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\begin{document}

\title[Lecture 4. $O(3)$\-/model: mass generation by instantons]{Lecture 4.\\$O(3)$\-/model: mass generation by instantons}
\date{}
\author{Michael Lashkevich}

\frame{\titlepage}

\begin{frame}
\frametitle{$O(3)$\-/model: topology of $\bn$\-/field}
\unpause

Consider the $O(3)$\-/model on the Euclidean plane:
\eq$$
S[\bn]={1\over2g}\int d^2x\,(\d_\mu\bn)^2,
\qquad
n_1^2+n_2^2+n_3^2=1.
\label{o3model}
$$
\unpause
We will be interested in the functions $\bn(x)$ with finite action. They must be constant at infinity:
\eq$$
\bn_0(x)\mathinner{\mathop\to\limits_{x\to\infty}}\bn_0.
\label{bcond}
$$
\unpause
It means that the function $\bn(x)$ lives on the \emph{topological sphere} $\R^2\cup\{\infty\}$.
\unpause
Hence, it realized a map
\eq$$
\bn:S^2\to S^{2\prime}.
$$
\unpause
Examples. Let $(\theta,\phi)$ are spherical coordinates on $S^2$ and $(\theta',\phi')$ are those on $S^{2\prime}$. Define the mapping
\eq$$
\theta'=\theta,
\qquad
\varphi'=q\varphi,
\qquad
q\in\Z.
$$
\unpause
The mappings with different values of $q$ cannot be deformed to each other continuously. Thus $q$ is a \emph{topological number}.
\unpause

In generic coordinates $(x^1,x^2)$ on $S^2$ and $(x^{\prime1},x^{\prime2})$ on $S^{2\prime}$. Define any metric $g'$ on the target sphere, such that
$$
S=\int_{S^{2\prime}}d^2x'\sqrt{g'}
$$
is finite.
\unpause
Then
$$
q={1\over S}\int_{x'(S^2)}d^2x'\,\sqrt{g'}={1\over S}\int_{S^2}d^2x{\d(x')\over\d(x)}\sqrt{g'}.
$$

\end{frame}

\begin{frame}
\frametitle{Topological charge: integral form}
\unpause

Assuming the spherical coordinates on $S^{2\prime}$ with the standard metric:
$$
q={1\over4\pi}\int^{2\pi}_0d\varphi\int^\pi_0d\theta\,
{\d(\theta',\varphi')\over\d(\theta,\varphi)}\sin\theta'
={1\over4\pi}\int d^2x\,
{\d(\theta',\varphi')\over\d(x^1,x^2)}\sin\theta'.
$$
\unpause
Put
\eq$$
\bn=(\sin\theta'\cos\varphi',\sin\theta'\sin\varphi',\cos\theta').
\label{jacsubst}
$$
\unpause
It can be checked by a direct calculation that
\eq$$
{1\over2}\bn\,(\d_\mu\bn\times\d_\nu\bn)\epsilon^{\mu\nu}
={\d(\theta',\varphi')\over\d(x^1,x^2)}\sin\theta'.
\label{jacobian}
$$
\unpause
Hence,
\eq$$
q={1\over8\pi}\int d^2x\,\bn\,(\d_\mu\bn\times\d_\nu\bn)\epsilon^{\mu\nu}.
\label{qint}
$$
\unpause
It can be proved in a simple way. $\boldsymbol a=\d_1\bn\,dx^1$ and $\boldsymbol b=\d_2\bn\,dx^2$ are small vectors on the sphere.
\unpause
The element $df$ of the surface in the parallelogram $(\boldsymbol a,\boldsymbol b)$ is
$$
df=\pm|\boldsymbol a\times\boldsymbol b|=\bn(\boldsymbol a\times\boldsymbol b)=\bn(\d_1\bn\,dx^1\times\d_2\bn\,dx^2)
={1\over2}\bn\,(\d_\mu\bn\times\d_\nu\bn)\epsilon^{\mu\nu}\,dx^1\,dx^2.
$$
$$
\begin{tikzpicture}
\draw[fill] (0,0) node[left] {$0$} circle (1pt);
\draw[->] (0,0) -- node[above] {$\bn$} (2,.5);
\fill[black!10] (2,.5) -- (2.3,.2) -- (2.3,-.4) -- (2,-.1) -- cycle;
\draw[->] (2,.5) -- node[above right] {$\boldsymbol b$} (2.3,.2);
\draw[->] (2,.5) -- node[left] {$\boldsymbol a$} (2,-.1);
\draw[-] (2.3,.2) -- (2.3,-.4);
\draw[-] (2,-.1) -- (2.3,-.4);
\node[outer sep=-2pt] (df) at (3,-.4) {$df$};
\draw[-{Stealth[length=1mm]}] (df) to[bend right] (2.15,.05);
\end{tikzpicture}
$$

