% File: cft2-prod.tex
% Section: CFT2
% Title: Operator products in CFT2
% Last modified: 07.11.2003
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\begin{document}

\begin{center}
\large\textbf{OPERATOR PRODUCTS IN CFT${}_2$}
\end{center}

\vspace{.1cm}

\begin{center}
\large\textbf{Algebras of chiral conformal fields}
\end{center}

Consider a typical set of local field operators in CFT${}_2$\,:
a primary chiral field $\varphi(z)$ of conformal dimension $\Delta$,
the $T_{zz}$\,-\,component (denoted by $T(z)$) of the energy-momentum
tensor ($\Delta=2$), and a chiral current $J^a(z)$ with $\Delta=1$\,.
The following operator-algebra relations are assumed to hold:

\begin{align}
  T(z)\,T(w) &= \frac{c}{2(z-w)^4}+\frac{2}{(z-w)^2}\,T(w)
+\frac{1}{z-w}\,\partial\, T(w) \label{1} \\
  T(z)\,\varphi(w) &= \frac{\Delta}{(z-w)^2}\,\varphi(w)+\frac{1}{z-w}\,
\partial \varphi(w) \label{2} \\
 J^a(z)\,J^b(w) &= \frac{\varkappa^{ab}K}{(z-w)^2}
+\frac{f_{c}^{ab}}{z-w}\,J^c(w) \label{3}
\end{align}
(regular terms are omitted).
Here the radially-ordered operator product is meant:
\begin{equation}
\label{rad}
A(z)B(w) = B(w)A(z) \equiv \mathcal{R}A(z)B(w) \doteq
\begin{cases}
A(z)B(w) & \qquad |z|>|w| \\ B(w)A(z) & \qquad |w|>|z|
\end{cases}
\end{equation}
To permute operators in (\ref{2}), one uses (\ref{rad})
and then expands in $z-w$ discarding regular terms:
\begin{equation}
\label{}
\varphi(z)\,T(w) = T(w)\,\varphi(z)
= \frac{\Delta}{(z-w)^2}\,\varphi(w)+\frac{\Delta-1}{z-w}\,
\partial \varphi(w)
\end{equation}

Now we define the corresponding modes (the contours surround zero):
\begin{align}
  T(z) &= \sum_{n}^{} \frac{L_n}{z^{n+2}} &
L_n &= \frac{1}{2\pi i}\oint dz\,z^{n+1}\,T(z) \\
 \varphi(z) &= \sum_{n}^{}z^{-n-\Delta}\,\varphi_n &
\varphi_n &= \frac{1}{2\pi i}\oint dz\,z^{n+\Delta-1}\,\varphi(z) \\
 J^a(z) &= \sum_{n}^{} z^{-n-1}\,J_{n}^{a} &
J_n^a &= \frac{1}{2\pi i}\oint dz\,z^n\,J^a(z)
\end{align}

Our aim is to derive the (centrally extended) Lie algebra relations
for the modes
\begin{align}
 [L_m,L_n] &=
(m-n)L_{m+n}+\frac{c}{12}\,m(m^2-1)\,\delta_{n,-m}\label{vir}\\
 [L_m,\varphi_n] &= (m(\Delta-1)-n)\,\varphi_{m+n}\label{lphi} \\
 [L_m,J_n^a] &= -n\,J_{m+n}^a \\
 [J_m^a,J_n^b] &=
f_{c}^{ab}\,J_{m+n}^c + m\,\delta_{n,-m}\,\varkappa^{ab}K
\end{align}
and fields (currents)
\begin{align}
 [T(z),\,T(w)] &= -\frac{c}{12}\,\delta'''(z-w)-2\delta'(z-w)\,T(w)
+ \delta(z-w)\,T'(w)\label{local1} \\
 [J^a(z),\,J^b(w)] &=
 -\varkappa^{ab}K\,\delta'(z-w)
+f_{c}^{ab}J^c(w)\,\delta(z-w)\label{local2}
\end{align}
from the operator product expansion (OPE) relations (\ref{1})--(\ref{3}).
It's important to note that when in commutators (inside square brackets)
an ordinary (not radially ordered!) product is assumed:
\begin{equation}
\label{}
[A(z)\,,B(w)] \neq \mathcal{R}A(z)B(w)-\mathcal{R}B(w)A(z) \equiv 0\,.
\end{equation}

