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\begin{document}

\title[Lecture 6. $O(N)$\-/model: integrability]{Lecture 6.\\$O(N)$\-/model: integrability and the exact $S$\-/matrix}
\date{}
\author{Michael Lashkevich}

\frame{\titlepage}

\begin{frame}
\frametitle{Conformal invariance of the action}
\unpause

Recall the action of the $O(N)$\-/model
$$
S[\bn,\omega]={1\over2g}\int d^2x\,((\d_\mu\bn)^2-\omega(\bn^2-1)).
$$
\unpause
The equation of motion is
$$
\d^\mu\d_\mu\bn+\omega\bn=0,
\qquad
\bn^2=1.
$$
\unpause
In the light cone coordinates $z$, $\bz$ we have
\eq$$
S[\bn,\omega]
=-{1\over g}\int dz\,d\bz\,
\left(\d\bn\bd\bn+{\omega\over4}(\bn^2-1)\right),
\label{ONaction}
$$
\unpause
and
\eq$$
4\d\bd\bn=\omega\bn,
\qquad
\bn^2=1.
\label{ONeqmotion}
$$
\unpause
The action is invariant with respect to {\it pseudoconformal\/} transformations
\eq$$
z\to f_1(z),
\qquad
\bz\to f_2(\bz),
\qquad
\omega\to{\omega\over f'_1(z)f'_2(\bz)}.
\label{pseudoconformal}
$$
\unpause
The transformations include, in particular, translations
$$
f_1(z)=z+c,
\qquad
f_2(\bz)=\bz+\bar c,
$$
\unpause
scaling and Lorentz transformations, for which
$$
f_1(z)=\lambda z,
\qquad
f_2(\bz)=\bar\lambda\bz,
$$
\unpause
and the inversion transformation
$$
f_1(z)=1/z,
\qquad
f_2(\bz)=1/\bz.
$$

\end{frame}

\begin{frame}
\frametitle{Classical conservation laws}
\unpause

Energy\-/momentum tensor:
$$
T_{zz}={1\over g}(\d\bn)^2,
\qquad
T_{\bz\bz}={1\over g}(\bd\bn)^2,
\qquad
T_{z\bz}=T_{\bz z}=-{\omega\over4g}(\bn^2-1).
$$
\unpause
On the equations of motion $T^\mu_\mu=-4T_{z\bz}=0$, which also expresses conformal invariance.
\unpause
The energy\-/momentum conservation:
\eq$$
\bd(\d\bn)^2=0,
\qquad
\d(\bd\bn)^2=0.
\label{im1classical}
$$
\unpause
There are higher spin conservation laws in the $O(N)$\-/model.
\unpause
Indeed, it is easy to check that
\eq$$
4\bd(\d^2\bn)^2=\d(\omega(\d\bn)^2)-3\d\omega(\d\bn)^2,
\quad
4\d(\bd^2\bn)^2=\bd(\omega(\bd\bn)^2)-3\bd\omega(\bd\bn)^2.
\label{im2classical}
$$
\unpause
Make a pseudoconformal transformation
$$
dz'=\left|\d\bn\over\d z\right|dz,
$$
\unpause
In the new coordinates $(\d\bn)^2=1$.
\unpause
Hence, there is a continuity equation
$$
\bd(\d\d\bn)^2=\d(2\d\bn\bd\bn).
$$
\unpause
Similarly
$$
\d(\bd\bd\bn)^2=\bd(2\bd\bn\d\bn).
$$

\end{frame}

\begin{frame}
\frametitle{Quantum conservation laws: heuristic derivation}
\unpause

On the quantum level anomalies appear that formally retain the Lorentz and scale invariance: $z\to\lambda z$, $\bz\to\bar\lambda\bz$.
\unpause
Hence, the r.h.s.\ of the energy\-/momentum conservation changes. It must be a $\d$\-/derivative.
\unpause
The only admissible form is
\eq$$
\bd(\d\bn)^2=-\beta\d\omega,
\label{im1anomaly}
$$
where $\beta$ is a constant.
\unpause
In fact, it breaks the scaling invariance, since it means that $T^\mu_\nu\ne0$.
\unpause

Consider $T_{zz}^2$. In classical case it satisfied $\bd T_{zz}^2=0$,
\unpause
but in quantum case has its own anomaly:
\eq$$
\bd(\d\bn)^4=-(2\beta+\alpha')(\d\bn)^2\d\omega+\d(\ldots).
\label{im12anomaly}
$$
\unpause
Similarly instead of (\ref{im2classical}) we have
\eq$$
\bd(\d^2\bn)^2=-(3+\alpha)(\d\bn)^2\d\omega+\d(\ldots).
\label{im2anomaly}
$$
\unpause
Hence,
\eq$$
\bd\left(
(\d^2\bn)^2-{3+\alpha\over2\beta+\alpha'}(\d\bn)^4
\right)=\d(\ldots).
\label{im2quantum}
$$
\unpause
We have two integrals of motion
So, there are at least two integrals of motion of spin~1 and spin~3:
\eq$$
I_1=\int dz{1\over2}(\d\bn)^2,
\qquad
I_3=\int dz\left(
{1\over2}(\d^2\bn)^2-{3+\alpha\over2(2\beta+\alpha')}(\d\bn)^4
\right),
\label{ims}
$$
which satisfy the equations $\bd I_1=0$, $\bd I_3=0$.

