The LATEX source file is followed by uuencoded postscript files of the four figures. BODY \documentstyle[12pt]{article} \if@twoside m \oddsidemargin 14truemm\evensidemargin 0mm \marginparwidth 85pt \else \oddsidemargin 7truemm\evensidemargin 7truemm \marginparwidth 68pt \fi \topmargin 5mm \headheight 0mm \headsep 0mm \textheight 9in \textwidth 165truemm \parindent=7mm \renewcommand{\baselinestretch}{1.} %\pagestyle{empty} \begin{document} \title{\bf A quantum pipette} \date{} \author{Pavel Exner} \maketitle \begin{center} Nuclear Physics Institute, Academy of Sciences \\25068 \v{R}e\v{z} near Prague \\ and Doppler Institute, Czech Technical University, \\ B\v rehov\'a 7, 11519 Prague, \\ Czech Republic \\ {\em exner@ujf.cas.cz} \end{center} \vspace{20mm} %DEFINITION OF THE MACROS USED \newcommand{\ie}{{\em i.e.}} \newcommand{\cf}{{\em cf. }} \newcommand{\R}{I\!\!R} \newcommand{\supp}{{\rm supp\,}} \newcommand{\Cf}{{\em Cf. }} \newcommand{\Ai}{{\rm Ai}} \newcommand{\Bi}{{\rm Bi}} \newcommand{\OO}{{\cal O}} \newcommand{\QED}{\mbox{\rule[-1.5pt]{6pt}{10pt}}} %END OF THE DEFINITION \begin{quotation} \noindent Curved quantum waveguides are known to bind particles. We show that a number of charged fermions in such a trap can be tuned by an external electrostatic field; if the latter is slowly increased, the bent duct can serve as a single particle ejector. \end{quotation} \section{Introduction} Though numerous quantum phenomena can be explained by usual semiclassical concepts, a complete theory --- even in the case of the nonrelativistic quantum mechanics --- is certainly deeper. A recent illustration was provided by properties of particles within bent tubes or other infinitely extended regions with Dirichlet boundaries (\ie, hard walls). It was demonstrated that such systems exhibit isolated energy eigenvalues \cite{ES,GJ,SRW} despite the absence of closed trajectories (apart of the obvious zero--measure set) in their classical counterparts. These bound states and related resonance effects in scattering \cite{DES} have attracted a considerable interest --- a list of references can be found in the review paper \cite{DE}. It is motivated not only by the mentioned theoretical reason, but rather by the fact that curved tubes (and more complicated regions constructed of them) can model some real physical systems. The most prominent among them are {\em quantum wires,} \ie, tiny strips of a very pure semiconductor material. Due to the purity and crystalic structure, an electron within the conductivity band can be regarded as a free particle of a certain effective mass. To replace the band by a halfline and to neglect the effective--mass dependence on the electron momentum is certainly a crude approximation; nevertheless, it is good enough to reproduce some properties of real quantum wires. There are other motivations to study Schr\"odinger equation in hard--wall tubes. A very recent one comes from the proposal to use {\em hollow optical fibres} as waveguides for transport of atoms or ions \cite{SMZ}; in view of the achievable widths of such ducts, quantum effects must again be taken into account. Despite numerous investigations of quantum waveguides during last