To begin with, let's perform the following computation.
\begin{multline}
\label{basic}
[L_m,\,\varphi(z)\,]
= \frac{1}{2\pi i}\left[\oint d\xi\,\xi^{m+1}\,
T(\xi),\,\varphi(z)\right]\\
= \frac{1}{2\pi i}\left(\oint_{C_1} d\xi\,\xi^{m+1}\,T(\xi)\right)\,\varphi(z)
 - \frac{\varphi(z)}{2\pi i}\oint_{C_2} d\xi\,\xi^{m+1}\,T(\xi) \\
= \frac{1}{2\pi i}\left(\oint_{C_1} - \oint_{C_2}\right)d\xi\,\xi^{m+1}
 \,\mathcal{R}\,T(\xi)\,\varphi(z) \\
= \frac{1}{2\pi i}\oint_z d\xi\,\xi^{m+1}\,
\left(\frac{\Delta}{(\xi-z)^2}\,\varphi(z)
+ \frac{1}{\xi-z}\,\partial\varphi(z)\right) \\
= (m+1)\Delta\,z^m\,\varphi(z) + z^{m+1}\,\partial\varphi(z)
\end{multline}
A point $z$ is assumed to be inside the contour $C_1$ but outside $C_2$
(and point 0 inside both).
All contours are followed counterclockwise. Standard formulas like
\begin{equation}
\label{}
\frac{1}{2\pi i}\oint_z \frac{d\xi f(\xi)}{(\xi-z)^{n+1}} = \frac{1}{n!}f^{(n)}(z)
\end{equation}
are widely used.

Now it is straightforward to obtain a commutator of the
corresponding modes:
\begin{multline}
[L_m,\varphi_n]
 = \frac{1}{2\pi i}\oint dz\,z^{n+\Delta-1}\,[L_m,\,\varphi(z)\,] \\
= \frac{1}{2\pi i}\oint dz\,z^{n+\Delta-1}
\left((m+1)\Delta\,z^m + z^{m+1}\,\partial_z\right)\varphi(z) \\
= \frac{1}{2\pi i}\oint dz\,z^{n+\Delta-1}\,
\sum_{k}^{}\left((m+1)\Delta-(k+\Delta)\right)\,
z^{m-k-\Delta}\,\varphi_k \\
= \frac{1}{2\pi i}\oint dz\,
\sum_{k}^{}z^{m+n-k-1}\,(m\Delta-k)\,\varphi_k
= (m(\Delta-1)-n)\,\varphi_{m+n}
\end{multline}

The same result can be derived as follows:
\begin{multline}
\sum_{n}^{}z^{-n-\Delta}\,[L_m,\,\varphi_n] \equiv [L_m,\,\varphi(z)\,]
=\left((m+1)\Delta\,z^m + z^{m+1}\,\partial_z\right)
\sum_{n}^{}z^{-n-\Delta}\,\varphi_n \\
= \sum_{n}^{}\,[(m+1)\Delta-(n+\Delta)]\,z^{m-n-\Delta}\,\varphi_n
= \sum_{n}^{}(m(\Delta-1)-n)\,z^{-n-\Delta}\,\varphi_{n+m}\,,
\end{multline}
and it remains to equate the coefficients of equal powers of $z$\,.

Quite analogously, we obtain the Virasoro algebra (\ref{vir})\,:
\begin{multline}
\sum_{n}^{}z^{-n-2}\,[L_m,\,L_n] \equiv [L_m,\,T(z)]
= \frac{1}{2\pi i}\left(\oint_{C_1} - \oint_{C_2}\right)d\xi
\,\xi^{m+1}\,T(\xi)\,T(z) \\
= \frac{1}{2\pi i}\oint_z d\xi\,\xi^{m+1}\,
\left(\frac{c}{2(\xi-z)^4}+\frac{2}{(\xi-z)^2}\,T(z)
+ \frac{1}{\xi-z}\,\partial T(z)\right) \\
= \frac{c}{12}\,m(m^2-1)\,z^{m-2}+2(m+1)\,z^m\,T(z)
+ z^{m+1}\,\partial_z T(z) \\
= \frac{c}{12}\,m(m^2-1)\,z^{m-2} + \sum_n(m-n)z^{-n-2}L_{m+n} \\
= \sum_n z^{-n-2}\,\left[(m-n)L_{m+n}
+ \frac{c}{12}\,m(m^2-1)\,\delta_{n,-m}\right]
\end{multline}

