mu}^\dagger(-\omega)L_{\mu}(-\omega)\rho}}\\
&=\sum_{\mu,\omega}\alpha_\mu(\omega)\bras{2L_\mu(\omega)\rho L_\mu^\dagger(\omega)-\brab{L_\mu^\dagger(\omega)L_\mu(\omega),\rho}}.
\end{align}
\end{subequations}

\subsection{Properties of the quantum Wasserstein metric}
Here we provide several useful properties regarding the metric defined in the main text.
\begin{lemma}
The inner product $\avg{\cdot,\mathcal{O}_\rho(\cdot)}$ satisfies the conjugate-symmetry condition $\avg{\xi,\mathcal{O}_\rho(\nu)}=\avg{\nu,\mathcal{O}_\rho(\xi)}^*$ for all operators $\nu,\xi\in\mathfrak{L}(\mathcal{H})$.
Here, $*$ denotes the complex conjugate.
\end{lemma}
\begin{proof}
For an arbitrary operator $X\in\mathfrak{L}(\mathcal{H})$ and $\theta\in\mathbb{R}$, we have
\begin{subequations}
\begin{align}
\avg{\xi,[X,[\rho]_{\theta}([X^\dagger,\nu])]}&=\Tr{\xi^\dagger[X,[\rho]_{\theta}([X^\dagger,\nu])]}\\
&=\Tr{[\xi^\dagger,X][\rho]_{\theta}([X^\dagger,\nu])}\\
&=\sum_{n,m}\Phi(e^{\theta/2}p_n,e^{-\theta/2}p_m)\Br{n}[X^\dagger,\nu]\Kt{m}\Br{m}[\xi^\dagger,X]\Kt{n},\label{eq:prop1}
\end{align}
\end{subequations}
where we have used Eq.~\eqref{eq:tilted.operator} in Eq.~\eqref{eq:prop1}.
Swapping $\xi$ and $\nu$, one obtains
\begin{subequations}
\begin{align}
\avg{\nu,[X,[\rho]_{\theta}([X^\dagger,\xi])]}^*&=\sum_{n,m}\Phi(e^{\theta/2}p_n,e^{-\theta/2}p_m)\Br{n}[X^\dagger,\xi]\Kt{m}^*\Br{m}[\nu^\dagger,X]\Kt{n}^*\\
&=\sum_{n,m}\Phi(e^{\theta/2}p_n,e^{-\theta/2}p_m)\Br{m}[\xi^\dagger,X]\Kt{n}\Br{n}[X^\dagger,\nu]\Kt{m}\\
&=\avg{\xi,[X,[\rho]_{\theta}([X^\dagger,\nu])]}.\label{eq:prop2}
\end{align}
\end{subequations}
As $\mathcal{O}_\rho(\nu)=\sum_{\mu,\omega}e^{-\beta\omega/2}\alpha_\mu(\omega)[L_{\mu}(\omega),[\rho]_{\beta\omega}([L_{\mu}^\dagger(\omega),\nu])]$, Eq.~\eqref{eq:prop2} implies that
\begin{equation}
\avg{\nu,\mathcal{O}_\rho(\xi)}^*=\avg{\xi,\mathcal{O}_\rho(\nu)}.\label{eq:consym1}
\end{equation}
\end{proof}

From Eq.~\eqref{eq:prop1}, one observes that
\begin{equation}
\avg{\xi,[X,[\rho]_{\theta}([X^\dagger,\xi])]}=\sum_{n,m}\Phi(e^{\theta/2}p_n,e^{-\theta/2}p_m)|\Br{n}[X^\dagger,\xi]\Kt{m}|^2\ge 0.
\end{equation}
Therefore, $\avg{\xi,\mathcal{O}_\rho(\xi)}\ge 0$ for an arbitrary operator $\xi$.
Equality is attained only when $[L_\mu^\dagger(\omega),\xi]=0$ for all $\mu$ and $\omega$.
When $\xi$ is a self-adjoint operator, i.e., $\xi^\dagger=\xi$, we have $\avg{\xi,\mathcal{K}_\rho(\xi)}=\avg{\xi,\mathcal{O}_\rho(\xi)}\ge 0$.

\begin{proposition}\label{proposition:vanish}
A self-adjoint operator $\nu$ satisfies $\mathcal{K}_\rho(\nu)=0$ if and only if $\nu$ is spanned by $\mathbb{I}_N$.
\end{proposition}
\begin{proof}
As $\mathcal{K}_\rho(\mathbb{I}_N)=0$, we need only show that if $\mathcal{K}_\rho(\nu)=0$, then $\nu$ is spanned by $\mathbb{I}_N$.
Noting that $0=\avg{\nu,\mathcal{K}_\rho(\nu)}=\avg{\nu,\mathcal{O}_\rho(\nu)}$, we find that $\avg{\nu,\mathcal{O}_\rho(\nu)}=0$ only when $[L_\mu^\dagger(\omega),\nu]=0$ for all $\mu$ and $\omega$.
As the dynamics of the quantum system are ergodic, this implies that $\nu$ is spanned by $\mathbb{I}_N$.

\end{proof}

\begin{proposition}\label{proposition:traceless}
$\mathcal{K}_\rho(\nu)$ is a traceless self-adjoint operator for all $\nu\in\mathfrak{H}(\mathcal{H})$.
\end{proposition}
\begin{proof}
The expression
\begin{equation}
\mathcal{K}_\rho(\nu)=i\beta^{-1}[\nu,\rho]+\mathcal{O}_\rho(\nu)=i\beta^{-1}[\nu,\rho]+\sum_{\mu,\omega}e^{-\beta\omega/2}\alpha_\mu(\omega)[L_{\mu}(\omega),[\rho]_{\beta\omega}([L_{\mu}^\dagger(\omega),\nu])]
\end{equation}
is a linear combination of commutators.
Therefore, $\Tr{\mathcal{K}_\rho(\nu)}=0$ is immediately derived.
Note that $(i\beta^{-1}[\nu,\rho])^\dagger=i\beta^{-1}[\nu,\rho]$, so we need only show that $\mathcal{O}_\rho(\nu)$ is self-adjoint.
Using the relations $[\rho]_\theta(X)^\dagger=[\rho]_{-\theta}(X^\dagger)$, $[X,Y]^\dagger=[Y^\dagger,X^\dagger]$, $e^{-\beta\omega/2}\alpha_\mu(\omega)=e^{\beta\omega/2}\alpha_\mu(-\omega)$, and $L_{\mu}^\dagger(\omega)=L_{\mu}(-\omega)$, we can prove that $\mathcal{O}_\rho(\nu)$ is self-adjoint as follows:
\begin{align}
\mathcal{O}_\rho(\nu)^\dagger&=\sum_{\mu,\omega}e^{-\beta\omega/2}\alpha_\mu(\omega)[L_{\mu}(\omega),[\rho]_{\beta\omega}([L_{\mu}^\dagger(\omega),\nu])]^\dagger\\
&=\sum_{\mu,\omega}e^{-\beta\omega/2}\alpha_\mu(\omega)[[\rho]_{\beta\omega}([L_{\mu}^\dagger(\omega),\nu])^\dagger,L_{\mu}(\omega)^\dagger]\\
&=\sum_{\mu,\omega}e^{\beta\omega/2}\alpha_\mu(-\omega)[[\rho]_{-\beta\omega}([\nu,L_{\mu}(\omega)]),L_{\mu}(-\omega)]\\
&=\sum_{\mu,\omega}e^{\beta\omega/2}\alpha_\mu(-\omega)[L_{\mu}(-\omega),[\rho]_{-\beta\omega}([L_{\mu}^\dagger(-\omega),\nu])]\\
&=\mathcal{O}_\rho(\nu).
\end{align}
\end{proof}

\begin{lemma}
For an arbitrary density operator $\rho$ and traceless self-adjoint operator $\vartheta$, there exists a unique traceless self-adjoint operator $\nu$ such that $\vartheta=\mathcal{K}_\rho(\nu)$.
\end{lemma}
\begin{proof}
Let $\mathcal{B}=\{\chi_{j,k}\}_{1\le j,k\le N}$ denote the set of generalized Gell-Mann matrices, which span the space of operators in the complex Hilbert space $\mathcal{H}$.
Specifically, $\chi_{j,k}$ can be expressed as follows:
\begin{equation}
\chi_{j,k}=\begin{cases}
E_{k,j}+E_{j,k}, & {\text{if}}~ j<k, \\
i(E_{k,j}-E_{j,k}), & {\text{if}}~ j>k, \\
\sqrt{\frac{2}{j(j+1)}}\bra{\sum_{l=1}^{j}E_{l,l}-jE_{j+1,j+1}}, & {\text{if}}~ j=k<N, \\
N^{-1}\mathbb{I}_N, & {\text{if}}~ j=k=N, \\
\end{cases}
\end{equation}
where $E_{j,k}$ denotes a matrix with $1$ in the $jk$-th entry and $0$ elsewhere.
In this construction, each $\chi_{j,k}$ is a Hermitian matrix and $\Tr{\chi_{j,k}}=\delta_{jN}\delta_{kN}$ for all $(j,k)$.
For convenience, we define a set $\overline{\mathcal{B}}:=\mathcal{B}\setminus\{\chi_{N,N}\}$.
For arbitrary traceless self-adjoint operator $X$, there exists real coefficients $c_{j,k}\in\mathbb{R}$ such that $X=\sum_{j,k}c_{j,k}\chi_{j,k}$.
Taking the trace of both sides of the equation, we obtain $0=\Tr{X}=\sum_{j,k}c_{j,k}\Tr{\chi_{j,k}}=c_{N,N}$.
This implies that $X$ can be expressed as a linear combination of matrices in $\overline{\mathcal{B}}$ with all real coefficients.

By propositions \ref{proposition:vanish} and \ref{proposition:traceless}, $\mathcal{K}_\rho(\chi_{j,k})$ is obviously a nonzero traceless self-adjoint operator for all $(j,k)\neq(N,N)$.
We now show that $\{\mathcal{K}_\rho(\chi_{j,k})\}_{(j,k)\neq(N,N)}$ is an independent set, i.e., $\sum_{(j,k)\neq(N,N)}c_{j,k}\mathcal{K}_\rho(\chi_{j,k})=0$ only when $c_{j,k}=0$ for all $j,k$.
Indeed, by the linearity of $\mathcal{K}_\rho$, we have
\begin{equation}
\sum_{(j,k)\neq(N,N)}c_{j,k}\mathcal{K}_\rho(\chi_{j,k})=\mathcal{K}_\rho\Big(\sum_{(j,k)\neq(N,N)}c_{j,k}\chi_{j,k}\Big)=0.
\end{equation}
Under proposition \ref{proposition:vanish}, $\sum_{(j,k)\neq(N,N)}c_{j,k}\chi_{j,k}$ must be spanned by $\mathbb{I}_N$ ($=N\chi_{N,N}$), i.e., $\sum_{(j,k)\neq(N,N)}c_{j,k}\chi_{j,k}=-c_{N,N}\chi_{N,N}$ for some $c_{N,N}$.
This is equivalent to $\sum_{1\le j,k\le N}c_{j,k}\chi_{j,k}=0$.
As $\mathcal{B}$ is a basis of $\mathcal{H}$, this equivalence requires that $c_{j,k}=0$ for all $j,k$.

Because $\{\mathcal{K}_\rho(\chi_{j,k})\}_{(j,k)\neq(N,N)}$ has $N^2-1$ elements, we can add another matrix $\phi$ to form a new basis of $\mathcal{H}$.
In terms of the elements of the new basis, $\mathbb{I}_N$ can then be expressed as
\begin{equation}
\mathbb{I}_N=z\phi+\sum_{(j,k)\neq(N,N)}c_{j,k}\mathcal{K}_\rho(\chi_{j,k}),\label{eq:Id.new.basis}
\end{equation}
where $z$ is some complex number.
Taking the trace of both sides of Eq.~\eqref{eq:Id.new.basis}, we obtain $N=z\,\Tr{\phi}$, which indicates that $z\neq 0$.
Therefore, $\phi$ can be expressed in terms of $\mathbb{I}_N$ and $\{\mathcal{K}_\rho(\chi_{j,k})\}_{(j,k)\neq(N,N)}$ as
\begin{equation}
\phi=z^{-1}\Big[\mathbb{I}_N-\sum_{(j,k)\neq(N,N)}c_{j,k}\mathcal{K}_\rho(\chi_{j,k})\Big].\label{eq:phi.new.basis}
\end{equation}
Equation \eqref{eq:phi.new.basis} implies that an arbitrary matrix can be expressed as a linear combination of elements in the following set:
\begin{equation}
\mathcal{S}:=\{\mathbb{I}_N\}\cup\{\mathcal{K}_\rho(\chi_{j,k})\}_{(j,k)\neq(N,N)}.
\end{equation}
Equivalently, $\mathcal{S}$ is a basis of $\mathcal{H}$.
Consequently, because $\mathcal{K}_\rho(\chi_{j,k})$ is traceless and self-adjoint, an arbitrary traceless self-adjoint operator $\vartheta$ can be expressed in terms of $\{\mathcal{K}_\rho(\chi_{j,k})\}_{(j,k)\neq(N,N)}$ with real coefficients $\{c_{j,k}\}$ as
\begin{equation}
\vartheta=\sum_{(j,k)\neq(N,N)}c_{j,k}\mathcal{K}_\rho(\chi_{j,k})=\mathcal{K}_\rho\Big(\sum_{(j,k)\neq(N,N)}c_{j,k}\chi_{j,k}\Big).
\end{equation}
Defining the traceless self-adjoint operator $\nu:=\sum_{(j,k)\neq(N,N)}c_{j,k}\chi_{j,k}$, one readily obtains $\vartheta=\mathcal{K}_\rho(\nu)$.
Finally, to prove the uniqueness of $\nu$, we assume two traceless self-adjoint operators $\nu_1$ and $\nu_2$ such that $\vartheta=\mathcal{K}_\rho(\nu_1)=\mathcal{K}_\rho(\nu_2)$, then $\mathcal{K}_\rho(\nu_1-\nu_2)=0$.
Applying the result in proposition \ref{proposition:vanish}, we have $\nu_1-\nu_2=z\mathbb{I}_N$ for some $z\in\mathbb{C}$.
Thus, $zN=\Tr{z\mathbb{I}_N}=\Tr{\nu_1-\nu_2}=0\Rightarrow z=0$, which implies the uniqueness of $\nu$.

\end{proof}

\begin{lemma}
Given an arbitrary traceless self-adjoint operator $\nu$, the equality $\avg{\nu+\lambda\mathbb{I}_N,\mathcal{K}_\rho(\nu+\lambda\mathbb{I}_N)}=\avg{\nu,\mathcal{K}_\rho(\nu)}$ holds for an arbitrary number $\lambda\in\mathbb{C}$.
\end{lemma}
\begin{proof}
Since $\mathcal{K}_\rho(\nu+\lambda\mathbb{I}_N)=\mathcal{K}_\rho(\nu)+\mathcal{K}_\rho(\lambda\mathbb{I}_N)=\mathcal{K}_\rho(\nu)$, we have
\begin{subequations}
\begin{align}
\avg{\nu+\lambda\mathbb{I}_N,\mathcal{K}_\rho(\nu+\lambda\mathbb{I}_N)}&=\avg{\nu+\lambda\mathbb{I}_N,\mathcal{K}_\rho(\nu)}\\
&=\avg{\nu,\mathcal{K}_\rho(\nu)}+\avg{\lambda\mathbb{I}_N,\mathcal{K}_\rho(\nu)}\\
&=\avg{\nu,\mathcal{K}_\rho(\nu)}+\lambda^*\Tr{\mathcal{K}_\rho(\nu)}\\
&=\avg{\nu,\mathcal{K}_\rho(\nu)},
\end{align}
\end{subequations}
where we have used the traceless property of $\mathcal{K}_\rho$ obtained in proposition \ref{proposition:traceless}.
\end{proof}

\subsection{Lower bound of the quantum Wasserstein distance in terms of the trace-like distance}
Here we derive the lower bound of the quantum Wasserstein distance $\mathcal{W}_{\rm q}(\rho_0,\rho_\tau)$ in terms of the trace-like distance.
From the definition of the quantum Wasserstein distance, given a fixed positive number $\delta>0$, there exists a smooth curve $\rho(t)$ with end points $\rho_0$ and $\rho_\tau$ such that
\begin{equation}
\tau\int_0^\tau\avg{\nu,\mathcal{K}_\rho(\nu)}dt\le\mathcal{W}_{\rm q}(\rho_0,\rho_\tau)^2+\delta.
\end{equation}
Here, $\nu(t)\in\mathfrak{H}(\mathcal{H})$ is a traceless self-adjoint operator that satisfies $\Dr{\rho}(t)=\mathcal{K}_\rho[\nu(t)]$.
Let $\rho(t)=\sum_np_n(t)\Mat{n(t)}$ be a spectral decomposition with an orthogonal basis $\langle n(t)|m(t)\rangle=\delta_{nm}$, and define the self-adjoint operator $\phi(t):=\sum_nc_n\Mat{n(t)}$, where $|c_n|\le 1$ are real constants to be determined later.
Evidently, $\phi(t)$ commutes with $\rho(t)$, i.e., $[\phi,\rho]=0$.
Now, using the relations $\Dr{\rho}=i\beta^{-1}[\nu,\rho]+\mathcal{O}_\rho(\nu)$ and $\avg{\phi,[\nu,\rho]}=0$, we have
\begin{subequations}
\begin{align}
\sum_nc_n[p_n(\tau)-p_n(0)]&=\Tr{\int_0^\tau\phi(t)\dot{\rho}(t)dt}\\
&=\int_0^\tau\avg{\phi,i\beta^{-1}[\nu,\rho]+\mathcal{O}_\rho(\nu)}dt\\
&=\int_0^\tau\avg{\phi,\mathcal{O}_\rho(\nu)}dt\\
&\le\bra{\int_0^\tau\avg{\phi,\mathcal{O}_\rho(\phi)}dt}^{1/2}\bra{\int_0^\tau\avg{\nu,\mathcal{O}_\rho(\nu)}dt}^{1/2}\\
&\le\bra{\tau^{-1}\int_0^\tau\avg{\phi,\mathcal{O}_\rho(\phi)}dt}^{1/2}\bra{\mathcal{W}_{\rm q}(\rho_0,\rho_\tau)^2+\delta}^{1/2}.\label{eq:trace.bound.1}
\end{align}
\end{subequations}
The first term in the last inequality \eqref{eq:trace.bound.1} can be rewritten as
\begin{subequations}
\begin{align}
\avg{\phi,\mathcal{O}_\rho(\phi)}&=\sum_{\mu,\omega}e^{-\beta\omega/2}\alpha_\mu(\omega)\avg{\phi,[L_{\mu}(\omega),[\rho]_{\beta\omega}([L_{\mu}^\dagger(\omega),\phi])]}\\
&=\sum_{\mu,\omega}e^{-\beta\omega/2}\alpha_\mu(\omega)\Tr{[\phi,L_{\mu}(\omega)][\rho]_{\beta\omega}([L_{\mu}^\dagger(\omega),\phi])}\\
&=\sum_{\mu,\omega}e^{-\beta\omega/2}\alpha_\mu(\omega)\avg{[L_{\mu}^\dagger(\omega),\phi],[\rho]_{\beta\omega}([L_{\mu}^\dagger(\omega),\phi])}.\label{eq:puri.term.1}
\end{align}
\end{subequations}
Before proceeding, we prove the following result.
\begin{proposition}\label{prop1}
For an arbitrary operator $X$, a real number $\theta$, and density operator $\rho$, the inequality
\begin{equation}
\avg{X,[\rho]_\theta(X)}\le\frac{1}{2}(e^{\theta/2}+e^{-\theta/2})\|X\|_\infty^2
\end{equation}
holds, where $\|X\|_\infty$ denotes the spectral norm of the operator $X$.
\end{proposition}
\begin{proof}
Using Eq.~\eqref{eq:tilted.operator}, we have
\begin{equation}
\avg{X,[\rho]_\theta(X)}=\sum_{n,m}\Phi(e^{\theta/2}p_n,e^{-\theta/2}p_m)\Br{n}X\Kt{m}\Br{m}X^\dagger\Kt{n}.
\end{equation}
Applying the inequality $\Phi(x,y)\le(x+y)/2$ and the relation $\sum_{n}\Mat{n}=\mathbb{I}_N$, we obtain
\begin{subequations}
\begin{align}
\avg{X,[\rho]_\theta(X)}&\le\frac{1}{2}\sum_{n,m}\bra{e^{\theta/2}p_n+e^{-\theta/2}p_m}\Br{n}X\Kt{m}\Br{m}X^\dagger\Kt{n}\\
&=\frac{1}{2}\sum_{n,m}e^{\theta/2}p_n\Br{n}X\Kt{m}\Br{m}X^\dagger\Kt{n}+\frac{1}{2}\sum_{m,n}e^{-\theta/2}p_m\Br{m}X^\dagger\Kt{n}\Br{n}X\Kt{m}\\
&=\frac{1}{2}\sum_{n}e^{\theta/2}p_n\Br{n}XX^\dagger\Kt{n}+\frac{1}{2}\sum_{m}e^{-\theta/2}p_m\Br{m}X^\dagger X\Kt{m}\label{eq:bound.tilted.op.c}\\
&\le\frac{1}{2}\sum_{n}e^{\theta/2}p_n\|X\|_\infty^2+\frac{1}{2}\sum_{m}e^{-\theta/2}p_m\|X\|_\infty^2\label{eq:bound.tilted.op.d}\\
&=\frac{1}{2}(e^{\theta/2}+e^{-\theta/2})\|X\|_\infty^2.\label{eq:bound.tilted.op.e}
\end{align}
\end{subequations}
Here we applied two facts: $\Br{n}XX^\dagger\Kt{n}\le\|X\|_\infty^2$ in Eq.~\eqref{eq:bound.tilted.op.d} and $\sum_np_n=1$ in Eq.~\eqref{eq:bound.tilted.op.e}.
\end{proof}

Returning to our problem, we apply proposition \ref{prop1} with $X=[L_{\mu}^\dagger(\omega),\phi]$ and $\theta=\beta\omega$, and hence obtain
\begin{equation}
\avg{[L_{\mu}^\dagger(\omega),\phi],[\rho]_{\beta\omega}([L_{\mu}^\dagger(\omega),\phi])}\le\frac{1}{2}(e^{-\beta\omega/2}+e^{\beta\omega/2})\|[L_{\mu}^\dagger(\omega),\phi]\|_\infty^2\le 2(e^{-\beta\omega/2}+e^{\beta\omega/2})\|L_{\mu}(\omega)\|_\infty^2.
\end{equation}
Here, we used the inequalities $\|[X,Y]\|_\infty\le\|XY\|_\infty+\|YX\|_\infty$ and $\|XY\|_\infty\le\|X\|_\infty\|Y\|_\infty$ for all $X,Y\in\mathfrak{L}(\mathcal{H})$.
Consequently, we have
\begin{equation}
\avg{\phi,\mathcal{O}_\rho(\phi)}\le 2\sum_{\mu,\omega}e^{-\beta\omega/2}\alpha_\mu(\omega)(e^{-\beta\omega/2}+e^{\beta\omega/2})\|L_{\mu}(\omega)\|_\infty^2=4\sum_{\mu,\omega}\alpha_\mu(\omega)\|L_{\mu}(\omega)\|_\infty^2.\label{eq:trace.bound.2}
\end{equation}
From Eqs.~\eqref{eq:trace.bound.1} and \eqref{eq:trace.bound.2}, we easily obtain the following inequality:
\begin{equation}
\mathcal{W}_{\rm q}(\rho_0,\rho_\tau)^2+\delta\ge\frac{(\sum_nc_n[p_n(\tau)-p_n(0)])^2}{4\tau^{-1}\int_0^\tau \sum_{\mu,\omega}\alpha_\mu(\omega)\|L_{\mu}(\omega)\|_\infty^2dt}.\label{eq:trace.bound.3}
\end{equation}
Setting $c_n={\rm sign}[p_n(\tau)-p_n(0)]$ and taking the limit $\delta\to 0$ in Eq.~\eqref{eq:trace.bound.3}, a lower bound of the quantum Wasserstein distance is obtained as
\begin{equation}
\mathcal{W}_{\rm q}(\rho_0,\rho_\tau)\ge\frac{\sum_n|p_n(\tau)-p_n(0)|}{2\sqrt{\tau^{-1}\int_0^\tau \sum_{\mu,\omega}\alpha_\mu(\omega)\|L_{\mu}(\omega)\|_\infty^2dt}}.\label{eq:trace.bound.4}
\end{equation}
From Eq.~\eqref{eq:trace.bound.4}, we wish to bound the Wasserstein distance by the trace-like distance $\mathsf{d}_{\rm T}(\rho_0,\rho_\tau)=\sum_{n=1}^{N}|a_n-b_n|$, where $a_1\le a_2\le\dots\le a_N$ and $b_1\le b_2\le\dots\le b_N$ are increasing eigenvalues of $\rho_0$ and $\rho_\tau$.
Given two arrays of real numbers, $\{x_n\}$ and $\{y_n\}$, one can prove that
\begin{equation}
\sum_n|x_n-y_n|\ge\sum_n|x_n-y_{\chi(n)}|,
\end{equation}
where $\{\chi(n)\}$ is a permutation of $\{n\}$ such that $y_{\chi(n)}\ge y_{\chi(m)}$ if $x_n\ge x_m$.
Therefore, $\sum_n|p_n(\tau)-p_n(0)|\ge \mathsf{d}_{\rm T}(\rho_0,\rho_\tau)$, so the bound in terms of the trace-like distance is written as
\begin{equation}
\mathcal{W}_{\rm q}(\rho_0,\rho_\tau)\ge\frac{\mathsf{d}_{\rm T}(\rho_0,\rho_\tau)}{2\sqrt{\tau^{-1}\int_0^\tau \sum_{\mu,\omega}\alpha_\mu(\omega)\|L_{\mu}(\omega)\|_\infty^2dt}}.
\end{equation}

