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\begin{document}

\title[Lecture 7. Thirring model: Bethe Ansatz]{Lecture 7\\Thirring model: solution by the Bethe Ansatz method}
\date{}

\frame{\titlepage}

\begin{frame}
\frametitle{Thirring model: the Hamiltonian formulation}
The action
\eq$$
S^{MT}[\psi,\bpsi]
=\int d^2x\,
\left(\bpsi(i\hat\d-m_0)\psi-{g\over2}(\bpsi\gamma^\mu\psi)^2\right)
\label{MT}
$$
with
\eq$$
\psi=\pMatrix{\psi_+\\\psi_-},
\qquad
\gamma^0=\pMatrix{&-i\\i&}=\sigma^2,
\qquad
\gamma^1=\pMatrix{&i\\i&}=i\sigma^1.
\label{gamma-matrices}
$$
\pause
The Hamiltonian
\eq$$
H=\int dx\,(-i\psi^+\sigma^3\d_x\psi+m_0\psi^+\sigma^2\psi
+2g\psi^+_+\psi_+\psi^+_-\psi_-)
\label{MTHam}
$$
\pause
with the commutation relations
\eq$$
\psi^+_{\alpha'}(x')\psi_\alpha(x)+\psi_\alpha(x)\psi^+_{\alpha'}(x')
=\delta_{\alpha'\alpha}\delta(x'-x).
\label{MTpsicommut}
$$
\pause
Conserved charges: the momentum $P$ and the particle number operator $Q$ are
\eq$$
P=-i\int dx\,\psi^+\d_x\psi,
\qquad
Q=\int dx\,\psi^+\psi.
\label{MTPQ}
$$

\end{frame}

\begin{frame}
\frametitle{Free fermion. Dirac's picture}

Consider the case $g=0$: free Dirac fermion.
\pause

Define the `reference' state $|\Omega\rangle$:
\eq$$
\psi_\alpha(x)|\Omega\rangle=0,
\qquad
\langle\Omega|\psi^+_\alpha(x)=0.
\label{Omegadef}
$$
\pause
``$N$\-/particle'' state:
\eq$$
|\chi_N\rangle
=\int d^Nx\,\chi^{\alpha_1\ldots\alpha_N}(x_1,\ldots,x_N)
\psi^+_{\alpha_1}(x_1)\ldots\psi^+_{\alpha_N}(x_N)|\Omega\rangle.
\label{Npart}
$$
\pause
\eq$$
Q|\chi_N\rangle=N|\chi_N\rangle
\quad\Rightarrow\quad
\cH\simeq\bigoplus^\infty_{N=0}\cH_N,
\qquad
\cH_N=\left\{v\in\cH\>|\>Qv=Nv\right\}.
$$
\pause
Define action $\hat H_N$ of $H$ on the wave function:
$$
H|\chi_N\rangle
=\int d^Nx\,(\hat H_N\chi)^{\alpha_1\ldots\alpha_N}(x_1,\ldots,x_N)
\psi^+_{\alpha_1}(x_1)\ldots\psi^+_{\alpha_N}(x_N)|\Omega\rangle.
$$
\pause
Explicitly, we have
$$
\hat H_N=\sum^N_{k=1}(-i\sigma^3_k\d_{x_k}+m_0\sigma^2_k),
$$
where $\sigma^i_k$ acts on the space of the $k$th particle:
$$
(\sigma^i_k\chi)^{\alpha_1\ldots\alpha_k\ldots\alpha_N}=\sum_{\alpha'_k}(\sigma^i)^{\alpha_k}_{\alpha'_k}\chi^{\alpha_1\ldots\alpha'_k\ldots\alpha_N}
$$