few years, many questions remain to be answered. This concerns, in particular, effects of external fields. Most attention has been paid to magnetic fields, either perpendicular to the waveguide plane (\cf \cite{E1,VOK} for further references) or threaded through the tube \cite{DJ}, or to the quantum Hall effect, while the influence of an electric field alone remained mostly untreated. One of our aims is to draw attention to the fact that Stark effect in non--straight tubes has a rich structure coming from combination of the curvature--induced attractive interaction and the electrostatic potential which is nonlinear along the tube even if the field is homogeneous. Instead of discussing it generally, however, we shall concentrate here on discussion of an interesting particular case. \section{Description of the model} We consider a particle whose motion is confined to a curved planar strip $\,\Omega\,$ of a constant width $\,d\,$ as sketched on Figure 1. Though we have in mind the systems mentioned in the introduction, in general we shall suppose only that the particle is a fermion of a nonzero charge $\,q\,$. We also assume that it is under influence of a homogeneous electric field of an intensity $\,E\;$; we denote $\,F:=qE\,$. Without loss of generality we shall suppose in the following that $\,F\ge 0\,$. The operator to study is the corresponding Hamiltonian, \begin{equation} \label{Hamiltonian} H_{\Omega}(F)\,:=\,-\Delta_D^{\Omega}+Fy\,, \end{equation} where $\,-\Delta_D^{\Omega}\,$ is the Dirichlet Laplacian on $\,L^2(\Omega)\,$ defined conventionally as in \cite[Sec.XIII.15]{RS}; for the sake of simplicity we put $\,\hbar=2m=1\,$. If the electric field is absent, $\,F=0\,$, the essential spectrum of $\,H_{\Omega}(F)\,$ starts at $\,\lambda_1:=(\pi/d)^2\,$ which is the lowest eigenvalue of the transverse Dirichlet problem. In addition, it has at least one eigenvalue below $\,\lambda_1\,$ whenever $\,\Omega\,$ is non--straight; a more precise formulation will be given below. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % %\caption{The model: a curved strip in an electric field} % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% What happens with these eigenvalues when the field is switched on depends, of course, substantially on the shape of $\,\Omega\,$. We restrict our attention here to the case when $\,\Omega\,$ is curved within a bounded region only, and outside it is perpendicular to the field direction. Moreover, we shall assume that the ``tilt'' of $\,\Omega\,$ is one--sided, so there are no field-induced bound states in the corresponding classical system. We have to put these assumptions on more mathematical terms. Following the standard procedure \cite{DE,ES} we choose the ``lower'' boundary of $\,\Omega\,$ as a reference curve; we call it $\,\Gamma\,$. We also introduce the usual curvilinear coordinates: $\,s\,$ is the arc length of $\,\Gamma\,$ and $\,u\,$ is the distance from $\,\Gamma\,$ (for points ``above'' the curve; it runs through the interval $\,[0,d]\,$). The Cartesian coordinates of the strip points are then given by \begin{equation} \label{parametrization} x\,=\,\xi(s)-u\dot\eta(s)\,, \quad y\,=\,\eta(s)+u\dot\xi(s)\,, \end{equation} where the functions $\,\xi,\eta\,$ satisfy the normalization condition $\,\dot\xi(s)^2\!