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use unicode math :P
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@ -4,18 +4,18 @@
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As in~\cite{Hartmann2017Dec} we choose
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\begin{equation}
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\label{eq:totalH}
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H = H_S + \underbrace{LB^\dagger + L^\dagger B}_{H_I} + H_B
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H = H_S + \underbrace{LB^†ger + L^†ger B}_{H_I} + H_B
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\end{equation}
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with the system hamiltonian \(H_S\), the bath hamiltonian
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\begin{equation}
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\label{eq:bathh}
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H_B = \sum_\lambda \omega_\lambda a^\dag a,
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H_B = ∑_λ ω_λ a^† a,
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\end{equation}
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the bath coupling system operator \(L\) and the bath coupling bath
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operator
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\begin{equation}
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\label{eq:bop}
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B=\sum_{\lambda} g_{\lambda} a_{\lambda}
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B=∑_{λ} g_{λ}^\ast a_{λ}
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\end{equation}
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which define the interaction hamiltonian \(H_I\).
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@ -35,25 +35,25 @@ Thus, we need to calculate
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\begin{eqnarray}
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\label{eq:calccomm}
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\begin{aligned}
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[H_B,H_I] &= [H_B, LB^\dag + L^\dag B] \\
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&= L[H_B, B^\dag ] + L^\dag [H_B, B] \\
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&= L[H_B, B^\dag ] - \hc.
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[H_B,H_I] &= [H_B, LB^† + L^† B] \\
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&= L[H_B, B^† ] + L^† [H_B, B] \\
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&= L[H_B, B^† ] - \hc.
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\end{aligned}
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\end{eqnarray}
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This checks out as the commutator has to be anti-hermitian due to
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\cref{eq:ehrenfest}.
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Using \([H_B, B^\dag ]=\sum_\lambda \omega_\lambda g^\ast_\lambda
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a^\dag_\lambda\) it follows that
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Using \([H_B, B^† ]=∑_λ ω_λ g_λ
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a^†_λ\) it follows that
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\begin{equation}
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\label{eq:expcomm}
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\begin{aligned}
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\ev{[H_B,H_I]} &= \sum_\lambda \omega_\lambda g^\ast_\lambda
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\ev{La^\dag_\lambda} - \cc
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= \sum_\lambda \omega_\lambda g^\ast_\lambda
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\ev{La^\dag_\lambda \eu^{\i \omega t}}_I - \cc\\
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&= \frac{1}{\i}\ev{L\partial_t{\sum_\lambda
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g^\ast_\lambda a^\dag_\lambda \eu^{\i \omega t}}}_I - \cc
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=\frac{1}{\i}\qty(\ev{L\dot{B}^\dag}_I + \cc)
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\ev{[H_B,H_I]} &= ∑_λ ω_λ g_λ
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\ev{La^†_λ} - \cc
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= ∑_λ ω_λ g^\ast_λ
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\ev{La^†_λ \eu^{\i ω t}}_I - \cc\\
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&= \frac{1}{\i}\ev{L\partial_t{∑_λ
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g^\ast_λ a^†_λ \eu^{\i ω t}}}_I - \cc
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=\frac{1}{\i}\qty(\ev{L\dot{B}^†}_I + \cc)
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\end{aligned}
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\end{equation}
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where we switched to the interaction picture with respect to \(H_B\)
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@ -62,7 +62,7 @@ In essence this is just the Heisenberg equation for \(H_I\). The
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expression for \(J\) follows
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\begin{equation}
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\label{eq:final_flow}
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J(t) = \ev{L^\dag\partial_t B(t) + L\partial_t B^\dag(t)}_I.
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J(t) = \ev{L^†\partial_t B(t) + L\partial_t B^†(t)}_I.
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\end{equation}
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From this point on, we will assume the interaction picture and drop the
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@ -70,55 +70,55 @@ From this point on, we will assume the interaction picture and drop the
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The two summands yield different expressions in terms of the NMQSD.
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For use with HOPS with the final goal of utilizing the auxiliary
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states the expression \(\ev{L^\dag\partial_t B(t)}\) should be
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states the expression \(\ev{L^†\partial_t B(t)}\) should be
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evaluated.