For illustration, the (affine Kac-Moody) algebra of the $J_m^a$ modes
can be calculated directly from OPE:
\begin{multline}
\label{direct}
[J_{m}^{a},\,J_n^b]
= \frac{1}{(2\pi i)^2} \left(\oint_{C_1}dz J^a(z)\oint_{C_2}dw J^b(w)
- \oint_{C_2}dw J^b(w)\oint_{C_1}dz J^a(z)\right)z^m\,w^n \\
= \frac{1}{(2\pi i)^2} \left(\oint_{C_1}dz\oint_{C_2}dw
- \oint_{C_2}dw\oint_{C_1}dz\right)z^mw^n\mathcal{R}J^a(z)\,J^b(w) \\
= \frac{1}{2\pi i}\oint dw\,w^n\,\frac{1}{2\pi i}\oint_w dz\,z^m
\left(\frac{\varkappa^{ab}K}{(z-w)^2}
+ \frac{f_{c}^{ab}}{z-w}\,J^c(w)\right) \\
= \frac{1}{2\pi i}\oint dw\,w^n\left(\varkappa^{ab}K\,m\,w^{m-1}
+ f_{c}^{ab}\,w^mJ^c(w)\right)
= \varkappa^{ab}K\,m\,\delta_{n,-m} + f_{c}^{ab}\,J^c_{m+n}
\end{multline}
In first two lines, the left contour is in both cases supposed
to embrace the right one. We see that the radial ordering
of operators is transformed into a specific ordering of the integration
contours.

To derive the formulas for local field commutators
 (\ref{local1}),(\ref{local2}), it is very
convenient to use the following representation of $\delta$-function
(corresponding to a complete set of functions $z^n$ on a circle):
\begin{equation}
\label{delta}
\delta(z-w) \doteq \frac{1}{w}\sum_{n}\left(\frac{z}{w}\right)^n
= \ldots+\frac{z}{w^2}+\frac{1}{w}+\frac{1}{z}+\frac{w}{z^2}+\ldots
= \delta(w-z)
\end{equation}
As usual, the principal relation with this
$\delta$-function is
\begin{equation}
\label{main}
\frac{1}{2\pi i}\oint dw\,f(w)\,\delta(z-w)
 = \frac{1}{2\pi i}\oint dw\,\sum_{mn}^{}f_m\,w^{m-n-1}z^{n}
 = \sum_n f_n\,z^n = f(z)
\end{equation}
(all integrations around zero).
Some other useful relations are easily derived, for example,
\begin{equation}
\label{}
f(z)\,\delta(z-w) = \sum_{mn}^{}f_m\,z^{m+n} w^{-n-1}
 = \sum_{mn}^{}f_m\,w^{m-n-1} z^{n} = f(w)\,\delta(z-w)
\end{equation}

\begin{equation}
\label{}
\delta'(z-w) = \frac{1}{zw}\sum_{n}n\left(\frac{z}{w}\right)^n\,, \qquad
\delta''(z-w) =
\frac{1}{z w^2}\sum_{n}n(n+1)\left(\frac{z}{w}\right)^n\,,
\end{equation}

\begin{equation}
\label{}
\delta'''(z-w) =
\frac{1}{z^2w^2}\sum_{n}n(n^2-1)\left(\frac{z}{w}\right)^n\,, \qquad
\frac{1}{2\pi i}\oint dw\,f(w)\,\delta'(z-w) = f'(z)\,.
\end{equation}

In a symbolic sense, the definition (\ref{delta}) may be formally
represented as
\begin{equation}
\label{diff}
\delta(z-w) = \left.\frac{1}{z-w}\,\right|_{|z|>|w|}
- \left.\frac{1}{z-w}\,\right|_{|z|<|w|}
\end{equation}
which is `almost' the Cauchy formula
\begin{equation}
\label{}
f(z) = \frac{1}{2\pi i} \oint_z\frac{dw\,f(w)}{w-z}
 = \frac{1}{2\pi i}\left(\oint_{C_1}-\oint_{C_2}\right)\frac{dw\,f(w)}{w-z}
\end{equation}
($C_1$ surrounds $z$ and 0 while $C_2$ only 0).
Differentiating (\ref{diff}), one obtains
\begin{equation}
\label{diffn}
\frac{(-)^n}{n!}\delta^{(n)}(z-w)
 = \left.\frac{1}{(z-w)^{n+1}}\,\right|_{|z|>|w|}
- \left.\frac{1}{(z-w)^{n+1}}\,\right|_{|z|<|w|}
\end{equation}

