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\begin{document}

\title[Lecture 3. Algebraic Bethe Ansatz]{Lecture 3\\Algebraic Bethe Ansatz}
\subtitle{A mini-course ``Solvable lattice models and Bethe Ansatz''\\(Ariel University, spring 2021)}
\author{Michael Lashkevich}
\institute{Landau Institute for Theoretical Physics,\\
Kharkevich Institute for Information Transmission Problems}
\date{}

\frame{\titlepage}

\begin{frame}
\frametitle{$L$ operators and reference state}
\unpause

Recall the definition of the $L$\-/operator:
\eq$$
L(u)=R_{0N}(u)\ldots R_{02}(u)R_{01}(u)=\pMatrix{A(u)&B(u)\\C(u)&D(u)}.
\label{Ludef}
$$
\unpause
These four operators can be represented graphically as
$$
\begin{tikzpicture}[scale=.5,>={Stealth[length=1mm]}]
\draw[-] (0,3.5) node[right] {\tiny$+$} -- (0,2.7); \draw[dashed] (0,2.7) -- (0,1.3);
  \draw[->] (0,1.3) -- (0,-.5) node[below] {\tiny$u$} node[right] {\tiny$+$} node[below=10pt] {$A(u)$};
\draw[->] (.5,3) -- (-.5,3) node[left] {\tiny$0$};
\draw[->] (.5,1) -- (-.5,1) node[left] {\tiny$0$};
\draw[->] (.5,0) -- (-.5,0) node[left] {\tiny$0$};
\end{tikzpicture}
\quad
\begin{tikzpicture}[scale=.5,>={Stealth[length=1mm]}]
\draw[-] (0,3.5) node[right] {\tiny$+$} -- (0,2.7); \draw[dashed] (0,2.7) -- (0,1.3);
  \draw[->] (0,1.3) -- (0,-.5) node[below] {\tiny$u$} node[right] {\tiny$-$} node[below=10pt] {$B(u)$};
\draw[->] (.5,3) -- (-.5,3) node[left] {\tiny$0$};
\draw[->] (.5,1) -- (-.5,1) node[left] {\tiny$0$};
\draw[->] (.5,0) -- (-.5,0) node[left] {\tiny$0$};
\end{tikzpicture}
\quad
\begin{tikzpicture}[scale=.5,>={Stealth[length=1mm]}]
\draw[-] (0,3.5) node[right] {\tiny$-$} -- (0,2.7); \draw[dashed] (0,2.7) -- (0,1.3);
  \draw[->] (0,1.3) -- (0,-.5) node[below] {\tiny$u$} node[right] {\tiny$+$} node[below=10pt] {$C(u)$};
\draw[->] (.5,3) -- (-.5,3) node[left] {\tiny$0$};
\draw[->] (.5,1) -- (-.5,1) node[left] {\tiny$0$};
\draw[->] (.5,0) -- (-.5,0) node[left] {\tiny$0$};
\end{tikzpicture}
\quad
\begin{tikzpicture}[scale=.5,>={Stealth[length=1mm]}]
\draw[-] (0,3.5) node[right] {\tiny$-$} -- (0,2.7); \draw[dashed] (0,2.7) -- (0,1.3);
  \draw[->] (0,1.3) -- (0,-.5) node[below] {\tiny$u$} node[right] {\tiny$-$} node[below=10pt] {$D(u)$};
\draw[->] (.5,3) -- (-.5,3) node[left] {\tiny$0$};
\draw[->] (.5,1) -- (-.5,1) node[left] {\tiny$0$};
\draw[->] (.5,0) -- (-.5,0) node[left] {\tiny$0$};
\end{tikzpicture}
\qquad
\uncover<4->{
\begin{tikzpicture}[scale=.5,>={Stealth[length=1mm]}]
\draw[-] (0,3.5) -- (0,2.7); \draw[dashed] (0,2.7) -- (0,1.3);
  \draw[-] (0,1.3) -- (0,-.5) node[below=10pt] {$|\Omega_+\rangle$};
\draw[->] (.5,3) node[above=-2pt] {\tiny$+$} -- (0,3);
\draw[->] (.5,1) node[above=-2pt] {\tiny$+$} -- (0,1);
\draw[->] (.5,0) node[above=-2pt] {\tiny$+$} -- (0,0);
\end{tikzpicture}
}
$$
\unpause
Notice that
$$
[S^z,A(u)]=[S^z,D(u)]=0,\quad[S^z,B(u)]=-B(u),\quad[S^z,C(u)]=C(u).
$$
\unpause
Consider also the pseudovacuum $|\Omega_+\rangle$.
\unpause
It can be defined as
$$
C(u)|\Omega_+\rangle=0.
$$
\unpause
How to flip spins?
\unpause
By means of $B(u)$ operators.

\end{frame}

\begin{frame}
\frametitle{$n=1$ case}
\unpause

Indeed, let
\eq$$
|u_1,u_2,\ldots,u_k\rangle=B(u_1)B(u_2)\ldots B(u_k)|\Omega_+\rangle.
\label{Bstates}
$$
Then
$$
S^z|u_1,\ldots,u_k\rangle=\left({N\over2}-k\right)|u_1,\ldots,u_k\rangle.
$$
\unpause
Consider the case $n=1$. You see that
$$
B(u)|\Omega_+\rangle
=\sum^N_{n=1}\begin{tikzpicture}[scale=.5,>={Stealth[length=1mm]},baseline=.5*2.5cm-2pt,color=red]
\draw[-] (0,6) node[right] {$+$} -- (0,4.7); \draw[dashed] (0,4.7) -- (0,3.3); \draw[-] (0,3.3) -- (0,3);
\draw[-,color=blue,thick] (1,3) node[above=-2pt] {$-$} -- (0,3) node[above right=-2pt] {\black$c$} -- (0,2.7); \draw[dashed,color=blue,thick] (0,2.7) -- (0,1.3);
  \draw[->,color=blue,thick] (0,1.3) -- (0,-1) node[right] {$-$};
\draw[->] (1,5) node[above=-2pt] {$+$} -- (0,5) node[above right=-2pt] {\black$a$} -- (-1,5) node[above=-2pt] {$+$};
\draw[->] (0,3) -- (-1,3) node[left] {$V_n$} node[above=-2pt] {$+$};
\draw[->] (1,1) node[above=-2pt] {$+$} -- (0,1) node[above right=-2pt] {\black$b$} -- (-1,1) node[above=-2pt] {$+$};
\draw[->] (1,0) node[above=-2pt] {$+$} -- (0,0) node[above right=-2pt] {\black$b$} -- (-1,0) node[above=-2pt] {$+$};
\end{tikzpicture}
\uncover<+->{=\sum_n b^{n-1}(u)c(u)a^{N-n}(u)|n\rangle}
\uncover<+->{\tikz[overlay,baseline=-2pt]{\draw (0,-1) node {$\displaystyle{}={a^N(u)c(u)\over b(u)}\sum_n\left(b(u)\over a(u)\right)^n|n\rangle.$}}}
\qquad\qquad\qquad
$$
\unpause
We see that
$$
B(u)|\Omega_+\rangle\sim\sum^N_{n=1}z^n(u)|n\rangle,
\qquad
z(u)={b(u)\over a(u)}.
$$
It is a Bethe wave function, if $z^N(u)=1$.

\end{frame}

\begin{frame}
\frametitle{$n=2$ case}
\unpause

Let $n=2$. We have $B(u_1)B(u_2)|\Omega_+\rangle={}$
\unpause
\vspace{-8pt}
\Multline*$$
\hspace{-8pt}\sum_{n_1<n_2}\Biggl\lgroup
\begin{tikzpicture}[scale=.5,>={Stealth[length=1mm]},baseline=.5*3.5cm-2pt,color=red]
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  \draw[-] (1,5.7) -- (1,5.5) node[above right=-2pt] {\black$b_1$} -- (1,5.3); \draw[dotted] (1,5.3) -- (1,2.7);
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\draw[->] (2,7) node[above=-2pt] {$+$} -- (0,7) -- (-1,7) node[above=-2pt] {$+$};
\draw[->] (0,5.5) -- (-1,5.5) node[left] {$V_{n_2}$} node[above=-2pt] {$+$};
\draw[->] (1,2.5) -- (-1,2.5) node[left] {$V_{n_1}$} node[above=-2pt] {$+$};
\draw[->] (2,1) node[above=-2pt] {$+$} -- (-1,1) node[above=-2pt] {$+$};
\draw[->] (2,0) node[above=-2pt] {$+$} -- (-1,0) node[above=-2pt] {$+$};
\begin{scope}[color=blue,thick]
  \draw[-] (2,5.5) node[above=-2pt] {$-$} -- (0,5.5) -- (0,5.3); \draw[dotted] (0,5.3) -- (0,2.7);
    \draw[-] (0,2.7) -- (0,2.5) node[above right=-2pt] {\black$b_2$} -- (0,2.3); \draw[dotted] (0,2.3) -- (0,1.2);
    \draw[->] (0,1.2) -- (0,1) node[above right=-2pt] {\black$b_2$} -- (0,0) node[above right=-2pt] {\black$b_2$} -- (0,-1) node[right] {$-$};
  \draw[-] (2,2.5) node[above=-2pt] {$-$} -- (1,2.5) -- (1,2.3); \draw[dotted] (1,2.3) -- (1,1.2);
    \draw[->] (1,1.2) -- (1,1) node[above right=-2pt] {\black$b_1$} -- (1,0) node[above right=-2pt] {\black$b_1$} -- (1,-1) node[right] {$-$};
\end{scope}
\end{tikzpicture}
\uncover<.(1)->{+
\begin{tikzpicture}[scale=.5,>={Stealth[length=1mm]},baseline=.5*3.5cm-2pt,color=red]
\draw[-] (0,8) node[right] {$+$} -- (0,7) node[above right=-2pt] {\black$a_2$} -- (0,6.8); \draw[dotted] (0,6.8) -- (0,5.7);
  \draw[-] (0,5.7) -- (0,5.5) node[above right=-2pt] {\black$a_2$} -- (0,5.3); \draw[dotted] (0,5.3) -- (0,2.7);
  \draw[-] (0,2.7) -- (0,2.5) node[above right=-2pt] {\black$c_2$};
\draw[-] (1,8) node[right] {$+$} -- (1,7) node[above right=-2pt] {\black$a_1$} -- (1,6.8); \draw[dotted] (1,6.8) -- (1,5.7);
  \draw[-] (1,5.7) -- (1,5.5) node[above right=-2pt] {\black$c_1$};
\draw[->] (2,7) node[above=-2pt] {$+$} -- (0,7) -- (-1,7) node[above=-2pt] {$+$};
\draw[->] (1,5.5) -- (-1,5.5) node[left] {$V_{n_2}$} node[above=-2pt] {$+$};
\draw[->] (0,2.5) -- (-1,2.5) node[left] {$V_{n_1}$} node[above=-2pt] {$+$};
\draw[->] (2,1) node[above=-2pt] {$+$} -- (-1,1) node[above=-2pt] {$+$};
\draw[->] (2,0) node[above=-2pt] {$+$} -- (-1,0) node[above=-2pt] {$+$};
\begin{scope}[color=blue,thick]
  \draw[-] (2,2.5) node[above=-2pt] {$-$} -- (0,2.5) -- (0,2.3); \draw[dotted] (0,2.3) -- (0,1.2);
    \draw[->] (0,1.2) -- (0,1) node[above right=-2pt] {\black$b_2$} -- (0,0) node[above right=-2pt] {\black$b_2$} -- (0,-1) node[right] {$-$};
  \draw[-] (2,5.5) node[above=-2pt] {$-$} -- (1,5.5) -- (1,5.3); \draw[dotted] (1,5.3) -- (1,2.7);
    \draw[-] (1,2.7) -- (1,2.5) node[above right=-2pt] {\black$a_1$} -- (1,2.3); \draw[dotted] (1,2.3) -- (1,1.2);
    \draw[->] (1,1.2) -- (1,1) node[above right=-2pt] {\black$b_1$} -- (1,0) node[above right=-2pt] {\black$b_1$} -- (1,-1) node[right] {$-$};
\end{scope}
\end{tikzpicture}
\Biggr\rgroup}
\uncover<.(2)->{+
\sum_{n_1<n<n_2}
\begin{tikzpicture}[scale=.5,>={Stealth[length=1mm]},baseline=.5*3.5cm-2pt,color=red]
\draw[-] (0,8) node[right] {$+$} -- (0,7) node[above right=-2pt] {\black$a_2$} -- (0,6.8); \draw[dotted] (0,6.8) -- (0,5.7);
  \draw[-] (0,5.7) -- (0,5.5) node[above right=-2pt] {\black$a_2$} -- (0,5.3); \draw[dotted] (0,5.3) -- (0,4.2);
  \draw[->] (0,4.2) --  (0,4) -- (-1,4) node[above=-2pt] {$+$} node[left] {$V_n$};
\draw[-] (1,8) node[right] {$+$} -- (1,7) node[above right=-2pt] {\black$a_1$} -- (1,6.8); \draw[dotted] (1,6.8) -- (1,5.7);
  \draw[-] (1,5.7) -- (1,5.5) node[above right=-2pt] {\black$c_1$} -- (-1,5.5) node[left] {$V_{n_2}$} node[above=-2pt] {$+$};
\draw[->] (2,7) node[above=-2pt] {$+$} -- (0,7) -- (-1,7) node[above=-2pt] {$+$};
\draw[-] (2,4) node[above=-2pt] {$+$} -- (1,4) -- (1,3.8); \draw[dotted] (1,3.8) -- (1,2.7);
  \draw[->] (1,2.7) -- (1,2.5) node[above right=-2pt] {\black$c_1$} -- (-1,2.5) node[left] {$V_{n_1}$} node[above=-2pt] {$+$};
\draw[->] (2,1) node[above=-2pt] {$+$} -- (-1,1) node[above=-2pt] {$+$};
\draw[->] (2,0) node[above=-2pt] {$+$} -- (-1,0) node[above=-2pt] {$+$};
\begin{scope}[color=blue,thick]
  \draw[-] (2,5.5) node[above=-2pt] {$-$} -- (1,5.5) -- (1,5.3); \draw[dotted] (1,5.3) -- (1,4.2);
    \draw[-] (1,4.2) -- (1,4) node[above right=-2pt] {\black$c_1$} -- (0,4) node[above right=-2pt] {\black$c_2$} -- (0,3.8);
   \draw[dotted] (0,3.8) -- (0,2.7); \draw[-] (0,2.7) -- (0,2.5) node[above right=-2pt] {\black$b_2$} -- (0,2.3); \draw[dotted] (0,2.3) -- (0,1.2);
    \draw[->] (0,1.2) -- (0,1) node[above right=-2pt] {\black$b_2$} -- (0,0) node[above right=-2pt] {\black$b_2$} -- (0,-1) node[right] {$-$};
  \draw[-] (2,2.5) node[above=-2pt] {$-$} -- (1,2.5) -- (1,2.3); \draw[dotted] (1,2.3) -- (1,1.2);
    \draw[->] (1,1.2) -- (1,1) node[above right=-2pt] {\black$b_1$} -- (1,0) node[above right=-2pt] {\black$b_1$} -- (1,-1) node[right] {$-$};
\end{scope}
\end{tikzpicture}}
\\
\uncover<.(3)->{{}={a_1^Na_2^Nc_1c_2\over b_1b_2}\sum_{n_1<n_2}\left({a_{21}\over b_{21}}z_1^{n_1}z_2^{n_2}+{a_{12}\over b_{12}}z_1^{n_2}z_2^{n_1}\right)|n_1,n_2\rangle,
}
$$
where $a_i=a(u_i)$, $a_{ij}=a(u_i-u_j)$ and so on, $z_i=b_i/a_i$.
\unpause\unpause\unpause\unpause
We see that
$$
B(u_1)B(u_2)|\Omega_+\rangle=B(u_2)B(u_1)|\Omega_+\rangle,
$$
and it is a Bethe vector.
\unpause
It can be checked that
$$
S(z_1,z_2)={a_{12}b_{21}\over b_{12}a_{21}}.
$$

\end{frame}

\begin{frame}
\frametitle{Commutation relations}
\unpause

We want to derive the Bethe equations in an alternative way.
\unpause
Since $T(u)=A(u)+D(u)$, we have to calculate the vectors
$$
T(u)|u_1,\ldots,u_k\rangle=(A(u)+D(u))B(u_1)B(u_2)\cdots B(u_k)|\Omega_+\rangle.
$$
What are the conditions, under which these vectors are proportional to $|u_1,\ldots,u_k\rangle$?
\unpause

First, find the commutation relations of the operators $A(u),\ldots,D(u)$. We have
$$
R_{12}(u_1-u_2)L_1(u_1)L_2(u_2)=L_2(u_2)L_1(u_1)R_{12}(u_1-u_2).
$$
\unpause
or
$$
\tikz[scale=.5,baseline=.5*1.5cm-2pt,>={Stealth[length=1mm]}]{
\draw[-] (0,4) node[left=-2pt] {{\uncover<4->{\tiny\red$+$}}} to[out=-90,in=90] node[above right=-2pt,pos=.9] {{\uncover<4->{\tiny\red$+$}}} (1,3) -- (1,2.7);
  \draw[dashed] (1,2.7) -- (1,1.3); \draw[->] (1,1.3) -- (1,-1) node[below] {$u_1$} node[right=-2pt] {{\uncover<4->{\tiny\blue$-$}}};
\draw[-] (1,4) node[right=-2pt] {{\uncover<4->{\tiny\red$+$}}} to[out=-90,in=90] node[above left=-2pt,pos=.9] {{\uncover<4->{\tiny\red$+$}}} (0,3) -- (0,2.7); 
  \draw[dashed] (0,2.7) -- (0,1.3); \draw[->] (0,1.3) -- (0,-1) node[below] {$u_2$} node[left=-2pt] {{\uncover<4->{\tiny\blue$-$}}};
\draw (.5,3.5) node[above] {{\uncover<4->{\tiny$a_{12}$}}};
\draw[->] (1.5,3) -- (-.6,3);
\draw[->] (1.5,1) -- (-.6,1);
\draw[->] (1.5,0) -- (-.6,0);
}
\quad=\quad
\tikz[scale=.5,baseline=.5*1.5cm-2pt,>={Stealth[length=1mm]}]{
\draw[-] (0,4) node[left=-2pt] {{\uncover<4->{\tiny\red$+$}}} -- (0,2.7); \draw[dashed] (0,2.7) -- (0,1.3);
  \draw[->] (0,1.3) -- (0,0) to[out=-90,in=90] node[below left=-2pt,pos=.1] {{\uncover<4->{\tiny\blue$-$}}}
  (1,-1) node[below] {$u_1$} node[right=-2pt] {{\uncover<4->{\tiny\blue$-$}}};
\draw[-] (1,4) node[right=-2pt] {{\uncover<4->{\tiny\red$+$}}} -- (1,2.7); \draw[dashed] (1,2.7) -- (1,1.3);
  \draw[->] (1,1.3) -- (1,0) to[out=-90,in=90] node[below right=-2pt,pos=.1] {{\uncover<4->{\tiny\blue$-$}}}
  (0,-1) node[below] {$u_2$} node[left=-2pt] {{\uncover<4->{\tiny\blue$-$}}};
\draw (.5,-.5) node[below] {{\uncover<4->{\tiny$a_{12}$}}};
\draw[->] (1.5,3) -- (-.6,3);
\draw[->] (1.5,1) -- (-.6,1);
\draw[->] (1.5,0) -- (-.6,0);
}
$$
\unpause
First, the ${}^{++}_{--}$\-/component of this relation gives
\eq$$
B(u_1)B(u_2)=B(u_2)B(u_1).
\label{BBcommut}
$$
It means that the states (\ref{Bstates}) are symmetric in~$u_i$.

\end{frame}

\begin{frame}
\frametitle{Commutation relations}
\unpause

To commute $T(u)$ with $B(u_i)$ we will need the commutations of operators $A(u)$ and $D(u)$ with $B(u)$:
\unpause
\Multline*$$
\only<2>{\vphantom{(}}
\only<3>{a(u_1-u_2)B(u_1)A(u_2)={}}
\only<4->{a(u_2-u_1)B(u_1)D(u_2)={}}
\\
\only<2>{
\tikz[scale=.5,baseline=.5*1.5cm-2pt,>={Stealth[length=1mm]}]{
\draw[-] (0,4) node[left=-2pt] {\tiny\red$\phantom+$} to[out=-90,in=90] node[above right=-2pt,pos=.9] {\tiny\red$\phantom+$} (1,3) -- (1,2.7);
  \draw[dashed] (1,2.7) -- (1,1.3); \draw[->] (1,1.3) -- (1,-1) node[below] {$u_1$} node[right=-2pt] {\tiny\blue$\phantom-$};
\draw[-] (1,4) node[right=-2pt] {\tiny\red$\phantom+$} to[out=-90,in=90] node[above left=-2pt,pos=.9] {\tiny\red$\phantom+$} (0,3) -- (0,2.7); 
  \draw[dashed] (0,2.7) -- (0,1.3); \draw[->] (0,1.3) -- (0,-1) node[below] {$u_2$} node[left=-2pt] {\tiny\red$\phantom+$};
\draw (.5,3.5) node[above] {\tiny$\phantom{a_{12}}$};
\draw[->] (1.5,3) -- (-.6,3);
\draw[->] (1.5,1) -- (-.6,1);
\draw[->] (1.5,0) -- (-.6,0);
}}
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\draw (.5,3.5) node[above] {\tiny$a_{12}$};
\draw[->] (1.5,3) -- (-.6,3);
\draw[->] (1.5,1) -- (-.6,1);
\draw[->] (1.5,0) -- (-.6,0);
}}
\only<4->{
\tikz[scale=.5,baseline=.5*1.5cm-2pt,>={Stealth[length=1mm]}]{
\draw[-] (0,4) node[left=-2pt] {\tiny\blue$-$} -- (0,2.7); \draw[dashed] (0,2.7) -- (0,1.3);
  \draw[->] (0,1.3) -- (0,0) to[out=-90,in=90] node[below left=-2pt,pos=.1] {\tiny\blue$-$}
  (1,-1) node[below] {$u_2$} node[right=-2pt] {\tiny\blue$-$};
\draw[-] (1,4) node[right=-2pt] {\tiny\red$+$} -- (1,2.7); \draw[dashed] (1,2.7) -- (1,1.3);
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  (0,-1) node[below] {$u_1$} node[left=-2pt] {\tiny\blue$-$};
\draw (.5,-.5) node[below] {\tiny$a_{21}$};
\draw[->] (1.5,3) -- (-.6,3);
\draw[->] (1.5,1) -- (-.6,1);
\draw[->] (1.5,0) -- (-.6,0);
}}
\quad=\quad
\only<2>{\tikz[scale=.5,baseline=.5*1.5cm-2pt,>={Stealth[length=1mm]}]{
\draw[-] (0,4) node[left=-2pt] {\tiny\red$\phantom+$} -- (0,2.7); \draw[dashed] (0,2.7) -- (0,1.3);
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  (1,-1) node[below] {$u_1$} node[right=-2pt] {\tiny\blue$\phantom-$};
\draw[-] (1,4) node[right=-2pt] {\tiny\red$\phantom+$} -- (1,2.7); \draw[dashed] (1,2.7) -- (1,1.3);
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  (0,-1) node[below] {$u_2$} node[left=-2pt] {\tiny\red$\phantom+$};
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\draw[->] (1.5,3) -- (-.6,3);
\draw[->] (1.5,1) -- (-.6,1);
\draw[->] (1.5,0) -- (-.6,0);
}}
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  \draw[->] (0,1.3) -- (0,0) to[out=-90,in=90] node[below left=-2pt,pos=.1] {\tiny\red$+$}
  (1,-1) node[below] {$u_1$} node[right=-2pt] {\tiny\blue$-$};
\draw[-] (1,4) node[right=-2pt] {\tiny\red$+$} -- (1,2.7); \draw[dashed] (1,2.7) -- (1,1.3);
  \draw[->] (1,1.3) -- (1,0) to[out=-90,in=90] node[below right=-2pt,pos=.1] {\tiny\blue$-$}
  (0,-1) node[below] {$u_2$} node[left=-2pt] {\tiny\red$+$};
\draw (.5,-.5) node[below] {\tiny$c_{12}$};
\draw[->] (1.5,3) -- (-.6,3);
\draw[->] (1.5,1) -- (-.6,1);
\draw[->] (1.5,0) -- (-.6,0);
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\tikz[scale=.5,baseline=.5*1.5cm-2pt,>={Stealth[length=1mm]}]{
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\quad+\quad
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\draw[-] (0,4) node[left=-2pt] {\tiny\red$\phantom+$} -- (0,2.7); \draw[dashed] (0,2.7) -- (0,1.3);
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  (0,-1) node[below] {$u_2$} node[left=-2pt] {\tiny\red$\phantom+$};
\draw (.5,-.5) node[below] {\tiny$\phantom{b_{12}}$};
\draw[->] (1.5,3) -- (-.6,3);
\draw[->] (1.5,1) -- (-.6,1);
\draw[->] (1.5,0) -- (-.6,0);
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\tikz[scale=.5,baseline=.5*1.5cm-2pt,>={Stealth[length=1mm]}]{
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\\
\only<2>{\vphantom{(}}
\only<3>{{}=c(u_1-u_2)B(u_2)A(u_1)+b(u_1-u_2)A(u_2)B(u_1)}
\only<4->{{}=c(u_2-u_1)B(u_2)D(u_1)+b(u_2-u_1)D(u_2)B(u_1)}
$$
\unpause\unpause\unpause
Finally we have
\Align$$
a(u_1-u_2)B(u_1)A(u_2)
&=c(u_1-u_2)B(u_2)A(u_1)+b(u_1-u_2)A(u_2)B(u_1),
\label{ABcommut}
\\
a(u_2-u_1)B(u_1)D(u_2)
&=c(u_2-u_1)B(u_2)D(u_1)+b(u_2-u_1)D(u_2)B(u_1).
\label{DBcommut}
$$

\end{frame}

\begin{frame}
\frametitle{Action of the transfer matrix: $n=1$}
\unpause

Apply these formulas to the action of the transfer matrix.
\unpause
Let us start from $n=1$:
\Align*$$
T(u)|u_1\rangle
&=(A(u)+D(u))B(u_1)|\Omega_+\rangle
\\
&=\biggl({a(u_1-u)\over b(u_1-u)}B(u_1)A(u)+{a(u-u_1)\over b(u-u_1)}B(u_1)D(u)
\\
&\quad-{c(u_1-u)\over b(u_1-u)}B(u)A(u_1)-{c(u-u_1)\over b(u-u_1)}B(u)D(u_1)\biggr)|\Omega_+\rangle.
$$
\unpause
Taking into account that $A(u)|\Omega_+\rangle=a^N(u)|\Omega_+\rangle$, $D(u)|\Omega_+\rangle=b^N(u)|\Omega_+\rangle$.%
\tikz[overlay,scale=.5,color=red,>={Stealth[length=1mm]},baseline=-.1cm,shift={($(current page.east)+(-4cm,-3cm)$)}]{
\draw[->] (0,3.5) node[right=-2pt] {$+$} node[above=2pt] {\tiny\black$A(u)|\Omega_+\rangle$} -- (0,-.5) node[right=-2pt] {$+$};
\draw[->] (.5,3) -- (-.5,3);
\draw[->] (.5,2) -- (-.5,2);
\draw[->] (.5,1) -- (-.5,1);
\draw[->] (.5,0) -- (-.5,0);
\begin{scope}[xshift=2.5cm]
  \draw[->,color=blue] (0,3.5) node[right=-2pt] {$-$} node[above=2pt] {\tiny\black$D(u)|\Omega_+\rangle$} -- (0,-.5) node[right=-2pt] {$-$};
  \draw[->] (.5,3) -- (-.5,3);
  \draw[->] (.5,2) -- (-.5,2);
  \draw[->] (.5,1) -- (-.5,1);
  \draw[->] (.5,0) -- (-.5,0);
\end{scope}
}
\unpause
We conclude that
\Align*$$
T(u)|u_1\rangle
&=\left({a(u_1-u)\over b(u_1-u)}a^N(u)+{a(u-u_1)\over b(u-u_1)}b^N(u)\right)|u_1\rangle
\\
&\quad-\left({c(u_1-u)\over b(u_1-u)}a^N(u_1)+{c(u-u_1)\over b(u-u_1)}b^N(u_1)\right)|u\rangle.
$$
\unpause
The vector $|u_1\rangle$ is an eigenvector of $T(u)$ it the second term vanishes:
$$
\left(b(u_1)\over a(u_1)\right)^N=-\left(c(u_1-u)b(u-u_1)\over c(u-u_1)b(u_1-u)\right)=1\quad\Leftrightarrow\quad z_1^N=1.
$$
\unpause
The corresponding eigenvector is
$$
\Lambda(u;u_1)=a^N(u){a(u_1-u)\over b(u_1-u)}+b^N(u){a(u-u_1)\over b(u-u_1)}.
$$

\end{frame}

\begin{frame}
\frametitle{Action of the transfer matrix: general case}
\unpause

\begin{block}{The Takhtajan\--Faddeev formulas:}
\vspace{-8pt}
\eq$$
\Aligned{
A(u)|u_1,\ldots,u_n\rangle
&=\alpha(u;u_1,\ldots,u_n)|u_1,\ldots,u_n\rangle
\\
&\quad
-\sum_{i=1}^n{c(u_i-u)\over b(u_i-u)}
\alpha(u_i;u_1,\ldots,\widehat{u_i},\ldots,u_n)
|u,u_1,\ldots,\widehat{u_i},\ldots,u_n\rangle,
\\
D(u)|u_1,\ldots,u_n\rangle
&=\delta(u;u_1,\ldots,u_n)|u_1,\ldots,u_n\rangle
\\
&\quad
-\sum_{i=1}^n{c(u-u_i)\over b(u-u_i)}
\delta(u_i;u_1,\ldots,\widehat{u_i},\ldots,u_n)
|u,u_1,\ldots,\widehat{u_i},\ldots,u_n\rangle,
}\label{ABaction}
$$
\vspace{-8pt}
where
\vspace{-2pt}
\eq$$
\alpha(u;u_1,\ldots,u_n)
=a^N(u)\prod_{i=1}^n{a(u_i-u)\over b(u_i-u)},
\quad
\delta(u;u_1,\ldots,u_n)
=b^N(u)\prod_{i=1}^n{a(u-u_i)\over b(u-u_i)}.
\label{alphadeltadef}
$$
\end{block}
\unpause

Thus the action of the transfer matrix is
$$
T(u)|u_1,\ldots,u_n\rangle=\Lambda(u;u_1,\ldots,u_n)|u_1,\ldots,u_n\rangle+\text{unwanted terms},
$$
where
$$
\Lambda(u;u_1,\ldots,u_n)=\alpha(u;u_1,\ldots,u_n)+\delta(u;u_1,\ldots,u_n)
$$

\end{frame}

\begin{frame}
\frametitle{Bethe equations and transfer matrix eigenvalue}
\unpause

Since ${c(u)\over b(u)}=-{c(-u)\over b(-u)}$, the unwanted terms in the r.h.s.\ have the form
$$
{c(u_i-u)\over b(u_i-u)}(\alpha(u_i;u_1,\ldots,\widehat{u_i},\ldots,u_n)-\delta(u_i;u_1,\ldots,\widehat{u_i},\ldots,u_n))
|u,u_1,\ldots,\widehat{u_i},\ldots,u_n\rangle.
$$
\unpause
They vanish, if the \emph{Bethe equations} are satisfied:
\vskip -8pt
$$
\alpha(u_i;u_1,\ldots,\widehat{u_i},\ldots,u_n)
=\delta(u_i;u_1,\ldots,\widehat{u_i},\ldots,u_n)
$$
\unpause
or
\vskip -8pt
\eq$$
\left(b(u_i)\over a(u_i)\right)^N=\prod_{\substack{j=1\\(j\ne i)}}^n{a(u_j-u_i)b(u_i-u_j)\over b(u_j-u_i)a(u_i-u_j)}.
\label{Betheeq}
$$
\unpause
Subject to these equations the vectors $|u_1,\ldots,u_n\rangle$ are eigenvectors with the eigenvalues
\eq$$
\Lambda(u;u_1,\ldots,u_n)=a^N(u)\prod_{i=1}^n{a(u_i-u)\over b(u_i-u)}+b^N(u)\prod_{i=1}^n{a(u-u_i)\over b(u-u_i)}.
\label{Lambda}
$$

\end{frame}

\begin{frame}
\frametitle{Explicit form of the Bethe equations}
\unpause

Let
$$
s(u)=\Cases{\sin u&\text{for $|\Delta|<1$};\\\sh u&\text{for $\Delta<-1$;}}
\quad
c(u)=\Cases{\cos u&\text{for $|\Delta|<1$};\\\ch u&\text{for $\Delta<-1$.}}
$$
\quad
The explicit form  of the Bethe equations:
\eq$$
\left(s(u_i)\over s(\lambda-u_i)\right)^N
=\prod_{\substack{j=1\\(j\ne i)}}^n
{s(u_i-u_j+\lambda)\over s(u_i-u_j-\lambda)}.
\label{Bethe-explicit}
$$
\unpause
Let
$$
u_i={\lambda\over2}+\i v_i,
\quad
e^{\i p(v)}={s(\lambda/2+\i v)\over s(\lambda/2-\i v)},
\quad
e^{\i\theta(v)}={s(\lambda+\i v)\over s(\lambda-\i v)}.
$$
The variables $v_i$ are defined in such a way that $|z_i|=1$ for real values of~$v_i$.
\unpause
Take logarithm of the Bethe equations:
$$
Np(v_i)=2\pi I_i+\sum_{j=1}^n\theta(v_i-v_j),
$$
where $I_i\in\Z+{1\over2}$ if $n\in2\Z$ and $I_i\in\Z$ if $n\in2\Z+1$.

\end{frame}

\end{document}