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\title{Analytical Solution for Quantum Brownian Motion with arbitrary BCF}
\author{Valentin Boettcher}
\date{\today}
\begin{document}
\maketitle
\tableofcontents

\section{Model}
\label{sec:model}
The model is given by the quadratic hamiltonian
\begin{equation}
  \label{eq:hamiltonian}
  H = \frac{\Omega}{4}\qty(p^2+q^2) + \frac{1}{2} q
  \sum_\lambda\qty(g_\lambda^\ast b_\lambda + g_\lambda
  b^\dag_\lambda)+\sum_\lambda\omega_\lambda b^\dag_\lambda b_\lambda,
\end{equation}
where \(a,a^\dag\) are the ladder operators of the harmonic
oscillator, \(q=a+a^\dag\) and \(p=\frac{1}{\iu}\qty(a-a^\dag)\) so
that \([q,p] = 2\iu\).

\section{Equations of Motion}
\label{sec:eqmot}

The Heisenberg equation yields
\begin{align}
  \dot{q} &=\Omega p \label{eq:qdot}\\
  \dot{p} &= -\Omega q - \int_0^t \Im[\alpha_0(t-s)] q(s)\dd{s} + W(t) \label{eq:pdot}
  \\
  \dot{b}_\lambda &= -\iu g_\lambda \frac{q}{2} - \iu\omega_\lambda b_\lambda
\end{align}
with the operator noise
\(W(t)=-\sum_\lambda \qty(g_\lambda^\ast b_\lambda(0)
\eu^{-\iu\omega_\lambda t } + g_\lambda b_\lambda^\dag(0)
\eu^{\iu\omega_\lambda t })\),
\(\ev{W(t)W(s)}=\alpha(t-s)\) and \(\alpha_0 \equiv \eval{\alpha}_{T=0}\).
The equations for \cref{eq:qdot} and \cref{eq:pdot} can be solved by
finding a matrix \(G(t)\) with \(G(0)=\id\) and
\begin{equation}
  \label{eq:eqmotprop}
  \dot{G}(t) = A G(t) - \int_0^t K(t-s) G(s)\dd{s},\quad A=\mqty(0 &
  \Omega \\ -\Omega & 0), \quad K(t)=\mqty(0 & 0\\ \Im[\alpha_0(t)] & 0).
\end{equation}
Then
\begin{equation}
  \label{eq:qpsol}
  \mqty(q(t)\\ p(t)) = G(t)\mqty(q(0)\\ p(0)) + \int_0^tG(t-s)
  \mqty(0\\ W(s))\dd{s}.
\end{equation}

Because we are only interested in solutions for \(t\geq 0\) and the
shape of the convolution in \cref{eq:eqmotprop} the solution may be
found by virtue of the Laplace transform.

Setting
\begin{equation}
  \label{eq:laplprop}
  \mathcal{L}\{G\}(z) = \int_0^\infty \eu^{-z\cdot t} G(t)
\end{equation}
leads to an algebraic formula
\begin{equation}
  \label{eq:galgebr}
  \mathcal{L}\{G\}(z) = \qty(z-A + \mathcal{L}\{K\}(z))^{-1}.
\end{equation}


\section{Solution}
\label{sec:solution}
We observe that
\begin{equation}
  \label{eq:mdef}
  M = z-A + \mathcal{L}\{K\}(z) = \mqty(z & -\Omega\\ \Omega +
  \mathcal{L}\{\Im[\alpha_0]\}(z) & z)
\end{equation}
and therefore
\begin{equation}
  \label{eq:minv}
  M^{-1} = \frac{1}{\Omega^2 + \Omega\mathcal{L}\{\Im[\alpha_0]\}(z) + z^2}
  \mqty(z & \Omega \\ -(\Omega + \mathcal{L}\{\Im[\alpha_0]\}(z)) & z).
\end{equation}
From this we can conclude that \(G_{11}=G_{22}\).
Because \(\ev{W(s)}=0\) for thermal initial states of the bath we have
\begin{equation}
  \label{eq:meanvals}
  \mqty(\ev{q(t)}\\ \ev{p(t)}) = G(t)\mqty(\ev{q(0)}\\ \ev{p(0)}).
\end{equation}
Knowing this, we can deduce from \(\ev{\dot{q}}= \Omega \ev{p}\) that
\begin{align}
  \label{eq:onlyoneneeded}
    G_{11} &= \frac{\dot{G}_{12}}{\Omega} & G_{21} &=\frac{\ddot{G}_{12}}{\Omega^2}.
\end{align}
Therefore it suffices if we concern ourselves with \(G_{12}\). We
nevertheless continue in full generality.

