\documentclass[fontsize=12pt,paper=a4,open=any, ,twoside=false,toc=listof,toc=bibliography, captions=nooneline,captions=tableabove,english,DIV=16,numbers=noenddot,final]{scrartcl} \usepackage{../hirostyle} \addbibresource{../references.bib} \synctex=1 \title{Calculating heat flows with HOPS} \author{Valentin Boettcher} \date{\today} \begin{document} \maketitle \tableofcontents \section{One Bath} \subsection{Linear NMQSD, Zero Temperature} As in~\cite{Hartmann2017Dec} we choose \begin{equation} \label{eq:totalH} H = H_S + \underbrace{LB^\dagger + L^\dagger B}_{H_I} + H_B \end{equation} with the system hamiltonian \(H_S\), the bath hamiltonian \begin{equation} \label{eq:bathh} H_B = \sum_\lambda \omega_\lambda a^\dag a, \end{equation} the bath coupling system operator \(L\) and the bath coupling bath operator \begin{equation} \label{eq:bop} B=\sum_{\lambda} g_{\lambda} a_{\lambda} \end{equation} which define the interaction hamiltonian \(H_I\). We define the heat flow out of the system as in~\cite{Kato2015Aug} through \begin{equation} \label{eq:heatflowdef} J = - \dv{\ev{H_B}}{t}. \end{equation} Working, for now, in the Schr\"odinger picture the Ehrenfest theorem can be employed to find \begin{equation} \label{eq:ehrenfest} \i\partial_t\ev{H_B} = \ev{[H_B,H]} = \ev{[H_B,H_I]}. \end{equation} Thus, we need to calculate \begin{eqnarray} \label{eq:calccomm} \begin{aligned} [H_B,H_I] &= [H_B, LB^\dag + L^\dag B] \\ &= L[H_B, B^\dag ] + L^\dag [H_B, B] \\ &= L[H_B, B^\dag ] - \hc. \end{aligned} \end{eqnarray} This checks out as the commutator has to be anti-hermitian due to \cref{eq:ehrenfest}. Using \([H_B, B^\dag ]=\sum_\lambda \omega_\lambda g^\ast_\lambda a^\dag_\lambda\) it follows that \begin{equation} \label{eq:expcomm} \begin{aligned} \ev{[H_B,H_I]} &= \sum_\lambda \omega_\lambda g^\ast_\lambda \ev{La^\dag_\lambda} - \cc = \sum_\lambda \omega_\lambda g^\ast_\lambda \ev{La^\dag_\lambda \eu^{\i \omega t}}_I - \cc\\ &= \frac{1}{\i}\ev{L\partial_t{\sum_\lambda g^\ast_\lambda a^\dag_\lambda \eu^{\i \omega t}}}_I - \cc =\frac{1}{\i}\qty(\ev{L\dot{B}^\dag}_I + \cc) \end{aligned} \end{equation} where we switched to the interaction picture with respect to \(H_B\) in keeping with the standard NMQSD formalism. In essence this is just the Heisenberg equation for \(H_I\). The expression for \(J\) follows \begin{equation} \label{eq:final_flow} J(t) = \ev{L^\dag\partial_t B(t) + L\partial_t B^\dag(t)}_I. \end{equation} From this point on, we will assume the interaction picture and drop the \(I\) subscript. The two summands yield different expressions in terms of the NMQSD. For use with HOPS with the final goal of utilizing the auxiliary states the expression \(\ev{L^\dag\partial_t B(t)}\) should be evaluated. We calculate \begin{equation} \label{eq:interactev} \ev{L^\dag\partial_t B(t)}=\ev{L^\dag\partial_t B(t)}{\psi(t)} = \int \braket{\psi(t)}{z}\mel{z}{L^\dag\partial_tB(t)}{\psi(t)}\frac{\dd[2]{z}}{\pi^N}, \end{equation} where \(N\) is the total number of environment oscillators and \(z=\qty(z_{\lambda_1}, z_{\lambda_2}, \ldots)\). To that