% !TeX spellcheck = en_US
% !TeX encoding = UTF-8
\documentclass[a4paper]{article}
\usepackage{amsmath,amssymb,fullpage,hyperref}
\usepackage[utf8]{inputenc}
\usepackage{tikz}
\usepackage[shortcuts]{extdash}
\usepackage{hyperref}
\usepackage[nosort]{cite}
\usepackage{color}

%\usepackage[notref,notcite]{showkeys}

\def\red{\color{red}}
\def\blue{\color{blue}}
\def\green{\color{green}}
\def\magenta{\color{magenta}}


\allowdisplaybreaks

\def\smallskip{\vskip\smallskipamount}
\def\medskip{\vskip\medskipamount}
\def\bigskip{\vskip\bigskipamount}

\let\{=\lbrace
\let\}=\rbrace
\let\[=\lbracket
\let\]=\rbracket

\long\def\items#1{\begin{itemize}\parsep=0pt\itemsep=0pt#1\end{itemize}}
\long\def\enum#1{\begin{enumerate}\parsep=0pt\itemsep=0pt#1\end{enumerate}}


\def\eq#1$$#2$${\begin{equation#1}#2\end{equation#1}}
\long\def\subeq#1{\begin{subequations}#1\end{subequations}}
\def\Split$$#1$${\begin{split}#1\end{split}}
\def\Align#1$$#2$${\begin{align#1}#2\end{align#1}}
\def\AlignAt#1$$#2$${\begin{alignat}{#1}#2\end{alignat}}
\def\Aligned#1{\begin{aligned}#1\end{aligned}}
\def\Gather#1$$#2$${\begin{gather#1}#2\end{gather#1}}
\def\Gathered#1{\begin{gathered}#1\end{gathered}}
\def\Multline#1$$#2$${\begin{multline#1}#2\end{multline#1}}
\def\Matrix#1{\begin{matrix}#1\end{matrix}}
\def\pMatrix#1{\begin{pmatrix}#1\end{pmatrix}}
\def\Cases#1{\begin{cases}#1\end{cases}}
\def\d{\partial}
\def\bd{{\bar\d}}
\def\i{{\rm i}}
\def\e{{\rm e}}
\def\bpsi{{\bar\psi}}
\def\bvarphi{{\bar\varphi}}
\def\bL{{\bar L}}
\def\bT{{\bar T}}
\def\Re{\mathop{\rm Re}\nolimits}
\def\Im{\mathop{\rm Im}\nolimits}
\def\sh{\mathop{\rm sh}\nolimits}
\def\ch{\mathop{\rm ch}\nolimits}
\def\tg{\mathop{\rm tg}\nolimits}
\def\ctg{\mathop{\rm ctg}\nolimits}
\def\tr{\mathop{\rm tr}\nolimits}
\def\arctg{\mathop{\rm arctg}\nolimits}
\def\bn{{\boldsymbol n}}
\def\bp{{\boldsymbol p}}
\def\br{{\boldsymbol r}}
\def\bR{{\boldsymbol R}}
\def\bS{{\boldsymbol S}}
\def\bJ{{\boldsymbol J}}
\def\bsigma{{\boldsymbol\sigma}}
\def\Z{{\mathbb Z}}
\def\R{{\mathbb R}}
\def\C{{\mathbb C}}
\def\cH{{\cal H}}
\def\cL{{\cal L}}
\def\cQ{{\cal Q}}
\def\ba{{\bar a}}
\def\balpha{{\bar\alpha}}
\def\bz{{\bar z}}
\def\bzeta{{\bar\zeta}}
\def\bw{{\bar w}}
\def\bw{{\bar w}}
\def\tphi{{\tilde\phi}}
\def\trho{{\tilde\rho}}
\def\ta{{\tilde a}}
\def\vq{{\vec q}}
\def\vx{{\vec x}}
\def\vy{{\vec y}}
\def\vJ{{\vec J}}
\def\va{{\vec a}}
\def\vb{{\vec b}}
\def\eff{{\rm eff}}
\def\const{{\rm \,const\,}}
\def\eF{{\epsilon_F}}
\def\Ech{{E_{\text{ch}}}}
\def\Esp{{E_{\text{sp}}}}
\def\Espel{{E_{\text{sp}}^{\text{el}}}}
\def\Eim{{E_{\text{im}}}}
\def\el{{\text{el}}}
\def\im{{\text{im}}}