\subsection{Lower bound of the entropy production in terms of the average energy-change distance}
Here we derive the lower bound of the entropy production $\Delta S_{\rm tot}$ in terms of the distance $\mathsf{d}_{\rm E}(\rho_0,\rho_\tau)=|\Tr{H(\rho_0-\rho_\tau)}|$.
The Lindblad master equation can be expressed as $\Dr{\rho}(t)=-i[H,\rho(t)]+\mathcal{O}_\rho[\phi(t)]$, where $\phi(t):=-\ln\rho(t)+\ln\rho^{\rm eq}$.
Using the relations $\Tr{H[H,\rho]}=0$ and $\Delta S_{\rm tot}=\int_0^\tau \avg{\phi,\mathcal{O}_\rho(\phi)}dt$, we obtain
\begin{subequations}
\begin{align}
|\Tr{H(\rho_0-\rho_\tau)}|&=\left|\Tr{H\int_0^\tau\dot{\rho}(t)dt}\right|\\
&=\left|\int_0^\tau\avg{H,\mathcal{O}_\rho(\phi)}dt\right|\\
&\le\bra{\int_0^\tau\avg{H,\mathcal{O}_\rho(H)}dt}^{1/2}\bra{\int_0^\tau\avg{\phi,\mathcal{O}_\rho(\phi)}dt}^{1/2}\\
&=\bra{\int_0^\tau\avg{H,\mathcal{O}_\rho(H)}dt}^{1/2}\sqrt{\Delta S_{\rm tot}}.\label{eq:energy.bound.1}
\end{align}
\end{subequations}
The first term in Eq.~\eqref{eq:energy.bound.1} can be rewritten as
\begin{subequations}
\begin{align}
\avg{H,\mathcal{O}_\rho(H)}&=\sum_{\mu,\omega}e^{-\beta\omega/2}\alpha_\mu(\omega)\avg{H,[L_{\mu}(\omega),[\rho]_{\beta\omega}([L_{\mu}^\dagger(\omega),H])]}\\
&=\sum_{\mu,\omega}e^{-\beta\omega/2}\alpha_\mu(\omega)\Tr{[H,L_{\mu}(\omega)][\rho]_{\beta\omega}([L_{\mu}^\dagger(\omega),H])}\\
&=\sum_{\mu,\omega}e^{-\beta\omega/2}\alpha_\mu(\omega)\avg{[L_{\mu}^\dagger(\omega),H],[\rho]_{\beta\omega}([L_{\mu}^\dagger(\omega),H])}.\label{eq:term.1}
\end{align}
\end{subequations}
Applying proposition \ref{prop1} with $X=[L_\mu^\dagger(\omega),H]$ and $\theta=\beta\omega$, one obtains
\begin{equation}
\avg{[L_\mu^\dagger(\omega),H],[\rho]_{\beta\omega}([L_\mu^\dagger(\omega),H])}\le\frac{1}{2}(e^{-\beta\omega/2}+e^{\beta\omega/2})\|[L_\mu^\dagger(\omega),H]\|_\infty^2=\frac{1}{2}(e^{-\beta\omega/2}+e^{\beta\omega/2})\omega^2\|L_\mu(\omega)\|_\infty^2.
\end{equation}
Consequently, we have
\begin{equation}
\avg{H,\mathcal{O}_\rho(H)}\le \frac{1}{2}\sum_{\mu,\omega}e^{-\beta\omega/2}\alpha_\mu(\omega)(e^{-\beta\omega/2}+e^{\beta\omega/2})\omega^2\|L_{\mu}(\omega)\|_\infty^2=\sum_{\mu,\omega}\alpha_\mu(\omega)\omega^2\|L_{\mu}(\omega)\|_\infty^2.\label{eq:energy.bound.2}
\end{equation}
From Eqs.~\eqref{eq:energy.bound.1} and \eqref{eq:energy.bound.2}, we readily obtain the following inequality:
\begin{equation}
\Delta S_{\rm tot}\ge\frac{\mathsf{d}_{\rm E}(\rho_0,\rho_\tau)^2}{\tau\sum_{\mu,\omega}\alpha_\mu(\omega)\omega^2\|L_{\mu}(\omega)\|_\infty^2}.\label{eq:energy.bound.3}
\end{equation}

A tighter bound in terms of the square of the heat current can be analogously obtained.
Applying the Cauchy--Schwarz inequality, we have
\begin{subequations}
\begin{align}
\int_0^\tau |\dot{Q}|dt &=\int_0^\tau\left|\Tr{H\dot{\rho}(t)}\right|dt\\
&=\int_0^\tau|\avg{H,\mathcal{O}_\rho(\phi)}|dt\\
&\le\bra{\int_0^\tau\avg{H,\mathcal{O}_\rho(H)}dt}^{1/2}\bra{\int_0^\tau\avg{\phi,\mathcal{O}_\rho(\phi)}dt}^{1/2}\\
&=\bra{\int_0^\tau\avg{H,\mathcal{O}_\rho(H)}dt}^{1/2}\sqrt{\Delta S_{\rm tot}}.
\end{align}
\end{subequations}
Subsequently, applying Eq.~\eqref{eq:energy.bound.2}, we obtain
\begin{equation}
\Delta S_{\rm tot}\ge\frac{\bra{\int_0^\tau |\dot{Q}|dt}^2}{\tau\sum_{\mu,\omega}\alpha_\mu(\omega)\omega^2\|L_{\mu}(\omega)\|_\infty^2}.
\end{equation}
This bound is a quantum version of the classical bound reported in Ref.~\cite{Shiraishi.2016.PRL}, in which the total entropy production is bounded from below by the square of the heat current.

Analogously, we can also derive a lower bound on $\Delta S_{\rm tot}$ in terms of the change of entropy $\Delta S$ of the system.
First, we can bound $\Delta S$ from above as
\begin{subequations}
\begin{align}
|\Delta S|&\le\int_0^\tau |\Tr{\Dr{\rho}(t)\ln\rho(t)}|dt\\
&=\int_0^\tau|\avg{\ln\rho,\mathcal{O}_\rho(\phi)}|dt\\
&\le\bra{\int_0^\tau\avg{\ln\rho,\mathcal{O}_\rho(\ln\rho)}dt}^{1/2}\bra{\int_0^\tau\avg{\phi,\mathcal{O}_\rho(\phi)}dt}^{1/2}\\
&=\sqrt{\Upsilon}\sqrt{\Delta S_{\rm tot}},
\end{align}
\end{subequations}
where $\Upsilon :=\int_0^\tau\avg{\ln\rho,\mathcal{O}_\rho(\ln\rho)}dt$.
Then, the total entropy production can be bounded from below as
\begin{equation}
\Delta S_{\rm tot}\ge \frac{|\Delta S|^2}{\Upsilon}.
\end{equation}
Equivalently, the heat dissipated to the environment can be bounded from below as
\begin{equation}
\Delta Q\ge \beta^{-1}\bra{-\Delta S+\frac{|\Delta S|^2}{\Upsilon}}.\label{eq:heat.bound}
\end{equation}
This bound is relevant to the inequality reported in Ref.~\cite{Timpanaro.2020.PRL}, $\Delta Q\ge\mathcal{Q}(\mathcal{S}^{-1}(-\Delta S))$.
However, the bound in Eq.~\eqref{eq:heat.bound} is not tight in the zero-temperature limit.

\subsection{Current-dissipation trade-off}
Here we derive a trade-off relation between the heat current and dissipation, i.e., we derive an upper bound on the ratio $J^2/\sigma_{\rm tot}$, where $J:=\Tr{H(t)\Dr{\rho}(t)}$ is the heat flow from the heat bath to the system and $\sigma_{\rm tot}$ is the total entropy production rate, which characterizes the irreversibility.
Using the relations $\Dr{\rho}(t)=-i[H(t),\rho(t)]+\mathcal{O}_\rho[\phi(t)]$ and $\sigma_{\rm tot}=\avg{\phi(t),\mathcal{O}_\rho[\phi(t)]}$ and applying the Cauchy--Schwarz inequality, we obtain
\begin{subequations}
\begin{align}
J^2=|\Tr{H(t)\Dr{\rho}(t)}|^2&=|\avg{H(t),\mathcal{O}_\rho[\phi(t)]}|^2\\
&\le\avg{H(t),\mathcal{O}_\rho[H(t)]}\avg{\phi(t),\mathcal{O}_\rho[\phi(t)]}\\
&=\avg{H(t),\mathcal{O}_\rho[H(t)]}\sigma_{\rm tot}.
\end{align}
\end{subequations}
Note that
\begin{subequations}
\begin{align}
\avg{H(t),\mathcal{O}_\rho[H(t)]}&=\sum_{\mu,\omega}e^{-\beta\omega/2}\alpha_\mu(\omega)\avg{[L_{\mu}^\dagger(\omega),H(t)],[\rho]_{\beta\omega}([L_{\mu}^\dagger(\omega),H(t)])}	\\
&=\sum_{\mu,\omega}e^{-\beta\omega/2}\alpha_\mu(\omega)\omega^2\avg{L_{\mu}^\dagger(\omega),[\rho]_{\beta\omega}(L_{\mu}^\dagger(\omega))}.
\end{align}
\end{subequations}
From Eq.~\eqref{eq:bound.tilted.op.c}, one can prove that
\begin{subequations}
\begin{align}
\avg{H(t),\mathcal{O}_\rho[H(t)]}&\le\frac{1}{2}\sum_{\mu,\omega}e^{-\beta\omega/2}\alpha_\mu(\omega)\omega^2\bras{e^{\beta\omega/2}\Tr{L_{\mu}^\dagger(\omega)L_{\mu}(\omega)\rho}+e^{-\beta\omega/2}\Tr{L_{\mu}(\omega)L_{\mu}^\dagger(\omega)\rho}}\\
&=\sum_{\mu,\omega}\alpha_\mu(\omega)\omega^2\Tr{L_{\mu}^\dagger(\omega)L_{\mu}(\omega)\rho}\\
&=\Tr{\mathsf{L}\rho},
\end{align}
\end{subequations}
where $\mathsf{L}:=\sum_{\mu,\omega}\alpha_\mu(\omega)\omega^2L_{\mu}^\dagger(\omega)L_{\mu}(\omega)$.
Decomposing $\rho=\rho_{\rm bd}+\rho_{\rm nd}$, where
\begin{subequations}
\begin{align}
\rho_{\rm bd}&=\sum_e\Pi_{e}\rho\Pi_{e},\\
\rho_{\rm nd}&=\sum_{e\neq e'}\Pi_{e}\rho\Pi_{e'},
\end{align}
\end{subequations}
and $\Pi_e$ is the projection to the eigenspace of $H$ with eigenvalue $e$.
As $[L_{\mu}^\dagger(\omega)L_{\mu}(\omega),H(t)]=0$, $[L_{\mu}^\dagger(\omega)L_{\mu}(\omega),\Pi_e]=0$ for all $e$.
Therefore, the coherence between eigenstates with different energies vanishes in $\Tr{\mathsf{L}\rho}$, i.e., $\Tr{\mathsf{L}\rho}=\Tr{\mathsf{L}\rho_{\rm bd}}$.
The trade-off relation between the heat current and dissipation is thus obtained as
\begin{equation}
\frac{J^2}{\sigma_{\rm tot}}\le\avg{H,\mathcal{O}_\rho(H)}\le\Tr{\mathsf{L}\rho_{\rm bd}}.
\end{equation}
This inequality, known as the current-dissipation trade-off relation \cite{Tajima.2020.arxiv}, implies that the ratio $J^2/\sigma_{\rm tot}$ is not enhanced by coherence between eigenstates with different energies, but is enhanced by coherence between degenerate energy eigenstates.

\subsection{Invalidity of the bound in terms of the relative entropy}
Here we prove that the total entropy production in thermalization processes cannot be bounded from below by the relative entropy between the initial and final states.
In thermalization processes, the dynamics of the density operator are governed by the Lindblad equation
\begin{equation}
\Dr{\rho}=-i[H,\rho]+\sum_{\mu,\omega}\alpha_\mu(\omega)\bras{2L_{\mu}(\omega)\rho L_{\mu}^\dagger(\omega)-\brab{L_{\mu}^\dagger(\omega)L_{\mu}(\omega),\rho}}.
\end{equation}
The total entropy production can be explicitly expressed as $\Delta S_{\rm tot}=S(\rho(0)||\rho^{\rm eq})-S(\rho(\tau)||\rho^{\rm eq})$, where $S(\rho_1 ||\rho_2):=\Tr{\rho_1(\ln\rho_1-\ln\rho_2)}$ is the relative entropy of $\rho_1$ with respect to $\rho_2$.
If the relative entropy satisfies the reverse triangle inequality:
\begin{equation}
S(\rho(0)||\rho^{\rm eq})\ge S(\rho(0)||\rho(\tau))+S(\rho(\tau)||\rho^{\rm eq}),
\end{equation}
then $\Delta S_{\rm tot}\ge S(\rho(0)||\rho(\tau))$ and the dissipation can be further bound by the quantum Fisher information and Wigner--Yanase metrics \cite{Mancino.2018.PRL}.
However, this inequality holds in the classical case \cite{Shiraishi.2019.PRL} but not in the general quantum case.
As a simple counterexample, consider that $\alpha_\mu(\omega)\to 0$ for all $\mu$ and $\omega$.
In this vanishing coupling limit, the total entropy production vanishes because the relative entropy is invariant under a unitary transform.
On the other hand, $S(\rho(0)||\rho(\tau))$ is always positive because $\rho(t)$ is changed under the internal dynamics; thus $\Delta S_{\rm tot}<S(\rho(0)||\rho(\tau))$.

