gaussian model solution in latex

tex/gaussian_model/main.tex

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+\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{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\eu^{\tilde{z}_l \cdot
+      t}\mqty(\tilde{z}_l & \Omega \\ \frac{\tilde{z}_l^2}{\Omega} & \tilde{z}_l) + \cc]
+\end{equation}
+with \(R_l={f_0(\tilde{z}_l)}/{p'(\tilde{z}_l)}\).
+\printbibliography{}
+\end{document}
+
+
+
+
+%%% Local Variables:
+%%% mode: latex
+%%% TeX-master: t
+%%% End: