Redfield Type Master Equation
In this appendix we discuss the derivation of the master equation (22): we follow the so-called projection technique [21, 22] applied to the interaction representation. We thus set
$$\begin{aligned} R_{\mathcal{S}\mathcal{B}}(t)&= \mathrm{e}^{it(H_\text {eff}+H_\mathcal{B})}\, {\varrho }_{\mathcal{S}\mathcal{B}}(t)\,\mathrm{e}^{-it(H_\text {eff}+H_\mathcal{B})}\end{aligned}$$
(49)
$$\begin{aligned} K_{\mathcal{S}\mathcal{B}}(t)&= \mathrm{e}^{it(H_\text {eff}+H_\mathcal{B})}\, {H}_{\mathcal{S}\mathcal{B}}(t)\,\mathrm{e}^{-it(H_\text {eff}+H_\mathcal{B})}, \end{aligned}$$
(50)
whence
$$\begin{aligned} \frac{\mathrm{d}R_{\mathcal{S}\mathcal{B}}(t)}{\mathrm{d}t}=\lambda \,\mathbb {K}_t[R_{\mathcal{S}\mathcal{B}}(t)], \quad \mathbb {K}_t[R_{\mathcal{S}\mathcal{B}}(t)]=-i, [K_{\mathcal{S}\mathcal{B}}(t),\,R_{\mathcal{S}\mathcal{B}}(t)]. \end{aligned}$$
(51)
Let \(\varrho _\beta \) be the bath Gibbs state at temperature \(T\) such that \([H_\mathcal{B},\,\varrho _\beta ]=0\); the following linear operators act as projectors on the states \( {\varrho }_{\mathcal{S}\mathcal{B}}(t)\) of the compound system \(\mathcal{S}+\mathcal{B}\):
$$\begin{aligned} \mathbb {P}[R_{\mathcal{S}\mathcal{B}}(t)]= (\mathrm{Tr}_\mathcal{B}(R_{\mathcal{S}\mathcal{B}}(t)))\otimes \varrho _\beta =R(t)\otimes \varrho _\beta ,\quad \mathbb {Q}=\mathrm{id}-\mathbb {P}, \end{aligned}$$
(52)
with
$$\begin{aligned} R(t)=\mathrm{Tr}_\mathcal{B}(R_{\mathcal{S}\mathcal{B}}(t))=\mathrm{e}^{itH_\text {eff}}\,\mathrm{Tr}_\mathcal{B}(\varrho _{\mathcal{S}\mathcal{B}}(t))\,\mathrm{e}^{-itH_\text {eff}} \end{aligned}$$
(53)
giving the time-evolving density matrix of the open quantum system \(\mathcal{S}\) in its own interaction representation.
Let \(\mathbb {K}^{PP}_t=\mathbb {P}\circ \mathbb {K}_t\circ \mathbb {P}\), \(\mathbb {K}^{PQ}_t=\mathbb {P}\circ \mathbb {K}_t\circ \mathbb {Q}\), \(\mathbb {K}^{QP}_t=\mathbb {Q}\circ \mathbb {K}_t\circ \mathbb {P}\) and \(\mathbb {K}^{QQ}_t=\mathbb {Q}\circ \mathbb {K}_t\circ \mathbb {Q}\) where \(\circ \) denotes the composition of maps. Then, (51) splits into the two coupled differential equations
$$\begin{aligned} \frac{\mathrm{d}\mathbb {P}[R_{\mathcal{S}\mathcal{B}}(t)]}{\mathrm{d}t}&= \lambda \,\mathbb {K}^{PP}_t\circ \mathbb {P}[R_{\mathcal{S}\mathcal{B}}(t)]+\lambda \,\mathbb {K}^{PQ}_t\circ \mathbb {Q}[R_{\mathcal{S}\mathcal{B}}(t)]\end{aligned}$$
(54)
$$\begin{aligned} \frac{\mathrm{d}\mathbb {Q}[R_{\mathcal{S}\mathcal{B}}(t)]}{\mathrm{d}t}&= \lambda \,\mathbb {K}^{QP}_t\circ \mathbb {P}[R_{\mathcal{S}\mathcal{B}}(t)]+\lambda \,\mathbb {K}^{QQ}_t\circ \mathbb {Q}[R_{\mathcal{S}\mathcal{B}}(t)]. \end{aligned}$$
(55)
The second equation is formally solved by
$$\begin{aligned} \mathbb {Q}[R_{\mathcal{S}\mathcal{B}}(t)]=\mathbb {W}_{t,0}^{QQ}\circ \mathbb {Q}[R_{\mathcal{S}\mathcal{B}}]+\lambda \,\int _0^t\mathrm{d}s\, \mathbb {W}_{t,s}^{QQ}\circ \mathbb {K}^{QP}_s\circ \mathbb {P}[R_{\mathcal{S}\mathcal{B}}(s)] \end{aligned}$$
(56)
with \(\mathbb {W}_{t,s}^{QQ}\) the time-ordered solution to
$$\begin{aligned} \frac{\mathrm{d}\mathbb {W}_{t,s}^{QQ}}{\mathrm{d}t}=\lambda \,\mathbb {K}^{QQ}_t\circ \mathbb {W}_{t,s}^{QQ}, \quad \mathbb {W}_{s,s}=\mathrm{id}. \end{aligned}$$
(57)
With initial condition \( {\varrho }_{\mathcal{S}\mathcal{B}}=\varrho \otimes \varrho _\mathcal{B}\), from \(\mathbb {Q}[R_{\mathcal{S}\mathcal{B}}]=\mathbb {Q}[ {\varrho }_{\mathcal{S}\mathcal{B}}]=0\) one gets
$$\begin{aligned} \mathbb {Q}[R_{\mathcal{S}\mathcal{B}}(t)]=\lambda \,\int _0^t\mathrm{d}s\, \mathbb {W}_{t,s}^{QQ}\circ \mathbb {K}^{QP}_s\circ \mathbb {P}[R_{\mathcal{S}\mathcal{B}}(s)] , \end{aligned}$$