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We calculate
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\begin{equation}
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\label{eq:interactev}
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\ev{L^\dag\partial_t B(t)}=\ev{L^\dag\partial_t B(t)}{\psi(t)} =
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\int \braket{\psi(t)}{z}\mel{z}{L^\dag\partial_tB(t)}{\psi(t)}\frac{\dd[2]{z}}{\pi^N},
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\ev{L^†\partial_t B(t)}=\ev{L^†\partial_t B(t)}{\psi(t)} =
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∫ \braket{\psi(t)}{z}\mel{z}{L^†\partial_tB(t)}{\psi(t)}\frac{\dd[2]{z}}{\pi^N},
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\end{equation}
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where \(N\) is the total number of environment oscillators and
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\(z=\qty(z_{\lambda_1}, z_{\lambda_2}, \ldots)\).
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\(z=\qty(z_{λ_1}, z_{λ_2}, \ldots)\).
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To that end,
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\begin{equation}
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\label{eq:nmqsdficate}
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\begin{aligned}
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\mel{z}{\partial_tB(t)}{\psi(t)} &= \sum_\lambda g_\lambda
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\qty(\partial_t \eu^{-\i\omega_\lambda
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t})\partial_{z^\ast_\lambda}\ket{\psi(z^\ast,t)} \\
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&= \int_0^t \sum_\lambda g_\lambda
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\qty(\partial_t \eu^{-\i\omega_\lambda
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t})\pdv{\eta_s^\ast}{z^\ast_\lambda}\fdv{\ket{\psi(z^\ast,t)}}{\eta^\ast_s}\dd{s}\\
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&= -\i\int_0^t\dot{\alpha}(t-s)\fdv{\ket{\psi(z^\ast,t)}}{\eta^\ast_s}\dd{s},
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\mel{z}{\partial_tB(t)}{\psi(t)} &= ∑_λ g_λ
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\qty(\partial_t \eu^{-\iω_λ
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t})\partial_{z^\ast_λ}\ket{\psi(z^\ast,t)} \\
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&= ∫_0^t ∑_λ g_λ
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\qty(\partial_t \eu^{-\iω_λ
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t})\pdv{\eta_s^\ast}{z^\ast_λ}\fdv{\ket{\psi(z^\ast,t)}}{\eta^\ast_s}\dd{s}\\
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&= -\i∫_0^t\dot{\alpha}(t-s)\fdv{\ket{\psi(z^\ast,t)}}{\eta^\ast_s}\dd{s},
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\end{aligned}
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\end{equation}
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where \(\eta^\ast_t\equiv -\i \sum_\lambda g^\ast_\lambda
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z^\ast_\lambda \eu^{\i\omega_\lambda t}\).
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where \(\eta^\ast_t\equiv -\i ∑_λ g^\ast_λ
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z^\ast_λ \eu^{\iω_λ t}\).
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With this we can write
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\begin{equation}
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\label{eq:steptoproc}
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\ev{L^\dag\partial_t B(t)} = -\i \mathcal{M}_{\eta^\ast}\bra{\psi(\eta,
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t)}L^\dag\int_0^t\dd{s} \dot{\alpha}(t-s)\fdv{\eta^\ast_s} \ket{\psi(\eta^\ast,t)}.
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\ev{L^†\partial_t B(t)} = -\i \mathcal{M}_{\eta^\ast}\bra{\psi(\eta,
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t)}L^†∫_0^t\dd{s} \dot{\alpha}(t-s)\fdv{\eta^\ast_s} \ket{\psi(\eta^\ast,t)}.
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\end{equation}
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Defining
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\begin{equation}
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\label{eq:defdop}
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D_t = \int_0^t\dd{s} \alpha(t-s)\fdv{\eta^\ast_s}
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D_t = ∫_0^t\dd{s} \alpha(t-s)\fdv{\eta^\ast_s}
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\end{equation}
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as in~\cite{Suess2014Oct} we can write
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\begin{equation}
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\label{eq:final_flow_nmqsd}
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J(t) = -\i \mathcal{M}_{\eta^\ast}\bra{\psi(\eta,
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t)}L^\dag\dot{D}_t\ket{\psi(\eta^\ast,t)} + \cc,
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t)}L^†\dot{D}_t\ket{\psi(\eta^\ast,t)} + \cc,
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\end{equation}
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where we've used that the integral in \(D_t\) can be expanded over the
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whole real axis. If we assume \(\alpha = \exp(-w t)\) then
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\begin{equation}
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\label{eq:hopsj}
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J(t) = \i \mathcal{M}_{\eta^\ast}\bra{\psi^{(0)}(\eta,
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t)}wL^\dag\ket{\psi^{(1)}(\eta^\ast,t)} + \cc.,
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t)}wL^†\ket{\psi^{(1)}(\eta^\ast,t)} + \cc.,
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\end{equation}
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where \(\ket{\psi^{(1)}(\eta^\ast,t)}\) is the first HOPS hierarchy
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state. This can be generalized to any BCF that is a sum of exponentials.
