Content-Type: multipart/mixed; boundary="-------------0404071042172" This is a multi-part message in MIME format. ---------------0404071042172 Content-Type: text/plain; name="04-102.keywords" Content-Transfer-Encoding: 7bit Content-Disposition: attachment; filename="04-102.keywords" superadiabaitc basis, exponential asymptotics ---------------0404071042172 Content-Type: application/x-tex; name="BetzTeufelArchive.tex" Content-Transfer-Encoding: 7bit Content-Disposition: inline; filename="BetzTeufelArchive.tex" \documentclass{article} \usepackage{amssymb, amsfonts} \usepackage{amsthm} \newtheorem{theorem}{Theorem} \newtheorem{lemma}{Lemma} \newtheorem{proposition}{Proposition} \newtheorem{corollary}{Corollary} \newtheorem*{assumption}{Assumption} \theoremstyle{definition} \newtheorem{definition}{Definition} \newtheorem{remark}{Remark} \newtheorem{notation}{Notation} \newtheorem{example}{Example} \addtolength{\topmargin}{-1cm} \addtolength{\textwidth}{2cm} \addtolength{\textheight}{2cm} \addtolength{\oddsidemargin}{-1cm} \newcommand{\qedi}{ } \newcommand {\Begin} [1] {\begin{#1} \setlength{\parskip}{0ex plus0ex minus0ex}} \newcommand {\kringel} [1] {\ensuremath{ \, \stackrel{\circ}{#1} \,}} \newcommand {\betrag} [1] {\ensuremath{ \left\vert #1 \right\vert } } \newcommand {\norm} [2] [] {\ensuremath{ \left\Vert #2 \right\Vert_{#1} } } \newcommand {\R} {\ensuremath{\mathbb{R}}} \newcommand {\Q} {\mathbb{Q}} \newcommand {\N} {\ensuremath{\mathbb{N}}} \newcommand {\B} {\ensuremath{\mathcal{B}}} \newcommand {\F} {\ensuremath {\mathcal{F}}} \newcommand {\Om} {\ensuremath{\Omega}} \newcommand {\om} {\omega} \newcommand {\WRaum} {\ensuremath{ (\Omega , \mathcal{S} , P) }} \newcommand {\Cov} {\mathrm{Cov}} \newcommand {\T} {\mathcal{T}} \newcommand {\tupel} [2] {\ensuremath{#1_1, \ldots,#1_#2}} \newcommand {\skalprod} [3] [] {\ensuremath{ \left\langle #2,#3 \right\rangle_{#1}}} \newcommand {\expval}[2] [] {\mathbb E_{#1}\left( #2 \right)} \newcommand {\spec} {\mathrm{spec}} \newcommand {\cKer} {\mathrm{Ker}} \newcommand {\W} {\mathcal{W}} \newcommand {\Z} {\mathbb{Z}} \newcommand {\supp} {\mathrm{supp}} \newcommand {\x} {\mathbf{x}} \newcommand {\y} {\mathbf{y}} \newcommand {\z} {\mathbf{z}} \newcommand {\muT} {\mu_{\scriptscriptstyle T}} \newcommand {\hoch}[2] {#1^{\scriptscriptstyle (#2)}} \newcommand {\cH} {\mathcal{H}} \newcommand {\mr} {\mathrm} \newcommand {\G} {\ensuremath {\mathcal{G}}} \newcommand {\g} {\ensuremath {\mathsf{G}}} \renewcommand {\S} {\mathcal{S}} \newcommand {\ldd}{:\!\!} \newcommand {\rdd}{\!\!:} \newcommand {\wick}[1]{\,\ldd{#1}\rdd\,} \newcommand {\mb} {\mathbf} \newcommand {\omitting}[1] {\backslash \hspace{-0.2 cm} #1} \newcommand {\sN} {\mathsf{N}} \renewcommand {\P} {\mathcal{P}} \newcommand {\sP} {\mathsf{P}} \newcommand {\C} {\mathbb C} \newcommand {\bS} {S} \renewcommand {\j} {\mathbf j} \newcommand{\cK} {\mathcal K} \newcommand{\cT} {\mathcal T} \newcommand{\commut}[2]{\left[ #1 , #2 \right]} \newcommand{\antcom}[2]{\left[ #1 , #2 \right]_+} \newcommand{\eps}{\varepsilon} \newcommand{\Or}{\mathcal{O}} \newcommand{\e}{\mathrm{e}} \newcommand{\cc}{ + \, {\mr{c.c.}}} \newcommand{\tRe}{\mathrm{Re}} \newcommand{\tIm}{\mathrm{Im}} \newcommand{\id}{\mathrm{id}} \newcommand{\D}{\mathrm{d}} \newcommand{\E}{\mathrm{e}} \newcommand{\I}{\mathrm{i}} \newcommand{\op}{{n_\eps}} \newcommand{\od}{{\rm}} \newcommand{\rom}{\renewcommand{\labelenumi}{{\rm(\roman{enumi})}}} \newcommand{\num}{\renewcommand{\labelenumi}{{\rm(\arabic{enumi})}}} \newcommand{\alp}{\renewcommand{\labelenumi}{{\rm(\alph{enumi})}}} \begin{document} \title{\LARGE \bf Precise coupling terms in adiabatic\\ quantum evolution} \author{Volker Betz\\ {\normalsize Institute for Biomathematics and Biometry, GSF Forschungszentrum,}\\ {\normalsize Postfach 1129, D-85758 Oberschlei{\ss}heim, Germany}\\{\normalsize volker.betz@gsf.de}\\[5mm] Stefan Teufel\\{\normalsize Mathematics Institute, University of Warwick,}\\ {\normalsize Coventry CV4 7AL, United Kingdom}\\{\normalsize teufel@maths.warwick.ac.uk}} \date{\today} \maketitle \begin{abstract} It is known that for multi-level time-dependent quantum systems one can construct superadiabatic representations in which the coupling between separated levels is exponentially small in the adiabatic limit. For a family of two-state systems with real-symmetric Hamiltonian we construct such a superadiabatic representation and explicitly determine the asymptotic behavior of the exponentially small coupling term. First order perturbation theory in the superadiabatic representation then allows us to describe the time-development of exponentially small adiabatic transitions. The latter result rigorously confirms the predictions of Sir Michael Berry for our family of Hamiltonians and slightly generalizes a recent mathematical result of George Hagedorn and Alain Joye. \end{abstract} \section{Introduction and main result} The decoupling of slow and fast degrees of freedom in the adiabatic limit is at the basis of many important approximations in physics, as, e.g., the Born-Oppenheimer approximation in molecular dynamics and the Peierls substitution in solid state physics. We refer to \cite{BMKNZ,Te} for recent reviews. Generically the decoupling is not exact and a coupling which is exponentially small in the adiabatic parameter remains. However, this small coupling has important physical consequences, as it makes possible, e.g., non-radiative decay to the ground state in molecules. Since Kato's proof from 1950 \cite{Ka} the adiabatic limit of quantum mechanics was considered also as a mathematical problem, with increased activity during the last 20 years. Some of the landmarks are \cite{Nen1,ASY,JoPf1,Nen2,HaJo}. We consider a two-state time-dependent quantum system described by the Schr\"o\-dinger equation \begin{equation} \label{H} \big( \I \eps \partial_t -H(t)\big) \psi(t) = 0 \end{equation} in the {\em adiabatic limit} $\eps\to 0$. For the moment we take the Hamiltonian $H(t)$ to be the real-symmetric $2 \times 2$-matrix \[%begin{equation} \label{hamiltonian2} H(t) = \rho(t)\left( \begin{array}{cc} \cos\theta(t) &\,\, \sin\theta(t) \\ \sin\theta(t) & \,-\hspace{-1pt} \cos\theta(t) \end{array} \right)\,. \]%end{equation} The eigenvalues of $H(t)$ are $\pm\rho(t)$ and we assume that the gap between them does not close, i.e.\ that $2\rho(t)\geq g>0$ for all $t\in\R$. As to be explained, even for this simple but prototypic problem there are open mathematical questions. In order to explain the concern of our work, namely the time-development of the exponentially small adiabatic transitions, let us briefly recall some important facts about (\ref{H}). Let $U_0(t)$ be the orthogonal matrix that diagonalizes $H(t)$, i.e. \begin{equation} \label{U_0} U_0(t) = \left( \begin{array}{cc} \cos(\theta(t)/2) & \sin(\theta(t)/2) \\ \sin(\theta(t)/2) & -\cos(\theta(t)/2) \end{array} \right)\,. \end{equation} Then the Schr\"odinger equation in the {\em adiabatic representation} becomes \[ U_0(t)\,\big( \I \eps \partial_t -H(t)\big)\,U_0^{\ast}(t)\, U_0(t) \psi(t) =: \big( \I\eps\partial_t - H^{\rm a}_\eps(t)\big) \psi^{\rm a}(t) = 0 \] with \[ H^{\rm a}_\eps(t) = \left(\begin{array}{cc} \rho(t)& \frac{\I\eps}{2}\theta'(t)\\[2mm]-\frac{\I\eps}{2}\theta'(t)&-\rho(t)\end{array}\right)\qquad \mbox{and}\qquad \psi^{\rm a} (t) = U_0(t) \psi(t)\,. \] Here and henceforth, primes denote time derivatives. First order perturbation theory in the adiabatic representation (cf.\ proof of Corollary~\ref{PropCor}) and integration by parts yields the adiabatic theorem \cite{BoFo,Ka}: The off-diagonal elements of the unitary propagator $K^{\rm a}(t,s)$ in the adiabatic basis, i.e.\ the solution of \[ \I\eps\partial_t K^{\rm a}_\eps(t,s) = H_\eps^{\rm a}(t) K_\eps^{\rm a}(t,s) \,,\qquad K_\eps^{\rm a}(s,s) = {\rm id}\,, \] vanish in the limit $\eps\to 0$. More precisely, let \begin{equation}\label{Pdef} P_+= \left(\begin{array}{cc}1&0\\0&0\end{array}\right)\,,\qquad P_-= \left(\begin{array}{cc}0&0\\0&1\end{array}\right)\,, \end{equation} which project onto the adiabatic subspaces in the adiabatic representation. Then \begin{equation}\label{AdiabaticThm} \|\, P_- \, K^{\rm a}_\eps(t,s)\, P_+ \| = \Or(\eps)\,. \end{equation} Therefore the transitions between the adiabatic subspaces are $\Or(\eps)$. This bound is optimal in the sense that in regions where $\theta(t)$ is not constant the leading order term in the asymptotic expansion of $P_- \, K^{\rm a}_\eps(t,s)\, P_+$ in powers of $\eps$ is proportional to $\eps$. However, if $\lim_{t\to\pm\infty}\theta'(t)=0$ then in the scattering limit the transitions between the adiabatic subspaces are much smaller: if the derivatives of $\theta\in C^\infty(\R)$ decay sufficiently fast, then for any $n\in\N$ \begin{equation}\label{Aeps} \mathcal{A}(\eps):=\lim_{t\to\infty}\|\, P_- \, K^{\rm a}_\eps(t,-t)\, P_+ \| = \Or(\eps^n)\,. \end{equation} If $\theta$ is analytic in a suitable neighborhood of the real axis, then transition amplitudes are even exponentially small, $\mathcal{A}(\eps) = \Or(\e^{-c/\eps})$ for some constant $c$ depending on the width of the strip of analyticity, see \cite{JoPf1,Ma}. It is well understood, see \cite{Le,Ga,Nen1}, how to reconcile the apparent contrariety between the smallness of the final amplitudes in (\ref{Aeps}) and the optimality of (\ref{AdiabaticThm}): the adiabatic basis is not the optimal basis for monitoring the transition process. For any $n\in \N$ there exist unitary transformations $U_\eps^n(t)$ such that the Hamiltonian in this $n^{\rm th}$ superadiabatic representation takes the form \begin{equation}\label{nthorderSAB} H^{n}_\eps(t) = \left(\begin{array}{cc} \rho^n_\eps(t)& c_\eps^n(t)\\ \overline{c}_\eps^n(t)&-\rho^n_\eps(t)\end{array}\right)\,\,\,\mbox{with}\,\,\, \rho^n_\eps(t) = \rho(t) +\Or(\eps^2)\,\,\,\mbox{and}\,\,\, |c_\eps^n(t)| = \Or(\eps^{n+1})\,. \end{equation} In the $n^{\rm th}$ superadiabatic basis the off-diagonal components of the propagator and hence also the transitions are of order $\Or(\eps^n)$, i.e.