end, \begin{equation} \label{eq:nmqsdficate} \begin{aligned} \mel{z}{\partial_tB(t)}{\psi(t)} &= \sum_\lambda g_\lambda \qty(\partial_t \eu^{-\i\omega_\lambda t})\partial_{z^\ast_\lambda}\ket{\psi(z^\ast,t)} \\ &= \int_0^t \sum_\lambda g_\lambda \qty(\partial_t \eu^{-\i\omega_\lambda t})\pdv{\eta_s^\ast}{z^\ast_\lambda}\fdv{\ket{\psi(z^\ast,t)}}{\eta^\ast_s}\dd{s}\\ &= -\i\int_0^t\dot{\alpha}(t-s)\fdv{\ket{\psi(z^\ast,t)}}{\eta^\ast_s}\dd{s}, \end{aligned} \end{equation} where \(\eta^\ast_t\equiv -\i \sum_\lambda g^\ast_\lambda z^\ast_\lambda \eu^{\i\omega_\lambda t}\). With this we can write \begin{equation} \label{eq:steptoproc} \ev{L^\dag\partial_t B(t)} = -\i \mathcal{M}_{\eta^\ast}\bra{\psi(\eta, t)}L^\dag\int_0^t\dd{s} \dot{\alpha}(t-s)\fdv{\eta^\ast_s} \ket{\psi(\eta^\ast,t)}. \end{equation} Defining \begin{equation} \label{eq:defdop} D_t = \int_0^t\dd{s} \alpha(t-s)\fdv{\eta^\ast_s} \end{equation} as in~\cite{Suess2014Oct} we can write \begin{equation} \label{eq:final_flow_nmqsd} J(t) = -\i \mathcal{M}_{\eta^\ast}\bra{\psi(\eta, t)}L^\dag\dot{D}_t\ket{\psi(\eta^\ast,t)} + \cc, \end{equation} where we've used that the integral in \(D_t\) can be expanded over the whole real axis. If we assume \(\alpha = \exp(-w t)\) then \begin{equation} \label{eq:hopsj} J(t) = \i \mathcal{M}_{\eta^\ast}\bra{\psi^{(0)}(\eta, t)}wL^\dag\ket{\psi^{(1)}(\eta^\ast,t)} + \cc., \end{equation} where \(\ket{\psi^{(1)}(\eta^\ast,t)}\) is the first HOPS hierarchy state. This can be generalized to any BCF that is a sum of exponentials. Interestingly one finds that \begin{equation} \label{eq:alternative} \ev{L\partial_t B^\dag(t)} = \i\int\frac{\dd[2]{z}}{\pi^N} \dot{\eta}_t^\ast \mel{\psi(\eta,t)}{L}{\psi(\eta^\ast,t)}. \end{equation} However, this seems numerically problematic because \(\eta^\ast\) is not differentiable in general. The previous expression has the advantage that we utilize the first hierarchy states that are already being calculated as a byproduct. In the language of~\cite{Hartmann2021Aug} we can generalize to \(\alpha(t) = \sum_i G_i \eu^{-W_i t}\) and thus \begin{equation} \label{eq:hopsflowrich} J(t) = \sum_\mu\frac{G_\mu W_\mu}{\bar{g}_\mu} \i\mathcal{M}_{\eta^\ast}\bra{\psi^{(0)}(\eta, t)}L^\dag\ket{\psi^{\vb{e}_\mu}(\eta^\ast,t)} + \cc, \end{equation} where \(\psi^{\vb{e}_\mu}\) is the \(\mu\)-th state of the first hierarchy and \(\bar{g}_\mu\) is an arbitrary scaling introduced in the definition of the hierarchy in~\cite{Hartmann2021Aug} to help with the scaling of the norm. \subsection{Nonlinear NMQSD, Zero Temperature} \label{sec:nonlin} In the spirit of the usual derivation of the nonlinear NMQSD we write \begin{equation} \label{eq:newb} \begin{aligned} \ev{L^\dag\dot{B(t)}} &= \int \frac{\dd[2]{z}}{\pi^N} \eu^{-\abs{z}^2} \braket{\psi}{z}\!\braket{z}{\psi} \frac{\braket{\psi(t)}{z}\!\mel{z}{L^\dag\dot{B(t)}}{\psi(t)}}{\braket{\psi}{z}\!\braket{z}{\psi}} \\ &= \int \frac{\dd[2]{z}}{\pi^N} \eu^{-\abs{z}^2} \frac{\mel{z(t)}{L^\dag\dot{B(t)}}{\psi(t)}}{\braket{\psi}{z(t)}\!