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@ -1,24 +1,27 @@
\chapter{An Analytical Solution for Quantum Brownian Motion Models} \chapter{An Analytical Solution for Quantum Brownian Motion Models}
\label{chap:analytsol} \label{chap:analytsol}
\begin{itemize}
\item \cref{chap:flow} is all nice and good, but need verification The results of \cref{chap:flow} are promising from a numerical
\item many actual numerical approximations to control besides theory perspective remain to be verified. Previous
verification work~\cite{Hartmann2017Dec,RichardDiss} has made it clear that the
\end{itemize} reduced system dynamics, but it is an open question whether bath
related quantities can be calculated to a similar degree of accuracy.
The best possible verification is the comparison with a soluble model,
ideally solved with a method completely different from the NMQSD. In
this chapter we will present the solution to the Heisenberg equations
of two quantum Brownian motion models for a single (\cref{sec:oneosc})
and two baths (\cref{sec:twoosc}) and time independent Hamiltonians.
These solutions will enable us to calculate the bath energy flow \(J\)
for one or two baths which will be compared with the numerical
solution in \cref{sec:hopsvsanalyt}.
\section{A Harmonic Oscillator coupled to a single Bath} \section{A Harmonic Oscillator coupled to a single Bath}
\label{sec:oneosc} \label{sec:oneosc}
\begin{itemize} A simple quadratic model that is soluble~\cite{Breuer2002Jun} and of
\item simplest model in the spirit of previous derivations the form \cref{eq:totalH} is given by
\item well known to be solvable, we will tease out the bath related
quantities in the heisenberg picture
\end{itemize}
The solution presented here is not entirely new, as the model is well
known. For reference see~\cite{Breuer2002Jun}. The trick with the
exponential expansion of the bath correlation function has been
arrived at independently, but may also be well known.
The model is given by the quadratic hamiltonian
\begin{equation} \begin{equation}
\label{eq:one_ho_hamiltonian} \label{eq:one_ho_hamiltonian}
H = \frac{Ω}{4}\qty(p^2+q^2) + \frac{1}{2} q H = \frac{Ω}{4}\qty(p^2+q^2) + \frac{1}{2} q
@ -29,14 +32,13 @@ where \(a,a^†\) are the ladder operators of the harmonic
oscillator, \(q=a+a^†\) and \(p=\frac{1}{\iu}\qty(a-a^†)\) so oscillator, \(q=a+a^†\) and \(p=\frac{1}{\iu}\qty(a-a^†)\) so
that \([q,p] = 2\iu\). that \([q,p] = 2\iu\).
The Heisenberg equation yields The Heisenberg equations for \cref{eq:one_ho_hamiltonian}
\begin{align} \begin{align}
\dot{q} &=Ω p \label{eq:qdot}\\ \dot{q} &=Ω p \label{eq:qdot}\\
\dot{p} &= -Ω q - \int_0^t \Im[α_0(t-s)] q(s)\dd{s} + W(t) \label{eq:pdot} \dot{p} &= -Ω q - \int_0^t \Im[α_0(t-s)] q(s)\dd{s} + W(t) \label{eq:pdot}
\\ \\
\dot{b}_λ &= -\iu g_λ \frac{q}{2} - \iu\omega_λ b_λ \dot{b}_λ &= -\iu g_λ \frac{q}{2} - \iu\omega_λ b_λ
\end{align} \end{align}
with the operator noise with the operator noise
\(W(t)=-\sum_λ \qty(g_λ^\ast b_λ(0) \(W(t)=-\sum_λ \qty(g_λ^\ast b_λ(0)
\eu^{-\iu\omega_λ t } + g_λ b_λ^†(0) \eu^{-\iu\omega_λ t } + g_λ b_λ^†(0)
@ -79,9 +81,12 @@ leads to an algebraic formula
\mathcal{L}\{G\}(z) = \qty(z-A + \mathcal{L}\{K\}(z))^{-1}. \mathcal{L}\{G\}(z) = \qty(z-A + \mathcal{L}\{K\}(z))^{-1}.
\end{equation} \end{equation}
\subsection{Solution} \subsection{Solution}
\label{sec:solution} \label{sec:solution}
To solve \cref{eq:eqmotprop} and find the propagator \(G\), we have to
find an explicit expression for \cref{eq:galgebr}, a simple matrix
inversion, and then apply the inverse transformation.
We observe that We observe that
\begin{equation} \begin{equation}
\label{eq:mdef} \label{eq:mdef}
@ -95,7 +100,7 @@ and therefore
\mqty(z & Ω \\ -(Ω + \mathcal{L}\{\Im[α_0]\}(z)) & z). \mqty(z & Ω \\ -(Ω + \mathcal{L}\{\Im[α_0]\}(z)) & z).
\end{equation} \end{equation}
From this we can conclude that \(G_{11}=G_{22}\). From this we can conclude that \(G_{11}=G_{22}\).
Because \(\ev{W(s)}=0\) for thermal initial states of the bath we have Because \(\ev{W(s)}=0\) holds for thermal initial states of the bath we have
\begin{equation} \begin{equation}
\label{eq:meanvals} \label{eq:meanvals}
\mqty(\ev{q(t)}\\ \ev{p(t)}) = G(t)\mqty(\ev{q(0)}\\ \ev{p(0)}). \mqty(\ev{q(t)}\\ \ev{p(t)}) = G(t)\mqty(\ev{q(0)}\\ \ev{p(0)}).
@ -107,11 +112,16 @@ Knowing this, we can deduce from \(\ev{\dot{q}}= Ω \ev{p}\) that
\end{align} \end{align}
These relations are true independent of the initial state of the These relations are true independent of the initial state of the
system. It therefore suffices if we concern ourselves with system. It therefore suffices if we concern ourselves with
\(G_{12}\). We nevertheless continue in full generality. \(G_{12}\).
We nevertheless continue in full generality and approach the inverse
Laplace transformation by expanding the BCF in terms of functions that
have a simple Laplace transform. As we also use an exponential
expansion in HOPS and are only interested in finite times, we may
choose \(α_0(t)=\sum_{n=1}^N G_n \eu^{-W_n t - \i \varphi_n}\) with
\(W_n=\gamma_n + \i\delta_n\) and
\(G_n, \varphi_n, \gamma_n,\delta_n\in\RR\) for \(t\geq 0\).
Assume that \(α_0(t)=\sum_{n=1}^N G_n \eu^{-W_n t - \i
\varphi_n}\) with \(W_n=\gamma_n + \i\delta_n\) and \(G_n,
\varphi_n, \gamma_n,\delta_n\in\RR\) for \(t\geq 0\).