\end{frame}

\begin{frame}
\frametitle{Free fermion: wave functions}

For $N=1$, the eigenfunction is
\eq$$
\chi_\lambda(x)=\pMatrix{e^{\lambda/2}\\ie^{-\lambda/2}}e^{ixm_0\sh\lambda}.
\label{1psolution}
$$
\pause
For general $N$ we have the Slater determinant
\eq$$
\chi^{\alpha_1\ldots\alpha_N}_{\lambda_1\ldots\lambda_N}(x_1,\ldots,x_N)
=\sum_\sigma(-1)^\sigma\prod^N_{k=1}
\chi^{\alpha_{\sigma_k}}_{\lambda_k}(x_{\sigma_k}).
\label{Npsolutions-free}
$$
The energy of the $N$\-/particle state is equal to
\eq$$
E_N(\lambda_1,\ldots,\lambda_N)=\sum^N_{k=1}\epsilon(\lambda_k),
\qquad
\epsilon(\lambda)=m_0\ch\lambda.
\label{ENfree}
$$
\pause
The periodic boundary condition
$$
\chi^{\alpha_1\alpha_2\ldots\alpha_N}(x_1+L,x_2,\ldots,x_N)=\chi^{\alpha_1\alpha_2\ldots\alpha_N}(x_1,x_2,\ldots,x_N)
$$
yields
\eq$$
e^{im_0L\sh\lambda_k}=1,
\qquad k=1,\ldots,N.
\label{thetaquant}
$$

\end{frame}

\begin{frame}
\frametitle{Free fermion: vacuum state}

Take the logarithm on the last equation
$$
\sh\lambda_k={2\pi n_k\over m_0L},
\qquad
n_k\in\Z.
$$
\pause
Hence,
$$
\lambda_k\in\Cases{\R&\Rightarrow\epsilon(\lambda_k)>0;\\\R+i\pi&\Rightarrow\epsilon(\lambda_k)<0.}
$$
In the ground state all one\-/particle states with $\epsilon(\lambda_k)<0$ must be filled up.
\pause
Let
$$
\lambda_k=i\pi+\xi_k,
\qquad
\xi_k\in\R.
$$
We have to cut the band from below:
\eq$$
\epsilon(\lambda_k)\ge-\Lambda
\quad\Rightarrow\quad
-\Theta<\xi_k<\Theta,
\qquad
\Theta\simeq\log{\Lambda\over m_0}.
\label{cutoff}
$$
\pause
The vacuum energy in the thermodynamic limit $L\to\infty$:
$$
E_0=-L\int^\Theta_{-\Theta}{d\xi\over2\pi}\,
\rho(\xi)m_0\ch\xi,
\qquad
\rho(\xi)={2\pi\over L}\left|dn\over d\xi\right|=m_0\ch\xi.
$$
Here $\rho(\xi)$ is the \emph{spectral density of states} of the valence band.

\end{frame}

\begin{frame}
\frametitle{Free fermion: excitations}

There are two types of excitations:
\begin{itemize}
\item Particles: $\lambda_k\in\R$;
\item Antiparticles or holes: absence of some $\lambda_k=i\pi+\xi_k$.
\end{itemize}
$$
\begin{tikzpicture}
\draw[->] (-3,0) -- (3,0) node[right] {$p=m_0\sh\lambda$};
\draw[->] (0,-3) -- (0,3) node[above] {$\epsilon=m_0\ch\lambda$};
\draw[dashed,thick,domain=-2:2] plot function{.5*(1+x**2)};
\draw[-,thick,domain=-2:2] plot function{-.5*(1+x**2)};
\filldraw (1,1) circle(3pt) node[below right] {particle};
\filldraw[color=white] (-1.5,-1.625) circle(3pt); \draw[-,thick] (-1.5,-1.625) circle(3pt) node[above left] {hole};
\draw[->] (2,-1) node[above] {Dirac sea} -- (1.6,-1.525);
\end{tikzpicture}
$$

\end{frame}

\begin{frame}
\frametitle{Interaction: Hamiltonian and wave function discontinuity}