+\dot\eta(s)^2=1\,$. One can use them to define the signed curvature of the reference curve, \begin{equation} \label{curvature} \gamma(s)\,:=\, \dot\eta(s)\ddot\xi(s)-\dot\xi(s)\ddot\eta(s)\,. \end{equation} The latter determines in turn the curve $\,\Gamma\,$ uniquely up to Euclidean transformations of the plane: we have \begin{equation} \label{inverse transformation} \xi(s)\,=\, \int_0^s \cos\beta(s_1)\,ds_1\,, \quad \eta(s)\,=\, \int_0^s \sin\beta(s_1)\,ds_1\,, \end{equation} where $\,\beta(s_2,s_1):= -\int_{s_1}^{s_2} \gamma(s)\,ds\,$ is the bending angle of $\,\Gamma\,$ between the points $\,s_1\,$ and $\,s_2\;$ (in distinction to \cite{E1,ES} we choose it to be anticlockwise positive), and $\,\beta(s):=\beta(s,0)\;$; the non--uniqueness has been removed by choosing the reference frame in such a way that $\,\xi(0)= \eta(0)= \dot\eta(0)=0\,$ and $\,\dot\xi(0)=1\,$. We adopt several general regularity assumptions, namely \begin{description} \item{\em (r1)} $\;\gamma\in L^1_{loc}(\R)\,$, \item{\em (r2)} $\;a\Vert\gamma\Vert_{\infty}<1\,$, \item{\em (r3)} $\;\Omega\,$ is not self--intersecting, \item{\em (r4)} $\;\gamma\,$ is piecewise $\,C^2\,$ with $\,\dot\gamma,\, \ddot\gamma\,$ bounded. \end{description} Among them, {\em (r2)} is needed to ensure that the other boundary of the strip is also smooth, while {\em (r4)} represents a strengthening of {\em (r1)}. Under {\em (r1)}--{\em (r3)}, one can rewrite $\,H_{\Omega}(0)\,$ as the Laplace--Beltrami operator on $\,L^2(\R\times[0,d],\, g^{1/2} ds\,du\,)\,$, where $\, g^{1/2}(s,u):= 1+u\gamma(s)\;$ \cite{DE,ES}; the potential part of (\ref{Hamiltonian}) can be easily expressed in terms of the curvilinear coordinates by means of (\ref{parametrization}) and (\ref{inverse transformation}). If the assumption {\em (r4)} is also valid, one can remove the Jacobian, \ie, to use the unitary operator $\,U:\:L^2(\Omega)\to L^2(\R\times[0,d])\,$ defined by \begin{equation} \label{straightening} (U\psi)(s,u)\,:=\, (1+u\gamma(s))^{1/2}\psi(x,y)\,. \end{equation} The operator resulting from this transformation, which by abuse of notation we will also denote as $\,H_{\Omega}(F)\,$, has the following form: \begin{equation} \label{Hamiltonian2} H_{\Omega}(F)\,=\,-\partial_s\, (1+u\gamma(s))^{-2}\, \partial_s\,-\, \partial_u^2 \,+\,V_F(s,u)\,, \end{equation} where \begin{eqnarray} \label{effective potential} V_F(s,u) &\!=\!& V_0(s,u)\,+\, F\int_0^s \sin\beta(s_1)\,ds_1\,+\, Fu\,\cos\beta(s) \nonumber \\ \\ V_0(s,u) &\!=\!& -\,\frac{\gamma(s)^2}{4(1+u\gamma(s))^2}\,+\, \frac{u\ddot\gamma(s)}{2(1+u\gamma(s))^3}\,-\, \frac{5}{4}\,\frac{u^2\dot\gamma(s)^2}{(1+u\gamma(s))^4}\,. \nonumber \end{eqnarray} After this preliminary, we can formulate the special assumptions of our model which we have sketched above: \begin{description} \item{\em (s1)} $\;\gamma\neq 0\,$ with $\,\supp\gamma\in [0,s_0]\,$ for some $\,s_0>0\,$, \item{\em (s2)} $\;\int_0^{s_0} \gamma(s)\,ds=0\,$, \item{\em (s3)} $\;\beta(s)\in [0,\pi]\,$ for $\,s\in[0,s_0]\,$. \end{description} \section{Existence of bound states} Let $\,N(F):= N(H_{\Omega}(F))\,$ be the number of bound states of $\,H_{\Omega}(F)\,$, \ie, the number of its isolated eigenvalues counting their multiplicity. Since $\,H_{\Omega}(F)\ge H_{\Omega}(F')\,$ holds obviously for $\,F\ge F'\,$, all eigenvalues are by the minimax principle nondecreasing functions of $\,F\,$. This does not mean automatically, however, that $\,N(\cdot)\,$ is monotonous, because we count the eigenvalues below $\,\inf\sigma_{ess}(H_{\Omega}(F))= \inf\sigma(h_u(F))\,$, where $\,h_u(F):= -\partial_u^2 +Fu\,$ with the Dirichlet condition at $\,u=0,d\,$, and the latter is also increasing. On the other hand, a strong enough field destroys all bound states. \vspace{3mm} \noindent {\bf Theorem:} \quad Assume {\em (r1)}--{\em (r3)} and {\em (s1)}--{\em (s3)}. Then \begin{description} \item{\em (a)} $\,N(0)\ge 1\,$. If, in addition {\em (r4)} is valid and $\;\int_0^{s_0} |\gamma(s)|\,ds\,$ is small enough, $\,H_{\Omega}(0)\,$ has just one bound state, $\,N(0)=1\,$. \item{\em (b)} There is a positive $\,F_0\,$ such that $\,N(F)=0\,$ for all $\,F\ge F_0\,$. \end{description} \vspace{3mm} \noindent {\em Proof:} \quad (a) \Cf \cite{GJ} and \cite[Secs.2, 4]{DE}. \\ (b) The idea is to estimate $\,H_{\Omega}(F)\,$ from below by an operator $\,\tilde H(F)\,$ in such a way that the threshold of the essential spectrum is preserved, $\,\inf\sigma_{ess}(H_{\Omega}(F)) =\inf\sigma_{ess}(\tilde H(F))\,$. In that case $\,N(F)\le \tilde N(F) := N(\tilde H(F))\,$ so it is sufficient to choose $\,\tilde H(F)\,$ which would have $\,\tilde N(F)=0\,$ for $\,F\,$ large enough. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % %\caption{The definition of $\,\tilde H(F)\,$. The full and dotted lines %represent Dirichlet and Neumann boundaries, respectively.} % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% The estimating operator is constructed in the way sketched on Figure 2. We cut the straight tube in left halfplane by an additional Neumann boundary. Furthermore, we deform $\,\Gamma\,$ to the right of the origin and close the obtained region by another Neumann boundary to a rectangle of sides $\,a,\,b\;$; this is always possible under the assumptions {\em (s1)}--{\em (s3)}. Finally, the upper boundary in the right halfplane is not important; we may remove it at all. We denote the operators $\,-\Delta+Fy\,$ in these regions with the appropriate boundary conditions as $\,H_j(F),\: j=1,2,3\,$, and define $\,\tilde H(F):= H_1(F)\oplus H_2(F)\oplus H_2(F)\,$. By the bracketing principle \cite[Sec.XIII.15]{RS}, $\;H_{\Omega}(F) \ge \tilde H(F)\,$, so it remains to check that $\,\tilde H(F)\,$ has the other required properties. Obviously, $\,\inf\sigma(H_3(F)) \ge Fb\,$. To estimate the bottom of the spectra of $\,H_1(F),\, H_2(F)\,$, we need to solve the transverse ($\,y$--direction) problem. Its eigenfunctions are linear combinations of the fundamental solutions \begin{equation} \label{fundamental solutions} u_{\lambda}(y)\,:=\, \Ai\left(F^{1/3}\left(y-{\lambda\over F} \right)\right)\,, \quad v_{\lambda}(y)\,:=\, \Bi\left(F^{1/3}\left(y-{\lambda\over F} \right)\right)\,, \end{equation} and the spectral conditions read \begin{equation} \label{spectral