\def\lcolon{\mathopen{\,:}}
\def\rcolon{\mathclose{:\,}}

\def\aux{{\tilde1}}



\begin{document}

\centerline{\bf Lecture~2}
\centerline{\bf Correlation functions and operator product expansions}
\bigskip

In the last lecture we found that
\eq$$
|p\rangle=\e^{\i q_0p}|0\rangle.
\label{pvac-0vac-rel}
$$
We want to use this fact to introduce a new class of operators. For this purpose rewrite (\ref{pvac-0vac-rel}) as follows
$$
|p\rangle=\e^{\i q_0p}|0\rangle
=\left.\lcolon\e^{\i p\varphi(\xi)}\rcolon|0\rangle\right|_{\xi^0\to-\infty}
=\lcolon\e^{\i p\varphi(x=0)}\rcolon|0\rangle.
$$
Indeed, the contribution of $\alpha_k$ ($k>0$) vanishes since they kill the vacuum, while the contribution of $\alpha_{-k}$ vanishes since it contains the factor $z^k\to0$.

The exponential operators are very important and deserve a special notation:
\eq$$
V_p(x)=\lcolon\e^{\i p\varphi(x)}\rcolon
=\e^{\i pQ}(z\bz)^{-2\i pP}
\exp\sum_{k>0}{-p(\alpha_{-k}z^k+\balpha_{-k}\bz^k)\over k}
\exp\sum_{k>0}{p(\alpha_kz^{-k}+\balpha_k\bz^{-k})\over k}.
\label{Vp-def}
$$

