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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.

\section{Applications}
\label{sec:applications}
Knowing \(G\) and \(\alpha\), we can calculate all observables of the
system. Simple closed form expressions of sums of exponentials can be
obtained by using an exponential expansions of \(\alpha\).

\subsection{Correlation Functions}
\label{sec:correl}

We proceed to calculate \(\ev{q(t)q(s)}\).

\printbibliography{}
\end{document}




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