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@ -126,7 +126,7 @@ state. This can be generalized to any BCF that is a sum of exponentials.
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Interestingly one finds that
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\begin{equation}
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\label{eq:alternative}
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\ev{L\partial_t B^\dag(t)} = \i\int\frac{\dd[2]{z}}{\pi^N}
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\ev{L\partial_t B^†(t)} = \i∫\frac{\dd[2]{z}}{\pi^N}
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\dot{\eta}_t^\ast \mel{\psi(\eta,t)}{L}{\psi(\eta^\ast,t)}.
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\end{equation}
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However, this seems numerically problematic because \(\eta^\ast\) is not
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@ -135,11 +135,11 @@ that we utilize the first hierarchy states that are already being
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calculated as a byproduct.
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In the language of~\cite{Hartmann2021Aug} we can generalize to
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\(\alpha(t) = \sum_i G_i \eu^{-W_i t}\) and thus
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\(\alpha(t) = ∑_i G_i \eu^{-W_i t}\) and thus
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\begin{equation}
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\label{eq:hopsflowrich}
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J(t) = \sum_\mu\frac{G_\mu W_\mu}{\bar{g}_\mu} \i\mathcal{M}_{\eta^\ast}\bra{\psi^{(0)}(\eta,
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t)}L^\dag\ket{\psi^{\vb{e}_\mu}(\eta^\ast,t)} + \cc,
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J(t) = ∑_\mu\frac{G_\mu W_\mu}{\bar{g}_\mu} \i\mathcal{M}_{\eta^\ast}\bra{\psi^{(0)}(\eta,
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t)}L^†\ket{\psi^{\vb{e}_\mu}(\eta^\ast,t)} + \cc,
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\end{equation}
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where \(\psi^{\vb{e}_\mu}\) is the \(\mu\)-th state of the first
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hierarchy and \(\bar{g}_\mu\) is an arbitrary scaling introduced in
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@ -154,30 +154,30 @@ In the spirit of the usual derivation of the nonlinear NMQSD we write
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\begin{equation}
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\label{eq:newb}
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\begin{aligned}
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\ev{L^\dag\dot{B(t)}} &= \int \frac{\dd[2]{z}}{\pi^N} \eu^{-\abs{z}^2}
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\ev{L^†\dot{B(t)}} &= ∫ \frac{\dd[2]{z}}{\pi^N} \eu^{-\abs{z}^2}
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\braket{\psi}{z}\!\braket{z}{\psi}
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\frac{\braket{\psi(t)}{z}\!\mel{z}{L^\dag\dot{B(t)}}{\psi(t)}}{\braket{\psi}{z}\!\braket{z}{\psi}}
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\frac{\braket{\psi(t)}{z}\!\mel{z}{L^†\dot{B(t)}}{\psi(t)}}{\braket{\psi}{z}\!\braket{z}{\psi}}
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\\
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&= \int \frac{\dd[2]{z}}{\pi^N} \eu^{-\abs{z}^2}
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\frac{\mel{z(t)}{L^\dag\dot{B(t)}}{\psi(t)}}{\braket{\psi}{z(t)}\!\braket{z(t)}{\psi}},
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&= ∫ \frac{\dd[2]{z}}{\pi^N} \eu^{-\abs{z}^2}
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\frac{\mel{z(t)}{L^†\dot{B(t)}}{\psi(t)}}{\braket{\psi}{z(t)}\!\braket{z(t)}{\psi}},
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\end{aligned}
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\end{equation}
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where \(z_{\lambda}^{*}(t)=z_{\lambda}^{*}+\i g_{\lambda} \int_{0}^{t}
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\dd{s} \eu^{-\i \omega_{\lambda} s}\ev{L^\dagger}_{s}\).