\ there are constants $C_n$ such that \begin{equation}\label{SuperAdiabaticThm} \|\, P_- \, K^{n}_\eps(t,s)\, P_+ \| \leq C_n \eps^{n}\,. \end{equation} In the scattering regime, where $\theta(t)$ becomes constant, the superadiabatic bases agree with the adiabatic basis, i.e.\ $\lim_{t\to\pm\infty} U_\eps^n(t) = U_0(t)$, and therefore the bound in (\ref{SuperAdiabaticThm}) basically yields (\ref{Aeps}). Typically $\lim_{n\to\infty}C_n\eps^n= \infty$ for all $\eps>0$, i.e.\ choosing $n$ larger while keeping $\eps$ fixed does not necessarily decrease the bound in (\ref{SuperAdiabaticThm}). However, one can choose $n_\eps=n(\eps)$ in such a way that $C_{n_\eps}\eps^{n_\eps}$ is minimal. If $\theta$ is analytic, one obtains the improved estimate \[ \|\, P_- \, K^{n_\eps}_\eps(t,s)\, P_+ \| = \Or(\e^{-c/\eps}) \] in the optimal superadiabatic basis $n_\eps$, see \cite{Nen2,JoPf2}. More interesting than bounds on $\mathcal{A}(\eps)$ is its actual value. Since $\mathcal{A}(\eps)$ is asymptotically smaller than any power of $\eps$, this question is beyond standard perturbation theory. For the case of analytic coupling $\theta$, asymptotic formulas of the type \begin{equation}\label{LZ} \mathcal{A}(\eps) = C\, \e^{-\frac{t_{\rm c}}{\eps}}\left(1 + \Or(\eps)\right) \end{equation} have been established, see e.g.\ \cite{JKP,Jo}, where the constants $C$ and $t_{\rm c}$ depend on the type and location of the complex singularities of $\theta'(t)/\rho(t)$. However, these results are obtained by solving (\ref{H}) not along the real axis but along a Stoke's line in the complex plane. As a consequence they give no information at all about the way in which the exponentially small final transition amplitude $\mathcal{A}(\eps)$ is build up in real time. This question of adiabatic transition histories is the concern of our paper. Berry \cite{Be} and, in a refined way Berry and Lim \cite{BeLi,LiBe}, gave an answer on a non-rigorous level and explicitly left a mathematically rigorous treatment as an interesting open problem. Only very recently Hagedorn and Joye \cite{HaJo} succeeded and confirmed Berry's results rigorously for a specific Hamiltonian. Although our work has been strongly motivated by the findings of Berry, our approach is slightly different. Let us first state our main result before we discuss its relation to the earlier ones. Without loss of generality we assume that $\rho(t)\equiv \frac{1}{2}$. It was observed in \cite{Be}, that this can always be achieved by transforming (\ref{H}) to the natural time scale $$ \tau(t) = 2 \int_0^t \varrho(s) \, \D s\,.$$ However, we can only treat a rather special class of Hamiltonians, since we must assume that in the natural time scale the coupling has the form \begin{equation} \label{thetadot} \theta'(t) = \I\gamma\left(\frac{1}{t+\I t_{\rm c}}-\frac{1}{t-\I t_{\rm c}}\right)= \frac{\gamma t_{\rm c}}{t^2 + t_{\rm c}^2} \end{equation} with $\gamma\in\R$ and $t_{\rm c}>0$. In other words we assume \begin{equation} \label{hamiltonian} H(t) = \frac{1}{2} \left( \begin{array}{cc} \cos\theta(t) &\,\, \sin\theta(t) \\ \sin\theta(t) &\,\, - \cos\theta(t) \end{array} \right)\qquad\mbox{with}\qquad \theta(t) = 2\,\gamma\arctan\left(\frac{t}{t_{\rm c}}\right)\,. \end{equation} We shall comment below on the meaning of this special choice and remark here that the Hamiltonian in \cite{HaJo} is (\ref{hamiltonian}) with $\gamma=\frac{1}{2}$. Our main result is the construction of an optimal superadiabatic basis in which the coupling term in the Hamiltonian is exponentially small and can be computed explicitly at leading order. This optimal basis is given as the $n_\eps^{\rm th}$ superadiabatic basis where $0\leq \sigma_\eps <2$ is such that \begin{equation}\label{ndef} n_\eps = \frac{t_{\rm c}}{\eps} -1+ \sigma_\eps\qquad\mbox{is an even integer.} \end{equation} \begin{theorem} \label{MainThm} Let $H(t)$ be as in (\ref{hamiltonian}) and $n_\eps$ as in (\ref{ndef}), and let $\eps_0>0$ be sufficiently small. Then for every $\eps\in (0,\eps_0]$ one can construct a family of unitary matrices $U_\eps^{n_\eps}(t)\in \C^{2\times 2}$, depending smoothly on $t\in\R$, such that \begin{equation}\label{UminusUnull} \|U_\eps^{n_\eps}(t) - U_0(t)\| =\Or\left(\frac{\eps^2}{1+t^2}\right) \end{equation} and \begin{equation}\label{Hop} U_\eps^{n_\eps}(t) \,\Big( \I \eps\partial_t - H(t)\Big)\,U_\eps^{n_\eps *}(t) = \I\eps\partial_t - \underbrace{\left(\begin{array}{cc} \rho_\eps^{n_\eps}(t)& c^{\op}_\eps(t) \\ \overline{c}^{\op}_\eps(t) & -\rho_\eps^{n_\eps}(t) \end{array}\right)}_{\displaystyle =:H^{\op}_\eps(t)}\,. \end{equation} Here \[ \rho_\eps^{n_\eps}(t)=\frac{1}{2} + \Or\left(\frac{\eps^2}{1+t^2}\right)\,, \] and for every $\alpha<\frac{3}{2}$ \begin{equation}\label{Hod} c^{\op}_\eps(t) = 2\I\,\sqrt{\frac{2\eps}{\pi t_{\rm c}}}\,\sin\left(\frac{\pi \gamma}{2}\right)\, \e^{-\frac{t_{\rm c}}{\eps}}\,\e^{-\frac{t^2}{2\eps t_{\rm c}}} \,\cos\left(\frac{t}{\eps}-\frac{t^3}{3\eps t_{\rm c}^2} + \frac{\sigma_\eps t}{t_{\rm c}} \right) + \Or\left(\phi^\alpha(\eps,t)\right)\,, \end{equation} with \begin{equation}\label{Phialpha} \phi^\alpha(\eps,t) = \left\{ \begin{array}{ll} \eps^\alpha \exp\left(-\frac{t_{\rm c}}{\eps} \left( 1 + \frac{t^2}{4 t_{\rm c}^2} \right) \right)\ &\,\, \mbox{if } |t| < t_{\rm c}, \\[3mm] {\displaystyle \frac{1}{1+t^2}} \exp \Big( -\frac{t_{\rm c}}{\eps} \Big(1+\frac{\ln 2}{2} \Big) \Big) &\,\, \mbox{if } |t| \geq t_{\rm c}. \end{array} \right. \end{equation} \end{theorem} \begin{remark} The explicit term in $c_\eps^{\op}$ is of order $\Or(\e^{-t_{\rm c}/\eps})$ only for times $|t|=\Or(\sqrt{\eps})$. For larger times all terms in $c_\eps^{\op}$ are exponentially small compared to the leading exponential $\e^{-t_{\rm c}/\eps}$. As a consequence, Taylor expansion of the cosine in $c_\eps^{\op}$ around $t/\eps$ for $|t|=\Or(\sqrt{\eps})$ shows that it can be replaced by $\cos(t/\eps)$ at the cost of lowering $\alpha$ to $\alpha<1$: for every $\alpha<1$ \[%begin{equation}\label{Hod2} c_\eps^{\op}(t) = 2\I\,\sqrt{\frac{2\eps}{\pi t_{\rm c}}}\,\sin\left(\frac{\pi \gamma}{2}\right)\, \e^{-\frac{t_{\rm c}}{\eps}}\,\e^{-\frac{t^2}{2\eps t_{\rm c}}} \,\cos\left(\frac{t}{\eps} \right) + \Or\left(\phi^\alpha(\eps,t)\right)\,. \]%end{equation} \end{remark} \begin{remark} The slow time decay of the error in (\ref{Phialpha}) for large times is due to the fact that $n_\eps$ is optimal for $t$ near $0$, but not for large $t$. \end{remark} \begin{remark} Taking $n_\eps$ defined in (\ref{ndef}) odd instead of even would yield slightly different off-diagonal elements in the effective Hamiltonian $H_\eps^{\op}(t)$. However, the resulting unitary propagator, cf.\ Corollary~\ref{PropCor}, would be the same at leading order. See the end of Section~5 for a discussion of this somewhat surprising fact. \end{remark} Let us shortly explain the idea of the proof of Theorem~\ref{MainThm} and at the same time the structure of our paper. First we construct the $n^{\rm th}$ order superadiabatic basis as in (\ref{nthorderSAB}) in two steps: in Section~2 we construct the projectors on the superadiabatic basis vectors and in Section~3 we construct the unitary basis transformation $U^n_\eps(t)$. We cannot use existing results here, e.g.\ \cite{Ga,Nen2}, since we need to keep carefully track of the exact form off the off-diagonal terms $c^n_\eps(t)$ of the superadiabatic Hamiltonian, and since we aim at a scalar recurrence relation instead of a matrix recurrence relation for the $c^n_\eps(t)$'s. The main mathematical challenge is the asymptotic analysis of the resulting recurrence relation, which is done in Section~4. This is also the only part where we have to assume the special form (\ref{thetadot}) for $\theta'$. Theorem~\ref{MainThm} then follows by choosing the order $n$ of the superadiabatic basis as in (\ref{ndef}), a choice which minimizes $c_\eps^{n}(t)$ near $t=0$. The details of this optimal truncation procedure and the proper proof of Theorem~\ref{MainThm} are given in Section~5. Finally in Section~6 we use first order perturbation theory in the optimal superadiabatic basis in order to obtain the following Corollary, in which we abbreviate \[ \Delta(t,s) := \arctan(t) -\arctan(s)\,. \] Also recall that erf$:\R\to (-1,1)$ with ${\rm erf}(x) = \frac{2}{\sqrt{\pi}} \int_0^x\e^{-x^2}\D x$ switches smoothly and monotonically from erf$(-\infty)=-1$ to erf$(\infty)=1$. \begin{corollary}\label{PropCor} The unitary propagator in the optimal superadiabatic basis \[ K_\eps^{\op}(t,s) = \left( \begin{array}{cc} k_\eps^+(t,s) & k_\eps^{\od}(t,s) \\\overline{k}_\eps^{\od}(t,s)& k_\eps^-(t,s) \end{array}\right)\,, \] i.e.\ the solution of \[ \I\eps\partial_t K_\eps^{\op}(t,s) = H_\eps^{\op}(t) K_\eps^{\op}(t,s) \,,\qquad K_\eps^{\op}(s,s) = {\rm id}\,, \] satisfies \begin{equation}\label{Kdiag} k_\eps^\pm(t,s) = \e^{\mp\frac{\I(t-s)}{2\eps}} + \Or(\eps\Delta(t,s)) \end{equation} and \begin{eqnarray}\nonumber k^{\od}_\eps(t,s) &=& \sin\left(\frac{\pi \gamma}{2}\right)\, \e^{-\frac{t_{\rm c}}{\eps}}\e^{-\frac{\I(t+s)}{2\eps}}\left({\rm erf}\left(\frac{t}{\sqrt{2\eps t_{\rm c}}}\right)-{\rm erf}\left(\frac{s}{\sqrt{2\eps t_{\rm c}}}\right)\right)\\&&+\,\Or\left(\sqrt{\eps}\e^{-\frac{t_{\rm c}}{\eps}}\Delta(t,s)\right)\,.