Consider the product
\Multline*$$
V_{p_1}(x')V_{p_2}(x)
=\e^{\i p_1Q}{\red\boxed{(z'\bz')^{-2\i p_1P}}}
\exp\sum_{k>0}{-p_1(\alpha_{-k}z^{\prime k}+\balpha_{-k}\bz^{\prime k})\over k}
{\blue\boxed{\exp\sum_{k>0}{p_1(\alpha_kz^{\prime-k}+\balpha_k\bz^{\prime-k})\over k}}}
\\
\times
{\red\boxed{\e^{\i p_2Q}}}(z\bz)^{-2\i p_2P}
{\blue\boxed{\exp\sum_{k>0}{-p_2(\alpha_{-k}z^k+\balpha_{-k}\bz^k)\over k}}}
\exp\sum_{k>0}{p_2(\alpha_kz^{-k}+\balpha_k\bz^{-k})\over k}.
$$
To render this to the normal ordered form the boxed parts must be commuted ({\red red} with {\red red}, {\blue blue} with {\blue blue}). By using the standard rule
$$
\e^f\e^g=\e^{[f,g]}\e^g\e^f,
$$
if the commutator $[f,g]$ is a $c$-number, we obtain
\eq$$
V_{p_1}(x')V_{p_2}(x)
=(z'-z)^{2p_1p_2}(\bz'-\bz)^{2p_1p_2}
\lcolon\e^{\i p_1\varphi(x')+\i p_2\varphi(x)}\rcolon.
\label{exp-exp-prod}
$$
More generally,
\eq$$
V_{p_N}(x_N)\lcolon\e^{\i\sum^{N-1}_{i=1}p_i\varphi(x_i)}\rcolon
=\prod^{N-1}_{i=1}(z_N-z_i)^{2p_Np_i}(\bz_N-\bz_i)^{2p_Np_i}
\times\lcolon\e^{\i\sum^N_{i=1}p_i\varphi(x_i)}\rcolon.
\label{expN-prod}
$$
Hence,
$$
\left\langle\prod^{\substack{\curvearrowleft\\N}}_{i=1}V_{p_i}(x_i)\right\rangle
=\prod^N_{i>j}(z_i-z_j)^{2p_ip_j}(\bz_i-\bz_j)^{2p_ip_j}
\times\langle\e^{\i Q\sum^N_{i=1}p_i}\rangle.
$$
We have nearly reached our destination, but we have to consider the last factor accurately. Let $p=\sum p_i$ and write
$$
\langle\e^{\i pQ}\rangle
=\langle0|\e^{\i pQ}|0\rangle
=\langle0|p\rangle.
$$
The r.h.s.\ must be zero, if $p\ne0$, and a nonzero constant (e.g.~1), if $p=0$. It is mathematically consistent, but the answer may look strange for a physicist. To advance in a more physical way, let us apply the rule
\eq$$
\langle\e^f\rangle=\e^{{1\over2}\langle f^2\rangle},
\label{exp-vev}
$$
if $f$ is linear in the basic oscillator operators. We have
\eq$$
\langle\e^{\i pQ}\rangle=\e^{-{p^2\over2}\langle Q^2\rangle}=R^{-2p^2}
\label{expipQ-calculation}
$$
according to the assumption of the last lecture. Therefore
\eq$$
\left\langle\prod^{\substack{\curvearrowleft\\N}}_{i=1}V_{p_i}(x_i)\right\rangle
=R^{-2\left(\sum^N_{i=1}p_i\right)^2}
\prod^N_{i>j}(z_i-z_j)^{2p_ip_j}(\bz_i-\bz_j)^{2p_ip_j}.
\label{exp-prod-gen}
$$
Since physically $R$ is the scale of the scrap on which the theory lives, the infinite plane corresponds $R\to\infty$. In this limit we reproduce the mathematical answer
\eq$$
\left\langle\prod^{\substack{\curvearrowleft\\N}}_{i=1}V_{p_i}(x_i)\right\rangle
=\prod^N_{i>j}(z_i-z_j)^{2p_ip_j}(\bz_i-\bz_j)^{2p_ip_j}\times\Cases{1,&\sum^N_{i=1}p_i=0;\\0,&\sum^N_{i=1}\ne0.}
\label{exp-prod}
$$

Let us reproduce this answer in yet more physical (and simple) way. Calculate $\left\langle\exp\i\sum p_i\varphi(x_i)\right\rangle$ by means of the functional integral. In fact, the only thing we need from the functional integral is~(\ref{exp-vev}). We obtain
$$
\left\langle\e^{\i\sum p_i\varphi(x_i)}\right\rangle
=\exp\left(-{1\over2}\sum^N_{i,j}p_ip_j\langle\varphi(x_i)\varphi(x_j)\rangle\right)
=\prod^N_{i,j}\left((z_i-z_j)(\bz_i-\bz_j)\over R^2\right)^{p_ip_j}.
$$
There are two types of terms in the exponent. For $i\ne j$ we may apply standard formula for pair correlation function, but if $i=j$ it formally gives infinity. Let us cut it at a small radius~$r_0$:
\eq$$
\langle\varphi(x')\varphi(x)\rangle
=\Cases{2\log{R^2\over(z'-z)(\bz'-\bz)},&\text{if $|x^2|>r_0^2$;}\\
2\log{R^2\over r_0^2},&\text{if $|x^2|\le r_0^2$.}}
\label{varphi-corr-cut}
$$
Then we have (if $|(x_i-x_j)^2|>r_0^2$)
\Multline*$$
\left\langle\e^{\i\sum p_i\varphi(x_i)}\right\rangle
=\left(r_0^2\over R^2\right)^{\sum p_i^2}
\prod^N_{i>j}\left((z_i-z_j)(\bz_i-\bz_j)\over R^2\right)^{2p_ip_j}
\\
=r_0^{2\sum p_i^2}R^{-2\left(\sum p_i\right)^2}
\prod^N_{i>j}(z_i-z_j)^{2p_ip_j}(\bz_i-\bz_j)^{2p_ip_j}.
$$
We see that the contribution of $r_0$ can be factorized between the exponents. Hence, the answer will coincide with (\ref{exp-prod-gen}), if we set
\eq$$
\e^{\i p\varphi(z)}=r_0^{2p^2}\lcolon\e^{\i p\varphi(z)}\rcolon.
\label{exp-normalexp}
$$
It means that the normal ordered exponent $\lcolon\e^{\i p\varphi(z)}\rcolon$ is a proper renormalization of the true exponent $\e^{\i p\varphi(z)}$. This argument also explains why the normal ordered exponent possesses the scaling dimension~$d=2p^2$: any correlation function on an infinite plane is invariant under the substitution
\eq$$
V_p(x)\to\lambda^{2p^2}V_p(x).
\label{exp-dim}
$$
In other words the scaling dimension $d$ is the eigenvalue of the dilation operator $D$ acting on the corresponding state~$|p\rangle$.

Now we will be interested in the energy\-/momentum tensor. According to the usual formula in the Minkowski space
$$
T^\mu_\nu=\d_\nu\varphi{\d\cL\over\d(\d_\mu\varphi)}-\delta^\mu_\nu\cL
$$
we obtain
\eq$$
T_{zz}\equiv-2\pi T={1\over8\pi}(\d\varphi)^2,
\qquad
T_{\bz\bz}\equiv-2\pi\bT={1\over8\pi}(\bd\varphi)^2,
\qquad
T_{z\bz}\equiv2\pi\Theta=0.
\label{em-tensor-def}
$$
The last equation reflects conformal invariance of the theory. The energy\-/momentum conservation law $\d_\mu T^\mu_\nu=0$ leads to
$$
\bd T=\d\Theta,
\qquad
\d\bT=\bd\Theta.
$$
Since $\Theta=0$ they reduce to
$$
\bd T=\d\bT=0.
$$
It means that the operator $T=T(z)$ is a function of the only variable~$z$, while $\bT=\bT(\bz)$ is a function of the only variable~$\bz$. On the Euclidean plane $T(z)$ is a complex analytic (holomorphic) function of~$z$, while $\bT(\bz)$ is a holomorphic function of~$\bz$ (it is usually called `antiholomorphic'). In quantum mechanics it is natural to define them as
\eq$$
T(z)=-{1\over4}\lcolon(\d\varphi)^2\rcolon,
\qquad
\bT(\bz)=-{1\over4}\lcolon(\bd\varphi)^2\rcolon.
\label{TbT-freeboson}
$$
On the cylinder it is natural to define the quantum energy\-/momentum tensor components according to
\eq$$
T_{\rm cyl}(\zeta)=-\left(2\pi\over L\right)^2\left(z^2T(z)-{1\over24}\right),
\qquad
\bT_{\rm cyl}(\bzeta)=-\left(2\pi\over L\right)^2\left(\bz^2\bT(z)-{1\over24}\right).
\label{TbTcyl-freeboson}
$$
The term $-1/24$ was added artificially to reproduce the term $-\pi/6L$ in the Hamiltonian $H$ being written in terms of the energy\-/momentum tensor:
\eq$$
H=\int^L_0d\xi^1\,T^{00}(\xi)=-\int^L_0{d\xi^1\over2\pi}(T_{\rm cyl}(\xi^1)+\bT_{\rm cyl}(\xi^1)).
\label{H-tensor}
$$
In one of the next lectures we will understand the true origin of this term.

It is convenient to introduce the Laurent components of the energy\-/momentum tensor:
\eq$$
\Aligned{
L_k
&=\oint{dz\over2\pi\i}z^{k+1}T(z)
=p^2\delta_{k0}+{1\over4}\sum_{l\in\Z\setminus\{0,k\}}\lcolon\alpha_l\alpha_{k-l}\rcolon,
\\
\bL_k
&=\oint{d\bz\over2\pi\i}\bz^{k+1}T(\bz)
=p^2\delta_{k0}+{1\over4}\sum_{l\in\Z\setminus\{0,k\}}\lcolon\balpha_l\balpha_{k-l}\rcolon.
}\label{LbL-def}
$$
The integration contours are defined in the Euclidean plane. They encircle coordinate origin $z=0$ ($\bz=0$) in the counter-clockwise direction of $z$ ($\bz$) and all points, where operators, which stand to the right of $T(z)$ ($\bT(\bz)$), are placed. For example, in the product $T(u)L_kT(w)$ it will enclose $w$, but not~$u$.

In these components (\ref{H-tensor}) can be rewritten as
\eq$$
H={2\pi\over L}\left(L_0+\bL_0-{1\over12}\right).
\label{H-L0bL0}
$$
Similarly, the momentum operator on the cylinder is given by
\eq$$
P={2\pi\over L}(L_0-\bL_0).
\label{P-L0bL0}
$$
On the plane the operator $D=L_0+\bL_0$ is the dilation operator, while $S=L_0-\bL_0$ is the angular momentum operator. Translations are generated by the operators $L_{-1},\bL_{-1}$. We may set
\eq$$
H_{\rm plane}=-\i(L_{-1}-\bL_{-1}),
\qquad
P_{\rm plane}=-\i(L_{-1}+\bL_{-1}).
\label{HP-plane}
$$

Consider the operator product
\Multline*$$
T(z')T(z)={1\over16}\lcolon(\d\varphi(x'))^2\rcolon\lcolon(\d\varphi(x))^2\rcolon
\\
={1\over16}\lcolon(\d\varphi(x'))^2(\d\varphi(x))^2\rcolon
+{1\over4}\langle\d\varphi(x')\,\d\varphi(x)\rangle\lcolon\d\varphi(x')\,\d\varphi(x)\rcolon
+{1\over8}\langle\d\varphi(x')\,\d\varphi(x)\rangle^2.
$$
We want to count singularities of the expression. The normal products are regular as $z'\to z$. Hence, the only singularity can stem from the correlation function
$$
\langle\d\varphi(x')\,\d\varphi(x)\rangle=-{\d^2\over\d z^{\prime2}}\langle\varphi(x')\varphi(x)\rangle
={2\over(z'-z)^2}.
$$
By expanding everything in powers of $z'-z$ we obtain the following operator product expansion (OPE):
\eq$$
T(z')T(z)={1/2\over(z'-z)^4}+{T(z)\over(z'-z)^2}+{\d T(z)\over z'-z}+O(1).
\label{TT-OPE}
$$
Consider now the commutator
$$
[L_k,L_l]=L_kL_l-L_lL_k
=\oint_{C_{\rm out}}{dz'\over2\pi\i}\oint_C{dz\over2\pi\i}z^{\prime k-1}z^{l-1}T(z')T(z)
-\oint_C{dz\over2\pi\i}\oint_{C_{\rm in}}{dz'\over2\pi\i}z^{\prime k-1}z^{l-1}T(z')T(z).
$$
The two terms differ in the order of contours. The contour $C_{\rm out}$ encloses the contour $C$, while the contour $C_{\rm in}$ is enclosed by~$C$. Hence,
$$
[L_k,L_l]
=\oint_C{dz\over2\pi\i}\oint_{C_z}{dz'\over2\pi\i}z^{\prime k-1}z^{l-1}T(z')T(z).
$$
Here $C_z$ is a small contour that encloses $z$ in a counter-clockwise direction. After substituting the OPE (\ref{TT-OPE}) and evaluating the residue we obtain
\eq$$
[L_k,L_l]=(k-l)L_{k+l}+{1\over12}k(k^2-1)\delta_{k+l,0}.
\label{Vir1-def}
$$
The algebra (\ref{Vir1-def}) is called the \emph{Virasoro algebra}. The same is valid for the left (antichiral) component $\bT(\bz)$, which produces another copy of the Virasoro algebra.

Similarly one can prove the OPE
\eq$$
T(z')V_p(x)={p^2V_p(x)\over(z'-z)^2}+{\d V_p(x)\over z'-z}+O(1).
\label{Texp-OPE}
$$
As a commutator one can rewrite it as
\eq$$
[L_k,V_p(x)]=p^2(k+1)z^kV_p(x)+z^{k+1}\d V_p(x).
\label{Lexp-commut}
$$
The sense of the above equations will be clarified later, when we will study conformal field theory.

\bigbreak
\bigskip
\centerline{\bf Problems}
\medskip

{\bf 1.}~Accurately prove the formulas (\ref{exp-exp-prod}),~(\ref{expN-prod}).

{\bf 2.}~Let us modify the energy\-/momentum tensor as follows:%
%
\footnote{We will see below that this form of the energy\-/momentum tensor is consistent with the Liouville theory, if $\i\cQ=b+b^{-1}$.}
%
\eq$$
T(z)=-{1\over4}(\d\varphi)^2+{\i\cQ\over2}\d^2\varphi,
\qquad
\bT(\bz)=-{1\over4}(\bd\varphi)^2+{\i\cQ\over2}\bd^2\varphi.
\label{TbT-Liouv}
$$
Prove that
$$
T(z')T(z)={c/2\over(z'-z)^4}+{T(z)\over(z'-z)^2}+{\d T(z)\over z'-z}+O(1),
\qquad
c=1-6\cQ^2,
$$
and
$$
[L_k,L_l]=(k-l)L_{k+l}+{c\over12}k(k^2-1)\delta_{k+l,0}.
$$

{\bf 3.}~Prove that for $T(z)$ from (\ref{TbT-Liouv}) the Virasoro algebra generators read
$$
L_k=p^2\delta_{k0}+{1\over4}\sum_{l\in\Z\setminus\{0,k\}}\lcolon\alpha_l\alpha_{k-l}\rcolon+{k+1\over2}\cQ\alpha_k.
$$

{\bf 4.}~For $T(z)$ from (\ref{TbT-Liouv}) prove the OPE
$$
T(z')V_p(x)={\Delta_pV_p(x)\over(z'-z)^2}+{\d V_p(x)\over z'-z}+O(1),
\qquad
\Delta_p=p(p-\cQ).
$$

{\bf 5.}~Define the so called correlation functions with a charge at infinity:
$$
\langle X\rangle_\cQ=\lim_{x_0\to\infty}(z_0\bz_0)^{2\cQ^2}\langle XV_{-\cQ}(x_0)\rangle.
$$
Prove that for exponential operators they read
\eq$$
\left\langle\prod^{\substack{\curvearrowleft\\N}}_{i=1}V_{p_i}(x_i)\right\rangle_\cQ
=\prod^N_{i>j}(z_i-z_j)^{2p_ip_j}(\bz_i-\bz_j)^{2p_ip_j}\times\Cases{1,&\sum^N_{i=1}p_i=\cQ;\\0,&\sum^N_{i=1}\ne0.}
\label{exp-corr-mod}
$$
Show the functions (\ref{exp-corr-mod}) to be invariant under the transformation
$$
V_p(z,\bz)\to(f'(z))^{\Delta_p}(\bar f'(\bz))^{\Delta_p}V_p(f(z),\bar f(\bz)),
$$
if $f(z)$ is a Möbius transformation
$$
f(z)={az+b\over cz+d}.
$$







\end{document}