For $g\ne0$ we have
\eq$$
\hat H_N=\sum^N_{k=1}(-i\sigma^3_k\d_{x_k}+m_0\sigma^2_k)
+g\sum^N_{k<l}\delta(x_k-x_l)
(1-\sigma^3_k\sigma^3_l).
\label{MTHN}
$$
\pause
Here
\eq$$
{1\over2}(1\otimes1-\sigma^3\otimes\sigma^3)
^{\alpha'_1\alpha'_2}_{\alpha_1\alpha_2}
=\delta^{\alpha'_1}_{\alpha_1}\delta^{\alpha'_2}_{\alpha_2}\delta_{\alpha_1,-\alpha_2}.
\label{sigmaid}
$$
\pause
The interaction term in the Hamiltonian (\ref{MTHN}) is poorly defined. Indeed, the wave function is \emph{discontinuous} at $x_k=x_l$, since the Schrödinger equation here is a \emph{first order} equation.
\pause
Consider a simple equation
$$
f'(x)-c\delta(x)f(x)=g(x,f(x))
$$
\pause
Let us regularize the delta\-/function in an arbitrary way:
\eq$$
f'(x)-c\delta_a(x)f(x)=g(x,f(x)),
\quad
\lim_{a\to0}\delta_a(x)=\delta(x),
\quad
\mathop{\rm supp}\delta_a=[-a,a].
$$
\pause
Let $\delta_a(x)=\epsilon'_a(x)$. Then
$$
{f'(x)\over f(x)}=c\epsilon'_a(x)+{\red\xcancel{\black g(x,f(x))\over f(x)}}
\quad\Rightarrow\quad
f(x)=\const e^{c\epsilon_a(x)}.
$$
\pause
\eq$$
f(+a)=e^cf(-a)
\quad\xRightarrow{\ a\to0\ }\quad
f(+0)=e^cf(-0).
$$

\end{frame}

\begin{frame}
\frametitle{Interaction: two-particle wave function}

Let
\eq$$
\chi^{\alpha_1\alpha_2}_{\lambda_1\lambda_2}(x_1,x_2)
=\Cases{
A_{12}\chi^{\alpha_1}_{\lambda_1}(x_1)\chi^{\alpha_2}_{\lambda_2}(x_2)
-A_{21}\chi^{\alpha_1}_{\lambda_2}(x_1)\chi^{\alpha_2}_{\lambda_1}(x_2)
\quad\text{for $x_1<x_2$,}\\
A_{21}\chi^{\alpha_1}_{\lambda_1}(x_1)\chi^{\alpha_2}_{\lambda_2}(x_2)
-A_{12}\chi^{\alpha_1}_{\lambda_2}(x_1)\chi^{\alpha_2}_{\lambda_1}(x_2)
\quad\text{for $x_1>x_2$.}
}
\label{2pAnsatz}
$$
\pause
It is antisymmetric:
$$
\chi^{\alpha_{\red1}\alpha_{\red2}}_{\lambda_1\lambda_2}(x_{\red1},x_{\red2})
=-\chi^{\alpha_{\red2}\alpha_{\red1}}_{\lambda_1\lambda_2}(x_{\red2},x_{\red1}).
$$
\pause
After applying the rule from the last slide we have
\eq$$
{A_{21}\over A_{12}}=R(\lambda_1-\lambda_2),
\qquad
R(\lambda)=e^{i\Phi(\lambda)}
={\ch{\lambda-ig\over2}\over\ch{\lambda+ig\over2}}.
\label{Phifunc}
$$
\pause
The function is periodic in $g$ with the period $2\pi$. Assume
\eq$$
-\pi<g<\pi.
\label{gregion}
$$
\pause
For $\Phi(\lambda)$ assume the skew symmetry
\eq$$
\Phi(-\lambda)=-\Phi(\lambda),
\label{Phiskewsym}
$$
with the cuts lie on the rays $(i(\pi-|g|),i\infty)$, $(-i(\pi-|g|),-i\infty)$.