(58)
whence, inserting (58) into (54),
$$\begin{aligned} \frac{\mathrm{d}\mathbb {P}[R_{\mathcal{S}\mathcal{B}}(t)]}{\mathrm{d}t}=\lambda \,\mathbb {K}^{PP}_t\circ \mathbb {P}[R_{\mathcal{S}\mathcal{B}}(t)]+\lambda ^2\,\int _0^t\mathrm{d}s\, \mathbb {K}^{PQ}_t\circ \mathbb {W}^{QQ}_{t,s}\circ \mathbb {K}_s^{QP}\circ \mathbb {P}[R_{\mathcal{S}\mathcal{B}}(s)]. \end{aligned}$$
(59)
The form of the interaction Hamiltonian in Eq. (21) and the fact that position operators have vanishing mean values with respect to Gibbs states yield \(\mathrm{Tr}_\mathcal{B}(\varrho _\mathcal{B}K_{\mathcal{S}\mathcal{B}}(t))=0\); then, \(\mathbb {P}[R_{\mathcal{S}\mathcal{B}}(t)]=R(t)\otimes \varrho _\beta \) implies
$$\begin{aligned} \frac{\mathrm{d}R(t)}{\mathrm{d}t}=-\,\lambda ^2 \int _0^t\mathrm{d}u\,\mathrm{Tr}_\mathcal{B}\Big (\Big [K_{\mathcal{S}\mathcal{B}}(t),\mathbb {Q}\circ \mathbb {W}^{QQ}_{t,u}\circ \mathbb {Q}\Big [K_{\mathcal{S}\mathcal{B}}(u),\,R(u)\otimes \varrho _\beta \Big ]\Big ]\Big ). \end{aligned}$$
(60)
The above equation depends on the history of the system state \(R(s)\) for all times \(0\le s\le t\); in order to eliminate this dependence, one takes into account the weak-coupling hypothesis \(\lambda \ll 1\) and looks at the dynamics as a function of a slow time parameter \(\tau =t\lambda ^2\). Firstly, by a change of integration variable \(s=t-u\), (60) is recast as
$$\begin{aligned} \frac{\mathrm{d}R(t)}{\mathrm{d}t}=-\,\lambda ^2 \int _0^t\mathrm{d}u\,\mathrm{Tr}_\mathcal{B}\Big (\Big [K_{\mathcal{S}\mathcal{B}}(t),\mathbb {Q}\circ \mathbb {W}^{QQ}_{t,t-u}\circ \mathbb {Q}\Big [K_{\mathcal{S}\mathcal{B}}(t-u),\,R_{t-u}\otimes \varrho _\beta \Big ]\Big ]\Big ). \end{aligned}$$
(61)
Then, letting \(\lambda \rightarrow 0\), \(\mathbb {W}^{QQ}_{t,s}\rightarrow \mathrm{id}\) as the right hand side of (57) vanishes, and
$$\begin{aligned}&\mathbb {Q}\circ \mathbb {W}^{QQ}_{t,s}\circ \mathbb {Q}\Bigg [\Big [K_{\mathcal{S}\mathcal{B}}(s),\,R(s)\otimes \varrho _\beta \Big ]\Bigg ]\,\rightarrow \, \mathbb {Q}\Bigg [\Big [K_{\mathcal{S}\mathcal{B}}(s),\, R(s)\otimes \varrho _\beta \Big ]\Bigg ]\\&\quad =\Big [K_{\mathcal{S}\mathcal{B}}(s),\, R(s)\otimes \varrho _\beta \Big ]. \end{aligned}$$
The last equality follows from \(\mathrm{Tr}_\mathcal{B}(\varrho _\beta K_{\mathcal{S}\mathcal{B}}(s))=0\), as explained before.
At this point, one usually sends the integration upper limit to \(+\infty \) and, in \(R(t-u)\), replaces \(t-u=\tau /\lambda ^2-u\) with \(t\); then, Eq. (61) reads
$$\begin{aligned} \frac{\mathrm{d}R(t)}{\mathrm{d}t}=-\lambda ^2 \int _0^{+\infty }\mathrm{d}u\,\mathrm{Tr}_\mathcal{B}\Big (\Big [K_{\mathcal{S}\mathcal{B}}(t),\,\Big [K_{\mathcal{S}\mathcal{B}}(t-u),\,R(t)\otimes \varrho _\beta \Big ]\Big ]\Big ). \end{aligned}$$
(62)
By going back from the interaction picture to the Schrödinger one, the following master equation for \(\varrho (t)\) is finally obtained,
$$\begin{aligned} \frac{\mathrm{d}\varrho (t)}{\mathrm{d}t}&= \mathbb {L}_t[\varrho (t)]=-i\,\Big [H_\text {eff},\, \varrho (t)\Big ]+\lambda ^2\, {\mathbb {N}}_t[\varrho (t)]\end{aligned}$$
(63)
$$\begin{aligned} {\mathbb {N}}_t[\varrho (t)]&= - \int _0^{+\infty }\mathrm{d}u\,\mathrm{Tr}_\mathcal{B}\Big (\Big [ {H}_{\mathcal{S}\mathcal{B}}(t),\, \Big [\mathrm{e}^{u(\mathbb {H}_\text {eff}+\mathbb {H}_\mathcal{B})}[ {H}_{\mathcal{S}\mathcal{B}}(t-u)],\, \varrho (t)\otimes \varrho _\beta \Big ]\Big ]\Big ), \end{aligned}$$
(64)
where
$$\begin{aligned} \mathrm{e}^{u(\mathbb {H}_\text {eff}+\mathbb {H}_\mathcal{B})}[X]=\mathrm{e}^{-iu(H_\text {eff}+H_\mathcal{B})}\,X\,\mathrm{e}^{iu(H_\text {eff}+H_\mathcal{B})}. \end{aligned}$$
(65)
Using (21) and the thermal state \(2\)-point functions (25) one finally obtains:
$$\begin{aligned} \nonumber {\mathbb {N}}_t[ \varrho (t)]&= -\sum _{\xi =1,3}\sum _n\lambda ^2_n \int _0^{+\infty }\mathrm{d}u\,\left\{ C(\omega _n,u)\Big [ {\sigma }_\xi (t),\, \mathrm{e}^{u\mathbb {H}_\text {eff}}[ {\sigma }_\xi (t-u)]\, \varrho (t)\Big ]\right. \\&\left. \quad +\,C^*(\omega _n,u)\Big [ \varrho (t)\,\mathrm{e}^{u\mathbb {H}_\text {eff}}[ {\sigma }_\xi (t-u)],\, {\sigma }_\xi (t)\Big ]\right\} , \end{aligned}$$
(66)
where
$$\begin{aligned} C(\omega _n,u)=2m\omega _n\mathrm{Tr}_\mathcal{B}\Big (\varrho _\beta \,q_{\xi ,n}\,\mathrm{e}^{-iuH_\mathcal{B}}\,q_{\xi ,n}\, \mathrm{e}^{iuH_\mathcal{B}}\Big ). \end{aligned}$$
(67)
Since the couplings \(\lambda _n\) do not depend on \(\xi =1,3\) the explicit dependence on time \(t\) disappears. Indeed, let
$$\begin{aligned} \mathcal{R}(t)=\begin{pmatrix} \cos \Omega t&{}\quad 0&{}\quad \sin \Omega t\\ 0&{}\quad 1&{}\quad 0\\ -\sin \Omega t&{}\quad 0&{}\quad \cos \Omega t \end{pmatrix} \end{aligned}$$
(68)
be the matrix which implements the rotation (21):
$$\begin{aligned} {\sigma }_\xi (t)=\sum _{\eta =1,2,3}\mathcal{R}_{\xi \eta }(t)\sigma _\eta , \quad \xi =1,3. \end{aligned}$$
(69)
Then, for generic \(2\times 2\) matrices \(A\) and \(B\) one finds
$$\begin{aligned} \sum _{\xi =1,3}A\, {\sigma }_\xi (t)\,B\, {\sigma }_\xi (t-u)&= \sum _{\xi =1,3}\sum _{\eta _1,\eta _2=1,2,3}\mathcal{R}_{\xi \eta _1}(t)\mathcal{R}_{\xi \eta _2}(t-u)\,A\, {\sigma }_{\eta _1}\,B\, {\sigma }_{\eta _2}\\&= \sum _{\eta _1,\eta _2=1,3}\mathcal{R}_{\eta _1\eta _2}(-u)\,A\, {\sigma }_{\eta _1}\,B\, {\sigma }_{\eta _2}= \sum _{\xi =1,3}A\,\sigma _\xi \,B\, {\sigma }_\xi (-u). \end{aligned}$$
Therefore, \( {\mathbb {N}}_t[ \varrho (t)]\) becomes time-independent and equals
$$\begin{aligned} \nonumber {\mathbb {N}}[ \varrho (t)]&= -\sum _{\xi =1,3}\sum _n\lambda ^2_n \int _0^{+\infty }\mathrm{d}u\,\left\{ C(\omega _n,u)\Big [\sigma _\xi ,\, \mathrm{e}^{-iu H_\text {eff}}\, {\sigma }_\xi (-u)\,\mathrm{e}^{iu H_\text {eff}} \varrho (t)\Big ]\right. \\&\left. \quad +\,C^*(\omega _n,u)\Big [ \varrho (t)\,\mathrm{e}^{-iu H_\text {eff}}\, {\sigma }_\xi (-u)\,\mathrm{e}^{iu H_\text {eff}},\,\sigma _\xi \Big ]\right\} , \end{aligned}$$
(70)
and, using (24) and (26), the master equation (63) reduces to (22).
Therefore, the generator \(\mathbb {L}_t\) in (63) becomes time-independent, too: \(\mathbb {L}_t=\mathbb {L}\). In order to recast it in Lindblad form as in (1) and (2), we first pass from the Pauli triple \(\sigma _{1,2,3}\) to the rotated one, \(\hat{\sigma }_{1,2,3}\), in (36):
$$\begin{aligned} \sigma _\xi =\sum _{j=1}^3\mathcal {V}_{j\xi }\hat{\sigma }_j,\quad \mathcal{V}=\frac{1}{\omega _{\text {eff}}}\begin{pmatrix}\omega _\text {eff}&{}\quad \! 0&{}\quad \! 0\\ 0&{}\quad \!\Delta &{}\quad \!\Omega \\ 0&{}\quad \!-\Omega &{}\quad \!\Delta \end{pmatrix}. \end{aligned}$$
(71)
Then, \(\displaystyle H_\text {eff}=\frac{\omega _{\text {eff}}}{2}\,\hat{\sigma }_3\) and
$$\begin{aligned} \mathrm{e}^{itH_\text {eff}}\hat{\sigma }_j\mathrm{e}^{-itH_\text {eff}}=\sum _{j,k=1}^3\mathcal{U}^{jk}_\text {eff}(t)\hat{\sigma }_k,\quad \mathcal{U}_{\text {eff}}(t)=\begin{pmatrix}\cos (\omega _{\text {eff}}t)&{}\quad -\sin (\omega _{\text {eff}}t)&{}\quad \!0\\ \sin (\omega _{\text {eff}}t)&{}\quad \cos (\omega _{\text {eff}}t)&{}\quad 0\\ 0&{}\quad 0&{}\quad 1\end{pmatrix}. \end{aligned}$$
(72)
One can thus rewrite the term \( {\mathbb {N}}[ \varrho (t)]\) as follows:
$$\begin{aligned} {\mathbb {N}}[ \varrho (t)] =-\sum _{j,k=1}^3\int _0^{+\infty }\mathrm{d}u\,\mathcal{Z}_{jk}(-u)\,\left\{ G(u)\,\Big [\hat{\sigma }_j,\, \hat{\sigma }_k\, \varrho (t)\Big ] +G^*(u)\,\Big [ \varrho (t)\,\hat{\sigma }_{k},\,\hat{\sigma }_j\Big ]\right\} , \end{aligned}$$
(73)
where \(G(u)\) is as in (24). By taking into account that \(\xi =1,3\) the coefficients \(\mathcal{Z}_{jk}(t)\) can be regrouped into the following matrix by introducing the projection \(\mathcal{P}=\text {diag}(1,0,1)\):
$$\begin{aligned} \mathcal{Z}(t)=\mathcal{V}\mathcal{P}\mathcal{R}(t)\mathcal{V}^T\mathcal{U}_\text {eff}(t)=\begin{pmatrix} cC+\frac{\Omega }{\omega _{\text {eff}}}sS &{}\quad \frac{\Omega }{\omega _{\text {eff}}}cS-sC &{}\quad \frac{\Delta }{\omega _{\text {eff}}}S\\ \frac{\Omega ^2}{\omega _{\text {eff}}^2}sC-\frac{\Omega }{\omega _{\text {eff}}}cS &{} \quad \frac{\Omega ^2}{\omega _{\text {eff}}^2}cC+\frac{\Omega }{\omega _{\text {eff}}}sS &{}\quad \frac{\Omega \Delta }{\omega _{\text {eff}}^2}C\\ \frac{\Omega \Delta }{\omega _{\text {eff}}^2}sC-\frac{\Omega }{\omega _{\text {eff}}}cS&{}\quad \frac{\Omega \Delta }{\omega _{\text {eff}}^2}cC+\frac{\Delta }{\omega _{\text {eff}}}sS &{}\quad \frac{\Delta ^2}{\omega _{\text {eff}}^2}C \end{pmatrix}, \end{aligned}$$
(74)
where \(c=\cos (\omega _{\text {eff}}t)\), \(s=\sin (\omega _{\text {eff}}t)\) and \(C=\cos (\Omega t)\), \(S=\sin (\Omega t)\).
Finally, by separating the purely dissipative contribution \(\mathbb {D}[\varrho (t)]\) to \( {\mathbb {N}}[ \varrho (t)]\) from the one corresponding to a Lamb-shift Hamiltonian \(H_{LS}\), one gets the right hand side of (63) as follows:
$$\begin{aligned} {\mathbb {L}}[ \varrho (t)]&= -i\,\Big [H_\text {eff}+\lambda ^2\,{H}_{LS},\, \varrho (t)\Big ]+\lambda ^2\,\mathbb {D}[\varrho (t)]\end{aligned}$$
(75)
$$\begin{aligned} \mathbb {D}[\varrho (t)]&= \sum _{j,k=1}^3{K}_{jk}\left( \hat{\sigma }_k\, \varrho (t)\,\hat{\sigma }_j-\frac{1}{2}\Big \{\hat{\sigma }_j\hat{\sigma }_k,\, \varrho (t)\Big \}\right) \end{aligned}$$
(76)
$$\begin{aligned} {K}_{jk}&= \int _0^{+\infty }\mathrm{d}u\,\Big (G(\tau )\,\mathcal{Z}_{jk}(-u)\,+\,G^*(\tau )\,\mathcal{Z}_{kj}(-u)\Big )= {K}^*_{kj}, \end{aligned}$$
(77)
where \(\displaystyle {H}_{LS}=\sum _{j,k=1}^3 {H}_{jk}\,\hat{\sigma }_j\hat{\sigma }_k\), with
$$\begin{aligned} {H}_{jk}=\frac{1}{2i}\int _0^{+\infty }\mathrm{d}\tau \,\Big (G(\tau )\,\mathcal{Z}_{jk}(-u)\,-\,G^*(\tau )\,\mathcal{Z}_{kj}(-u)\Big )= {H}^*_{kj}. \end{aligned}$$
(78)
In order to recast the action of \( {\mathbb {L}}[ \varrho (t)]\) as that of a \(4\times 4\) matrix \(-2\mathcal{L}\) on the Bloch vector \((1, {r}_1(t), {r}_2(t), {r}_3(t))\) as in (37), one considers the linear action of \( {\mathbb {L}}\) on the Pauli matrices \(\hat{\sigma }_{1,2,3}\) and on the identity \(\hat{\sigma }_0=1\): \( {\mathbb {L}}[\hat{\sigma }_\mu ]=\sum _{j=1}^3 {\mathcal{L}}_{j\mu }\hat{\sigma }_j\). With \( {r}_0(t)=1\) because of trace conservation, this gives
$$\begin{aligned} {\mathbb {L}}[ \varrho (t)]&= \frac{1}{2}\left( {\mathbb {L}}[1]+\sum _{j=1}^3 {r}_j(t) {\mathbb {L}}[\hat{\sigma }_j]\right) = -\sum _{j=1}^3\left( \sum _{\mu =0}^3 {\mathcal{L}}_{j\mu } {r}_\mu (t)\right) \hat{\sigma }_j\end{aligned}$$
(79)
$$\begin{aligned} {\mathcal{L}}&= {\mathcal{H}}_\text {eff}\,+\lambda ^2\, {\mathcal{H}}_{LS}\,+\lambda ^2\, {\mathcal{D}}=\begin{pmatrix}0&{}\quad 0&{}\quad 0&{}\quad 0\\ {\mathcal{L}}_{10}&{}\quad {\mathcal{L}}_{10}&{} \quad {\mathcal{L}}_{10}&{}\quad {\mathcal{L}}_{10}\\ {\mathcal{L}}_{20}&{} \quad {\mathcal{L}}_{21}&{}\quad {\mathcal{L}}_{22}&{}\quad {\mathcal{L}}_{23}\\ {\mathcal{L}}_{30}&{} \quad {\mathcal{L}}_{31}&{}\quad {\mathcal{L}}_{32}&{}\quad {\mathcal{L}}_{33} \end{pmatrix}. \end{aligned}$$
(80)