\label{Koffdiag} \end{eqnarray} Outside the transition region, more precisely for $|t|>\eps^\beta$ and $|s|>\eps^\beta$ for some $\beta<\frac{1}{2}$, (\ref{Koffdiag}) holds with the error term replaced by $\Or(\eps^\alpha\e^{-\frac{t_{\rm c}}{\eps}}\Delta(t,s))$ for every $\alpha<1$. \end{corollary} Corollary~\ref{PropCor} immediately implies the existence of solutions to (\ref{H}) of the form \begin{equation}\label{approxsol} \psi(t) = U_\eps^*(t)\,\left(\begin{array}{c}\e^{-\frac{\I t}{2\eps}}\\\sin\left(\frac{\pi \gamma}{2}\right)\, \e^{-\frac{t_{\rm c}}{\eps}}\e^{\frac{\I t}{2\eps}}\left({\rm erf}\left(\frac{t}{\sqrt{2\eps t_{\rm c}}}\right)+1\right) \end{array}\right)+\,\Or\left(\sqrt{\eps}\e^{-\frac{t_{\rm c}}{\eps}}\right)\,. \end{equation} They start at large negative times in the positive energy adiabatic subspace and smoothly and monotonically develop the exponentially small component in the negative energy adiabatic subspace in a $\sqrt{\eps}$-neighborhood of $t=0$. Berry and Lim \cite{Be,BeLi} argue that this behavior is universal: whenever $\theta'$ has the form \[ \theta'(t) = \frac{\pm\I\gamma}{t\pm\I t_{\rm c}}+ \Or(|t\pm\I t_{\rm c}|^\alpha)\qquad\mbox{for some}\,\,\alpha>-1 \] near its singularities $\pm\I t_{\rm c}$ closest to the real axis, then (\ref{approxsol}) should hold. For the Landau-Zener Hamiltonian, which describes the generic situation, one finds $\gamma=\frac{1}{3}$ and $\alpha=-\frac{1}{3}$. Hagedorn and Joye \cite{HaJo} proved (\ref{approxsol}) for the Hamiltonian (\ref{hamiltonian}) with $\gamma=\frac{1}{2}$. In the approach of Berry and, slightly modified, of Hagedorn and Joye, the optimal superadiabatic basis vectors are obtained through optimal truncation of an asymptotic expansion of the true solution of (\ref{H}) in powers of $\eps$. In contrast, in our approach the optimal superadiabatic basis is constructed by approximately diagonalizing the Hamiltonian. The main advantage of ``transforming the Hamiltonian'' over ``expanding the solutions'' is that the former approach can be applied, at least heuristically, to more general adiabatic problems, cf.\ \cite{Te}, as for example the Born-Oppenheimer approximation. While we cannot control the asymptotics for the Born-Oppen\-heimer model rigorously yet, the heuristic application of the idea yields new physical insight into adiabatic transition histories and new expressions for the exponentially small off-diagonal elements of the $S$-matrix for simple Born-Oppenheimer type models, cf.\ \cite{BeTe}. Therefore we see the rigorous results obtained in this paper also as a first attempt to justify the application of analogous ideas to more complicated but also more relevant systems. Furthermore, the concept of an adiabatically renormalized Hamiltonian was used to derive a criterion for selecting possible transition sequences in multi-level problems \cite{WiMo}. For the specific problem (\ref{H}) the knowledge of two linearly independent solutions is of course equivalent to the knowledge of the propagator and the effective Hamiltonian in the optimal superadiabatic basis. Therefore we shortly explain which aspects of our result constitute an improvement compared to \cite{HaJo}: Most importantly, our proof does not rely on the a priori knowledge of the scattering amplitude $\mathcal{A}(\eps)$. Indeed, our result yields for the first time a proof of (\ref{LZ}) based on superadiabatic evolution, as expressed in Corollary~\ref{SmatrixCor}. Moreover, we allow for a slightly larger class of Hamiltonians and obtain more detailed error estimates, which, in particular, give rise to close to optimal error bounds in the expansion of the $S$-matrix, cf.\ Corollary~\ref{SmatrixCor}. Finally, we also get explicitly the next order correction in (\ref{Koffdiag}) resp.\ (\ref{approxsol}), cf.\ Section~6. It should be noted, however, that the improved error estimates and the next order corrections could have been obtained also based on the proof in \cite{HaJo}. We finally turn to the scattering limit. Let $K_\eps^0(t,s)$ denote the propagator in the original basis and define the scattering matrix in the adiabatic basis by \[ S_\eps^{\rm a} := \lim_{t\to\infty} \e^{\frac{\I H_0 t}{\eps}}\, U_0(t)\,K_\eps^0(t,-t)\,U_0^*(-t)\, \e^{\frac{\I H_0 t}{\eps}}\,, \quad\mbox{where}\,\, H_0 = \left(\begin{array}{cc}\frac{1}{2}&0\\0&-\frac{1}{2}\end{array}\right)\,. \] Since, according to (\ref{UminusUnull}), for large negative and positive times the optimal superadiabatic basis agrees with the adiabatic basis, $S_\eps^{\rm a}$ can be computed with help of the optimal superadiabatic propagator from Corollary~\ref{PropCor}. \begin{corollary} \label{SmatrixCor} For $\beta<1$ we have \[ S_\eps^{\rm a} = \left( \begin{array}{cc} 1 +\Or(\eps) & 2\,\sin\left(\frac{\pi \gamma}{2}\right)\, \e^{-\frac{t_{\rm c}}{\eps}}\,(1+\Or(\eps^\beta))\\ 2\,\sin\left(\frac{\pi \gamma}{2}\right)\, \e^{-\frac{t_{\rm c}}{\eps}}\,(1+\Or(\eps^\beta))& 1+\Or(\eps) \end{array} \right)\,. \] \end{corollary} \begin{proof} According to (\ref{UminusUnull}) we have \[ S_\eps^{\rm a} = \lim_{t\to\infty} \e^{\frac{\I H_0 t}{\eps}}\, U_\eps^{n_\eps}(t)\,K_\eps^0(t,-t)\,U_\eps^{n_\eps\,\ast}(-t)\, \e^{\frac{\I H_0 t}{\eps}} = \lim_{t\to\infty} \e^{\frac{\I H_0 t}{\eps}}\,K_\eps^{\op}(t,-t)\, \e^{\frac{\I H_0 t}{\eps}}\,. \] Now the claim follows from inserting (\ref{Kdiag}) and (\ref{Koffdiag}) with the improved error estimate outside of the transition region. \qedi \end{proof} From Corollary~\ref{SmatrixCor} we conclude that the transition amplitude is given by \[ \mathcal{A}(\eps) = \|P_- S_\eps^{\rm a} P_+\| = 2\,\sin\left(\frac{\pi \gamma}{2}\right)\, \e^{-\frac{t_{\rm c}}{\eps}}\,\left(1+\Or(\eps^\beta)\right)\,,\qquad\mbox{for any} \,\,\beta<1\,, \] which agrees with the results of \cite{Jo}, as explained in \cite{BeLi}. We conclude the introduction with two recommendations for further reading: The numerical results of Berry and Lim \cite{LiBe} beautifully illustrate the idea of optimal superadiabatic bases and universal adiabatic transition histories. The introduction of the paper of Hagedorn and Joye \cite{HaJo} gives a slightly different viewpoint on the problem and, in particular, a short discussion on how exponential asymptotics for the Schr\"odinger equation (\ref{H}) fit into the broader field of exponential asymptotics for ordinary differential equations. \medskip \noindent{\bf Acknowledgements:} We are grateful to Alain Joye and George Hagedorn for many helpful discussions. \section{Superadiabatic projections} For the present and the following section we assume that $H(t)$ has the form (\ref{hamiltonian}), but with some arbitrary $\theta\in C^\infty(\R)$. The first aim is to construct time-dependent matrices $\pi^{(n)} \in \R^{2 \times 2}$ with \begin{eqnarray} & & (\pi^{(n)})^2 - \pi^{(n)} = \Or(\varepsilon^{n+1}), \label{pi1}\\ & & \commut{\I\eps \partial_t - H}{\pi^{(n)}} = \Or(\varepsilon^{n+1}). \label{pi2} \end{eqnarray} Here, $\commut{A}{B} = AB - BA$ denotes the commutator two operators $A$ and $B$. Likewise, we will later use $\antcom{A}{B} = AB + BA$ to denote the anti-commutator of $A$ and $B$. Equation (\ref{pi1}) says that $\pi^{(n)}$ is a projection up to errors of order $\eps^{n+1}$, while (\ref{pi2}) implies that $\pi^{(n)}(t)$ is approximately equivariant, i.e.\ \[ K_\eps^0(t,s)\,\pi^{(n)}(s) = \pi^{(n)}(t)\,K_\eps^0(t,s) + \Or(\eps^{n})\,. \] Recall the $K_\eps^0(t,s)$ is the unitary propagator for (\ref{H}). Hence $\pi^{(n)}(t)$ is an almost projector onto an almost equivariant subspace. We construct $\pi^{(n)}$ inductively starting from the Ansatz \begin{equation} \label{ansatz} \pi^{(n)} = \sum_{k=0}^n \pi_k \varepsilon^k\,. \end{equation} By (\ref{hamiltonian}), $H$ has two eigenvalues $\pm1/2$. Let $\pi_0$ be the projection onto the eigen\-space corresponding to $+ 1/2$, and $\pi^{(0)} = \pi_0$ according to (\ref{ansatz}). It is easily checked that (\ref{pi1}) and (\ref{pi2}) are fulfilled for $n=0$. In order to construct $\pi_n$ for $n > 0$, let us write $G_n(t)$ for the term of order $\eps^{n+1}$ in (\ref{pi2}), i.e. \begin{equation} \label{G def} (\pi^{(n)})^2 - \pi^{(n)} = \eps^{n+1} G_{n+1} + \Or(\eps^{n+2})\,. \end{equation} Obviously, \begin{equation} \label{G calculated} G_{n+1} = \sum_{j=1}^{n} \pi_j \pi_{n+1-j}. \end{equation} \begin{proposition} \label{general scheme} Assume that $\pi^{(n)}$ given by (\ref{ansatz}) fulfills (\ref{pi1}) and (\ref{pi2}). Then a unique matrix $\pi_{n+1}$ exists such that $\pi^{(n+1)}$ defined as in (\ref{ansatz}) fulfills (\ref{pi1}) and (\ref{pi2}). $\pi_{n+1}$ is given by \begin{equation}\label{recursion1} \pi_{n+1} = G_{n+1} - \pi_0G_{n+1} - G_{n+1}\pi_0 - \I \commut{\pi'_n}{\pi_0}. \end{equation} Furthermore $\pi'_n$ is off-diagonal with respect to $\pi_0$, i.e. \begin{equation} \label{offdiag} \pi_0 \pi'_n \pi_0 = (1-\pi_0) \pi'_n (1-\pi_0) = 0, \end{equation} and $G_{n}$ is diagonal with respect to $\pi_0$, i.e. \begin{equation} \label{diag} \pi_0 G_{n+1} (1-\pi_0) = (1-\pi_0) G_{n+1} \pi_0 = 0. \end{equation} \end{proposition} \begin{remark} The fact that the superadiabatic projections are unique answers the question raised in \cite{Be} to which extend the superadiabatic basis constructed there is uniquely determined. \end{remark} \begin{remark} Our construction can be seen as a special case of the construction in \cite{EmWe}. It was applied in the same context in \cite{PST,Te}. The role and the importance of the superadiabatic subspaces as opposed to the superadiabatic evolution have been emphasized by Nenciu \cite{Nen2}. He constructs the superadiabatic projections for much more general time-dependent Hamiltonians. However, Nenciu's construction is less suitable for the explicit computations we need to perform. \end{remark} \begin{proof} Let $\pi^{(n+1)}$ be given by (\ref{ansatz}) and suppose $\pi^{(n)}$ fulfills (\ref{pi1}) and (\ref{pi2}). Let $\tilde\pi_{n+1}$ be an arbitrary matrix, and define $\tilde\pi^{(n+1)} = \pi^{(n)} + \eps^{n+1} \tilde\pi_{n+1}$. Then $$ (\tilde \pi^{(n+1)})^2 - \tilde \pi^{(n+1)} = (\pi^{(n)})^2 - \pi^{(n)} + \eps^{n+1} \left( \antcom{\tilde \pi^{(n+1)}}{\tilde \pi_{n+1}} - \tilde \pi_{n+1} \right).