1$, as mentioned before, not all zero-energy ground states are translation invariant. \begin{enumerate} \item There are two translation invariant zero-energy ground states, $\omega_\uparrow$ and $\omega_\downarrow$, respectively determined by \begin{equation} \omega_\uparrow(S^3_x)=S,\mbox{ and } \omega_\downarrow(S^3_x)=-S, \quad\mbox{ for all }x\in\Ir. \end{equation} \item There is an infinite family of pure {\it kink} ground states, which are all obtained as thermodynamic limits of the finite-volume ground states of the Hamiltonians ${\tilde H}^{+-}_\Lambda$, described in the previous paragraphs. For a more detailed description of the resulting space of states see \cite{GW}. The kink ground states satisfy \begin{equation} \omega(h^{+-}_{x,x+1})=0 . \end{equation} Note that the translation invariant ground states have zero energy for each of the interactions $h^{\alpha\beta}_{x,x+1}$. \item There is an infinite family of pure {\it antikink} ground states which can be obtained from the kink ground states either by reflection (interchanging left and right), or by a rotation over $\pi$ about an axis in the XY plane (spin flip, interchanging up and down). All antikink ground states satisfy \begin{equation} \omega(h^{-+}_{x,x+1})=0 . \end{equation} \end{enumerate} \end{itemize} Gottstein and Werner \cite{GW} also prove that if $\omega$ is a pure zero-energy ground state of the XXZ chains with local Hamiltonians $\tilde{H}_\Lambda^{-+}$, then exactly one of the following must be true: \begin{itemize} \item Either $\omega$ is translation invariant, and it is then either $\omega_\uparrow$ or $\omega_\downarrow$, \item or $\omega$ is represented by a unit vector in the GNS Hilbert space of the state $\omega_0$ obtained by the following thermodynamic limit: \be \omega_0(A)=\lim_{b\to +\infty}\lim_{a\to -\infty}\psi^{-+}_{[a,b]}(A) \ee which describes an antikink centered at the origin. Here $\psi_{[a,b]}^{-+}$ is given by (\ref{defkinkpsi}). For a description of the GNS representation $\pi_{\omega_0}$ of $\omega_0$, and the GNS Hilbert space $\HH_{\omega_0}$, see \cite{GW} or \cite{KN1}. So, in this case there is a vector $\psi\in\HH_{\omega_0}$, such that for all $A\in\A$ \be \omega(A)=\langle \psi,\pi_{\omega_0}(A)\psi\rangle \; . \ee \end{itemize} Moreover, $\psi$ belongs to the kernel of the GNS Hamiltonian. Together with Theorem \ref{thm:mainXXZbis}, this result (and its analogue for kink states) also proves that all infinite-volume ground states of the XXZ chains are thermodynamic limits of finite-volume ground states of local Hamiltonians with one-site boundary terms as, e.g., the $\tilde{H}_\Lambda^{\alpha\beta}$ defined in (\ref{Ham_mp}--\ref{Ham_ppmm}). %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section[PTXXZ]{Proof of Theorem \ref{thm:mainXXZ}} \label{sec:proofXXZ} \setcounter{equation}{0}% \setcounter{theorem}{0}% We begin with three short lemmas that will be used in the proof of Theorem \ref{thm:mainXXZ} and Theorem \ref{thm:mainXXX}. \begin{lemma}\label{lem:finite_energy} Let ${\tilde H}_{[a,b]}=\sum_{x=a-1}^bh_{x,x+1}$ be local Hamiltonians with a translation invariant interaction satisfying $0\leq h_{x,x+1}\leq h\idty$, and suppose there exists a state $\eta$ of $\A$ such that $\eta(h_{x,x+1})=0$, for all $x\in\Ir$. Then, if $\omega$ is a ground state, i.e., $\omega$ satisfies \eq{gs}, we have \be 0\leq\omega({\tilde H}_{[a,b]})\leq 2h \label{energy_bound}\ee for all $a1$. If there exists \be \lim_n \omega\circ\tau_{a_n}=\eta, \label{limitn} \ee then there exists $t\in [0,1]$ such that \begin{equation} \eta=t\omega_\uparrow + (1-t)\omega_\downarrow, \end{equation} where $\omega_\uparrow$ and $\omega_\downarrow$ are the ``up'' and ``down'' ground states defined in \eq{up_and_down}. \end{lemma} \begin{proof} By Lemma \ref{lem:finite_energy} we have that $\eta$ is a zero-energy ground state with \begin{equation} \eta(h_{x,x+1})=0,\mbox{ for all } x\in\Ir. \end{equation} This implies that $\eta$ is a convex combination of the two translation invariant zero-energy ground states $\omega_\uparrow$ and $\omega_\downarrow$. \end{proof} We will use the following elementary lemma in our estimates. \begin{lemma}\label{lem:state_bound} Let $\omega$ be a state of a C*-algebra $\A$, and suppose $P\in\A$ is an orthogonal projection. Then, for all $A\in \A$ we have \be \left\vert \omega(A)-\omega(PAP)\right\vert\leq 2\sqrt{1-\omega(P)} \Vert A\Vert. \ee \end{lemma} \begin{proof} Note that \begin{equation} \omega(A)-\omega(PAP)=\omega(A(\idty-P))+\omega((\idty-P) A P). \end{equation} The two terms in the right-hand side can be estimated using the Cauchy-Schwarz inequality for the positive sesqui-linear form $(X,Y)\mapsto\omega(X^*Y)$: \bea \vert \omega(A(\idty-P))\vert^2 &\leq& \omega(\idty-P)\omega(AA^*),\\ \vert \omega((\idty-P)AP)\vert^2 &\leq& \omega(\idty-P)\omega(PAA^*P). \eea As $\omega(AA^*)$ and $\omega(PAA^*P)$ are both bounded by $\Vert A\Vert^2$, the result follows. \end{proof} The set of all states of $\A$ is $w*$-compact. Therefore, for any state $\omega$ there are sequences $a_n$ of integers, with $\lim_n a_n=+\infty$, and such that the $w*$-limits of $\omega\circ\tau_{a_n}$ and $\omega\circ\tau_{-a_n}$ exist. The following theorem is then sufficient to prove Theorem \ref{thm:mainXXZ}. \begin{theorem} \label{thm:mainXXZbis} Let $\omega$ be any ground state of the spin-S XXZ ferromagnetic chain with anisotropic coupling $\Delta>1$, and let $[a_n,b_n]$ be a sequence of intervals tending to $\Ir$ (i.e., $a_n\to-\infty$ and $b_n\to+\infty$), such that \begin{equation} w^\ast\mbox{--}\lim_{n\rightarrow\infty}\omega\circ\tau_{-a_n}\equiv \omega_{-\infty}=t_{-\infty}\omega_\uparrow+(1-t_{-\infty})\omega_\downarrow, \label{asymp1} \end{equation} and \begin{equation} w^\ast\mbox{--}\lim_{n\rightarrow\infty}\omega\circ\tau_{b_n}\equiv \omega_{+\infty}=t_{+\infty}\omega_\uparrow+(1-t_{+\infty})\omega_\downarrow. \label{asymp2}\end{equation} Here $t_{\pm\infty}\in[0,1]$. Then the following properties hold: \par\noindent i) $\omega$ has well-defined asymptotics, i.e., \begin{equation} w^\ast\mbox{--}\lim_{n\rightarrow\infty}\omega\circ\tau_{\pm n} =\omega_{\mp\infty} \end{equation} ii) $\omega$ has a convex decomposition as follows \begin{equation} \omega= t^{++}\omega_\uparrow + t^{--}\omega_\downarrow + t^{+-}\varphi^{+-} + t^{-+}\varphi^{-+} \label{convexdecomp}\end{equation} where the states $\varphi^{+-}$ and $\varphi^{-+}$ satisfy, for all $x\in\Ir$, \begin{equation} \varphi^{+-}(h^{+-}_{x,x+1}) = 0,\quad \varphi^{-+}(h^{-+}_{x,x+1})=0. \end{equation} \end{theorem} Note that the complete list of such states $\varphi^{+-}$ and $\varphi^{-+}$ is known due to the work of Gottstein and Werner described in the previous section. The state $\varphi^{+-}$ is a convex combination of the translations of a kink state, and $\varphi^{-+}$ is a convex combination of the translations of an antikink state. The following relations between the convex combination coefficients are obvious: \begin{equation} t_{-\infty} = t^{++} + t^{+-},\quad t_{+\infty} = t^{-+} + t^{++}. \end{equation} \begin{proof} We use the Bratteli-Kishimoto-Robinson characterization of ground states stated in Theorem \ref{thm:BKR}. A suitable choice for the boundary terms is: \be W_{[a,b]}:= -2S A(\Delta) \left(P^+_{a-1}P^-_{b+1}+P^-_{a-1}P^+_{b+1}\right), \label{defW} \ee where $P_x^{\pm}$ are the projections onto the state with $S_x^{(3)}=S$ at the site $x$ for the plus sign, and the state with $S_x^{(3)}=-S$ for the minus sign. The corresponding local Hamiltonian is \begin{equation} {\tilde H}_{[a,b]}=\sum_{x=a-1}^b h_{x,x+1}+W_{[a,b]}. \end{equation} First, we will show that for any ground state $\omega$ satisfying the assumptions of the theorem the quantity $E_{\Lambda}(\omega)$ defined by \be E_{\Lambda}(\omega)=\inf\{\omega^\prime(\tilde H_\Lambda) \mid \omega^\prime_{\Lambda^c}=\omega_{\Lambda^c}\} \label{E_vanishes} \ee tends to zero as $\Lambda=[a_n,b_n]\uparrow\Ir$ by showing that $\liminf_n E_{[a_n,b_n]}(\omega) \geq 0$ (part a), and $\limsup_n E_{[a_n,b_n]}(\omega)\leq 0$ (part b). This fact can then be combined with the assumed asymptotics (\ref{asymp1} -\ref{asymp2}) to complete the proof of the theorem (see part c). \medskip {\it a) Proof of $\liminf_n E_{[a_n,b_n]}(\omega)\geq 0$.