The matrix \( {\mathcal{L}}\) consists of an antisymmetric Hamiltonian contribution \( {\mathcal{H}}_\text {eff}+\lambda ^2\, {\mathcal{H}}_{LS}\), where
$$\begin{aligned} {\mathcal{H}}_\text {eff}&= \begin{pmatrix}0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad \omega _{\text {eff}}/2&{}\quad 0\\ 0&{}\quad -\omega _{\text {eff}}/2&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad -0&{}\quad 0 \end{pmatrix}\end{aligned}$$
(81)
$$\begin{aligned} {\mathcal{H}}_{LS}&= \begin{pmatrix}0&{}\quad \!0&{}\quad \!0&{}\quad \!0\\ 0&{}\quad 0&{}\quad {\mathcal{H}}_{12}&{}\quad {\mathcal{H}}_{13}\\ 0&{}\quad - {\mathcal{H}}_{12}&{}\quad 0&{}\quad {\mathcal{H}}_{23}\\ 0&{}\quad - {\mathcal{H}}_{13}&{}\quad - {\mathcal{H}}_{23}&{}\quad 0 \end{pmatrix}, \quad \left\{ \begin{matrix} {\mathcal{H}}_{12}=&{}\quad 2\,\mathcal{I}m( {H}_{21})\\ {\mathcal{H}}_{13}=&{}\quad 2\,\mathcal{I}m( {H}_{31})\\ {\mathcal{H}}_{23}=&{}\quad 2\,\mathcal{I}m( {H}_{32}) \end{matrix} \right. , \end{aligned}$$
(82)
plus a purely dissipative term
$$\begin{aligned} {\mathcal{D}}&= \begin{pmatrix}0&{}\quad \!0&{}\quad \!0&{}\quad \!0\\ {\mathcal{K}}_{10}&{}\quad {\mathcal{K}}_{11}&{} \quad {\mathcal{K}}_{12}&{}\quad {\mathcal{K}}_{13}\\ {\mathcal{K}}_{20}&{}\quad {\mathcal{K}}_{12}&{}\quad {\mathcal{K}}_{22}&{}\quad {\mathcal{K}}_{23}\\ {\mathcal{K}}_{30}&{}\quad {\mathcal{K}}_{13}&{}\quad {\mathcal{K}}_{23}&{}\quad {\mathcal{K}}_{33}\\ \end{pmatrix},\quad \left\{ \begin{matrix} {\mathcal{K}}_{10}&{}\quad =\mathcal{I}m( {K}_{23})\\ {\mathcal{K}}_{20}&{}\quad =\mathcal{I}m( {K}_{31})\\ {\mathcal{K}}_{30}&{}\quad =\mathcal{I}m( {K}_{12}) \end{matrix}\right. \end{aligned}$$
(83)
$$\begin{aligned}&\left\{ \begin{matrix} {\mathcal{K}}_{11}&{}\quad = {K}_{22}+ {K}_{33}\\ {\mathcal{K}}_{22}&{}\quad = {K}_{11}+ {K}_{33}\\ {\mathcal{K}}_{33}&{}\quad = {K}_{11}+ {K}_{22} \end{matrix}\right. ,\quad \left\{ \begin{matrix} {\mathcal{K}}_{12}&{}\quad =-\mathcal{R}e( {K}_{12})\\ {\mathcal{K}}_{13}&{}\quad =-\mathcal{R}e( {K}_{13})\\ {\mathcal{K}}_{23}&{}\quad =-\mathcal{R}e( {K}_{23}) \end{matrix}\right. . \end{aligned}$$
(84)
Using the expressions in (74), (77) and (78), the matrix entries explicitly read
$$\begin{aligned} {\mathcal{K}}_{10}&= \frac{\Delta }{\omega _{\text {eff}}}\int _0^{+\infty }\mathrm{d}u\int _0^{+\infty }\mathrm{d}\omega \,J(\omega )\,\sin (\omega u)\,\bigg (\sin (\omega _{\text {eff}}u)\sin (\Omega u)\nonumber \\&\quad +\,\frac{\Omega }{\omega _{\text {eff}}}\Big (\cos (\omega _{\text {eff}}u)-1\Big )\cos (\Omega u) \bigg ) \end{aligned}$$
(85)
$$\begin{aligned} {\mathcal{K}}_{20}&= \frac{\Delta }{\omega _{\text {eff}}}\int _0^{+\infty }\mathrm{d}u\int _0^{+\infty }\mathrm{d}\omega \,J(\omega )\,\sin (\omega u)\,\bigg (-\Big (1+\cos (\omega _{\text {eff}}u)\Big )\sin (\Omega u) \nonumber \\&\quad +\,\frac{\Omega }{\omega _{\text {eff}}}\sin (\omega _{\text {eff}}u)\cos (\Omega u) \bigg )\end{aligned}$$
(86)
$$\begin{aligned} {\mathcal{K}}_{30}&= \int _0^{+\infty }\mathrm{d}u\int _0^{+\infty }\mathrm{d}\omega \,J(\omega )\,\sin (\omega u)\,\bigg (-\frac{2\Omega ^2+\Delta ^2}{\omega _{\text {eff}}^2}\sin (\omega _{\text {eff}}u)\cos (\Omega u)\nonumber \\&\quad +\,2\frac{\Omega }{\omega _{\text {eff}}}\cos (\omega _{\text {eff}}u)\sin (\Omega u)\bigg ) \end{aligned}$$
(87)
$$\begin{aligned} {\mathcal{K}}_{11}&= 2\int _0^{+\infty }\mathrm{d} u\int _0^{+\infty }\mathrm{d}\omega \,J(\omega )\,\cos (\omega u)\coth \left( \frac{\beta \omega }{2}\right) \,\left( \frac{\Omega }{\omega _{\text {eff}}}\sin (\omega _{\text {eff}}u)\sin (\Omega u)\right. \nonumber \\&\left. \quad +\,\frac{\Omega ^2}{\omega _{\text {eff}}^2}\cos (\omega _{\text {eff}}u)\cos (\Omega u)+\frac{\Delta ^2}{\omega _{\text {eff}}^2}\cos (\Omega u) \right) \end{aligned}$$
(88)
$$\begin{aligned} {\mathcal{K}}_{12}&= -\frac{\Delta ^2}{\omega _{\text {eff}}^2}\int _0^{+\infty }\mathrm{d}u\int _0^{+\infty }\mathrm{d}\omega \,J(\omega )\,\cos (\omega u)\, \coth \left( \frac{\beta \omega }{2}\right) \,\sin (\omega _{\text {eff}}u)\cos (\Omega u)\end{aligned}$$
(89)
$$\begin{aligned} {\mathcal{K}}_{13}\!&= \!\frac{\Delta }{\omega _\text {eff}}\int _0^{+\infty }\mathrm{d}u\int _0^{+\infty }\mathrm{d}\omega \,J(\omega )\,\cos (\omega u)\coth \coth \left( \frac{\beta \omega }{2}\right) \,\left( \!-\frac{\Omega }{\omega _{\text {eff}}}\sin (\omega _{\text {eff}}u)\cos (\Omega u)\right. \nonumber \\&\left. \quad +\,\sin (\Omega u)\Big (\cos (\omega _{\text {eff}}u)-1\Big )\right) \end{aligned}$$
(90)
$$\begin{aligned} {\mathcal{K}}_{22}&= 2\int _0^{+\infty }\mathrm{d}u\int _0^{+\infty }\mathrm{d}\omega \,J(\omega )\,\cos (\omega u)\coth \left( \frac{\beta \omega }{2}\right) \,\left( \cos (\omega _{\text {eff}}u)\cos (\Omega u)\right. \nonumber \\&\left. \quad +\,\frac{\Omega }{\omega _{\text {eff}}}\sin (\omega _{\text {eff}}u)\sin (\Omega u)+\frac{\Delta ^2}{\omega _{\text {eff}}^2}\cos (\Omega u) \right) \end{aligned}$$
(91)
$$\begin{aligned} {\mathcal{K}}_{23}&= -\frac{\Delta }{\omega _{\text {eff}}}\int _0^{+\infty }\mathrm{d}u\int _0^{+\infty }\mathrm{d}\omega \,J(\omega )\,\cos (\omega u)\coth \left( \frac{\beta \omega }{2}\right) \,\left( \sin (\omega _{\text {eff}}u)\sin (\Omega u)\right. \nonumber \\&\left. \quad +\,\frac{\Omega }{\omega _{\text {eff}}}\Big (1+\cos (\omega _{\text {eff}}u)\Big )\cos (\Omega u)\right) \end{aligned}$$
(92)
$$\begin{aligned} {\mathcal{K}}_{33}&= 2\int _0^{+\infty }\mathrm{d}u\int _0^{+\infty }\mathrm{d}\omega \,J(\omega )\,\cos (\omega u)\coth \left( \frac{\beta \omega }{2}\right) \,\left( \frac{2\Omega ^2+\Delta ^2}{\omega _{\text {eff}}^2}\cos (\omega _{\text {eff}}u)\cos (\Omega u)\right. \nonumber \\&\left. \quad +\,2\frac{\Omega }{\omega _{\text {eff}}}\sin (\omega _{\text {eff}}u)\sin (\Omega u)\right) ; \end{aligned}$$
(93)
$$\begin{aligned} {\mathcal{H}}_{12}&= \int _0^{+\infty }\mathrm{d} u\int _0^{+\infty }\mathrm{d}\omega \,J(\omega )\,\cos (\omega u)\coth \left( \frac{\beta \omega }{2}\right) \,\left( \frac{2\Omega ^2+\Delta ^2}{\omega _{\text {eff}}^2}\sin (\omega _{\text {eff}}u)\cos (\Omega u)\right. \nonumber \\&\left. \quad -\,2\frac{\Omega }{\omega _{\text {eff}}}\cos (\omega _{\text {eff}}u)\sin (\Omega u)\right) \end{aligned}$$
(94)
$$\begin{aligned} {\mathcal{H}}_{13}&= \frac{\Delta }{\omega _{\text {eff}}}\int _0^{+\infty }\mathrm{d}u\int _0^{+\infty }\mathrm{d}\omega \,J(\omega )\,\cos (\omega u)\, \coth \left( \frac{\beta \omega }{2}\right) \,\left( \frac{\Omega }{\omega _{\text {eff}}}\sin (\omega _{\text {eff}}u)\cos (\Omega u)\right. \nonumber \\&\left. \quad -\,\sin (\Omega u)\Big (1+\cos (\omega _{\text {eff}}u)\Big )\right) \end{aligned}$$
(95)
$$\begin{aligned} {\mathcal{H}}_{23}&= \frac{\Delta }{\omega _{\text {eff}}}\int _0^{+\infty }\mathrm{d}u\int _0^{+\infty }\mathrm{d}\omega \,J(\omega )\,\cos (\omega u)\coth \left( \frac{\beta \omega }{2}\right) \,\left( -\sin (\omega _{\text {eff}}u)\sin (\Omega \tau )\right. \nonumber \\&\left. \quad +\,\frac{\Omega }{\omega _{\text {eff}}}\Big (1-\cos (\omega _{\text {eff}}u)\Big )\cos (\Omega u)\right) . \end{aligned}$$
(96)
In the main text, we have added a superscript “\(\text {Red}\)” to \(\mathcal {L}\), \(\mathcal {H}_{LS}\) and \(\mathcal {D}\) in order to distinguish them from the analogous expressions pertaining to a completely positive dynamics which are obtained in the next appendix.
Completely Positive Master Equation
A physically consistent reduced dynamics can be obtained by a more careful treatment; it leads to a completely positive time-evolution, thus avoiding all the inconsistencies of the Redfield dynamics used in [19]. We again use the projection technique but we follow the analysis of [8] without passing to the interaction representation. By repeating the arguments of the previous appendix, one arrives at the following analog of equation (60):
$$\begin{aligned} \frac{\mathrm{d} \varrho (t)}{\mathrm{d}t}&=-i\,[H_\text {eff},\, \varrho (t)]\nonumber \\&\quad -\,\lambda ^2 \int _0^t\mathrm{d}u\,\mathrm{Tr}_\mathcal{B}\Big (\Big [ {H}_{\mathcal{S}\mathcal{B}}(t),\,\mathbb {Q}\circ \mathbb {U}^{QQ}_{t,s}\circ \mathbb {Q}\Big [\Big [ {H}_{\mathcal{S}\mathcal{B}}(u),\, \varrho (t) \otimes \varrho _\mathcal{B}\Big ]\Big ]\Big ). \end{aligned}$$
(97)
As shown in [8], a sounder strategy than the one that led to equation (62) in the previous appendix consists firstly in formally integrating (97), yielding
$$\begin{aligned} \varrho (t)&= \mathrm{e}^{t\mathbb {H}_\text {eff}}[ {\varrho }] -\lambda ^2 \int _0^t\mathrm{d}u\,\int _0^u\mathrm{d}v\,\mathrm{e}^{(t-u)\mathbb {H}_\text {eff}} \nonumber \\&\quad \times \, \Bigg [ \mathrm{Tr}_\mathcal{B}\Big (\Big [ {H}_{\mathcal{S}\mathcal{B}}(u),\mathbb {Q}\circ \mathbb {U}^{QQ}_{u,v}\circ \mathbb {Q}\Big [ {H}_{\mathcal{S}\mathcal{B}}(v),\, {\varrho }(v) \otimes \varrho _\mathcal{B}\Big ]\Big ]\Big )\Bigg ]. \end{aligned}$$
(98)
Secondly, in changing the double integral into
$$\begin{aligned}&\int _0^t\mathrm{d}v\,\int _v^t\mathrm{d}u\,\mathrm{e}^{(t-u)\mathbb {H}_\text {eff}}\,\Bigg [\mathrm{Tr}_\mathcal{B}\Big (\Big [ {H}_{\mathcal{S}\mathcal{B}}(u),\mathbb {Q}\circ \mathbb {U}^{QQ}_{u,v}\circ \mathbb {Q}\Big [{H}_{\mathcal{S}\mathcal{B}}(v), \, \varrho (v)\otimes \varrho _\mathcal{B}\Big ]\Big ]\Big )\Bigg ]\\&\quad =\int _0^t\mathrm{d}v\,\int _0^{t-v}\mathrm{d}w\,\mathrm{e}^{(t-v-w)\mathbb {H}_\text {eff}}\,\\&\qquad \times \Bigg [ \mathrm{Tr}_\mathcal{B}\Big (\Big [ {H}_{\mathcal{S}\mathcal{B}}(v+w),\,\mathbb {Q}\circ \mathbb {U}^{QQ}_{v+w,v}\,\circ \, \mathbb {Q}\Big [ {H}_{\mathcal{S}\mathcal{B}}(v),\, \varrho (v) \otimes \varrho _\mathcal{B}\Big ]\Big ]\Big )\Bigg ], \end{aligned}$$
and, finally, in going to the slow time-scale \(\tau =t\lambda ^2\), \(\lambda \ll 1\) where, using (65), one replaces
$$\begin{aligned} \mathbb {Q}\circ \mathbb {U}^{QQ}_{v+w,v}\circ \mathbb {Q}\Big [\Big [ {H}_{\mathcal{S}\mathcal{B}}(v),\, \varrho (v) \otimes \varrho _\mathcal{B}\Big ]\Big ]\quad \hbox {by}\quad \mathrm{e}^{w(\mathbb {H}_\text {eff}+\mathbb {H}_\mathcal{B})}\Bigg [\Big [ {H}_{\mathcal{S}\mathcal{B}}(v),\, \varrho (v) \otimes \varrho _\mathcal{B}\Big ]\Bigg ], \end{aligned}$$
so that the second integral with respect to \(\mathrm{d}w\) becomes
$$\begin{aligned} \overline{\mathbb {N}}_v[ \varrho (v)]&= \int _0^{+\infty }\mathrm{d}w\, \mathrm{e}^{-w\mathbb {H}_\text {eff}}\nonumber \\&\times \,\Bigg [\mathrm{Tr}_\mathcal{B}\Big (\Big [ {H}_{\mathcal{S}\mathcal{B}}(v+w),\, \mathrm{e}^{w(\mathbb {H}_\text {eff}+\mathbb {H}_\mathcal{B})}\,\Big [\Big [ {H}_{\mathcal{S}\mathcal{B}}(v),\, {\varrho }_{v} \otimes \varrho _\mathcal{B}\Big ]\Big ]\Big ]\Big )\Bigg ]. \end{aligned}$$
(99)
This yields
$$\begin{aligned} \varrho (t)=\mathrm{e}^{t\mathbb {H}_\text {eff}}[ {\varrho }]-\lambda ^2\int _0^t\mathrm{d}v\,\overline{\mathbb {N}}_v[ \varrho (v)] \end{aligned}$$
that solves the master equation:
$$\begin{aligned} \frac{\mathrm{d} \varrho (t)}{\mathrm{d}t}&= -i\,\Big [H_\text {eff},\, \varrho (t)\Big ]+ \lambda ^2\overline{\mathbb {N}}_t[ \varrho (t)]. \end{aligned}$$
(100)
By proceeding as in the previous appendix, one recasts (100) in the time-independent form
$$\begin{aligned} \frac{\mathrm{d} \varrho (t)}{\mathrm{d}t}&= \overline{\mathbb {L}}[ \varrho (t)]=-i\,\Big [H_\text {eff},\, \varrho (t)\Big ]+\lambda ^2\, \overline{\mathbb {N}}[ \varrho (t)]\end{aligned}$$
(101)
$$\begin{aligned} \overline{\mathbb {N}}[ \varrho (t)]&= -i\,\Big [\overline{H}_{LS},\, \varrho (t)\Big ]+\lambda ^2\,\sum _{j,k=1}^3 {K}_{jk}\Big (\hat{\sigma }_k\, \varrho (t)\,\hat{\sigma }_j-\frac{1}{2}\Big \{\hat{\sigma }_j\hat{\sigma }_k,\, \varrho (t)\Big \}\Big ), \end{aligned}$$
(102)
where \(\overline{H}_{LS}=\sum _{j,k=1}^3 {H}_{jk}\hat{\sigma }_j\hat{\sigma }_k\) and, with respect to (77) and (78), the coefficients now read
$$\begin{aligned} {H}_{jk}&= \frac{1}{2i}\int _0^{+\infty }\mathrm{d}u\,\Big (G(u)\,\mathcal{Z}_{kj}(u)\,-\,G^*(u)\,\mathcal{Z}_{jk}(u)\Big )= {H}^*_{kj}\end{aligned}$$
(103)
$$\begin{aligned} {K}_{jk}&= \int _0^{+\infty }\mathrm{d}u\,\Big (G(u)\,\mathcal{Z}_{kj}(u)\,+\,G^*(u)\,\mathcal{Z}_{jk}(u)\Big )= {K}^*_{kj}. \end{aligned}$$
(104)
Remark 5
In the Bloch representation, the generator \(\overline{\mathbb {L}}[ \varrho (t)]\) corresponds to the action on the Bloch vector \((1, {r}_1(t), {r}_2(t), {r}_3(t))\) of a \(4\times 4\) matrix \(\overline{\mathcal{L}}= {\mathcal{H}}_\text {eff}\,+\lambda ^2\,\overline{\mathcal{H}}_{LS}\,+\lambda ^2\,\overline{\mathcal{D}}\), whose coefficients are all equal to the ones in (85)–(96) apart from \( {\mathcal{K}}_{10}\), \( {\mathcal{K}}_{12}\), \( {\mathcal{K}}_{13}\) and \( {\mathcal{H}}_{23}\) which have opposite signs.