This leads to a mathematically simple expression for the Laplace This leads to a mathematically simple expression for the Laplace
transform transform
\begin{equation} \begin{equation}
@ -166,10 +176,10 @@ of degree \(2(N+1)\) where we \textbf{assume that there are
no roots with multiplicity greater than one}. no roots with multiplicity greater than one}.
With this we can now calculate the inverse laplace transform of With this we can now calculate the inverse laplace transform of
expressions of the form \(\frac{f_0(z)g(z)}{p(z)}\) where \(g(z)\) is expressions of the form \(\frac{f_0(z)g(z)}{p(z)}\) where \(g(z)\) is
any holonome function so that \(\frac{f_0(z)g(z)}{p(z)} \eu^{z\cdot any holonomic function so that \(\frac{f_0(z)g(z)}{p(z)} \eu^{z\cdot
t}\) falls off fast enough for \(t\geq 0\), t}\) falls off fast enough for \(t\geq 0\),
\(\Re(z)>\max_l{\Re(\tilde{z}_l)}=\Delta\) and \(\Re(z) \rightarrow \(\Re(z)>\max_l{\Re(\tilde{z}_l)}=\Delta\) and \(\Re(z) \rightarrow
-\infty\). With this we can close the contour of the inverse Laplace -\infty\). Now the contour of the inverse Laplace
transform transform
\begin{equation} \begin{equation}
\label{eq:invlap} \label{eq:invlap}
@ -177,7 +187,7 @@ transform
\frac{1}{2\pi\i}\int_{\Delta - \i\infty}^{\Delta + \i\infty} \frac{f_0(z)g(z)}{p(z)} \eu^{z\cdot \frac{1}{2\pi\i}\int_{\Delta - \i\infty}^{\Delta + \i\infty} \frac{f_0(z)g(z)}{p(z)} \eu^{z\cdot
t}\dd{z} t}\dd{z}
\end{equation} \end{equation}
to the left to obtain can be closed to the left to obtain
\begin{equation} \begin{equation}
\label{eq:simpleinvtrans} \label{eq:simpleinvtrans}
\mathcal{L}^{-1}\qty{\frac{f_0(z)g(z)}{p(z)}}(t) \mathcal{L}^{-1}\qty{\frac{f_0(z)g(z)}{p(z)}}(t)
@ -192,7 +202,8 @@ all our purposes. For completeness we give
p'(z) = 2\mu\sum_{k=1}^{N+1}\qty[(z-\Re(\tilde{z}_k))\prod_{\substack{n=1\\ p'(z) = 2\mu\sum_{k=1}^{N+1}\qty[(z-\Re(\tilde{z}_k))\prod_{\substack{n=1\\
n\neq k}}^{N+1}(z-\tilde{z}_n)(z-\tilde{z}^\ast_n)]. n\neq k}}^{N+1}(z-\tilde{z}_n)(z-\tilde{z}^\ast_n)].
\end{equation} \end{equation}
We can immediately conclude that all elements of \(G\) are sums of
It can be immediately concluded that all elements of \(G\) are sums of
exponentials, just like the BCF. In particular exponentials, just like the BCF. In particular
\begin{equation} \begin{equation}
\label{eq:gfinal} \label{eq:gfinal}
@ -201,7 +212,17 @@ exponentials, just like the BCF. In particular
\end{equation} \end{equation}
with \(R_l={f_0(\tilde{z}_l)}/{p'(\tilde{z}_l)}\). with \(R_l={f_0(\tilde{z}_l)}/{p'(\tilde{z}_l)}\).
It may be noted that this solution does not contain any notion of
temperature, as we are working in the Heisenberg picture.
\subsubsection{Negative Times} \subsubsection{Negative Times}
For completeness, it may be of interest to find a solution for
negative times. This solution is relatively unphysical, as the initial
condition of a product state plays a pivotal role in open system
dynamics~\cite{Rivas2012}. Therefore a system that starts out in some
entangled state just to reach the perfect product state at \(t=0\) is
not something that is likely to be applicable to physical questions.
The solution detailed above is only valid for positive times. Because The solution detailed above is only valid for positive times. Because
we strive to employ the same formalism again for negative times, we we strive to employ the same formalism again for negative times, we
will concern ourselves with the transformed quantities will concern ourselves with the transformed quantities
@ -236,13 +257,20 @@ and
\subsection{Applications} \subsection{Applications}
\label{sec:applications} \label{sec:applications}
\begin{itemize} Having found an expression for \(G\), we have in principle solved the
\item the actual ``meat'' that is of interest to verify HOPS model. It remains however to apply that solution in a way that is
\end{itemize} contributing towards our goal of validating the results of
\cref{chap:flow}.
Knowing \(G\) and \(α\), we can calculate all observables of the Knowing \(G\) and \(α\), we can calculate all observables of the
system. Simple closed form expressions of sums of exponentials can be system with the ultimate goal of finding an expression for
obtained by using an exponential expansions of \(α\). Throughout, \(J=-∂_{t}\ev{H_{B}}\). Simple closed form expressions of sums of
we assume a thermal bath initial state so that \(\ev{W(t)}=0\). exponentials will be obtained by using an exponential expansion of
\(α\).
Throughout, we assume a thermal bath initial state so that
\(\ev{W(t)}=0\).
\subsubsection{Correlation Functions} \subsubsection{Correlation Functions}
\label{sec:correl} \label{sec:correl}
@ -259,7 +287,7 @@ We proceed to calculate \(\ev{q(t)q(s)}\). For brevity we set
\end{aligned} \end{aligned}
\end{equation} \end{equation}
For a pure harmonic oscillator initial state \(\ket{n}\) we have For a Fock type initial state \(\ket{n}\) we have
\begin{equation} \begin{equation}
\label{eq:hoexp} \label{eq:hoexp}
\begin{aligned} \begin{aligned}
@ -326,73 +354,17 @@ For further evaluation of \cref{eq:bathderiv_1} we have to calculate
\biggr] \biggr]
\end{aligned} \end{aligned}
\\ \\
&= ∫_0^t\dd{r} g_1(t-r)\qty[g_2(t-r) + g_3(t,r)] = Λ_1(t) + Λ_2(t) &= ∫_0^t\dd{r} g_1(t-r)\qty[g_2(t-r) + g_3(t,r)] = Λ_1(t) + Λ_2(t),
\end{aligned} \end{aligned}
\end{equation} \end{equation}
where \(Λ(t,s)\) was defined in \cref{eq:qcorrel}.
This expression now only uses \(α(t)\) for \(t\geq 0\) so that we can This expression now only uses \(α(t)\) for \(t\geq 0\) so that we can
once again employ the exponential expansion for \(α\). In fact, all once again employ the exponential expansion for \(α\).
other quantities in \cref{eq:lambdafold} have exponential expansion so
that we can now define\footnote{Note that this is inconsistent with
\cref{sec:solution}.}
\begin{equation}
\label{eq:expansions}
\begin{aligned}
α_0&=∑_k U_k\eu^{-Q_k t} & \dot{α}_0&=∑_k P_k\eu^{-L_k t} & α(t)
&= ∑_nG_n\eu^{-W_n t} \\
A(t) &= ∑_l A_l\eu^{-C_l t} & B(t) &= ∑_l B_l\eu^{-C_l t}.
\end{aligned}
\end{equation}
With this we can calculate, We will arrive at expressions that are weighted sums of exponentials
\begin{align} whose detailed calculation is quite tedious and can be found
\label{eq:lambdaintegrals} in\cref{sec:explicit_flow}.
∫_r^t\dd{s}B(s-r)\dot{α}_0(t-s)
&=\sum_{m,k}\underbrace{\frac{B_mP_k}{L_k-C_m}}_{\equiv
Γ^1_{mk}}\qty[\eu^{-C_m(t-r)}-\eu^{-L_k(t-r)}]=g_1(t-r)\\
∫_0^{t-r}\dd{u}B(t-r-u)α(u)
&=\sum_{n,l}\underbrace{\frac{B_nG_l}{C_n-W_l}}_{\equiv
Γ^2_{nl}}\qty[\eu^{-W_l(t-r)}-\eu^{-C_n(t-r)}]=g_2(t-r)\\
∫_0^{r}\dd{u}B(t-r+u)α^\ast(u)
&=\sum_{n,l}\underbrace{\frac{B_nG_l^\ast}{C_n+W_l^\ast}}_{\equiv
Γ^3_{nl}}\qty[\eu^{-C_n(t-r)}-\eu^{-W_l^\ast r-C_n t}]=g_3(t,r)
\end{align}
and
\begin{align}
\label{eq:finalsummands}
Λ_1(t)&= ∑_{m,k,n,l}Γ^1_{mk}Γ^2_{nl}\qty[\frac{1-\eu^{-(C_m+W_l)t}}{C_m+W_l}
-
\frac{1-\eu^{-(C_m+C_n)t}}{C_m+C_n}-
\frac{1-\eu^{-(L_k+W_l)t}}{L_k+W_l}
+
\frac{1-\eu^{-(L_k+C_n)t}}{L_k+C_n}]\\
Λ_2(t)&=
\begin{aligned}[t]
∑_{m,k,n,l}Γ^1_{mk}Γ^3_{nl}\biggl[\frac{1-\eu^{-(C_m+C_n)t}}{C_m+C_n}
&-\frac{1-\eu^{-(L_k+C_n)t}}{L_k+C_n}
\\&-\frac{\eu^{-(C_n+W_l^\ast)t}-\eu^{-(C_m+C_n)t}}{C_m-W_l^\ast}
+\frac{\eu^{-(C_n+W_l^\ast)t}-\eu^{-(L_k+C_n)t}}{L_k-W^\ast_l}\biggr]
\end{aligned}
\end{align}
Also required for \cref{eq:bathderiv_1} are
\begin{align}
\label{eq:ABconv}
∫_0^t\dd{s}A(s)\dot{α}_0(t-s) &= ∑_{n,m}\underbrace{\frac{A_nP_m}{L_m-C_n}}_{\equiv
Γ^A_{nm}}\qty[\eu^{-C_n t}-\eu^{-L_m t}]\\
∫_0^t\dd{s}B(s)\dot{α}_0(t-s) &= ∑_{n,m}Γ^1_{nm}\qty[\eu^{-C_n t}-\eu^{-L_m t}]
\end{align}
and
\begin{multline}
\label{eq:nonzerotemplim}
∫_0^t\dd{s}A(s)\qty(α(s)-α_0(s)) =\\
∑_{m,n}\frac{A_nG_m}{C_n+W_m}\qty(1-\eu^{-(C_n+W_m)t}) - ∑_{m,n}\frac{A_nU_m}{C_n+Q_m}\qty(1-\eu^{-(C_n+Q_m)t}).
\end{multline}
This concludes the calculation. A possible measure of simplification
would be to write \cref{eq:bathderiv_1} as a sum of exponentials and
give explicit expressions for the coefficients and exponents. This is
not required for now. Code implementing this can be found under
\url{https://github.com/vale981/hopsflow}.
\section{Two coupled Harmonic Oscillators coupled to two Baths}% \section{Two coupled Harmonic Oscillators coupled to two Baths}%
\label{sec:twoosc} \label{sec:twoosc}
@ -403,16 +375,19 @@ not required for now. Code implementing this can be found under
not trivially additive and we can reuse the method from not trivially additive and we can reuse the method from
\cref{sec:oneosc} without alteration \cref{sec:oneosc} without alteration
\end{itemize} \end{itemize}
As we would like to verify our method also for more than one bath, a
model with two baths is required.
The considerations of~\cref{sec:oneosc} can be straight forwardly The considerations of~\cref{sec:oneosc} can be straight forwardly
generalized to the case of two coupled oscillators coupled in turn to generalized to the case of two coupled oscillators coupled in turn to
a bath each. a bath each. This construction is chosen so that the previous results
can be reused and the coupling to the baths is not trivial.
We will not give explicit formulas for the results in terms of sums of We will not give explicit formulas for the results in terms of sums of
exponentials, as they are quite extensive and easily obtained via the exponentials, as they are quite extensive and easily obtained via the
use of a computer algebra system or the aforementioned code. use of a computer algebra system or the aforementioned code.
The model is again given by a quadratic hamiltonian The model is again given by a quadratic Hamiltonian
\begin{equation} \begin{equation}
\label{eq:hamiltonian_two_bath} \label{eq:hamiltonian_two_bath}
\begin{aligned} \begin{aligned}
@ -427,28 +402,27 @@ The \(a_i^{(i)},a_i^{†}\) are the ladder operators of the harmonic
oscillators and \(q_i=a_i+a_i^†\) and \(p=\frac{1}{\iu}\qty(a_i-a_i^†)\) so oscillators and \(q_i=a_i+a_i^†\) and \(p=\frac{1}{\iu}\qty(a_i-a_i^†)\) so
that \([q_i,p_j] = 2\iuδ_{ij}\) and \([q_i,q_j] = [p_i,p_j] = 0\). that \([q_i,p_j] = 2\iuδ_{ij}\) and \([q_i,q_j] = [p_i,p_j] = 0\).