\end{frame}

\begin{frame}
\frametitle{IMs and elasticity of scattering}
\unpause

By taking into account both components and both chiralities, we obtain four integrals of motion (\emph{IMs}): $I_{\pm1}$, $I_{\pm3}$, which satisfy the equations $\dot I_s=0$. We have $I_1\sim p_z$, $I_{-1}\sim p_\bz$.
\unpause
Let $|\theta_1,\ldots,\theta_n\rangle$ be a common eigenstate of IMs with $n$ particles with rapidities $\theta_i$, i.e.\ $p_{iz}=-{m\over2}e^{\theta_i}$, $p_{i\bz}={m\over2}e^{-\theta_i}$.
\unpause
The eigenvalue of $I_s$ must be a homogeneous Laurent polynomial of $e^{\theta_i}$ of the power $s$.
\unpause

\emph{Locality}. Local IMs are space integrals, so that if particles are far from each other, their eigenvalues must be sums of eigenvalues on one\-/particle states.
\unpause
Hence
$$
I_s|\theta_1,\ldots,\theta_n\rangle=\const\sum^n_{i=1}e^{s\theta_i}|\theta_1,\ldots,\theta_n\rangle.
$$
\unpause
Consider a $2\to n$ particle scattering. Due to integrals of motion we have
$$
e^{s\theta_1}+e^{s\theta_2}
=\sum^n_{i=1}e^{s\theta'_i}
\qquad(s=-3,-1,1,3).
$$
\unpause
Let us fix the rapidities $\theta'_1,\ldots,\theta'_n$. We have \emph{four equations} for \emph{two variables} $\theta_1,\theta_2$, which depend on \emph{$n$ parameters}.
\unpause
If $n\ge3$, these equations only have solutions for special values of the parameters $\theta'_i$.
\unpause
But amplitudes must be analytic in $\theta'_i$.
\unpause
$\Rightarrow$ Amplitudes of these processes are \emph{identically zero}.
\unpause
$\Rightarrow$ Only processes with $n=2$ are allowed and \emph{$\theta'_i=\theta_i$} (up to an inessential permutation).
\unpause

In fact, the model has integrals of motion with all $s\in2\Z+1$.
\unpause
In this case only \emph{$n\to n$} processes are allowed with \emph{$\theta'_i=\theta_i$}. It is the \emph{ideally elastic scattering} characteristic for \emph{integrable models}.

\end{frame}

\begin{frame}
\frametitle{Factorizable scattering}
\unpause

For integrable models we have

\begin{block}{Factorized scattering assumption}
The scattering amplitude of $n$ particles into $n$ particles factorizes into the product of all pairwise scattering amplitudes in any order with summation over the internal states of the intermediate particles.
\end{block}

\unpause
Graphically it can be depicted as
\eq$$
\tikz[baseline=-2pt]{
\draw[->] (-1.333,-1.5) node[below left] {$p_1\!\!\!$} -- (1.619,1.5) node[above right] {$\!\!p_1$};
\draw[->] (-0.572,-1.5) node[below] {$p_2$} -- (1.048,1.5) node[above] {$p_2$};
\draw[->] (0.095,-1.5) node[below] {$p_3$} -- (-0.667,1.5) node[above] {$p_3$};
\draw[->] (1.429,-1.5) node[below right] {$\!\!p_4$} -- (-1.619,1.5) node[above left] {$p_4\!\!\!$};
\draw[-,fill=white] (0,0) circle (.9);
}
\quad=\quad
\tikz[baseline=-2pt]{
\draw[->] (-1.333,-1.5) node[below left] {$p_1\!\!\!$} -- (1.619,1.5) node[above right] {$\!\!p_1$};
\draw[->] (-0.572,-1.5) node[below] {$p_2$} -- (1.048,1.5) node[above] {$p_2$};
\draw[->] (0.095,-1.5) node[below] {$p_3$} -- (-0.667,1.5) node[above] {$p_3$};
\draw[->] (1.429,-1.5) node[below right] {$\!\!p_4$} -- (-1.619,1.5) node[above left] {$p_4\!\!\!$};
}
\label{factscat}
$$

\end{frame}

\begin{frame}
\frametitle{Asymptotic $n$\-/particle wave function}
\unpause

Suppose there is some characteristic distance $R$ beyond which virtual particles are not born. Then on large distances $|x_i-x_j|\gg R$ the wave  eigenfunction is indistinguishable from an $n$\-/particle wave function.
\unpause
Due to an infinite number of IMs all particles have constant momenta $p_i$ up to a permutation, and the wave function is a combination of the same plain waves for any order of $x_i$:
\Multline$$
\psi_{\beta_1p_1,\ldots,\beta_np_n}(\alpha_1x_1,\ldots,\alpha_nx_n)
=\sum_{\tau\in S_n}
A^{\alpha_{\sigma_1}\ldots\alpha_{\sigma_n}}
_{\beta_{1\,\hphantom\sigma}\ldots\beta_n}[\tau]
e^{i\sum^n_{i=1}p_{\tau_i}x_{\sigma_i}}
\\
\text{for}\quad
x_{\sigma_1}<x_{\sigma_2}<\cdots<x_{\sigma_n},
\quad
|x_i-x_j|\gg R.
\label{psifunction}
$$
Here $\alpha_i$ is an internal space of states of the particle located at $x_i$.
\unpause
The function is symmetric with respect to the permutations $\alpha_ix_i\leftrightarrow\alpha_jx_j$.
\unpause

The labels $\beta_i$ can be defined, e.g. by the requirement
$$
A^{\alpha_1\ldots\alpha_n}_{\beta_1\ldots\beta_n}[\id]
=\prod^n_{i=1}\delta^{\alpha_i}_{\beta_i}.
$$
For $p_1>p_2>\cdots>p_n$, the parameters $\beta_i$ naturally describe the internal states of the \emph{incoming} particles.
\unpause