}\newline We write \begin{equation} P_x=P_x^++P_x^-. \end{equation} The assumed asymptotic behavior of $\omega$ implies that for any $\epsilon>0$, there exists $N$ such that, for all $n\geq N$, we have \bea \omega(\idty-P_{a_n-1})&\leq&\epsilon,\label{Pan1}\\ \omega(P^\alpha_{a_n-1}(\idty-P^\alpha_{a_n}))&\leq&\epsilon,\label{Pan2}\\ \omega(P^{\alpha}_{a_n-1}A_{a_n-1}P^{\beta}_{a_n-1})&\leq&\epsilon, \quad\mbox{ if }\alpha\neq\beta \quad \mbox{for any} \ A_{a_n-1}\in {\cal A}_{\{a_n-1\}}\label{Pan3} \eea and similar inequalities hold with $a_n-1$ replaced by $b_n+1$. Clearly, for $\Lambda=[a_n,b_n],n\geq N$, any $\omega^\prime$ such that $\omega^\prime_{\Lambda^c}=\omega_{\Lambda^c}$ satisfies the same inequalities. In order to estimate the finite-volume energy of such $\omega^\prime$, first introduce resolutions of the identity at the site $a_n-1$ as follows \begin{eqnarray} \omega^\prime(\tilde{H}_{[a_n,b_n]}) &=&\omega^\prime(P_{a_n-1}\tilde{H}_{[a_n,b_n]}P_{a_n-1}) +\omega^\prime(P_{a_n-1}\tilde{H}_{[a_n,b_n]}(\idty-P_{a_n-1}))\ret &+&\omega^\prime((\idty-P_{a_n-1})\tilde{H}_{[a_n,b_n]}P_{a_n-1}) +\omega^\prime((\idty-P_{a_n-1})\tilde{H}_{[a_n,b_n]}(\idty-P_{a_n-1})).\ret \label{decom1} \end{eqnarray} {From} (\ref{Pan1}), the non-vanishing terms are the first and the fourth ones. The fourth term is non-negative. In fact it is written as \begin{equation} \omega'((\idty-P_{a_n-1})\tilde{H}_{[a_n,b_n]}(\idty-P_{a_n-1})) =\sum_{x=a_n-1}^{b_n}\omega'((\idty-P_{a_n-1})h_{x,x+1}(\idty-P_{a_n-1}))\ge 0 \end{equation} because the contribution from the boundary term $W_{[a_n,b_n]}$ of (\ref{defW}) is vanishing. We decompose the first term in the right-hand side of (\ref{decom1}) as \begin{eqnarray} \omega^\prime(P_{a_n-1}\tilde{H}_{[a_n,b_n]}P_{a_n-1}) &=&\omega^\prime(P_{a_n-1}^+\tilde{H}_{[a_n,b_n]}P_{a_n-1}^+) +\omega^\prime(P_{a_n-1}^+\tilde{H}_{[a_n,b_n]}P_{a_n-1}^-)\ret &+&\omega^\prime(P_{a_n-1}^-\tilde{H}_{[a_n,b_n]}P_{a_n-1}^+) +\omega^\prime(P_{a_n-1}^-\tilde{H}_{[a_n,b_n]}P_{a_n-1}^-). \end{eqnarray} {From} (\ref{Pan3}), the only non-vanishing terms are of the form $\omega^\prime(P_{a_n-1}^\alpha\tilde{H}_{[a_n,b_n]}P_{a_n-1}^\alpha)$. Since similar properties hold for the site $b_n+1$, we have only to treat the forms \begin{equation} \omega^\prime(P_{a_n-1}^\alpha P_{b_n+1}^\beta\tilde{H}_{[a_n,b_n]} P_{a_n-1}^\alpha P_{b_n+1}^\beta) \end{equation} as the rest of non-vanishing contributions. We show that these are non-negative. To do this we consider the following decompositions of the Hamiltonian: \be {\tilde H}_{[a,b]}={\tilde H}_{[a,b]}^{\alpha\beta} +\delta W^{\alpha\beta}_{[a,b]} \label{decompham} \ee with \begin{equation} \delta W^{\alpha\beta}_{[a,b]}:=A(\Delta) \left[\frac{1}{2}(\alpha-\beta)\left(S^{(3)}_{a-1}-S^{(3)}_{b+1}\right) -2S\left(P^+_{a-1}P^-_{b+1}+P^-_{a-1}P^+_{b+1}\right)\right]. \label{deltaW} \end{equation} Using this expression, one can easily show \be P^\alpha_{a-1}P^\beta_{b+1}\tilde H_{[a,b]}P^\alpha_{a-1}P^\beta_{b+1} = P^\alpha_{a-1}P^\beta_{b+1}{\tilde H}^{\alpha\beta}_{[a,b]} P^\alpha_{a-1}P^\beta_{b+1}\geq 0. \ee Clearly the corresponding terms in the expectation are non-negative. Thus, we have shown that \be \omega(\tilde{H}_{[a_n,b_n]})\geq -C\times\epsilon, \ee where $C$ is a positive constant and as $\epsilon>0$ is arbitrary, this proves $\liminf_n E_{[a_n,b_n]}(\omega)\geq 0$. \medskip {\it b) Proof of $\limsup_n E_{[a_n,b_n]}(\omega)\leq 0$.