\subsection{Equations of Motion} The Heisenberg equations for \cref{eq:hamiltonian_two_bath} are
\label{sec:eqmot_two}
The Heisenberg equation yields
\begin{align} \begin{align}
\dot{q}_i &= Ω_i p_i \label{eq:qidot}\\ \dot{q}_i &= Ω_i p_i \label{eq:qidot}\\
\dot{p}_i &= -(Ω_i+γ) q_i - \int_0^t \Im[α^{(i)}_0(t-s)] q_i(s)\dd{s} + W_i(t) \label{eq:pidot} \dot{p}_i &= -(Ω_i+γ) q_i - \int_0^t \Im[α^{(i)}_0(t-s)] q_i(s)\dd{s} + W_i(t) \label{eq:pidot}
\\ \\
\dot{b}^{(i)}_λ &= -\iu g^{(i)}_λ \frac{q_i}{2} - \iu\omega^{(i)}_λ b^{(i)}_λ \dot{b}^{(i)}_λ &= -\iu g^{(i)}_λ \frac{q_i}{2} - \iu\omega^{(i)}_λ b^{(i)}_λ,
\end{align} \end{align}
with the operator noise with the operator noise
\(W_i(t)=-\sum_λ \qty(g_λ^{(i),\ast} b^{(i)}_λ(0) \(W_i(t)=-\sum_λ \qty(g_λ^{(i),\ast} b^{(i)}_λ(0)
\eu^{-\iu\omega^{(i)}_λ t } + g_λ^{(i)} b_λ^{(i),†}(0) \eu^{-\iu\omega^{(i)}_λ t } + g_λ^{(i)} b_λ^{(i),†}(0)
\eu^{\iu\omega^{(i)}_λ t })\) satisfying \(\ev{W_i(s)}=0\) and \eu^{\iu\omega^{(i)}_λ t })\) satisfying \(\ev{W_i(s)}=0\) and
\(\ev{W_i(t)W_j(s)}=δ_{ij}α^{(i)}(t-s)\). We introduced \(α^{(i)}_0 \(\ev{W_i(t)W_j(s)}=δ_{ij}α^{(i)}(t-s)\). We introduced
\equiv \eval{α^{(i)}}_{T=0}\). \(α^{(i)}_0 \equiv \eval{α^{(i)}}_{T=0}\).
We have given most quantities an extra index and accounted for the We have given most quantities an extra index and accounted for the
coupling between the two oscillators. Apart from this, the equations coupling between the two oscillators. Apart from this, the equations
of motion have the same structure as in \cref{seq:eqmot}. of motion have the same structure as in \cref{sec:oneosc}.
Again, we obtain \subsection{Solution}
\label{sec:eqmot_two}
With the same general program as before, we we first obtain
\begin{equation} \begin{equation}
\label{eq:bsoltwo} \label{eq:bsoltwo}
b^{(i)}_λ(t) = b^{(i)}_λ(0) \eu^{-\iu ω^{(i)}_λ t} - \frac{\iu g^{(i)}_λ}{2}∫_0^t b^{(i)}_λ(t) = b^{(i)}_λ(0) \eu^{-\iu ω^{(i)}_λ t} - \frac{\iu g^{(i)}_λ}{2}∫_0^t
@ -501,7 +475,7 @@ This can be done easily\footnote{We have use a computer algebra
\begin{equation} \begin{equation}
\label{eq:minvtwo} \label{eq:minvtwo}
M^{-1}(z) = \frac{1}{\det[M](z)} \tilde{M}(z) M^{-1}(z) = \frac{1}{\det[M](z)} \tilde{M}(z)
\end{equation} \end{equation}
where \(\tilde{M}\) is a matrix containing only polynomials of \(z\) where \(\tilde{M}\) is a matrix containing only polynomials of \(z\)
and of the Laplace transforms of the bath correlation functions. and of the Laplace transforms of the bath correlation functions.
@ -509,15 +483,16 @@ The numerator is
\begin{equation} \begin{equation}
\label{eq:numerator} \label{eq:numerator}
\begin{aligned} \begin{aligned}
\det[M](z)=a& b \Lambda \Omega +a \left(\gamma \det[M](z)=a(z)& b(z) \Lambda \Omega +a(z) \left(\gamma
\Lambda \Omega +\Lambda ^2 \Omega +\Omega z^2\right) \Lambda \Omega +\Lambda ^2 \Omega +\Omega z^2\right)
+b +b(z)
\left(\gamma \Lambda \Omega +\Lambda \Omega ^2+\Lambda z^2\right)\\ \left(\gamma \Lambda \Omega +\Lambda \Omega ^2+\Lambda z^2\right)\\
&+\gamma \Lambda ^2 \Omega +\gamma \Lambda &+\gamma \Lambda ^2 \Omega +\gamma \Lambda
\Omega ^2+\Lambda ^2 \Omega ^2+\gamma \Lambda z^2+\gamma \Omega z^2+\Lambda ^2 z^2+\Omega ^2 z^2+z^4, \Omega ^2+\Lambda ^2 \Omega ^2+\gamma \Lambda z^2+\gamma \Omega z^2+\Lambda ^2 z^2+\Omega ^2 z^2+z^4,
\end{aligned} \end{aligned}
\end{equation} \end{equation}
where \(a=\mathcal{L}\{\Im[α^{(1)}_0]\}\) and \(b=\mathcal{L}\{\Im[α^{(2)}_0]\}\). where \(a(z)=\mathcal{L}\{\Im[α^{(1)}_0]\}(z)\) and
\(b(z)=\mathcal{L}\{\Im[α^{(2)}_0]\}(z)\).
Using the same approach as in \cref{sec:solution}, we arrive at an Using the same approach as in \cref{sec:solution}, we arrive at an
expression similar to \cref{eq:prefactorrational} for expression similar to \cref{eq:prefactorrational} for
@ -534,17 +509,17 @@ exponentials
G(t) = \sum_{l=1}^{2+N_1+N_2}\qty[R_l \tilde{M}(\tilde{z}_l)\eu^{\tilde{z}_l \cdot G(t) = \sum_{l=1}^{2+N_1+N_2}\qty[R_l \tilde{M}(\tilde{z}_l)\eu^{\tilde{z}_l \cdot
t} + \cc] t} + \cc]
\end{equation} \end{equation}
with \(R_l={f_0(\tilde{z}_l)}/{p'(\tilde{z}_l)}\). with \(R_l={f_0(\tilde{z}_l)}/{p'(\tilde{z}_l)}\) as defined in
\cref{sec:solution}.
\subsection{Applications} \subsection{Applications}
\subsubsection{Correlation Functions} \subsubsection{Correlation Functions}
\label{sec:correltwo} \label{sec:correltwo}
\begin{itemize}
\item same game as in \cref{sec:applications}
\end{itemize}
We can now proceed to calculate the correlation functions We can now proceed to calculate the correlation functions
\(C(t,s) = \ev{x_i(t)x_j(s)}\) where the \(x_i\) are the phase space operators \(C(t,s) = \ev{x_i(t)x_j(s)}\) where the \(x_i\) are the phase space operators
of the two harmonic oscillators. of the two harmonic oscillators. This will enable use to calculate the
system energies of the two oscillators (omitted here) and again the
bath energy flows of the two baths.