\end{frame}

\begin{frame}
\frametitle{Self\-/duality equations}
\unpause

From the identity
\eq$$
\int d^2x\,(\d_\mu\bn+\epsilon_{\mu\nu}\bn\times\d^\nu\bn)^2
=2\int d^2x\,(\d_\mu\bn)^2
-2\int d^2x\,\bn\,(\d_\mu\bn\times\d_\nu\bn)\epsilon^{\mu\nu}
\label<.>{intid}
$$
\unpause
we obtain
\eq$$
S[\bn]
={4\pi q\over g}
+{1\over4g}\int d^2x\,(\d_\mu\bn+\epsilon_{\mu\nu}\bn\times\d^\nu\bn)^2.
\label<.>{Sqplus}
$$
\unpause
By the substitution $\bn\to-\bn$, $q\to-q$ we have
\eq$$
S[\bn]
=-{4\pi q\over g}
+{1\over4g}\int d^2x\,(\d_\mu\bn-\epsilon_{\mu\nu}\bn\times\d^\nu\bn)^2.
\label<.>{Sqminus}
$$
\unpause
Thus
\eq$$
S[\bn]\ge{4\pi|q|\over g}.
\label{Sqmin}
$$
\unpause
The equality in (\ref{Sqmin}) is achieved, if one of the \emph{self\-/duality equations} is satisfied:
\Align$$
\d_\mu\bn
&=-\epsilon_{\mu\nu}\bn\times\d^\nu\bn
\qquad(q>0),
\label{sdpos}
\\
\d_\mu\bn
&=\epsilon_{\mu\nu}\bn\times\d^\nu\bn
\hphantom{-}\qquad(q<0).
\label{sdneg}
$$
\unpause
These are first\-/order differential equations. Every their solution is a solution to the equations of motion, but not vice versa.

\end{frame}

\begin{frame}
\frametitle{Solutions to the self\-/duality equations}
\unpause

The stereographic projection:
\Gather$$
\begin{tikzpicture}
\draw[-] (-2,1) -- (5,1);
\fill (0,0) circle (1pt) node[below left=-2pt] {$O$};
\fill (0,-1) circle (1pt) node[below,overlay] {S}; \fill (0,1) circle (1pt) node[above,overlay] {N};
\draw[-,name path=circle] (0,0) circle (1cm);
\draw[-,name path=line] (0,-1) -- (4,1) node[above] {$w$}; \fill (4,1) circle (1pt);
\path[name intersections={of=circle and line}];
\draw[->,thick] (0,0) -- node[sloped,above=-1pt] {$\bn$} (intersection-2);
\end{tikzpicture}
\notag
\\
\uncover<+->{n_1+in_2={2w\over1+|w|^2},
\qquad
n_3={1-|w|^2\over1+|w|^2}.}
\label{stereogr}
$$
\unpause
The complex parameter $w$ becomes a coordinate on the sphere.
\unpause
Substituting it to the self\-/duality equations we obtain
\Align$$
\bd w
&=0
\qquad(q>0),
\label{sdcpos}
\\
\d w
&=0
\qquad(q<0).
\label{sdcneg}
$$
\unpause
Take, for example, $q>0$.
\unpause
A regular solution around $\bn=\boldsymbol e_3$ corresponds to a \emph{simple node} of $w\simeq{n_1+in_2\over2}$,
\unpause
while a regular solution around $\bn=-\boldsymbol e_3$ corresponds to a \emph{simple pole} of $w\simeq{2\over n_1-in_2}$.
\unpause
Hence, the general solution is
\eq$$
w(n,\va,\vb,c;z)=c\prod^n_{j=1}{z-a_j\over z-b_j},
\label{solution-pos}
$$
\unpause
The values $a_j,b_j\in\C\cup\{\infty\}$, but $a_i\ne b_j$ ($\forall i,j$).

\end{frame}

\begin{frame}
\frametitle{Solutions to the self\-/duality equations}
\unpause

Let $w_0$ be large enough and generic. Then the equation $w(n,\va,\vb,c;z)=w_0$ has exactly $n$ solutions.
\unpause
$\Rightarrow$ It has exactly $n$ solution un to multiplicity for any $w_0$.
\unpause
$\Rightarrow$
\eq$$
n=q.
\tag{\ref{solution-pos}a}
$$
\unpause
Similarly, for $q<0$ we have
\eq$$
w(q,\va,\vb,c;\bz)=c\prod^{-q}_{j=1}{\bz-a_j\over\bz-b_j},
\label{solution-neg}
$$

\end{frame}

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

Discuss the calculation of the functional integral.
\unpause
Let $w(z,\bz)$ be an arbitrary function. Then the action reads