\subsection{Quantum Otto heat engine}
Consider a quantum Otto heat engine consisting of a two-level atom, whose energy levels (the excited state $\Kt{e}$ and the ground state $\Kt{g}$) are changed by an external controller.
The atom is alternatively coupled with two heat baths at different inverse temperatures $\beta_h<\beta_c$, and undergoes two isochoric and two adiabatic processes.
The system Hamiltonian is given by $H(t)=\omega(t)\sigma_z/2$, where $\sigma_z=\Mat{e}-\Mat{g}$ is the Pauli matrix in the $z$ direction.
The heat engine is cyclically operated as follows:
\begin{itemize}
	\item[1.] During adiabatic expansion in time $\tau_a$, the frequency changes from $\omega_h$ to $\omega_c$, and work is produced due to the change in internal energy. Here, the word {\it adiabatic} means that the system is isolated from the heat baths and there is no heat exchange during the process, although jumps between energy eigenstates may occur.
	\item[2.] During the cold isochore in time $\tau_c$, the atom is in contact with the cold bath and the frequency $\omega_c$ remains unchanged. In this process, heat $Q_c$ is transferred from the working medium to the cold bath.
	\item[3.] During adiabatic compression in time $\tau_a$, the frequency is reversed from $\omega_c$ to $\omega_h$ and work is done on the medium.
	\item[4.] During the hot isochore in time $\tau_h$, the atom is in contact with the hot bath and the frequency $\omega_h$ is fixed. In this process, heat $Q_h$ is extracted from the hot bath by the working medium.
\end{itemize}
During the adiabatic process, the dynamics of the density matrix are governed by the von Neumann equation
\begin{equation}
\Dr{\rho}(t)=-i[H(t),\rho(t)].
\end{equation}
During an isochoric process $k=h$ or $c$, the atom is thermalized and the time evolution of the density matrix follows the Lindblad master equation \cite{Breuer.2002}:
\begin{equation}
\Dr{\rho}(t)=-i[H_k,\rho(t)]+\mathcal{D}_k[\rho(t)],\label{eq:Lin.iso}
\end{equation}
where the dissipator $\mathcal{D}_k[\cdot]$ is defined by
\begin{equation}
\mathcal{D}_k[\rho]=\alpha_k\bar{n}(\omega_k)(2\sigma_+\rho\sigma_--\{\sigma_-\sigma_+,\rho\})+\alpha_k(\bar{n}(\omega_k)+1)(2\sigma_-\rho\sigma_+-\{\sigma_+\sigma_-,\rho\}).
\end{equation}
Here, $H_k=\omega_k\sigma_z/2$, $\sigma_{\pm}=(\sigma_x\pm i\sigma_y)/2$, $\alpha_k$ is a positive damping rate, and $\bar{n}(\omega_k)=(e^{\beta_k\omega_k}-1)^{-1}$ is the Planck distribution.

\subsubsection{Analytical solution of the density matrix}
In the stationary state, the density matrix can be analytically calculated as
\begin{equation}
\rho(t)=(e^{\lambda(t)}+e^{-\lambda(t)})^{-1}e^{\lambda(t)\sigma_z},\label{eq:den.mat.ans}
\end{equation}
where $\lambda(t)$ is a periodic function satisfying $\lambda(t+\tau)=\lambda(t)$ with $\tau=2\tau_a+\tau_h+\tau_c$.
In the following analysis, we determine the analytical form of $\lambda(t)$.
For any operators $X$ and $Y$, one can prove that
\begin{equation}
e^{-\lambda X}Ye^{\lambda X}=\sum_{n=0}^{\infty}\frac{(-\lambda)^n}{n!}[X,Y]_n,
\end{equation}
where the nested commutator is recursively defined as $[X,Y]_n=[X,[X,Y]_{n-1}]$ and $[X,Y]_0=Y$.
Using the relations $[\sigma_z,\sigma_{+}]=2\sigma_{+}$ and $[\sigma_z,\sigma_{-}]=-2\sigma_{-}$, one readily obtains
\begin{subequations}
\begin{align}
[\sigma_z,\sigma_{+}]_n&=(+2)^n\sigma_{+},\\
[\sigma_z,\sigma_{-}]_n&=(-2)^n\sigma_{-}.
\end{align}
\end{subequations}
Subsequently,
\begin{subequations}
\begin{align}
e^{-\lambda\sigma_z}\sigma_{+}e^{\lambda\sigma_z}&=\sum_{n=0}^{\infty}\frac{(-\lambda)^n(2)^n}{n!}\sigma_{+}=e^{-2\lambda}\sigma_{+}\Leftrightarrow e^{\lambda\sigma_z}\sigma_{+}=e^{2\lambda}\sigma_{+}e^{\lambda\sigma_z},\\
e^{-\lambda\sigma_z}\sigma_{-}e^{\lambda\sigma_z}&=\sum_{n=0}^{\infty}\frac{(-\lambda)^n(-2)^n}{n!}\sigma_{-}=e^{2\lambda}\sigma_{-}\Leftrightarrow e^{\lambda\sigma_z}\sigma_{-}=e^{-2\lambda}\sigma_{-}e^{\lambda\sigma_z}.
\end{align}
\end{subequations}
Noting that $\sigma_+\sigma_-=(\mathbb{I}_2+\sigma_z)/2$ and $\sigma_-\sigma_+=(\mathbb{I}_2-\sigma_z)/2$, the dissipator term can be calculated as
\begin{subequations}\label{eq:dis.term}
\begin{align}
\mathcal{D}_k[e^{\lambda\sigma_z}]&=\alpha_k\bar{n}(\omega_k)(2\sigma_+e^{\lambda\sigma_z}\sigma_--\{\sigma_-\sigma_+,e^{\lambda\sigma_z}\})+\alpha_k(\bar{n}(\omega_k)+1)(2\sigma_-e^{\lambda\sigma_z}\sigma_+-\{\sigma_+\sigma_-,e^{\lambda\sigma_z}\})\\
&=2\bras{\alpha_k\bar{n}(\omega_k)(e^{-2\lambda}\sigma_+\sigma_--\sigma_-\sigma_+)+\alpha_k(\bar{n}(\omega_k)+1)(e^{2\lambda}\sigma_-\sigma_+-\sigma_+\sigma_-)}e^{\lambda\sigma_z}\\
&=\alpha_k\bras{\bar{n}(\omega_k)(e^{-2\lambda}-1)+(\bar{n}(\omega_k)+1)(e^{2\lambda}-1)+\brab{\bar{n}(\omega_k)(e^{-2\lambda}+1)-(\bar{n}(\omega_k)+1)(e^{2\lambda}+1)}\sigma_z}e^{\lambda\sigma_z}
\end{align}
\end{subequations}
Inserting Eqs.~\eqref{eq:den.mat.ans} and \eqref{eq:dis.term} into Eq.~\eqref{eq:Lin.iso}, we obtain
\begin{equation}
\bras{-\frac{e^{\lambda}-e^{-\lambda}}{e^{\lambda}+e^{-\lambda}}+\sigma_z}\Dr{\lambda}(t)=\alpha_k\bras{\bar{n}(\omega_k)(e^{-2\lambda}-1)+(\bar{n}(\omega_k)+1)(e^{2\lambda}-1)+\brab{\bar{n}(\omega_k)(e^{-2\lambda}+1)-(\bar{n}(\omega_k)+1)(e^{2\lambda}+1)}\sigma_z},
\end{equation}
which is satisfied if $\lambda(t)$ obeys the following differential equation:
\begin{equation}
\Dr{\lambda}(t)=\alpha_k\bras{\bar{n}(\omega_k)(e^{-2\lambda}+1)-(\bar{n}(\omega_k)+1)(e^{2\lambda}+1)}.
\end{equation}
This equation can be analytically solved for $\lambda(t)$.
The result is
\begin{equation}
\lambda(t)=\begin{cases}
\lambda(\tau_a), & {\rm if}~0\le t<\tau_a,\\
\frac{1}{2}\ln\bras{\frac{\bar{n}(\omega_c)e^{2\alpha_c(2\bar{n}(\omega_c)+1)(t-\tau_a)}-z_c}{(\bar{n}(\omega_c)+1)e^{2\alpha_c(2\bar{n}(\omega_c)+1)(t-\tau_a)}+z_c}}, & {\rm if}~\tau_a\le t<\tau_a+\tau_c,\\
\lambda(2\tau_a+\tau_c), & {\rm if}~\tau_a+\tau_c\le t<2\tau_a+\tau_c,\\
\frac{1}{2}\ln\bras{\frac{\bar{n}(\omega_h)e^{2\alpha_h(2\bar{n}(\omega_h)+1)(t-2\tau_a-\tau_c)}-z_h}{(\bar{n}(\omega_h)+1)e^{2\alpha_h(2\bar{n}(\omega_h)+1)(t-2\tau_a-\tau_c)}+z_h}}, & {\rm if}~2\tau_a+\tau_c\le t< \tau,
\end{cases}
\end{equation}
where the constant $z_k$ can be explicitly determined through the boundary conditions as
\begin{subequations}
\begin{align}
z_c&=[2\bar{n}(\omega_c)+1]\bra{\frac{\bar{n}(\omega_c)}{2\bar{n}(\omega_c)+1}-\frac{\bar{n}(\omega_h)}{2\bar{n}(\omega_h)+1}}\Bigg/\bras{1+\frac{1-e^{-2\alpha_c(2\bar{n}(\omega_c)+1)\tau_c}}{e^{2\alpha_h(2\bar{n}(\omega_h)+1)\tau_h}-1}},\\
z_h&=[2\bar{n}(\omega_h)+1]\bra{\frac{\bar{n}(\omega_h)}{2\bar{n}(\omega_h)+1}-\frac{\bar{n}(\omega_c)}{2\bar{n}(\omega_c)+1}}\Bigg/\bras{1+\frac{1-e^{-2\alpha_h(2\bar{n}(\omega_h)+1)\tau_h}}{e^{2\alpha_c(2\bar{n}(\omega_c)+1)\tau_c}-1}}.
\end{align}
\end{subequations}

\subsubsection{Thermodynamics and efficiency}
For each $1\le i\le 4$, let $\rho_i$ denote the density matrix at the beginning of process $i$.
Note that the density matrix is unchanged during the adiabatic processes, i.e., $\rho_1=\rho_2$ and $\rho_3=\rho_4$.
During an isochoric process, a heat quantity $Q_c=\Tr{H_c(\rho_2-\rho_3)}$ is transferred to the cold bath, or a heat quantity $Q_h=\Tr{H_h(\rho_1-\rho_4)}$ is extracted from the hot bath.
The total work $W$ extracted from the working medium is
\begin{equation}
-W=\int_0^{\tau_a}\Tr{\partial_tH(t)\rho(t)}dt+\int_{\tau_a+\tau_c}^{\tau-\tau_h}\Tr{\partial_tH(t)\rho(t)}dt=\Tr{\rho_1(H_c-H_h)}+\Tr{\rho_3(H_h-H_c)}.
\end{equation}
By conservation of energy, we have $-W+Q_h-Q_c=0$.
The efficiency $\eta$ is then defined as
\begin{equation}
\eta:=\frac{W}{Q_h}=1-\frac{Q_c}{Q_h}.
\end{equation}
The total entropy produced during the isochoric processes is
\begin{subequations}
\begin{align}
\Delta S_{\rm tot}^{h}&=\Delta S_{h}-\beta_hQ_h\ge 0,\\
\Delta S_{\rm tot}^{c}&=\Delta S_{c}+\beta_cQ_c\ge 0,
\end{align}
\end{subequations}
where $\Delta S_{h}=\Tr{\rho_4\ln\rho_4}-\Tr{\rho_1\ln\rho_1}$ and $\Delta S_{c}=\Tr{\rho_2\ln\rho_2}-\Tr{\rho_3\ln\rho_3}$ are the changes in the von Neumann entropy during the hot and cold isochoric processes, respectively.
As $\Delta S_{h}+\Delta S_{c}=0$, $\beta_cQ_c-\beta_hQ_h\ge 0$ follows from the second law of thermodynamics.
Using this inequality, one can prove that $\eta$ cannot exceed the Carnot efficiency, given by
\begin{equation}
\eta\le 1-\frac{\beta_h}{\beta_c}=:\eta_{\rm C}.
\end{equation}
In the following, we tighten the bound on the efficiency.
According to Eqs.~(5)~ and (6)~ in the main text, the total entropy productions during the isochoric processes are bounded from below by the distances $\mathsf{d}_{\rm T}(\cdot,\cdot)$ and $\mathsf{d}_{\rm E}(\cdot,\cdot)$ as follows:
\begin{align}
\Delta S_{\rm tot}^{h}&=\Delta S_{h}-\beta_hQ_h\ge\max\brab{\frac{\mathsf{d}_{\rm T}(\rho_1,\rho_4)^2}{4\tau_h\mathcal{A}_{\rm T}^h},\frac{\mathsf{d}_{\rm E}(\rho_1,\rho_4)^2}{\tau_h\mathcal{A}_{\rm E}^h}},\label{eq:bound.hot}\\
\Delta S_{\rm tot}^{c}&=\Delta S_{c}+\beta_cQ_c\ge\max\brab{\frac{\mathsf{d}_{\rm T}(\rho_2,\rho_3)^2}{4\tau_c\mathcal{A}_{\rm T}^c},\frac{\mathsf{d}_{\rm E}(\rho_2,\rho_3)^2}{\tau_c\mathcal{A}_{\rm E}^c}}.\label{eq:bound.cold}
\end{align}
Here, $\mathcal{A}_{\rm T}^k:=\alpha_k(2\bar{n}(\omega_k)+1)$ and $\mathcal{A}_{\rm E}^k:=\omega_k^2\alpha_k(2\bar{n}(\omega_k)+1)$ for $k=h$ or $c$.
From Eqs.~\eqref{eq:bound.hot} and \eqref{eq:bound.cold}, we obtain
\begin{equation}
\beta_cQ_c-\beta_hQ_h\ge\max\brab{\frac{\mathsf{d}_{\rm T}(\rho_1,\rho_4)^2}{4\tau_h\mathcal{A}_{\rm T}^h},\frac{\mathsf{d}_{\rm E}(\rho_1,\rho_4)^2}{\tau_h\mathcal{A}_{\rm E}^h}}+\max\brab{\frac{\mathsf{d}_{\rm T}(\rho_2,\rho_3)^2}{4\tau_c\mathcal{A}_{\rm T}^c},\frac{\mathsf{d}_{\rm E}(\rho_2,\rho_3)^2}{\tau_c\mathcal{A}_{\rm E}^c}}=:\mathfrak{g}.
\end{equation}
Consequently, a tighter bound on $\eta$ is obtained as
\begin{equation}
\eta\le\eta_{\rm C}-\frac{\mathfrak{g}}{\beta_cQ_h}=:\eta_{\rm G}.
\end{equation}

\section{Classical Markov jump processes}
\subsection{Alternative expression of the classical master equation}