$$ Using (\ref{G def}), we see that terms of order $\eps^{n+1}$ vanish if and only if \begin{equation} \label{eq001} G_{n+1} = \tilde \pi_{n+1} - \antcom{\pi_0}{\tilde \pi_{n+1}} = (1-\pi_0) \tilde \pi_{n+1}(1-\pi_0) - \pi_0 \tilde \pi_{n+1} \pi_0. \end{equation} %Now (\ref{diag}) follows immediately. Multiplying (\ref{eq001}) with $(1 - \pi_0)$ and with $\pi_0$ on both sides and subtracting the results, we find that $\tilde \pi_{n+1}$ must fulfill \begin{equation} \label{eq002} (1-\pi_0) \tilde \pi_{n+1}(1-\pi_0) + \pi_0 \tilde \pi_{n+1} \pi_0 = G_{n+1} - \antcom{G_{n+1}}{\pi_0}. \end{equation} Similarly $$ \commut{\I \eps \partial_t - H}{\tilde \pi^{(n+1)}} = \commut{\I \eps \partial_t - H}{\pi^{(n)}} + \eps^{n+1} \commut{\I \eps \partial_t - H}{\tilde \pi_{n+1}}. $$ Again terms of order $\eps^{n+1}$ vanish if and only if \begin{equation} \label{offdiag explicit} \I \pi'_n = \commut{H}{\tilde \pi_{n+1}}. \end{equation} Since $\pi_0$ is the projector onto the eigenspace of $H$, we have $\pi_0 H = H \pi_0 = E\pi_0$, where $E = 1/2$ is the positive eigenvalue of $H$, and similarly $(1-\pi_0)H = H(1-\pi_0) = -E(1-\pi_0)$. When we multiply (\ref{offdiag explicit}) first with with $\pi_0$ from the left and with $1-\pi_0$ from the right, then the other way round, and finally subtract the second result from the first, we get \begin{equation} \label{eq003} 2 E (\pi_0 \tilde \pi_{n+1} (1-\pi_0) + (1-\pi_0) \tilde \pi_{n+1} \pi_0 ) = -\I \commut{\pi'_n}{\pi_0}. \end{equation} Now we divide (\ref{eq003}) by $2E$ and add (\ref{eq002}) to find \begin{equation} \label{eq004} \tilde \pi_{n+1} = G_{n+1} - \antcom{G_{n+1}}{\pi_0} - \frac{\I}{2E} \commut{\pi'_n}{\pi_0}. \end{equation} Thus $\tilde\pi_{n+1}$ is uniquely determined by the requirement that $\tilde\pi^{(n+1)}$ should fulfill (\ref{pi1}) and (\ref{pi2}). On the other hand, $\commut{H}{G_{n+1} - \antcom{G_{n+1}}{\pi_0}} = 0$ and \[\pi_0 \commut{\pi'_n}{\pi_0} \pi_0 = (1-\pi_0) \commut{\pi'_n}{\pi_0} (1-\pi_0) = 0\,, \] and thus $\pi_{n+1}$ given by the right hand side of (\ref{eq004}) indeed fulfills (\ref{eq002}) and (\ref{offdiag explicit}). This shows existence. (\ref{diag}) and (\ref{offdiag}) now follow directly from (\ref{eq001}) and (\ref{offdiag explicit}). \qedi \end{proof} The calculation of $\pi^{(n)}$ via the matrix recurrence relation (\ref{recursion1}) and (\ref{G calculated}) is now possible in principle, but extremely cumbersome. In order to make more explicit calculations possible, we introduce a special basis of $\R^{2 \times 2}$. Recall that $U_0(t)$ as defined in (\ref{U_0}) is the unitary transformation into the basis consisting of the eigenvectors of $H$, i.e. the adiabatic basis, and let $V_0(t) = \frac{2}{\theta'(t)}U'_0(t)$. With $P=P_+$ as in (\ref{Pdef}) we then have $U_0^2 = V_0^2 = \id$ and $PU_0V_0P = PV_0U_0P = 0$, and $\pi_0 = U_0PU_0$. Moreover, since $G_1 = 0$ by (\ref{G calculated}), (\ref{recursion1}) implies \begin{equation} \label{pi_1} \pi_1 = -\frac{\I}{2} \theta' (V_0PU_0 - U_0PV_0). \end{equation} Motivated by this, we put \begin{eqnarray*} X = V_0PU_0 -U_0PV_0\,, & \quad& Y = V_0PV_0 - U_0PU_0\,, \\ Z = V_0PU_0 + U_0PV_0\,, & \quad & W = V_0PV_0 + U_0PU_0\,. \end{eqnarray*} It is immediate that this is a basis of $\R^{2 \times 2}$ for all $t$, and in fact $$ X = \left(\begin{array}{cc}0 & -1 \\ 1 & 0 \end{array}\right), \quad W = \left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right), \quad Y = -2 H, \quad Z = \frac{-1}{\theta'} Y'.$$ Our reason for representing $X$ through $Z$ via $U_0$ and $V_0$ is that the following important relations now follow without effort: \begin{eqnarray} && X' = 0, \quad Y' = - \theta' Z, \quad Z' = \theta' Y, \label{derivatives} \\ && \antcom XY = \antcom XZ = \antcom YZ = 0,\quad -X^2 = Y^2 = Z^2 = W, \label{products} \\ && \commut{X}{\pi_0} = Z, \quad \commut{Y}{\pi_0} = 0, \quad \commut{Z}{\pi_0} = X, \label{commutators} \\ && W - \antcom{W}{\pi_0} = Y. \label{W equation} \end{eqnarray} These relations show that this basis behaves extremely well under the operations involved in the recursion (\ref{recursion1}). This enables us to obtain \begin{proposition} \label{function recursion} For all $n \in \N$, $\pi_n$ is of the form \begin{equation} \label{xyz ansatz} \pi_n = x_nX + y_nY + z_nZ, \end{equation} where the functions $x_n, y_n $ and $z_n$ satisfy the differential equations \begin{eqnarray} x'_n & = & \I z_{n+1}, \label{E1a}\\ y'_n & = & - \theta' z_n, \label{E1b}\\ z'_n & = & \I x_{n+1} + \theta' y_n. \label{E1c} \end{eqnarray} Moreover, \begin{equation} \label{functions recursion start} x_1(t) = -\frac{\I}{2}\theta'(t), \quad y_1(t) = z_1(t) = 0. \end{equation} \end{proposition} \begin{remark} Hence, for all even $n$, $x_n = 0$, while for all odd $n$, $y_n = z_n = 0$. \end{remark} \begin{proof} (\ref{functions recursion start}) was already noticed in (\ref{pi_1}). Now suppose $\pi_n$ is given by (\ref{xyz ansatz}). By (\ref{products}) and (\ref{W equation}), $G_{n+1} - \antcom{G_{n+1}}{\pi_0}$ is proportional to $Y$ with a prefactor given through (\ref{G calculated}), and by (\ref{recursion1}), (\ref{derivatives}) and (\ref{commutators}), \begin{equation} \label{true recursion} \pi_{n+1} = \sum_{j=1}^n (- x_j x_{n+1-j} + y_j y_{n+1-j} + z_j z_{n+1-j}) Y + \I(\theta'y_n - z'_n) X - \I x'_nZ. \end{equation} Comparing with (\ref{xyz ansatz}) shows (\ref{E1a}) and (\ref{E1c}). To show (\ref{E1b}), we use (\ref{offdiag}). This gives $$ 0 = \pi_0 \pi'_n \pi_0 = (y_n' + \theta' z_n) \pi_0Y\pi_0 + (z_n' - \theta'y_n)\pi_0Z\pi_0 + x'_n \pi_0X\pi_0.$$ Since $\pi_0Z\pi_0 = \pi_0X\pi_0 = 0$ and $\pi_0Y\pi_0 = \pi_0$, the claim follows. \qedi \end{proof} \begin{remark} From (\ref{E1a}) through (\ref{E1c}) we may derive recursions for calculating $x_n$ or $z_n$, e.g. \begin{equation} \label{zn recursion with integration constant} z_{n+2}(t) = - \frac{\D}{\D t} \left( z'_n(t) + \theta'(t) \left( \int \theta'(t) z_n(t) \, \D t + C \right) \right). \end{equation} The constant of integration $C$ must (and in some cases can) be determined by comparison with (\ref{true recursion}). \end{remark} Using (\ref{E1a})--(\ref{E1c}), we can give very simple expressions for the quantities appearing in (\ref{pi1}) and (\ref{pi2}). As for (\ref{pi2}), we use (\ref{derivatives}) and the differential equations to find \begin{eqnarray} \commut{\I\eps \partial_t - H}{\pi^{(n)}} & = & \I \eps^{n+1} \pi_n' = \I \eps^{n+1} \left(x_n' X + (y_n' + \theta' z_n) Y + (z_n' - \theta' y_n) Z\right) = \nonumber \\ & = & - \eps^{n+1} (z_{n+1} X + x_{n+1} Z) \label{pi_n'}. \end{eqnarray} Now we turn to $(\pi^{(n)})^2 - \pi^{(n)}$, the term by which $\pi^{(n)}$ fails to be a projector. Let us write \begin{equation} \label{gnk} (\pi^{(n)})^2 - \pi^{(n)} = \sum_{k=1}^{n} \eps^{n+k} G_{n+1,k}. \end{equation} With our earlier convention, $G_{n+1,1} = G_{n+1}$. Explicitly, (\ref{ansatz}) and (\ref{gnk}) give \begin{equation} \label{Gnk shape} G_{n+1,k} = \antcom{\pi_k}{\pi_n} + \antcom{\pi_{k+1}}{\pi_{n-1}} + \ldots = \sum_{j=0}^{n-k} \pi_{j+k} \pi_{n-j}. \end{equation} \begin{proposition} \label{proj correction} For each $n \in \N$, there exist functions $g_{n+1,k}, k \leq n$ with \begin{equation} \label{projector error} ((\pi^{(n)})^2 - \pi^{(n)})(t) = \left(\sum_{k=1}^{n} \eps^{n+k} g_{n+1,k}(t)\right) W. \end{equation} For each $k \leq n$, $$g'_{n+1,k} = 2 \I ( x_k z _{n+1} - z_k x_{n+1}).$$ \end{proposition} \begin{proof} By (\ref{products}), each $G_{n+1,k}$ is proportional to $W$. Using (\ref{xyz ansatz}) additionally, we find $\antcom{\pi_k}{\pi_m} = 2 (- x_k x_m + y_k y_m + 2 z_k z_m)W$, and thus (\ref{Gnk shape}) yields $$ g_{n+1,k} = \sum_{j=0}^{n-k} - x_{j+k} x_{n-j} + y_{j+k} y_{n-j} + z_{j+k} z_{n-j}.