Exchange of two neighboring particles means their scattering. Therefore
\eq$$
A^{\alpha_1\ldots\alpha_{i+1}\alpha_i\ldots\alpha_n}
_{\beta_1\ldots\beta_i\,\beta_{i+1}\ldots\beta_n}[\tau s^i]
=\sum_{\alpha'_i\alpha'_{i+1}}
S^{\alpha_i\alpha_{i+1}}_{\alpha'_i\alpha'_{i+1}}(p_{\tau_i},p_{\tau_{i+1}})
A^{\alpha_1\ldots\alpha'_i\alpha'_{i+1}\ldots\alpha_n}
_{\beta_1\ldots\beta_i\,\beta_{i+1}\ldots\beta_n}[\tau].
\label{Arelation}
$$
Here $s^i$ is the permutation of numbers $i$ and $i+1$. The matrix $S(p_1,p_2)$ is the \emph{two\-/particle $S$\-/matrix}.
\unpause
This defines the coefficients $A$ and proves the factorization assumption.

\end{frame}

\begin{frame}
\frametitle{Consistency requirements: three particle permutation}
\unpause

Due to the factorization the two\-/particle $S$\-/matrix satisfy a set of equations.
\unpause

Let us inverse the order of three consecutive particles, say $1,2,3$. We may do it in two ways:
$$
\Matrix{
            &        &132&\to&312&        &\\
            &\nearrow&            &   &            &\searrow&\\
123&        &            &   &            &        &321.\\
            &\searrow&            &   &            &\nearrow&\\
            &        &213&\to&231&        &
}
$$

\end{frame}

\begin{frame}
\frametitle{Consistency requirements: three particle permutation}
\unpause

The first way $123\to132\to312\to321$ leads to the relation
\Multline*$$
A^{\alpha_3\alpha_2\alpha_1\ldots}_{\cdots}[321\ldots]
\\
=\sum_{\beta_1,\beta_2,\beta_3}\left(\sum_{\gamma_1,\gamma_2,\gamma_3}
S^{\alpha_1\alpha_2}_{\gamma_1\gamma_2}(p_1,p_2)
S^{\gamma_1\alpha_3}_{\beta_1\gamma_3}(p_1,p_3)
S^{\gamma_2\gamma_3}_{\beta_2\beta_3}(p_2,p_3)\right)
A^{\beta_1\beta_2\beta_3\ldots}_{\cdots}[123\ldots]
$$
\unpause
or graphically
$$
\sum_{\text{internal lines}}
\quad
\tikz[baseline=-2pt,yscale=1.2]{
  \draw[->] (1,-1) node[below right] {$\!\!p_3$} --
    node[pos=.1, above right=-3pt] {$\beta_3$}
    node[pos=.35, below left=-3pt] {$\gamma_3$}
    node[pos=.7, below left=-3pt] {$\alpha_3$} (-1,1);
  \draw[->] (.5,-1) node[below] {$p_2$} --
    node[pos=.1, left=-2pt] {$\beta_2$}
    node[pos=.5, right=-2pt] {$\gamma_2$}
    node[pos=.9, left=-1pt] {$\alpha_2$} (.5,1);
  \draw[->] (-1,-1) node[below left] {$p_1\!\!\!$} --
    node[pos=.25, above left=-3pt] {$\beta_1$}
    node[pos=.65, above left=-3pt] {$\gamma_1$}
    node[pos=.9, below right=-3pt] {$\alpha_1$} (1,1);
}
$$
\unpause
We will write it as
$$
A_{321\ldots}
=S_{12}(p_1,p_2)S_{13}(p_1,p_3)S_{23}(p_2,p_3)
A_{123\ldots},
$$

\end{frame}

\begin{frame}
\frametitle{Consistency requirements: three particle permutation}
\unpause

The second way $123\to213\to231\to321$ leads to the relation
\Multline*$$
A^{\alpha_3\alpha_2\alpha_1\ldots}_{\cdots}[321\ldots]
\\
=\sum_{\beta_1,\beta_2,\beta_3}\left(\sum_{\gamma_1,\gamma_2,\gamma_3}
S^{\alpha_2\alpha_3}_{\gamma_2\gamma_3}(p_2,p_3)
S^{\alpha_1\gamma_3}_{\gamma_1\beta_3}(p_1,p_3)
S^{\gamma_1\gamma_2}_{\beta_1\beta_2}(p_1,p_2)\right)
A^{\beta_1\beta_2\beta_3\ldots}_{\cdots}[123\ldots],
$$
\unpause
or graphically
$$
\sum_{\text{internal lines}}
\quad
\tikz[baseline=-2pt,yscale=1.2]{
  \draw[->] (1,-1) node[below right] {$\!\!p_3$} --
    node[pos=.25, above right=-3pt] {$\beta_3$}
    node[pos=.65, above right=-3pt] {$\gamma_3$}
    node[pos=.85, below left=-3pt] {$\alpha_3$} (-1,1);
  \draw[->] (-.5,-1) node[below] {$p_2$} --
    node[pos=.1, right=-1pt] {$\beta_2$}
    node[pos=.5, left=-2pt] {$\gamma_2$}
    node[pos=.9, right=1pt] {$\alpha_2$} (-.5,1);
  \draw[->] (-1,-1) node[below left] {$p_1\!\!\!$} --
    node[pos=.15, above left=-2pt] {$\beta_1$}
    node[pos=.35, below right=-2pt] {$\gamma_1$}
    node[pos=.75, below right=-2pt] {$\alpha_1$} (1,1);
}
$$
\unpause
We will write it as
$$
A_{321\ldots}
=S_{23}(p_2,p_3)S_{13}(p_1,p_3)S_{12}(p_1,p_2)
A_{123\ldots},
$$

\end{frame}

\begin{frame}
\frametitle{Consistency requirements: three particle permutation}
\unpause

The result of these two permutation processes must be the same. Hence, we have the

\begin{block}{Yang\--Baxter equation}
\Multline$$
\sum_{\gamma_1,\gamma_2,\gamma_3}
S^{\alpha_1\alpha_2}_{\gamma_1\gamma_2}(p_1,p_2)
S^{\gamma_1\alpha_3}_{\beta_1\gamma_3}(p_1,p_3)
S^{\gamma_2\gamma_3}_{\beta_2\beta_3}(p_2,p_3)
\\
=
\sum_{\gamma_1,\gamma_2,\gamma_3}
S^{\alpha_2\alpha_3}_{\gamma_2\gamma_3}(p_2,p_3)
S^{\alpha_1\gamma_3}_{\gamma_1\beta_3}(p_1,p_3)
S^{\gamma_1\gamma_2}_{\beta_1\beta_2}(p_1,p_2),
\label{YBlong}
$$
\unpause
or, shorter,
\eq$$
S_{12}(p_1,p_2)S_{13}(p_1,p_3)S_{23}(p_2,p_3)
=S_{23}(p_2,p_3)S_{13}(p_1,p_3)S_{12}(p_1,p_2),
\label{YB}
$$
\unpause
or, graphically,
\eq$$
\tikz[baseline=-2pt,yscale=1.2]{
  \draw[->] (1,-1) node[below right] {$\!\!p_3$} --
    node[pos=.1, above right=-3pt] {$\beta_3$}
    node[pos=.7, below left=-3pt] {$\alpha_3$} (-1,1);
  \draw[->] (.5,-1) node[below] {$p_2$} --
    node[pos=.1, left=-2pt] {$\beta_2$}
    node[pos=.9, left=-1pt] {$\alpha_2$} (.5,1);
  \draw[->] (-1,-1) node[below left] {$p_1\!\!\!$} --
    node[pos=.25, above left=-3pt] {$\beta_1$}
    node[pos=.9, below right=-3pt] {$\alpha_1$} (1,1);
}
\quad=\quad
\tikz[baseline=-2pt,yscale=1.2]{
  \draw[->] (1,-1) node[below right] {$\!\!p_3$} --
    node[pos=.25, above right=-3pt] {$\beta_3$}
    node[pos=.85, below left=-3pt] {$\alpha_3$} (-1,1);
  \draw[->] (-.5,-1) node[below] {$p_2$} --
    node[pos=.1, right=-1pt] {$\beta_2$}
    node[pos=.9, right=1pt] {$\alpha_2$} (-.5,1);
  \draw[->] (-1,-1) node[below left] {$p_1\!\!\!$} --
    node[pos=.15, above left=-2pt] {$\beta_1$}
    node[pos=.75, below right=-2pt] {$\alpha_1$} (1,1);
}
\label{YBgraph}
$$

\end{block}

\end{frame}

\begin{frame}
\frametitle{Consistency requirements: inversability}
\unpause

Now demand that two consequent permutation of the same two particle leads to identical map: $12\to21\to12$. This implies the

\begin{block}{Unitarity condition}

\eq$$
\sum_{\gamma_1,\gamma_2}S^{\alpha_1\alpha_2}_{\gamma_1\gamma_2}(p_1,p_2)
S^{\gamma_2\gamma_1}_{\beta_2\beta_1}(p_2,p_1)
=\delta^{\alpha_1}_{\beta_1}\delta^{\alpha_2}_{\beta_2},
\label{unitarity-long}
$$
\unpause
or
\eq$$
S_{12}(p_1,p_2)S_{21}(p_2,p_1)=1,
\label{unitarity}
$$
\unpause
or
\eq$$
\tikz[baseline=-2pt]{
  \draw[->] (.5,-1) node[below] {$\beta_2$} to[out=150,in=-150,looseness=1.5] (.5,1) node[above] {$\alpha_2$};
  \draw[->] (-.5,-1) node[below] {$\beta_1$} to[out=30,in=-30,looseness=1.5] (-.5,1) node[above] {$\alpha_1$};
}
\quad=\quad
\tikz[baseline=-2pt]{
  \draw[->] (.5,-1) node[below] {$\beta_2$} -- (.5,1) node[above] {$\alpha_2$};
  \draw[->] (-.5,-1) node[below] {$\beta_1$} -- (-.5,1) node[above] {$\alpha_1$};
}
\label{unitarity-graph}
$$

\end{block}

\end{frame}

\begin{frame}
\frametitle{Relativistic condition: crossing symmetry}
\unpause

The last condition is due to the theory is relativistic.

\begin{block}{Crossing symmetry}

\eq$$
S^{\alpha_1\alpha_2}_{\beta_1\beta_2}(p_1,p_2)
=\sum_{\alpha'_1\beta'_1}C_{\beta_1\beta'_1}S^{\alpha_2\beta'_1}_{\beta_2\alpha'_1}(p_2,-p_1)C_{\alpha'_1\alpha_1}.
$$
The momenta $p_1$, $p_2$ are assumed here as space\-/time (2D), and $C$ is the charge conjugation matrix.
\unpause
More concisely
$$
S_{12}(p_1,p_2)=C_1S_{2\tilde1}(p_2,-p_1)C_1,
$$
where $\tilde1$ means transposition in this space.
\unpause
Graphically
\eq$$
\tikz[baseline=-2pt,scale=.7]{
  \draw[->] (1,-1) node[below right=-1pt] {$p_2$} --
    node[pos=.15,above right=-3pt] {$\beta_2$}
    node[pos=.85,below left=-3pt] {$\alpha_2$} (-1,1);
  \draw[->] (-1,-1) node[below left=-1pt] {$p_1\!$} --
      node[pos=.15,above left=-3pt] {$\beta_1$}
      node[pos=.85,below right=-3pt] {$\alpha_1$} (1,1);
}
\quad=\quad
\tikz[baseline=-2pt,scale=.7]{
  \draw[->] (1,-1) node[below right=-2pt] {$p_2$} --
    node[pos=.15,above right=-3pt] {$\beta_2$}
    node[pos=.85,below left=-3pt] {$\alpha_2$} (-1,1);
  \draw[<-] (-1,-1) --
      node[pos=.15,above left=-3pt] {$\bar\beta_1$}
      node[pos=.85,below right=-3pt] {$\bar\alpha_1$} (1,1) node[above right=-3pt] {$\!\!-p_1$};
}
\label{cross-graph}
$$