We now show that the master equation $\Dr{\mbm{p}}=\mathsf{R}\mbm{p}$ can be expressed as $\Dr{\mbm{p}}=\mathsf{K}_p\mbm{f}$, where $\mathsf{R}=[R_{mn}]$ with $R_{nn}=-\sum_{m(\neq n)}R_{mn}$, $\mathsf{K}_p=\sum_{1\le n<m\le N}R_{nm}p^{\rm eq}_m\Phi\bra{\frac{p_n}{p^{\rm eq}_n},\frac{p_m}{p^{\rm eq}_m}}\mathsf{E}_{nm}$, and $\mbm{f}=-\nabla_{\mbm{p}} D(\mbm{p}||\mbm{p}^{\rm eq})$.
Here, $\nabla_{\mbm{p}}:=[\partial_{p_1},\dots,\partial_{p_N}]^\top$.
Specifically, we need to show that
\begin{equation}
(\mathsf{K}_p\mbm{f})_n=\sum_{m(\neq n)}[R_{nm}p_m-R_{mn}p_n]\label{eq:mastereq.row}
\end{equation}
holds for all $n$.
Indeed, using the relations $f_n=-(\ln p_n-\lnp^{\rm eq}_n - 1)$ and $R_{nm}p^{\rm eq}_m=R_{mn}p^{\rm eq}_n$, Eq.~\eqref{eq:mastereq.row} can be verified as follows:
\begin{subequations}
\begin{align}
(\mathsf{K}_p\mbm{f})_n&=\sum_{m(\neq n)}R_{nm}p^{\rm eq}_m\Phi\bra{\frac{p_n}{p^{\rm eq}_n},\frac{p_m}{p^{\rm eq}_m}}(\mathsf{E}_{nm}\mbm{f})_n\\
&=\sum_{m(\neq n)}R_{nm}p^{\rm eq}_m\Phi\bra{\frac{p_n}{p^{\rm eq}_n},\frac{p_m}{p^{\rm eq}_m}}(f_n-f_m)\\
&=\sum_{m(\neq n)}R_{nm}p^{\rm eq}_m\frac{p_n/p^{\rm eq}_n-p_m/p^{\rm eq}_m}{\ln p_n-\lnp^{\rm eq}_n-\ln p_m+\lnp^{\rm eq}_m}(\ln p_m-\lnp^{\rm eq}_m-\ln p_n+\lnp^{\rm eq}_n)\\
&=\sum_{m(\neq n)}R_{nm}p^{\rm eq}_m\bra{\frac{p_m}{p^{\rm eq}_m}-\frac{p_n}{p^{\rm eq}_n}}\\
&=\sum_{m(\neq n)}[R_{nm}p_m-R_{mn}p_n].
\end{align}
\end{subequations}

\subsection{Properties of the matrix $\mathsf{K}_p$}
The matrix $\mathsf{K}_p$ is symmetric and positive semi-definite.
Its properties are given below.
\begin{lemma}
For an arbitrary distribution $\mbm{p}$ satisfying $p_n>0$ for all $n$, ${\rm ker}(\mathsf{K}_p)=\{\mbm{v}\in\mathbb{R}^{N\times 1}~|~\mbm{v}\propto\mbm{1}\}$.
\end{lemma}
\begin{proof}
As the system is irreducible, there exists a set of $N-1$ unordered pairs, $\mathcal{Y}=\{(i,j)\,|\,R_{ij}\neq 0\}$, such that for arbitrary states $n\neq m$, there is a path $n=i_0\to i_1\to\dots\to i_k=m$ and $(i_{l},i_{l+1})\in\mathcal{Y}$ for all $0\le l\le k-1$.
Assuming $\mbm{v}\in{\rm ker}(\mathsf{K}_p)$, we have
\begin{equation}
0=\avg{\mbm{v},\mathsf{K}_p\mbm{v}}=\sum_{1\le n<m\le N}R_{nm}p^{\rm eq}_m\Phi\bra{\frac{p_n}{p^{\rm eq}_n},\frac{p_m}{p^{\rm eq}_m}}(v_m-v_n)^2.
\end{equation}
This expression means that $v_i-v_j=0$ for all $(i,j)\in\mathcal{Y}$, or equivalently, $\mbm{v}\propto\mbm{1}$.
\end{proof}

\begin{lemma}
There exists a vector $\mbm{v}$ for which $\Dr{\mbm{p}}=\mathsf{K}_p\mbm{v}$.
Such a vector is unique under the condition $\avg{\mbm{1},\mbm{v}}=0$.
\end{lemma}
\begin{proof}
For any $\mbm{v}$ satisfying $\mathsf{K}_p\mbm{v}=0$ [i.e., $\mbm{v}\in {\rm ker}(\mathsf{K}_p)$], then $\mbm{v}\propto\mbm{1}\Rightarrow\mbm{v}^\top\Dr{\mbm{p}}=0$; equivalently, $\Dr{\mbm{p}}\in{\rm ker}(\mathsf{K}_p)^\perp$.
According to the Fredholm alternative, the equation $\Dr{\mbm{p}}=\mathsf{K}_p\mbm{v}$ always has a nonzero solution $\mbm{v}$.
Defining $\overline{\mbm{v}}:=\mbm{v}-N^{-1}\avg{\mbm{1},\mbm{v}}\mbm{1}$, we can write $\Dr{\mbm{p}}=\mathsf{K}_p\overline{\mbm{v}}$ and $\avg{\mbm{1},\overline{\mbm{v}}}=0$.
Assume that there exist two solutions $\mbm{v}_1$ and $\mbm{v}_2$ satisfying $\avg{\mbm{1},\mbm{v}_1}=\avg{\mbm{1},\mbm{v}_2}=0$.
We then have $\mathsf{K}_p(\mbm{v}_1-\mbm{v}_2)=0\Rightarrow\mbm{v}_1-\mbm{v}_2=c\mbm{1}$ for some $c\in\mathbb{R}$.
Moreover, $\avg{\mbm{1},\mbm{v}_1-\mbm{v}_2}=0\Rightarrow Nc=0\Rightarrow c=0$, which proves the uniqueness of $\mbm{v}$.
\end{proof}

\subsection{Geodesic equation of the modified Wasserstein distance}
We here derive the geodesic equation that determines the shortest path between two distributions $\mbm{p}_0$ and $\mbm{p}_\tau$.
To this end, we consider the following functional, which is minimized along the geodesic path $\{\mbm{p}(t)\}_{0\le t\le\tau}$:
\begin{equation}
\mathcal{J}[\mbm{p}(t)]=\int_0^\tau\avg{\mbm{v}(t),\mathsf{K}_p\mbm{v}(t)}dt,
\end{equation}
where $\mbm{v}(t)$ and $\mbm{p}(t)$ are related through $\Dr{\mbm{p}}(t)=\mathsf{K}_p\mbm{v}(t)$.
Consider an arbitrary perturbation path $\{\mbm{q}(t)\}_{0\le t\le \tau}$ that satisfies $\mbm{q}(0)=\mbm{q}(\tau)=0$ and $\sum_nq_n(t)=0$ for all $0\le t\le \tau$.
Because the functional $\mathcal{J}[\mbm{\gamma}(t)]$ is minimized when $\mbm{\gamma}=\mbm{p}$, the function $\Theta(\epsilon)=\mathcal{J}[\mbm{p}(t)+\epsilon \mbm{q}(t)]$ has a minimum at $\epsilon=0$, so $\Theta'(0)=0$.
The functional evaluated at $\mbm{\gamma}=\mbm{p}+\epsilon\mbm{q}$ can be written as
\begin{align}
\mathcal{J}[\mbm{p}(t)+\epsilon \mbm{q}(t)]=\int_0^\tau\avg{\mbm{\vartheta}(t),\mathsf{K}_{p+\epsilon q}\mbm{\vartheta}(t)}dt,\label{eq:functional}
\end{align}
where $\mbm{\vartheta}(t)$ is determined from $\Dr{\mbm{p}}(t)+\epsilon\Dr{\mbm{q}}(t)=\mathsf{K}_{p+\epsilon q}\mbm{\vartheta}(t)$.
From Eq.~\eqref{eq:functional}, we have
\begin{equation}
0=\Theta'(0)=\int_0^\tau\bras{\avg{\partial_\epsilon\mbm{\vartheta}(t),\mathsf{K}_p\mbm{v}(t)}+\avg{\mbm{v}(t),\partial_\epsilon\mathsf{K}_{p+\epsilon q}\mbm{v}(t)}+\avg{\mbm{v}(t),\mathsf{K}_p\partial_\epsilon\mbm{\vartheta}(t)}}_{\epsilon=0}dt.\label{eq:pardiff.eps}
\end{equation}
Hereafter, we omit the notation of evaluating at $\epsilon=0$ for conciseness.
The first and third terms in Eq.~\eqref{eq:pardiff.eps} are equal by symmetry of $\mathsf{K}_p$; that is, $\avg{\partial_\epsilon\mbm{\vartheta}(t),\mathsf{K}_p\mbm{v}(t)}=\avg{\mbm{v}(t),\mathsf{K}_p\partial_\epsilon\mbm{\vartheta}(t)}$.
Taking the partial derivative of both sides of $\Dr{\mbm{p}}(t)+\epsilon\Dr{\mbm{q}}(t)=\mathsf{K}_{p+\epsilon q}\mbm{\vartheta}(t)$ with respect to $\epsilon$ and evaluating at $\epsilon=0$, we obtain
\begin{equation}
\Dr{\mbm{q}}(t)=\partial_\epsilon\mathsf{K}_{p+\epsilon q}\mbm{v}(t)+\mathsf{K}_p\partial_\epsilon\mbm{\vartheta}(t)\Rightarrow\avg{\mbm{v}(t),\mathsf{K}_p\partial_\epsilon\mbm{\vartheta}(t)}=\avg{\mbm{v}(t),\Dr{\mbm{q}}(t)}-\avg{\mbm{v}(t),\partial_\epsilon\mathsf{K}_{p+\epsilon q}\mbm{v}(t)}.\label{eq:pardiff.master}
\end{equation}
From Eqs.~\eqref{eq:pardiff.eps} and \eqref{eq:pardiff.master}, we have
\begin{equation}
0=\int_0^\tau\bras{2\avg{\mbm{v}(t),\Dr{\mbm{q}}(t)}-\avg{\mbm{v}(t),\partial_\epsilon\mathsf{K}_{p+\epsilon q}\mbm{v}(t)}}dt=-\int_0^\tau\bras{2\avg{\Dr{\mbm{v}}(t),\mbm{q}(t)}+\avg{\mbm{v}(t),\partial_\epsilon\mathsf{K}_{p+\epsilon q}\mbm{v}(t)}}dt.\label{eq:tmp1}
\end{equation}
Since
\begin{equation}
\partial_\epsilon\mathsf{K}_{p+\epsilon q}=\sum_{1\le n<m\le N}R_{nm}p^{\rm eq}_m\partial_\epsilon\Phi\bra{\frac{p_n+\epsilon q_n}{p^{\rm eq}_n},\frac{p_m+\epsilon q_m}{p^{\rm eq}_m}}\mathsf{E}_{nm},
\end{equation}
we have
\begin{equation}
\avg{\mbm{v}(t),\partial_\epsilon\mathsf{K}_{p+\epsilon q}\mbm{v}(t)}=\sum_{m,n}R_{mn}[v_m(t)-v_n(t)]^2\Psi\bra{\frac{p_n(t)}{p^{\rm eq}_n(t)},\frac{p_m(t)}{p^{\rm eq}_m(t)}}q_n(t)=\avg{\mbm{r}(t),\mbm{q}(t)},\label{eq:tmp2}
\end{equation}
where $\Psi(x,y)=[x-\Phi(x,y)]/[x(\ln x-\ln y)]$ and $r_n(t):=\sum_{m}R_{mn}[v_m(t)-v_n(t)]^2\Psi\bra{p_n(t)/p^{\rm eq}_n(t),p_m(t)/p^{\rm eq}_m(t)}$.
From Eqs.~\eqref{eq:tmp1} and \eqref{eq:tmp2}, we have
\begin{equation}
\int_0^\tau\avg{2\Dr{\mbm{v}}(t)+\mbm{r}(t),\mbm{q}(t)}dt=0.
\end{equation}
Because $\{\mbm{q}(t)\}_{0\le t\le\tau}$ is an arbitrary perturbation path, the term in the inner product must be zero, i.e., $\Dr{\mbm{v}}(t)+\mbm{r}(t)/2=0$.
Finally, the geodesic equation that determines the shortest path between states $\mbm{p}_0$ and $\mbm{p}_\tau$ is obtained as follows:
\begin{equation}
\begin{cases}
\Dr{\mbm{p}}(t)-\mathsf{K}_p\mbm{v}(t)=0,\\
\Dr{\mbm{v}}(t)+\displaystyle\frac{1}{2}\mbm{r}(t)=0,\label{eq:Wasserstein.geodesic.equ}
\end{cases}
\end{equation}
with boundary conditions $\mbm{p}(0)=\mbm{p}_0$ and $\mbm{p}(\tau)=\mbm{p}_\tau$.