$$ Thus by using Proposition \ref{function recursion}, \begin{eqnarray*} g'_{n+1,k} & = & \sum_{j=0}^{n-k} \I ( z_{j+k+1} x_{n-j} + x_{j+k} z_{n-j+1} ) - ( \theta' z_{j+k} y_{n-j} + \theta' y_{j+k} z_{n-j}) + \\ && + \theta y_{j+k} z_{n-j} + \theta' y_{j+k} z_{n-j} - \I ( x_{j+k+1} z_{n-j} + z_{j+k} x_{n-j+1} ) = \\ & = & \I \sum_{j=0}^{n-k} ( ( z_{j+k+1} x_{n-j} - z_{j+k} x_{n-j+1}) + ( x_{j+k} z_{n-j+1} - x_{j+k+1} z_{n-j}) = \\ & = & 2 \I ( x_k z_{n+1} - z_k x_{n+1} ). \end{eqnarray*} The last equality follows because the sum is a telescopic sum. \qedi \end{proof} Since $W = \id$ is independent of $t$, Proposition \ref{proj correction} gives the derivative of the correction $(\pi^{(n)})^2 - \pi^{(n)}$ to a projector. As above, this gives an easy way for estimating the correction itself provided we have some clue how to choose the constant of integration. \section{Construction of the unitary} \label{unitary} We now proceed to construct the unitary transformation $U_\eps^n$ into the $n^{\rm th}$ superadiabatic basis. By (\ref{ansatz}) and (\ref{recursion1}), $\pi^{(n)}$ is self-adjoint. Thus it has two orthonormal eigenvectors $v_n$ and $w_n$. Let $$v_0 = \left( \begin{array}{c} \cos(\theta/2) \\ \sin(\theta/2) \end{array} \right), \quad w_0 = \left( \begin{array}{c} \sin(\theta/2) \\ -\cos(\theta/2) \end{array} \right)$$ be the eigenvectors of $\pi_0$, and write \begin{equation} \label{v_n ansatz} v_n = \alpha v_0 + \beta w_0, \quad w_n = \overline{\alpha} w_0 - \overline{\beta} v_0 \qquad (\alpha, \beta \in \C). \end{equation} We make this representation unique by requiring $0 \leq \alpha \in \R$. Let $U_\eps^n$ be the unitary operator taking $(v_n,w_n)$ to the standard basis $(e_1,e_2)$ of $\R^2$ , i.e. \begin{equation} \label{U} U_\eps^n = e_1 v_n^{\ast} + e_2 w_n^{\ast}, \end{equation} where all vectors are column vectors. Note that the definition (\ref{U_0}) of $U_0$ is consistent with (\ref{U}) for $n=0$. $U_\eps^n$ diagonalizes $\pi^{(n)}$, thus \begin{equation} \label{diag pi_n} U_\eps^n \pi^{(n)} U_\eps^{n\, \ast} = D \equiv \left( \begin{array}{cc} \lambda_1 & 0 \\ 0 & \lambda_2 \end{array} \right), \end{equation} where $\lambda_{1,2}$ are the eigenvalues of $\pi^{(n)}$. Although $\alpha, \beta$ and $\lambda_{1,2}$ depend on $n$, $\eps$ and $t$, we suppress this from the notation. \begin{lemma} \label{U errors} $$ U_0 U_\eps^{n\,\ast} = \left( \begin{array}{cc} \alpha & -\overline{\beta} \\ \beta & \alpha \end{array} \right), \quad \mbox{and} \quad U_0 U_\eps^{n\,\ast '} = \left( \begin{array}{cc} \alpha' + \beta & \,\,\alpha -\overline{\beta}' \\ \beta' - \alpha & \,\,\alpha' + \overline{\beta} \end{array} \right). $$ \end{lemma} \begin{proof} The calculations are straightforward and we only show the second equality. First note that $v_0' = - w_0$ and $w_0' = v_0$. Thus $$ U_\eps^{n\,\ast '} = ((\alpha' + \beta) v_0 + (\beta' - \alpha) w_0) e_1^{\ast} + (( \alpha - \overline{\beta}')v_0 + (\alpha' + \overline{\beta})w_0)e_2^\ast,$$ and using the orthogonality of $v_0$ and $w_0$ yields the claim, $$ U_0 U_\eps^{n\,\ast '} = e_1 (\alpha' + \beta) e_1^\ast + e_1 ( \alpha - \overline{\beta}') e_2^\ast + e_2 (\beta' - \alpha) e_1^\ast + e_2 ( \alpha' +\overline{\beta})e_2^{\ast}\,.\qquad\qedi$$ \end{proof} It will turn out that $\beta, \alpha' \alpha$, and $\beta'$ are small quantities, $\lambda'_1, \lambda'_2$, and $\lambda_2$ are even much smaller, while $\alpha^2$ and $\lambda_1$ are large, i.e. of order $1 + \Or(\eps)$. This motivates the form in which we present the following result. \begin{proposition} \label{general transformed matrix} Suppose $\lambda_1 \neq \lambda_2$. Then for each $n \in \N$, $$ U_\eps^n (\I \eps \partial_t - H) U_\eps^{n\,\ast} = \I \eps \partial_t - \left( \begin{array}{cc} \frac{1}{2} & \frac{\alpha^2 \eps^{n+1}}{\lambda_1 - \lambda_2} (x_{n+1} - z_{n+1}) \\ \frac{\alpha^2 \eps^{n+1}}{\lambda_1 -\lambda_2} (-x_{n+1} - z_{n+1}) & -\frac{1}{2} \end{array} \right) + R,$$ with $$ R = {\footnotesize\left( \begin{array}{cc} \eps \tIm (\overline{\beta}(2 \alpha + \beta')) + |\beta|^2 & - \frac{\eps^{n+1} \overline{\beta}^2}{\lambda_1 - \lambda_2} (x_{n+1} + z_{n+1}) \\ \frac{\eps^{n+1} \beta^2}{\lambda_1 -\lambda_2} (x_{n+1} - z_{n+1}) & - \eps \tIm (\overline{\beta}(2 \alpha + \beta')) - |\beta|^2 \end{array} \right)}.$$ \end{proposition} \begin{proof} Let us write $U_\eps^n (\I \eps \partial_t - H) U_\eps^{n\,\ast} = (M_{i,j})$, $i,j \in \{1,2\}$. $M_{1,1}$ and $M_{2,2}$ are calculated in a straightforward manner, using Lemma \ref{U errors} together with the fact $U_0 H U_0^{\ast} = {\scriptsize \left( \begin{array}{cc} 1/2 & 0 \\ 0 & -1/2 \end{array} \right)}$: \begin{eqnarray*} \lefteqn{U_\eps^n (\I \eps \partial_t - H) U_\eps^{n\,\ast} = \I \eps \partial_t + \I \eps U_\eps^n U_0^{\ast} U_0 U_\eps^{n\,\ast '} - U_\eps^n U_0^{\ast} U_0 H U_0^{\ast} U_0 U_\eps^{n\,\ast} = }\\ &=& \I \eps \partial_t + \I \eps {\footnotesize \left( \begin{array}{cc} \alpha & \overline{\beta} \\ -\beta & \alpha \end{array} \right) \left( \begin{array}{cc} \alpha' + \beta & \alpha -\overline{\beta}' \\ \beta' - \alpha & \alpha' + \overline{\beta} \end{array} \right) - \frac{1}{2} \left( \begin{array}{cc} \alpha & \overline{\beta} \\ -\beta & \alpha \end{array} \right) \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right) \left( \begin{array}{cc} \alpha & -\overline{\beta} \\ \beta & \alpha \end{array} \right)}. \end{eqnarray*} Carrying out the matrix multiplication yields \begin{equation} \label{the diag elements} M_{1,1} = - M_{2,2} = \I \eps \partial_t + \I \eps ((\alpha(\alpha' + \beta) + \overline{\beta}(\beta' - \alpha)) - \frac{1}{2} (\alpha^2 - |\beta|^2). \end{equation} We now use $\alpha^2 + |\beta|^2 = 1$ to obtain $ 0 = 2 \alpha \alpha' + \beta' \overline{\beta} + \overline{\beta}' \beta = 2 \tRe(\alpha \alpha' + \overline{\beta}\beta')$ and $\alpha^2 -|\beta|^2 = 1 - 2 |\beta|^2$. Plugging these into (\ref{the diag elements}) gives the diagonal coefficients of $M$. Although we could get expressions for the off-diagonal coefficients by the same method, these would not be useful later on. Instead we use (\ref{diag pi_n}), i.e. $U_\eps^{n\,\ast} D = \pi^{(n)} U_\eps^{n\,\ast}$ together with (\ref{pi_n'}) and obtain \begin{equation} \label{commut trick} U_\eps^n (\I \eps \partial_t - H) U_\eps^{n\,\ast}D = DU_\eps^n (\I \eps \partial_t - H) U_\eps^{n\,\ast} - \eps^{n+1} U_\eps^n (z_{n+1} X + x_{n+1} Z) U_\eps^{n\,\ast}. \end{equation} By multiplying (\ref{commut trick}) with $e_j e_j^{\ast}$ from the left and by $e_k e_k^{\ast}$ from the right $(j,k \in \{1,2\})$ and rearranging, we obtain \begin{eqnarray} \label{commut trick 2} \lefteqn{(\lambda_k - \lambda_j) \,e_j \,e_j^{\ast} \,U_\eps^n (\I \eps \partial_t - H)\, U_\eps^{n\,\ast} e_k \,e_k^{\ast} = } \\ & = & - \eps^{n+1} e_j \,e_j^{\ast} \,U_\eps^n (z_{n+1} X + x_{n+1} Z) \,U_\eps^{n\,\ast} e_k\, e_k^{\ast} - \I \delta_{k,j} \,\eps \,\lambda'_j \,e_j\, e^{\ast}_j\,. \nonumber \end{eqnarray} From the equalities $ U_0 X U_0^{\ast} = { \left( \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right)}$, $U_0 Z U_0^{\ast} = { \left( \begin{array}{cc} 0 & -1 \\ -1 & 0 \end{array} \right)}$ and Lemma \ref{U errors} we obtain $$ U_\eps^n X U_\eps^{n\,\ast} = {\footnotesize \left( \begin{array}{cc} \alpha(\beta-\overline\beta) & \alpha^2 + \overline\beta^2 \\ -(\alpha^2+\beta^2) & -\alpha(\beta - \overline{\beta}) \end{array} \right)}, \qquad U_\eps^n Z U_\eps^{n\,\ast} = {\footnotesize \left( \begin{array}{cc} -\alpha(\beta+\overline\beta) & -(\alpha^2 - \overline\beta^2) \\ -(\alpha^2-\beta^2) & \alpha(\beta + \overline{\beta}) \end{array} \right)}.$$ The expressions for $M_{1,2}$ and $M_{2,1}$ follow by taking $k \neq j$ in (\ref{commut trick 2}). \qedi \end{proof} We now use our results from the previous section to express $\alpha, \beta$ and $\lambda_{1,2}$ in terms of $x_k, y_k$ and $z_k$, $k \leq n$. Let us define \begin{eqnarray} \label{xi} \xi & \equiv & \xi(n,\eps,t) = {\textstyle \sum_{k=1}^n} \eps^k x_k(t), \\ \label{eta} \eta & \equiv & \eta(n,\eps,t) = {\textstyle\sum_{k=1}^n} \eps^k y_k(t), \\ \label{zeta} \zeta & \equiv & \zeta(n,\eps,t) = {\textstyle\sum_{k=1}^n} \eps^k z_k(t). \end{eqnarray} Moreover, let \begin{equation} \label{g} g \equiv g(n,\eps,t) = {\textstyle\sum_{k=1}^{n}} \eps^{n+k} g_{n+1,k}(t) \end{equation} be the quantity appearing in (\ref{projector error}). \begin{lemma} \label{eigenvalues} The eigenvalues of $\pi^{(n)}$ solve the quadratic equation $$ \lambda_{1,2}^2 - \lambda_{1,2} - g = 0.$$ \end{lemma} \begin{proof} By (\ref{diag pi_n}) and Proposition \ref{proj correction} we obtain $$ \left(\begin{array}{cc} \lambda_1^2 - \lambda_1 & 0 \\ 0 & \lambda_2^2 - \lambda_2 \end{array} \right) = U_\eps^n ((\pi^{(n)})^2 - \pi^{(n)}) \,U_\eps^{n\,\ast} = U_\eps^n g W U_\eps^{n\,\ast} = \left( \begin{array}{cc} \,g\, & \,0\, \\[2pt] \,0\, & \,g\, \end{array} \right)\,.\qquad\qedi $$ \end{proof} \begin{lemma} \label{alpha beta} $$ \alpha^2 (\lambda_1 - \lambda_2) = 1 - \eta - \lambda_2, \qquad \mbox{and } \qquad \alpha \beta (\lambda_1 - \lambda_2) = - \xi - \zeta.$$ \end{lemma} \begin{proof} We have \begin{equation} \label{no1} \pi^{(n)} = \lambda_1 v_n v_n^{\ast} + \lambda_2 w_n w_n^{\ast}. \end{equation} Plugging in (\ref{v_n ansatz}), we obtain \begin{eqnarray*} \pi^{(n)} v_0 &=& \lambda_1 \alpha v_n - \lambda_2 \beta w_n = (\lambda_1\alpha^2 + \lambda_2 |\beta|^2) v_0 + (\lambda_1 - \lambda_2)\alpha \beta w_0 = \\ &=& (\alpha^2(\lambda_1 - \lambda_2) + \lambda_2) v_0 + (\lambda_1 - \lambda_2)\alpha \beta w_0. \end{eqnarray*} In the last step, we used $|\beta|^2 + \alpha^2 = 1$. On the other hand, from (\ref{ansatz}) and (\ref{recursion1}) we have \begin{equation}\label{p0minuspi} \pi^{(n)} = \pi_0 + \sum_{k=1}^n \eps^k (x_k X + y_kY + z_kZ)\,, \end{equation} and since $Xv_0 = Zv_0 = -w_0$, $\pi_0 v_0 = v_0$ and $Y v_0 = - v_0$, we find $$ \pi^{(n)} v_0 = (1 - \eta) v_0 - (\xi + \zeta) w_0.$$ Comparing coefficients finishes the proof. \qedi \end{proof} \begin{theorem} \label{general diag} Let $\eps_0>0$ be sufficiently small. For $\eps\in(0,\eps_0]$ assume there is a bounded function $q$ on $\R$ such that $\xi(t)$, $\eta(t)$, $\zeta(t)$ and their derivatives $\xi'(t), \eta'(t), \zeta'(t)$ are all bounded in norm by $\eps q(t)$. Then \begin{eqnarray} \label{general offdiag eq} \lefteqn{U_\eps^n (\I \eps \partial_t - H) U_\eps^{n\,\ast} =}\\&=& \I \eps \partial_t- \left( \begin{array}{cc} \frac{1}{2} + \Or(\eps^2 q^2) & \eps^{n+1} (x_{n+1} - z_{n+1})\, (1 + \Or(\eps q)) \\[2mm] \eps^{n+1} (-x_{n+1} - z_{n+1})\, (1 + \Or(\eps q)) & -\frac{1}{2} + \Or(\eps^2 q^2) \end{array} \right)\,.