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where \(z_{λ}^{*}(t)=z_{λ}^{*}+\i g_{λ} ∫_{0}^{t}
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\dd{s} \eu^{-\i ω_{λ} s}\ev{L^†ger}_{s}\).
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We find that next steps are the same as in \cref{sec:nonlin} by noting
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\begin{equation}
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\label{eq:deriv_trick}
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\eval{\partial_{z^\ast_\lambda}}_{z^\ast=z_\lambda^\ast(t)} =
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\int_0^t\dd{s}\eval{\pdv{\eta^\ast}{z^\ast_\lambda}}_{z^\ast=z^\ast_\lambda(t)}
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\eval{\partial_{z^\ast_λ}}_{z^\ast=z_λ^\ast(t)} =
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∫_0^t\dd{s}\eval{\pdv{\eta^\ast}{z^\ast_λ}}_{z^\ast=z^\ast_λ(t)}
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\fdv{}{\eta^\ast_s(z^\ast=z^\ast(t))} =
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\int_0^t\dd{s}\eval{\pdv{\eta^\ast}{z^\ast_\lambda}}_{z^\ast=0}
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∫_0^t\dd{s}\eval{\pdv{\eta^\ast}{z^\ast_λ}}_{z^\ast=0}
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\fdv{}{\eta^\ast_s(z^\ast=z^\ast(t))},
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\end{equation}
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which does alter the definition of \(D_t\) but results in the same
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HOPS equations.
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The shifted process \(\tilde{\eta}^\ast=
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\eta^\ast(z^\ast(t),t)=\eta^\ast(t) +
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\int_0^t\dd{s}\alpha^\ast(t-s)\ev{L^\dag}_{\psi_s}\) appears directly
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∫_0^t\dd{s}\alpha^\ast(t-s)\ev{L^†}_{\psi_s}\) appears directly
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in the NMQSD equation but results only in a slight change in the
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functional derivative. Note however that
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\begin{equation}
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@ -196,7 +196,7 @@ Therefore,
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\label{eq:newbcontin}
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J(t) =
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-\i
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\mathcal{M}_{\tilde{\eta}^\ast}\frac{\mel{\psi(\tilde{\eta},t)}{L^\dag\dot{\tilde{D}}_t}{\psi(\tilde{\eta}^\ast,t)}}{\braket{\psi(\tilde{\eta},t)}{\psi(\tilde{\eta}^\ast,t)}}
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\mathcal{M}_{\tilde{\eta}^\ast}\frac{\mel{\psi(\tilde{\eta},t)}{L^†\dot{\tilde{D}}_t}{\psi(\tilde{\eta}^\ast,t)}}{\braket{\psi(\tilde{\eta},t)}{\psi(\tilde{\eta}^\ast,t)}}
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+ \cc,
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\end{equation}
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where the dependence on \(\tilde{\eta}\) is symbolic and to be
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@ -206,9 +206,9 @@ Again we express the result in the language of~\cite{Hartmann2021Aug}
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to obtain
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\begin{equation}
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\label{eq:hopsflowrich}
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J(t) = \sum_\mu\frac{G_\mu W_\mu}{\bar{g}_\mu}
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J(t) = ∑_\mu\frac{G_\mu W_\mu}{\bar{g}_\mu}
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\i\mathcal{M}_{\eta^\ast}\frac{\bra{\psi^{(0)}(\eta,
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t)}L^\dag\ket{\psi^{\vb{e}_\mu}(\eta^\ast,t)}}{\bra{\psi^{(0)}(\eta,
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t)}L^†\ket{\psi^{\vb{e}_\mu}(\eta^\ast,t)}}{\bra{\psi^{(0)}(\eta,
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t)}\ket{\psi^{0}(\eta^\ast,t)}} + \cc.
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\end{equation}
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@ -225,7 +225,7 @@ method described in~\cite{Hartmann2017Dec}.
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The shift operator
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\begin{equation}
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\label{eq:shiftop}
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\vb{D}(y) = \bigotimes_\lambda \eu^{y_\lambda a_\lambda^\dag-y^\ast_\lambda a_\lambda}
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\vb{D}(y) = \bigotimes_λ \eu^{y_λ a_λ^†-y^\ast_λ a_λ}
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\end{equation}
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the ground state of the environment into an arbitrary
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coherent state
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@ -240,55 +240,55 @@ thermal initial bath as
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\begin{equation}
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\label{eq:shiftbath}
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\rho =
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\prod_\lambda\qty(\int\dd[2]{y_\lambda}
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\frac{\eu^{-\abs{y_\lambda}^2\bar{n}_\lambda}}{\pi\bar{n}_\lambda})
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U(t)\vb{D}(y)\ketbra{\psi}\otimes\ketbra{0}\vb{D}(y)^\dag U(t)^\dag.