Assume that \(\alpha_0(t)=\sum_{n=1}^N G_n \eu^{-W_n t - \i
  \varphi_n}\) with \(W_n=\gamma_n + \i\delta_n\) and \(G_n,
\varphi_n, \gamma_n,\delta_n\in\RR\).
This leads to a mathematically simple expression for the Laplace
transform
\begin{equation}
  \label{eq:laplace_alpha}
  \mathcal{L}\qty{\Im[\alpha_0]}(z) = -\sum_n G_n\qty[\frac{(z+\gamma_n)\sin\varphi_n+\delta_n\cos\varphi_n}{\delta_n^2+(\gamma_n+z)^2}].
\end{equation}
Because \(\mathcal{L}\{\Im[\alpha_0]\}\) appears in the denominator of
\cref{eq:minv} it is desirable to write \cref{eq:laplace_alpha} with a
common denominator. Introducing \(s_n = \sin\varphi_n,\, c_n =
\cos\varphi_n\) and \(z_n= -W_k\) we arrive
at
\begin{equation}
  \label{eq:laplace_alpha_better}
  \begin{aligned}
  \mathcal{L}\qty{\Im[\alpha_0]}(z) &= - \sum_n
  G_n\frac{(z_n+\gamma_n)s_n+ \delta_nc_n}{(z-z_n)(z-z_n^\ast)} \\
  &= -\frac{\sum_n G_n \qty((z_n+\gamma_n)s_n+
    \delta_nc_n)\prod_{k\neq
      n}(z-z_k)(z-z_k^\ast)}{\prod_{k}(z-z_k)(z-z_k^\ast)} \\
  &= \frac{\sum_n f_n(z)\prod_{k\neq n}(z-z_k)(z-z_k^\ast)}{\prod_{k}(z-z_k)(z-z_k^\ast)}
  \end{aligned}
\end{equation}
with the polynomials of first degree
\(f_n(z)=-G_n \qty((z_n+\gamma_n)s_n+\delta_nc_n)\).  Because the
above expression is a rational function, the components of
\cref{eq:minv} are rational functions for which the Laplace transform
is particularly simple to invert using the residue theorem. With this
in mind we now calculate
\begin{equation}
  \label{eq:prefactorrational}
      \frac{1}{\Omega^2 + \Omega\mathcal{L}\{\Im[\alpha_0]\}(z) + z^2}
      % =\frac{\prod_{k}(z-z_k)(z-z_k^\ast)}{\qty[(z+\i\Omega)(z-\i\Omega)]\prod_{k}(z-z_k)(z-z_k^\ast)
      %   + \sum_n\Omega f_n(z)\prod_{k\neq n}(z-z_k)(z-z_k^\ast)}\\
      =\frac{f_0(z)}{p_1(z) + \sum_n q_n(z)}
      =
      \frac{f_0(z)}{\mu\prod_{n=1}^{N+1}(z-\tilde{z}_l)(z-\tilde{z}^\ast_l)}
      = \frac{f_0(z)}{p(z)}
\end{equation}
where
\begin{align}
  f_0(z) &= \prod_{k}(z-z_k)(z-z_k^\ast) \\
  p_1(z) &= \qty[(z+\i\Omega)(z-\i\Omega)]\prod_{k}(z-z_k)(z-z_k^\ast) \\
  q_n(z) &= \Omega f_n(z)\prod_{k\neq n}(z-z_k)(z-z_k^\ast)
\end{align}
and \(\mu\in\RR\). The \(\tilde{z}_l\) are the roots of the real
polynomial
\begin{equation}
  \label{eq:polyp}
  p(z) = p_1(z) + \sum_{n=1}^{N}q_n(z)
\end{equation}
of degree \(2(N+1)\) where we \textbf{assume that there are
  no roots with multiplicity greater than one}.
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
any holonome function so that \(\frac{f_0(z)g(z)}{p(z)} \eu^{z\cdot
  t}\) falls off fast enough for \(t\geq 0\),
\(\Re(z)>\max_l{\Re(\tilde{z}_l)}=\Delta\) and \(\Re(z) \rightarrow
-\infty\). With this we can close the contour of the inverse Laplace
transform
\begin{equation}
  \label{eq:invlap}
  \mathcal{L}^{-1}\qty{\frac{f_0(z)g(z)}{p(z)}}(t) =
  \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}
\end{equation}
to the left to obtain
\begin{equation}
  \label{eq:simpleinvtrans}
  \mathcal{L}^{-1}\qty{\frac{f_0(z)g(z)}{p(z)}}(t)
  =
  \sum_{l=1}^{N+1}\qty[\frac{f_0(\tilde{z}_l)g(\tilde{z}_l)}{p'(\tilde{z}_l)}
  \eu^{\tilde{z}_l \cdot t} + \cc]
\end{equation}
where we assumed that \(g(z)^\ast=g(z^\ast)\) which is the case for
all our purposes. For completeness we give
\begin{equation}
  \label{eq:pderiv}
  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)].
\end{equation}
We can immediately conclude that all elements of \(G\) are sums of
exponentials, just like the BCF. In particular
\begin{equation}
  \label{eq:gfinal}
  G(t) = \sum_{l=1}^{N+1}\qty[R_l \mqty(\tilde{z}_l & \Omega \\ \frac{\tilde{z}_l^2}{\Omega} & \tilde{z}_l)\eu^{\tilde{z}_l \cdot
    t} + \cc]
\end{equation}
with \(R_l={f_0(\tilde{z}_l)}/{p'(\tilde{z}_l)}\). Note that we have
assumed a thermal initial state which lead to a particularity simple
expression for \(G_{21}\) as remarked earlier.
\printbibliography{}
\end{document}




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