\end{frame}

\begin{frame}
\frametitle{Interaction: Bethe Ansatz and Bethe equations}

The $N$\-/particle solution (\emph{Bethe Ansatz}):
\eq$$
\chi^{\alpha_1\ldots\alpha_N}_{\lambda_1\ldots\lambda_N}(x_1,\ldots,x_N)
=\sum_\tau(-1)^{\sigma\tau}
A_\tau\prod^N_{k=1}\chi^{\alpha_{\sigma_k}}_{\lambda_{\tau_k}}(x_{\sigma_k})
\quad\text{for $x_{\sigma_1}<\ldots<x_{\sigma_N}$}.
\label{BetheAnsatz}
$$
\pause
The coefficients $A$ satisfy the relations
\eq$$
A_{\ldots,i+1,i,\ldots}
=R(\lambda_{i}-\lambda_{i+1})
A_{\ldots,i,i+1,\ldots}.
\label{Arelations}
$$
\pause
The periodic boundary condition
$$
\chi^{\alpha_1\alpha_2\ldots\alpha_N}(x_1+L,x_2,\ldots,x_N)=\chi^{\alpha_1\alpha_2\ldots\alpha_N}(x_1,x_2,\ldots,x_N)
$$
imposes on $\lambda_k$ the system of \emph{Bethe equations}:
\eq$$
e^{im_0L\sh\lambda_k}\prod^N_{\scriptstyle l=1\atop\scriptstyle l\ne k}R(\lambda_k-\lambda_l)=1.
\label{BetheEqs}
$$
\pause
A set $\{\lambda_1,\ldots,\lambda_N\}$ that satisfy (\ref{BetheEqs}) is called a \emph{solution} to the Bethe equations, while each element is called a \emph{root} of the Bethe equations. All roots are \emph{different}:
\eq$$
\lambda_k\ne\lambda_l,
\quad\text{if $k\ne l$.}
\label{nnne}
$$

\end{frame}

\begin{frame}
\frametitle{Ground state}

Take logarithm of the Bethe equations:
\eq$$
m_0L\sh\lambda_k+\sum^N_{l=1}\Phi(\lambda_k-\lambda_l)=2\pi n_k,
\qquad
n_k\in\Z.
\label{BetheLog}
$$
\pause
Energy and momentum:
\eq$$
E_N(\lambda_1,\ldots,\lambda_N)=m_0\sum^N_{i=1}\ch\lambda_i,
\qquad
P_N(\lambda_1,\ldots,\lambda_N)=m_0\sum^N_{i=1}\sh\lambda_i.
\label{EP}
$$
\pause
\textbf{Conjecture:} all negative energy one\-/particle states are filled up in the \emph{ground (vacuum) state}. Then
$$
\lambda_k=i\pi+\xi_k,
\qquad
n_k=k_0-k.
$$
\pause
Hence,
\eq$$
m_0L\sh\xi_k=2\pi(k-k_0)+\sum^N_{l=1}\Phi(\xi_k-\xi_l).
\label{BetheLog-vac}
$$

\end{frame}

\begin{frame}
\frametitle{Ground state: thermodynamic limit}

Take the difference for neighboring $k$ and divide by $L(\xi_{k+1}-\xi_k)$:
$$\smash{
m_0{\sh\xi_{k+1}-\sh\xi_k\over\xi_{k+1}-\xi_k}
={2\pi\over L(\xi_{k+1}-\xi_k)}+{1\over L}\sum^N_{l=1}{\Phi(\xi_{k+1}-\xi_l)-\Phi(\xi_k-\xi_l)\over\xi_{k+1}-\xi_k}.}
$$
\pause
In the limit $L\to\infty$ the values $\xi_k$ become \emph{dense}. Then define
\eq$$
\rho(\xi)={2\pi\over L(\xi_{k+1}-\xi_k)}={2\pi\over L}{dk\over d\xi}.
\label{rhodef}
$$
\pause
Substituting the sum by and integral, we obtain
\eq$$
m_0\ch\xi=\rho(\xi)+\int^\Theta_{-\Theta}{d\xi\over2\pi}'\,\Phi'(\xi-\xi')\rho(\xi').
\label{BetheTL}
$$
We again need the ultraviolet cutoff $\Theta$.
\pause