Now, with the use of $\delta$-function (\ref{delta}), we compute
\begin{multline}
\label{locj}
 [J^a(z),\,J^b(w)] = \sum_{mn}z^{-m-1}w^{-n-1}[J^a_m,\,J^b_n] \\
=  \sum_{mn}z^{-m-1}w^{-n-1}(f_{c}^{ab}\,J_{m+n}^c
 + m\,\delta_{n,-m}\,\varkappa^{ab}K) \\
= -\varkappa^{ab}K\sum_n \frac{n}{zw}\left(\frac{z}{w}\right)^n
+ f_{c}^{ab}\sum_m z^{-m-1}w^m\sum_n w^{-m-n-1}J_{m+n}^{c} \\
= -\varkappa^{ab}K\,\delta'(z-w)+f_{c}^{ab}J^c(w)\,\delta(z-w)\,,
\end{multline}

\begin{multline}
[T(z),\,T(w)] = \sum_{mn}z^{-m-2}w^{-n-2}[L_m,\,L_n] \\
= \sum_{mn}z^{-m-2}w^{-n-2}\left((m-n)L_{m+n}
+\frac{c}{12}\,m(m^2-1)\,\delta_{n,-m}\right) \\
=  -\frac{c}{12}\,\delta'''(z-w)
+ \sum_{mn}z^{-m-2}w^{-n-2}(2(m+1)-m-n-2)L_{m+n} \\
= -\frac{c}{12}\,\delta'''(z-w)
+2\frac{1}{zw}\sum_m (m+1)\left(\frac{w}{z}\right)^{m+1}
\sum_n w^{-m-n-2}L_{m+n} \\
-\frac{1}{z}\sum_m\left(\frac{w}{z}\right)^{m+1}
\sum_n (m+n+2)w^{-m-n-3}L_{m+n} \\
= -\frac{c}{12}\,\delta'''(z-w)-2\delta'(z-w)\,T(w)+\delta(z-w)\,T'(w)\,.
\end{multline}

Such formulas are also `derived' by taking into account
the following formal representation (similar to (\ref{diff})
for the local commutators
\begin{align}
[T(z),\,T(w)] &= \mathcal{R}T(z)T(w)
\left(\bigl|_{|z|>|w|}\quad - \quad\bigr|_{|z|<|w|}\right) \\
[J^a(z),\,J^b(w)] &= \mathcal{R}J^a(z)J^b(w)
\left(\bigl|_{|z|>|w|}\quad - \quad\bigr|_{|z|<|w|}\right)
\end{align}
and then transforming the differences into $\delta$-functions by
eq. (\ref{diffn}).

