2$ be the smallest integer for which $\gamma_m\neq \epsilon_m^{(2s)}$. As $\gamma_n \leq \epsilon_n^{(2s)}$, for all $n$, this means $\gamma_m < \epsilon_m^{(2s)}$. Therefore, i) from above cannot hold with $n+1=m$, and hence ii) must hold, i.e., $\gamma_m\geq\gamma_{m-1}$. We assumed that $m$ was the {\it smallest\/} integer such that $\gamma_m \neq \epsilon_m^{(2s)}$, hence $\gamma_{m-1} =\epsilon_{m-1}^{(2s)}$. We conclude that $\epsilon_m^{(2s)} > \gamma_m\geq \gamma_{m-1} =\epsilon_{m-1}^{(2s)}$, which is in contradiction with the assumption \eq{strict}. Therefore, no $m \geq 2$ such that $\gamma_m\neq \epsilon_m^{(2s)}$ exists and the lemma is proved. \end{proof} We now return to the spin-1/2 XXZ Heisenberg chain for the computation of $\epsilon_n^{(1)}$. \begin{Lem}\label{lem:epsilonn} For the $SU_q(2)$ invariant spin-1/2 ferromagnetic XXZ chain with $\Delta \geq 1$, $\epsilon_n^{(1)}$ defined in \eq{epsilonJ} is given by \be \epsilon_n^{(1)}=1-\Delta^{-1}\cos (\pi/n) \ee In particular one has $\epsilon_{n+1}^{(1)} < \epsilon_n^{(1)}$. \end{Lem} \begin{proof} %In order to show the statements of Lemma \ref{lem:epsilonn}, We calculate the eigenvalues ${\cal E}$ of the Hamiltonian $H^{(q)}_L$ of (\ref{hammu}) in the one down spin sector by using a transfer matrix method. An arbitrary vector $\psi$ in that subspace can be written as \begin{equation} \psi=\sum_{x=1}^L a_x D_x , \end{equation} where $D_x$ denotes the basis vector with all spins up except at the site $x$ where the spin is down. %, and $a_x$ are %coefficients to be determined. For $\psi$ to be an eigenvector the coefficients $a_x$ must satisfy $\langle D_y|H^q_L\psi\rangle={\cal E} a_y$ where ${\cal E}$ is the eigenvalue, which amounts to \begin{equation} a_{y+1}=2\Delta(1 -{\cal E})a_y-a_{y-1} \quad \mbox{for \quad $2\le y \le L-1$}, \label{naka} \end{equation} \begin{equation} a_2=2\Delta[1/2+A(\Delta)-{\cal E}]a_1,\quad a_{L-1}=2\Delta[1/2-A(\Delta)-{\cal E}]a_L. \label{edge1} \end{equation} The equations (\ref{naka}) can be rewritten as \begin{equation} \left( \begin{array}{c} a_{y+1}\\ a_y \end{array} \right)=T \left( \begin{array}{c} a_y \\ a_{y-1} \end{array} \right)\quad,\quad \mbox{with} \quad T= \left( \begin{array}{cc} 2\Delta(1-{\cal E})& -1\\ 1&0 \end{array} \right). \label{transfer} \end{equation} %in terms of the transfer matrix %\begin{equation} %T= %\left( %\begin{array}{cc} %2\Delta(1-{\cal E})& -1\\ %1&0 %\end{array} %\right). %\label{T} %\end{equation} By using (\ref{transfer}) repeatedly, we get \begin{equation} \left( \begin{array}{c} a_L\\ a_{L-1} \end{array} \right)=T^{L-2} \left( \begin{array}{c} a_2\\ a_1 \end{array} \right). \end{equation} Combining this with (\ref{edge1}), we have \begin{equation} a_L \left( \begin{array}{c} 1\\ 2\Delta{[1/2-A(\Delta)-{\cal E}]} \end{array} \right)=a_1T^{L-2} \left( \begin{array}{c} 2\Delta{[1/2+A(\Delta)-{\cal E}]}\\ 1 \end{array} \right). \label{Teq} \end{equation} This equation can be solved in terms of the eigenvalues and -vectors of the transfer matrix $T$. % (\ref{transfer}). The eigenvalues $\lambda$ of $T$ are given by the roots of the equation \begin{equation} \lambda^2-2\Delta(1-{\cal E})\lambda+1=0, \label{chara} \end{equation} given by \begin{equation} \lambda_{\pm}=\Delta(1-{\cal