We find We find
\begin{equation} \begin{equation}
@ -556,6 +531,18 @@ We find
The matrix \(Θ_{ij}\) contains the bath-induced correlations and can The matrix \(Θ_{ij}\) contains the bath-induced correlations and can
be calculated as in the single-oscillator case. be calculated as in the single-oscillator case.
For two oscillators that are initially in Fock states
\(\ket{ψ_{0}}=\ket{n}\otimes\ket{m}\) we have
\begin{equation}
\label{eq:initial_corr}
C(0,0) =
\begin{pmatrix}
1 + 2 n & \i & 0 & 0 \\
-\i & 1 + 2 n & 0 & 0 \\
0 & 0 & 1 + 2 m & \i \\
0 & 0 & -\iu & 1 + 2m
\end{pmatrix}.
\end{equation}
\subsubsection{Bath Enery Derivative} \subsubsection{Bath Enery Derivative}
\label{sec:bathflowtwo} \label{sec:bathflowtwo}
@ -568,6 +555,10 @@ Similar to the calculations in \cref{sec:bathflow} we find
∑_{k=1,2}G_{2n-1,2k}∂_s\qty(α^{(k)}(s)-α_0^{(k)}(s)). ∑_{k=1,2}G_{2n-1,2k}∂_s\qty(α^{(k)}(s)-α_0^{(k)}(s)).
\end{equation} \end{equation}
This can be evaluated using the exponential expansions and yields This can be evaluated using the exponential expansions and yields yet
yet another sum of exponentials. The steady state flow can then be another sum of exponentials. The steady state flow can then be found
found be setting all exponentials to zero. be setting all exponentials to zero, although care has to be taken, as
an exponential fit of the BCF may be only valid for finite times.
This concludes the calculation. Python code implementing the solution
can be found under \url{https://github.com/vale981/hopsflow}.

255
src/flow.tex
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@ -159,7 +159,7 @@ state. This can be generalized to any BCF that is a sum of exponentials.
Interestingly one finds that Interestingly one finds that
\begin{equation} \begin{equation}
\label{eq:alternative} \label{eq:alternative}
\ev{L∂_t B^†(t)} = \i∫\frac{\dd[2]{z}}{\pi^N} \ev{L∂_t B^†(t)} = \i \mathcal{M}_{η^\ast}
\dot{η}_t^\ast \mel{\psi(η,t)}{L}{\psi(η^\ast,t)}. \dot{η}_t^\ast \mel{\psi(η,t)}{L}{\psi(η^\ast,t)}.
\end{equation} \end{equation}
This expression is undesirable as it does not exist for all bath This expression is undesirable as it does not exist for all bath
@ -201,7 +201,8 @@ Following the usual derivation of the nonlinear NMQSD we write
\end{aligned} \end{aligned}
\end{equation} \end{equation}
where \(\tilde{z}_{\lambda}^{*}(t)=z_{\lambda}^{*}+\i g_{\lambda} ∫_{0}^{t} where \(\tilde{z}_{\lambda}^{*}(t)=z_{\lambda}^{*}+\i g_{\lambda} ∫_{0}^{t}
\dd{s} \eu^{-\i ω_{\lambda} s}\ev{L^†}_{s}\). \dd{s} \eu^{-\i ω_{\lambda} s}\ev{L^†}_{s}\) and \(\ev{L^\dag}_{t}=ψ(\tilde{η}_{t}^\ast)_{t}^\dag L^\dag
ψ(\tilde{η}_{t}^\ast)_{t}\) as in \cref{sec:nmqsd_basics}.
It has to be shown now, that the term It has to be shown now, that the term
\({\braket{\psi}{\tilde{\vb{z}}(t)}\!\braket{\tilde{\vb{z}}(t)}{\psi}}\) \({\braket{\psi}{\tilde{\vb{z}}(t)}\!\braket{\tilde{\vb{z}}(t)}{\psi}}\)
@ -438,26 +439,102 @@ This is an expression that we can easily evaluate with the HOPS
method. We will however refrain from doing so, as it turns out in method. We will however refrain from doing so, as it turns out in
\cref{sec:hopsvsanalyt} that consistent results can be obtained using \cref{sec:hopsvsanalyt} that consistent results can be obtained using
the derivative of the stochastic process \(ξ\). the derivative of the stochastic process \(ξ\).
\section{Generalization to Multiple Baths}
\label{sec:multibath}
Another requirement for thermodynamic application is the ability to
couple to multiple baths of possibly different structure and
temperature.
Due to the structure of the NMQSD and HOPS the results above can be
generalized in straight-forward manner to models of the form
\begin{equation}
\label{eq:multimodel}
H = H_\sys + ∑_{n=1}^N \qty[H_\bath\nth + \qty(L_n^†B_n + \hc)],
\end{equation}
where \(N\) is the number of baths, \(H_\sys\) is the system
Hamiltonian, \(H_\bath\nth = ∑_λω_λ\nth a_λ^{(n),†}a_λ\nth\),
\(B_n=∑_{λ} g_λ\nth a_λ\nth\) and the \(L_n={(\vb{L})}_n\) are
arbitrary operators in the system Hilbert space.
Note that this models a situation where each bath couples with the
system through exactly one spectral density and is therefore not fully
general.
We refer to \cref{sec:hops_multibath} for an review of the NMQSD
theory and HOPS method for multiple baths.
Because the bath energy change is being calculated directly and not
through energy conservation as in~\cite{Kato2016Dec}, we find
\begin{equation}
\label{eq:general_n_flow}
J_n=-\dv{\ev{H_\bath^{(n)}}}{t} = \iu\ev{[H_\bath^{(n)},
H_\inter^{(n)}]}
\end{equation}
regardless of the (non-) commutativity\footnote{For example, the
three-level model used in \cite{Uzdin2015Sep,Klatzow2019Mar} has
non-commuting couplings.} of the interaction
Hamiltonians. Therefore, we can apply the formalism of the previous
sections almost unchanged, by taking care that all quantities involved
in the expression of \(J_n\) refer to the \(n\)th bath denoted by sub
and superscripts.
This can be achieved by making the replacements
\begin{equation}
\label{eq:replacements}
\begin{aligned}
D_t \rightarrow D_t^{(n)} &\equiv
∫_0^t\dd{s}α_n(t-s)\fdv{η^\ast_n(s)} \\
ξ(t) \rightarrow ξ_n(t)&\equiv∑_{\lambda} g^{(n)}_{\lambda}
y_{\lambda} \eu^{-\mathrm{i} ω^{(n)}_{\lambda} t}
\end{aligned}
\end{equation}
in the previous sections, where the quantities involved are as in
\cref{sec:hops_multibath} and \cref{eq:xiproc}.