\nonumber \end{eqnarray} \end{theorem} \begin{proof} From (\ref{p0minuspi}) and our assumptions it follows that $\pi^{(n)} - \pi_0 = \Or(\eps q)$. Thus $\lambda_1 = 1 + \Or(\eps q)$ and $\lambda_2 = \Or(\eps q)$, and from Lemma \ref{eigenvalues} we infer $g = \Or(\eps)$ and $$ \lambda_1 = {\textstyle\frac{1}{2}} \left( 1 + \sqrt{1 + 4 g} \right)\,,\qquad\lambda_2 = {\textstyle\frac{1}{2}} \left( 1 - \sqrt{1 + 4 g} \right)\,.$$ Since $\lambda_1 - \lambda_2 \neq 0$, Lemma \ref{alpha beta} yields $$ \alpha^2 = \frac{1 + \sqrt{1+4g} - 2 \eta}{2 \sqrt{1 + 4g}}, \quad \beta = \frac{-\xi - \zeta}{\sqrt{1+4g} \alpha}\,.$$ Hence $ \alpha^2 = 1 + \Or(\eps q),$ and $\beta, \beta' $ and $ \alpha \alpha' = (\alpha^2)' /2$ are all $\Or(\eps q)$. Plugging these into the matrix $R$ in Proposition \ref{general transformed matrix} shows the claim. \qedi \end{proof} \section{Solving the recursion: a pair of simple poles} In order to make further progress, we need to understand the asymptotic behavior of the off-diagonal elements of the effective Hamiltonian in the $n^{\rm th}$ superadiabatic basis for large $n$. According to (\ref{general offdiag eq}) this amounts to the asymptotics of $x_n$ and $z_n$ as given by the recursion from Proposition \ref{function recursion}. It is clear that the function $\theta'$ alone determines the behavior of this recursion. We will study here the special case \begin{equation} \label{theta ansatz} \theta'(t) = \frac{\I \gamma}{t+\I t_{\rm c}} - \frac{ \I \gamma}{t-\I t_{\rm c}} = \frac{\gamma t_{\rm c}}{t^2 + t_{\rm c}^2}\,. \end{equation} The reason lies in the intuition that the poles of $\theta'$ closest to the real axis determine the superadiabatic transitions, and that these transitions are of universal form whenever these poles are of order one, see \cite{Be,BeLi} for details. As in \cite{HaJo}, we have to restrict to the special case that $\theta'$ has no contribution besides these poles in order solve the recursion. We now have two parameters left in $\theta'$. The distance $t_{\rm c}$ of the poles from the real axis determines the exponential decay rate in the in the off-diagonal elements of the Hamiltonian and the strength of the residue $\gamma$ determines the pre-factor in front of the exponential. As is done in \cite{HaJo}, we could get rid of the parameter $t_{\rm c}$ by rescaling time, but we choose not to do so because $t_{\rm c}$ plays a nontrivial role in optimal truncation and the error bounds obtained therein, and keeping this parameter will make things more transparent. We use (\ref{zn recursion with integration constant}) in order to determine the asymptotics of $z_n$. From Proposition \ref{function recursion} together with (\ref{true recursion}) it is clear that $y_n$ must go to zero as $t \to \pm \infty$. This fixes the constant of integration in (\ref{zn recursion with integration constant}), and we arrive at the linear two-step recursion \begin{equation} \label{zn recursion} z_{n+2}(t) = - \frac{\D}{\D t} \left( z'_n(t) + \theta'(t) \int_{-\infty}^t \theta'(s) z_n(s) \, \D s \right). \end{equation} The fact that the recursion is linear will make its analysis simpler than the one of the nonlinear recursion in \cite{HaJo}. We rewrite $\theta'$ as $$ \theta'(t) = \frac{\gamma}{t_{\rm c}} (f + \overline f) \qquad \mbox{with}\qquad f(t) = \frac{\I t_{\rm c}}{t + \I t_{\rm c}}\,.$$ For $z_n$, we will make an Ansatz as a sum of powers of $f$ and $\overline{f}$. The reason for the success of this approach is the fact that this representation is stable under differentiation and integration, and also under multiplication with $\theta'$ through the partial fraction expansion. More explicitly, we have \begin{lemma} \label{identities} For each $m \geq 1$, \begin{eqnarray} \theta' \tIm(f^m) &=& \frac{\gamma}{t_{\rm c}} \sum_{k=0}^{m-1} 2^{-k} \tIm(f^{m+1-k}), \label{Im mult} \\ \theta' \tRe(f^m) &=& \frac{\gamma}{t_{\rm c}} \left( \sum_{k=0}^{m-1} 2^{-k} \tRe(f^{m+1-k}) + 2^{-m} \theta' \right), \label{Re mult} \\ \tIm(f^m)' &=& -\frac{m}{t_{\rm c}} \tRe(f^{m+1}), \label{Im diff}\\ \tRe(f^m)' &=& \frac{m}{t_{\rm c}} \tIm(f^{m+1}). \label{Re diff} \end{eqnarray} \end{lemma} \begin{proof} We have $ f + \overline{f} = \frac{2t_{\rm c}^2}{t^2+t_{\rm c}^2} = 2 f\overline{f}$, and thus $$ f^k \overline f = \frac{1}{2} f^{k-1}(f+\overline f) = \frac{1}{2} (f^k + f^{k-1}\overline f)$$ and \begin{equation} \label{multiplication rule} \theta' f^{n-j} = \frac{\gamma}{t_{\rm c}} \left( \sum_{k=0}^{n-j} 2^{-k} f^{n+1 - (j+k)} + 2^{-n+j}\overline f \right). \end{equation} Taking the complex conjugate of (\ref{multiplication rule}) and adding it to resp.\ subtracting it from (\ref{multiplication rule}), we arrive at (\ref{Im mult}) and (\ref{Re mult}). To prove (\ref{Im diff}) and (\ref{Re diff}), it suffices to use that $(f^k)' = k f^{k+1} / (\I t_{\rm c})$ along with the complex conjugate equation. \qedi \end{proof} %The preceding simple Lemma is all we need to obtain \begin{proposition} \label{z_n and recursion} For each even $n \in \N$ and $j = 0, \ldots, n-1$, let the numbers $a_j^{(n)}$ be recursively defined through \begin{eqnarray} a_0^{(2)} & = & 1, \qquad a_1^{(2)} = 0\,, \label{a_n rekursion start}\\ a_j^{(n+2)} &=& \frac{n+1-j}{(n+1)\,n} \left( (n-j)\,a_j^{(n)} - \gamma^2 \sum_{k=0}^j \frac{1}{n-k} \sum_{m=0}^k a_m^{(n)} \right) \quad (j 0$ such that for all $n \in \N$ $$ |a_1^{(n)}| \leq C_1 \frac{\ln n}{n-1}\,.$$ \item For each $p>1$ there exists $C_2 > 0$ such that for all $n \in \N$ $$ \sup_{j \geq 2} p^{-j} |a_j^{(n)}| \leq \frac{C_2}{n-1}\,.$$ \end{enumerate} \end{proposition} \begin{proof} (a) By (\ref{a_n rekursion start}), $a_0^{(2)} = 1$, and $$ a_0^{(n+2)} = a_0^{(n)}\left( 1 - \frac{\gamma^2}{n^2} \right).$$ Comparing with the product representation of the sine function, $$ \sin(\pi x) = \pi x \prod_{n=1}^{\infty} \left( 1 - \frac{x^2}{n^2} \right)\,,$$ we arrive at (a). \\ (b) Put $\alpha_n = (n-1) a_1^{(n)}$. Then by (\ref{a_n rekursion step}), $$ \alpha_{n+2} = \alpha_n \left( 1 - \frac{\gamma^2}{(n-1)^2} \right) - \gamma^2 \left( \frac{1}{n} + \frac{1}{n-1} \right) a_0^{(n)}.$$ thus for $n-1 > \gamma$, we have $$ |\alpha_{n+2}| \leq |\alpha_n| + \gamma^2 \left( \frac{1}{n} + \frac{1}{n-1} \right) \max_{m \in \N} |a_0^{(m)}|,$$ which shows (b). \\ (c) Put $c_j^{(n)} = (n-1) p^{-j} a_j^{(n)}$, and $c^{(n)} = \max_{j \geq 2} |c_j^{(n)}|$. We will show that the sequence $c^{(n)}$ is bounded. We have \begin{eqnarray} c_j^{(n+2)} & = & \frac{n+1-j}{n(n-1)} \left( (n-j) c_j^{(n)} - \gamma^2 \sum_{k=2}^j \frac{1}{n-k} \sum_{m=2}^k p^{-j+m} c_m^{(n)} - \right. \nonumber \\ && \left. -(n-1) p^{-j} \gamma^2 \left( a_0^{(n)} \sum_{k=0}^j \frac{1}{n-k} + a_1^{(n)} \sum_{k=1}^j \frac{1}{n-k} \right) \right). \label{bigformula} \end{eqnarray} Now \begin{equation} \label{term1} \left| \sum_{k=2}^j \frac{1}{n-k} \sum_{m=2}^k p^{-j+m} c_m^{(n)} \right| \leq c^{(n)} \frac{1}{n-j}\frac{p^2}{(p-1)^2}, \end{equation} and \begin{equation} \label{term2} p^{-j} \sum_{k=0}^j \frac{1}{n-k} \leq \frac{(j+1) p^{-j}}{n-j} \leq \frac{1}{(n-j)\ln p}. \end{equation} We plug these results into (\ref{bigformula}) and obtain \begin{eqnarray*} %\label{estimate of bigformula} |c_j^{(n+2)}| &\leq& c^{(n)} \left( \frac{(n+1-j)(n-j)}{n(n-1)} + \frac{(n+1-j)\,\gamma^2 p^2}{(n-j)(p-1)^2} \frac{1}{n(n-1)} \right) +\\ & & + \frac{1}{n} \frac{(n+1-j)\,p^2\,\gamma^2}{(n-j) (p-1)^2 \ln p}(|a_0^{(n)}|+|a_1^{(n)}|). \end{eqnarray*} By (a) and (b), $a_0^{(n)}$ and $a_1^{(n)}$ are bounded. Taking the supremum over $j \geq 2$ above, we see that there exist constants $B_1$ and $B_2$ with $$ c^{(n+2)} \leq c^{(n)} \left( \frac{n-2}{n} + \frac{B_1}{n(n-1)} \right) + \frac{B_2}{n},$$ hence $$ c^{(n+2)} - c^{(n)} \leq \frac{1}{n} \left(\left( -2 + \frac{B_1}{n-1}\right) c^{(n)} + B_2 \right).$$ Now let $n-1 > B_1$. Then for $c^{(n)} > B_2$, the above inequality shows $c^{(n+2)} < c^{(n)}$, while for $c^{(n)} \leq B_2$, $c^{(n+2)} \leq c^{(n)} + B_2/n \leq B_2 (1 + 1/n).$ Thus $c^{(n)}$ is a bounded sequence. \qedi \end{proof} \begin{remark} We will make no use of the fact that the logarithmic correction to the $1/n$-decay of the higher coefficients occurs only in the coefficient $a_1^{(n)}$. We chose to include this in the statement of the preceding theorem anyway, because this gives some insight into the nature of the recursion and is not hard to prove. \end{remark} \begin{remark} Numerical calculations of the first few thousand $a_j^{(n)}$ suggest that (c) above continues to be true if we choose $p=1$, but this seems to be much harder to prove. However, the estimate above is more than good enough for us. \end{remark} \begin{remark} The constants appearing in the proof of Proposition~\ref{a_n limit} (b) and (c) are not optimal, and could be improved by more careful arguments. This is unimportant for our purposes, and for the sake of brevity and readability we chose to use the simple estimates given. \end{remark} \begin{corollary} \label{b_n limit} Let $b_j^{(n)}$ be given by (\ref{bj in proof}). Then for each $p > 1$, there exists $C_3>0$ such that $$ \sup_{j \geq 0} p^{-j} b_j^{(n)} \leq \frac{C_3}{n-1}.