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\begin{document}

\title[Lecture 9. Ice model]{Lecture 9\\Ice model and commuting transfer matrices}
\date{}

\frame{\titlepage}

\begin{frame}
\frametitle{Ice model: configurations}

The `ice model' (\tikz[baseline=-2pt,scale=.5]{\draw[fill=black] (0,0) circle (2mm);} is Oxygen, \tikz[baseline=-2pt,scale=.5]{\draw[-,fill=white] (0,0) circle (1mm);} is Hydrogen):
$$
\begin{tikzpicture}[scale=.5,baseline=.5*1.5cm-2pt]
\draw[-] (0,-.5) -- (0,3.5);
\draw[-] (1,-.5) -- (1,3.5);
\draw[-] (2,-.5) -- (2,3.5);
\draw[-] (3,-.5) -- (3,3.5);
\draw[-] (-.5,0) -- (3.5,0);
\draw[-] (-.5,1) -- (3.5,1);
\draw[-] (-.5,2) -- (3.5,2);
\draw[-] (-.5,3) -- (3.5,3);
\foreach \x in {0,...,3} {
  \foreach \y in {0,...,3} {
    \draw[fill=black] (\x,\y) circle (2mm);
  }
}
\draw[-,fill=white,yshift=0cm] (0,.35) circle (1mm);
\draw[-,fill=white,yshift=1cm] (0,.65) circle (1mm);
\draw[-,fill=white,yshift=2cm] (0,.65) circle (1mm);
\draw[-,fill=white,yshift=3cm] (0,.35) circle (1mm);
%
\begin{scope}[xshift=1cm]
  \draw[-,fill=white,yshift=-1cm] (0,.65) circle (1mm);
  \draw[-,fill=white,yshift=0cm] (0,.65) circle (1mm);
  \draw[-,fill=white,yshift=1cm] (0,.65) circle (1mm);
  \draw[-,fill=white,yshift=2cm] (0,.65) circle (1mm);
\end{scope}
%
\begin{scope}[xshift=2cm]
  \draw[-,fill=white,yshift=-1cm] (0,.65) circle (1mm);
  \draw[-,fill=white,yshift=0cm] (0,.35) circle (1mm);
  \draw[-,fill=white,yshift=1cm] (0,.35) circle (1mm);
  \draw[-,fill=white,yshift=2cm] (0,.35) circle (1mm);
  \draw[-,fill=white,yshift=3cm] (0,.35) circle (1mm);
\end{scope}
%
\begin{scope}[xshift=3cm]
  \draw[-,fill=white,yshift=0cm] (0,.35) circle (1mm);
  \draw[-,fill=white,yshift=1cm] (0,.35) circle (1mm);
  \draw[-,fill=white,yshift=2cm] (0,.35) circle (1mm);
\end{scope}
%
\draw[-,fill=white,xshift=0cm] (.35,0) circle (1mm);
\draw[-,fill=white,xshift=1cm] (.35,0) circle (1mm);
\draw[-,fill=white,xshift=2cm] (.65,0) circle (1mm);
%
\begin{scope}[yshift=1cm]
  \draw[-,fill=white,xshift=-1cm] (.65,0) circle (1mm);
  \draw[-,fill=white,xshift=0cm] (.35,0) circle (1mm);
  \draw[-,fill=white,xshift=1cm] (.35,0) circle (1mm);
  \draw[-,fill=white,xshift=2cm] (.65,0) circle (1mm);
\end{scope}
%
\begin{scope}[yshift=2cm]
  \draw[-,fill=white,xshift=0cm] (.35,0) circle (1mm);
  \draw[-,fill=white,xshift=1cm] (.35,0) circle (1mm);
  \draw[-,fill=white,xshift=2cm] (.35,0) circle (1mm);
  \draw[-,fill=white,xshift=3cm] (.35,0) circle (1mm);
\end{scope}
%
\begin{scope}[yshift=3cm]
  \draw[-,fill=white,xshift=0cm] (.65,0) circle (1mm);
  \draw[-,fill=white,xshift=1cm] (.65,0) circle (1mm);
  \draw[-,fill=white,xshift=2cm] (.65,0) circle (1mm);
  \draw[-,fill=white,xshift=3cm] (.35,0) circle (1mm);
\end{scope}
\end{tikzpicture}
\uncover<3->{
\quad=\quad
\begin{tikzpicture}[scale=.5,baseline=.5*1.5cm-2pt]
% vertical lines
\begin{scope}[decoration={markings,mark=at position .7 with {\arrow{>}}},xshift=0cm]
  \draw[-,postaction=decorate,yshift=-1cm] (0,1) -- (0,0);
  \draw[-,postaction=decorate,yshift=0cm] (0,1) -- (0,0);
  \draw[-,postaction=decorate,yshift=1cm] (0,0) -- (0,1);
  \draw[-,postaction=decorate,yshift=2cm] (0,0) -- (0,1);
  \draw[-,postaction=decorate,yshift=3cm] (0,1) -- (0,0);
\end{scope}
%
\begin{scope}[decoration={markings,mark=at position .7 with {\arrow{>}}},xshift=1cm]
  \draw[-,postaction=decorate,yshift=-1cm] (0,0) -- (0,1);
  \draw[-,postaction=decorate,yshift=0cm] (0,0) -- (0,1);
  \draw[-,postaction=decorate,yshift=1cm] (0,0) -- (0,1);
  \draw[-,postaction=decorate,yshift=2cm] (0,0) -- (0,1);
  \draw[-,postaction=decorate,yshift=3cm] (0,0) -- (0,1);
\end{scope}
%
\begin{scope}[decoration={markings,mark=at position .7 with {\arrow{>}}},xshift=2cm]
  \draw[-,postaction=decorate,yshift=-1cm] (0,0) -- (0,1);
  \draw[-,postaction=decorate,yshift=0cm] (0,1) -- (0,0);
  \draw[-,postaction=decorate,yshift=1cm] (0,1) -- (0,0);
  \draw[-,postaction=decorate,yshift=2cm] (0,1) -- (0,0);
  \draw[-,postaction=decorate,yshift=3cm] (0,1) -- (0,0);
\end{scope}
%
\begin{scope}[decoration={markings,mark=at position .7 with {\arrow{>}}},xshift=3cm]
  \draw[-,postaction=decorate,yshift=-1cm] (0,1) -- (0,0);
  \draw[-,postaction=decorate,yshift=0cm] (0,1) -- (0,0);
  \draw[-,postaction=decorate,yshift=1cm] (0,1) -- (0,0);
  \draw[-,postaction=decorate,yshift=2cm] (0,1) -- (0,0);
  \draw[-,postaction=decorate,yshift=3cm] (0,0) -- (0,1);
\end{scope}
%
% horizontal lines
\begin{scope}[decoration={markings,mark=at position .7 with {\arrow{>}}},yshift=0cm]
  \draw[-,postaction=decorate,xshift=-1cm] (1,0) -- (0,0);
  \draw[-,postaction=decorate,xshift=0cm] (1,0) -- (0,0);
  \draw[-,postaction=decorate,xshift=1cm] (1,0) -- (0,0);
  \draw[-,postaction=decorate,xshift=2cm] (0,0) -- (1,0);
  \draw[-,postaction=decorate,xshift=3cm] (0,0) -- (1,0);
\end{scope}
%
\begin{scope}[decoration={markings,mark=at position .7 with {\arrow{>}}},yshift=1cm]
  \draw[-,postaction=decorate,xshift=-1cm] (0,0) -- (1,0);
  \draw[-,postaction=decorate,xshift=0cm] (1,0) -- (0,0);
  \draw[-,postaction=decorate,xshift=1cm] (1,0) -- (0,0);
  \draw[-,postaction=decorate,xshift=2cm] (0,0) -- (1,0);
  \draw[-,postaction=decorate,xshift=3cm] (0,0) -- (1,0);
\end{scope}
%
\begin{scope}[decoration={markings,mark=at position .7 with {\arrow{>}}},yshift=2cm]
  \draw[-,postaction=decorate,xshift=-1cm] (1,0) -- (0,0);
  \draw[-,postaction=decorate,xshift=0cm] (1,0) -- (0,0);
  \draw[-,postaction=decorate,xshift=1cm] (1,0) -- (0,0);
  \draw[-,postaction=decorate,xshift=2cm] (1,0) -- (0,0);
  \draw[-,postaction=decorate,xshift=3cm] (1,0) -- (0,0);
\end{scope}
%
\begin{scope}[decoration={markings,mark=at position .7 with {\arrow{>}}},yshift=3cm]
  \draw[-,postaction=decorate,xshift=-1cm] (1,0) -- (0,0);
  \draw[-,postaction=decorate,xshift=0cm] (0,0) -- (1,0);
  \draw[-,postaction=decorate,xshift=1cm] (0,0) -- (1,0);
  \draw[-,postaction=decorate,xshift=2cm] (0,0) -- (1,0);
  \draw[-,postaction=decorate,xshift=3cm] (1,0) -- (0,0);
\end{scope}
%
\end{tikzpicture}
}
\uncover<4->{
\quad=\quad
\begin{tikzpicture}[scale=.5,baseline=.5*1.5cm-2pt,>={Stealth[length=1mm]}]
% vertical lines
\begin{scope}[xshift=0cm]
  \draw[<-,yshift=-1cm] (0,.5) node[left=-2pt] {\tiny$-$} -- (0,1);
  \draw[-,yshift=0cm] (0,0) -- node[left=-2pt] {\tiny$-$} (0,1);
  \draw[-,yshift=1cm] (0,0) -- node[left=-2pt] {\tiny$+$} (0,1);
  \draw[-,yshift=2cm] (0,0) -- node[left=-2pt] {\tiny$+$} (0,1);
  \draw[-,yshift=3cm] (0,0) -- (0,.5) node[left=-2pt] {\tiny$-$};
\end{scope}
%
\begin{scope}[xshift=1cm]
  \draw[<-,yshift=-1cm] (0,.5) node[left=-2pt] {\tiny$+$} -- (0,1);
  \draw[-,yshift=0cm] (0,0) -- node[left=-2pt] {\tiny$+$} (0,1);
  \draw[-,yshift=1cm] (0,0) -- node[left=-2pt] {\tiny$+$} (0,1);
  \draw[-,yshift=2cm] (0,0) -- node[left=-2pt] {\tiny$+$} (0,1);
  \draw[-,yshift=3cm] (0,0) -- (0,.5) node[left=-2pt] {\tiny$+$};
\end{scope}
%
\begin{scope}[xshift=2cm]
  \draw[<-,yshift=-1cm] (0,.5) node[left=-2pt] {\tiny$+$} -- (0,1);
  \draw[-,yshift=0cm] (0,0) -- node[left=-2pt] {\tiny$-$} (0,1);
  \draw[-,yshift=1cm] (0,0) -- node[left=-2pt] {\tiny$-$} (0,1);
  \draw[-,yshift=2cm] (0,0) -- node[left=-2pt] {\tiny$-$} (0,1);
  \draw[-,yshift=3cm] (0,0) -- (0,.5) node[left=-2pt] {\tiny$-$};
\end{scope}
%
\begin{scope}[xshift=3cm]
  \draw[<-,yshift=-1cm] (0,.5) node[left=-2pt] {\tiny$-$} -- (0,1);
  \draw[-,yshift=0cm] (0,0) -- node[left=-2pt] {\tiny$-$} (0,1);
  \draw[-,yshift=1cm] (0,0) -- node[left=-2pt] {\tiny$-$} (0,1);
  \draw[-,yshift=2cm] (0,0) -- node[left=-2pt] {\tiny$-$} (0,1);
  \draw[-,yshift=3cm] (0,0) -- (0,.5) node[left=-2pt] {\tiny$+$};
\end{scope}
%
% horisontal lines
\begin{scope}[yshift=0cm]
  \draw[<-,xshift=-1cm] (.5,0) node[above=-2pt] {\tiny$-$} -- (1,0);
  \draw[-,xshift=0cm] (0,0) -- node[above=-2pt] {\tiny$-$} (1,0);
  \draw[-,xshift=1cm] (0,0) -- node[above=-2pt] {\tiny$-$} (1,0);
  \draw[-,xshift=2cm] (0,0) -- node[above=-2pt] {\tiny$+$} (1,0);
  \draw[-,xshift=3cm] (0,0) -- (.5,0) node[above=-2pt] {\tiny$+$};
\end{scope}
%
\begin{scope}[yshift=1cm]
  \draw[<-,xshift=-1cm] (.5,0) node[above=-2pt] {\tiny$+$} -- (1,0);
  \draw[-,xshift=0cm] (0,0) -- node[above=-2pt] {\tiny$-$} (1,0);
  \draw[-,xshift=1cm] (0,0) -- node[above=-2pt] {\tiny$-$} (1,0);
  \draw[-,xshift=2cm] (0,0) -- node[above=-2pt] {\tiny$+$} (1,0);
  \draw[-,xshift=3cm] (0,0) -- (.5,0) node[above=-2pt] {\tiny$+$};
\end{scope}
%
\begin{scope}[yshift=2cm]
  \draw[<-,xshift=-1cm] (.5,0) node[above=-2pt] {\tiny$-$} -- (1,0);
  \draw[-,xshift=0cm] (0,0) -- node[above=-2pt] {\tiny$-$} (1,0);
  \draw[-,xshift=1cm] (0,0) -- node[above=-2pt] {\tiny$-$} (1,0);
  \draw[-,xshift=2cm] (0,0) -- node[above=-2pt] {\tiny$-$} (1,0);
  \draw[-,xshift=3cm] (0,0) -- (.5,0) node[above=-2pt] {\tiny$-$};
\end{scope}
%
\begin{scope}[yshift=3cm]
  \draw[<-,xshift=-1cm] (.5,0) node[above=-2pt] {\tiny$-$} -- (1,0);
  \draw[-,xshift=0cm] (0,0) -- node[above=-2pt] {\tiny$+$} (1,0);
  \draw[-,xshift=1cm] (0,0) -- node[above=-2pt] {\tiny$+$} (1,0);
  \draw[-,xshift=2cm] (0,0) -- node[above=-2pt] {\tiny$+$} (1,0);
  \draw[-,xshift=3cm] (0,0) -- (.5,0) node[above=-2pt] {\tiny$-$};
\end{scope}
\end{tikzpicture}
}
$$
\uncover<2->{Each oxygen atom has two hydrogen atom next to it.}
\uncover<4->{Small arrows on the right figure define the orientation of the lattice lines and vertices, which will be important later.}

\Align*$$
\begin{tikzpicture}[scale=.5,baseline=-2pt]
\draw[-] (0,0) -- (1,0); \draw[fill=black] (0,0) circle (2mm); \draw[fill=black] (1,0) circle (2mm); \draw[-,fill=white] (.65,0) circle (1mm);
\end{tikzpicture}
&\uncover<3->{
\quad=\quad
\begin{tikzpicture}[scale=.5,baseline=-2pt]
\draw[-,postaction=decorate,decoration={markings,mark=at position .7 with {\arrow{>}}}] (0,0) -- (1,0);
\end{tikzpicture}
}
\uncover<4->{
\quad=\quad
\begin{tikzpicture}[scale=.5,baseline=-2pt,>={Stealth[length=1mm]}]
\draw[<-] (0,0) -- node[above=-2pt] {\tiny$+$} (1,0);
\end{tikzpicture}
}
\\
\begin{tikzpicture}[scale=.5,baseline=-2pt]
\draw[-] (0,0) -- (1,0); \draw[fill=black] (0,0) circle (2mm); \draw[fill=black] (1,0) circle (2mm); \draw[-,fill=white] (.35,0) circle (1mm);
\end{tikzpicture}
&\uncover<3->{
\quad=\quad
\begin{tikzpicture}[scale=.5,baseline=-2pt]
\draw[-,postaction=decorate,decoration={markings,mark=at position .7 with {\arrow{>}}}] (1,0) -- (0,0);
\end{tikzpicture}
}
\uncover<4->{
\quad=\quad
\begin{tikzpicture}[scale=.5,baseline=-2pt,>={Stealth[length=1mm]}]
\draw[<-] (0,0) -- node[above=-2pt] {\tiny$-$} (1,0);
\end{tikzpicture}
}
\\
\begin{tikzpicture}[scale=.5,baseline=.25cm-2pt]
\draw[-] (0,0) -- (0,1); \draw[fill=black] (0,0) circle (2mm); \draw[fill=black] (0,1) circle (2mm); \draw[-,fill=white] (0,.65) circle (1mm);
\end{tikzpicture}
&\uncover<3->{
\quad=\quad
\begin{tikzpicture}[scale=.5,baseline=.25cm-2pt]
\draw[-,postaction=decorate,decoration={markings,mark=at position .7 with {\arrow{>}}}] (0,0) -- (0,1);
\end{tikzpicture}
}
\uncover<4->{
\quad=\quad
\begin{tikzpicture}[scale=.5,baseline=.25cm-2pt,>={Stealth[length=1mm]}]
\draw[<-] (0,0) -- node[left=-2pt] {\tiny$+$} (0,1);
\end{tikzpicture}
}
\\
\begin{tikzpicture}[scale=.5,baseline=.25cm-2pt]
\draw[-] (0,0) -- (0,1); \draw[fill=black] (0,0) circle (2mm); \draw[fill=black] (0,1) circle (2mm); \draw[-,fill=white] (0,.35) circle (1mm);
\end{tikzpicture}
&\uncover<3->{
\quad=\quad
\begin{tikzpicture}[scale=.5,baseline=.25cm-2pt]
\draw[-,postaction=decorate,decoration={markings,mark=at position .7 with {\arrow{>}}}] (0,1) -- (0,0);
\end{tikzpicture}
}
\uncover<4->{
\quad=\quad
\begin{tikzpicture}[scale=.5,baseline=.25cm-2pt,>={Stealth[length=1mm]}]
\draw[<-] (0,0) -- node[left=-2pt] {\tiny$-$} (0,1);
\end{tikzpicture}
}
$$

\end{frame}

\begin{frame}
\frametitle{Ice model: Boltzmann weights}

\emph{Six\-/vertex model}: the Boltzmann weights are associated with vertices:
$$
Z=\sum_{\substack{\text{configu-}\\\text{rations}}}\prod_\text{vertices}R^{\ve'_1\ve'_2}_{\ve_1\ve_2},
\quad
R^{\ve'_1\ve'_2}_{\ve_1\ve_2}
=
\begin{tikzpicture}[scale=.5,baseline=-2pt,>={Stealth[length=1mm]}]
\draw[<-] (-.6,0) node[left] {\tiny$\ve_2$} -- (.5,0) node[right] {\tiny$\ve'_2$};
\draw[<-] (0,-.6) node[below] {\tiny$\ve_1$} -- (0,.5) node[above] {\tiny$\ve'_1$};
\end{tikzpicture},
\quad
{\red\mathop{\boxed{\black\ve'_1+\ve'_2=\ve_1+\ve_2}}\limits_\text{\normalsize\strut Ice condition}}.
$$
\pause
We have six vertices
\Align*$$
R^{++}_{++}=a
&=
\begin{tikzpicture}[scale=.5,baseline=-2pt]
\draw[-] (-.5,0) -- (.5,0);
\draw[-] (0,-.5) -- (0,.5);
\draw[fill=black] (0,0) circle (2mm); \draw[-,fill=white] (0,-.35) circle (1mm); \draw[-,fill=white] (-.35,0) circle (1mm);
\end{tikzpicture}
=
\begin{tikzpicture}[scale=.5,baseline=-2pt]
\draw[>->] (-.5,0) -- (.5,0);
\draw[>->] (0,-.5) -- (0,.5);
\end{tikzpicture}
=
\begin{tikzpicture}[scale=.5,baseline=-2pt,>={Stealth[length=1mm]}]
\draw[<-] (-.6,0) node[left] {\tiny$+$} -- (.5,0) node[right] {\tiny$+$};
\draw[<-] (0,-.6) node[below] {\tiny$+$} -- (0,.5) node[above] {\tiny$+$};
\end{tikzpicture},
&R^{--}_{--}=a'
&=
\begin{tikzpicture}[scale=.5,baseline=-2pt]
\draw[-] (-.5,0) -- (.5,0);
\draw[-] (0,-.5) -- (0,.5);
\draw[fill=black] (0,0) circle (2mm); \draw[-,fill=white] (0,.35) circle (1mm); \draw[-,fill=white] (.35,0) circle (1mm);
\end{tikzpicture}
=
\begin{tikzpicture}[scale=.5,baseline=-2pt]
\draw[<-<] (-.5,0) -- (.5,0);
\draw[<-<] (0,-.5) -- (0,.5);
\end{tikzpicture}
=
\begin{tikzpicture}[scale=.5,baseline=-2pt,>={Stealth[length=1mm]}]
\draw[<-] (-.6,0) node[left] {\tiny$-$} -- (.5,0) node[right] {\tiny$-$};
\draw[<-] (0,-.6) node[below] {\tiny$-$} -- (0,.5) node[above] {\tiny$-$};
\end{tikzpicture}
\\
R^{+-}_{+-}=b
&=
\begin{tikzpicture}[scale=.5,baseline=-2pt]
\draw[-] (-.5,0) -- (.5,0);
\draw[-] (0,-.5) -- (0,.5);
\draw[fill=black] (0,0) circle (2mm); \draw[-,fill=white] (0,-.35) circle (1mm); \draw[-,fill=white] (.35,0) circle (1mm);
\end{tikzpicture}
=
\begin{tikzpicture}[scale=.5,baseline=-2pt]
\draw[<-<] (-.5,0) -- (.5,0);
\draw[>->] (0,-.5) -- (0,.5);
\end{tikzpicture}
=
\begin{tikzpicture}[scale=.5,baseline=-2pt,>={Stealth[length=1mm]}]
\draw[<-] (-.6,0) node[left] {\tiny$-$} -- (.5,0) node[right] {\tiny$-$};
\draw[<-] (0,-.6) node[below] {\tiny$+$} -- (0,.5) node[above] {\tiny$+$};
\end{tikzpicture},
&R^{-+}_{-+}=b'
&=
\begin{tikzpicture}[scale=.5,baseline=-2pt]
\draw[-] (-.5,0) -- (.5,0);
\draw[-] (0,-.5) -- (0,.5);
\draw[fill=black] (0,0) circle (2mm); \draw[-,fill=white] (0,.35) circle (1mm); \draw[-,fill=white] (-.35,0) circle (1mm);
\end{tikzpicture}
=
\begin{tikzpicture}[scale=.5,baseline=-2pt]
\draw[>->] (-.5,0) -- (.5,0);
\draw[<-<] (0,-.5) -- (0,.5);
\end{tikzpicture}
=
\begin{tikzpicture}[scale=.5,baseline=-2pt,>={Stealth[length=1mm]}]
\draw[<-] (-.6,0) node[left] {\tiny$+$} -- (.5,0) node[right] {\tiny$+$};
\draw[<-] (0,-.6) node[below] {\tiny$-$} -- (0,.5) node[above] {\tiny$-$};
\end{tikzpicture}
\\
R^{-+}_{+-}=c
&=
\begin{tikzpicture}[scale=.5,baseline=-2pt]
\draw[-] (-.5,0) -- (.5,0);
\draw[-] (0,-.5) -- (0,.5);
\draw[fill=black] (0,0) circle (2mm); \draw[-,fill=white] (0,-.35) circle (1mm); \draw[-,fill=white] (0,.35) circle (1mm);
\end{tikzpicture}
=
\begin{tikzpicture}[scale=.5,baseline=-2pt]
\draw[<->] (-.5,0) -- (.5,0);
\draw[>-<] (0,-.5) -- (0,.5);
\end{tikzpicture}
=
\begin{tikzpicture}[scale=.5,baseline=-2pt,>={Stealth[length=1mm]}]
\draw[<-] (-.6,0) node[left] {\tiny$-$} -- (.5,0) node[right] {\tiny$+$};
\draw[<-] (0,-.6) node[below] {\tiny$+$} -- (0,.5) node[above] {\tiny$-$};
\end{tikzpicture},
&R^{+-}_{-+}=c'
&=
\begin{tikzpicture}[scale=.5,baseline=-2pt]
\draw[-] (-.5,0) -- (.5,0);
\draw[-] (0,-.5) -- (0,.5);
\draw[fill=black] (0,0) circle (2mm); \draw[-,fill=white] (-.35,0) circle (1mm); \draw[-,fill=white] (.35,0) circle (1mm);
\end{tikzpicture}
=
\begin{tikzpicture}[scale=.5,baseline=-2pt]
\draw[>-<] (-.5,0) -- (.5,0);
\draw[<->] (0,-.5) -- (0,.5);
\end{tikzpicture}
=
\begin{tikzpicture}[scale=.5,baseline=-2pt,>={Stealth[length=1mm]}]
\draw[<-] (-.6,0) node[left] {\tiny$+$} -- (.5,0) node[right] {\tiny$-$};
\draw[<-] (0,-.6) node[below] {\tiny$-$} -- (0,.5) node[above] {\tiny$+$};
\end{tikzpicture}
$$
\pause
$$
R=\pMatrix{a&&&\\&b&c&\\&c'&b'&\\&&&a'}\
\text{in the basis $(++),(+-),(-+),(--)$.}
$$

\end{frame}

\begin{frame}
\frametitle{Six-vertex models: solvable case}

The six\-/vertex model is \emph{solvable}, if
$$
R^{-\ve'_1\>-\ve'_2}_{-\ve_1\>-\ve_2}=R^{\ve'_1\ve'_2}_{\ve_1\ve_2}
$$
or
$$
a'=a,
\quad
b'=b
\quad
c'=c.
$$
\pause
The transfer matrix
\eq$$
T^{\ve'_1\ldots\ve'_N}_{\ve_1\ldots\ve_N}
=\sum_{\mu_1\ldots\mu_N}R^{\mu_2\ve'_1}_{\mu_1\ve_1}
R^{\mu_3\ve'_2}_{\mu_2\ve_2}\ldots R^{\mu_1\ve'_N}_{\mu_N\ve_N}.
\label{transfermatrixdef}
$$
\pause
Let us consider the matrix $R$ as an operator in the tensor product of two two\-/dimensional spaces:
$$
R:\C^2\otimes\C^2\to\C^2\otimes\C^2,
\qquad
v_{\ve_1}\otimes v_{\ve_2}\mapsto
R^{\ve_1\ve_2}_{\ve'_1\ve'_2}v_{\ve'_1}\otimes v_{\ve'_2}.
$$
Here $v_\ve$ is the natural basis in $V=\C^2$.
\pause
Consider the tensor product $V_1\otimes V_2\otimes\cdots\otimes V_k$ of identical spaces $V_i\simeq V$. Let $R_{ij}$ is the $R$ matrix acting on $V_i\otimes V_j$.
\pause

Then the transfer matrix can be written as
\eq$$
T=\tr_{V_0}(R_{0N}\ldots R_{02}R_{01}):\
V_1\otimes V_2\otimes\cdots\otimes V_N\to V_1\otimes V_2\otimes\cdots\otimes V_N.
\label{transfermatrixtr}
$$
The space $V_1\otimes\cdots\otimes V_N$ is called \emph{quantum space}, while the space $V_0$ is called \emph{auxiliary space}.

\end{frame}

\begin{frame}
\frametitle{Six-vertex models: $L$ operator}

The operator under the trace is
\eq$$
L=R_{0N}\ldots R_{02}R_{01}:\ V_0\otimes V_1\otimes\cdots\otimes V_N\to V_0\otimes V_1\otimes\cdots\otimes V_N.
\label{Ldef}
$$
\pause
We will consider it as an operator in the quantum space and a matrix in the auxiliary space
$$
L=\pMatrix{A&B\\ C&D},
\quad
A,B,C,D:V_1\otimes V_2\otimes\cdots\otimes V_N\to V_1\otimes V_2\otimes\cdots\otimes V_N.
$$
\pause
Then
\eq$$
T=\tr_{V_0}L=A+D.
\label{transfermatrixAplusD}
$$

\end{frame}

\begin{frame}
\frametitle{Commuting transfer matrices and Yang\--Baxter equation}

Integrability demands the existence of extra commuting integrals of motion $I_n$:
$$
[T,I_n]=0,
\quad
[I_m,I_n]=0.
$$
How to construct them?
\pause

Let use search for the operators $T'=\tr_{V_0}L'$, $L'=R'_{0N}\ldots R'_{02}R'_{01}$ with some matrix $R'$.
\pause

\begin{block}{Theorem}
If there exist nondegenerate matrices $R'$, $R''$ such that
\eq$$
R''_{12}R'_{13}R_{23}=R_{23}R'_{13}R''_{12},
\label{YBE}
$$
or, graphically
\eq$$
\begin{tikzpicture}[baseline=-2pt]
\draw[->,name path=l1] (-.25,1) -- (.75,-1);
\draw[->,name path=l2] (.25,1) -- (-.75,-1);
\draw[->,name path=l3] (1,-.5) -- (-1,-.5);
\begin{scope}[name intersections={of=l1 and l2,by=R11},name intersections={of=l1 and l3,by=R1},name intersections={of=l2 and l3,by=R}]
\draw[-] (R) node[below] {$\quad R$};
\draw[-] (R1) node[below] {$R'\quad$};
\draw[-] (R11) node[left] {$R''$};
\end{scope}
\end{tikzpicture}
\quad=\quad
\begin{tikzpicture}[baseline=-2pt]
\draw[->,name path=l1] (-.75,1) -- (.25,-1);
\draw[->,name path=l2] (.75,1) -- (-.25,-1);
\draw[->,name path=l3] (1,.5) -- (-1,.5);
\begin{scope}[name intersections={of=l1 and l2,by=R11},name intersections={of=l1 and l3,by=R1},name intersections={of=l2 and l3,by=R}]
\draw[-] (R) node[above] {$R\quad$};
\draw[-] (R1) node[above] {$\quad R'$};
\draw[-] (R11) node[left] {$R''$};
\end{scope}
\end{tikzpicture}
\tag{$\ref{YBE}'$}
$$
then
\eq$$
[T,T']=0
\label{TT'-commut}
$$
\end{block}

\end{frame}

\begin{frame}
\frametitle{Commuting transfer matrices: a proof}

A graphical proof:
\Multline$$
T'T=
\\
\begin{tikzpicture}[scale=.5,baseline=1cm-2pt]
\draw[-] (0,4.5) -- (0,2.5); \draw[dashed] (0,2.5) -- (0,.5); \draw[->] (0,.5) -- (0,-.6) node[below] {$L$};
\draw[-] (1,4.5) -- (1,2.5); \draw[dashed] (1,2.5) -- (1,.5); \draw[->] (1,.5) -- (1,-.6) node[below] {$L'$};
\draw[->] (1.5,4) -- (-.6,4);
\draw[->] (1.5,3) -- (-.6,3);
\draw[->] (1.5,0) -- (-.6,0);
\end{tikzpicture}
\ =\
\begin{tikzpicture}[scale=.5,baseline=1cm-2pt]
\draw[-] (0,6) to[out=-90,in=90] (1,5) to[out=-90,in=90] (0,4); \draw[-] (.5,5.5) node[right=4pt] {\rlap{$R^{\prime\prime-1}$}};
\draw[-] (1,6) to[out=-90,in=90] (0,5) to[out=-90,in=90] (1,4); \draw[-] (.5,4.5) node[right=4pt] {\rlap{$R''$}};
\draw[-] (0,4) -- (0,2.5); \draw[dashed] (0,2.5) -- (0,.5); \draw[->] (0,.5) -- (0,-.6) node[below] {$L$};
\draw[-] (1,4) -- (1,2.5); \draw[dashed] (1,2.5) -- (1,.5); \draw[->] (1,.5) -- (1,-.6) node[below] {$L'$};
\draw[->] (1.5,4) -- (-.6,4);
\draw[->] (1.5,3) -- (-.6,3);
\draw[->] (1.5,0) -- (-.6,0);
\end{tikzpicture}
\ =\
\begin{tikzpicture}[scale=.5,baseline=1cm-2pt]
\draw[-] (0,5) to[out=-90,in=90] (1,4) to[out=-90,in=90] (0,3); \draw[-] (.5,4.5) node[right=4pt] {\rlap{$R^{\prime\prime-1}$}};
\draw[-] (1,5) to[out=-90,in=90] (0,4) to[out=-90,in=90] (1,3); \draw[-] (.5,3.5) node[right=4pt] {\rlap{$R''$}};
\draw[-] (0,3) -- (0,2.5); \draw[dashed] (0,2.5) -- (0,.5); \draw[->] (0,.5) -- (0,-.6);
\draw[-] (1,3) -- (1,2.5); \draw[dashed] (1,2.5) -- (1,.5); \draw[->] (1,.5) -- (1,-.6);
\draw[->] (1.5,4) -- (-.6,4);
\draw[->] (1.5,3) -- (-.6,3);
\draw[->] (1.5,0) -- (-.6,0);
\end{tikzpicture}
\ =\
\begin{tikzpicture}[scale=.5,baseline=1cm-2pt]
\draw[-] (0,5) to[out=-90,in=90] (1,4); \draw[-] (1,4) -- (1,3) to[out=-90,in=90] (0,2); \draw[-] (.5,4.5) node[right=4pt] {\rlap{$R^{\prime\prime-1}$}};
\draw[-] (1,5) to[out=-90,in=90] (0,4); \draw[-] (0,4) -- (0,3) to[out=-90,in=90] (1,2); \draw[-] (.5,2.5) node[right=4pt] {\rlap{$R''$}};
\draw[dashed] (0,2) -- (0,.5); \draw[->] (0,.5) -- (0,-.6);
\draw[dashed] (1,2) -- (1,.5); \draw[->] (1,.5) -- (1,-.6);
\draw[->] (1.5,4) -- (-.6,4);
\draw[->] (1.5,3) -- (-.6,3);
\draw[->] (1.5,0) -- (-.6,0);
\end{tikzpicture}
\ =\
\begin{tikzpicture}[scale=.5,baseline=1cm-2pt]
\draw[-] (0,5) to[out=-90,in=90] (1,4);
\draw[-] (1,5) to[out=-90,in=90] (0,4);
\draw[-] (0,4) -- (0,2.5); \draw[dashed] (0,2.5) -- (0,.5); \draw[-] (0,.5) -- (0,0) node[above left] {$L'$};
  \draw[->] (0,0) to[out=-90,in=90] (1,-1) -- (1,-1.3);
\draw[-] (1,4) -- (1,2.5); \draw[dashed] (1,2.5) -- (1,.5); \draw[-] (1,.5) -- (1,0) node[above right] {$L$};
  \draw[->] (1,0) to[out=-90,in=90] (0,-1) -- (0,-1.3);
\draw[-] (.5,4.5) node[right=4pt] {\rlap{$R^{\prime\prime-1}$}};
\draw[-] (.5,-.5) node[right=4pt] {\rlap{$R''$}};
\draw[->] (1.5,4) -- (-.6,4);
\draw[->] (1.5,3) -- (-.6,3);
\draw[->] (1.5,0) -- (-.6,0);
\end{tikzpicture}
\ =\
\begin{tikzpicture}[scale=.5,baseline=1cm-2pt]
\draw[-] (0,6) to[out=-90,in=90] (1,5) to[out=-90,in=90] (0,4); \draw[-] (.5,5.5) node[right=4pt] {\rlap{$R''$}};
\draw[-] (1,6) to[out=-90,in=90] (0,5) to[out=-90,in=90] (1,4); \draw[-] (.5,4.5) node[right=4pt] {\rlap{$R^{\prime\prime-1}$}};
\draw[-] (0,4) -- (0,2.5); \draw[dashed] (0,2.5) -- (0,.5); \draw[->] (0,.5) -- (0,-.6) node[below] {$L'$};
\draw[-] (1,4) -- (1,2.5); \draw[dashed] (1,2.5) -- (1,.5); \draw[->] (1,.5) -- (1,-.6) node[below] {$L$};
\draw[->] (1.5,4) -- (-.6,4);
\draw[->] (1.5,3) -- (-.6,3);
\draw[->] (1.5,0) -- (-.6,0);
\end{tikzpicture}
\ =\
\begin{tikzpicture}[scale=.5,baseline=1cm-2pt]
\draw[-] (0,4.5) -- (0,2.5); \draw[dashed] (0,2.5) -- (0,.5); \draw[->] (0,.5) -- (0,-.6) node[below] {$L'$};
\draw[-] (1,4.5) -- (1,2.5); \draw[dashed] (1,2.5) -- (1,.5); \draw[->] (1,.5) -- (1,-.6) node[below] {$L$};
\draw[->] (1.5,4) -- (-.6,4);
\draw[->] (1.5,3) -- (-.6,3);
\draw[->] (1.5,0) -- (-.6,0);
\end{tikzpicture}
\\
=TT'.
$$
\pause
A more conventional proof is based on the relation
$$
R''_{12}L'_1L_2=L_2L'_1R''_{12},
$$
which is proved by induction.
\pause
Then
$$
\Aligned{
T'T
&=\tr_{V_1\otimes V_2}(L'_1L_2)
=\tr_{V_1\otimes V_2}((R''_{12})^{-1}R''_{12}L'_1L_2)
=\tr_{V_1\otimes V_2}((R''_{12})^{-1}L_2L'_1R''_{12})
\\
&=\tr_{V_1\otimes V_2}(R''_{12}(R''_{12})^{-1}L_2L'_1)
=\tr_{V_1\otimes V_2}(L_2L'_1)=TT'.
}
$$

\end{frame}

\begin{frame}
\frametitle{Yang\--Baxter equation: solution for the six\-/vertex model}

The solution can be found in the form
\eq$$
\Aligned{
R
&=R(\lambda,u_2-u_3),
\\
R'
&=R(\lambda,u_1-u_3),
\\
R''
&=R(\lambda,u_1-u_2)
}\label{YBEsolution}
$$
with a given matrix\-/valued function $R(\lambda,u)$.
\pause
Since the common factor of $a,b,c$ is arbitrary, assume $a=1$. Trigonometric solution(s):
$$
\Aligned{
b(u)
&={\sin u\over\sin(\lambda-u)},
&b(u)
&={\sh u\over\sh(\lambda-u)},
\\
c(u)
&={\sin\lambda\over\sin(\lambda-u)}
&c(u)
&={\sh\lambda\over\sh(\lambda-u)}
\\
\span(a<b+c,\ b<a+c,\ c<a+b);\quad
&\span(c>a+b).
}
$$
\pause
The cases $a>b+c$ and $b>a+c$ and not interesting from the thermodynamic point of view and will be discussed later.
\pause
The parameter $\lambda$ is the same for $R,R',R''$ and can be expressed as
\eq$$
\left.\Matrix{-\cos\lambda\\-\ch\lambda}\right\}=\Delta\equiv{a^2+b^2-c^2\over2ab}.
\label{Deltadef}
$$
\pause
Thus we will omit the parameter $\lambda$ from now on:
$$
R(u)\equiv R(\lambda,u),\ a(u)\equiv a(\lambda,u)\text{ etc.}
$$

\end{frame}

\begin{frame}
\frametitle{Yang\--Baxter equation: spectral parameter}

The spectral parameters can be associated to lines:
$$
R(\lambda,u-v)^{\ve_3\ve_4}_{\ve_1\ve_2}
=\begin{tikzpicture}[baseline=-2pt]
\draw[<-] (-.6,0) node[left] {$\ve_2$} -- node[pos=.25,above] {$v$} (.5,0) node[right] {$\ve_4$};
\draw[<-] (0,-.6) node[below] {$\ve_1$} -- node[pos=.25,right] {$u$} (0,.5) node[above] {$\ve_3$};
\end{tikzpicture}
$$
\pause
This $R$ matrix is the solution to the Yang\--Baxter equation in the form
\Multline$$
R_{12}(\lambda,u_1-u_2)R_{13}(\lambda,u_1-u_3)R_{23}(\lambda,u_2-u_3)
\\
=R_{23}(\lambda,u_2-u_3)R_{13}(\lambda,u_1-u_3)R_{12}(\lambda,u_1-u_2).
\label{YBrel}
$$
Graphically:
\eq$$
\begin{tikzpicture}[baseline=-2pt]
\draw[->,name path=l1] (-.25,1) -- (.75,-1) node[below] {$\ u_1$};
\draw[->,name path=l2] (.25,1) -- (-.75,-1) node[below] {$u_2$};
\draw[->,name path=l3] (1,-.5) -- (-1,-.5) node[left] {$u_3$};
\end{tikzpicture}
\quad=\quad
\begin{tikzpicture}[baseline=-2pt]
\draw[->,name path=l1] (-.75,1) -- (.25,-1) node[below] {$\ u_1$};
\draw[->,name path=l2] (.75,1) -- (-.25,-1) node[below] {$u_2$};
\draw[->,name path=l3] (1,.5) -- (-1,.5) node[left] {$u_3$};
\end{tikzpicture}
\tag{$\ref{YBrel}'$}
$$
\pause
Besides, the $R$ matrix satisfy the relations
\eq$$
b(u)R(\lambda-u)^{\ve_3\ve_4}_{\ve_1\ve_2}=R(u)^{\ve_2\>-\ve_3}_{\ve_4\>-\ve_1},
\qquad
R_{12}(u)R_{21}(-u)=1,
\qquad
R(0)=P=\tikz[baseline=-2pt,>={Stealth[length=1mm]},scale=.5]{
  \draw[->] (0,.5) -- (0,.2) to[out=-90,in=0] (-.2,0) -- (-.5,0);
  \draw[->] (.5,0) -- (.2,0) to[out=180,in=90] (0,-.2) -- (0,-.5);
}\ .
\label{addrels}
$$

\end{frame}

\begin{frame}
\frametitle{Integrals of motion}

We have
\eq$$
[T(u),T(u')]=0
\quad\forall u,u'.
\label{commfam}
$$
But not all the integrals of motion $T(u)$ are independent.
\pause

First of all, $T(0)$ is nothing but the shift operator:
\eq$$
T(0)=\
\begin{tikzpicture}[scale=.5,baseline=1cm-2pt]
\draw[-] (0,4.8) -- (0,2.5); \draw[dashed] (0,2.5) -- (0,0.5); \draw[->] (0,.5) -- (0,-.8) node[below] {$u$};
\draw[->] (.8,4) -- (-.8,4) node[left] {$u$};
\draw[->] (.8,3) -- (-.8,3) node[left] {$u$};
\draw[->] (.8,0) -- (-.8,0) node[left] {$u$};
\end{tikzpicture}
\ =\
\begin{tikzpicture}[scale=.5,baseline=1cm-2pt]
\draw[->] (0,4.8) -- (0,4.2) to[out=-90,in=0] (-.2,4) -- (-.8,4);
\draw[->] (.8,4) -- (.2,4) to[out=180,in=90] (0,3.8) -- (0,3.2) to[out=-90,in=0] (-.2,3) -- (-.8,3);
\draw[-] (.8,3) -- (.2,3) to[out=180,in=90] (0,2.8) -- (0,2.5); \draw[dashed] (0,2.5) -- (0,0.5);
  \draw[->] (0,.5) -- (0,.2) to[out=-90,in=0] (-.2,0) -- (-.8,0);
\draw[->] (.8,0) -- (.2,0) to[out=180,in=90] (0,-.2) -- (0,-.8);
\end{tikzpicture}
\label{T0-shift}
$$
\pause

Then decompose the product $T^{-1}(0)T(u)$ in~$u$:
\eq$$
T^{-1}(0)T(u)=1-\sum_{n=1}^\infty{H_nu^n\over n!}.
\label{T-decom}
$$
\pause
Hamiltonians $H_n$ commute with $T(u)$ and mutually commute:
\eq$$
[T(0),H_n]=[H_m,H_n]=0
\quad\forall m,n.
\label{Hn-commut}
$$
The set $T(0)$, $H_1,\ldots,H_{N-1}$ form a set of independent integrals of motion.
\pause
Operators $H_n$ are \emph{local} in the sense that each of them is a sum of term, which involves a finite number ($n+1$) of neighboring nodes.

\end{frame}

\begin{frame}
\frametitle{Six\-/vertex model and XXZ Heisenberg chain}

Let us find the Hamiltonian $H_1$ explicitly:
\Multline*$$
-H_1=T^{-1}(0)T'(0)=
\\
\left.d\over du\right|_{u=0}\begin{tikzpicture}[scale=.5,baseline=1cm-2pt]
\draw[<-] (1,4.8) node[above] {$0$} -- (1,4.2) to[out=-90,in=180] (1.2,4) -- (1.8,4);
\draw[<-] (-.8,4) node[left] {$0$} -- (.8,4) to[out=0,in=90] (1,3.8) -- (1,3.2) to[out=-90,in=180] (1.2,3) -- (1.8,3);
\draw[<-] (-.8,3) node[left] {$0$} -- (.8,3) to[out=0,in=90] (1,2.8) -- (1,2.5); \draw[dashed] (1,2.5) -- (1,0.5);
  \draw[-] (1,.5) -- (1,.2) to[out=-90,in=180] (1.2,0) -- (1.8,0);
\draw[<-] (-.8,0) node[left] {$0$} -- (.8,0) to[out=0,in=90] (1,-.2) -- (1,-.8);
\draw[-] (0,4.8) -- (0,2.5); \draw[dashed] (0,2.5) -- (0,0.5); \draw[->] (0,.5) -- (0,-.8) node[below] {$u$};
\end{tikzpicture}
=\left.d\over du\right|_{u=0}\sum^N_{n=1}
\vbox spread -1cm{\vss\hbox{\begin{tikzpicture}[scale=.5,baseline=1.5cm-2pt]
\draw[<-] (-.8,6) node[above] {\tiny$V_N$} -- (-.2,6) to[out=0,in=-90] (0,6.2) -- (0,6.8) to[out=90,in=180] (.2,7) -- (.8,7) to[out=0,in=90]
  (1,6.8) -- (1,6.2) to[out=-90,in=180] (1.2,6) -- (1.8,6);
\draw[<-] (-.8,4) node[above] {\tiny$V_{n+2}$} -- (-.2,4) to[out=0,in=-90] (0,4.2) -- (0,4.5); \draw[dashed] (0,4.5) -- (0,5.5);
  \draw[-] (0,5.5) -- (0,5.8) to[out=90,in=180] (.2,6) -- (.8,6) to[out=0,in=-90] (1,5.8) -- (1,5.5); \draw[dashed] (1,5.5) -- (1,4.5);
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\draw[<-] (-.8,3) node[above] {\tiny$V_{n+1}$} -- (.8,3) to[out=0,in=90] (1,2.8) -- (1,2.2) to[out=-90,in=180] (1.2,2) -- (1.8,2);
\draw[<-] (-.8,2) node[above] {\tiny$V_n$} -- (-.2,2) to[out=0,in=-90] (0,2.2) -- (0,3.8) to[out=90,in=180] (.2,4)
  -- (.8,4) to[out=0,in=90] (1,3.8) -- (1,3.2) to[out=-90,in=180] (1.2,3) -- (1.8,3);
\draw[<-] (-.8,0) node[above] {\tiny$V_1$} -- (-.2,0) to[out=0,in=-90] (0,.2) -- (0,.5); \draw[dashed] (0,.5) -- (0,1.5);
  \draw[-] (0,1.5) -- (0,1.8) to[out=90,in=180] (.2,2) -- (.8,2) to[out=0,in=90] (1,1.8) -- (1,1.5); \draw[dashed] (1,1.5) -- (1,.5);
  \draw[-] (1,0.5) -- (1,0.2) to[out=-90,in=180] (1.2,0) -- (1.8,0);
\end{tikzpicture}}}
=\left.d\over du\right|_{u=0}\sum^N_{n=1}
\begin{tikzpicture}[scale=.5,baseline=1.5cm-2pt]
\draw[<-] (-.8,6) node[above] {\tiny$V_N$} -- (1.8,6);
\path (.5,6) -- node[pos=.4] {$\vdots$} (.5,4);
\draw[<-] (-.8,4) node[above] {\tiny$V_{n+2}$} -- (1.8,4);
\draw[<-] (-.8,3) node[above] {\tiny$V_{n+1}$} -- (0,3) to[out=0,in=180] (1,2) -- (1.8,2);
\draw[<-] (-.8,2) node[above] {\tiny$V_n$} -- (0,2) to[out=0,in=180] (1,3) -- (1.8,3);
\draw[-] (.5,2.5)+(240:.3) arc (240:120:.3) node[midway,left=-2pt] {$u$};
\path (.5,2) -- node[pos=.4] {$\vdots$} (.5,0);
\draw[<-] (-.8,0) node[above] {\tiny$V_1$} -- (1.8,0);
\end{tikzpicture}
\\
=\sum^N_{n=1}\check R'_{n,n+1}(0),
$$
\pause
where
$$
\Aligned{
\check R(u)
&\arraycolsep=2pt\def\arraystretch{.9}=PR(u)
=\pMatrix{a(u)&&&\\&c(u)&b(u)&\\&b(u)&c(u)&\\&&&a(u)}
=1+{u\over\sin\lambda}\pMatrix{0&&&\\&\cos\lambda&1&\\&1&\cos\lambda&\\&&&0}
+O(u^2)
\\
&=1-{u\over\sin\lambda}\left(h-{\cos\lambda\over2}\right)+O(u^2),
}
$$

\end{frame}

\begin{frame}
\frametitle{Six\-/vertex model and XXZ Heisenberg chain}

Here
$$
h=-{1\over2}(\sigma^x\otimes\sigma^x+\sigma^y\otimes\sigma^y-\cos\lambda\>\sigma^z\otimes\sigma^z).
$$
Hence
$$
H_1\sin\lambda=H_{\rm XXZ}+{N\Delta\over2},
$$
where $H_{\rm XXZ}$ is the Hamiltonian of the \emph{XXZ Heisenberg} chain:
\eq$$
H_{\rm XXZ}=-{1\over2}\sum_{n=1}^N(\sigma^x_n\sigma^x_{n+1}+\sigma^y_n\sigma^y_{n+1}+\Delta\sigma^z_n\sigma^z_{n+1})
\label{XXZdef}
$$
with $\Delta$ given by~(\ref{Deltadef}):
$$
\Delta={a^2+b^2-c^2\over2ab}=\Cases{-\cos\lambda\\-\ch\lambda}.
$$

\end{frame}

\begin{frame}
\frametitle{XXZ Heisenberg chain: pseudovacuums}

Due to the ice condition the $z$ component of total spin
$$
S^z={1\over2}\sum^N_{i=1}\sigma^z_n
$$
is a conserved charge:
\eq$$
[T(u),S^z]=[H_{\rm XXZ},S^z]=0.
\label{Szcommut}
$$
Thus the space of states is split into the sum over eigenvalues of~$S^z$.
\pause

Define the \emph{pseudovacuums}
\eq$$
|\Omega_\pm\rangle
=\underbrace{v_\pm\otimes v_\pm\otimes\ldots\otimes v_\pm}_N.
\label{Omegadef}
$$
\pause
\rlap{\hskip\textwidth\llap{\red
\begin{tikzpicture}[overlay,baseline=0cm,scale=.4,>={Stealth[length=1mm]}]
\draw[<-] (-.5,-1) node[left] {\tiny$+$} -- (.5,-1) node[right] {\tiny$+$};
\draw[<-] (-.5,-2) node[left] {\tiny$+$} -- (.5,-2) node[right] {\tiny$+$};
\draw[<-] (-.5,-3) node[left] {\tiny$+$} -- (.5,-3) node[right] {\tiny$+$};
\draw[<-] (-.5,-4) node[left] {\tiny$+$} -- (.5,-4) node[right] {\tiny$+$};
\draw[->] (0,-.5) node[above] {\tiny$\pm$} -- (0,-4.5) node[below] {\tiny$\pm$};
\draw (0,1) node {$T(u)|\Omega_+\rangle$};
\end{tikzpicture}
}}%
Evidently,
\Gather*$$
S^z|\Omega_\pm\rangle=\pm{N\over2}|\Omega_\pm\rangle,
\qquad
T(u)|\Omega_\pm\rangle=(a^N(u)+b^N(u))|\Omega_\pm\rangle,
\\
H_{\it XXZ}|\Omega_\pm\rangle=-{N\Delta\over2}|\Omega_\pm\rangle.
$$

\end{frame}

\begin{frame}
\frametitle{XXZ Heisenberg chain: pseudoparticles}

States of fixed spin $S^z=N/2-k$ are linear combinations of the states
\eq$$
|n_1,\ldots,n_k\rangle
=\sigma^-_{n_1}\ldots\sigma^-_{n_k}|\Omega_+\rangle,
\qquad
\sigma^{\pm}={\sigma^x\pm i\sigma^y\over2}.
\label{basicvectors}
$$
\pause
Consider $k=1$. The state
\eq$$
|\Psi_1(z)\rangle
=\sum_n z^n|n\rangle.
\label{Psi1}
$$
\pause
is an eigenvector of the Hamiltonian,
\eq$$
H_{\it XXZ}|\Psi_1(z)\rangle
=\left(-{N\Delta\over2}+\epsilon(z)\right)|\Psi_1(z)\rangle,
\qquad
\epsilon(z)=2\Delta-z-z^{-1},
\label{HPsi1}
$$
if $z^N=1$.
\pause
Three regimes:
\begin{itemize}
\item $\Delta>1$: $\epsilon(z)>0$ $\forall z$. The states $|\Omega_\pm\rangle$ are ground states. The excitation are physical excitations (\emph{magnons}).
\pause
\item $\Delta<-1$: $\epsilon(z)<0$ $\forall z$. The states $|\Omega_\pm\rangle$ are states of the \emph{highest} energy. The ground state corresponds to $S^z=0$ or $\pm{1\over2}$, and excited states separated by an \emph{energy gap}.
\pause
\item $-1<\Delta<1$: $\epsilon(z)$ does not have definite sign. The ground state corresponds to $S^z=0$ or $\pm{1\over2}$. \emph{No energy gap}.
\pause

\end{itemize}

\end{frame}

\begin{frame}
\frametitle{Six\-/vertex model: three regimes}

1.~\emph{Ferroelectric} regime: $\Delta>0$. Let $a>b+c$. \only<1>{Ground configurations:}\only<2>{Excitations?}\only<3->{Excitations:}
$$
\begin{tikzpicture}[scale=.6,>={Stealth[length=1mm]},baseline=.6*1.5cm-2pt]
\foreach \i in {0,...,3} {
  \draw[->] (\i,3.5) node[right=-2pt] {\tiny$+$} -- (\i,3) -- node[right=-2pt] {\tiny$+$} (\i,2) -- node[right=-2pt] {\tiny$+$} (\i,1)
    -- node[right=-2pt] {\tiny$+$} (\i,0) -- (\i,-.5) node[right] {\tiny$+$};
}
\foreach \i in {0,...,3} {
  \draw[->] (3.5,\i) node[above=-2pt] {\tiny$+$} -- (3,\i) -- node[above=-2pt] {\tiny$+$} (2,\i) -- node[above=-2pt] {\tiny$+$} (1,\i)
    -- node[above=-2pt] {\tiny$+$} (0,\i) -- (-.5,\i) node[above=-2pt] {\tiny$+$};
}
\uncover<2->{
  \draw[fill,color=white] (1.5,1cm+5pt) circle (5pt);
  \draw[-,thick,color=red] (1,1) -- node[above=-2pt] {\tiny$-$} (2,1);
}
\uncover<3->{
  \draw[fill,color=white] (.5,1cm+5pt) circle (5pt);
  \draw[-,thick,color=red] (0,1) -- node[above=-2pt] {\tiny$-$} (1,1);
  \draw[fill,color=white] (5pt,.5) circle (5pt);
  \draw[-,thick,color=red] (0,0) -- node[right=-2pt] {\tiny$-$} (0,1);
  \draw[fill,color=white] (-.5,5pt) circle (5pt);
  \draw[<-,thick,color=red] (-.5,0) node[above=-2pt] {\tiny$-$} -- (0,0);
  \draw[fill,color=white] (2cm+5pt,1.5) circle (5pt);
  \draw[-,thick,color=red] (2,1) -- node[right=-2pt] {\tiny$-$} (2,2);
  \draw[fill,color=white] (2.5,2cm+5pt) circle (5pt);
  \draw[-,thick,color=red] (2,2) -- node[above=-2pt] {\tiny$-$} (3,2);
  \draw[fill,color=white] (3cm+5pt,2.5) circle (5pt);
  \draw[-,thick,color=red] (3,2) -- node[right=-2pt] {\tiny$-$} (3,3);
  \draw[fill,color=white] (3cm+5pt,3.5) circle (5pt);
  \draw[-,thick,color=red] (3,3) -- (3,3.5) node[right=-2pt] {\tiny$-$};
}
\end{tikzpicture}
\quad\text{and}\quad
\begin{tikzpicture}[scale=.6,>={Stealth[length=1mm]},baseline=.6*1.5cm-2pt]
\foreach \i in {0,...,3} {
  \draw[->] (\i,3.5) node[right=-2pt] {\tiny$-$} -- (\i,3) -- node[right=-2pt] {\tiny$-$} (\i,2) -- node[right=-2pt] {\tiny$-$} (\i,1)
    -- node[right=-2pt] {\tiny$-$} (\i,0) -- (\i,-.5) node[right] {\tiny$-$};
}
\foreach \i in {0,...,3} {
  \draw[->] (3.5,\i) node[above=-2pt] {\tiny$-$} -- (3,\i) -- node[above=-2pt] {\tiny$-$} (2,\i) -- node[above=-2pt] {\tiny$-$} (1,\i)
    -- node[above=-2pt] {\tiny$-$} (0,\i) -- (-.5,\i) node[above=-2pt] {\tiny$-$};
}
\end{tikzpicture}
$$
\uncover<4->{On a large lattice any excitations have vanishing weight. $\Rightarrow$ \emph{Frozen order}.}

\uncover<5->{
2.~\emph{Antiferroelectric} regime: $\Delta<-1$, $c>a+b$. \only<5>{Ground configurations:}\only<6>{Excitations?}\only<7->{Excitations:}
$$
\begin{tikzpicture}[scale=.6,>={Stealth[length=1mm]},baseline=.6*1.5cm-2pt]
\foreach \i in {3,1} {
  \draw[->] (\i,3.5) node[right=-2pt] {\tiny$+$} -- (\i,3) -- node[right=-2pt] {\tiny$-$} (\i,2) -- node[right=-2pt] {\tiny$+$} (\i,1)
    -- node[right=-2pt] {\tiny$-$} (\i,0) -- (\i,-.5) node[right] {\tiny$+$};
}
\foreach \i in {2,0} {
  \draw[->] (\i,3.5) node[right=-2pt] {\tiny$-$} -- (\i,3) -- node[right=-2pt] {\tiny$+$} (\i,2) -- node[right=-2pt] {\tiny$-$} (\i,1)
    -- node[right=-2pt] {\tiny$+$} (\i,0) -- (\i,-.5) node[right] {\tiny$-$};
}
\foreach \i in {3,1} {
  \draw[->] (3.5,\i) node[above=-2pt] {\tiny$-$} -- (3,\i) -- node[above=-2pt] {\tiny$+$} (2,\i) -- node[above=-2pt] {\tiny$-$} (1,\i)
    -- node[above=-2pt] {\tiny$+$} (0,\i) -- (-.5,\i) node[above=-2pt] {\tiny$-$};
}
\foreach \i in {2,0} {
  \draw[->] (3.5,\i) node[above=-2pt] {\tiny$+$} -- (3,\i) -- node[above=-2pt] {\tiny$-$} (2,\i) -- node[above=-2pt] {\tiny$+$} (1,\i)
    -- node[above=-2pt] {\tiny$-$} (0,\i) -- (-.5,\i) node[above=-2pt] {\tiny$+$};
}
\uncover<6->{
  \draw[fill,color=white] (1.5,1cm+5pt) circle (5pt);
  \draw[-,thick,color=red] (1,1) -- node[above=-2pt] {\tiny$+$} (2,1);
}
\uncover<7->{
  \draw[fill,color=white] (1cm+5pt,1.5) circle (5pt);
  \draw[-,thick,color=red] (1,1) -- node[right=-2pt] {\tiny$-$} (1,2);
  \draw[fill,color=white] (1.5,2cm+5pt) circle (5pt);
  \draw[-,thick,color=red] (1,2) -- node[above=-2pt] {\tiny$-$} (2,2);
  \draw[fill,color=white] (2cm+5pt,1.5) circle (5pt);
  \draw[-,thick,color=red] (2,1) -- node[right=-2pt] {\tiny$+$} (2,2);
}
\end{tikzpicture}
\quad\text{and}\quad
\begin{tikzpicture}[scale=.6,>={Stealth[length=1mm]},baseline=.6*1.5cm-2pt]
\foreach \i in {3,1} {
  \draw[->] (\i,3.5) node[right=-2pt] {\tiny$-$} -- (\i,3) -- node[right=-2pt] {\tiny$+$} (\i,2) -- node[right=-2pt] {\tiny$-$} (\i,1)
    -- node[right=-2pt] {\tiny$+$} (\i,0) -- (\i,-.5) node[right] {\tiny$-$};
}
\foreach \i in {2,0} {
  \draw[->] (\i,3.5) node[right=-2pt] {\tiny$+$} -- (\i,3) -- node[right=-2pt] {\tiny$-$} (\i,2) -- node[right=-2pt] {\tiny$+$} (\i,1)
    -- node[right=-2pt] {\tiny$-$} (\i,0) -- (\i,-.5) node[right] {\tiny$+$};
}
\foreach \i in {3,1} {
  \draw[->] (3.5,\i) node[above=-2pt] {\tiny$+$} -- (3,\i) -- node[above=-2pt] {\tiny$-$} (2,\i) -- node[above=-2pt] {\tiny$+$} (1,\i)
    -- node[above=-2pt] {\tiny$-$} (0,\i) -- (-.5,\i) node[above=-2pt] {\tiny$+$};
}
\foreach \i in {2,0} {
  \draw[->] (3.5,\i) node[above=-2pt] {\tiny$-$} -- (3,\i) -- node[above=-2pt] {\tiny$+$} (2,\i) -- node[above=-2pt] {\tiny$-$} (1,\i)
    -- node[above=-2pt] {\tiny$+$} (0,\i) -- (-.5,\i) node[above=-2pt] {\tiny$-$};
}
\end{tikzpicture}
$$
}%
\uncover<8->{The excitations have finite weight. $\Rightarrow$ \emph{Nontrivial thermodynamics}.}

\uncover<9->{
3.~\emph{Disordered} regime: $|\Delta|<1$. No ground configurations. It turns out that this regime is always \emph{critical}.
}

\end{frame}

\begin{frame}
\frametitle{Heisenberg chain: two\-/particle states}

Consider the case $k=2$. Let us search for an eigenstate in the form
\eq$$
|\Psi_2(z_1,z_2)\rangle
=\sum_{n_1<n_2}(A_{12}z_1^{n_1}z_2^{n_2}+A_{21}z_2^{n_1}z_1^{n_2})
|n_1,n_2\rangle.
\label{Psi2}
$$
\pause
The action of the Hamiltonian moves $n_i$ by $\pm1$. Thus, the action on the contributions with $n_2-n_1>1$ does not differ from the action on the one\-/particle state. Hence, \emph{if} the state is an eigenstate, we have
$$
H_{\it XXZ}|\Psi_2(z_1,z_2)\rangle
=\left(-{N\Delta\over2}+\epsilon(z_1)+\epsilon(z_2)\right)
|\Psi_2(z_1,z_2)\rangle.
$$
\pause
When is it the case? First, check the action on the terms with $n_2-n_1=1$. We obtain
\eq$$
{A_{21}\over A_{12}}=S(z_1,z_2)
\equiv-{1+z_1z_2-2\Delta z_2\over 1+z_1z_2-2\Delta z_1}.
\label{S12def}
$$
\pause
Second, we have to impose the periodicity condition:
\eq$$
z_1^NS(z_1,z_2)=1,
\quad
z_2^NS(z_2,z_1)=1.
\label{k2betheeqs}
$$

\end{frame}

\begin{frame}
\frametitle{Heisenberg chain: Bethe Ansatz}

Consider general $k$. The \emph{Bethe Ansatz} is
$$
|\Psi_k(z_1,\ldots,z_k)\rangle=\sum_{n_1<\ldots<n_k}\sum_{\sigma\in S_k}
A_{\sigma_1\ldots\sigma_k}\prod_{j=1}^k z_{\sigma_j}^{n_j}
|n_1,\ldots,n_k\rangle.
$$
\pause
It is an eigenvector of the Hamiltonian, if (1)
\eq$$
A_{\ldots ji\ldots}/A_{\ldots ij\ldots}=S(z_i,z_j)
\label{AjiAijratio}
$$
\pause
and (2) the \emph{Bethe equations}
\eq$$
z_i^N\prod_{j,\ j\ne i}S(z_i,z_j)=1
\label{periodicity}
$$
are satisfied.
\pause
The eigenvalue is given by
\eq$$
H_{\it XXZ}|\Psi_k(z_1,\ldots,z_k)\rangle
=\left(-{N\Delta\over2}+\sum_{i=1}^k\epsilon(z_i)\right)
|\Psi_k(z_1,\ldots,z_k)\rangle,
\label{HPsik}
$$
\pause
Next time we rederive the Bethe equations in a different way and solve them for the ground state. We also will find the corresponding eigenvalue of the transfer matrix.

\end{frame}

\begin{frame}
\frametitle{Seminar}



\end{frame}

%\begin{frame}
%\frametitle{Seminar: Operator product expansions in CFT}
%\setcounter{equation}1
%
%Recall the Virasoro algebra
%\eq$$
%[L_m,L_n]=(m-n)L_{m+n}+{c\over12}m(m^2-1)\delta_{m+n,0},
%\label{Vir}
%$$
%and commutation with primary operators:
%\eq$$
%[L_n,\Phi_\Delta(z,\bar z)]=(n+1)\Delta z^n\Phi_\Delta(z,\bar z)+z^{n+1}\d\Phi_\Delta(z,\bar z).
%\label{LPhi}
%$$
%The highest weight vectors $|\Delta\rangle$:
%\eq$$
%L_n|\Delta\rangle=0\quad(n>0),
%\qquad
%L_0|\Delta\rangle=\Delta|\Delta\rangle.
%\label{hwv}
%$$
%Analogously for $\bar L_n$ and $\bar z$.
%
%The operator product expansion
%\eq$$
%\Phi_{\Delta'}(z,\bar z)|\Delta\rangle=\sum_k C_k(z\bar z)^{\Delta_k-\Delta-\Delta'}\varphi_k(z)\bar\varphi_k(\bar z)|\Delta_k\rangle.
%\label{Phi-OPE}
%$$
%Here
%\eq$$
%\varphi_k(z)|\Delta_k\rangle=\left(1+z\beta_1L_{-1}+z^2(\beta_2L_{-2}+\beta_{11}L_{-1}^2)+\cdots\right)|\Delta_k\rangle.
%\label{varphi-OPE}
%$$
%On the other hand from (\ref{LPhi}) and (\ref{hwv}) we have
%\eq$$
%z^{\Delta_k-\Delta-\Delta'}L_n\varphi_k(z)|\Delta_k\rangle=\left((n+1)\Delta'z^n+z^{n+1}\d\right)z^{\Delta_k-\Delta-\Delta'}\varphi_k(z)|\Delta_k\rangle
%\quad(n>0).
%\label{Lvarphi}
%$$
%\emph{The task:} Write equations for the coefficients $\beta_1$, $\beta_2$, $\beta_{11}$.
%
%
%\end{frame}


\end{document}