\subsection{Lower bound of the modified Wasserstein distance in terms of the total variation distance}
Here we derive the lower bound of the Wasserstein distance in terms of the total variation distance, $\mathsf{d}_{\rm V}(\mbm{p},\mbm{q})=\sum_n|p_n-q_n|$.
In variational form, the distance $\mathsf{d}_{\rm V}(\mbm{p},\mbm{q})$ can be expressed as
\begin{equation}
\mathsf{d}_{\rm V}(\mbm{p},\mbm{q})=\max_{\|\mbm{w}\|_\infty\le 1}\,\{\mbm{w}^\top(\mbm{p}-\mbm{q})\}=\max_{\|\mbm{w}\|_\infty\le 1}\,\avg{\mbm{w},\mbm{p}-\mbm{q}},
\end{equation}
where the maximum is taken over all real vectors $\mbm{w}=[w_1,\dots,w_N]^\top$ and $\|\mbm{w}\|_\infty:=\max_n|w_n|$.
Equality is attained when $w_n={\rm sign}(p_n-q_n)$.
Here, the sign function ${\rm sign}(x)$ of $x$ is defined as ${\rm sign}(x)=1$ for $x\ge 0$ and $-1$ otherwise.
By definition of the modified Wasserstein distance, given a fixed positive number $\delta>0$, there exists a smooth curve $\mbm{p}(t)$ with end points $\mbm{p}_0$ and $\mbm{p}_\tau$ such that
\begin{equation}
\tau\int_0^\tau\avg{\mbm{v},\mathsf{K}_p\mbm{v}}dt\le\mathcal{W}_{\rm c}(\mbm{p}_0,\mbm{p}_\tau)^2+\delta.
\end{equation}
Here, $\mbm{v}(t)\in\mathbb{R}^{N\times 1}$ is determined from $\Dr{\mbm{p}}(t)=\mathsf{K}_p\mbm{v}(t)$.
For an arbitrary vector $\mbm{w}$ with $\|\mbm{w}\|_\infty\le 1$, we have
\begin{subequations}\label{eq:bound.tot.var.dist}
\begin{align}
\avg{\mbm{w},\mbm{p}_\tau-\mbm{p}_0}&=\int_0^\tau\avg{\mbm{w},\mathsf{K}_p\mbm{v}}dt\\
&\le\bra{\int_0^\tau\avg{\mbm{w},\mathsf{K}_p\mbm{w}}dt}^{1/2}\bra{\int_0^\tau\avg{\mbm{v},\mathsf{K}_p\mbm{v}}dt}^{1/2}\\
&\le\bra{\tau^{-1}\int_0^\tau\avg{\mbm{w},\mathsf{K}_p\mbm{w}}dt}^{1/2}\bra{\mathcal{W}_{\rm c}(\mbm{p}_0,\mbm{p}_\tau)^2+\delta}^{1/2}.\label{eq:L1dist.bound.tmp1}
\end{align}
\end{subequations}
To further bound the first term in Eq.~\eqref{eq:L1dist.bound.tmp1}, we apply the inequalities $\Phi(x,y)\le(x+y)/2$ and $(w_n-w_m)^2\le 4$, and obtain
\begin{subequations}
\begin{align}
\avg{\mbm{w},\mathsf{K}_p\mbm{w}}&=\sum_{m>n}R_{nm}p^{\rm eq}_m\Phi\bra{\frac{p_n}{p^{\rm eq}_n},\frac{p_m}{p^{\rm eq}_m}}\avg{\mbm{w},\mathsf{E}_{nm}\mbm{w}}\\
&=\sum_{m>n}R_{nm}p^{\rm eq}_m\Phi\bra{\frac{p_n}{p^{\rm eq}_n},\frac{p_m}{p^{\rm eq}_m}}(w_n-w_m)^2\label{eq:class.temp1} \\
&\le 2\sum_{m>n}R_{nm}p^{\rm eq}_m\bra{\frac{p_n}{p^{\rm eq}_n}+\frac{p_m}{p^{\rm eq}_m}}\\
&=2\sum_{m>n}[R_{nm}p_m+R_{mn}p_n].
\end{align}
\end{subequations}
Consequently, we have
\begin{equation}
\mathcal{W}_{\rm c}(\mbm{p}_0,\mbm{p}_\tau)^2+\delta\ge\frac{\avg{\mbm{w},\mbm{p}_0-\mbm{p}_\tau}^2}{2\tau^{-1}\int_0^\tau\sum_{m>n}[R_{nm}(t)p_m(t)+R_{mn}(t)p_n(t)]dt}.
\end{equation}
Taking the maximum over all $\mbm{w}$ and the limit $\delta\to 0$, we obtain
\begin{equation}
\mathcal{W}_{\rm c}(\mbm{p}_0,\mbm{p}_\tau)^2\ge\frac{\mathsf{d}_{\rm V}(\mbm{p}_0,\mbm{p}_\tau)^2}{2\mathcal{A}_{\rm V}},\label{eq:ClaWasDisLowBou}
\end{equation}
where $\mathcal{A}_{\rm V}:=\tau^{-1}\int_0^\tau\sum_{m\neq n}R_{mn}(t)\gamma_n(t)dt$ is the average dynamical activity along the geodesic path $\{\mbm{\gamma}(t)\}_{0\le t\le \tau}$.
The dynamical activity characterizes the time scale of the system.
As it indicates the time-symmetric changes in the system, it plays important roles in nonequilibrium phenomena \cite{Maes.2020.PR}.
From Eqs.~(11)~and \eqref{eq:ClaWasDisLowBou}, the classical speed limits of the state transformation are obtained as
\begin{equation}
\tau\ge\frac{\mathcal{W}_{\rm c}(\mbm{p}(0),\mbm{p}(\tau))^2}{\Delta S_{\rm tot}}\ge\frac{\mathsf{d}_{\rm V}(\mbm{p}(0),\mbm{p}(\tau))^2}{2\Delta S_{\rm tot}\mathcal{A}_{\rm V}}.\label{eq:ClaSpeLim}
\end{equation}
These inequalities imply a trade-off relation between the time needed to transform the system state and the physical quantities such as entropy production and dynamical activity.
Specifically, a fast transformation necessitates high dissipation and frenesy.
The last bound in the inequality \eqref{eq:ClaSpeLim} is analogous to, but distinct from, a bound derived in Ref.~\cite{Shiraishi.2018.PRL}.
In the earlier study, $\mathcal{A}_{\rm V}$ is replaced by the average dynamical activity along the path described by the time evolution of the system.

We can derive a bound that is tighter than Eq.~\eqref{eq:ClaWasDisLowBou}.
First, we divide the set of states $\mathcal{N}=\{1,\dots,N\}$ into two subsets $\mathcal{N}=\mathcal{X}_-\cup\mathcal{X}_+$, where
\begin{align}
\mathcal{X}_-&:=\{n~|~1\le n\le N,~ p_n(0)<p_n(\tau)\},\\
\mathcal{X}_+&:=\{n~|~1\le n\le N,~ p_n(0)\ge p_n(\tau)\}.
\end{align}
For convenience, we define ${\rm par}(n):=-1$ if $n\in\mathcal{X}_-$ and ${\rm par}(n):=1$ if $n\in\mathcal{X}_+$.
Then, $\mathsf{d}_{\rm V}(\mbm{p}_0,\mbm{p}_\tau)=\avg{\mbm{w},\mbm{p}_0-\mbm{p}_\tau}$, where $w_n={\rm par}(n)$.
From Eq.~\eqref{eq:class.temp1}, noticing that $(w_n-w_m)^2=0$ if $n$ and $m$ belong to the same subset $\mathcal{X}_-$ or $\mathcal{X}_+$, and $(w_n-w_m)^2=4$ if $n$ and $m$ belong to different subsets, we have
\begin{subequations}
\begin{align}
\avg{\mbm{w},\mathsf{K}_p\mbm{w}}&\le 2\sum_{m>n~{\&}~{\rm par}(m)\neq{\rm par}(n)}R_{nm}p^{\rm eq}_m\bra{\frac{p_n}{p^{\rm eq}_n}+\frac{p_m}{p^{\rm eq}_m}}\\
&=2\sum_{m\in\mathcal{X}_-,\,n\in\mathcal{X_+}}[R_{mn}p_n+R_{nm}p_m]=2\sum_{{\rm par}(m)\neq{\rm par}(n)}R_{mn}p_n.
\end{align}
\end{subequations}
Subsequently, we obtain a tighter bound as
\begin{equation}
\mathcal{W}_{\rm c}(\mbm{p}_0,\mbm{p}_\tau)^2\ge\frac{\mathsf{d}_{\rm V}(\mbm{p}_0,\mbm{p}_\tau)^2}{2\mathcal{A}_{\rm V}^{\rm par}},
\end{equation}
where $\mathcal{A}_{\rm V}^{\rm par}:=\tau^{-1}\int_0^\tau\sum_{{\rm par}(m)\neq{\rm par}(n)}R_{mn}(t)\gamma_n(t)dt$ is the average of the {\it partial} dynamical activity along the geodesic path $\{\mbm{\gamma}(t)\}_{0\le t\le \tau}$.
Since $\mathcal{A}_{\rm V}^{\rm par}$ involves only transition rates between states in $\mathcal{X}_-$ and $\mathcal{X}_+$, $\mathcal{A}_{\rm V}^{\rm par}\le \mathcal{A}_{\rm V}$.

Following the same approach (i.e., applying Eq.~\eqref{eq:bound.tot.var.dist} for the path described by the system dynamics), we obtain
\begin{subequations}
\begin{align}
\avg{\mbm{w},\mbm{p}_\tau-\mbm{p}_0}&=\int_0^\tau\avg{\mbm{w},\mathsf{K}_p\mbm{h}}dt\\
&\le\bra{\int_0^\tau\avg{\mbm{w},\mathsf{K}_p\mbm{w}}dt}^{1/2}\bra{\int_0^\tau\avg{\mbm{h},\mathsf{K}_p\mbm{h}}dt}^{1/2}\\
&=\bra{\int_0^\tau\avg{\mbm{w},\mathsf{K}_p\mbm{w}}dt}^{1/2}\sqrt{\Delta S_{\rm tot}}.
\end{align}
\end{subequations}
Analogously, we can prove that
\begin{equation}
\tau\ge\frac{\mathsf{d}_{\rm V}(\mbm{p}(0),\mbm{p}(\tau))^2}{2\Delta S_{\rm tot}\mathcal{A}^{\rm par}},\label{eq:tighter.csl}
\end{equation}
where $\mathcal{A}^{\rm par}:=\tau^{-1}\int_0^\tau\sum_{{\rm par}(m)\neq{\rm par}(n)}R_{mn}(t)p_n(t)dt$ is the average of the {\it partial} dynamical activity along the path described by the time evolution of the system.
Obviously, Eq.~\eqref{eq:tighter.csl} is tighter than the bound reported in Ref.~\cite{Shiraishi.2018.PRL}.
The newly obtained bound indicates that the speed of the state transformation is not constrained by the total dynamical activity, but by the partial dynamical activity induced by transitions between states in $\mathcal{X}_-$ and $\mathcal{X}_+$.

\subsection{Thermalization process of a three-level system}
Here we illustrate the derived bound on the thermalization process of a three-level system.
The transition rates are time-independent and equal to
\begin{equation}
R_{mn}=w_{mn}e^{\beta(\mathcal{E}_n-\mathcal{E}_m)/2}{\rm sech}[\beta(\mathcal{E}_n-\mathcal{E}_m)/2],
\end{equation}
where $w_{mn}=w_{nm}$ are nonnegative constants.
Evidently, the transition rates satisfy the detailed balance conditions $R_{mn}p^{\rm eq}_n=R_{nm}p^{\rm eq}_m$.
According to Eq.~(11)~in the main text, the entropy production is bounded from below by the modified Wasserstein distance as
\begin{equation}
\Delta S_{\rm tot}\ge \frac{\mathcal{W}_{\rm c}(\mbm{p}(0),\mbm{p}(\tau))^2}{\tau}.
\end{equation}
The total entropy production can be explicitly expressed as $\Delta S_{\rm tot}=D(\mbm{p}(0)||\mbm{p}^{\rm eq})-D(\mbm{p}(\tau)||\mbm{p}^{\rm eq})$.
In thermalization processes satisfying the detailed balance conditions, Ref.~\cite{Shiraishi.2019.PRL} proved that the relative entropy satisfies the reverse triangle inequality:
\begin{equation}
D(\mbm{p}(0)||\mbm{p}^{\rm eq})\ge D(\mbm{p}(0)||\mbm{p}(\tau))+D(\mbm{p}(\tau)||\mbm{p}^{\rm eq}).
\end{equation}
Subsequently, the entropy production during thermalization processes is bounded from below by an information-theoretical quantity of the initial and final states, $\Delta S_{\rm tot}\ge D(\mbm{p}(0)||\mbm{p}(\tau))$.
For fixed transition rates, Fig.~\ref{fig:three.level} plots the entropy production, modified Wasserstein distance, and relative entropy as functions of time $\tau$.
In this figure, the distance term $\mathcal{W}_{\rm c}^2(\mbm{p}(0),\mbm{p}(\tau))/\tau$ and the relative entropy always lie below the entropy production $\Delta S_{\rm tot}$.
The modified Wasserstein distance is tight in the short-time regime, whereas the relative entropy saturates in the long-time limit.
Therefore, these two bounds complementarily characterize the irreversibility in thermalization processes.
\begin{figure}
\centering
	\includegraphics[width=0.8\linewidth]{FIG_Supp1.eps}
	\protect\caption{Numerical verification of the derived bound. $\Delta S_{\rm tot}$ (solid line), $\mathcal{W}_{\rm c}(\mbm{p}(0),\mbm{p}(\tau))^2/\tau$ (dashed line), and $D(\mbm{p}(0)||\mbm{p}(\tau))$ (dash-dotted line) during the thermalization process of a three-level system. Parameters are set as $\beta=1,w_{12}=1,w_{23}=2,w_{13}=0,\mathcal{E}_1=3,\mathcal{E}_2=-0.5,\mathcal{E}_3=6$, and $\mbm{p}(0)=[0.1,0.1,0.8]^\top$.}\label{fig:three.level}
\end{figure}

redpajama_set_name: RedPajamaArXiv


text:
David Kevin Woolf
Associate Professor, School of Energy, Geoscience, Infrastructure and Society
Associate Professor, School of Energy, Geoscience, Infrastructure and Society, Institute for Life and Earth Sciences
Associate Professor, Research Centres and Themes, Energy Academy
Associate Professor, Research Centres and Themes, International Centre for Island Technology
EmailD.K.Woolf@hw.ac.uk
Willing to speak to media
1 Entry for encyclopedia/dictionary
Chapter (peer-reviewed)
Bubbles and their role in gas exchange
Woolf, D. K., 1997, The Sea Surface and Global Change. Liss, P. S. & Duce, R. A. (eds.). Cambridge, UK: Cambridge University Press, p. 173-206 34 p.
gas exchange
breaking wave
Report Group 1 - Physical processes in the microlayer and the air-sea exchange of trace gases
Liss, P. S., Watson, A. J., Bock, E. J., Jahne, B., Asher, W. E., Frew, N. M., Hasse, L., Korenowski, G. M., Merlivat, L., Phillips, L. F., Schluessel, P. & Woolf, D. K., 1997, The Sea Surface and Global Change. Liss, P. S. & Duce, R. A. (eds.). Cambridge University Press, p. 1-33 33 p.
Radar altimetry: introduction and application to air-sea interaction
Woolf, D. K. & Gommenginger, C., 2008, Remote sensing of the European seas. Barale, V. & Gade, M. (eds.). Springer, p. 283-294 12 p.
radar altimetry
air-sea interaction
altimetry
Characterisation of a tidal energy site: hydrodynamics and seabed structure
Easton, M., Woolf, D. K., Harendza, A. & Jackson, A., 2011, 9th European Wave and Tidal Energy Conference (EWTEC 2011). Sustainable Energy Research Group, University of Southampton, 10 p.
sand wave
environmental constraint
site characterization

redpajama_set_name: RedPajamaCommonCrawl


text:
RegistroBR
==========

Registro.br offers a fully-functional python client, available from ftp://ftp.registro.br/pub/isavail/ . This is a modified library version.