$$ \end{corollary} \begin{proof} For $j \leq n-1$, we have $n-1 \leq j(n-j)$, and thus Proposition~\ref{a_n limit} (c) gives \begin{eqnarray*} p^{-j} b_j^{(n)} & \leq & \frac{p^{-j}}{n-j} \left( (a_0^{(n)} + a_1^{(n)}) + \frac{C_2}{n-1} \frac{p^2}{p-1} (p^{j-1}-1) \right) \leq \\ & \leq & p^{-j} \frac{j(a_0^{(n)} + a_1^{(n)})}{n-1} + \frac{p\, C_2}{p-1} \frac{1}{n-1} \leq \frac{C_3}{n-1}\,.\qquad\qedi \end{eqnarray*} \end{proof} Having good control over the coefficients $a_j^{(n)}$, we can now derive relatively sharp estimates on the functions $x_n, y_n$ and $z_n$. Let us fix $\alpha < 1$ and define $$R_n^\alpha(t) = \frac{1}{(n-1)^\alpha} \max \left\{ \left| \frac{t_{\rm c}}{t+\I t_{\rm c}} \right|^n, \left( \frac{1}{\sqrt{2}} \right)^{n-2} \left| \frac{t_{\rm c}}{t+\I t_{\rm c}} \right|^2 \right\}.$$ Obviously, for $t \leq t_{\rm c}$ the first function in the maximum above dominates, for $t>t_{\rm c}$ the second one does. For families of functions $g_n(t), G_n(t)$ we write $$ g_n(t) = \Or(G_n(t))$$ if there exists $C>0$ such that $|g_n(t)| \leq C |G_n(t)|$ for all $n \in \N$ and all $t \in \R$. \begin{theorem} \label{main estimates} For $n>1$ and $\alpha < 1$, we have \begin{eqnarray} x_n(t) & = & \I \frac{(n-1)!}{t_{\rm c}^n} \left( \frac{2\sin(\gamma\pi/2)}{\pi} \tRe \left(\left( 1 - \I \frac{t}{t_{\rm c}} \right)^{-n}\right) + \Or(R_n^\alpha(t)) \right), \label{AA}\\ y_n(t) & = & \frac{(n-1)!}{t_{\rm c}^n} \,\,\Or(R_n^\alpha(t)), \label{BB}\\ z_n(t) & = & - \frac{(n-1)!}{t_{\rm c}^n} \left( \frac{2\sin(\gamma\pi/2)}{\pi} \tIm \left(\left( 1 - \I \frac{t}{t_{\rm c}} \right)^{-n}\right) + \Or(R_n^\alpha(t)) \right), \label{AB}\ \end{eqnarray} \end{theorem} \begin{proof} With the definition of $f$ and Proposition~\ref{a_n limit} (a) we get $$ a_0^{(n)} \tIm(f^n) = \frac{2 \sin(\gamma\pi/2)}{\pi\gamma} \tIm \left(\left( 1 - \I \frac{t}{t_{\rm c}} \right)^{-n}\right) + O\left( \frac{1}{n^2} \left|\frac{t_{\rm c}}{t+\I t_{\rm c}} \right|^n\right)$$ when $n$ is even, and a similar formula for $a_0^{(n+1)} \tRe(f^n)$ when $n$ is odd. This covers the $j=0$ terms in (\ref{x_n recursion step}) and (\ref{z_n recursion step}). For the remaining terms, let $$c_j^{(n)} = \left\{ \begin{array}{ll} a_j^{(n)} & \,\,\mbox{if } n \mbox{ is even,} \\ n a_j^{(n)}/(n-j) &\,\, \mbox{if } n \mbox{ is odd.} \end{array} \right.$$ Now $n/(n-j) \leq j$ for $j < n$, and thus by Proposition~\ref{a_n limit} (b) and (c) for each $p>1$ we can find $C>0$ such that $$c_j^{(n)} \leq j p^j \frac{C}{(n-1)^{\alpha}} $$ for all $j \geq 1$. (For $j \geq 2$, we may even choose $\alpha = 1$, but we will not exploit this.) For $|t| \leq t_{\rm c}$, we have $|t_{\rm c}/(t+\I t_{\rm c})|^{-j} \leq 2^{j/2}$, so we get $$ \left| \sum_{j=2}^{n-1} \frac{c_j^{(n)}}{2^j} \left( \frac{\I t_{\rm c}}{t+\I t_{\rm c}}\right)^{n-j} \right| \leq \left( \frac{C}{(n-1)^\alpha} \sum_{j=2}^{n-1} j \left(\frac{p}{\sqrt{2}} \right)^j \right) \left| \frac{t_{\rm c}}{t+\I t_{\rm c}} \right|^n.$$ If we choose $p < \sqrt{2}$, the sum on the right hand sided is bounded uniformly in $n$. Combining this with our above calculations, (\ref{AA}) and (\ref{AB}) are proved for $|t| < t_{\rm c}$. For $|t| > t_{\rm c}$, we have $|t_{\rm c} / (t+\I t_{\rm c})| \leq 1/\sqrt{2}$, and thus $$ \left| \sum_{j=2}^{n-2} \frac{c_j^{(n)}}{2^j} \left( \frac{\I t_{\rm c}}{t+\I t_{\rm c}}\right)^{n-j} \right| \leq \frac{C}{(n-1)^\alpha} \left| \frac{t_{\rm c}}{t+\I t_{\rm c}} \right|^2 \sum_{j=2}^{n-2} j \left(\frac{p}{2}\right)^j \left( \frac{1}{\sqrt{2}} \right)^{n-2-j}. $$ If we choose again $p < \sqrt{2}$, the sum on the right hand side is bounded by $\tilde{C}(1/\sqrt{2})^{n-2}$ uniformly in $n$. For the term with $j = n-1$, this does not work since then $n-2-j < 0$. But for $n$ even, this term vanishes since then $c_{n-1}^{(n)} = 0$, and for $n$ odd, it equals $$ \frac{c_{n-1}^{(n)}}{2^{n}} \tRe\left( \frac{\I t_{\rm c}}{t+\I t_{\rm c}} \right) = \frac{n a_{n-1}^{(n+1)}}{2^n} \frac{t_{\rm c}^2}{t^2+t_{\rm c}^2} \leq \frac{\tilde C}{n-1} \left(\frac{1}{\sqrt{2}}\right)^{n-2} \left| \frac{t_{\rm c}}{t+\I t_{\rm c}} \right|^2.$$ This proves (\ref{AA}) and (\ref{AB}) for $|t| \geq t_{\rm c}$. The proof of (\ref{BB}) is similar and uses Corollary \ref{b_n limit}. \qedi \end{proof} %{$t_{\rm c}$ as switching point is a little bit arbitrary. %Everything smaller than $\sqrt{3} t_{\rm c}$ (I think...) does it, %but then other constants get worse. Maybe here we make a remark %about this.} \section{Optimal truncation} By the results of the previous section $\pi_k$ grows like $(k-1)!/t_{\rm c}^k$. Hence, the sum $\pi^{(n)} = \sum_{k=0}^n \eps^k \pi_k$ does not converge to an exactly equivariant projection $\pi^{(\infty)}$ as $n \to \infty$. This is the reason why we see exponentially small transitions. The basis in which these transitions develop smoothly is the optimal superadiabatic basis: since we cannot go all the way to infinity with $n$, we fix $\eps$ and choose $n=n(\eps)$ such that the off-diagonal elements in (\ref{general diag}) become minimal. Using Stirling's formula and (\ref{AA}) resp.\ (\ref{AB}), it is easy to see that the place to truncate is at $n(\eps) = t_{\rm c}/ \eps$. This $n(\eps)$ is in general not a natural number, but we will find that a change of $n$ which is of order one does not change the results. Before we go into more details, we need a preliminary result. \begin{lemma} \label{power to exponential} Uniformly in $x \in [0,1]$ and for $k > 0$, we have $$ (1+x)^{-k} = \e^{-k x} + \e^{-kx/2} \Or{\textstyle \left(\frac{1}{k}\right)}.$$ \end{lemma} \begin{proof} We start with the equality \begin{equation} \label{diff1} (1+x)^{-k} - \e^{-kx} = \e^{-kx} \left(\e^{k(x-\ln(1+x))} -1 \right). \end{equation} At first consider $x > \sqrt{1/k}$. There we use the inequality $(x-\ln(1+x)) \leq x/3$, valid for $0 \leq x \leq 1$, in (\ref{diff1}) and obtain $$ | (1+x)^{-k} - \e^{-kx} | \leq \e^{-kx} \left( \e^{kx/3} - 1 \right) = \e^{-kx/2} \left( \e^{-kx/6} - \e^{-kx/2}\right). $$ For $x> \sqrt{1/k}$, the term in the last bracket above is $\Or(1/k)$, and we are done in this case. For $x\leq\sqrt{1/k}$, we use $(x-\ln(1+x)) \leq x^2/2$ and rearrange (\ref{diff1}) to get $$ \e^{kx/2} ((1+x)^{-k} - \e^{-kx}) = \e^{-kx/2} \left(\e^{kx^2/2} -1 \right) =: f(x,k). $$ To find out where $f(x,k)$ is maximal, we calculate $$ \frac{\D}{\D x}f(x,k) = \frac{k}{2} \e^{-kx/2} \left( 1 + \e^{kx^2/2}(2x-1)\right).$$ The derivative is zero exactly at the solutions of the equation \begin{equation} \label{nullst} \ln(1-2x)/x^2 = -k/2. \end{equation} Now $\ln(1-2x)/x^2 = -2/x + R(x)$, where $R(x)$ is a power series in $x$, convergent for $x < 1/2$. Thus for $x < \sqrt{1/k}$ and $k$ sufficiently large, there exists exactly one solution $x^{\ast}(k)$ of (\ref{nullst}), and $x^{\ast}(k) < C/k$ uniformly in $k$ for some $C>0$. Since $\frac{\D}{\D x}f(x,k) > 0$ for $x < 1/k^2$, $f(x,k)$ has a maximum at $x^{\ast}(k)$. Thus $$ f(x,k) \leq f(x^{\ast}(k),k) \leq \e^{-C/2k}-1 = \Or(1/k)$$ for $x < \sqrt{1/k}$, and the claim is proved. \qedi \end{proof} Lemma \ref{power to exponential} immediately yields \begin{equation} \label{R1neu} \left(1+\frac{a}{k}\right)^{-k} = \e^{-a} \,\left(1+\Or\left({\textstyle \frac{1}{k}}\right)\right) \end{equation} uniformly on compact intervals of $a$ by taking $x = a/k$. We now turn to the proof of Theorem \ref{MainThm}, which we deduce from Theorems \ref{general diag} and \ref{main estimates}. As stated already in (\ref{ndef}), we will use \begin{equation}\label{ndef2} n_\eps = \frac{t_{\rm c}}{\eps} -1+ \sigma_\eps, \end{equation} where $\sigma_\eps \in [0,2[$ is such that $n_{\eps}$ is an even integer. The advantage of this convention about $\sigma_\eps$ is that now the off-diagonal components in (\ref{general offdiag eq}) are always given by $\eps^{n_\eps+1}x_{n_\eps+1}$ since $z_{n+1}=0$ for even $n$. Of course we could as well consider the asymptotic behavior of $\eps^{n+1}z_{n+1}$ for odd $n$ and one would expect to end up with the same result. However, it is obvious from (\ref{AA}) and (\ref{AB}) that $x_{n+1}$ is purely imaginary and $z_{n+1}$ is real at leading order. Thus the large $n$ asymptotics of the off-diagonal component of the effective Hamiltonian do depend on whether we consider even or odd superadiabatic bases. On the other hand, the asymptotics of the propagator must be independent of the exact choice of basis. We will discuss this point after giving the proof of Theorem~\ref{MainThm} based on the above convention. \begin{proof}[Proof of Theorem~\ref{MainThm}] We want to apply Theorem \ref{general diag} and thus have to check that $\xi, \eta$ and $\zeta$ defined in (\ref{xi})--(\ref{zeta}) together with their derivatives are $\Or(\eps \theta')$. From Proposition~\ref{z_n and recursion} together with Proposition~\ref{a_n limit} we infer that there exists $C>0$ such that $|x_k(t)| \leq C \theta'(t) (k-1)!/ t_{\rm c}^{k}$ for each $k$. The same is true for $y_n$ and $z_n$. Using the differential equations (\ref{E1a})--(\ref{E1c}), we find that there is $C' > 0$ with $|x_n'(t)| \leq C' \theta'(t) n!/ t_{\rm c}^{n+1}.$ This means that $$ |\xi'(t)| \leq \eps C' \theta'(t) \sum_{k=1}^n \eps^k t_{\rm c}^{-k-1} k! \,\eps^{k-1},$$ with similar expressions for the other quantities. Now taking $\eps = t_{\rm c}/(n_\eps - \sigma_\eps)$, we find \[ \sum_{k=0}^{n_\eps} t_{\rm c}^{-k} \eps^k (k+1)! = \sum_{k=0}^{n_\eps} \frac{(k+1)!}{(n_\eps-\sigma_\eps)^k} = \left( 1 + \frac{2}{n_\eps-\sigma_\eps} + \frac{3!