We have lost one equation. Thus substitute it by the normalization condition
\eq$$
\int^\Theta_{-\Theta}{d\xi\over2\pi}\,\rho(\xi)={N\over L}.
\label{rhonorm}
$$
\pause
Approximate solution
\eq$$
\rho(\xi)=\rho_0\ch{\pi\xi\over\pi+g},
\qquad
\rho_0\sim m_0e^{g\Theta\over\pi+g}.
\label{rhosolution}
$$
$\rho_0$ is finite $\Rightarrow$ \emph{renormalization} of $m_0$.

\end{frame}

\begin{frame}
\frametitle{Excitations in the thermodynamic limit: holes}

Consider the Dirac sea with \emph{holes}: $n_k$ do not cover a segment of $\Z$. Then
\eq$$
m_0L\sh\xi_k=-2\pi n_k+\sum^N_{l=1}\Phi(\xi_k-\xi_l).
\label{BetheLog-holes}
$$
\pause
Define $\xi(n)$ by the equation
\eq$$
m_0L\sh\xi(n)=-2\pi n+\sum^N_{l=1}\Phi(\xi(n)-\xi_l)
\label{BetheLog-holes-beta(n)}
$$
Evidently $\xi(n_k)=\xi_k$. In this case the state is filled. Otherwise it is empty.
\pause
Define two spectral densities:
\begin{itemize}
\item Density of states: $\rho(\xi(n))={2\pi\over L|\xi(n+1)-\xi(n)|}\simeq{2\pi\over L}\left|dn\over d\xi(n)\right|$.
\pause
\item Density of particles $\rho^\bullet(\xi)=\left\langle{2\pi\over L|\xi_{k+1}-\xi_k|}\right\rangle_{\xi_k\simeq\xi}
=\left\langle{2\pi\over L}\left|dk\over d\xi_k\right|\right\rangle_{\xi_k\simeq\xi}$.
\end{itemize}
There difference $\rho(\xi)-\rho^\bullet(\xi)=\rho^\circ(\xi)$ is the \emph{density of holes}.

\end{frame}

\begin{frame}
\frametitle{Integral equation for holes}

Integral equation
\eq$$
m_0\ch\xi
=\rho(\xi)+\int^\Theta_{-\Theta}{d\xi'\over2\pi}\,\Phi'(\xi-\xi')(\rho(\xi')-\rho^\circ(\xi')).
\label{p-eq2}
$$
\pause
Let $\rho_\text{vac}(\xi)$ is the ground state (vacuum) density of states~(\ref{rhosolution}), which solves
\eq$$
m_0\ch\xi=\rho_\text{vac}(\xi)+\int^\Theta_{-\Theta}{d\xi'\over2\pi}\,\Phi'(\xi-\xi')\rho_\text{vac}(\xi').
\tag{\ref{BetheTL}}
$$
Let
$$
\delta\rho(\xi)=\rho(\xi)-\rho_\text{vac}(\xi).
$$
\pause
Take the difference of (\ref{p-eq2}) and (\ref{BetheTL}):
\eq$$
\delta\rho(\xi)
+\int^\Theta_{-\Theta}{d\xi'\over2\pi}\,\Phi'(\xi-\xi')\delta\rho(\xi')
=\int^\Theta_{-\Theta}{d\xi'\over2\pi}\,\Phi'(\xi-\xi')\rho^\circ(\xi').
\label{deltarhoeq}
$$
Let us solve it in the limit $\Theta\to\infty$.