Foreshadowing slightly it may be states that in the light of
\cref{sec:general_obs} it might be an interesting question what impact
mixed bath hierarchy states have. For a cyclic machine with long
strokes, where only one bath is coupled to the system at a time, it
might be efficient to truncate the hierarchy in a way that discards
mixed bath states more readily than single bath hierarchy states as
the correlations between the baths are expected to be small.
\section{Generalization to Time Dependent Hamiltonians}
\label{sec:timedep}
To extract energy from a quantum thermal machine without an explicit
work reservoir, external modulation is required.
The above discussion is based on the model \cref{eq:multimodel} which
did not include explicit time modulations of \(H_\sys\) or \(L\). As
we did not calculate any explicit time derivatives of those two
operators, the results of the previous sections remain valid when we
substitute \(H_\sys\rightarrow H_\sys(t)\) and \(L\rightarrow L(t)\).
For the total power we find
\begin{equation}
\label{eq:power}
\dv{\ev{H}}{t} = \ev{\pdv{H_\inter}{t}} + \ev{\pdv{H_\sys}{t}},
\end{equation}
which can be evaluated as we will find in \cref{sec:intener} by
replacing \(L(t)\) with \(\dot{L}(t)\).
\section{General Collective Bath Observables} \section{General Collective Bath Observables}
\label{sec:general_obs} \label{sec:general_obs}
\begin{itemize} Now that we have introduced the formalism using the example of the
\item now we have all the basic tools ready we can discuss what the bath energy flow \(J\) in
most general observables are that we can calculate \cref{sec:flow_lin,sec:nonlin_flow,sec:lin_finite,sec:multibath,sec:timedep},
\item elucidates the meaning of the hierarchy states we may proceed to more general observables of the form can be
\end{itemize} generalized to calculate expectation values (and thus moments) of
The results obtained in \cref{sec:flow_lin,sec:nonlin_flow,sec:lin_finite} arbitrary observables of the form
can be generalized to calculate expectation values (and thus moments)
of arbitrary observables of the form
\begin{equation} \begin{equation}
\label{eq:collective_obs} \label{eq:collective_obs}
O = f(B^†, B) = ∑_{α}F_α\qty(B^†)^{α_1}B^{α_2} O = f(B^†, B) = ∑_{α}F_α\qty(B^†)^{α_1}B^{α_2}
\end{equation} \end{equation}
where \(α\) is a multi-index, \(B\) is as in~\cref{eq:totalH} and the where \(α\) is a two-dimensional multi-index, \(B\) is as
\(F_α\) are general observables acting on the system only. Note that in~\cref{eq:totalH} and the \(F_α\) are general observables acting on
\(f\) is already normal-ordered. We will restrict the discussion to the system only. Note that \(f\) is already normal-ordered. We will
the case of a single bath, as the generalization to multiple baths is restrict the discussion to the case of a single bath, as the
straight-forward. generalization to multiple baths is straight-forward.
To evaluate \(\ev{O}\) we have to find the value of To evaluate \(\ev{O}\) we have to find the value of
\(\ev{\qty(B^†)^a B^b}\). This can be achieved by interjecting the \(\ev{\qty(B^†)^a B^b}\). This can be achieved by interjecting the
@ -475,12 +552,12 @@ For zero temperature, we find following the procedures of
\sqrt{\frac{G^{\vb{k}}}{\vb{k}!}}ψ^{\vb{k}}\\ \sqrt{\frac{G^{\vb{k}}}{\vb{k}!}}ψ^{\vb{k}}\\
\label{eq:bdagmel}\mel{ψ}{\qty(B^†)^a}{z} &= \label{eq:bdagmel}\mel{ψ}{\qty(B^†)^a}{z} &=
\begin{aligned}[t] \begin{aligned}[t]
\qty(\mel{z}{B^a}{ψ})^†&= \qty((-\iu D_t)^b\ket{ψ(η^\ast,t)})^\dag\\ \qty(\mel{z}{B^a}{ψ})^†&= \qty((-\iu D_t)^a\ket{ψ(η^\ast,t)})^\dag\\
&= (\iu)^a∑_{\abs{\vb{k}}=a}\binom{a}{\vb{k}} (-\iu)^{\vb{k}} &= (\iu)^a∑_{\abs{\vb{k}}=a}\binom{a}{\vb{k}} (-\iu)^{\vb{k}}
\sqrt{\frac{\qty(G^{\vb{k}})^\ast}{\vb{k}!}}\qty(ψ^{\vb{k}})^\dag \sqrt{\frac{\qty(G^{\vb{k}})^\ast}{\vb{k}!}}\qty(ψ^{\vb{k}})^\dag,
\end{aligned} \end{aligned}
\end{align} \end{align}
in ``fock-space'' normalization where \(\vb{k}! = k_1!k_2!\ldots\) and where \(\vb{k}! = k_1!k_2!\ldots\) and
\(G^{\vb{k}}=G_1^{k_1}G_2^{k_2}\ldots\) following the usual \(G^{\vb{k}}=G_1^{k_1}G_2^{k_2}\ldots\) following the usual
conventions of multi-indices. Thus, expressions involving the bath conventions of multi-indices. Thus, expressions involving the bath
operator \(B\) to the \(b\)th power lead to expressions involving the operator \(B\) to the \(b\)th power lead to expressions involving the
@ -513,7 +590,7 @@ involving the HOPS hierarchy states this reduces to dividing by the
norm of the zeroth hierarchy state. norm of the zeroth hierarchy state.
The generalization to multiple baths may be performed in the same The generalization to multiple baths may be performed in the same
manner as will be discussed in \cref{sec:multibath}. This allows to manner as was discussed in \cref{sec:multibath}. This allows to
calculate the expectation value involving multiple bath operators calculate the expectation value involving multiple bath operators
\(B^{(n)}\). Interestingly, the generalization of \cref{eq:bmel} to \(B^{(n)}\). Interestingly, the generalization of \cref{eq:bmel} to
multiple baths immediately links hierarchy states of the form multiple baths immediately links hierarchy states of the form
@ -529,7 +606,7 @@ and inserting the coherent state resolution of unity we find terms of
the form the form
\begin{equation} \begin{equation}
\label{eq:with_process} \label{eq:with_process}
\mel{z}{\qty(B^\dag)^b}{ψ} \sim \qty(η^\ast(t))^b\ket{ψ(η^\ast,t)}. \mel{z}{\qty(B^\dag)^b}{ψ} \sim \qty(η^\ast_{t})^b\ket{ψ(η^\ast,t)}.