}\newline In order to obtain an upper bound implying that also $\limsup E_{\Lambda}(\omega) =0$, we choose a trial state $\omega^\prime$ defined as follows: \begin{equation} \omega^\prime=\sum_{\alpha,\beta=\pm}\xi^{\alpha\beta}+\xi' \end{equation} with \begin{equation} \xi^{\alpha\beta}=\psi^{\alpha\beta}_{[a_n+1,b_n-1]} \otimes\omega_{[a_n+1,b_n-1]^c}(P^\alpha_{a_n}P^\beta_{b_n}(\ \cdots\ ) P^\alpha_{a_n}P^\beta_{b_n}) \end{equation} and \begin{equation} \xi'=\psi^{++}_{[a_n+1,b_n-1]}\otimes\omega_{[a_n+1,b_n-1]^c} ((\idty-P_{a_n}P_{b_n})(\cdots)(\idty-P_{a_n}P_{b_n})), \end{equation} where $\psi^{-+}_{[a_n+1,b_n-1]}$ is given by (\ref{defkinkpsi}), i.e., the antikink centered at the origin, $\psi_{[a_n+1,b_n-1]}^{+-}$ is the kink which is the reflection of the kink, and $\psi^{\alpha\alpha}$ is the state with all spins $S_x^{(3)}=\alpha S$ for $\alpha=\pm$. Before making the energy estimates, we check that $\omega'(A)=\omega(A)$ for all $A\in{\cal A}_{[a_n,b_n]^c}$. This is verified as follows: \begin{eqnarray} \omega'(A)&=&\sum_{\alpha,\beta=\pm}\xi^{\alpha\beta}(A)+\xi'(A)\ret &=&\sum_{\alpha,\beta=\pm}\omega(P_{a_n}^\alpha P_{b_n}^\beta A) +\omega((\idty-P_{a_n}P_{b_n})A)\ret &=&\omega(P_{a_n}P_{b_n}A)+\omega((\idty-P_{a_n}P_{b_n})A)=\omega(A) \end{eqnarray} for $A\in{\cal A}_{[a_n,b_n]^c}$. To compute the energy of each term we again use the decomposition (\ref{decompham}). The interior terms of the Hamiltonian, i.e., ${\tilde H}^{\alpha\beta}_{[a_n+2,b_n-2]}$, are identically zero. Therefore we have \begin{eqnarray} \xi^{\alpha\beta}\left({\tilde H}_{[a_n,b_n]}\right) &=&\xi^{\alpha\beta}\left(h_{a_n-1,a_n}^{\alpha\beta} +h_{a_n,a_n+1}^{\alpha\beta} +h_{b_n-1,b_n}^{\alpha\beta}+h_{b_n,b_n+1}^{\alpha\beta}\right)\ret &+&\xi^{\alpha\beta}\left(\delta W_{[a_n,b_n]}^{\alpha\beta}\right), \label{xialphabetaest1} \end{eqnarray} and \begin{eqnarray} \xi'\left({\tilde H}_{[a_n,b_n]}\right)&=& \xi'\left(h_{a_n-1,a_n}+h_{a_n,a_n+1}+h_{b_n-1,b_n}+h_{b_n,b_n+1}\right)\ret &-&2SA(\Delta)\xi'\left(P_{a_n-1}^+P_{b_n+1}^-+P_{a_n-1}^-P_{b_n+1}^+\right). \label{xipest} \end{eqnarray} Owing to (\ref{Pan1}) and \begin{equation} \idty-P_{a_n}P_{b_n}=(\idty-P_{a_n})P_{b_n}+P_{a_n}(\idty-P_{b_n}) +(\idty-P_{a_n})(\idty-P_{b_n}), \end{equation} the right-hand side of (\ref{xipest}) tends to zero as $n\to\infty$. By definition, we can write $\xi^{\alpha\beta}\left({\tilde H}_{[a_n,b_n]}\right)$ of (\ref{xialphabetaest1}) as \begin{eqnarray} \xi^{\alpha\beta}\left({\tilde H}_{[a_n,b_n]}\right) &=&\omega\left(P_{b_n}^\beta P_{a_n}^\alpha h_{a_n-1,a_n}^{\alpha\beta} P_{a_n}^\alpha\right) +\xi^{\alpha\beta}(h_{a_n,a_n+1}^{\alpha\beta}) +\xi^{\alpha\beta}(h_{b_n-1,b_n}^{\alpha\beta})\ret&+& \omega\left(P_{a_n}^\alpha P_{b_n}^\beta h_{b_n-1,b_n}^{\alpha\beta} P_{b_n}^\beta\right) +\omega\left(P_{a_n}^\alpha P_{b_n}^\beta\delta W_{[a_n,b_n]}^{\alpha\beta}\right). \label{xialphabetaest} \end{eqnarray} The first term in the right-hand side is written as \begin{eqnarray} & &\omega\left(P_{b_n}^\beta P_{a_n}^\alpha h_{a_n-1,a_n}^{\alpha\beta} P_{a_n}^\alpha\right)\ret &=&\omega\left(P_{b_n}^\beta P_{a_n}^\alpha h_{a_n-1,a_n}^{\alpha\beta} P_{a_n}^\alpha(\idty-P_{a_n-1}^\alpha)\right) +\omega\left(P_{b_n}^\beta P_{a_n}^\alpha h_{a_n-1,a_n}^{\alpha\beta} P_{a_n}^\alpha P_{a_n-1}^\alpha\right). \end{eqnarray} This first term in the right-hand side is vanishing as $a_n\rightarrow -\infty$ from (\ref{Pan2}), and the second term also is vanishing because $h_{a_n-1,a_n}^{\alpha\beta}P_{a_n}^\alpha P_{a_n-1}^\alpha=0$. Therefore the first term $\omega\left(P_{b_n}^\beta P_{a_n}^\alpha h_{a_n-1,a_n}^{\alpha\beta} P_{a_n}^\alpha\right)$ of (\ref{xialphabetaest}) is vanishing. Similarly the fourth term in the right-hand side of (\ref{xialphabetaest}) is vanishing as $b_n\rightarrow +\infty$. The fifth term in the right-hand side of (\ref{xialphabetaest}) tends to zero as $n\to\infty$. To show this, we introduce a resolution of the identity \begin{equation} \idty=P_{a_n-1}^\alpha P_{b_n+1}^\beta+P_{a_n-1}^\alpha(\idty-P_{b_n+1}^\beta) +(\idty-P_{a_n-1}^\alpha)P_{b_n+1}^\beta+(\idty-P_{a_n-1}^\alpha) (\idty-P_{b_n+1}^\beta). \end{equation} Owing to this and (\ref{Pan2}), we have only to show that \begin{equation} \omega\left(P_{a_n}^\alpha P_{b_n}^\beta \delta W_{[a_n,b_n]}^{\alpha\beta} P_{a_n-1}^\alpha P_{b_n+1}^\beta\right) \end{equation} is vanishing in the limit $n\to\infty$. But this is identically zero for all $\alpha,\beta=\pm$. The rest are the second and the third terms in the right-hand side of (\ref{xialphabetaest}). We treat only the second term because the third one is treated in the same way. To begin with, we note the following: {From} Lemma~\ref{lem:probnotS} and Lemma~\ref{lem:state_bound} we obtain \begin{equation} \left\Vert\psi^{\alpha\beta}_{[a_n+1,b_n-1]}(P^\alpha_{a_n+1} P^\beta_{b_n-1}(\ \cdots\ ) P^\alpha_{a_n+1}P^\beta_{b_n-1})-\psi^{\alpha\beta}_{[a_n+1,b_n-1]}(\cdots) \right\Vert \leq\mu\exp\left[-\nu\min\{-a_n,b_n\}\right] \label{psiasym} \end{equation} for some positive constants $\mu, \nu$, and $\min\{-a_n,b_n\}$ represents the distance of the position of the center of the kink to the origin. Combining (\ref{psiasym}) with \begin{equation} P^\alpha_{a_n}P^\alpha_{a_n+1}h^{\alpha\beta}_{a_n,a_n+1}P^\alpha_{a_n} P^\alpha_{a_n+1}=0 \quad \mbox{for all} \ \alpha,\beta=\pm, \end{equation} we get the desired result $\xi^{\alpha\beta}(h_{a_n,a_n+1}^{\alpha\beta})\rightarrow 0$ as $a_n\rightarrow -\infty$. This concludes the proof of $\limsup_n E_{[a_n,b_n]}(\omega)\leq 0$. \medskip %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% {\it c)} We can now complete the proof as follows. {From} the asymptotics it follows that there is a sequence $\epsilon_n \downarrow 0$ such that \begin{equation} \bigl\vert \omega_{[a_n,b_n]}(A)-\sum_{\alpha,\beta=\pm} \omega(P^\alpha_{a_n -1}P^\beta_{b_n+1} A)\bigr\vert\leq \epsilon_n \Vert A\Vert \end{equation} for all $A\in\A_{[a_n,b_n]}$. This implies that there are four subsequences of states defined by \begin{equation} \omega^{\alpha\beta}_k(A) =\frac{\omega(P^\alpha_{a_{n_k} -1}P^\beta_{b_{n_k}+1}A P^\alpha_{a_{n_k} -1}P^\beta_{b_{n_k}+1})}{\omega(P^\alpha_{a_{n_k} -1}P^\beta_{b_{n_k}+1})} \end{equation} with the following properties: \begin{eqnarray} &&w^\ast\mbox{--}\lim_{k\rightarrow\infty}\omega^{\alpha\beta}_k \equiv\omega^{\alpha\beta}\ \mbox{exists}\label{property1},\\ &&w^\ast\mbox{--}\lim_{k\rightarrow\infty}\left\vert\omega -\sum_{\alpha,\beta=\pm} t^{\alpha\beta} \omega^{\alpha\beta}_k\right\vert =0, \quad \mbox{with} \ t^{\alpha\beta}\ge 0, \label{property3}\\ &&\lim_{k\rightarrow\infty}\omega^{\alpha\beta}_k(h_{x,x+1}^{\alpha\beta})=0 \quad \mbox{for all $x\in \Ir$}.\label{property2} \end{eqnarray} The last property (\ref{property2}) follows from the above results in parts a and b. To see this, we note that \begin{eqnarray} \omega_k^{\alpha\beta}\left({\tilde H}_{[a_{n_k},b_{n_k}]}\right) &=&\omega_k^{\alpha\beta}\left({\tilde H}_{[a_{n_k},b_{n_k}]}^{\alpha\beta} \right) +\omega_k^{\alpha\beta}\left(\delta W_{[a_{n_k},b_{n_k}]}^{\alpha\beta}\right), \end{eqnarray} where we have used the decompositions (\ref{decompham}) of the Hamiltonian ${\tilde H}_{[a,b]}$. The second term in the right-hand side is vanishing as $a_{n_k}\rightarrow -\infty, b_{n_k}\rightarrow +\infty$ from the asymptotics and the definition (\ref{deltaW}) of $\delta W_{[a,b]}^{\alpha\beta}$. Combining these observations with (\ref{property3}) and with the results of parts a and b, i.e., $\omega({\tilde H}_{[a_n,b_n]})\rightarrow 0$ as $a_n\rightarrow -\infty, b_n\rightarrow +\infty$, we have \begin{equation} \omega_k^{\alpha\beta}({\tilde H}_{[a_{n_k},b_{n_k}]}^{\alpha\beta}) =\sum_{x=a_{n_k}-1}^{b_{n_k}} \omega_k^{\alpha\beta}(h_{x,x+1}^{\alpha\beta}) \rightarrow 0 \quad \mbox{as} \ a_{n_k}\rightarrow -\infty, b_{n_k}\rightarrow +\infty. \end{equation} This implies (\ref{property2}) because $h_{x,x+1}^{\alpha\beta}\ge 0$ for all $\alpha,\beta=\pm$ and for all $x\in \Ir$. As the convex decomposition (\ref{convexdecomp}), if it exists, is unique, its terms must be proportional to the $\omega^{\alpha\beta}$ because of (\ref{property3}). The property (\ref{property2}) identifies the $\omega^{\alpha\beta}$ as the known zero-energy states, which proves the statement ii) of Theorem~\ref{thm:mainXXZbis}. The statement i) follows from ii). \end{proof} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section[PTXXX]{Proof of Theorem \ref{thm:mainXXX}} \label{sec:proofXXX} \setcounter{equation}{0}% \setcounter{theorem}{0}% The proof of the absence of non-translation-invariant ground