master-thesis/tex/energy_transfer/main.tex

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\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)^\dag + 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}{L^\dag\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)} + c.c.,
\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)} + c.c.,
\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.

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\end{document}




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