E})\pm \sqrt{\Delta^2(1-{\cal E})^2-1}. \label{lam} \end{equation} Consider first the case $\Delta(1-{\cal E})\ne \pm1$. Then the eigenvectors are determined by \begin{equation} u_\pm= \left( \begin{array}{c} \lambda_\pm\\ 1 \end{array} \right). \end{equation} %for the eigenvalue $\lambda_\pm$. In terms of %these vectors $u_\pm$, the vector with $A(\Delta)=\sqrt{1-\Delta^{-2}}/2$ in (\ref{Teq}) can be rewritten as \begin{equation} \left( \begin{array}{c} 2\Delta{[1/2+A(\Delta)-{\cal E}]}\\ 1 \end{array} \right) = \left( \begin{array}{c} \Delta+\sqrt{\Delta^2-1}-2\Delta{\cal E} \\ 1 \end{array} \right)=\alpha_+ u_+ +\alpha_- u_- \label{vec1} \end{equation} with \begin{equation} \alpha_\pm = {1 \over 2}\left[1\pm {1 \over \sqrt{\Delta^2(1-{\cal E})^2-1}} (\sqrt{\Delta^2-1}-\Delta{\cal E})\right] . \label{alpha} \end{equation} Similarly the vector in the left-hand side of (\ref{Teq}) can be rewritten as \begin{equation} \left( \begin{array}{c} 1\\ \Delta-\sqrt{\Delta^2-1}-2\Delta{\cal E} \end{array} \right) =\beta_+ \left( \begin{array}{c} 1\\ \lambda_- \end{array} \right) +\beta_- \left( \begin{array}{c} 1\\ \lambda_+ \end{array} \right) =\beta_+\lambda_- u_+ +\beta_-\lambda_+ u_- \label{vec2} \end{equation} with \begin{equation} \beta_\pm = {1 \over 2}\left[1\pm {1 \over \sqrt{\Delta^2(1-{\cal E})^2-1}} (\sqrt{\Delta^2-1}+\Delta{\cal E})\right] , \label{beta} \end{equation} where we have used $\lambda_+\lambda_-=1$. %which follows from (\ref{chara}). Substituting (\ref{vec1}) and (\ref{vec2}) into (\ref{Teq}), we have \begin{equation} ({a_L/a_1})\left[\beta_+\lambda_-u_++ \beta_-\lambda_+u_-\right] = T^{L-2}\left(\alpha_+ u_++\alpha_- u_-\right) =\alpha_+\lambda_+^{L-2} u_+ +\alpha_-\lambda_-^{L-2} u_-. \end{equation} Here we have assumed $a_1\ne 0$. Actually $a_1=0$ implies $\psi=0$. Since the vectors $u_+$ and $u_-$ are independent of each other, we get \begin{equation} (a_L/a_1)\beta_+=\alpha_+\lambda_+^{L-1} \label{lam+}\quad \mbox{ and}\quad (a_L/a_1)\beta_-=\alpha_-\lambda_-^{L-1}. \end{equation} %where we have used $\lambda_+\lambda_-=1$. If $\alpha_-=0$, we get ${\cal E}=0$, $\beta_-=0$, $\alpha_+=\beta_+=1$ and $\lambda_+=\Delta+\sqrt{\Delta^2-1}$ from (\ref{alpha}) and (\ref{beta}). % This solution with The eigenvalue ${\cal E}=0$ is %corresponding to the ground state in the one down spin sector. When $\alpha_-\ne 0$, $\alpha_\pm,\beta_\pm$ are all non-vanishing. Therefore, from (\ref{lam+}), we have \begin{equation} \lambda_+^{2L-2}= {\alpha_- \over \alpha_+}\times {\beta_+ \over \beta_-} , \label{lam+2} \end{equation} where we have used $\lambda_+\lambda_-=1$. % Note that \begin{equation} {\alpha_- \over \alpha_+} ={\Delta -\sqrt{\Delta^2-1}\over \lambda_+}\times {\lambda_+-(\Delta+\sqrt{\Delta^2-1}) \over \lambda_+-(\Delta-\sqrt{\Delta^2-1})} \end{equation} and \begin{equation} {\beta_+ \over \beta_-} ={\Delta +\sqrt{\Delta^2-1}\over \lambda_+}\times {\lambda_+-(\Delta-\sqrt{\Delta^2-1}) \over \lambda_+-(\Delta+\sqrt{\Delta^2-1})} \end{equation} from (\ref{alpha}), (\ref{beta}) and (\ref{lam}). Combining these with the above (\ref{lam+2}), we have $\lambda_+^{2L}=1$. This implies that $\lambda_+=e^{i\pi \ell /L}$, with $\ell$ an integer. From (\ref{lam}), we get the energy eigenvalues \begin{equation} {\cal E}_L(\ell)=1-{\lambda_++\lambda_-\over 2\Delta} =1-\Delta^{-1}\cos(\pi \ell /L). \label{sol} \end{equation} Here $\ell =1,2,\ldots,L-1$ because $\lambda_\pm\ne \pm 1$ which are the degenerate roots of (\ref{chara}) when $\Delta(1-{\cal E})=\pm 1$. % %Consequently, we have the solution ${\cal E}=0$ corresponding %to one of the ground states, and the solutions (\ref{sol}). Since we have found $L$ distinct eigenvalues, we obtained the complete set of eigenvalues. In particular this implies that there are no solutions with $\Delta(1-{\cal E})=\pm 1$ (except when $\Delta =1$). \end{proof} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \masection{The infinite chain}\label{sec:infinite} Before we can prove rigorous statements about the spectrum of the infinite chain we need to introduce the mathematical objects that define the infinite system. Although all interesting properties of the infinite chain can be expressed as results for limits of quantities defined for finite chains, the converse is not true. Not all limits of finite chain quantities give interesting or even sensible statements about the infinite chain. %By using a clean definition of the infinite %system we will have no difficulty in sorting out the relevant %statements about the thermodynamic limit of the XXZ chain. Let the symbols $\up\up,\up\down,\down\up,\down\down$ denote the four superselection sectors of the infinite XXZ chain with $\Delta>1$, corresponding to {\it up, kinks, antikinks\/}, and {\it down\/} respectively. We can describe the GNS Hilbert spaces \cite{BraRo} of these four superselection sectors as the so-called incomplete tensor products \cite{Gui} $\Hs_{\alpha\beta}$, for $\alpha$ and $\beta=\up$ or $\down$, defined by \begin{equation} \Hs_{\alpha\beta}=\overline{\bigcup_\Lambda \left(\bigotimes_{x\in\Lambda} \Cx^2\otimes\bigotimes_{y\in\Lambda^c}\Omega_{\alpha\beta}(y)\right)} \label{Hab} \end{equation} where \begin{equation} \Omega_{\alpha\beta}(y)= \cases{\ket{\alpha}&if $y\leq 0$\cr \ket{\beta}&if $y> 0$\cr} \label{Oaby} \end{equation} We also define the vectors $\Omega_{\alpha\beta}$ as the infinite product vectors \begin{equation} \Omega_{\alpha\beta}=\bigotimes_{y\in\Ir}\Omega_{\alpha\beta}(y) \in\Hs_{\alpha\beta} \label{Oab} \end{equation} Let $\A_\Lambda$ denote the local observables acting non-trivially only on the sites in the finite set $\Lambda$. Local observables $X\in\A_\Lambda$ act on $\Hs_{\alpha\beta}$ in the obvious way, e.g., the spin matrices at the site $x$ act on the $x^{th}$ factor of the tensor product \eq{Hab}. From the definitions above it is clear that vectors $\psi$ of the form \begin{equation} \psi=X\Omega_{\alpha\beta},\qquad X\in \bigcup_\Lambda \A_\Lambda \label{psi} \end{equation} form a dense subspace of $\Hs_{\alpha\beta}$. Note that if $\alpha\neq\beta$, $\Omega_{\alpha\beta}$ is {\it not\/} the GNS vector representing one of the kink (or antikink) ground states. Let $\Omega^{GNS}_{\alpha\beta}$ denote the GNS vector in the $\alpha\beta$ sector, or one of the GNS vectors in the case of kinks or antikinks. That $\Omega^{GNS}_{\alpha\beta}\in\Hs_{\alpha\beta}$ follows from the explicit expansion \cite{GW} $$ \Omega^{GNS}_{\alpha\beta}=Z(q)^{-1}\sum_{k=0}^{\infty} \sum_{x_1