\end{block}

\end{frame}

\begin{frame}
\frametitle{Bootstrap equations for the $S$\-/matrix}
\unpause

Finally, we obtain a set of equations, which are called \emph{bootstrap equations} for the $S$\-/matrix. Repeat them in terms of rapidities:

\begin{block}{Bootstra equations}

\begin{enumerate}
\item<+->{\it Yang\--Baxter equation}
\eq$$
S_{12}(\theta_1-\theta_2)S_{13}(\theta_1-\theta_3)S_{23}(\theta_2-\theta_3)
=S_{23}(\theta_2-\theta_3)S_{13}(\theta_1-\theta_3)S_{12}(\theta_1-\theta_2).
\label{YBtheta}
$$
\item<+->{\it Unitarity}
\eq$$
S_{12}(\theta)S_{21}(-\theta)=1.
\label{unitarity-theta}
$$
\item<+->{\it Crossing symmetry}
\eq$$
S_{12}(\theta)=C_1S_{2\tilde1}(i\pi-\theta)C_1.
\label{cross-theta}
$$
\end{enumerate}

\end{block}
\unpause
The physical sheet:
\eq$$
0\le\Im\theta<\pi,
\label{physsheet}
$$
\unpause
The Mandelstam variable
$$
s=(p_1^\mu+p_2^\mu)^2=m_1^2+m_2^2+2m_1m_2\ch\theta.
$$
\unpause
The line $\Im\theta=0$ corresponds to the cut $[(m_1+m_2)^2,+\infty)$.
\unpause

The line $\Im\theta=\pi$ corresponds to the cut $(-\infty,(m_1-m_2)^2]$.
\unpause

The $S$\-/matrix is real on imaginary axis: $S(iu)\in\R\qquad\text{for}\quad u\in\R$. The only poles on the physical sheet is on the imaginary axis.
\unpause

A pole $iu_0\in[0,i\pi]$ corresponds to a bound state, if $\Res_{u=u_0}S^{\alpha\beta}_{\alpha\beta}(iu)>0$.

\end{frame}

\begin{frame}
\frametitle{Bootstrap equations for the $S$\-/matrix of the $O(N)$ model}
\unpause