redpajama_set_name: RedPajamaGithub


text:
require 'spec_helper'
require 'ruby_util/hash_with_method_access_shared'
require 'tempfile'
require 'fileutils'
require 'ruby_util/string'

describe Xenuti::Report do
  let(:report) { Xenuti::Report.new }
  let(:tmp) do
    tmp = Dir.mktmpdir
    at_exit do
      FileUtils.rm_rf(tmp)
    end
    tmp
  end

  # This does not work, as constructor initializes the hash
  # It is fine though, since method access is tested on other places anyway
  # it_behaves_like 'hash with method access', Xenuti::Report.new

  describe '::prev_report' do
    it 'should return the previous report' do
      Dir.mkdir tmp + '/reports'
      newer = '2014-05-30T15:38:04+02:00'
      older = '2014-05-30T15:37:04+02:00'

      Dir.mkdir tmp + '/reports/' + newer
      File.open(tmp + '/reports/' + newer + '/report.yml', 'w+') do |file|
        file.write <<-EOF.unindent
        --- !ruby/hash:Xenuti::Report
        scan_info:
          version: 0.0.1
          start_time: 2014-05-30 15:37:04.001 +02:00
        scanner_reports: []
        config: {}
        name: :new
        EOF
      end

      Dir.mkdir tmp + '/reports/' + older
      File.open(tmp + '/reports/' + older + '/report.yml', 'w+') do |file|
        file.write <<-EOF.unindent
        --- !ruby/hash:Xenuti::Report
        scan_info:
          version: 0.0.1
          start_time: 2014-05-27 15:37:04.002 +02:00
        scanner_reports: []
        config: {}
        name: :old
        EOF
      end

      config = Xenuti::Config.from_hash('general' => { 'workdir' => tmp })
      expect(Xenuti::Report.prev_report(config).name).to eq(:new)
    end

    it 'should return nil when directory does not contain any report yet' do
      config = Xenuti::Config.from_hash('general' => { 'workdir' => FIXTURES })
      expect(Xenuti::Report.prev_report(config)).to be_eql(nil)
    end
  end

  describe '::diff' do
    it 'should diff with older report correctly' do
      # TODO
      # fail 'Implement this test'
    end
  end

  describe '::reports_dir' do
    it 'output should be the same across multiple calls' do
      config = Xenuti::Config.from_hash('general' => { 'workdir' => tmp })
      reports_dir1 = report.report_dir(config)
      sleep 1
      reports_dir2 = report.report_dir(config)
      expect(reports_dir1).to be_eql(reports_dir2)
    end
  end

  describe '#save and ::load' do
    it 'report should be identical after saving and loading again' do
      config = Xenuti::Config.from_hash('general' => { 'workdir' => tmp })
      report['config'] = config
      report.scan_info.start_time = Time.now
      report.save(config)
      latest = Xenuti::Report.prev_report(config)
      expect(latest).to be_eql(report)
    end
  end

  describe '#duration' do
    it 'should compute duration correctly' do
      report.scan_info.start_time = Time.new(2008, 6, 21, 13, 30, 1.1)
      report.scan_info.end_time = Time.new(2008, 6, 21, 13, 30, 2.3)
      expect(report.duration).to be_eql(1.2)
    end
  end
end


The Football Association have defended their anti-doping rules after Channel 4's Dispatches programme revealed names of several players who have failed drugs tests.
It is the FA's policy to keep private the names of players who fail tests for social drugs out of competition, although punishments of up to six months for first-time offenders can be handed out.
The FA want players to be allowed privacy to seek help for their problems, where required, and stress there is no requirement on them to identify those who have failed tests for recreational drugs away from game time.
Players with Premier League experience, who are no longer active in the top flight, were named in the Dispatches programme as having tested positive for cocaine.
The FA stress they are working to eliminate illegal substances of all types, and underlined that players who test positive for social drugs will be punished, even if their names are not disclosed.
Anyone who fails a test for a performance-enhancing drug is named, whether they are tested in or out of competition.
In a statement, the FA said: 'The FA operates a comprehensive anti-doping programme which is the largest of any sport in the UK. The FA prohibits all the doping offences listed in the World Anti-Doping Agency code and applies all the sanctions laid down in the code for those offences.
'In addition, the FA, supported by all the football stakeholders, recognise the issues that social drugs may cause and choose to go beyond the WADA Code by proactively testing all samples for social drugs, irrespective of whether the tests are conducted in or out of competition.
'Football is one of the only sports in the UK that ban social drugs at all times, and were the first to do so.
'Any player who tests positive for a social drug out of competition is charged and subject to a sanction which ordinarily includes a suspension from all football activity for a period of up to six months for a first time offence.
'They are also subject to target testing for a period of two years. The FA do not report the name of the player as this offence is not a WADA Code offence and privacy allows for the player to undergo any necessary rehabilitation and counselling.
'All England representative teams are subject to UEFA and FIFA regulations whilst Premier and Football League clubs and players are subject to strict FA whereabouts regulations.
'Players are drug tested on a no-advance notice basis. Testing can take place at matches, training grounds, players' home addresses and locations abroad. 70 per cent of all FA drug tests are conducted out of competition, as recognised by WADA as the most effective means of deterring and detecting doping.
'Any club or player in breach of the FA whereabouts regulations are subject to disciplinary processes.
David Howman, director general of the World Anti Doping Agency, believes the FA should consider naming all players who are found to have used banned drugs, no matter what they are found to have taken or when they are tested.
Howman, a sports lawyer who also held senior positions in rugby and tennis in his native New Zealand before joining WADA eight years ago, suggested the FA would do well to make their doping detection programme fully transparent.
He told the Daily Telegraph: 'Why does there have to be so much secrecy? When sports people get involved with substances that are ordinarily part of a criminal justice programme, what is so special about sport that it ought not to come into the public arena?
'There will of course be times when revealing a player's identity might seriously compromise his rehabilitation, particularly if he has a serious drug addiction or a medical condition. I accept the need for confidentiality in such cases.


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Can Prostate Cancer Screening Improve Men's Lives?
PSA Test: Overall Gain in Quality-of-Life Years May Not Apply to Individual Men
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Reviewed by Louise Chang, MD
Aug. 15, 2012 -- A European study suggests that men who undergo PSA prostate cancer screening will gain years of high-quality life.
But it's not quite that simple.
Those extra years of high-quality life showed up across big groups of screened men. Individual men may get different results -- especially if screening-associated risks such as impotence and incontinence bother them more than the possible benefit of a life free of prostate cancer.
Whether to get a PSA test is a tricky decision.
Urologists, who see men dying of late-stage prostate cancer, tend to favor PSA screening for all men as they reach middle age. They want to catch the cancers while they are still treatable.
Testing experts, including the U.S. Preventive Services Task Force, calculate that the real harms of prostate cancer over-diagnosis and over-treatment outweigh potential benefits. It's an ongoing debate.
The new study, published today in the New England Journal of Medicine, "moves the ball down the field quite a lot," according to medical decision-making expert Harold C. Sox, MD, professor of medicine at Dartmouth's Geisel School of Medicine.
Instead of trying to compare prostate cancer to treatment side effects, the new study looks at the quality of life in men who do and don't get PSA tests.
PSA Pluses and Minuses
Harry J. de Koning, MD, professor of public health and screening evaluation at the Netherlands' Erasmus Medical Center, led a team that analyzed data from a large European clinical trial of PSA screening. They translated all outcomes of regular PSA testing into quality life years gained or lost.
"I think prostate cancer screening has been shown to be effective, but there are a lot of negative side effects," de Koning tells WebMD. "What we now say is that a man age 55 to 69, if he is screened every four years, will gain something like 8.5 years of life. But we say you should diminish this by 20%, to about 6.7 years, because then you get healthy life years. It is a longer life, but a mix of good things and bad things."
De Koning and colleagues arrived at their calculation by assigning life-quality values to each possible outcome of PSA screening.
"So a person who just got a biopsy thinks, 'Today is 10% worse than my average day in life.' Maybe some say 20% or 40%. That is what we try to calculate," de Koning says. "Or if a man is getting palliative care for metastatic cancer and you ask how much worse this state is than normal, he may say it's 50% worse than two years ago when he was healthy."
These values, while reasonable and based on previous quality-of-life research, clearly would vary from person to person. They calculate that 1,000 men who receive PSA screening would gain 56 years of high-quality life. But they note that if slightly different values were used, these 1,000 men might lose as many as 21 quality life years or gain as many as 97.
That gap makes it impossible to recommend for or against PSA screening for all men, says Sox, whose editorial comments accompany the de Koning report in the NEJM.
"If a significant number of men could end up losing quality life years as a result of their feelings about the kind of side effects that can happen with prostate cancer treatment, a significant number of patients also would gain quality life years by not worrying about dying of prostate cancer," Sox tells WebMD. "So you can't have a general recommendation either for or against PSA testing. You have to talk to the patient and find out what his feelings are."
PSA Decision: Case by Case
That's almost exactly the current recommendation of the American Cancer Society. Unlike the U.S. Preventive Services Task Force, which based on its own risk/benefit analysis flatly advises men not to get the test, the ACS advises men to sit down with a medical professional and discuss all the risks and all the benefits.
That's not very satisfying to men who want protection from prostate cancer, but who don't want to gamble on a test that might lead to severe side effects from unnecessary treatment.
And there's the rub. De Koning and colleagues note that in the European study, 42% of cancers detected by PSA screening were "over-diagnosed" -- that is, they never would have done harm.
"Strategies to reduce over-diagnosis would seem to be necessary before screening can be generally advocated," they write.
Until this happens, Sox recommends improving our "meager" understanding of how men feel about the quality of their lives after treatment for prostate cancer.
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SOURCES: Harold C. Sox Jr., MD, professor of medicine, Geisel School of Medicine at Dartmouth, N.H. Harry de Koning, MD, professor of public health and screening evaluation, Erasmus Medical Center, Netherlands. Heijnsdijk, E. New England Journal of Medicine, Aug. 19, 2012. Sox, H.C. New England Journal of Medicine, Aug. 19, 2012.
©2012 WebMD, LLC. All Rights Reserved.
Screening Tests Every Man Should Have See Slideshow
Prostate Cancer Symptoms, PSA Test, Treatments
Prostate cancer is the most common cancer in men. Learn the signs and symptoms of prostate cancer, along with causes and treatments. Know the stages, survival rates and lower your risk of prostate cancer.
Read more: Prostate Cancer Symptoms, PSA Test, Treatments