Due to (\ref{main}), the local field commutators similar to
(\ref{local1}),(\ref{local2}) should be understood in the sense
of integration over a closed contour. Really, if we define
for some function $\varepsilon(z)$ the quantities
\begin{equation}
\label{}
T_\varepsilon = \frac{1}{2\pi i}\oint d\xi\,\varepsilon(\xi)\,T(\xi)
\qquad\qquad
J^a_\varepsilon = \frac{1}{2\pi i}\oint d\xi\,\varepsilon(\xi)\,J^a(\xi)
\end{equation}
and then integrate (\ref{local1}) and (\ref{local2}) multiplied by
$\varepsilon(z)$ over a closed contour, the following commutators
are derived,
\begin{align}
[T_\varepsilon,\,T(z)] &= \frac{c}{12}\varepsilon'''(z)
+ 2\varepsilon'(z)T(z) + \varepsilon(z)T'(z) \label{teps} \\
[J^a_\varepsilon,\,J^b(z)] &= \varkappa^{ab}K\varepsilon'(z)
+ f_{c}^{ab}\varepsilon(z)J^c(z)
\end{align}
which can be also calculated directly, in complete analogy
with the proof of (\ref{basic}). Here is an example
of such a computation:
\begin{multline}
\label{}
[T_\varepsilon,\,\varphi(z)\,]
= \frac{1}{2\pi i}\left[\oint d\xi\,\varepsilon(\xi)\,T(\xi),\,
\varphi(z)\right]
= \frac{1}{2\pi i}\left(\oint_{C_1} - \oint_{C_2}\right)
d\xi\,\varepsilon(\xi)\,T(\xi)\,\varphi(z) \\
= \frac{1}{2\pi i}\oint_z d\xi\,\varepsilon(\xi)\,T(\xi)\,\varphi(z)
= \frac{1}{2\pi i}\oint_z d\xi\,\varepsilon(\xi)\,
\left(\frac{\Delta}{(\xi-z)^2}\,\varphi(z)
+ \frac{1}{\xi-z}\,\partial\varphi(z)\right) \\
=\Delta\,\varepsilon'(z)\,\varphi(z)+\varepsilon(z)\,\partial\varphi(z)
\end{multline}
(as in (\ref{basic}), $z$ is inside $C_1$ but outside $C_2$). We see that
$T_\varepsilon$ generates conformal transformations
\begin{equation}
\label{}
z\rightarrow z' = z + \varepsilon(z)\,, \ \ \
\varphi'(z') = \left(\frac{dz'}{dz}\right)^{-\Delta}\,\varphi(z)
\end{equation}
because of
\begin{equation}
\label{}
\delta_{\varepsilon}\varphi(z) \doteq \varphi'(z)-\varphi(z)
 = -(\Delta\varepsilon'+\varepsilon\partial)\varphi(z)
 = -[T_\varepsilon\,,\varphi(z)]\,.
\end{equation}
Variations $\delta_\varepsilon \doteq -\varepsilon\partial_z-\Delta\varepsilon'$
obey the Lie algebra
\begin{equation}
\label{}
[\delta_{\varepsilon_1}\,,\delta_{\varepsilon_2}]
 = \delta_{[\varepsilon_1,\varepsilon_2]}\,, \ \ \
[\varepsilon_1\,,\varepsilon_2] \doteq \varepsilon_1'\varepsilon_2
 - \varepsilon_1\varepsilon_2'\,.
\end{equation}
From (\ref{teps}) one sees that in case of $T_\varepsilon$ such algebra
is centrally extended:
\begin{equation}
\label{}
[T_{\varepsilon_1}\,,T_{\varepsilon_2}] = T_{[\varepsilon_1,\varepsilon_2]}
 + \frac{c}{24}\oint\frac{dz}{2\pi i}
(\varepsilon_1'''\varepsilon_2 - \varepsilon_1\varepsilon_2''')\,.
\end{equation}

\newpage

\begin{center}
\large\textbf{Normal products}
\end{center}

In its full form, the operator product expansion looks like
\begin{equation}
\label{ope}
A(z)B(w) \equiv \mathcal{R}A(z)B(w) \doteq \sum_n (z-w)^n (AB)_n(w)\,.
\end{equation}
Its singular part (which is the only part of OPE
commonly written down) is given by
\begin{equation}
\label{sing}
\underbrace{A(z)B(w})\, \doteq \sum_{n<\,0} (z-w)^n (AB)_n(w)
= -\frac{1}{2\pi i} \oint_w\frac{dx}{x-z}A(x)B(w)
\end{equation}
($z$ outside the contour)\,. However, the $n=0$ term in (\ref{ope}),
a `normal product',
\begin{equation}
\label{norm}
:\!A(w)B(w)\!:\ \equiv \ :\!AB\!:\!(w) \doteq (AB)_0(w)
 = \frac{1}{2\pi i} \oint_w\frac{dx}{x-w}A(x)B(w)
\end{equation}
is also widely used for building composite local operators.
Such operators are also subject to OPE procedure due to `Wick formula'
\begin{equation}
\label{wick}
\underbrace{A(z):\!\!BC\!:\!(w}) = \frac{1}{2\pi i} \oint_w\frac{dx}{x-w}
\left(\underbrace{A(z)B(x})C(w) + B(x)\underbrace{A(z)C(w})\right)\,.
\end{equation}
The proof
\begin{multline}
\underbrace{A(z):\!\!BC\!:\!(w}) = -\frac{1}{(2\pi i)^2}\oint_{C_1}\frac{dy\,A(y)}{y-z}
\oint_w\frac{dx}{x-w}B(x)C(w) \\
= -\frac{1}{(2\pi i)^2}\oint_w\frac{dx}{x-w}\oint_x\frac{dy\,A(y)B(x)}{y-z}C(w)
-\frac{1}{(2\pi i)^2}\oint_w\frac{dx\,B(x)}{x-w}\oint_{C_2}\frac{dy\,A(y)C(w)}{y-z}
\end{multline}
reduces to deforming contour $C_1$ which is chosen to embrace
(while $C_2$ is embraced by) another integration contour (around $w$);
$z$ is outside all contours, and $w$ is inside both $C_1$ and $C_2$.