\braket{z(t)}{\psi}}, \end{aligned} \end{equation} where \(z_{\lambda}^{*}(t)=z_{\lambda}^{*}+\i g_{\lambda} \int_{0}^{t} \dd{s} \eu^{-\i \omega_{\lambda} s}\ev{L^\dagger}_{s}\). We find that next steps are the same as in \cref{sec:nonlin} by noting \begin{equation} \label{eq:deriv_trick} \eval{\partial_{z^\ast_\lambda}}_{z^\ast=z_\lambda^\ast(t)} = \int_0^t\dd{s}\eval{\pdv{\eta^\ast}{z^\ast_\lambda}}_{z^\ast=z^\ast_\lambda(t)} \fdv{}{\eta^\ast_s(z^\ast=z^\ast(t))} = \int_0^t\dd{s}\eval{\pdv{\eta^\ast}{z^\ast_\lambda}}_{z^\ast=0} \fdv{}{\eta^\ast_s(z^\ast=z^\ast(t))}, \end{equation} which does alter the definition of \(D_t\) but results in the same HOPS equations. The shifted process \(\tilde{\eta}^\ast= \eta^\ast(z^\ast(t),t)=\eta^\ast(t) + \int_0^t\dd{s}\alpha^\ast(t-s)\ev{L^\dag}_{\psi_s}\) appears directly in the NMQSD equation but results only in a slight change in the functional derivative. Note however that \begin{equation} \label{eq:fdvclarification} \fdv{}{\eta^\ast_s(z^\ast=z^\ast(t))} \neq \fdv{}{\tilde{\eta}^\ast_s} \end{equation} which is not problematic as we have (implicit in~\cite{Diosi1998Mar}) \begin{equation} \label{eq:fdvhops} \fdv{}{\eta^\ast_s(z^\ast=z^\ast(t))} \ket{\psi(z^\ast)} = \fdv{}{\eta^\ast_s}\ket{\psi(z^\ast(t, z^\ast_0), t)} \end{equation} so that the usual HOPS hierarchy follows. Note \(z^\ast_0 = z^\ast(0)\). Therefore, \begin{equation} \label{eq:newbcontin} J(t) = -\i \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)}} + \cc, \end{equation} where the dependence on \(\tilde{\eta}\) is symbolic and to be understood in the context of \cref{eq:fdvhops}. Again we express the result in the language of~\cite{Hartmann2021Aug} to obtain \begin{equation} \label{eq:hopsflowrich} J(t) = \sum_\mu\frac{G_\mu W_\mu}{\bar{g}_\mu} \i\mathcal{M}_{\eta^\ast}\frac{\bra{\psi^{(0)}(\eta, t)}L^\dag\ket{\psi^{\vb{e}_\mu}(\eta^\ast,t)}}{\bra{\psi^{(0)}(\eta, t)}\ket{\psi^{0}(\eta^\ast,t)}} + \cc. \end{equation} \subsection{Linear Theory, Finite Temperature} The finite temperature case needs some additional considerations as the previous sections dealt explicitly with mean values in a pure state. The Ehrenfest theorem still holds in mixed states, but we would like to recover the usual pure state zero temperature formalism. There are multiple methods for dealing with a thermal initial such as the thermofield method (see~\cite{Diosi1998Mar}), but because the results discussed here are to be applied with the HOPS method we shall use the method described in~\cite{Hartmann2017Dec}. The shift operator \begin{equation} \label{eq:shiftop} \vb{D}(y) = \bigotimes_\lambda \eu^{y_\lambda a_\lambda^\dag-y^\ast_\lambda a_\lambda} \end{equation} the ground state of the environment into an arbitrary coherent state \begin{equation} \label{eq:shiftwork} \vb{D}(y)\ket{0} = \ket{y} \end{equation} where \(y=(y_1,y_2,\ldots)\) as usual. This allows us to write the density matrix of the system with a thermal initial bath as \begin{equation} \label{eq:shiftbath} \rho = \prod_\lambda\qty(\int\dd[2]{y_\lambda} \frac{\eu^{-\abs{y_\lambda}^2\bar{n}_\lambda}}{\pi\bar{n}_\lambda}) U(t)\vb{D}(y)\ketbra{\psi}\otimes\ketbra{0}\vb{D}(y)^\dag U(t)^\dag. \end{equation} The usual step is now to insert \(\id =\vb{D}(y)\vb{D}^\dag(y)\) to