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\prod_λ\qty(∫\dd[2]{y_λ}
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\frac{\eu^{-\abs{y_λ}^2\bar{n}_λ}}{\pi\bar{n}_λ})
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U(t)\vb{D}(y)\ketbra{\psi}\otimes\ketbra{0}\vb{D}(y)^† U(t)^†.
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\end{equation}
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The usual step is now to insert \(\id =\vb{D}(y)\vb{D}^\dag(y)\) to
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The usual step is now to insert \(\id =\vb{D}(y)\vb{D}^†(y)\) to
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arrive at a new time translation operator
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\begin{equation}
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\label{eq:utilde}
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\tilde{U}(t) = \vb{D}^\dag(y)U(t)\vb{D}(y)
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\tilde{U}(t) = \vb{D}^†(y)U(t)\vb{D}(y)
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\end{equation}
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and to interpret the integral in \cref{eq:shiftbath} in a monte-carlo
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sense which leads to a stochastic contribution to the system Hamiltonian
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\begin{equation}
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\label{eq:thermalh}
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H_{\mathrm{sys}}^{\mathrm{shift}}=L \xi^{*}(t)+L^{\dagger} \xi(t)
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H_{\mathrm{sys}}^{\mathrm{shift}}=L \xi^{*}(t)+L^{†ger} \xi(t)
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\end{equation}
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with the stochastic process
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\begin{equation}
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\label{eq:xiproc}
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\xi(t):=\sum_{\lambda} g_{\lambda} y_{\lambda} \eu^{-\mathrm{i} \omega_{\lambda} t}
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\xi(t):=∑_{λ} g_{λ} y_{λ} \eu^{-\mathrm{i} ω_{λ} t}
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\end{equation}
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with corresponding moments \(\mathcal{M}(\xi(t))=0=\mathcal{M}(\xi(t) \xi(s))\) and
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\[
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\mathcal{M}\left(\xi(t) \xi^{*}(s)\right)=\frac{1}{\pi} \int_{0}^{\infty} \mathrm{d} \omega \bar{n}(\beta \omega) J(\omega) e^{-\mathrm{i} \omega(t-s)}.
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\mathcal{M}\left(\xi(t) \xi^{*}(s)\right)=\frac{1}{\pi} ∫_{0}^{\infty} \mathrm{d} ω \bar{n}(\beta ω) J(ω) e^{-\mathrm{i} ω(t-s)}.
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\]
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Remember that we want to calculate
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\begin{equation}
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\label{eq:whatreallymatters}
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\begin{aligned}
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\ev{L^\dag\dot{B}(t)} &= \tr[L^\dag\dot{B}(t)\rho(t)] \\
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&=\prod_\lambda\qty(\int\dd[2]{y_\lambda}
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\frac{\eu^{-\abs{y_\lambda}^2\bar{n}_\lambda}}{\pi\bar{n}_\lambda})\tr[L^\dag\dot{B}(t)
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U(t)\vb{D}(y)\ketbra{\psi}\otimes\ketbra{0}\vb{D}(y)^\dag U(t)^\dag] .
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\ev{L^†\dot{B}(t)} &= \tr[L^†\dot{B}(t)\rho(t)] \\
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&=\prod_λ\qty(∫\dd[2]{y_λ}
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\frac{\eu^{-\abs{y_λ}^2\bar{n}_λ}}{\pi\bar{n}_λ})\tr[L^†\dot{B}(t)
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U(t)\vb{D}(y)\ketbra{\psi}\otimes\ketbra{0}\vb{D}(y)^† U(t)^†] .