As an example of implementing these techniques let's consider two
primary fields $b(z)$ with $\Delta = \lambda$ and $c(z)$ with
$\Delta = 1-\lambda$\,. A parameter $\nu$ is to distinguish
the cases of $b$ and $c$ being fermions $(\nu=1)$ or bosons
$(\nu=-1)$\,: $b(z)\,c(w)=-\nu c(w)\,b(z)$\,. Reparametrization
ghosts in conformally invariant string theory correspond to
$\lambda=2, \ \nu=1$\,.

The following OPE relations are assumed to be our starting point:
\begin{equation}
\label{cb}
c(z)\,b(w) = \frac{1}{z-w}\,, \ \ \ \ b(z)\,c(w) = \frac{\nu}{z-w}\,,
\ \ \ \ c(z)\,c(w) = b(z)\,b(w) = 0
\end{equation}
The corresponding mode expansions are chosen to be
\begin{equation}
\label{bcmode}
b(z) = \sum_{n}z^{-n-\lambda}\,b_n\,, \ \ \ \ \
c(z) = \sum_{n}z^{-n-1+\lambda}\,c_n\,.
\end{equation}
Proceeding analogously to (\ref{direct}) we obtain
\begin{equation}
\label{cmbn}
c_m b_n + \nu b_n c_m = \delta_{m,-n}\,, \ \ \ \
b_m b_n + \nu b_n b_m = c_m c_n + \nu c_n c_m = 0\,.
\end{equation}
Really,
\begin{multline}
\label{}
c_m b_n + \nu b_n c_m = \frac{1}{(2\pi i)^2}
\left(\oint_{C_1}dz\oint_{C_2}dw- \oint_{C_2}dw\oint_{C_1}dz\right)
z^{m-\lambda} w^{n-1+\lambda} c(z) b(w) \\
= \frac{1}{2\pi i}\oint dw\,w^{n-1+\lambda}\,
 \frac{1}{2\pi i}\oint_w dz\,z^{m-\lambda}\frac{1}{z-w}
= \frac{1}{2\pi i}\oint dw\,w^{n+m-1}
= \delta_{n,-m}
\end{multline}

Now we introduce two composite operators build from $b$ and $c$ --
the `ghost stress tensor' $t$ (with $\Delta=2$)
\begin{equation}
\label{t}
t(z) = \nu\lambda :\!\partial c\;b\!: +(1-\lambda):\!\partial b\;c\!:
\ = -\lambda :\!b\,\partial c\!: +(1-\lambda):\!\partial b\;c\!:
\ = \sum_{n}z^{-n-2}\,t_n
\end{equation}
and the ghost number current $j$ (with $\Delta=1$)
\begin{equation}
\label{j}
j(z) = :\!bc\!: \ = \ -\nu:\!cb\!: \ = \ \sum_{n}z^{-n-1}\,j_n\,.
\end{equation}