\end{equation} \end{equation}
The corresponding version of~\cref{eq:f_ex_zero} would only depend on The corresponding version of~\cref{eq:f_ex_zero} would only depend on
the zeroth order state and the stochastic processes. It has been the zeroth order state and the stochastic processes. It has been
@ -541,6 +618,18 @@ their average dynamics whereas the stochastic process fluctuates
around zero and does not contain much information about the actual around zero and does not contain much information about the actual
dynamics. dynamics.
% The process \(\pqty{η^\ast_{t}}^{b}\) has the autocorrelation function
% \(\ev{\pqty{η_{t}}^{b}\pqty{η^\ast_{s}}^{b}}=b! (α(t-s))^{b}\). Now
% for a decaying function
% \begin{equation}
% \label{eq:bcf_exponentiated}
% \pqty{\frac{α(τ)}{α(0)}}^{b}\xrightarrow{b\to ∞}
% \begin{cases}
% 1 & τ = 0 \\
% 0 & τ \neq 0,
% \end{cases}
% \end{equation}
% so that we end up with a process that is some approximation of white noise.
Also, this alternative method could be used convergence and Also, this alternative method could be used convergence and
consistency check, as expressions of the form~\cref{eq:with_process} consistency check, as expressions of the form~\cref{eq:with_process}
only involve the hierarchy cutoff and the exponential expansion of the only involve the hierarchy cutoff and the exponential expansion of the
@ -553,9 +642,10 @@ BCF in an indirect manner.
\item can also be used to quantify how ``strong'' the coupling is \item can also be used to quantify how ``strong'' the coupling is
\item simple application of the above formalism, even simpler than flow \item simple application of the above formalism, even simpler than flow
\end{itemize} \end{itemize}
A simple application of the formalism discussed To access all contributions to the total energy \(\ev{H}\) a way to
in~\cref{sec:general_obs} is the expectation value of the interaction calculate the expectation value of the interaction energy
Hamiltonian. \(\ev{H_{\inter}}\) is required. This is a application of the
formalism discussed in~\cref{sec:general_obs}.
For zero temperature and the nonlinear method we arrive at For zero temperature and the nonlinear method we arrive at
\begin{equation} \begin{equation}
@ -579,7 +669,7 @@ In HOPS terms \cref{eq:intexp} corresponds to
For nonzero temperature an extra term For nonzero temperature an extra term
\begin{equation} \begin{equation}
\label{eq:interexptherm} \label{eq:interexptherm}
\mathcal{M}_{\tilde{η}^\ast}\frac{\mel{\psi(\tilde{η},t)}{L^†ξ(t)}{\psi(\tilde{η}^\ast,t)}}{\braket{\psi(\tilde{η},t)}{\psi(\tilde{η}^\ast,t)}} \mathcal{M}_{\tilde{η}^\ast,ξ}{\mel{\psi(\tilde{η},t)}{L^†ξ(t)}{\psi(\tilde{η}^\ast,t)}}
+ \cc + \cc
\end{equation} \end{equation}
has to be added to \cref{eq:intexp}, where \(ξ\) is the thermal has to be added to \cref{eq:intexp}, where \(ξ\) is the thermal
@ -587,10 +677,12 @@ stochastic process.
\subsection{Higher Orders of the Coupling Hamiltonian} \subsection{Higher Orders of the Coupling Hamiltonian}
\label{sec:higher_order_coupling} \label{sec:higher_order_coupling}
\begin{itemize} In this section, the question of how many hierarchy orders have to be
\item a slight dive into the meaning and importance of the hierarchy included in the simulation to consistently calculate the expectation
states for convergence value of powers of the interaction Hamiltonian. Being nonessential for
\end{itemize} the understanding of the rest of the work, this section may be
skipped.
For self adjoint coupling operators \(L=L^\dag\) we can use Wick's For self adjoint coupling operators \(L=L^\dag\) we can use Wick's
theorem to find a normally ordered expression for \(H_\inter^n=L^n(B^\dag + theorem to find a normally ordered expression for \(H_\inter^n=L^n(B^\dag +
B)^n\). B)^n\).
@ -687,100 +779,15 @@ depends strongly on hierarchy states of order \(\sim \sqrt{n}\).
\begin{figure}[h] \begin{figure}[h]
\centering \centering
\plot{interaction_orders/k_weights} \plot{interaction_orders/k_weights}
\caption{\label{fig:kdist}The unnormalized distribution of the coefficients in \caption{\label{fig:kdist}The unnormalized distribution of the
\cref{eq:interactionnormal} with respect to \(k\) for different coefficients in \cref{eq:interactionnormal} with respect to \(k\)
\(n\). As a particular \(k\) can appear multiple times in the sum, for different \(n\). As a particular \(k\) can appear multiple
only the maximal coefficient for each \(k\) is being considered in times in the sum, only the maximal coefficient for each \(k\) is
the lines with the triangle markers. The maximum of this being considered in the lines with the triangle markers. The
distribution is given by \cref{eq:finalk}. The lines with the maximum of this distribution is given by \cref{eq:finalk}. The
circle markers show the full distribution. The dotted lines lines with the circle markers show the full distribution and the
correspond to binomial distributions centered at \(k_m\).} lines with the triangle markers show only the normalized
distribution of the maximum of \(l! 2^l k! (n-2l-k)!\) over \(l\).
The dotted lines correspond to binomial distributions centered at
\(k_m\).}
\end{figure} \end{figure}
\section{Generalization to Multiple Baths}
\label{sec:multibath}
\begin{itemize}
\item for thermodynamic machines: need multiple baths
\item due to the structure of NMQSD/HOPS generalization is
straightforward
\end{itemize}
The results above can be generalized in straight-forward manner to
models of the form
\begin{equation}
\label{eq:multimodel}
H = H_\sys + ∑_{n=1}^N \qty[H_\bath\nth + \qty(L_n^†B_n + \hc)],
\end{equation}
where \(N\) is the number of baths, \(H_\sys\) is the (possibly time
dependent) system Hamiltonian,
\(H_\bath\nth = ∑_λω_λ\nth a_λ^{(n),†}a_λ\nth\),
\(B_n=∑_{λ} g_λ\nth a_λ\nth\) and the \(L_n={(\vb{L})}_n\) are
arbitrary operators in the system Hilbert space (again possibly time
dependent). This models a situation where each bath couples with the
system through exactly one spectral density and is therefore not fully
general.
We refer to \cref{sec:hops_multibath} for an review of the NMQSD
theory and HOPS method for multiple baths.
Because the bath energy change is being calculated directly and not
through energy conservation as in~\cite{Kato2016Dec}, we find
\begin{equation}
\label{eq:general_n_flow}
J_n=-\dv{\ev{H_\bath^{(n)}}}{t} = \iu\ev{[H_\bath^{(n)},
H_\inter^{(n)}]}
\end{equation}
regardless of the (non-) commutativity\footnote{For example, the
three-level model used in \cite{Uzdin2015Sep,Klatzow2019Mar} has
non-commuting couplings.} of the interaction
Hamiltonians. Therefore, we can apply the formalism of the previous
sections almost unchanged, by taking care that all quantities involved
in the expression of \(J_n\) refer to the \(n\)th bath denoted by sub
and superscripts.