}{(n_\eps-\sigma_\eps)^2} + \ldots \right). \] Each of the $n_\eps+1$ terms in the sum above is bounded by $\mathrm{const}/(n_\eps-\sigma_\eps)$ except the first which is $1$. This shows $$ |\xi'(t)| \leq \eps \theta' C' {\textstyle\left(1 + \frac{n_\eps}{n_\eps-\sigma_\eps}\right)}\,,$$ and Theorem \ref{general diag} gives (\ref{Hop}) with $c_\eps^{\op}(t) = \eps^{n_\eps+1}x_{n_\eps+1}(t)(1+\Or(\eps \theta'(t))$. Recall that $z_{n_\eps+1}(t) = 0$ due to our convention. It remains to determine the leading order asymptotics of $\eps^{n_\eps+1}x_{n_\eps+1}$. For convenience of the reader let us rewrite (\ref{AA}) as \begin{equation}\label{epsx} \eps^{n_\eps+1}x_{n_\eps+1}(t) = \I \frac{\eps^{n_\eps+1}n_\eps!}{t_{\rm c}^{n_\eps+1}} {\textstyle\left[ \frac{2\sin(\gamma\pi/2)}{\pi} \,\tRe \left(\hspace{-2pt}\left( 1 - \I \frac{t}{t_{\rm c}} \right)^{-(n_\eps+1)}\right) + \Or\left(R_{n_\eps+1}^\beta(t)\right)\hspace{-1pt} \right]}. \end{equation} \begin{lemma} \label{stirling} With (\ref{ndef2}), we have $$ \frac{\eps^{n_\eps+1}n_\eps!}{ t_{\rm c}^{n_\eps+1}}= \sqrt{\frac{2 \pi \eps}{t_{\rm c}}} \e^{-\frac{t_{\rm c}}{\eps}}(1 + \Or(\eps)).$$ \end{lemma} \begin{proof} Stirling's formula for $(n+1)!$ implies $$ n! = \frac{1}{n+1}\left( \frac{n+1}{\e} \right)^{n+1} \sqrt{n+1} \sqrt{2 \pi}\,\left(1+\Or\left({\textstyle \frac{1}{n+1}}\right)\right)\,.$$ Together with (\ref{R1neu}) this yields \begin{eqnarray*} \eps^{n_\eps+1} n_\eps! & = & t_{\rm c}^{n_\eps+1}\, \e^{-(n_\eps+1)} \left( 1 - \frac{\sigma_\eps}{n_\eps+1} \right)^{-(n_\eps+1)} \sqrt{\frac{2\pi}{n_\eps+1}}\,\left(1+\Or\left({\textstyle \frac{1}{n_\eps+1}}\right)\right) = \\ & = & t_{\rm c}^{n_\eps+1}\, \e^{-(n_\eps+1)}\,\e^{\sigma_\eps} \,\sqrt{\frac{2\pi}{n_\eps+1}}\, \left(1+\Or\left({\textstyle \frac{1}{n_\eps+1}}\right)\right) = \\ & = & t_{\rm c}^{n_\eps+1}\, \e^{-\frac{t_{\rm c}}{\eps}}\, \sqrt{\frac{2 \pi \eps}{t_{\rm c} + \eps\sigma_\eps }}\,(1 + \Or(\eps)). \end{eqnarray*} Finally, $$\sqrt{\frac{2 \pi \eps}{t_{\rm c} + \eps\sigma_\eps }} = \sqrt{\frac{2 \pi \eps}{t_{\rm c}}} \left(1+ \frac{\eps\sigma_\eps}{t_{\rm c}} \right)^{-1/2} = \sqrt{\frac{2 \pi \eps}{t_{\rm c}}}\,(1+\Or(\eps))\,.\qquad\qedi $$ \end{proof} Lemma~\ref{stirling} takes care of the first factor in (\ref{epsx}). Turning to the terms inside the square brackets in (\ref{epsx}), let us first note that for $|t| \geq t_c$, both terms are $$\Or(2^{-(n_\eps-1)/2}/(1+t^2) )= \Or(\exp(-t_{\rm c} \ln 2/(2 \eps))/(1+t^2))\,,$$ proving the theorem in this case. For $|t| < t_c$, we investigate the modulus and the phase separately. Let $0<\beta<1$. From Lemma \ref{power to exponential} it follows that \[ {\textstyle\left| 1 + \I \frac{t}{t_{\rm c}} \right|}^{n_{\eps}+1} = {\textstyle\left(1+ \frac{t^2}{t_{\mr c}^2}\right)}^{(t_{\mr c}/ \eps + \sigma_\eps)/2} ={\textstyle \left(1+ \frac{t^2}{t_{\mr c}^2}\right)}^{\sigma_\eps/2} \left( \e^{-\frac{t^2}{2 t_{\mr c} \eps}} + \Or \left( \eps \e^{-\frac{t^2}{4 t_{\mr c} \eps}} \right) \right). \] For $|t| \geq \eps^{\beta/2}$, $\exp(-t^2/(2 t_{\mr c} \eps)) = \Or(\eps \exp(-t^2/(4 t_{\mr c} \eps)))$. Thus neither the prefactor involving $\sigma_\eps$ above nor the phase play any role in this region. For $|t| < \eps^{\beta/2}$, $(1+ t^2/t_{\mr c}^2)^{\sigma_\eps/2} = 1 + \Or(\sigma_\eps \eps^\beta)$ and therefore $$ {\textstyle\left| 1 + \I \frac{t}{t_{\rm c}} \right|}^{n_{\eps}+1} = \e^{-\frac{t^2}{2 t_{\mr c} \eps}} + \Or \left( \eps^\beta \e^{-\frac{t^2}{4 t_{\mr c} \eps}} \right).$$ The same reasoning applies to $R_{n_\eps+1}^\beta$ and gives $$ R_{n_\eps+1}^\beta(t) \leq \eps^\beta \left( \e^{-\frac{t^2}{2 t_{\rm c} \eps}} + \Or\left(\eps^\beta \e^{-\frac{t^2}{4 t_{\rm c} \eps}} \right)\right). $$ Turning to the phase in the region $|t| < \eps^{\beta/2}$, we find \begin{eqnarray*} \e^{\I (n_\eps+1)\arctan (t/t_{\rm c})} &=& \exp \left( \I \left({\textstyle \frac{t_{\rm c}}{\eps}} + \sigma_\eps \right) \left( (t/t_{\rm c}) -{\textstyle \frac{1}{3}}(t/t_{\rm c})^3 + \Or((t/t_{\rm c})^5)\right) \right) = \\ & = & \exp \left( \I \left( {\textstyle \frac{t}{\eps}} - {\textstyle \frac{t^3}{3\eps t_{\rm c}^2} } + {\textstyle\frac{\sigma_\eps t}{t_{\rm c}}}\right)+ \Or(t_{\rm c} (t/t_{\rm c})^5/\eps) + \Or(\sigma_\eps (t/t_{\rm c})^3) \right)\\ & =& \exp \left( \I \left( {\textstyle \frac{t}{\eps}} - {\textstyle\frac{t^3}{3\eps t_{\rm c}^2}} + {\textstyle\frac{\sigma_\eps t}{t_{\rm c}}}\right)\right) \left(1 + \Or(\eps^{5\beta /2 - 1}) + \Or(\eps^{3\beta/2})\right). \end{eqnarray*} Now we just have to collect all the pieces and add the complex conjugate. \qedi \end{proof} Let us now see what of the above would have changed for $n_{\eps}$ odd. Then $x_{n_\eps+1} = 0$, and (\ref{AB}) together with Lemma \ref{power to exponential} and \ref{stirling} yields \begin{eqnarray} \label{zeta eps} c_\eps^{n_\eps}(t) &=& -\eps^{n_\eps+1}z_{n_\eps+1}(t)\, (1 + \Or(\eps\theta'))\\& =& 2\,\sqrt{\frac{2\eps}{\pi t_{\rm c}}}\,\sin\left(\frac{\pi \gamma}{2}\right)\, \e^{-\frac{t_{\rm c}}{\eps}}\,\e^{-\frac{t^2}{2\eps t_{\rm c}}} \,\sin\left(\frac{t}{\eps}-\frac{t^3}{3\eps t_{\rm c}^2} + \frac{\sigma_\eps t}{t_{\rm c}} \right)+ \Or\left(\phi^\alpha(\eps,t)\right)\,.\nonumber \end{eqnarray} At first, this looks like an important difference, since now the off-diagonal elements in the transformed Hamiltonian are purely real-valued in leading order, while in the other case they were purely imaginary. However, in the computation of the propagator, another factor of $\exp(\pm \I t/\eps)$ from the dynamical phase appears, cf.\ (\ref{oscint}). At leading order only the resonant term of the Hamiltonian survives, which is the same for odd and even $n_\eps$. \section{First order perturbation in the optimal superadiabatic basis} In this section we prove Corollary~\ref{PropCor}. Since we use standard first order perturbation theory, we stay sketchy in some parts. After splitting $H_\eps^{\op}(t)$, see (\ref{Hop}), as \[ H_\eps^{\op}(t) =\left( \begin{array}{cc}\frac{1}{2}&0\\0&-\frac{1}{2}\end{array}\right) + V_\eps(t) =: H_0 + V_\eps(t)\,, \] Dyson expansion in the interaction picture (cf.\ \cite{ReSi2}, Thm.\ X.69) yields \begin{eqnarray*} K_\eps^{\op}(t,s) &=& \e^{-\frac{\I t H_0}{\eps}} \left( {\rm id} -\frac{\I}{\eps}\int_s^t\e^{\frac{\I \tau H_0}{\eps}}\,V_\eps(\tau)\,\e^{-\frac{\I \tau H_0}{\eps}}\,\D \tau\right)\e^{\frac{\I s H_0}{\eps}}\\ &&+ \,\left( \begin{array}{cc} \Or(\eps^2) & \Or(\eps\e^{-\frac{t_{\rm c}}{\eps}}) \\\Or(\eps\e^{-\frac{t_{\rm c}}{\eps}})&\Or(\eps^2) \end{array}\right) \Delta(t,s)\,. \end{eqnarray*} Thus we only need to evaluate the integral \begin{eqnarray*}\nonumber -\frac{\I}{\eps}\int_s^t\e^{\frac{\I \tau H_0}{\eps}}\,V_\eps(\tau)\,\e^{-\frac{\I \tau H_0}{\eps}}\,\D \tau &=&-\frac{\I}{\eps}\int_s^t \left(\begin{array}{cc}0&\e^{\frac{\I\tau}{\eps}} c_\eps^{\op}(\tau)\\ \e^{-\frac{\I\tau}{\eps}} \overline{c}_\eps^{\op}(\tau) &0 \end{array}\right)\,\D \tau\\&&\, +\left(\begin{array}{cc}\Or(\eps)&0\\0&\Or(\eps)\end{array}\right)\Delta(t,s)\,. \end{eqnarray*} Inserting (\ref{Hod}) and using (\ref{Phialpha}) gives \begin{eqnarray}\nonumber\lefteqn{ -\frac{\I}{\eps}\int_s^t\e^{\frac{\I\tau}{\eps}} c_\eps^{\op}(\tau)\D\tau=}\\ \nonumber&=& \sqrt{\frac{2}{\eps\pi t_{\rm c}}}\,\sin\left(\frac{\pi \gamma}{2}\right)\, \e^{-\frac{t_{\rm c}}{\eps}} \int_s^t\e^{\frac{\I\tau}{\eps}} \,\e^{-\frac{\tau^2}{2\eps t_{\rm c}}} \,\left(\e^{-\frac{\I \tau}{\eps} +\frac{\I\tau^3}{3\eps t_{\rm c}^2} - \frac{\I \sigma \tau}{t_{\rm c}} }+ \e^{\frac{\I \tau}{\eps} -\frac{\I\tau^3}{3\eps t_{\rm c}^2} + \frac{\I \sigma \tau}{t_{\rm c}} } \right) \D\tau \\ & & +\, \Or\left(\eps^\alpha\e^{-\frac{t_{\rm c}}{\eps}}\Delta(t,s)\right) = (*)\,,\label{oscint} \end{eqnarray} for each $\alpha<1$. Now we replace the exponentials $\e^{\pm(\frac{\I\tau^3}{3\eps t_{\rm c}^2} - \frac{\I \sigma \tau}{t_{\rm c}} )}$ by $1\pm(\frac{\I\tau^3}{3\eps t_{\rm c}^2} - \frac{\I \sigma \tau}{t_{\rm c}} ) $. Using $|\e^{\I\varphi}-1-\I\varphi|\leq \varphi^2$, we conclude that the resulting error is bounded by a constant times \[ \eps^{-\frac{1}{2}} \e^{-\frac{t_{\rm c}}{\eps}} \int_{-\infty}^\infty \,\e^{-\frac{\tau^2}{2\eps t_{\rm c}}} \left(\frac{\tau^6}{\eps^2}+\frac{\tau^4}{\eps} +\tau^2 \right)\D\tau= \Or(\eps\e^{-\frac{t_{\rm c}}{\eps}} )\,. \] Hence we obtain \begin{eqnarray}\nonumber (*) &=& {\textstyle\sqrt{\frac{2}{\eps\pi t_{\rm c}}}\,\sin\left(\frac{\pi \gamma}{2}\right)\, \e^{-\frac{t_{\rm c}}{\eps}}}\hspace{-1pt} \int_s^t \hspace{-1pt}\e^{-\frac{\tau^2}{2\eps t_{\rm c}}} \,\left(1+{\textstyle\frac{\I\tau^3}{3\eps t_{\rm c}^2} - \frac{\I \sigma \tau}{t_{\rm c}}}+ \e^{\frac{2\I \tau}{\eps} } {\textstyle \left(1-\frac{\I\tau^3}{3\eps t_{\rm c}^2} + \frac{\I \sigma \tau}{t_{\rm c}}\right)}\right)\hspace{-1pt} \D\tau\\&& +\, \Or\left(\eps^\alpha\e^{-\frac{t_{\rm c}}{\eps}}\Delta(t,s)\right)\label{oscint2} \end{eqnarray} with $\alpha<1$, where the first summand in the integrand gives rise to the explicit term in (\ref{Koffdiag}). The remaining terms can be integrated explicitly as well, most conveniently using Maple or Mathematica. They are all of order $\Or(\sqrt{\eps}\e^{-\frac{t_{\rm c}}{\eps}}\Delta(t,s))$ uniformly in $t$ and $s$ resp.\ of order $\Or(\eps^\alpha\e^{-\frac{t_{\rm c}}{\eps}}\Delta(t,s))$ for $|t|$ and $|s|$ larger than $\eps^\beta$ for some $\beta<\frac{1}{2}$. To illustrate the reasoning note that \[ \int_s^t \e^{-\frac{\tau^2}{2\eps t_{\rm c}}}\tau\,\D\tau = \eps t_{\rm c}\left(\e^{-\frac{s^2}{2\eps t_{\rm c}}}-\e^{-\frac{t^2}{2\eps t_{\rm c}}}\right)\,. \] This is uniformly of order $\Or(\eps)$, but of order $\Or(\e^{-\eps^{2\beta-1}})$ for $|t|$ and $|s|$ larger than $\eps^\beta$. 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