The formal solutions to (101) are given by
$$\begin{aligned} \varrho (t)=\mathrm{e }^{t\overline{\mathbb {L}}}[\varrho ]=\mathrm{e}^{t\mathbb {H}_\text {eff}}[\varrho ]+\lambda ^2\int _0^t\mathrm{d}s\,\mathrm{e}^{t\mathbb {H}_\text {eff}}\circ \overline{\mathbb {N}}[ {\varrho }_s] \end{aligned}$$
(105)
with \(\mathrm{e}^{t\mathbb {H}_\text {eff}}[\varrho ]=\exp (-it\, H_\text {eff})\,\varrho \,\exp (it\, H_\text {eff})\).
On the slow time scale \(\tau =\lambda ^2\,t\), one rewrites
$$\begin{aligned} \mathrm{e}^{-t\mathbb {H}_\text {eff}}\circ \mathrm{e}^{t\overline{\mathbb {L}}}[\varrho ]=\varrho +\int _0^\tau \mathrm{d}u\, \Big \{\mathrm{e}^{-u/\lambda ^2\mathbb {H}_\text {eff}}\circ \overline{\mathbb {N}}\circ \mathrm{e}^{u/\lambda ^2\mathbb {H}_\text {eff}}\Big \}\Big [\mathrm{e}^{-u\mathbb {H}_\text {eff}}\circ \mathrm{e}^{u\overline{\mathbb {L}}}[\varrho ]\Big ]; \end{aligned}$$
(106)
when \(\lambda \rightarrow 0\) the fast oscillations in the term in curly brackets average to zero. This allows one to replace that term by its ergodic average
$$\begin{aligned} \mathbb {N}=\lim _{T\rightarrow +\infty }\frac{1}{2T}\int _{-T}^T\mathrm{d}s\,\mathrm{}^{-s/\lambda ^2\mathbb {H}_\text {eff}}\circ \overline{\mathbb {N}}\circ \mathrm{}^{s/\lambda ^2\mathbb {H}_\text {eff}}, \end{aligned}$$
(107)
which fulfils \(\mathbb {N}\circ \mathbb {H}_\text {eff}=\mathbb {H}_\text {eff}\circ \mathbb {N}\), so that the resulting master equation is
$$\begin{aligned} \frac{\mathrm{d} \varrho (t)}{\mathrm{d}t}= {\mathbb {L}}[ \varrho (t)]=-i\,\Big [H_\text {eff}+\lambda ^2\, {H}_{LS},\, \varrho (t)\Big ] +\lambda ^2 {\mathbb {D}}[ \varrho (t)] \end{aligned}$$
(108)
In the Bloch representation the action of \(\mathbb {N}\) corresponds to that of the \(4\times 4\) matrix
$$\begin{aligned} {\mathcal{N}}=\lim _{T\rightarrow +\infty }\frac{1}{2T}\int _0^T\mathrm{d}s\,\mathcal{U}_\text {eff}(-s)\,\overline{\mathcal{N}}\,\mathcal{U}_\text {eff}(s), \end{aligned}$$
(109)
with \(\overline{\mathcal{N}}\) the \(4\times 4 \) matrix corresponding to \(\overline{\mathbb {N}}\) and \(\mathcal{U}_\text {eff}(s)\) the \(4\times 4\) matrix with \(1,0,0,0\) in the first row and column and, in the rest, the \(3\times 3\) matrix in (72). Then, the action of \( {\mathbb {L}}\) can be represented by means of the \(4\times 4\) matrix \( {\mathcal{L}}=\mathcal{H}_\text {eff}+\lambda ^2\, {\mathcal{H}}_{LS}+\lambda ^2\, {\mathcal{D}}\), where \(\mathcal{H}_\text {eff}\) is as in (81), while
$$\begin{aligned} {\mathcal{H}}_{LS}=\begin{pmatrix}0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad {\mathcal{H}}_{12}&{}\quad 0\\ 0&{}\quad - {\mathcal{H}}_{12}&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0 \end{pmatrix},\quad {\mathcal{D}}=\begin{pmatrix}0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad \mathcal {K}_{11}+ \mathcal {K}_{22}&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad \mathcal {K}_{11}+ \mathcal {K}_{22}&{}\quad 0\\ \mathcal {K}_{30}&{}\quad 0&{}\quad 0&{}\quad \mathcal {K}_{33} \end{pmatrix}.\nonumber \\ \end{aligned}$$
(110)