This can be achieved by making the replacements
\begin{equation}
\label{eq:replacements}
\begin{aligned}
D_t \rightarrow D_t^{(n)} &\equiv
∫_0^t\dd{s}α_n(t-s)\fdv{η^\ast_n(s)} \\
ξ(t) \rightarrow ξ_n(t)&\equiv∑_{\lambda} g^{(n)}_{\lambda}
y_{\lambda} \eu^{-\mathrm{i} ω^{(n)}_{\lambda} t}
\end{aligned}
\end{equation}
in the previous sections, where the quantities involved are as in
\cref{sec:hops_multibath} and \cref{eq:xiproc}.
In the light of \cref{sec:general_obs} it might be an interesting
question what impact mixed bath hierarchy states have. For a cyclic
machine with long strokes, where only one bath is coupled to the
system at a time, it might be efficient to truncate the hierarchy in a
way that discards mixed bath states more readily than single bath
hierarchy states as the correlations between the baths are expected to
be small.
\section{Generalization to Time Dependent Hamiltonians}
\label{sec:timedep}
\begin{itemize}
\item second important ingredient: external modulation, at least for
our purposes here
\item again here -> no magic, straight forward
\end{itemize}
The above discussion is based on the model \cref{eq:totalH} which did
not include explicit time modulations of \(H_\sys\) or \(L\). As we
did not calculate any explicit time derivatives of those two
operators, the results of the previous sections remain valid when we
substitute
\begin{align}
\label{eq:timedepsusbs}
H_\sys&\rightarrow H_\sys(t) & L\rightarrow L(t).
\end{align}
For the total power we find
\begin{equation}
\label{eq:power}
\dv{\ev{H}}{t} = \ev{\pdv{H_\inter}{t}} + \ev{\pdv{H_\sys}{t}},
\end{equation}
which can be evaluated as in \cref{sec:intener} by replacing \(L(t)\)
with \(\dot{L}(t)\).

67
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@ -437,3 +437,70 @@ The polynomial expressions for the smoothstep functions are
&{\text{if }}1\leq x\\ &{\text{if }}1\leq x\\
\end{cases}}. \end{cases}}.
\end{equation} \end{equation}
\subsection{Explicit Expressions for the Bath Energy Flow of the QBM
Model}
\label{sec:explicit_flow}
Here we detail the rest of the calculations omitted in
\cref{sec:bathflow}.
All quantities in \cref{eq:lambdafold} have exponential expansion so
that we can now define\footnote{Note that this is inconsistent with
\cref{sec:solution}.}
\begin{equation}
\label{eq:expansions}
\begin{aligned}
α_0&=∑_k U_k\eu^{-Q_k t} & \dot{α}_0&=∑_k P_k\eu^{-L_k t} & α(t)
&= ∑_nG_n\eu^{-W_n t} \\
A(t) &= ∑_l A_l\eu^{-C_l t} & B(t) &= ∑_l B_l\eu^{-C_l t}.
\end{aligned}
\end{equation}
With this we can calculate,
\begin{align}
\label{eq:lambdaintegrals}
∫_r^t\dd{s}B(s-r)\dot{α}_0(t-s)
&=\sum_{m,k}\underbrace{\frac{B_mP_k}{L_k-C_m}}_{\equiv
Γ^1_{mk}}\qty[\eu^{-C_m(t-r)}-\eu^{-L_k(t-r)}]=g_1(t-r)\\
∫_0^{t-r}\dd{u}B(t-r-u)α(u)
&=\sum_{n,l}\underbrace{\frac{B_nG_l}{C_n-W_l}}_{\equiv
Γ^2_{nl}}\qty[\eu^{-W_l(t-r)}-\eu^{-C_n(t-r)}]=g_2(t-r)\\
∫_0^{r}\dd{u}B(t-r+u)α^\ast(u)
&=\sum_{n,l}\underbrace{\frac{B_nG_l^\ast}{C_n+W_l^\ast}}_{\equiv
Γ^3_{nl}}\qty[\eu^{-C_n(t-r)}-\eu^{-W_l^\ast r-C_n t}]=g_3(t,r)
\end{align}
and
\begin{align}
\label{eq:finalsummands}
Λ_1(t)& =
\begin{aligned}[t]
∑_{m,k,n,l}Γ^1_{mk}Γ^2_{nl}\biggl[\frac{1-\eu^{-(C_m+W_l)t}}{C_m+W_l}
&-
\frac{1-\eu^{-(C_m+C_n)t}}{C_m+C_n}
\\&-
\frac{1-\eu^{-(L_k+W_l)t}}{L_k+W_l}
+
\frac{1-\eu^{-(L_k+C_n)t}}{L_k+C_n}\biggr]
\end{aligned}\\
Λ_2(t)&=
\begin{aligned}[t]
∑_{m,k,n,l}Γ^1_{mk}&Γ^3_{nl}\biggl[\frac{1-\eu^{-(C_m+C_n)t}}{C_m+C_n}
-\frac{1-\eu^{-(L_k+C_n)t}}{L_k+C_n}
\\&-\frac{\eu^{-(C_n+W_l^\ast)t}-\eu^{-(C_m+C_n)t}}{C_m-W_l^\ast}
+\frac{\eu^{-(C_n+W_l^\ast)t}-\eu^{-(L_k+C_n)t}}{L_k-W^\ast_l}\biggr]
\end{aligned}
\end{align}
Also required for \cref{eq:bathderiv_1} are
\begin{align}
\label{eq:ABconv}
∫_0^t\dd{s}A(s)\dot{α}_0(t-s) &= ∑_{n,m}\underbrace{\frac{A_nP_m}{L_m-C_n}}_{\equiv
Γ^A_{nm}}\qty[\eu^{-C_n t}-\eu^{-L_m t}]\\
∫_0^t\dd{s}B(s)\dot{α}_0(t-s) &= ∑_{n,m}Γ^1_{nm}\qty[\eu^{-C_n t}-\eu^{-L_m t}]
\end{align}
and
\begin{multline}
\label{eq:nonzerotemplim}
∫_0^t\dd{s}A(s)\qty(α(s)-α_0(s)) =\\
∑_{m,n}\frac{A_nG_m}{C_n+W_m}\qty(1-\eu^{-(C_n+W_m)t}) - ∑_{m,n}\frac{A_nU_m}{C_n+Q_m}\qty(1-\eu^{-(C_n+Q_m)t}).
\end{multline}