conditions} u_{\lambda}(0)v_{\lambda}(d)- u_{\lambda}(d)v_{\lambda}(0)=0\,, \quad u_{\lambda}(0)v'_{\lambda}(b)- u'_{\lambda}(b)v_{\lambda}(0)=0 \end{equation} in the first and the second region, respectively. Introducing the parameters $\,\eta:= F^{-1/3}\,$ and $\,\delta:=a_1+\lambda F^{-2/3}\,$, where $\,a_1\approx -2.33\,$ is the first zero of $\,\Ai\,$ and using asymptotic properties of the Airy functions \cite{AS}, we find from (\ref{fundamental solutions}) and (\ref{spectral conditions}) that the lowest eigenvalue is in the first case given by \begin{equation} \label{lambda} \lambda_1(F)\,=\, F^{2/3}\, \left\lbrack -\,a_1\,+\, c_1\, e^{-(4/3)d^{3/2}\sqrt F} (1+\OO(F^{-1/2})) \right\rbrack\,, \end{equation} where $\,c_1:=\,-\,{\Bi(a_1)\over 2\Ai'(a_1)}\,\approx 0.325\,$. The spectrum of $\,H_1(F)\,$ is then purely continuous and starts from $\,\lambda_1(F)\,$. On the other hand, the spectrum of $\,H_2(F)\,$ is purely discrete, the lowest eigenvalue being obtained in a similar way as \begin{equation} \label{mu} \mu_1(F)\,=\,\left(\pi\over 2a\right)^2\,+\, F^{2/3}\, \left\lbrack -\,a_1\,-\, c_1\, e^{-(4/3)b^{3/2}\sqrt F} (1+\OO(F^{-1/2})) \right\rbrack\,. \end{equation} Hence for all $\,F\,$ large enough, $\,\lambda_1(F)\le \min\{\mu_1(F), Fb\}\,$, so the bottom of the spectrum of $\,\tilde H(F)\,$ is determined by that of $\,H_1(F)\,$. Since the latter has no eigenvalues, we arrive at the sought conclusion. \quad \QED \section{Thin strips: a semiclassical estimate} The above general result yields only a very rough estimate of the critical field strength $\,F_0\,$ and it says nothing about the behaviour of the function $\,N(\cdot)\,$ in the interval $\,[0,F_0]\,$. To get a better idea, we shall discuss in this section on a heuristic level the case of a thin strip, $\,d\|\gamma\|_{\infty}\ll 1\,$. We shall suppose that the curvature $\,\gamma\,$ is smooth enough, so that we can replace $\,H_{\Omega}(F)\,$ by the unitarily equivalent operator (\ref{Hamiltonian2}). If the strip is thin, the problem can be then reduced to discussion of the one--dimensional Schr\"odinger operator \begin{equation} \label{thin Hamiltonian} H(F)\,:=\,-\partial^2_s\,+\,V_F(s)\,:=\, -\partial^2_s\,-\,{1\over4}\,\gamma(s)^2\,+\, F\int_0^s \sin\beta(s_1)\,ds_1\, \end{equation} on $\,L^2(\R)\,$, which is the leading term in the projection of the operator $\,H_{\Omega}(F)\,$ onto the lowest transverse mode with the corresponding contribution $\,\lambda_1:=(\pi/d)^2\,$ to the energy subtracted --- for details see \cite[Sec.5]{DE}. The number of bound states in the curved strip can be estimated semiclassically: denoting $\,\tilde V_F(s):= \min\{0,V_F(s)\}\,$ and $\,\tilde p_F(s):= \sqrt{-\tilde V_F(s)}\,$, we have \begin{equation} \label{semiclassical N} N(F)\,\approx\, {1\over\pi}\, \int_0^{s_0} \tilde p(s)\,ds\,. \end{equation} The estimate makes certainly sense as long as the phase space allowed for the classical motion is large enough. We know that it fails if the field is absent and the strip is only slightly curved, because then $\,N(0)=1\,$ for an arbitrarily small non--zero $\,\gamma\,$. On the other hand, if $\,N(0)\gg 1\,$ we can still use it to assess the critical field value $\,F_0\,$, because the property of one--dimensional Schr\"odinger operators on which the above claim is based requires the potential to decay fast enough at infinity \cite{BGS} which is certainly not true for $\,F\neq 0\,$ when the asymptotics on the two sides are different. The right side of the relation (\ref{semiclassical N}) is a decreasing function of $\,F\,$ and $\,\lim_{F\to\infty} N(F)=0\,$ in accordance with the general result discussed above. As an illustration, consider the following particular case: %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % %\caption{A strip with alternative bends for $\,\beta=\pi\,$ and %$\,N=2\,$.} % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \noindent {\bf Example:} \quad Suppose that the waveguide has $\,2N\,$ bends of alternating orientation, each of them of radius $\,R\,$ and angle $\,\beta\;$ (\cf Figure 3). Then $\,s_0=2N\beta R\,$ and $\,V_0(s)= -(2R)^{-2}$, so $\,N(0)\approx {N\beta\over\pi}\,$ and the estimate can be used as long as the number of bends $\,N\gg {\pi\over\beta}\,$. Using the parametrization (\ref{parametrization}), we find easily $$ y(s)= \left\lbrace \begin{array}{lll} 2nR(1\!-\!\cos\beta)+R\left(1\!-\!\cos\left(s-2n\beta R\over R \right)\right) & \dots & s\in(2n\beta R,(2n\!+\!1)\beta R) \\ \\ 2(n\!+\!1)R(1\!-\!\cos\beta)-R\left(1\!-\!\cos\left(2(n+1)\beta R-s\over R \right)\right) & \dots & s\in((2n\!+\!1)\beta R,(2n\!+\!2)\beta R) \\ \end{array}\!\!\! \right. $$ Hence $\,y(\cdot)\,$ is a sum of two functions, a linear and a periodic one. For the purpose of an estimate we neglect the oscillating part; this yields $$ V_F(s)\,\approx\, {2Fs\over\beta}\, \sin^2{\beta\over 2}\,-\, {1\over 4R^2} $$ so performing the integration in (\ref{semiclassical N}), we get \begin{equation} \label{N example} N(F)\,\approx\, \left\lbrace \begin{array}{lll} {\beta\over 24\pi FR^3 \sin^2{\beta\over 2}} \left\lbrack 1-\left(1\!-\!16NFR^3 \sin^2{\beta\over 2}\right)^{3/2} \right\rbrack & \dots & 0? @ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_ begin 644 figpip M)24E)24E)24E)24E)24E)24E)24E)24E)24E)24E)24E)24E)24E)24E)24Ez M)24E)24E)24E)24E)24E)24E)24E)24E#0HE)24E)24E)24E)24E)24E)24Ey M)24E)24E)24E)24E)24E)24E)24E)24E)24E)24E)24E)24E)24E)24E)24Ex M)24E)24-"E!3($9)3$4@3T8@1DE'55)%(#$-"B4E)24E)24E)24E)24E)24Ew M)24E)24E)24E)24E)24E)24E)24E)24E)24E)24E)24E)24E)24E)24E)24Ev M)24E)24E)0T*)24E)24E)24E)24E)24E)24E)24E)24E)24E)24E)24E)24Eu M)24E)24E)24E)24E)24E)24E)24E)24E)24E)24E)24E#0H-"B4A4%,M061Ot M8F4M,BXP( T*)25#6%N($UA9V5N=&$@665Lp M;&]W($)L86-K#0HE)5!A9V5S.B H871E;F0I#0HE)4)O=6YD:6YG0F]X.B Ho M871E;F0I#0HE)45N9$-O;6UE;G1S#0H-"B4E0F5G:6Y00T*+TUA=&A7;W)K&1E9B![k M97AC:"!D969](&)D968-"B]XV5X8V@@W-E=&-M>6MC;VQOF4@>'-T;W)E"0DEW!O< T*+U=I;F1O=W-,871I;C%%;F-O9&EN9R R-38@87)Rp M87D@8F1E9@T*25-/3&%T:6XQ16YC;V1I;F<@5VEN9&]WPT*97AC:"!D=7 @=VAEe MW!O<"!3=&%N9&%R9$5N8V]D:6YG?2!I9F5L" O1DE$(&YE('MD969]>W!O<"!P;W!](&EFb M96QS92!](&9OPT*,R Q(')O;&P@,2!I;F1E> T*9'5P(&ESx MW!O<"!P;W!](&EF96QS90T*97AC:"!