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\begin{document}

\title[Lecture 1. $O(2)$\-/model and BKT transition]{Lecture 1.\\$O(2)$\-/model and Berezinskii\--Kosterlitz\--Thouless transition}
\date{}
\author{Michael Lashkevich}

\frame{\titlepage}

\begin{frame}
\frametitle{$O(2)$\-/model}
\unpause

We will often consider the models in two\-/dimensional space\-/time with the action
\eq$$
S[\bn]={1\over2g}\int d^2x\,(\d_\mu\bn)^2,
\qquad
\bn^2\equiv\sum^N_{i=1}n^2_i=1,
\label<.>{O(N)-model}
$$
which are called \emph{$\bn$\-/field models} or \emph{$O(N)$\-/models}.
\unpause

In this lecture we consider the \emph{$N=2$} case.
\unpause
Let
$$
n_1=\cos\varphi,
\qquad
n_2=\sin\varphi.
$$
\unpause
Then
\Gather$$
S[\varphi]={1\over2g}\int d^2x\,(\d_\mu\varphi)^2,
\label<.>{varphiaction}
\\
\uncover<+->{\varphi(x)\sim\varphi(x)+2\pi.}
\label<.>{varphiequiv}
$$
\unpause
It looks like a free field with $\langle\varphi(x')\varphi(x)\rangle=-{g\over4\pi}\log(-(x'-x)^2)$.
\unpause
If it were the case, we would have for operators consistent with (\ref{varphiequiv}) the power\-/like behavior:
\eq$$
\langle
e^{im\varphi(x')}e^{in\varphi(x)}
\rangle
\sim
\left(-(x'-x)^2\right)^{{g\over4\pi}mn},
\qquad
m,n\in\Z.
\label{masslesscorr}
$$
\unpause
What can break this behavior?

\end{frame}

\begin{frame}
\frametitle{Vortices in the Euclidean plane}
\unpause

Consider the classical equation of motion in the Euclidean plane
\eq$$
\nabla^2\varphi=0.
\label<.>{eqmotion}
$$
\unpause
It admits the solutions
\eq$$
\varphi_{\vq\vx}(x)
=\sum^n_{a=1}q_a\Im\log(z-z_a)
=\sum^n_{a=1}{q_a\over2i}\log{z-z_a\over\bz-\bz_a},
\qquad
q_a\in\Z.
\label<.>{vortices}
$$
\tikz[overlay] \node[draw,red,text width=.3\textwidth,above left=5pt] at (current page.0) {\vspace{-1em}$$
\Aligned{
z
&=x^1+ix^2=x^1-x^0,
\\
\bz
&=x^1-ix^2=x^1+x^0.
}
$$\vspace{-1em}};
\unpause
For $n=1$ assuming $x-x_1=re^{i\theta}$ we have
$$
\varphi_{q_1x_1}(x)=q_1\theta,
$$
which is a \emph{vortex} at the point $x_1$.
\unpause

The solution (\ref{vortices}) is a solution to (\ref{eqmotion}) even at the points $x=x_a$.
\unpause
Indeed,
$$
\d_\mu\d^\mu{1\over2i}\log{z\over\bz}=\d_\mu\d^\mu\arctg{x^2\over x^1}
=-\epsilon_{\mu\nu}\d_\mu{x^\nu\over r^2}
=\epsilon^{\mu\nu}\d_\mu\d_\nu\log{1\over r}.
$$
\unpause
Then for any smooth, bounded and decreasing fast enough function $\varphi(x)$ we have
$$
\int d^2x\,\varphi(x)\d_\mu\d^\mu{1\over2i}\log{z\over\bz}
=\int d^2x\,(\epsilon^{\mu\nu}\d_\mu\d_\nu\varphi(x))\log{1\over r}=0,
$$
since the integral of $\log r$ converges at $x=0$.
\unpause
We immediately obtain
\eq$$
\int d^2x\,\d^\mu\varphi\,\d_\mu\varphi_{\vq\vx}=0.
\label{varphi-varphiqx}
$$

\end{frame}

\begin{frame}
\frametitle{Classical action of vertices}
\unpause

Let us calculate the classical action on the vertex solution:
\Align*$$
S[\varphi_{\vq\vx}]
&={2\over g}\int d^2x\,\d\varphi_{\vq\vx}\,\bd\varphi_{\vq\vx}
\tikz[overlay] \node[draw,red,text width=.25\textwidth,below=4em,left=5pt] at (current page.north east) {\vspace{-1em}$$
\d={\d\over\d z},\quad\bd={\d\over\d\bz}
$$\vspace{-1em}};
\uncover<+->{={1\over2g}\int d^2x\sum_{a,b}{q_aq_b\over(z-z_a)(\bz-\bz_b)}}
\notag
\\
&\uncover<+->{{}={1\over2g}\left(
\sum_aq^2_a\int{d^2x\over|z-z_a|^2}
\uncover<+->{{}+\sum_{a<b}q_aq_b\int d^2x\,
{(z-z_a)(\bz-\bz_b)+(\bz-\bz_a)(z-z_b)\over|z-z_a|^2|z-z_b|^2}}
\right).}
$$
\unpause
The first integral is
$$
\int{d^2x\over|z-z_a|^2}
\simeq2\pi\int^R_{r_0}{dr\over r}
=2\pi\log{R\over r_0},
$$
where $R$ and $r_0$ are infrared and ultraviolet cutoff parameters.
\unpause
The second integral is
$$
\int d^2x\,{(z-z_a)(\bz-\bz_b)+(\bz-\bz_a)(z-z_b)\over|z-z_a|^2|z-z_b|^2}=2\pi\log{R^2\over|z_a-z_b|^2}.
$$
\unpause
Hence
\Align$$
S[\varphi_{\vq\vx}]
&={1\over2g}\left(
\pi\sum_aq^2_a\log{R^2\over r^2_0}
+2\pi\sum_{a<b}q_aq_b\log{R^2\over|z_a-z_b|^2}
\right)
\label{vortices-int2}
\\
&={\pi\over2g}\left(\sum_a q_a\right)^2\log R^2
-{\pi\over2g}\sum_a q^2_a\log r^2_0
+{1\over2g}\sum_{a<b}q_aq_b\,2\pi\log{1\over|z_a-z_b|^2}.
\label{claction}
$$

\end{frame}

\begin{frame}
\frametitle{Classical action of vertices}
\unpause

So we have
\eq$$
S[\varphi_{\vq\vx}]={\pi\over2g}\left(\sum_a q_a\right)^2_{\tikz[overlay] \node[left] (sumqa) {};}\log R^2
-{\pi\over2g}\sum_a q^2_a\tikz[overlay] \node[left] (sumqa2) {};\log r^2_0
+{1\over2g}\sum_{a<b}q_aq_b\,2\pi\log{1\over|z_a-z_b|^2}.
\tag{\ref{claction}}
$$
\unpause
Since $R\to\infty$, the action is only finite if the first\tikz[overlay] \node[above left] (firstterm) {}; term vanishes:%
\only<.>{\tikz[overlay,red] \draw[->] (firstterm) to[out=145,in=-25] (sumqa);}
\eq$$
\sum_a q_a=0.
\label{neutrality}
$$
\unpause
The second\tikz[overlay] \node[above left] (secondterm) {}; term is the sum of `energies' of the cores of vortices.%
\only<.-.(2)>{\tikz[overlay,red] \draw[->] (secondterm) to[out=80,in=-100] (sumqa2);}
\unpause
It becomes finite in a regularized version of the theory, e.g.\ the $|\phi|^4$ model:
$$
S[\phi]
=\int d^2x\,\left(
|\d_\mu\phi|^2-{\lambda\over4}(|\phi|^2-\phi^2_0)^2
\right),
$$
\unpause
We will see that the behavior of the gas of vortices depends on $g$ rather than on $r_0$.

\end{frame}

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

Now we want to calculate the functional integral over $\varphi$. We split it into a sum over vortex configurations:
\eq$$
Z[J]
=\sum^\infty_{n\tikz[overlay] \node[left] (nvar) {};=0}{\tikz[overlay] \node[right] (r0powers) {};r_0^{-2n}\over n\tikz[overlay] \node[left] (nfactorial) {};!}
\sum_{q_1,\ldots,q_n\ne0\atop q_1+\cdots\tikz[overlay] \node[left] (qvar) {};+q_n=0}
\int d^2x_1\cdots\tikz[overlay] \node[left] (xvar) {}; d^2x_n\,
\int D\chi\tikz[overlay] \node[left] (chivar) {};\,
e^{-S[\chi+\varphi_{\vq\vx}]-(J,\chi+\varphi_{\vq\vx})}.
\tikz[overlay] \node[draw,red,above=2em,left=-4em] {$(f,g)=\int d^2x\,f(x)g(x)$};
\label{zj}
$$
\unpause
Here $n\tikz[overlay] \node[above left] (ndescr) {};$ is the number over vortices,%
\only<.>{\tikz[overlay,red] \draw[->] (ndescr) -- (nvar);}
\unpause
$q_i\tikz[overlay] \node[above left] (qdescr) {};$ are vorticities,%
\only<.>{\tikz[overlay,red] \draw[->] (qdescr) -- (qvar);}
\unpause
$x_i\tikz[overlay] \node[above left] (xdescr) {};$ are position of vortices,%
\only<.>{\tikz[overlay,red] \draw[->] (xdescr) -- (xvar);}
\unpause
and the field $\chi\tikz[overlay] \node[above left] (chidescr) {};$ runs all configurations \emph{without} the identification $\red\xcancel{\black\chi\sim\chi+2\pi}$.%
\only<.>{\tikz[overlay,red] \draw[->] (chidescr) -- (chivar);}
\unpause
The factor $1/n!\tikz[overlay] \node[above left] (nfactorialdescr) {};$ is caused by the fact that configurations of vortices are permutation invariant.%
\only<.>{\tikz[overlay,red] \draw[->] (nfactorialdescr) -- (nfactorial);}
\unpause
The factor $\tikz[overlay] \node[above right] (r0powersdescr) {};r_0^{-2n}$ is necessary to nondimensionolize the integrals.%
\only<.>{\tikz[overlay,red] \draw[->] (r0powersdescr) -- (r0powers);} We may think that every vortex lives in a cell of size $r_0$.
\unpause

The action is given by
$$
S[\chi+\varphi_{\vq\vx}]=S[\varphi_{\vq\vx}]+S[\chi]
+{\alt<.(1)->{\transparent1}{\transparent0}\red\xcancel{\transparent1\black{1\over g}\int d^2x\,\d^\mu\chi\,\d_\mu\varphi_{\vq\vx}}}.
$$
\unpause
We have seen that the last term vanishes.
\unpause
Hence the generating function factorizes:
\Align$$
Z[J]
&=Z_0[J]\sum^\infty_{n=0}{1\over n!}
\sum_{q_1,\ldots,q_n\atop q_1+\cdots+q_n=0}
r_0^{{\pi\over g}\sum q_a^2-2n}\times
\notag
\\
&\quad\times
\int d^2x_1\cdots d^2x_n\,
{\red\left({\black\prod_{a<b}|z_a-z_b|^{2{\pi\over g}q_aq_b}}\right)^{\tikz[overlay] \node (expScl-explicit) {};}}
\tikz[overlay,red] {\node[right,xshift=.5cm,yshift=1cm] (expScl) {$e^{-S[\varphi_{\vq\vx}]}\times\const$};
  \draw[-,double distance=1.5pt,shorten <=-2pt,shorten >=-4pt] ([yshift=2pt]expScl.south west) -- (expScl-explicit);}
\,e^{-(J,\varphi_{\vq\vx})},
\label{ZJ-factorization}
\\
Z_0[J]
&=\int D\chi\,e^{-S[\chi]-(J,\chi)}.
\label{Z0J-def}
$$

\end{frame}

\begin{frame}
\frametitle{Berezinskii\--Kosterlitz\--Thouless (BKT) transition}
\unpause

The source $J(x)$ is not arbitrary due to the identification $\varphi\sim\varphi+2\pi$. It must have the form
\eq$$
J_{\vJ\vy}(x)=-i\sum^k_{j=1}J_j\delta(x-y_j),
\qquad
J_i\in\Z.
\label<.>{JJy-def}
$$
\unpause
Then
$$
e^{-(J,\varphi)}=\exp i\sum^k_{j=1}J_j\varphi(y_j)
$$ 
is unique\-/valued.
\unpause
We have
\Align$$
Z[J_{\vJ\vy}]
&=\sum^\infty_{n=0}{1\over n!}
\sum_{q_1,\ldots,q_n\atop q_1+\cdots+q_n=0}
r_0^{{\pi\over g}\sum_a q_a^2+{g\over4\pi}\sum_j J^2_j-2n}
\int d^2x_1\cdots d^2x_n\,
\notag
\\
&\hskip -2em\times
\prod_{a<b}|z_a-z_b|^{{2\pi\over g}q_aq_b}
\prod_{a,j}\left(w_j-z_a\over\bw_j-\bz_a\right)^{q_aJ_j/2}\,
\prod_{j<j'}|w_j-\tikz[overlay] \node[above left=2pt] (wuse) {};w_{j'}|^{{g\over2\pi}J_jJ_{j'}}.
\tikz[overlay,red] {\node[left=5pt,draw] (wdef) at ([yshift=1cm]current page.0) {$w_j=y^1_j+iy^2_j$};
  \draw[->] (wdef.south west) to[bend right] (wuse.north);}
\label{ZJfin}
$$
\unpause
This looks as a partition function of a two\-/dimensional plasma.
\unpause
At high `temperature' $g$ the plasma is `ionized', vortices are separated and correlation functions decrease exponentially due to the Debye\-/type screening.
\unpause
Al low `temperature' the vortices of opposite vorticities attract and neutralize each other.
\unpause
In contrast to the usual plasma here these regimes are switched at a definite value of $g$. It is called the \emph{Berezinskii\--Kosterlitz\--Thouless (BKT) transition}.

\end{frame}

\begin{frame}
\frametitle{Critical value of $g$}
\unpause

Now let us find the critical value $g_\text{BKT}$ of the `temperature' $g$ corresponding to the BKT transition.
\unpause

The plasma phase corresponds to infrared divergent integrals, while the neutral phase corresponds to infrared convergent integrals.
\unpause
One integral is always divergent due to the translation invariance, so that we ignore it.
\unpause
Then the divergence index is
$$
I_n=2(n-1)+{2\pi\over g}\sum_{a<b}q_aq_b.
$$
If $I_n<0$, the $(n-1)$\-/tuple integral converges at large $x_a-x_b$.
\unpause
Estimate the index $I_n$. It depends on the sum
$$
\sum_{a<b}q_aq_b={1\over2}\sum_{a\ne b}q_aq_b
={1\over2}\left(\sum_a q_a\right)^2-{1\over2}\sum_a q_a^2
=-{1\over2}\sum_a q_a^2\le-{n\over2}.
$$
\unpause
Thus all integrals converge, if
$$
I_n\le2(n-1)+2{\pi\over g}\left(-{n\over2}\right)<0.
$$
\unpause
For large $n$ it gives the exact bound: if this condition is satisfied, all integrals are convergent, while if it is not satisfied, there exists a configuration $\{q_a\}$ for any given $n$, for which the integral diverges.
\unpause
By taking $n\to\infty$ we obtain the critical value
\eq$$
g_\text{BKT}={\pi\over2}.
\label{gKT}
$$

\end{frame}

\begin{frame}
\frametitle{Critical value of $g$}
\unpause

\begin{itemize}
\item<.->For $g>g_\text{BKT}$ the correlation length $\xi\sim r_0f(g)$. The excitations are massive with the mass $m\sim\xi^{-1}$.
\unpause
In the limit $r_0\to0$ the exponent $e^{-S}\to0$ for vortex solutions, but effectively the volume of the phase space grows, so that the contribution of the vortices is always of the same order.

\item<+->For $g<g_\text{BKT}$ the theory is massless and for $r\gg r_0$ coincides with a reduction of the free massless boson theory compatible with the identification $\varphi\sim\varphi+2\pi$.
\end{itemize}

\end{frame}

\begin{frame}
\frametitle{Free massless field}
\unpause

Consider a free massless field $\phi(x)$ with the action
\eq$$
S_0[\phi]={1\over8\pi}\int d^2x\,(\d_\mu\phi)^2.
\label{S0phi}
$$
\unpause
The classical equation of motion is
$$
\d_\mu\d^\mu\phi=0.
$$
\unpause
Define the \emph{dual} field $\tilde\phi(x)$ as a solution to the equation
\eq$$
\d_\mu\tphi=\epsilon_{\mu\nu}\,\d^\nu\phi,
\qquad
\epsilon_{01}=-\epsilon_{10}=1,
\label{phidual}
$$
\unpause
or
\eq$$
\d\tphi=\d\phi,
\qquad
\bd\tphi=-\bd\phi.
$$
\unpause
We rewrite it as follows
\eq$$
\Aligned{
\phi(x)
&=\phi_R(z)+\phi_L(\bz),
\\
\tphi(x)
&=\phi_R(z)-\phi_L(\bz).
}
\label{phiLR}
$$
\unpause
This decomposition (up to some subtleties) is valid in the quantum case. The correlation functions
\eq$$
\langle\phi_R(z)\phi_R(z')\rangle_0
=\log{R\over z-z'},
\quad
\langle\phi_L(\bz)\phi_L(\bz')\rangle_0
=\log{R\over\bz-\bz'},
\quad
\langle\phi_R(z)\phi_L(\bz')\rangle_0
=0
\label{LRcorr}
$$
are consistent with the theory.

\end{frame}

\begin{frame}
\frametitle{Exponential operators}
\unpause

Consider the exponents $e^{i\alpha\phi_{R,L}(x)}$ of the fields. Their correlation functions diverge.
\unpause
In the functional integral manner we can derive them as follows:
\vspace{-8pt}
\Multline*$$
\hspace{-8pt}\left\langle e^{i\alpha_1\phi_R(z_1)}\cdots e^{i\alpha_n\phi_R(z_n)}\right\rangle_0
=\left\langle e^{i\sum^n_{a=1}\alpha_a\phi_R(z_a)}\right\rangle_0
\uncover<+->{{}=\exp\left\langle-{1\over2}\left(\sum^n_{a=1}\alpha_a\phi_R(z_a)\right)^2\right\rangle_0}
\\
\hspace{-8pt}\uncover<+->{{}=\exp\left(-{1\over2}\sum^n_{a=1}\alpha_a^2{\red\overbrace{\black\langle\phi_R^2\rangle_0}^{\vrule width 0pt depth 4pt\smash{\log{R\over r_0}}}}
-\sum_{a<b}\alpha_a\alpha_b{\red\overbrace{\black\langle\phi_R(z_a)\phi(z_b)\rangle_0}^{\vrule width 0pt depth 4pt\smash{\log{R\over z_a-z_b}}}}\right)}
\uncover<+->{{}=\left(r_0\over R\right)^{{1\over2}\sum_a\alpha_a^2}\prod^n_{a<b}\left(z_a-z_b\over R\right)^{\alpha_a\alpha_b}}
\\
\uncover<+->{{}=r_0^{{1\over2}\sum_a\alpha_a^2}R^{-{1\over2}\left(\sum_a\alpha_a\right)^2}\prod^n_{a<b}(z_a-z_b)^{\alpha_a\alpha_b}.}
$$
\unpause
Here we assumed the $T$ ordered averages such that $z_a$ is assumed `later' than $z_{a+1}$.
\unpause
We see that the renormalization
\eq$$
e^{i\alpha\phi_{R,L}}=r_0^{\alpha^2/2}\lcolon e^{i\alpha\phi_{R,L}}\rcolon,
\qquad
e^{i\alpha\phi}=r_0^{\alpha^2}\lcolon e^{i\alpha\phi}\rcolon,
\qquad
e^{i\alpha\tphi}=r_0^{i\alpha^2}\lcolon e^{i\alpha\tphi}\rcolon.
\label{expredef}
$$
makes the operators $\lcolon\cdots\rcolon$ finite.
\unpause
We have
\eq$$
\Aligned{
\langle\lcolon e^{i\alpha_1\phi_R(z_1)}\rcolon\cdots\lcolon e^{i\alpha_n\phi_R(z_n)}\rcolon\rangle_0
&=R^{-{1\over2}\left(\sum_a\alpha_a\right)^2}
\prod_{a<b}(z_a-z_b)^{\alpha_a\alpha_b},
\\
\langle\lcolon e^{i\alpha_1\phi_L(\bz_1)}\rcolon\cdots\lcolon e^{i\alpha_n\phi_L(\bz_n)}\rcolon\rangle_0
&=R^{-{1\over2}\left(\sum_a\alpha_a\right)^2}
\prod_{a<b}(\bz_a-\bz_b)^{\alpha_a\alpha_b}.
}
\label{expLRcorr}
$$

\end{frame}

\begin{frame}
\frametitle{Scaling transformation}
\unpause

The renormalized exponents $\lcolon e^{i\alpha\phi_{R,L}}\rcolon$ are no more dimensionless and have the dimensions $\alpha^2/2$ in mass (inverse length) units.
\unpause
These dimensions coincide with the \emph{scaling dimensions} of the operators.
\unpause
A system of operators $O_i(x)$ possesses dimensions $d_i$, if all correlation function are invariant under simultaneous transformations
$$
O_i(x)\to s^{d_i}O_i(sx).
$$
\unpause
Indeed, in the limit $R\to\infty$ we have
\eq$$
\Aligned{
\langle\lcolon e^{i\alpha_1\phi_R(z_1)}\rcolon\cdots\lcolon e^{i\alpha_n\phi_R(z_n)}\rcolon\rangle_0
&=\Cases{\prod_{a<b}(z_a-z_b)^{\alpha_a\alpha_b},&\sum_a\alpha_a=0;\\0&\text{otherwise,}}
\\
\langle\lcolon e^{i\alpha_1\phi_L(\bz_1)}\rcolon\cdots\lcolon e^{i\alpha_n\phi_L(\bz_n)}\rcolon\rangle_0
&=\Cases{\prod_{a<b}(\bz_a-\bz_b)^{\alpha_a\alpha_b},&\sum_a\alpha_a=0;\\0&\text{otherwise.}}
}
\label{expLRcorr-inf}
$$
These correlation functions are invariant under the scaling transformation.
\unpause
Then we have
\Multline$$
\left\langle\prod^k_{j=1}e^{i\beta_j\tphi(y_j)}
\prod^n_{a=1}e^{i\alpha_a\phi(x_a)}\right\rangle_0
=r_0^{\sum_a\alpha_a^2+\sum_j\beta_j^2}\prod_{a<b}|z_a-z_b|^{2\alpha_a\alpha_b}
\times
\\*
\times
\prod_{j<j'}|w_j-w_{j'}|^{2\beta_a\beta_b}
\prod_{a,j}\left(w_j-z_a\over\bw_j-\bz_a\right)^{\alpha_a\beta_j}
\times\Cases{1,&\sum\alpha_a=\sum\beta_j=0;\\
  0&\text{otherwise.}}
\label{tphiphicorr}
$$

\end{frame}

\begin{frame}
\frametitle{Partition function in terms of the free boson}
\unpause

This coincides with the integrand of $Z[J]$ if
\eq$$
\alpha_a
=\sqrt{\pi\over g}\,q_a,
\qquad
\beta_j
=\sqrt{g\over4\pi}\,J_j.
\label{alpha-beta-def}
$$
\unpause
Then we have
\vspace{-8pt}
\Multline*$$
Z[J_{\vJ\vy}]
=\sum^\infty_{n=0}{1\over n!}
\sum_{q_1,\ldots,q_n\atop q_1+\cdots+q_n=0}
r_0^{-2n}
\int d^2x_1\cdots d^2x_n\,
\\
\times
\left\langle
\prod^k_{j=1}e^{i\sqrt{g\over4\pi}\,J_j\tphi(y_j)}
\prod^n_{a=1}e^{i\sqrt{\pi\over g}\,q_a\phi(x_a)}
\right\rangle_0.
\label{ZJphi}
$$
\unpause
The integrand is remarkably symmetric with respect to the replacements
$$
g\leftrightarrow (2\pi)^2g^{-1},
\qquad
k\leftrightarrow n,
\qquad
q_a\leftrightarrow J_j,
\qquad
\phi(x)\leftrightarrow\tphi(x).
$$
\unpause
Moreover, the Lagrangian of the free field is written identically in terms of both the fields $\phi$ and~$\tphi$. Thus we can identify
\eq$$
\varphi(x)=\sqrt{g\over4\pi}\tphi(x).
\label{varphi-tphi}
$$

\end{frame}

\begin{frame}
\frametitle{Sine\-/Gordon theory}
\unpause

Since $r_0^{{\pi\over g}q^2}\ll r_0^{q{\pi\over g}}$, we may neglect the contribution of a $q$\-/vortex compared to the contribution of $q$ instances of a $1$\-/vortex.
\unpause
It means that we may restrict the sum over vorticities to $q_a=\pm1$.
\unpause
Hence
\Align$$
Z[J_{\vJ\vy}]
&=\sum^\infty_{n=0}{r_0^{-4n}\over(2n)!}\int d^2x_1\cdots d^2x_{2n}\,\sum_{q_1,\ldots,q_{2n}=\pm1}
\left\langle\prod^k_{j=1}e^{i\sqrt{g\over4\pi}\,J_j\tphi(y_j)}\prod^{2n}_{a=1}e^{iq_a\sqrt{\pi\over g}\phi(x_a)}\right\rangle_0
\nonumber
\\
&\hskip -2em\uncover<+->{{}=\sum^\infty_{n=0}{r_0^{-4n}\over(2n)!}\int d^2x_1\cdots d^2x_{2n}\,
\left\langle\prod^k_{j=1}e^{i\sqrt{g\over4\pi}\,J_j\tphi(y_j)}
\prod^{2n}_{a=1}\left(e^{i\sqrt{\pi\over g}\phi(x_a)}+e^{-i\sqrt{\pi\over g}\phi(x_a)}\right)\right\rangle_0}
\nonumber
\\
&\uncover<+->{{}=\left\langle\prod^k_{j=1}e^{i\sqrt{g\over4\pi}\,J_j\tphi(y_j)}
\exp\left(2r_0^{-2}\int d^2x\,\cos\sqrt{\pi\over g}\phi(x)\right)\right\rangle_0}
\nonumber
\\
&\uncover<+->{{}=\int D\phi\,e^{-S_{\rm SG}[\phi]}\prod^k_{j=1}e^{i\sqrt{g\over4\pi}\,J_j\tphi(y_j)},}
\label{Z-SG}
$$
\unpause
where
\eq$$
S_{\rm SG}[\phi]
=\int d^2x\left({(\d_\mu\phi)^2\over8\pi}-\mu\lcolon\cos\beta\phi\rcolon\right)
\label{SGdef}
$$
is the action of the sine\-/Gordon model with the parameters
\eq$$
\beta=\sqrt{\pi\over g},
\qquad
\mu=2r_0^{{\pi\over g}-2}.
\label{SGparams}
$$

\end{frame}

\begin{frame}
\frametitle{Scaling dimension of the perturbation term}
\unpause

The sine\-/Gordon model is a perturbation of the free massless fermion model with the perturbation term $\sim\lcolon\cos\beta\phi$ in the Lagrangian with the scaling dimension
$$
\Delta_\text{p}=\beta^2={\pi\over g}.
$$
There are three regimes:
\begin{enumerate}
\item<+->$\Delta_\text{p}<2$ ($g>g_\text{BKT}$). The perturbation is \emph{relevant} and \emph{superrenormalizable}. It does not change the ultraviolet behavior of the theory, but essentially changes the infrared behavior.
\item<+->$\Delta_\text{p}>2$ ($g<g_\text{BKT}$). The perturbation is \emph{irrelevant} and \emph{nonrenormalizable}. It changes the infrared behavior breaking the perturbation theory beyond the leading (tree) contributions. The infrared behavior remains free\-/fermion\-/like.
\item<+->$\Delta_\text{p}=2$ ($g=g_\text{BKT}$). The perturbation is \emph{marginal}. In the case of the sine\-/Gordon theory it is also \emph{renormalizable}. Nevertheless it changes both infrared and ultraviolet behavior.
\end{enumerate}


\end{frame}

\begin{frame}
\frametitle{Seminar: free massless boson}
\unpause

1. Define
\Gather*$$
\varphi(z)=Q-iP\log z+\sum_{k\ne0}{a_k\over ik}z^{-k},
\\
[P,Q]=-i,
\qquad
[a_k,a_l]=k\delta_{k+l,0},
\\
P|0\rangle=a_k|0\rangle=0,\quad\langle0|a_{-k}=0\quad(k>0).
\label{vac-def}
$$%
\unpause
Calculate
$$
\langle\varphi(z')\varphi(z)\rangle={}
\uncover<.>{\rlap{\red??}}\uncover<+->{\langle Q^2\rangle+\log{1\over z'-z}.}
$$%
\unpause
2. Define
$$
\Aligned{
e^{i\alpha\varphi(r_0,z)}
&=\exp\left(i\alpha Q+\alpha P\log z+\alpha\sum_{k>0}\left({a_k\over k}z^{-k}-{a_{-k}\over k}(z-r_0)^k\right)\right),
\\
\lcolon e^{i\alpha\varphi(r_0,z)}\rcolon
&=e^{i\alpha Q}z^{\alpha P}\exp\left(-\alpha\sum_{k>0}{a_{-k}\over k}(z-r_0)^k\right)
\exp\left(\alpha\sum_{k>0}{a_k\over k}z^{-k}\right).
}
$$%
\unpause
Calculate the coefficient:
$$
e^{i\alpha\varphi(r_0,z)}
=\uncover<.>{\rlap{\red??}}
\uncover<+->{r_0^{\alpha^2/2}}
\lcolon e^{i\alpha\varphi(r_0,z)}\rcolon.
$$


\end{frame}

\begin{frame}
\frametitle{Seminar: free massless boson}
\unpause
3. Define
$$
\lcolon e^{i\alpha\varphi(z)}\rcolon=\lcolon e^{i\alpha\varphi(0,z)}\rcolon
=e^{i\alpha Q}z^{\alpha P}\exp\left(-\alpha\sum_{k>0}{a_{-k}\over k}z^k\right)
\exp\left(\alpha\sum_{k>0}{a_k\over k}z^{-k}\right).
$$%
\unpause
Calculate the coefficient
$$
\lcolon e^{i\alpha_1\varphi(z')}\rcolon\lcolon e^{i\alpha_2\varphi(z)}\rcolon
=\uncover<.>{\rlap{\red??}}
\uncover<+->{(z'-z)^{\alpha_1\alpha_2}}
\lcolon  e^{i\alpha_2\varphi(z')+i\alpha_1\varphi(z)}\rcolon.
$$%
\unpause
4. Calculate
$$
\langle\lcolon e^{i\sum^N_{j=1}\alpha_i\varphi(z_i)}\rcolon\rangle
=\uncover<.>{\rlap{\red??}}
\uncover<+->{\Cases{1,&\sum_j\alpha_j=0;\\0&\text{otherwise}.}}
$$%
\unpause
5. Calculate
$$
\langle\lcolon e^{i\alpha_1\varphi(z_1)}\rcolon\cdots\lcolon e^{i\alpha_N\varphi(z_N)}\rcolon\rangle
=\uncover<.>{\rlap{\red??}}
\uncover<+->{\prod^N_{i<j}(z_i-z_j)^{\alpha_i\alpha_j}\times\Cases{1,&\sum_j\alpha_j=0;\\0&\text{otherwise}.}}
$$%
\unpause
6. Prove that this correlation function is invariant under the transformation
$$
\lcolon e^{i\alpha_i\varphi(z_i)}\rcolon\to\lambda^{\alpha^2/2}\lcolon e^{i\alpha_i\varphi(\lambda z_i)}\rcolon.
$$%
\unpause
7. Prove that this correlation function is invariant under the transformation
$$
\lcolon e^{i\alpha_i\varphi(z_i)}\rcolon\to z_i^{-\alpha^2}\lcolon e^{i\alpha_i\varphi(-z_i^{-1})}\rcolon.
$$


\end{frame}



\end{document}
