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\begin{document}

\title[Lecture 5. $O(N)$\-/model: $1/N$\-/expansion]{Lecture 5.\\$O(N)$\-/model: $1/N$\-/expansion}
\date{}
\author{Michael Lashkevich}

\frame{\titlepage}

\begin{frame}
\frametitle{Functional integral}
\unpause

Consider the general $O(N)$\-/model in the Minkowski space:
\eq$$
S[\bn]={1\over2g}\int d^2x\,(\d_\mu\bn)^2,
\qquad
\bn^2=1.
\label<.>{ONmodel}
$$
\unpause
Instead of the restriction $\bn^2=1$, let us introduce an auxiliary field $\omega$:
\eq$$
S[\bn,\omega]={1\over2g}\int d^2x\,((\d_\mu\bn)^2-\omega(\bn^2-1)).
\label{ONmodelOmega}
$$
\unpause
The generating functional of correlation functions:
\eq$$
Z[J]=\int D\omega\,D\bn\,e^{iS[\bn,\omega]+ig^{-1/2}\int d^2x\,\bJ\bn}.
$$
\unpause
The integral over $\bn$ is Gaussian. Indeed, the exponent looks like
$$
iS[\bn,\omega]+ig^{-1/2}\int d^2x\,\bJ\bn
=-{1\over2}\left(
{n_i\over g^{1/2}},K(\omega)\delta_{ij}{n_j\over g^{1/2}}
\right)
+\left(iJ_i,{n_i\over g^{1/2}}\right)+i\int d^2x\,{\omega\over2g},
$$
where
$$
K(\omega)=i(\d^2_\mu+\omega).
$$
\unpause
Thus we obtain
$$
Z[J]=\int D\omega\,
(\det(\d^2_\mu+\omega))^{-N/2}
\exp\left(i\int d^2x\,{\omega\over2g}
-{1\over2}\int d^2x\,d^2x'\,J_i(x)G(x,x'|\omega)J_i(x')
\right),
$$
where $G(x,x'|\omega)$ is the solution of the equation
\eq$$
i(\d^2_\mu+\omega(x))G(x,x'|\omega)=\delta(x-x').
\label{Gdef}
$$

\end{frame}

\begin{frame}
\frametitle{Stationary point approximation as $N\to\infty$}
\unpause

Rewrite it as
\Gather$$
Z[J]=\int D\omega\,\exp\left(iS_\eff[\omega]-{1\over2}\int d^2x\,d^2x'\,J_i(x)G(x,x'|\omega)J_i(x')\right),
\label{ZJomega}
\\
S_\eff[\omega]=i{N\over2}\tr\log(\d^2_\mu+\omega)+\int d^2x\,{\omega\over2g}.
\label{Seff}
$$
\unpause
Find the saddle point as $N\to\infty$. Suppose that the saddle point is
$$
\omega(x)=\const=\omega_0.
$$
\unpause
Then
\Align$$
\tr\log(\d^2_\mu+\omega_0)
&=V\int{d^2k\over(2\pi)^2}\log(\omega_0-k^2-i0)\uncover<+->{{}=iV\int_E {d^2k\over(2\pi)^2}\log(\omega_0+k^2)}
\nonumber
\\
&\uncover<+->{{}={iV\over2\pi}\int^\Lambda_0dk\,k\log(\omega_0+k^2)}
\uncover<+->{{}={iV\over4\pi}\int^{\omega_0+\Lambda^2}_{\omega_0}du\,\log u}
\uncover<+->{{}={iV\over4\pi}\left[u\log{u\over e}\right]^{\omega_0+\Lambda^2}_{\omega_0}}
\nonumber
\\
&\uncover<+->{{}={iV\over4\pi}\left((\omega_0+\Lambda^2)\log{\omega_0+\Lambda^2\over e}-\omega_0\log{\omega_0\over e}\right)}
\nonumber
\\
&\uncover<+->{{}={iV\over4\pi}\left(\omega_0\log{\Lambda^2\over\omega_0}+\Lambda^2\log{\omega_0+\Lambda^2\over e}\right).}
\label{trlogK}
$$
where $\Lambda$ is an ultraviolet cutoff parameter.

\end{frame}

\begin{frame}
\frametitle{Stationary point approximation as $N\to\infty$}
\unpause

The stationary point equation is
$$
0={dS[\omega_0]\over d\omega_0}=V\left(-{N\over8\pi}\log{\Lambda^2\over\omega_0}+{1\over2g}\right),
$$
\unpause
which results in
\eq$$
\omega_0=m^2=\Lambda^2\exp\left(-{4\pi\over Ng}\right).
\label<.>{omega0}
$$
\unpause
The parameter $\omega_0=m^2$ is \emph{dimensional} and \emph{finite}. We will see that it is the mass of particles.
\unpause
Instead, the parameter $g$ depends on the UV cutoff $\Lambda$ and is non\-/physical. Its dependence can be described by the RG equation
\eq$$
{dg\over d\log\Lambda}=\beta(g)=-{N\over2\pi}g^2.
\label<.>{beta}
$$
\unpause
This phenomenon is called the \emph{dynamic mass generation} or \emph{dimensional transmutation}.

\end{frame}

\begin{frame}
\frametitle{$1/N$ expansion}
\unpause

Let
\eq$$
\omega(x)=m^2+(2/N)^{1/2}\rho(x).
\label{omega-expansion}
$$
\unpause
Assuming $G(x,x')=G(x,x'|m^2)$ we have
\Align$$
S_\eff[\omega]
&=\const
+i{N\over2}\tr\log\left(1+(2/N)^{1/2}\rho(\d^2_\mu+m^2)^{-1}\right)
+{1\over(2N)^{1/2}g}\tr\rho
\nonumber
\\
&\uncover<+->{{}=\const+i{N\over2}\tr\log(1+i(2/N)^{1/2}\rho G)+{1\over(2N)^{1/2}g}\tr\rho}
\nonumber
\\
&\uncover<+->{{}=\const+{\red\alt<.(1)->{\transparent1}{\transparent0}
\xcancel{\black\transparent1\left({1\over(2N)^{1/2}g}\tr\rho-\left(N\over2\right)^{1/2}\tr\rho G\right)}}
-i{N\over2}\sum^\infty_{n=2}{(-i)^n(2/N)^{n/2}\over n}\tr(\rho G)^n.}
\nonumber
$$
\unpause
The first parenthesis vanishes.
\unpause
Indeed,
$$
\Aligned{
\tr\rho
&=\int d^2x\,\rho(x),
\\
\unpause
\tr\rho G
&=\int d^2x\,\rho(x)G(x,x)\uncover<+->{{}=G(0,0)\int d^2x\,\rho(x)}
\uncover<+->{{}=G(0,0)\tr\rho}
\\
&\uncover<+->{{}=V\int{d^2k\over(2\pi)^2}{i\over k^2-m^2+i0}\,\tr\rho}
\uncover<+->{{}={V\over4\pi}\log{\Lambda^2\over m^2}\,\tr\rho}
\uncover<+->{{}=(gN)^{-1}\tr\rho.}
}
$$
\unpause
Finally we have
\eq$$
S_\eff[\omega]=\const-i{N\over2}\sum^\infty_{n=2}{(-i)^n\over n}\left(2\over N\right)^{n/2}
\int d^{2n}x\,\rho(x_1)G(x_1,x_2)\ldots\rho(x_n)G(x_n,x_1).
\label{Seff-decomp}
$$

\end{frame}

\begin{frame}
\frametitle{$1/N$ expansion: the $D$ propagator}
\unpause

The term with $n=2$ is a quadratic form in $\rho$:
$$
{i\over2}\int d^2x_1\,d^2x_2\,\rho(x_1)G(x_1,x_2)\rho(x_2)G(x_2,x_1).
$$
\unpause
Hence we will consider the inverse $D(x_1,x_2)$ of its kernel
$$
D^{-1}(x_1,x_2)=G(x_1,x_2)G(x_2,x_1)
$$
as a propagator of the field $\rho(x)$.
\unpause
In the momentum space
\eq$$
D(k)=
\
\tikz[baseline=-2pt]{
  \draw[dashed] (-.5,0) -- node[anchor=south] {$k$} (.5,0);
}
\
=-\left(
\int{d^2q\over(2\pi)^2}\,
{1\over(q^2-m^2+i0)((q+k)^2-m^2+i0)}
\right)^{-1}.
\label{Dk}
$$
\unpause
Explicitly, we have
\eq$$
D^{-1}(k)={i\over2\pi k^2}
{1\over\sqrt{1-{4m^2\over k^2}}}
\log{\sqrt{1-{4m^2\over k^2}}+1\over\sqrt{1-{4m^2\over k^2}}-1}.
\label{Dkexplicit}
$$
\unpause
This cumbersome formula becomes quite elementary in the appropriate parameterization:
\eq$$
D(k)=4\pi im^2\>{\sh\theta\over\theta},
\quad
k^2=-4m^2\sh^2{\theta\over2}.
\label{Dktheta}
$$

\end{frame}

\begin{frame}
\frametitle{$1/N$ expansion: the $G$ propagator and the vertex}
\unpause

Now expand the function $G(x,x'|\omega)$:
\Align$$
G[\omega]
&={1\over G^{-1}+i(2/N)^{1/2}\rho}=\sum^\infty_{n=0}(-i)^n{\textstyle\left(2\over N\right)^{n/2}}G(\rho G)^n,
\nonumber
\\
\uncover<+->{G(x_1,x_2|\omega)}
&\uncover<.->{{}=\sum^\infty_{n=0}(-i)^n\left(2\over N\right)^{n/2}\int d^{2n}y\,G(x_1,y_1)\rho(y_1)G(y_1,y_2)\ldots\rho(y_n)G(y_n,x_2).}
\nonumber
$$
\unpause
Thus the expression for $S_\text{eff}$ and for $G[\omega]$ can be written in terms of the propagator
\eq$$
G_{ij}(p)=
\quad
\tikz[baseline=-2pt]{
  \draw[-] (-1,0) node[anchor=east] {$i$} -- node[anchor=south] {$p$} (1,0) node[anchor=west] {$j$};
}
\quad
=G(p)\delta_{ij}={i\delta_{ij}\over p^2-m^2+i0}
\label{Gp}
$$
\unpause
and the vertex
\eq$$
\tikz[baseline=-2pt] {
  \draw[-] (-.5,0) node[anchor=south] {$i$} -- (.5,0) node[anchor=south] {$j$};
  \draw[dashed] (0,0) -- (0,1);
}
\quad
=-i\left(2\over N\right)^{1/2}\delta_{ij}.
\label{vertex}
$$

\end{frame}

\begin{frame}
\frametitle{$1/N$ expansion: Feynman rules}
\unpause

We can formulate Feynman rules:
\begin{enumerate}

\item<+-> A diagram consists of dashed lines (\ref{Dk}), solid lines (\ref{Gp}) and vertices (\ref{vertex}).

\item<+-> The outer lines of a diagram can only be solid lines corresponding to the massive particles $\varphi_i=g^{-1/2}n_i$.

\item<+->\label{loops} Closed loops of solid lines must contain at least three vertices.

\end{enumerate}
\unpause
The relation between $g$, $m$ and $\Lambda$ is given by
$$
\left\langle\sum^N_{i=1}\varphi^2_i(x)\right\rangle={1\over g}.
$$
\unpause
For example, in the order $1/N$ one can obtain
\eq$$
m^2=\Lambda^2\exp\left(-{4\pi\over(N-2)g'}\right),
\qquad
{1\over g'}={1\over g}+{\Lambda^2\over4\pi m^2\log(\Lambda^2/m^2)}.
\label{omega0corr}
$$
\unpause
The quadratic divergence originates in the rigid constraint $\bn^2=1$. Adding to the action a term $\alpha\int d^2x\,\omega^2$ mitigates it to a logarithmic one.

\end{frame}

\begin{frame}
\frametitle{Two\-/particle scattering}
\unpause

We have particles $\varphi_i$, $i=1,\ldots,N$ of the mass $m$. Consider the scattering process $\varphi_i+\varphi_j\to\varphi_{i'}+\varphi_{j'}$. Let $p_1=m\sh\theta_1$, $p_2=m\sh\theta_2$ be the momenta of the incoming particles, and $p'_1=m\sh\theta'_1$, $p'_2=m\sh\theta'_2$ be the momenta of the outgoing particles. The $\theta$ variables are called \emph{rapidities} of particles.
\unpause
The energy and momentum conservation equations
$$
\Aligned{
m\ch\theta_1+m\ch\theta_2
&=m\ch\theta'_1+m\ch\theta'_2,
\\
m\sh\theta_1+m\sh\theta_2
&=m\sh\theta'_1+m\sh\theta'_2
}
$$
\unpause
have two solutions: $\theta'_1=\theta_1$, $\theta'_2=\theta_2$ and $\theta'_1=\theta_2$, $\theta'_2=\theta_1$.
\unpause
Thus the $S$ matrix can be written as
\Multline*$$
S^{i'j'}_{ij}(\theta_1,\theta_2;\theta'_1,\theta'_2)
=(2\pi)^2\delta(p'_1-p_1)\delta(p'_2-p_2)S^{i'j'}_{ij}(\theta_1-\theta_2)
\\
+(2\pi)^2\delta(p'_2-p_1)\delta(p'_1-p_2)S^{j'i'}_{ij}(\theta_1-\theta_2).
$$
\unpause
We can rewrite the delta\-/function in terms of the delta\-/function over space\-/time momenta:
\Multline*$$
S^{i'j'}_{ij}(\theta_1,\theta_2;\theta'_1,\theta'_2)
=(2\pi)^2\delta^{(2)}(P'-P){\sh(\theta_1-\theta_2)\over\ch\theta_1\ch\theta_2}S^{i'j'}_{ij}(\theta_1-\theta_2)
\\
=(2\pi)^2\delta^{(2)}(P'-P){4m^2\sh(\theta_1-\theta_2)\over4\varepsilon_1\varepsilon_2}S^{i'j'}_{ij}(\theta_1-\theta_2),
$$
where $P^\mu=p_1^\mu+p_2^\mu$, $P^{\prime\mu}=p_1^{\prime\mu}+p_2^{\prime\mu}$.

\end{frame}

\begin{frame}
\frametitle{Two\-/particle scattering}
\unpause

Hence
$$
S^{i'j'}_{ij}(\theta_1-\theta_2)
=\delta^{i'}_i\delta^{j'}_j+{M^{i'j'}_{ij}(\theta_1-\theta_2)\over4m^2\sh(\theta_1-\theta_2)}.
$$
\unpause
The compatibility condition with the $O(N)$\-/symmetry gives
\eq$$
S^{i'j'}_{ij}(\theta)=\delta_{i'j'}\delta_{ij}S_1(\theta)+\delta_{i'i}\delta_{j'j}S_2(\theta)+\delta_{j'i}\delta_{i'j}S_3(\theta).
\label{S1S2S3}
$$

\end{frame}

\begin{frame}
\frametitle{Two\-/particle scattering: $1/N$ contribution}
\unpause

Calculate the $S$ matrix in the order $1/N$. We will use the formula
With the formula
\eq$$
D(k)=4\pi im^2\>{\sh\vartheta\over\vartheta},
\quad
k^2=-4m^2\sh^2{\vartheta\over2}.
\tag{\ref{Dktheta}}
$$
\unpause
Then
\Align$$
4m^2\sh\theta\,S_1(\theta)
&=\quad
\tikz[baseline=-2pt]{
\draw[-] (-1.0,0.5) node[anchor=east] {$p_1$} -- (-0.5,0.0);
\draw[-] (-1.0,-0.5) node[anchor=east] {$p_2$} -- (-0.5,0.0);
\draw[-] (0.5,0.0) -- (1.0,0.5) node[anchor=west] {$p_1$};
\draw[-] (0.5,0.0) -- (1.0,-0.5) node[anchor=west] {$p_2$};
\draw[dashed] (-0.5,0.0) -- node[above=3pt] {$\vartheta=i\pi-\theta$} (0.5,0.0);
}
\quad,
&S_1(\theta)
&=-{2\pi i\over N(i\pi-\theta)},
\nonumber
\\
\uncover<+->{4m^2\sh\theta\,(S_2(\theta)-1)
&=\quad
\tikz[baseline=-2pt] {
  \draw[-] (-1,.5) node[anchor=east] {$p_1$} -- (1,.5) node[anchor=west] {$p_1$};
  \draw[-] (-1,-.5) node[anchor=east] {$p_2$} -- (1,-.5) node[anchor=west] {$p_2$};
  \draw[dashed] (0,-.5) -- node[right] {$\vartheta=0$} (0,.5);
}
\quad,
&S_2(\theta)
&=1-{2\pi i\over N\sh\theta},}
\nonumber
\\
\uncover<+->{4m^2\sh\theta\,S_3(\theta)
&=
\quad
\tikz[baseline=-2pt] {
  \draw[-] (-1,.5) node[anchor=east] {$p_1$} -- (1,.5) node[anchor=west] {$p_2$};
  \draw[-] (-1,-.5) node[anchor=east] {$p_2$} -- (1,-.5) node[anchor=west] {$p_1$};
  \draw[dashed] (0,-.5) -- node[right] {$\vartheta=\theta$} (0,.5);
}
\quad,
&S_3(\theta)
&=-{2\pi i\over N\theta}.}
\nonumber
$$


\end{frame}

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






\end{frame}



\end{document}