&35,-w M"GT@8F1E9@T*#0HO8W-M('L-"@DQ(&1P:3)P;VEN="!D:78@+3$@9'!I,G!Ov M:6YT(&1I=B!S8V%L90T*"6YE9R!T7!ER!;+C4@9'!I,G!O:6YT(&UU;" T(&1P:3)Pr M;VEN="!M=6Q=(# @PT*"6QI;F5T;PT*"7-TPT*"7MR;&EN971O?2!R97!E870-"@E]k M(&)D968-"B]04"![#0H)8VQO'-T"!Md M871R:7@@9&5F#0HO36%K94]V86P@>PT*"6YE=W!A=&@-"@ET36%T"!S971M871R:7@-"@E](&)D968-"B]&3R![#0H)a M36%K94]V86P-"@ES=')O:V4-"@E](&)D968-"B]03R![#0H)36%K94]V86P-z M"@EF:6QL#0H)?2!B9&5F#0H-"B]01"![#0H),B!C;W!Y(&UO=F5T;R!L:6YEy M=&\@#H@(" Q-C,@(" @u M-S$@(" T.#$@(" V.#D-"DUA=&A7;W)K0T*+V,P('L@," P(# @p MR Q(# @,"!So MR Q(#$@,"!S"D@#H@(" Q-C,@(" @-S$@(" T.#$@(" V.#D-y M"B4E4&%G97,Z(# P,0T*)25%3T8-"@T*)24E)24E)24E)24E)24E)24E)24Ex M)24E)24E)24E)24E)24E)24E)24E)24E)24E)24E)24E)24E)24E)24E)24Ew M)24E#0HE)24E)24E)24E)24E)24E)24E)24E)24E)24E)24E)24E)24E)24Ev M)24E)24E)24E)24E)24E)24E)24E)24E)24E)24-"E!3($9)3$4@3T8@1DE'u M55)%(#(-"B4E)24E)24E)24E)24E)24E)24E)24E)24E)24E)24E)24E)24Et M)24E)24E)24E)24E)24E)24E)24E)24E)24E)24E)0T*)24E)24E)24E)24Es M)24E)24E)24E)24E)24E)24E)24E)24E)24E)24E)24E)24E)24E)24E)24Er M)24E)24E)24E)24E#0H-"B4A4%,M061O8F4M,BXP( T*)25#6%N($UA9V5N=&$@665L;&]W($)L86-K#0HE)5!A9V5S.B Hm M871E;F0I#0HE)4)O=6YD:6YG0F]X.B H871E;F0I#0HE)45N9$-O;6UE;G1Sl M#0H-"B4E0F5G:6Y00T*k M+TUA=&A7;W)K&1E9B![97AC:"!D969](&)D968-"B]XV5X8V@@W-E=&-M>6MC;VQOF4@>'-T;W)E"0DEq MW!O< T*+U=I;F1On M=W-,871I;C%%;F-O9&EN9R R-38@87)R87D@8F1E9@T*25-/3&%T:6XQ16YCm M;V1I;F<@5VEN9&]WPT*97AC:"!D=7 @=VAEW!O<"!3b M=&%N9&%R9$5N8V]D:6YG?2!I9F5L" Oz M1DE$(&YE('MD969]>W!O<"!P;W!](&EF96QS92!](&9OPT*v M,R Q(')O;&P@,2!I;F1E> T*9'5P(&ESW!Ou M<"!P;W!](&EF96QS90T*97AC:"!&35,-"GT@8F1E9@T*#0HO8W-M('L-"@DQt M(&1P:3)P;VEN="!D:78@+3$@9'!I,G!O:6YT(&1I=B!S8V%L90T*"6YE9R!Ts M7!ER!;72 P('-E=&1AR!;p M+C4@9'!I,G!O:6YT(&UU;" T(&1P:3)P;VEN="!M=6Q=(# @PT*k M"6QI;F5T;PT*"7-TPT*"7MR;&EN971O?2!R97!E870-"@E](&)D968-"B]04"![#0H)8VQO'-T"!M871R:7@@9&5F#0HO36%K94]V86P@a M>PT*"6YE=W!A=&@-"@ET36%T"!S971My M871R:7@-"@E](&)D968-"B]&3R![#0H)36%K94]V86P-"@ES=')O:V4-"@E]x M(&)D968-"B]03R![#0H)36%K94]V86P-"@EF:6QL#0H)?2!B9&5F#0H-"B]0w M1"![#0H),B!C;W!Y(&UO=F5T;R!L:6YE=&\@0T*+V,P('L@," P(# @R Q(# @,"!SR Q(#$@,"!Sk M"D@C,N<',-"B4E0W)E871Ip M;VY$871E.B P,R\S,2\Y-2 @,30Z,C4Z-#4-"B4E1&]C=6UE;G1.965D961&o M;VYT#H@*&%T96YD*0T*)25%;F1#;VUM96YTV)I;F0@9&5F?2!B:6YD(&1E9@T*+VQD968@>VQO860@9&5F?2!B:6YDi M(&1E9@T*+WAD968@>V5X8V@@9&5F?2!B9&5F#0HO>'-T;W)E('ME>&-H('-Th M;W)E?2!B9&5F#0H-"B4@;W!E7-T96T@;6%P<&EN9W,-"B]Ds M<&DR<&]I;G0@,"!D968-"@T*)2!F;VYT(&-O;G1R;VP-"B]&;VYT4VEZ92 Pr M(&1E9@T*+T9-4R![#0H)+T9O;G13:7IE('AS=&]R90D))7-A=F4@F4@p M;F5G(# @,%T-"@EM86ME9F]N= T*"7-E=&9O;G0-"@E]8F1E9@T*#0HO25-/o M3&%T:6XQ16YC;V1I;F<@=VAER]7:6YD;W=S3&%T:6XQ16YC;V1I;F<@d M4W1A;F1A&-H#0ID=7 @,R Q(')O;&P-"F9I;F1F;VYT(&1Ua M<"!L96YG=&@@9&EC="!B96=I;@T*("![(#$@:6YD97@@+T9)1"!N92![9&5Fz M?7MP;W @<&]P?2!I9F5LPT*9FEN9&9O;G0@+T-H87)3=')I;F=S(&=E= T*w M+T%GR!;-B!D<&DR<&]I;G0@;75L72 P('-E=&1AR!;n M+C4@9'!I,G!O:6YT(&UU;" T(&1P:3)P;VEN="!M=6P@-B!D<&DR<&]I;G0@m M;75L(#0@9'!I,G!O:6YT(&UU;%T@,"!S971D87-H('T@8F1E9@T*#0HE(&UAl M8W)OW)L:6YE=&]](')E<&5A= T*"7T@8F1E9@T*+T%0('L-"@E[PT*"6-L;W-E<&%T:"!F:6QL#0H)h M?2!B9&5F#0HO1% @>PT*"6-L;W-E<&%T:"!S=')O:V4-"@E](&)D968-"B]-g M4B![#0H)-" M,B!R;VQL(&UO=F5T;PT*"61U<" @,"!E>&-H(')L:6YE=&\-f M"@EE>&-H(# @&-H(')L:6YE=&\-"@EC;&]Se M97!A=&@-"@E](&)D968-"B]&4B![#0H)35(@PT*"4U2(&9I;&P-"@E](&)D968-"B],,6D@>PT*"7L@8W5R"!C=7)R96YT;6%TPT*"4UA:V5/=F%L#0H)PT*"4UA:V5/=F%L#0H)9FEL; 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