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You can withdraw consent by refusing the test, we won't force you," Demello Rice said.* * *The law is undergoing a sea change in response to public pressure for greater protections Even if the machine turns out to be positive, this does not mean that you can be convicted for DWI. Houston has previously testified that, "There is no margin of error with the breath test because it is a singular event without any ability to determine error"Really, really? RFIRadio frequency interference (RFI) is electromagnetic interference that affects radio frequency bands from 100kHz to 5 GHz.
Flood, 523 So.2d 1180 (Fla. This is due in large part to a person holding their breath prior to blowing and then producing more volume and warmer breath. The first time you try them they are often impossible regardless of whether you have been drinking. Significance of Vatiation in Blood-Breath Partition Coefficient of Alcohol, British Medical Journal, 1976; Mason, M.
The New Hampshire regulations require that the simulator temperature be +/- .2 degrees Celsius again emphasizing the importance of temperature on breath testing. If you blow before you peak, then your breath alcohol level can be as much as 50% higher than your blood alcohol level, because breath alcohol comes from your arterial blood This means that for every particle of alcohol in your lungs, it assumes there are 2100 particles in your blood. Intoxilyzer 8000EN and the Common Challenges it Raises in Court In pre-trial motion hearings, the common goal is to request that the judge remove any mention of the breath test reading
What's the matter with Kansas? The email below is form Ms. We encourage our juries to try these tests back in the jury room so they can see how easy they are to fail. 5. In California, thousands of cases were thrown into confusion because of defective breathalyzers; Ventura County in southern California ended up dismissing at least 64 cases because of a glitch.The main culprit
The machine was malfunctioning, but gave a false test result without alerting the operator to the problem. Breath tests have an integral role in protecting the public from wanton drunk drivers who kill millions each year. A real DUI lawyer will have won at least a few jury trials in the last year to six months. This is the most common source of police error in DUI cases.
State, 210 Ga. Even if the Intoxilyzer machine is working perfectly when an individual blows into it, the results may not be accurate. You Can Beat a Breath or Blood Test Result DUI - It is Not Hopeless: Because there are many opportunities for a strong defense in cases with State-administered results, we regularly It is necessary to contact an experienced Las Vegas DUI defense lawyer if you had to blow into the Intoxilyzer 5000EN and received a positive test result.
Related Explore posts in the same categories: Uncategorized This entry was posted on July 9, 2012 at 3:39 am and is filed under Uncategorized. The CMI, Inc. If the operator fails to perform a proper 20 minute observation period, change a mouthpiece, label the independent sample tubes, types in the wrong simulator serial number or any other countless Back to top Las Vegas DUI Attorney Explains Problems with Intoxilyzer 5000EN The breath test machine currently used in the of Nevada is the Intoxilyzer 5000EN which is manufactured by a
Public distrust of police has continued to swell in recent years as controversial policies, such as stop and frisk, have made headlines for being discriminatory or violating basic privacy rights. Jones. App. 150, 345 S.E.2d 670 (1986); Oxley v. States now all use a .08 percent blood alcohol content reading as the legal intoxication limit, but there have been cases where drivers are arrested for blowing much lower than the legal
Breath Tests and DUI Defense Lawyer If you are arrested and have submitted to a breath alcohol test, contact a Las Vegas DUI Attorney to discuss possible defenses. State, 234 Ga. Second, was there enough evidence - probable cause - to arrest you for DUI in the first place? They are about getting in and out of court in as little time as possible.
Since the state failed to proffer an officer could rely on the suspect's statement and still comply with the protocol, the Court held the State had failed to meet its burden Expert testimony is especially important when other evidence such as the performance of the FST's is insufficient to help the jury come to a clear verdict. This is not science fiction. Breathing PatternsIn New Hampshire the police are trained to tell subject to blow when told and to keep blowing while there is a tone made by the machine.
For every breath degree over 34, the Intoxilyzer 5000 overestimates the breath alcohol content by about 6.8%. Machine ErrorNo machine is ever perfect and even the manufacturer CMI, Inc has brochures and sales materials available on line that detail that detail the Intoxilyzer 5000EN "is better than federal I would argue that most courts would say it is. Since then, refusal rates have dropped in 11 states and increased in 12 others, with refusals averaging between 19 percent and 25 percent.
Before you plea, make sure your lawyer has reviewed the police report and video tape. 10. When the prosecutor is unable to make such a showing for a particular instrument or a particular breath alcohol reading, then the defense attorney must be prepared to file and litigate Perry 118 P.3d 178, 2005 WL 2076580 (2005) ***UNPUBLISHED*** The Court of Appeals held the district court did not err in suppressing the results of a breath test based upon the Google it yourself. 4.
All Rights Reserved. 102 E 26th St Bryan, Texas 77803 979-823-9111 This publication and the information included in it are not intended to serve as a substitute for consultation with an Not all persons partition alcohol from their blood based upon this ratio. Likewise, the factory can make other repairs from the factory in Ohio via phone line. [Mr. Carey offers a free initial case evaluation, and will go over the facts of your case, your history, your rights and options.
See Department of Public Safety v. Alobaidi, T.A.A & Payne, J.P. Orr, 122 Ga. Field Sobriety Tests Are Not Reliable Indicators of Alcohol Impairment: The three most commonly used field sobriety evaluations are the Horizontal Gaze Nystagmus Test, or eye jerking test, the nine step
The roadside test is only "to give the officer an indication" of a driver's intoxication, Britta Demello Rice, a prosecutor for North Dakota's Burleigh County, told ThinkProgress. Studies show partition ratios as low as 834:1. The Intoxilyzer assumes the temperature of the person's exhaled breath is 34 degrees Centigrade. Intoxilyzers are used in more than 35 states.
If you submitted to a breath test machine after a DUI arrest in Clark County, Nevada or a surrounding area, then contact an experienced Las Vegas DUI attorney to discuss possible When a person makes an effort to blow (as evidenced by puffed cheeks and getting red in the face) a strong argument to exclude evidence of a refusal can be made. Mr. It is rarely known whether the person is still absorbing alcohol from their last drink when the breath test is administered. 4.
The Court also decided that drivers could only challenge the validity of a breathalyzer machine, not the supporting science.The legal argument is that drivers implied consent when they got a license, Coupled with the increase in police militarization exemplified by the Ferguson, Missouri protests after unarmed teen Michael Brown was shot and killed by a police officer, the public is questioning how If they never have or if its been more than a few years, you should be worried.
intoxilyzer margin of error
Perry 118 P.3d 178, 2005 WL 2076580 (2005) ***UNPUBLISHED*** The Court of Appeals held the district court did not err in suppressing the results of a breath test based upon the Houston's comments about no margin of error) for them then to weigh in the determination of a person?s innocence or guilt in a criminal trial. �To allow anything less would critically Indeed the "very same" case in one county, town or city may result in a very different outcome in another New York county, town, or c...
intoxilyzer 8000 margin of error
The machine operator may not know how to conduct the test. In Arizona, law agencies use the Intoxilyzer 8000 device. Professor Rolla N. Generated Wed, 19 Oct 2016 05:26:47 GMT by s_wx1126 (squid/3.5.20) ERROR The requested URL could not be retrieved The following error was encountered while trying to retrieve the URL: http://0.0.0.8/ Connection Pete Tarpon Springs South Pinellas Beaches West Pasco MORE NEIGHBORHOODS Editorials & letters Perspective Submit a letter to the editor Editori...
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Since it takes between 45 minutes and 3 hours for alcohol to be absorbed into the system, an individual's BAC may continue to rise for some time after he is stopped The testimony in this case clearly demonstrates that the Intoxilyzer 5000 EN does fall within this constitutional proscription. Physiological Aspect of Breath-Alcohol Measurement, Alcohl, Drugs & Driving, Vol.6 No. 2, A.W. In part that is because we are dealing with human beings. As will be discussed later, breath machines ...
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The key to the hotel's character is its commitment to the French Paradox, a lifestyle based on the suggested benefits of French living, including a study on the positive effects of wine consumption in fighting cardiovascular disease. The atmosphere is consequently one of simultaneous revelry and relaxation; it's likely to be a rare opportunity to eat, drink and be pampered with the tranquil beauty of a vineyard as an inviting backdrop.
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The majority of the 61 rooms or suites have breathtaking views of the grapevines from their balconies or verandahs, making a scroll, cycle or even row through your grounds all the more appealing. Even the various hotel buildings beautifully blend into the landscape, despite being contemporary, as they were designed using recycled, local materials.
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Here's a list of solar energy terms to help you navigate through technical texts. If we left something out, let us know!
Alternating current (AC) – AC is the electric current that your home uses for powering electrical appliances. It is an electric current in which the flow of electric charge periodically reverses direction, whereas in direct current (DC, also dc), the flow of electric charge is only in one direction.
Amorphous semiconductor – a semiconductor that is not made out of crystalline silicon. It is used to make some thin film solar panels. It's less efficient than traditional crystalline semiconductors.
Array – a set of photovoltaic panels. A home solar electric system might include several arrays on different parts of the roof.
Azimuth – It is a horizontal angle measured clockwise from any fixed reference plane or easily established base direction line an important measurement (usually in degrees) used to optimize the placement of photovoltaic panels. The azimuth angle is the compass direction from which the sunlight is coming, and therefore where solar panels should optimally face for the most solar power production. A positive azimuth angle generally indicates the sun is east of south, and a negative azimuth angle generally indicates the sun is west of south.
Balance of System (BOS) – The balance of system (BOS) encompasses all components of a photovoltaic system other than the photovoltaic panels. This includes wiring, switches, a mounting system, one or many solar inverters, a battery bank and battery charger. It is basically all the parts of a solar electric system excluding the solar panels and the inverter. Balance of system usually includes items like racking, wires, conduit, and safety shut-offs. The balance of system might account for 15% of the total cost of a solar electric system.
Base load – the amount of electric power a utility must supply constantly to meet the demand for energy. It is the minimum level of demand on an electrical supply system over 24 hours. Base load power sources are those plants, which can generate dependable power to consistently meet demand.
Battery – a battery back up for a solar electric system stores the extra power the solar system makes. A home can use that power when the solar system isn't producing enough energy, at night, or in a power outage.
BIPV panel – Building-integrated photovoltaics (BIPV) are photovoltaic materials that are used to replace conventional building materials in parts of the building envelope such as the roof, skylights, or facades. It is also a type of solar electric panel that uses silicon as a semiconductor and acts like a part of your roof. They're also known as solar roof shingles. BIPV can be done on new or existing roofs.
Cell – the smallest part of a solar panel that converts light into solar electricity.
Direct Current (DC) – a type of low voltage electrical current. DC electricity is produced by solar cells and must be converted into AC before it's usable in a house.
Gigawatt – one billion watts.
Grid – a system of high-tension cables in a region that distributes electricity to homes, businesses, and other buildings.
Grid-tied/Grid connected – a solar system that is connected to the power grid and uses the grid as a backup source of power.
Ground mount – a solar electrical system that is mounted on the ground instead of on a roof.
Interconnection – The physical and bureaucratic process of getting your system connected to the energy grid. Governed by sometimes nonsensical rules imposed by state governments and utility companies.
Inverter – A piece of equipment that converts DC electricity generated by solar panels to AC electricity used by your home.
IRR – Internal Rate of Return. Used to show the relative performance of an investment. Allows you to compare the potential return on solar to other investments.
Kilowatt (kW) – A unit of measurement equal to one thousand watts. Solar panels are rated according to how many kW they produce at any given moment.
Kilowatt-hour (kWh) – A unit of measurement of electricity generation equal to one kilowatt of generation (or consumption) operating for one hour.
Megawatt (MW) – A unit of measurement equal to one million watts.
Microinverter – A type of inverter that changes DC electricity to AC electricity on a per-panel basis. The output from several microinverters is combined and often fed to the electrical grid.
Module – a solar panel, or a group of solar cells.
Monocrystalline panel – a solar panel that's made from a large, single silicon crystal and has a patchwork pattern. Monocrystalline panels are more expensive and more efficient than multi- or poly- crystalline panels.
Nameplate Rating – Also known as STC, or Standard Test Conditions – Nameplate rating refers to the performance of a piece of electricity-generating equipment under ideal laboratory conditions.
Photon – a "packet" of light energy
Photovoltaic – The generation of electric energy from light.
Polycrystalline panel – (also multicrystalline) a solar panel that's made from small silicon crystals oriented in lots of different directions. Polycrystalline panels are less expensive and less efficient than monocrystalline panels.
Power purchase agreement (PPA) – a contract between a power producer and a power consumer, which states that the consumer will purchase a certain amount of power from the producer. In this type of solar agreement, a company installs a system on a homeowner's property and then sells the energy to the homeowner at sub-market prices for a defined period of time.
PTC – Performance Test Conditions – This type of measurement of electrical output from a solar panel accounts for real-world conditions, including energy losses from wiring, and more.
Rebate – Usually a direct, lump sum payment given to a homeowner or installation company when they install solar panels on a home. Utility companies most often pay rebates to homeowners because the utilities can count the homeowners' generation as clean energy toward their goals under an RPS law.
RPS Law – Renewable Portfolio Standard. An RPS law (often simply called an "RPS") sets goal for renewable energy generation as a percentage of total output in a state. Most states have enacted RPS laws to spur development of renewable energy, and an RPS is the primary tool used to mandate rebates, tax exemptions, and tax credits for solar and other renewables.
Semiconductor – a material that has a limited ability to conduct electric current. Semiconductors used in different types of solar panels include copper indium diselenide, cadmium telluride gallium arsenide, and silicon.
Silicon – a dark gray, semi-metallic, chemical element. Silicon is the material most commonly used semiconductor used in solar cells and computer chips.
Solar constant – the average amount of solar radiation that reaches the earth's upper atmosphere, equal to 1353 watts per square meter. Solar Lease – a financial agreement, in which a company installs a solar generation system on a homeowner's property, then leases the panels to the homeowner over a given period. At the end of the lease's term, the homeowner is often given the option to renew the lease or purchase the panels for the remainder of the cost, minus interest.
Solar noon – the time of day when the sun reaches its highest point in the sky. This time divides the daylight hours for that day exactly in half. Solar noon may be quite a bit different from 'clock' noon.
Solar Panels – The part of a solar power generation system that takes in photons from the sun and turns it into electrical energy, to be sent to the inverter and used in the home or transmitted to the utility company.
Stand alone/off grid – a solar electric system that is not connected to a power grid and which may or may not have a battery.
STC – Standard Test Conditions – Laboratory conditions under which solar panels are tested
Thin Film – A type of solar cell that uses a chemical layer instead of a silicon crystal as the method of converting solar energy to electrical energy. Thin-film is, as the name implies, thinner and less bulky than traditional solar cells, but not necessarily cheaper on a per-watt-generated basis.
Tilt – The degree of elevation of a solar panel relative to the ground. Tilt can either be fixed or adjustable by manual or automatic means. Tilt is one of two necessary orientation variables used to get maximum possible production from solar panels.
Tracking – A system that adjust the azimuth and tilt of solar panels to maximize the production of energy from solar panels.
Watt – A unit of energy equal to one joule per second.
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