\Align$$
S[w,\bw]
&={4\pi q\over g}+{8\over g}\int d^2x\,{\bd w\d\bw\over(1+|w|^2)^2}
\label{Sqwplus}
\\
&=-{4\pi q\over g}+{8\over g}\int d^2x\,{\d w\bd\bw\over(1+|w|^2)^2}.
$$
\unpause
Suppose $q\ge0$. Let
\eq$$
S_q[\varphi,\bvarphi]
=S[w(q,\va,\vb,c;z)(1+\varphi(z,\bz)),
  w^*(q,\va,\vb,c;z)(1+\bvarphi(z,\bz))].
\label{Sqactiondef}
$$
\unpause
It is easy to see that
\eq$$
S_q[\varphi,\bvarphi]
={4\pi q\over g}
+{8\over g}\int d^2x\,{|w|^2\over(1+|w|^2)^2}\bd\varphi\d\bvarphi
$$
in the quadratic approximation.
\unpause
The $q$\-/instanton action is
\eq$$
\Aligned{
Z_q
&={e^{-4\pi q/g}\over(q!)^2}
\int d\mu(\va,\vb,c)\,
Z[w(q,\va,\vb,c;z)],
\\
Z[w]
&=\int D\varphi\,D\bvarphi\,\exp\left(
-{8\over g}\int d^2x\,{|w|^2\over(1+|w|^2)^2}\bd\varphi\d\bvarphi
\right)}
\label{Zqdef}
$$
with a conformal invariant measure $\mu$.

\end{frame}

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

The only conformal invariant measure is
\eq$$
\mu(\va,\vb,c)
=k^q{d^2c\over|c|^2}\prod^q_{j=1}d^2a_j\,d^2b_j\,
\prod_{i<j}|a_i-a_j|^4|b_i-b_j|^4\prod_{i,j}|a_i-b_j|^{-4}
\label{instmeasure}
$$
with a certain constant~$k$.
\unpause
The partition function $Z[w]$ is $g$\-/independent,
\unpause
but due to the UV and IR divergences is not literally conformal invariant.
\unpause
Instead,
$$
Z[w]\to Z[w']\prod^q_{j=1}\left|da'_j\over da_j\right|^{2\alpha}
\left|db'_j\over db_j\right|^{2\alpha}
$$
with a certain exponent $\alpha$.
\unpause
Hence,
$$
Z[w]\sim f(c)\prod_{i<j}|a_i-a_j|^{-4\alpha}|b_i-b_j|^{-4\alpha}
\prod_{i\ne j}|a_i-b_j|^{4\alpha}.
$$
\unpause
A rather complex calculation results in $\alpha=1/2$ and $f(c)=|c|^2/(1+|c|^2)^2$, so the integral over $c$ gives just a finite factor.
\unpause
Thus we have
\eq$$
Z_q\sim{\lambda^q\over(q!)^2}\int\prod^q_{j=1}d^2a_j\,d^2b_j\,
\prod_{i<j}|a_i-a_j|^2|b_i-b_j|^2\prod_{i,j}|a_i-b_j|^{-2},
\label{Zqfin}
$$
where $\lambda\sim e^{-4\pi/g}$.

\end{frame}

\begin{frame}
\frametitle{Total partition function}
\unpause

The total partition function $Z=\sum_{q\in\Z}Z_q$ formally coincides with the partition function of the sine\-/Gordon model with $\beta^2=1$, i.e.\ with the partition function of massive free fermions.
\unpause
Literally it is \emph{not correct}.
\unpause
The problem is the we cannot find contributions of solutions that behave like multi\-/instanton in certain regions of space and like multi\-/antisoliton in other regions.
\unpause
The only correct conclusion is that the total theory is \emph{massive}.
\unpause
We will see below that the $O(N)$\-/models with $N>3$, where there are no solitons, are massive too. It is not related to solitons.
\unpause
Moreover, the $O(3)$\-/model looks falls in to the general pattern $N\ge3$ very well.
\unpause

What is the real consequence of the solitons?
\unpause
In the case of solitons we may modify the action:
\eq$$
S_\theta(\bn)=S(\bn)+i\theta q
={1\over2g}\int d^2x\,(\d_\mu\bn)^2
+i{\theta\over8\pi}
\int d^2x\,\bn(\d_\mu\bn\times\d_\nu\bn)\epsilon^{\mu\nu}.
\label{thetaaction}
$$
\unpause
The second \emph{theta\-/term} is purely topological, but it essentially modifies the theory. The partition function reads
$$
Z=\sum^\infty_{q=-\infty}e^{i\theta q}Z_q.
$$
\unpause
In particular, it is known that for $\theta=\pi$ the $O(3)$\-/model is massless, but not scale\-/invariant.
\unpause
For $\theta\ne0,\pi\pmod{2\pi}$ the theory does not seem to be exactly solvable.


\end{frame}

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






\end{frame}



\end{document}