arrive at a new time translation operator \begin{equation} \label{eq:utilde} \tilde{U}(t) = \vb{D}^\dag(y)U(t)\vb{D}(y) \end{equation} and to interpret the integral in \cref{eq:shiftbath} in a monte-carlo sense which leads to a stochastic contribution to the system Hamiltonian \begin{equation} \label{eq:thermalh} H_{\mathrm{sys}}^{\mathrm{shift}}=L \xi^{*}(t)+L^{\dagger} \xi(t) \end{equation} with the stochastic process \begin{equation} \label{eq:xiproc} \xi(t):=\sum_{\lambda} g_{\lambda} y_{\lambda} \eu^{-\mathrm{i} \omega_{\lambda} t} \end{equation} with corresponding moments \(\mathcal{M}(\xi(t))=0=\mathcal{M}(\xi(t) \xi(s))\) and \[ \mathcal{M}\left(\xi(t) \xi^{*}(s)\right)=\frac{1}{\pi} \int_{0}^{\infty} \mathrm{d} \omega \bar{n}(\beta \omega) J(\omega) e^{-\mathrm{i} \omega(t-s)}. \] Remember that we want to calculate \begin{equation} \label{eq:whatreallymatters} \begin{aligned} \ev{L^\dag\dot{B}(t)} &= \tr[L^\dag\dot{B}(t)\rho(t)] \\ &=\prod_\lambda\qty(\int\dd[2]{y_\lambda} \frac{\eu^{-\abs{y_\lambda}^2\bar{n}_\lambda}}{\pi\bar{n}_\lambda})\tr[L^\dag\dot{B}(t) U(t)\vb{D}(y)\ketbra{\psi}\otimes\ketbra{0}\vb{D}(y)^\dag U(t)^\dag] . \end{aligned} \end{equation} To recover the zero temperature formulation of this expectation value we again insert a \(\id\), but have to commute \(\vb{D}(y)^\dag\) past \(\dot{B}(t)\). This leads to the expression \begin{equation} \label{eq:pureagain} \begin{aligned} \ev{L^\dag\dot{B}(t)} &=\prod_\lambda\qty(\int\dd[2]{y_\lambda} \frac{\eu^{-\abs{y_\lambda}^2\bar{n}_\lambda}}{\pi\bar{n}_\lambda})\\ &\qquad\times\tr[L^\dag(\dot{B}(t) + \dot{\xi}(t)) \vb{D}^\dag(y) U(t)\vb{D}(y)\ketbra{\psi}\otimes\ketbra{0}\vb{D}^\dag(y)U(t)^\dag\vb{D}(y)] \\ &=\prod_\lambda\qty(\int\dd[2]{y_\lambda} \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]. \end{aligned} \end{equation} which returns us to the zero temperature formalism with a transformed Hamiltonian and the replacement \begin{eqnarray} \label{eq:breplacement} B(t) \rightarrow B(t) + \xi(t) \end{eqnarray} which plausibly corresponds to the \(L^\dag\) part of \(H_I + H_{\mathrm{sys}}^{\mathrm{shift}}\). The appearance of \(\dot{\xi}(t)\) may cause concern. However, for twice differentiable \(\mathcal{M}(\xi(t)\xi^\ast(s))\) the sample trajectories are smooth. Alternatively we can calculate \begin{equation} \label{eq:gettingarounddot} \begin{aligned} \ev{\dot{H}_{\mathrm{sys}}^{\mathrm{shift}}} &= \dv{\ev{H_{\mathrm{sys}}^{\mathrm{shift}}}}{t} - \frac{1}{\iu}\qty(\ev{H_{\mathrm{sys}}^{\mathrm{shift}}H} -\ev{H H_{\mathrm{sys}}^{\mathrm{shift}}}) \\ &=\dv{\ev{H_{\mathrm{sys}}^{\mathrm{shift}}}}{t} - \frac{1}{\iu}\ev{[H_{\mathrm{sys}}^{\mathrm{shift}}, H]} \\ &=\dv{\ev{H_{\mathrm{sys}}^{\mathrm{shift}}}}{t} - \frac{1}{\iu}\ev{[H_{\mathrm{sys}}^{\mathrm{shift}}, H_I]}. \end{aligned} \end{equation} 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 \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}. \end{equation} This is an expression that we can easily evaluate with the HOPS method. \printbibliography{} \end{document} %%% Local Variables: %%% mode: latex %%% TeX-master: t %%% End: