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\begin{document}

\title[Lecture 4. Solving Bethe equations]{Lecture 4\\Solving Bethe equations}
\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{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}

\begin{frame}
\frametitle{Solving the Bethe equations for the ground state}

Let us obtain the ground state, i.e.\ the state with the largest eigenvalue of the transfer matrices in the thermodynamic limit.
\unpause
Note that the XXZ one\-/particle energy
$$
\epsilon(v)=2\Delta-2\cos p(v)
$$
is an even function, $\epsilon(-v)=\epsilon(v)$ with an absolute minimum at $v=0$ and monotonous for $0\le v<\infty$ if $|\Delta|<1$ and for $0\le v\le{\pi\over2}$ for $\Delta<-1$. It means that the `Dirac sea' must be symmetric.
\unpause

Thus formulate the conjectures:
\begin{enumerate}
\item In the ground state all Bethe roots $v_i$ are real and, in the thermodynamic limit, densely fill a region $-v_F<v<v_F$.
\unpause
\item In the ground state all values of $I_i$ are consecutive.
\unpause
\item In the ground state $S^z/N\to0$ as $N\to\infty$.
\end{enumerate}
\unpause
Taking the thermodynamic limit in a usual way, we obtain the integral equations
\eq$$
p'(v)=\rho(v)+\int^{v_F}_{-v_F}{dv'\over2\pi}\,\theta'(v-v')\rho(v'),
\qquad
\int^{v_F}_{-v_F}{dv\over2\pi}\,\rho(v)={n\over N},
\label{inteq}
$$
where $\rho(v)={2\pi dI\over Ndv}$ is the density of particles = density of states.
\unpause

We have
$$
p'(v)={s(\lambda)\over s({\lambda\over2}+\i v)s({\lambda\over2}-\i v)},
\quad
\theta'(v)={2s(2\lambda)\over s(\lambda+\i v)s(\lambda-\i v)}.
$$

\end{frame}

\begin{frame}
\frametitle{Solving the Bethe equations for the ground state}
\unpause

For which values $\bar v_F$ of $v_F$ the equation is solvable analytically?
\unpause
\begin{itemize}
\item If $|\Delta|<1$ the functions $p'(v),\theta'(v)\sim e^{-2|v|}$ as $v\to\pm\infty$. Hence, $\bar v_F=\infty$.
\unpause
Hence,
\eq$$
\rho(v)=\int^\infty_{-\infty} dk\,\rho_ke^{-ikv},\ldots
\label{Fourier-int}
$$
\unpause
\item If $\Delta<-1$ the functions $p'(v),\theta'(v)$ are periodic with the period $\pi$. Hence, $\bar v_F={\pi\over2}$.
\unpause
Hence,
\eq$$
\rho(v)=2\sum_{k\in2\Z}\rho_ke^{-ikv},\ldots
\label{Fourier-series}
$$
\end{itemize}
\unpause
Then
$$
\rho_k=p'_k-\theta'_k\rho_k,
$$
\unpause
We have
$$
\Aligned{
p'_k
&={\sh{(\pi-\lambda)k\over2}\over\sh{\pi k\over2}},
\quad
&\theta'_k&={\sh{(\pi-2\lambda)k\over2}\over\sh{\pi k\over2}}
\quad
&(|\Delta|<1);
\\
p'_k
&=e^{-\lambda|k|/2},
\quad
&\theta'_k&=e^{-\lambda|k|}
\quad
&(\Delta<-1).
}
$$
\unpause
We have for the density
$$
\rho_k={p'_k\over1+\theta'_k}={1\over2\ch{\lambda k\over2}}
$$
in both cases.
\unpause
Then
$$
{n\over N}=\int^{\bar v_F}_{-\bar v_F}{dv\over2\pi}\,\rho(v)=\rho_0={1\over2}
\quad\Rightarrow\quad
{S^z\over N}\to0.
$$

\end{frame}

\begin{frame}
\frametitle{Free energy}
\unpause

Recall the expression for the eigenvalue
\eq$$
\Lambda(u;u_1,\ldots,u_n)=a^N(u)\prod_{i=1}^n{a(u_i-u)\over b(u_i-u)}+b^N(u)\prod_{i=1}^n{a(u-u_i)\over b(u-u_i)}.
\tag{\ref{Lambda}}
$$
\unpause
Now let us calculate the free energy per vertex of the six\-/vertex model:
\Align*$$
f=-\lim_{N\to\infty}{\log\Lambda_\text{max}(u)\over N}
=-\max\biggl(\log a(u)
&+\int^{v_F}_{-v_F}{dv\over2\pi}\,\rho(v)\only<2>{\log{a(iv-u+\lambda/2)\over b(iv-u+\lambda/2)}}\only<3->{{\red(-\i)p(\i u+v)}},
\\
\log b(u)
&+\int^{v_F}_{-v_F}{dv\over2\pi}\,\rho(v)\only<2>{\log{a(u-iv-\lambda/2)\over b(u-iv-\lambda/2)}}\only<3->{\rlap{$\red(-\i)p(\i(\lambda-u)+v)$}
\hphantom{{}\log{a(u-iv-\lambda/2)\over b(u-iv-\lambda/2)}}}
\biggr).
$$
\unpause\unpause
For $v_F=\bar v_F$ we can use the Fourier transform. For $|\Delta|<1$ we have
$$
f=\min\left(-\log a(u)-\int{dk\over k}\rho_{-k}p'_ke^{ku},
 -\log b(u)-\int{dk\over k}\rho_kp'_ke^{k(\lambda-u)}\right).
$$
\unpause
By symmetrizing the we find that the two alternatives coincide, so that
\Align$$
f
&=-\log a(u)-\int^\infty_0{dk\over k}\,
{\sh uk\sh{\pi-\lambda\over2}k\over\sh{\pi\over2}k\ch{\lambda\over2}k}
\notag
\\
&=-\log b(u)-\int^\infty_0{dk\over k}\,
{\sh(\lambda-u)k\sh{\pi-\lambda\over2}k\over\sh{\pi\over2}k\ch{\lambda\over2}k}.
\label{feglfin}
$$

\end{frame}

\begin{frame}
\frametitle{Free energy}
\unpause

In the case $\Delta<-1$ the free energy reads
$$
f=\min\left(-\log a(u)-{1\over\pi}\sum_{k\in2Z}{1\over k}\rho_{-k}p'_ke^{ku},
 -\log b(u)-{1\over\pi}\sum_{k\in2Z}{1\over k}\rho_kp'_ke^{k(\lambda-u)}\right).
$$
\unpause
Finally, we have
\Align$$
f
&=-\log a(u)-u-\sum_{m=1}^\infty{e^{-\lambda m}\sh2um\over m\ch\lambda m}
\notag
\\
&=-\log b(u)-(\lambda-u)
-\sum_{m=1}^\infty{e^{-\lambda m}\sh2(\lambda-u)m\over m\ch\lambda m}.
\label{feaffin}
$$
\unpause
\emph{Why are these two cases so different?}
\unpause
Because in the case $|\Delta|<1$ there is a \emph{gapless} spectrum, while in the case $\Delta<-1$ there is a \emph{gap} between the two largest eigenvalues of $T(u)$ and all other eigenvalues.
\unpause

\emph{What if $v_F<\bar v_F$?}
\unpause
This case corresponds to general homogeneous six\-/vertex model with arbitrary $a,a',b,b',c,c'$. The ratio $c/c'$ is inessential, but nonunit rations $a/a'$, $b/b'$ correspond to an \emph{external field}. They can be related to $v_F$. The integral equations do not have an analytic solution, but can be solved numerically. The two alternatives for the free energy are different.



\end{frame}



\end{document}