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\end{aligned}
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\end{equation}
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To recover the zero temperature formulation of this expectation value we
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again insert a \(\id\), but have to commute \(\vb{D}(y)^\dag\) past
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again insert a \(\id\), but have to commute \(\vb{D}(y)^†\) past
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\(\dot{B}(t)\). This leads to the expression
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\begin{equation}
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\label{eq:pureagain}
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\begin{aligned}
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\ev{L^\dag\dot{B}(t)} &=\prod_\lambda\qty(\int\dd[2]{y_\lambda}
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\frac{\eu^{-\abs{y_\lambda}^2\bar{n}_\lambda}}{\pi\bar{n}_\lambda})\\
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&\qquad\times\tr[L^\dag(\dot{B}(t) + \dot{\xi}(t))
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\vb{D}^\dag(y) U(t)\vb{D}(y)\ketbra{\psi}\otimes\ketbra{0}\vb{D}^\dag(y)U(t)^\dag\vb{D}(y)] \\
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&=\prod_\lambda\qty(\int\dd[2]{y_\lambda}
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\frac{\eu^{-\abs{y_\lambda}^2\bar{n}_\lambda}}{\pi\bar{n}_\lambda})\tr[L^\dag\qty{\dot{B}(t) + \dot{\xi}(t)}
|
||||
\tilde{U}(t)\ketbra{\psi}\otimes\ketbra{0} \tilde{U}(t)^\dag].
|
||||
\ev{L^†\dot{B}(t)} &=\prod_λ\qty(∫\dd[2]{y_λ}
|
||||
\frac{\eu^{-\abs{y_λ}^2\bar{n}_λ}}{\pi\bar{n}_λ})\\
|
||||
&\qquad\times\tr[L^†(\dot{B}(t) + \dot{\xi}(t))
|
||||
\vb{D}^†(y) U(t)\vb{D}(y)\ketbra{\psi}\otimes\ketbra{0}\vb{D}^†(y)U(t)^†\vb{D}(y)] \\
|
||||
&=\prod_λ\qty(∫\dd[2]{y_λ}
|
||||
\frac{\eu^{-\abs{y_λ}^2\bar{n}_λ}}{\pi\bar{n}_λ})\tr[L^†\qty{\dot{B}(t) + \dot{\xi}(t)}
|
||||
\tilde{U}(t)\ketbra{\psi}\otimes\ketbra{0} \tilde{U}(t)^†].
|
||||
\end{aligned}
|
||||
\end{equation}
|
||||
which returns us to the zero temperature formalism with a transformed
|
||||
|
|
@ -297,7 +297,7 @@ Hamiltonian and the replacement
|
|||
\label{eq:breplacement}
|
||||
B(t) \rightarrow B(t) + \xi(t)
|
||||
\end{eqnarray}
|
||||
which plausibly corresponds to the \(L^\dag\) part of \(H_I + H_{\mathrm{sys}}^{\mathrm{shift}}\).
|
||||
which plausibly corresponds to the \(L^†\) part of \(H_I + H_{\mathrm{sys}}^{\mathrm{shift}}\).
|
||||
|
||||
The appearance of \(\dot{\xi}(t)\) may cause concern. However, for
|
||||
twice differentiable \(\mathcal{M}(\xi(t)\xi^\ast(s))\) the sample
|
||||
|
|
@ -322,13 +322,13 @@ Alternatively we can calculate
|
|||
Now,
|
||||
\begin{equation}
|
||||
\label{eq:hshcomm}
|
||||
[H_{\mathrm{sys}}^{\mathrm{shift}}, H_I] = \xi(t) [L^\dag, L]
|
||||
B(t)^\dag + \xi^\ast(t) [L, L^\dag] B
|
||||
[H_{\mathrm{sys}}^{\mathrm{shift}}, H_I] = \xi(t) [L^†, L]
|
||||
B(t)^† + \xi^\ast(t) [L, L^†] B
|
||||
\end{equation}
|
||||
and therefore
|
||||
\begin{equation}
|
||||
\label{eq:finalex}
|
||||
\ev{[H_{\mathrm{sys}}^{\mathrm{shift}}, H_I]} = -i \mathcal{M}_{\eta^\ast}\mel{\psi}{\xi(t)^\ast[L,L^\dag]D_t}{\psi}.
|
||||
\ev{[H_{\mathrm{sys}}^{\mathrm{shift}}, H_I]} = -i \mathcal{M}_{\eta^\ast}\mel{\psi}{\xi(t)^\ast[L,L^†]D_t}{\psi}.
|
||||
\end{equation}
|
||||
This is an expression that we can easily evaluate with the HOPS
|
||||
method.
|
||||
|
|
|
|||
Loading…
Add table
Reference in a new issue