Recall that for the $O(N)$ ($N\ge3$) model we have
\eq$$
S^{\alpha'\beta'}_{\alpha\hphantom{'}\beta}(\theta)
=\delta^{\alpha'\beta'}\delta_{\alpha\beta}
\uncover<.>{\tikz[overlay,yshift=-.7cm] {\draw[red,-] (-.5,-.5) -- (0,0) -- (.5,-.5); \draw[blue,<->] (-.5,.5) -- (0,0) -- (.5,.5);}}
S_1(\theta)
+\delta^{\alpha'}_\alpha\delta^{\beta'}_\beta
\uncover<.>{\tikz[overlay,yshift=-.7cm] {\draw[red,->] (-.5,-.5) -- (.5,.5); \draw[blue,->] (.5,-.5) -- (0,0) -- (-.5,.5);}}
S_2(\theta)
+\delta^{\beta'}_\alpha\delta^{\alpha'}_\beta
\uncover<.>{\tikz[overlay,yshift=-.7cm] {\draw[red,->] (-.5,-.5) -- (0,0) -- (-.5,.5); \draw[blue,->] (.5,-.5) -- (0,0) -- (.5,.5);}}
S_3(\theta).
\label{S1S2S3}
$$
\unpause
The Yang\--Baxter equations takes the form
The Yang\--Baxter equation for it takes the form
\Align$$
&S_2(\theta)S_3(\theta+\theta')S_3(\theta')
+S_3(\theta)S_3(\theta+\theta')S_2(\theta')
\notag
\\
&\uncover<.>{\tikz[overlay,yshift=-1cm,scale=.7]{
  \draw[red,->] (-1,-1) -- (0,0) -- (-1,1);
  \draw[blue,->] (.5,-1) -- (.5,-.5) -- (0,0) -- (1,1);
  \draw[green,->] (1,-1) -- (.5,-.5) -- (.5,1);
  \draw (1.5,0) node {$+$};
  \begin{scope}[xshift=3cm]
    \draw[red,->] (-1,-1) -- (0,0) -- (-1,1);
    \draw[blue,->] (.5,-1) -- (.5,.5) -- (1,1);
    \draw[green,->] (1,-1) -- (0,0) -- (.5,.5) -- (.5,1);
    \draw (2,0) node {$=$};
  \end{scope}
  \begin{scope}[xshift=7cm]
    \draw[red,->] (-1,-1) -- (-.5,-.5) -- (-.5,.5) -- (-1,1);
    \draw[blue,->] (-.5,-1) -- (-.5,-.5) -- (1,1);
    \draw[green,->] (1,-1) -- (-.5,.5) -- (-.5,1);
  \end{scope}
}}
\qquad=S_3(\theta)S_2(\theta+\theta')S_3(\theta'),
\label{yb1}
\\
\uncover<+->{&S_2(\theta)S_1(\theta+\theta')S_1(\theta')
+S_3(\theta)S_2(\theta+\theta')S_1(\theta')}
\notag
\\
&\uncover<.>{\tikz[overlay,yshift=-1cm,scale=.7]{
  \draw[red,->] (-1,-1) -- (0,0) -- (.5,-.5) -- (.5,1);
  \draw[blue,-] (.5,-1) -- (.5,-.5) -- (1,-1);
  \draw[green,<->] (-1,1) -- (0,0) -- (1,1);
  \draw (1.5,0) node {$+$};
  \begin{scope}[xshift=3cm]
    \draw[red,->] (-1,-1) -- (.5,.5) -- (.5,1);
    \draw[blue,-] (.5,-1) -- (.5,-.5) -- (1,-1);
    \draw[green,<->] (-1,1) -- (.5,-.5) -- (.5,.5) -- (1,1);
    \draw (2,0) node {$=$};
  \end{scope}
  \begin{scope}[xshift=7cm]
    \draw[red,->] (-1,-1) -- (-.5,-.5) -- (-.5,1);
    \draw[blue,->] (-.5,-1) -- (-.5,-.5) -- (0,0) -- (1,-1);
    \draw[green,<->] (-1,1) -- (-.5,.5) -- (0,0) -- (1,1);
  \end{scope}
}}
\qquad
\uncover<.->{{}=S_3(\theta)S_1(\theta+\theta')S_2(\theta'),}
\label{yb2}
\\
&\uncover<+->{NS_1(\theta)S_3(\theta+\theta')S_1(\theta')
+S_1(\theta)S_3(\theta+\theta')(S_2(\theta')+S_3(\theta'))}
\notag
\\
&\qquad\quad
\uncover<.->{{}+(S_2(\theta)+S_3(\theta))S_3(\theta+\theta')S_1(\theta')}
\notag
\\
&\qquad\quad
\uncover<.->{{}+S_1(\theta)(S_1(\theta+\theta')+S_2(\theta+\theta'))S_1(\theta')}
\notag
\\
&\uncover<.>{\tikz[overlay,xshift=-1cm,yshift=-1cm,scale=.7]{
  \draw[red,->] (-1,-1) -- (0,0) -- (-1,1);
  \draw[blue,-] (.5,-1) -- (.5,-.5) -- (1,-1);
  \draw[green,<->] (.5,1) -- (.5,.5) -- (1,1);
  \draw[yellow,-] (0,0) -- (.5,-.5) -- (.5,.5) -- cycle;
  \draw (1.5,0) node {$+$};
  \begin{scope}[xshift=3cm]
    \draw[red,->] (-1,-1) -- (0,0) -- (-1,1);
    \draw[blue,-] (.5,-1) -- (.5,.5) -- (0,0) -- (1,-1); \draw[blue] (.5,-.5) node[right] {\tiny$S_2+S_3$};
    \draw[green,<->] (.5,1) -- (.5,.5) -- (1,1);
    \draw (1.5,0) node {$+$};
  \end{scope}
  \begin{scope}[xshift=6cm]
    \draw[red,->] (-1,-1) -- (0,0) -- (-1,1);
    \draw[blue,-] (.5,-1) -- (.5,-.5) -- (1,-1);
    \draw[green,<->] (.5,1) -- (.5,-.5) -- (0,0) -- (1,1); \draw[green] (.5,.5) node[right] {\tiny$S_2+S_3$};
    \draw (1.5,0) node {$+$};
  \end{scope}
  \begin{scope}[xshift=9cm]
    \draw[red,->] (-1,-1) -- (.5,.5) -- (.5,-.5) -- (-1,1); \draw[red] (0,0) node[left] {\tiny$S_1+S_2$};
    \draw[blue,-] (.5,-1) -- (.5,-.5) -- (1,-1);
    \draw[green,<->] (.5,1) -- (.5,.5) -- (1,1);
    \draw (2,0) node {$=$};
  \end{scope}
  \begin{scope}[xshift=13cm]
    \draw[red,->] (-1,-1) -- (-.5,-.5) -- (-.5,.5) -- (-1,1);
    \draw[blue,-] (-.5,-1) -- (-.5,-.5) -- (0,0) -- (1,-1);
    \draw[green,<->] (-.5,1) -- (-.5,.5) -- (0,0) -- (1,1);
  \end{scope}
}}
\qquad
\uncover<.->{=S_3(\theta)S_1(\theta+\theta')S_3(\theta').}
\label{yb3}
$$
\unpause

Let us solve these equations.

\end{frame}

\begin{frame}
\frametitle{Bootstrap equations for the $S$\-/matrix of the $O(N)$ model}
\unpause

Solve the first equation
\eq$$
S_2(\theta)S_3(\theta+\theta')S_3(\theta')
+S_3(\theta)S_3(\theta+\theta')S_2(\theta')
=S_3(\theta)S_2(\theta+\theta')S_3(\theta').
\tag{\ref{yb1}}
$$
\unpause
Let $h(\theta)=S_2(\theta)/S_3(\theta)$. Then it takes the form
$$
h(\theta)+h(\theta')=h(\theta+\theta').
$$
\unpause
Therefore, $h(\theta)\sim\theta$ and
\eq$$
S_3(\theta)=-i{\lambda\over\theta}S_2(\theta).
\label{S32ratio}
$$
\unpause
Solve the second equation
\eq$$
S_2(\theta)S_1(\theta+\theta')S_1(\theta')
+S_3(\theta)S_2(\theta+\theta')S_1(\theta')
=S_3(\theta)S_1(\theta+\theta')S_2(\theta').
\tag{\ref{yb2}}
$$
\unpause
Let $g(\theta)=S_2(\theta)/S_1(\theta)$. Then
$$
g(\theta+\theta')-g(\theta')
={\theta\over i\lambda}.
$$
\unpause
This equation has a solution
$$
g(\theta)={\theta-i\kappa\over i\lambda}.
$$
\unpause
Substituting it into the third equation (\ref{yb3}), we get
$$
\kappa={N-2\over2}\lambda.
$$
\unpause
It meas that
\eq$$
S_1(\theta)=-{i\lambda\over i(N-2)\lambda/2-\theta}S_2(\theta).
\label{S12ratio}
$$

\end{frame}

\begin{frame}
\frametitle{Bootstrap equations for the $S$\-/matrix of the $O(N)$ model}
\unpause

Now substitute it into the crossing symmetry equation
\Align$$
S_2(\theta)
&=S_2(i\pi-\theta),
\label{cross22}
\\
S_1(\theta)
&=S_3(i\pi-\theta).
\label{cross13}
$$
We obtain
\eq$$
\lambda={2\pi\over N-2}.
\label{lambdavalue}
$$
\unpause
Now impose the unitarity condition:
\Gather$$
\uncover<.>{
\tikz[overlay,yshift=-1cm,scale=.5]{
\draw[red,->] (-.3,-1) to[out=30,in=50-180] (0,-.75) to[out=50,in=-50] (0,.75) to[out=180-50,in=-30] (-.3,1) -- +(180-30:.2);
\draw[blue,->] (.3,-1) to[out=180-30,in=-50] (0,-.75) to[out=180-50,in=50-180] (0,.75) to[out=50,in=30-180] (.3,1) -- +(30:.2);
\draw (1,0) node {$+$};
\begin{scope}[xshift=2cm]
  \draw[red,->] (-.3,-1) to[out=30,in=50-180] (0,-.75) to[out=180-50,in=50-180] (0,.75) to[out=180-50,in=-30] (-.3,1) -- +(180-30:.2);
  \draw[blue,->] (.3,-1) to[out=180-30,in=-50] (0,-.75) to[out=50,in=-50] (0,.75) to[out=50,in=30-180] (.3,1) -- +(30:.2);
  \draw (1.5,0) node {$=$};
\end{scope}
\begin{scope}[xshift=5cm]
  \draw[red,->] (-.3,-1) -- (-.3,1);
  \draw[blue,->] (.3,-1) -- (.3,1);
\end{scope}
}}
S_2(\theta)S_2(-\theta)+S_3(\theta)S_3(-\theta)=1,
\label{unit1}
\\
\uncover<+->{
\uncover<.>{
\tikz[overlay,yshift=-1cm,scale=.5]{
\draw[red,->] (-.3,-1) to[out=30,in=50-180] (0,-.75) to[out=180-50,in=50-180] (0,.75) to[out=50,in=30-180] (.3,1) -- +(30:.2);
\draw[blue,->] (.3,-1) to[out=180-30,in=-50] (0,-.75) to[out=50,in=-50] (0,.75) to[out=180-50,in=-30] (-.3,1) -- +(180-30:.2);
\draw (1,0) node {$+$};
\begin{scope}[xshift=2cm]
  \draw[red,->] (-.3,-1) to[out=30,in=50-180] (0,-.75) to[out=50,in=-50] (0,.75) to[out=50,in=30-180] (.3,1) -- +(30:.2);
  \draw[blue,->] (.3,-1) to[out=180-30,in=-50] (0,-.75) to[out=180-50,in=50-180] (0,.75) to[out=180-50,in=-30] (-.3,1) -- +(180-30:.2);
  \draw (1.5,0) node {$=$};
\end{scope}
\begin{scope}[xshift=5cm]
  \draw (0,0) node {$0$};
\end{scope}
}}
S_2(\theta)S_3(-\theta)+S_3(\theta)S_2(-\theta)=0,}
\label{unit2}
\\
\uncover<+->{
NS_1(\theta)S_1(-\theta)
\uncover<.>{
\tikz[overlay,xshift=1cm,yshift=-1cm,scale=.5]{
\draw[red,-] (-.3,-1) to[out=30,in=50-180] (0,-.75) to[out=-50,in=180-30] (.3,-1);
\draw[blue,->] (-.3,1) to[out=-30,in=180-50] (0,.75) to[out=50,in=30-180] (.3,1) -- +(30:.2); \draw[blue,->] (-.3,1) -- +(180-30:.2);
\draw[yellow,-] (0,.75) to[out=-50,in=50] (0,-.75) to[out=180-50,in=50-180] cycle;
\draw (1,0) node {$+$};
\begin{scope}[xshift=2cm]
\draw[red,-] (-.3,-1) to[out=30,in=50-180] (0,-.75) to[out=-50,in=180-30] (.3,-1);
\draw[blue,->] (-.3,1) to[out=-30,in=180-50] (0,.75) to[out=-50,in=50] (0,-.75) to[out=180-50,in=50-180] (0,.75) to[out=50,in=30-180] (.3,1)
  -- +(30:.2); \draw[blue,->] (-.3,1) -- +(180-30:.2); \draw[blue] (0,.75) node[right=2pt] {\tiny$S_2+S_3$};
\draw (1.5,0) node {$+$};
\end{scope}
\begin{scope}[xshift=5cm]
\draw[red,-] (-.3,-1) to[out=30,in=50-180] (0,-.75) to[out=50,in=-50] (0,.75) to[out=50-180,in=180-50] (0,-.75) to[out=-50,in=180-30] (.3,-1);
  \draw[red] (0,-.75) node[right] {\tiny$S_2+S_3$};
\draw[blue,->] (-.3,1) to[out=-30,in=180-50] (0,.75) to[out=50,in=30-180] (.3,1)
  -- +(30:.2); \draw[blue,->] (-.3,1) -- +(180-30:.2);
  \draw (2,0) node {$=$};
\end{scope}
\begin{scope}[xshift=8.5cm]
  \draw (0,0) node {$0$};
\end{scope}
}}
+S_1(\theta)(S_2(-\theta)+S_3(-\theta))
+(S_2(\theta)+S_3(\theta))S_1(-\theta)
=0}
\label{unit3}
$$
\unpause
By substituting the solution of the YB equation to the crossing symmetry and unitarity equations, we obtain
\eq$$
S_2(\theta)=S_2(i\pi-\theta),
\qquad
S_2(\theta)S_2(-\theta)={\theta^2\over\theta^2+\lambda^2}.
\label{unit22}
$$
\unpause
There are many solutions to these equations (the \emph{CDD (Castillejo\--Dalitz\--Dyson) ambiguity}). If we take any solution and multiply it by a factor
$$
{\sh\theta+i\sin\alpha\over\sh\theta-i\sin\alpha},
$$
we will again have a solution.
\unpause
We will search the `minimal' solution, which has the least number of poles and zeros on the physical sheet.