\end{frame}

\begin{frame}
\frametitle{The limit $\Theta\to\infty$. Fourier transform}

For $\Theta\to\infty$ we may apply the Fourier transform:
$$
\tilde X(\omega)
=\int^\infty_{-\infty}{d\xi\over2\pi}\,X(\xi)e^{i\xi\omega},
\quad
X=\Phi',\delta\rho,\rho^\circ,\ldots
$$
\pause
Equation (\ref{deltarhoeq}) takes the form
$$
\delta\tilde\rho(\omega)+\tilde\Phi'(\omega)\delta\tilde\rho(\omega)=\Phi'(\omega)\tilde\rho^\circ(\omega).
$$
\pause
It is easy to check that
\eq$$
\tilde\Phi'(\omega)=-{\sh g\omega\over\sh\pi\omega},
\qquad
\delta\tilde\rho(\omega)=-{\sh g\omega\over2\sh{\pi-g\over2}\omega\ch{\pi+g\over2}\omega}\tilde\rho^\circ(\omega).
\label{Phi-deltarho-omega}
$$

\end{frame}

\begin{frame}
\frametitle{Fractional charge of a hole}


Effect of a cutoff: \emph{fractional charge}. Formal charge of a hole is $-1$, but when holes are inserted, particles are pulled into the region $-\Theta\le\xi\le\Theta$ or pushed off it. We have two quantities:
\eq$$
\Delta N
=-L\int^\Theta_{-\Theta}{d\xi\over2\pi}\,\rho^\circ(\xi)=-\tilde\rho^\circ(0),
\quad
\Delta Q
=L\int^\Theta_{-\Theta}{d\xi\over2\pi}\,(\delta\rho(\xi)-\rho^\circ(\xi)).
\label{DeltaN-DeltaQ-def}
$$
\pause
We obtain
\eq$$
\Delta Q=-\int^\Theta_{-\Theta}{d\xi\over2\pi}\,\int^\infty_{-\infty}d\omega\,e^{-i\omega\xi}
{\sh\pi\omega\over2\sh{\pi-g\over2}\omega\ch{\pi+g\over2}\omega}\tilde\rho^\circ(\omega)
\simeq{\pi\over\pi-g}\Delta N.
\label{DeltaQ-fin}
$$
\pause
Then the charge of one hole is
\eq$$
z^\circ=-{\Delta Q\over\Delta N}=-{\pi\over\pi-g}.
\label{charge-zh-fin}
$$
\pause
It means that the quantity $Q$ does not coincide with the physical number of particles. Hence,
\eq$$
Q=|z^\circ|\int dx\,\psi_\phys^+\psi_\phys+\const
\quad\Rightarrow\quad
\psi=|z^\circ|^{1/2}\psi_\phys.
\label{Q-psiphys}
$$
\pause
It corresponds to the renormalization of the coupling constant $g$:
\eq$$
g_\phys=g|z^\circ|={g\over1-g/\pi}
\quad\Leftrightarrow\quad
{1\over g_\phys}={1\over g}-{1\over\pi}.
\label{g-gphys}
$$
\pause
$$
-\pi<g<\pi\quad\Leftrightarrow\quad-{\pi\over2}<g_\phys<\infty
$$
in consistency with the boson\--fermion correspondence.

\end{frame}

\begin{frame}
\frametitle{Mass of a hole}

Calculate the energy and momentum of a state with holes:
$$
\Aligned{
E[\rho^\circ]-E[0]
&=m_0L\int^\Theta_{-\Theta}{d\xi\over2\pi}\,(\rho^\circ(\xi)-\delta\rho(\xi))\ch\xi,
\\
P[\rho^\circ]
&=m_0L\int^\Theta_{-\Theta}{d\xi\over2\pi}\,(\rho^\circ(\xi)-\delta\rho(\xi))\sh\xi.
}
$$
\pause
Let $g<0$. The integrals converge as $\Theta\to\infty$, but at $\Theta=\infty$ we have $E[\rho^\circ]-E[0]=P[\rho^\circ]=0$. Thus $m_0$ must be \emph{renormalized to infinity}.
\pause

Leg $g>0$. The integrals diverge as $\Theta\to\infty$. Thus $m_0$ must be \emph{renormalized to zero}.
\pause

Assuming $\Theta$ large but finite, we can obtain
\eq$$
E[\rho^\circ]-E[0]=L\int^\infty_{-\infty}{d\xi\over2\pi}\,\epsilon(\xi)\rho^\circ(\xi),
\qquad
P[\rho^\circ]=L\int^\infty_{-\infty}{d\xi\over2\pi}\,p(\xi)\rho^\circ(\xi)
\label{EP-fin}
$$
where
\eq$$
\epsilon(\lambda)=m\ch{\pi\lambda\over\pi+g},
\qquad
p(\lambda)=m\sh{\pi\lambda\over\pi+g},
\qquad
m={M\over g}\ctg\left({\pi\over2}{\pi-g\over\pi+g}\right)
\label{spectrum}
$$
and
\eq$$
m_0=M\exp\left(-{g\over\pi+g}\Theta\right)
\sim M\left(m_0\over\Lambda\right)^{g/(\pi+g)}.
\label{m0div}
$$

\end{frame}

\begin{frame}
\frametitle{Masses and scales}

Thus, the particles have a relativistic spectrum with the mass
$$
m\sim m_0\exp\left({g\over\pi+g}\Theta\right)
$$
and the rapidity
$$
\theta={\pi\xi\over\pi+g}.
\label{theta-beta-rel}
$$
\pause

Comparing with the estimation
$$
m_0\sim m^{2-\beta^2},
$$
where $\beta$ is the coupling constant of the sine\-/Gordon model, we obtain
$$
{g\over\pi}=1-\beta^2.
$$
\pause
In terms of the physical coupling constant $g_\phys$ it reads
$$
{g_\phys\over\pi}=\beta^{-2}-1,
$$
in consistency with the bosonization.

\end{frame}

\begin{frame}
\frametitle{Model system}

Consider the model system of `spinless' fermions with the $S$ matrix $S(\theta)=e^{i\Psi(\theta)}$, $\Psi(-\theta)=-\Psi(\theta)$. If particles are far from each other we may apply the Bethe Ansatz to them and obtain the Bethe equations
\eq$$
e^{imL\sh\theta_k}\prod^N_{l=1\atop l\ne k}S(\theta_l-\theta_k)=1,
\label{auxBetheEqs}
$$
where $\theta_k$ are physical rapidities of the particles and $m$ is their physical mass.
\pause
By taking logarithm we obtain the equations
\eq$$
mL\sh\theta_k+\sum^N_{l=1}\Psi(\theta_k-\theta_l)=2\pi n_k.
\label{auxBetheLog}
$$
\pause
Define $\theta(n)$ by the equation
\eq$$
mL\sh\theta(n)=\sum^N_{l=1}\Psi(\theta(n)-\theta_l)=2\pi n.
\label{auxTheta(n)}
$$
\pause
Introduce the densities:
\eq$$
\Aligned{
\rho_*(\theta(n))
&={2\pi\over L|\theta(n+1)-\theta(n)|}\simeq{2\pi\over L}\left|dn\over d\theta(n)\right|,
\\
\rho^\bullet_*(\theta)
&=\left\langle{2\pi\over L|\theta_{k+1}-\theta_k|}\right\rangle_{\theta_k\simeq\theta}
=\left\langle{2\pi\over L}\left|dk\over d\theta_k\right|\right\rangle_{\theta_k\simeq\theta}.
}\label{aux-rho-rhof-def}
$$

\end{frame}

\begin{frame}
\frametitle{Model system: integral equations}

In the same way we obtain the integral equation
\eq$$
m\ch\theta+\int^\infty_{-\infty}{d\theta'\over2\pi}\,\Psi'(\theta-\theta')\rho^\bullet_*(\theta')=2\pi\rho_*(\theta).
\label{aux-p-eq2}
$$
\pause
If $\rho^\bullet_*(\theta)=0$, we immediately obtain $\rho_{*\text{vac}}(\theta)=m\ch\theta$. With $\delta\rho_*=\rho_*-\rho_{*\text{vac}}$ we have
\eq$$
\delta\rho_*(\theta)=\int^\infty_{-\infty}{d\theta'\over2\pi}\,\Psi'(\theta-\theta')\rho^\bullet_*(\theta').
\label{aux-deltarhoeq}
$$
\pause
For the Fourier transforms we have
\eq$$
\delta\tilde\rho_*(\omega)=\tilde\Psi'(\omega)\rho^\bullet_*(\omega).
\label{aux-p-eq2-fourier}
$$
\pause
Identify
\eq$$
\rho^\circ(\xi)={d\theta\over d\xi}\rho^\bullet_*(\theta),
\quad
\delta\rho(\xi)={d\theta\over d\xi}\delta\rho_*(\theta),
\quad
\theta={\pi\over\pi+g}\xi.
\label{rho*-rho-rel}
$$
\pause
In the Fourier transforms
\eq$$
\tilde\rho^\circ\left({\pi\over\pi+g}\omega\right)=\tilde\rho^\bullet_*(\omega),
\qquad
\delta\tilde\rho\left({\pi\over\pi+g}\omega\right)=\delta\tilde\rho_*(\omega).
\label{rho*-rho-omega-rel}
$$

\end{frame}

\begin{frame}
\frametitle{Thirring model: $S$ matrix}

Comparing it with~(\ref{Phi-deltarho-omega}) we obtain
\Align$$
\Psi(\theta)
&=i\int^\infty_{-\infty}{d\omega\over\omega}
{\sh{\pi\omega\over2}\sh{\pi(p-1)\omega\over2}\over\sh\pi\omega\sh{\pi p\omega\over2}}e^{-i\theta\omega}
\notag
\\
&=2\int^\infty_0{d\omega\over\omega}
{\sh{\pi\omega\over2}\sh{\pi(p-1)\omega\over2}\over\sh\pi\omega\sh{\pi p\omega\over2}}\sin\theta\omega,
\quad
\beta^2=1-{g\over\pi}=2{p\over p+1}.
\label{Psi-fin}
$$
This identifies the function $S(\theta)=e^{i\Psi(\theta)}$ with the matrix element $S(\theta)^{--}_{--}$ of two antifermions in the Thirring model.
\pause

Other matrix elements can be found by means of the Yang\--Baxter equations and in the basis $(++,+-,-+,--)$ are given by
\eq$$
S(\theta)
=\left(S^{\alpha_1\alpha_2}_{\beta_1\beta_2}(\theta)\right)
=\pMatrix{a(\theta)&&&\\
&b(\theta)&c(\theta)&\\
&c(\theta)&b(\theta)&\\
&&&a(\theta)},
$$
where $a(\theta)=e^{i\Psi(\theta)}$, and
\eq$$
{b(\theta)\over a(\theta)}
={\sh{\theta\over p}\over\sh{i\pi-\theta\over p}},
\qquad
{c(\theta)\over a(\theta)}
={\sh{i\pi\over p}\over\sh{i\pi-\theta\over p}}.
\label{ba-ca-ratios}
$$

\end{frame}

\begin{frame}
\frametitle{Thirring model: bound states}

For $g<0$ (or $g_\phys<0$, $\beta^2<1$, $p<1$) the elements $b(\theta)$ and $c(\theta)$ have poles on the physical sheet at the points
\eq$$
\theta_n=i\pi-i\pi pn,
\qquad n=1,2,\ldots,\left\lfloor1\over p\right\rfloor,
\label{theta-SG-bound}
$$
which correspond to the neutral bound states with the masses
\eq$$
M_n=2m\sin{\pi pn\over2}
\label{masses-SG-bound}
$$
In the sine\-/Gordon model these bound states correspond to the breather excitations, and in the classical limit $\beta^2\to0$ their spectrum becomes continuous in consistency with the classical field theory.


\end{frame}

\begin{frame}
\frametitle{Seminar}

\end{frame}



\end{document}