states in the isotropic case is similar to the proof in the anisotropic case (Theorem \ref{thm:mainXXZbis}). It differs from it in two points. The first difference is that in the isotropic case all ground states turn out to be zero energy ground states for the {\em same} local Hamiltonians. This simplifies the proof. The second difference, however, makes the proof more subtle than in the anisotropic case. This is due to the broken continuous rotation symmetry. The possible asymptotics of pure ground states now depend on two continuous parameters, which can be taken to be two angles: $(\theta,\phi)=\Omega\in S^2$. For the same reason the excitation spectrum is gapless and, therefore, the method of proof followed in \cite{Mat} cannot be adapted to the isotropic case. The state of a single spin pointing in the direction $\Omega$ is represented by the vector \begin{equation} \ket{\Omega}:=U(\Omega)\ket{S}, \end{equation} where \be U(\Omega):=e^{-i\phi S^{(3)}}e^{-i\theta S^{(2)}}, \ee and $\ket{S}$ is the normalized eigenvector of $S^{(3)}$ satisfying $S^{(3)}\ket{S}=S\ket{S}$. It follows that the vectors \begin{equation} U(\Omega)\ket{S},\quad \Omega\in S^2 \end{equation} span the $(2S+1)$-dimensional irreducible unitary representation of SU(2). For the same reason the vectors \begin{equation} \ket{\Omega}_\Lambda:=\bigotimes_{x\in\Lambda}(\ket{\Omega})_x \end{equation} span the maximum total spin subspace for any finite volume $\Lambda\subset\Ir$. It is also straightforward to check that the orthogonal projection onto the subspace of maximal total spin, $P_\Lambda$, can be written as \cite{KlaSka} \be P_\Lambda:=\frac{2kS+1}{4\pi}\int d\Omega\,Q_\Lambda(\Omega), \label{defP} \ee where, as usual, $d\Omega=\sin \theta d\theta d\phi$, and \be Q_\Lambda(\Omega)=\bigotimes_{x\in\Lambda}(\ket{\Omega}\bra{\Omega})_x. \label{defQ}\ee It is obvious from the form of the isotropic Hamiltonian (\eq{Ham_ppmm} with $\Delta =1$) that zero-energy states are supported by the maximum total spin subspace on each finite volume, which is permutation invariant. It follows immediately that all zero-energy ground states of this Hamiltonian are translation invariant. Therefore, for the proof of Theorem \ref{thm:mainXXX} it is sufficient to show that all ground states are zero-energy states for this Hamiltonian. By Theorem \ref{thm:BKR}, and the fact that the local interactions are non-negative, this will follow if we prove that for any ground state \be \lim_{\Lambda\uparrow\Ir}E_\Lambda(\omega)=0, \label{energy_XXX_vanishes}\ee where $E_\Lambda(\omega)$ is defined in \eq{E_vanishes}. By Lemma~\ref{lem:finite_energy} we have to do this for any $\omega$ with left and right asymptotics such that \be \lim_{x\to\pm\infty}\omega(\idty-P_{\Lambda +x})=0 \label{asymXXX} \ee for all finite $\Lambda\subset\Ir$. To do this we follow the same strategy as in the proof of Theorem \ref{thm:mainXXZ}: we need to construct local modifications of an arbitrary ground state $\omega$ that have arbitrarily low energy. This will be done by inserting a long-wavelength spin-wave state that gradually turns the spin from $\Omega^-$ to $\Omega^+$, conditioned upon the asymptotic orientation to the left and to the right being $\Omega^-$ and $\Omega^+$ respectively. The sum over the possible asymptotic behaviors appearing in the proof of Theorem \ref{thm:mainXXZ} will here become an integral over $\Omega^-$ and $\Omega^+$. Some care has to be taken with the conditioning in order not to spoil the energy estimates, which need to be done carefully, too. A quick estimate produces useless bounds. Now, we fill in the technical details of the proof sketched above. \vspace{.5truecm} \noindent {\bf Proof of Theorem \ref{thm:mainXXX}:}\newline \noindent As explained above we want to show \eq{energy_XXX_vanishes}. Due to the non-negative interaction we only need to show \be \limsup_{\Lambda\uparrow\Ir}E_\Lambda(\omega)\leq 0. \label{limsup_XXX}\ee Let $\omega$ be a ground state. We will construct a trial state $\omega^\prime$ which coincides with $\omega$ outside a finite interval $\Lambda=[a,b]$. We will use a corridor of $m$ sites to perform the conditioning. Therefore, let $m\geq 1$ be such that $b-a+1>2m$. The state $\omega^\prime$ is then defined by \begin{equation} \omega':=\int d\Omega^{+}\int d\Omega^{-} \eta_{[a+m,b-m]}^{\Omega^{+},\Omega^{-}}\otimes \omega_{{[a+m,b-m]}^c}^{\Omega^{+},\Omega^{-}}+\omega^\pprime, \label{defomegap} \end{equation} where $\eta_{[a+m,b-m]}^{\Omega^{+},\Omega^{-}}$, $\omega_{{[a+m,b-m]}^c}^{\Omega^{+},\Omega^{-}}$, and $\omega^\pprime$ are non-negative functionals on $\A_{[a+m,b-m]}$, \hfill\break $\A_{{[a+m,b-m]}^c}$, and $\A$, respectively, defined as follows. For convenience put ${\hat \Lambda}=[a+m,b-m]$. \begin{equation} \eta_{\hat \Lambda}^{\Omega^{+},\Omega^{-}}(\,\cdots\,):= \omega_{\uparrow,{\hat \Lambda}}\left(\left( V_{\hat \Lambda}^{\Omega^{+},\Omega^{-}}\right)^* (\,\cdots\,)V_{\hat \Lambda}^{\Omega^{+},\Omega^{-}}\right), \label{defvarphi} \end{equation} where $\omega_{\uparrow,\hat{\Lambda}}$ is the state with $S^{(3)}_x=S$ for all $x\in\hat\Lambda$, and $V_{\hat \Lambda}^{\Omega^+,\Omega^-}$ is the unitary defined by \begin{eqnarray} V_{\hat \Lambda}^{\Omega^+,\Omega^-}&:=& \exp\left[-i\phi^-\sum_{x=a+m}^{b-m}S_x^{(3)}-\frac{i}{\hat L} \left(\phi^+-\phi^-\right) \sum_{x=a+m}^{b-m}(x-a-m)S_x^{(3)}\right] \nonumber \\ &\times& \exp\left[-i\theta^-\sum_{x=a+m}^{b-m}S_x^{(2)}-\frac{i}{\hat L} \left(\theta^+-\theta^-\right)\sum_{x=a+m}^{b-m}(x-a-m) S_x^{(2)}\right] \end{eqnarray} with ${\hat L}=b-a-2m$, and $\Omega^\pm=(\theta^\pm,\phi^\pm)$. \bea &&\omega_{{\hat \Lambda}^c}^{\Omega^{+},\Omega^{-}}(\cdots) :=\left(\frac{2mS+1}{4\pi}\right)^2\nonumber\\ &&\times\omega_{{\hat \Lambda}^c} \left(Q_{[a,a+m-1]}(\Omega^-) Q_{[b-m+1,b]}(\Omega^+)(\,\cdots\,) Q_{[a,a+m-1]}(\Omega^-) Q_{[b-m+1,b]}(\Omega^+)\right), \label{omegapro}\eea where $Q_\Lambda(\Omega)$ is defined in \eq{defQ}. \begin{equation} \omega''(\,\cdots\,):=\omega_{\uparrow, \hat \Lambda}(\,\cdots\,)\otimes \omega_{{\hat \Lambda}^c}\left((\idty-P_{[a,a+m-1]}P_{[b-m+1,b]}) (\,\cdots\,)(\idty-P_{[a,a+m-1]}P_{[b-m+1,b]}\right), \label{defomegapp} \end{equation} with $P_{[k,\ell]}$ defined in \eq{defP}. Note that in the state $\eta_{\hat \Lambda}^{\Omega^{+},\Omega^{-}}$ the orientation of spins gradually rotate from $\Omega^{-}$ on the left to $\Omega^{+}$ on the right. Due to the known asymptotics of any ground state, the state $\omega_{{\hat \Lambda}^c}^{\Omega^{+},\Omega^{-}}$, up to normalization, represents $\omega$ conditioned upon the spins having orientation $\Omega^-$ on the sites $a,\ldots, a+m-1$ and $\Omega^+$ on the sites $b-m+1,\ldots,b$. In finite volume all this is approximate, and the term $\omega''$ is exactly the correction needed to reproduce $\omega$ on the complement of $\Lambda$. Before making the energy estimates, we first verify this fact, i.e., check that $\omega'(A)=\omega(A)$ for all $A\in {\cal A}_{\Lambda^c}$. Note that the projection operators $P_{[a,a+m-1]}$, $P_{[b-m+1,b]}$, \hfill\break $Q_{[a,a+m-1]}(\Omega^-)$, and $Q_{[b-m+1,b]}(\Omega^+)$ are elements of the algebra ${\cal A}_{\Lambda \cap {\hat \Lambda}^c}$, and therefore commute with any $A\in \A_{\Lambda^c}$ by the definitions. This fact and the definitions given above suffice to check the following for all $A\in\A_{\Lambda^c}$: \begin{eqnarray} \omega'(A)&=&\int d\Omega^+\int d\Omega^-\, \eta_{\hat \Lambda}^{\Omega^+,\Omega^-}(\idty) \omega_{{\hat \Lambda}^c}^{\Omega^{+},\Omega^{-}}(A) +\omega''(A)\ret &=&\left(\frac{2mS+1}{4\pi}\right)^2\int d\Omega^+\int d\Omega^-\, \omega(Q_{[a,a+m-1]}(\Omega^-) Q_{[b-m+1,b]}(\Omega^+)A)\ret &+&\omega((\idty-P_{[a,a+m-1]}P_{[b-m+1,b]})A)\ret &=&\omega(P_{[a,a+m-1]}P_{[b-m+1,b]}A) +\omega((\idty-P_{[a,a+m-1]}P_{[b-m+1,b]})A)=\omega(A). \end{eqnarray} Next, we estimate the energy of $\omega^\prime$. By definition, we have \begin{eqnarray} \omega'(H_{\Lambda\cup\partial\Lambda})&=&\int d\Omega^+\int d\Omega^- \eta_{\hat \Lambda}^{\Omega^+,\Omega^-} \left(H_{\hat \Lambda}\right)\nonumber \\ &+&\int d\Omega^+\int d\Omega^- \left[\omega_{{\hat \Lambda}^c}^{\Omega^+,\Omega^-}(h_{a-1,a}) +\omega_{{\hat \Lambda}^c}^{\Omega^+,\Omega^-}(h_{b,b+1})\right] \nonumber \\ &+&\sum_{x=b-m}^b\omega''(h_{x,x+1})+\sum_{x=a-1}^{a+m-1} \omega''(h_{x,x+1}). \label{omegapenergy} \end{eqnarray} First, we estimate $\eta_{\hat \Lambda}^{\Omega^+,\Omega^-} \left(H_{\hat\Lambda}\right)$. Due to the rotation invariance this quantity depends only on the angle between $\Omega^-$ and $\Omega^+$. Therefore we can consider the case $\theta^+=\theta,\theta^-=\phi^+=\phi^-=0$. The energy is then given by the usual spin wave energy: \begin{eqnarray} \eta_{\hat \Lambda}^{\Omega^+,\Omega^-}\left(H_{\hat \Lambda}\right) &=&\sum_{x=a+m}^{b-m-1}\omega_\uparrow \left(\left(V_{\hat \Lambda}^{\Omega^+,\Omega^-}\right)^\ast h_{x,x+1}V_{\hat \Lambda}^{\Omega^+,\Omega^-}\right) \nonumber \\ &=&S^2{\hat L}\left[1-\cos(\theta/{\hat L})\right] . \end{eqnarray} This implies \be \eta_{\hat \Lambda}^{\Omega^+,\Omega^-}\left(H_{\hat \Lambda}\right) \rightarrow 0, \mbox{ as }{\hat L}=b-a-2m \rightarrow \infty . \label{est1}\ee Therefore the first integral in the right-hand side of (\ref{omegapenergy}) vanishes in the limit ${\hat \Lambda}\uparrow\Ir$. Next consider the second integral in the right-hand side of (\ref{omegapenergy}). Using the definitions (\ref{defP}), (\ref{defQ}), and (\ref{omegapro}), we have \begin{eqnarray} & &\int d\Omega^+\int d\Omega^- \omega_{{\hat \Lambda}^c}^{\Omega^+,\Omega^-}(h_{b,b+1})\nonumber\\ &=&\frac{2mS+1}{4\pi}\int d\Omega^+ \omega(Q_{[b-m+1,b]}(\Omega^+) h_{b,b+1}Q_{[b-m+1,b]}(\Omega^+)P_{[a,a+m-1]}) \nonumber\\ &\le&\omega\left(\frac{2mS+1}{4\pi}\int d\Omega^+ Q_{[b-m+1,b]}(\Omega^+) h_{b,b+1}Q_{[b-m+1,b]}(\Omega^+)\right). \label{term2}\end{eqnarray} The operator \begin{equation} \frac{2mS+1}{4\pi}\int d\Omega^+ Q_{[b-m+1,b]}(\Omega^+) h_{b,b+1}Q_{[b-m+1,b]}(\Omega^+) \end{equation} commutes with SU(2) rotations. Therefore, as any weak limit $\omega \circ\tau_b$, $b\to\infty$, is supported by the highest spin irreducible representation of SU(2), the limit $b\to\infty$ of \eq{term2} is given by \begin{eqnarray} & &\frac{2mS+1}{4\pi}\int d\Omega \,\omega^{(+\infty)} \left(Q_{[1,m]}(\Omega)h_{m,m+1}Q_{[1,m]}(\Omega)\right)\nonumber\\ &=&\frac{2mS+1}{4\pi}\int d\Omega \,\omega_\uparrow \left(Q_{[1,m]}(\Omega)h_{m,m+1}Q_{[1,m]}(\Omega)\right). \label{est2} \end{eqnarray} Here again $\omega_\uparrow$ is the state determined by $\omega_\uparrow(S_x^{(3)})=S$, for all $x\in \Ir$. The right-hand side of \eq{est2} can easily be calculated: \begin{eqnarray} & &\frac{2mS+1}{4\pi}\int d\Omega \,\omega_\uparrow \left(Q_{[1,m]}(\Omega)h_{m,m+1}Q_{[1,m]}(\Omega)\right)\nonumber\\ &=&\frac{2mS+1}{4\pi}\int d\Omega \ S^2 \left(\cos \frac{\theta}{2}\right)^{4mS}(1-\cos \theta)\nonumber\\ &=&\frac{S^2(2mS+1)}{2}\int_0^\pi d\theta \sin \theta \left(\frac{1+\cos\theta}{2}\right)^{2mS}(1-\cos \theta)= \frac{S^2}{mS+1}. \label{cosint} \end{eqnarray} Clearly this quantity vanishes as $m\to \infty$. The contribution in the second term of \eq{omegapenergy} coming form the left asymptotics can be estimated in the same way. Therefore, we have shown \begin{equation} \lim_{m \rightarrow \infty} \lim_{b \rightarrow \infty} \lim_{a \rightarrow -\infty} \int d\Omega^+\int d\Omega^- \left[\omega_{{\hat \Lambda}^c}^{\Omega^+,\Omega^-}(h_{a-1,a}) +\omega_{{\hat \Lambda}^c}^{\Omega^+,\Omega^-}(h_{b,b+1})\right]=0 . \label{cosint2} \end{equation} Finally, we consider the two summations in the right-hand side of \eq{omegapenergy}. Note that \begin{eqnarray} \idty-P_{[a,a+m-1]}P_{[b-m+1,b]} &=&P_{[a,a+m-1]}(\idty-P_{[b-m+1,b]})+(\idty-P_{[a,a+m-1]})P_{[b-m+1,b]}\ret &+&(\idty-P_{[a,a+m-1]})(\idty-P_{[b-m+1,b]}). \end{eqnarray} Combining this with \eq{asymXXX} and \eq{defomegapp}, we can conclude \begin{equation} \lim_{b\rightarrow \infty}\lim_{a\rightarrow -\infty} \sum_{x=b-m}^b\omega''(h_{x,x+1})=0 ,\mbox{ and } \lim_{b\rightarrow \infty}\lim_{a\rightarrow -\infty} \sum_{x=a-1}^{a+m-1}\omega''(h_{x,x+1})=0 \label{est3}\end{equation} for a fixed $m$. 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