\end{frame}

\begin{frame}
\frametitle{Bootstrap equations for the $S$\-/matrix of the $O(N)$ model}
\unpause

We have the equations
\eq$$
S_2(\theta)=S_2(i\pi-\theta),
\qquad
S_2(\theta)S_2(-\theta)={\theta^2\over\theta^2+\lambda^2}.
\tag{\ref{unit22}}
$$
\unpause
Unitarity equation $\Rightarrow$ Simple zero $\theta=0$
\unpause
$\stackrel{\text{\red crossing}}\Rightarrow$ Simple zero $\theta=i\pi$
\unpause
$\stackrel{\text{\blue unitarity}}\Rightarrow$ Simple pole $\theta=-i\pi$
\unpause
$\stackrel{\text{\red crossing}}\Rightarrow$ Simple pole $\theta=2i\pi$ $\stackrel{\text{\blue unitarity}}\Rightarrow$ \dots
\unpause

We have series of zeros and poles:
\eq$$
\Aligned{{}
&\text{Zeros:}\quad
&&\theta=-2\pi in,i\pi+2\pi in,
\quad&&
\\
&\text{Poles:}\quad
&&\theta=-i\pi-2\pi in,2\pi i+2\pi in,\quad&&n=0,1,2,\ldots
}\label{zp1}
$$
\unpause
Unitarity equation $\Rightarrow$ Simple pole either at $\theta=-i\lambda$ ($S^{(+)}$ solution) or at $\theta=i\lambda$ ($S^{(-)}$ solution).
\unpause
Similarly we obtain for $S^{(\pm)}$:
\eq$$
\Aligned{{}
&\text{Zeros:}\quad
&&\theta=\mp i\lambda-i\pi-2\pi in,\pm i\lambda+2\pi i+2\pi in,
\quad&&
\\
&\text{Poles:}\quad
&&\theta=\mp i\lambda-2\pi in,\pm i\lambda+i\pi+2\pi in,\quad
&&n=0,1,2,\ldots
}\label{zp2}
$$
\unpause
The function that has poles and zeros at (\ref{zp1}) and (\ref{zp2}) is
\eq$$
S^{(\pm)}_2(\theta)
=Q^{(\pm)}(\theta)Q^{(\pm)}(i\pi-\theta),
\quad
Q^{(\pm)}(\theta)
={\Gamma\left(\pm{\lambda\over2\pi}-i{\theta\over2\pi}\right)
\Gamma\left({1\over2}-i{\theta\over2\pi}\right)
\over
\Gamma\left({1\over2}\pm{\lambda\over2\pi}-i{\theta\over2\pi}\right)
\Gamma\left(-i{\theta\over2\pi}\right)}.
\label{S2pm}
$$

\end{frame}

\begin{frame}
\frametitle{$S$\-/matrix: $N\to\infty$ behavior and final choice}
\unpause

Take the limit $N\to\infty$. We obtain
\Align$$
S^{(\pm)}_1(\theta)
&=-{2\pi i\over N(i\pi-\theta)},
\label{S1expand}
\\
S^{(\pm)}_2(\theta)
&=1\mp{2\pi i\over N\sh\theta},
\label{S2expand}
\\
S^{(\pm)}_3(\theta)
&=-{2\pi i\over N\theta}.
\label{S3expand}
$$
\unpause
By comparing with the $1/N$\-/expansion we conclude that the solution $S^{(+)}(\theta)$ is the $S$\-/matrix of the $O(N)$\-/model.
\unpause
The solution $S^{(-)}(\theta)$ is the $S$\-/matrix of the $N$\-/component fermion Neveu\--Schwartz model (see a problem to the last lecture).
\unpause

Notice, that
\eq$$
S^{(\pm)}_{12}(0)=\mp P_{12},
\label{S0sign}
$$
where $P_{12}:a\times b\mapsto b\times a$ is the permutation operator of the spaces 1 and~2.
\unpause
This means that for the particles in the $O(N)$\-/model a kind of the Pauli principle applies, although we considered the particles to be bosons. Two particles cannot have the same momentum.



\end{frame}

\begin{frame}
\frametitle{Seminar}
\unpause






\end{frame}



\end{document}