The following OPE relations involving $t$ and $j$ can be obtained
by the use of Wick's formula:
\begin{align}
t(z)\,b(w) &= \frac{\lambda}{(z-w)^2}\,b(w)
+ \frac{1}{z-w}\,\partial b(w) \label{tb} \\
t(z)\,c(w) &= \frac{1-\lambda}{(z-w)^2}\,c(w)
+ \frac{1}{z-w}\,\partial c(w) \label{tc} \\
t(z)\,j(w) &= \frac{\nu\,(2\lambda-1)}{(z-w)^3}
+ \frac{1}{(z-w)^2}\,j(w)
+ \frac{1}{z-w}\,\partial j(w) \label{tj} \\
t(z)\,t(w) &= \frac{-\nu\,(6\lambda^2-6\lambda+1)}{(z-w)^4}
+ \frac{2}{(z-w)^2}\,t(w)
+ \frac{1}{z-w}\,\partial t(w) \label{tt} \\
j(z)\,b(w) &= \frac{b(w)}{z-w} \label{jb} \\
j(z)\,c(w) &= \frac{-c(w)}{z-w} \label{jc} \\
j(z)\,j(w) &= \frac{\nu}{(z-w)^2} \label{jj}\,.
\end{align}
To prove (\ref{tb}) we begin with
\begin{multline}
\label{}
t(w)\,b(z) = b(z)\,t(w) = \frac{1}{2\pi i} \oint_w\frac{dx}{x-w}\,
[ -\lambda\underbrace{b(z)\,b(x})\,\partial c(w) \\
 + (1-\lambda)\underbrace{b(z)\,\partial b(x})\,c(w)
 + \nu\lambda \,b(x)\underbrace{b(z)\,\partial c(w})
 - \nu(1-\lambda)\,\partial b(x)\underbrace{b(z)\,c(w})\, ] \\
= \frac{1}{2\pi i} \oint_w\frac{dx}{x-w}\,
[ \nu\lambda\, b(x)\,\partial_w\frac{\nu}{z-w}
 - \nu(1-\lambda)\,\partial b(x)\,\frac{\nu}{z-w}\, ] \\
= \frac{\lambda}{(z-w)^2}\,b(w) - \frac{1-\lambda}{z-w}\,\partial b(w)\,.
\end{multline}
Therefore,
\begin{equation}
\label{}
t(z)\,b(w) = \frac{\lambda}{(z-w)^2}\,b(z)
 + \frac{1-\lambda}{z-w}\,\partial b(z)
= \frac{\lambda}{(z-w)^2}\,b(w) + \frac{1}{z-w}\,\partial b(w)+\ldots\,.
\end{equation}
As another example, we derive (\ref{tj}). Note the use of the first
terms of the expansion
\begin{equation}
\label{}
b(x)\,c(w) = \frac{\nu}{x-w} + j(w) + f(w)(x-w) +\ldots\,.
\end{equation}
at intermediate stages of the calculation:
\begin{multline}
\label{}
t(z)\,j(w) \equiv t(z)\,:\!bc\!:(w)
= \frac{1}{2\pi i} \oint_w\frac{dx}{x-w}\,
[ \underbrace{t(z)\,b(x})\,c(w) + b(x)\underbrace{t(z)\,c(w})\,] \\
= \frac{1}{2\pi i} \oint_w\frac{dx}{x-w}\,
\left(\frac{\lambda}{(z-x)^2}b(x)\,c(w)
 + \frac{1}{z-x}\partial b(x)\,c(w)
 + \frac{1-\lambda}{(z-w)^2}b(x)\,c(w) \right. \\
 + \left.\frac{1}{z-w}b(x)\,\partial c(w) \right)
= \frac{1}{2\pi i} \oint_w\frac{dx}{x-w}\,
\left(\frac{\nu\lambda}{(z-x)^2(x-w)} + \frac{\lambda j(w)}{(z-x)^2}
 - \frac{\nu}{(z-x)(x-w)^2} \right.\\
 + \frac{f(w)}{z-x}
 + \left.\frac{\nu(1-\lambda)}{(z-w)^2(x-w)} + \frac{(1-\lambda)j(w)}{(z-w)^2}
 + \frac{\nu}{(z-w)(x-w)^2} + \frac{\partial j(w)}{z-w}
  - \frac{f(w)}{z-w}\right) \\
= \frac{2\nu\lambda}{(z-w)^3} + \frac{\lambda j(w)}{(z-w)^2}
 - \frac{\nu}{(z-w)^3} + \frac{(1-\lambda)j(w)}{(z-w)^2}
 + \frac{\partial j(w)}{z-w} \\
= \frac{\nu\,(2\lambda-1)}{(z-w)^3} + \frac{j(w)}{(z-w)^2}
 + \frac{\partial j(w)}{z-w}\,.
\end{multline}

We see from (\ref{tt}) that $t(z)$ is an honest stress tensor with
\begin{equation}
\label{}
c = -2\nu\,(6\lambda^2-6\lambda+1) = \nu(1-3Q^2)\,, \ \
Q = \nu(2\lambda-1)\,,
\end{equation}
its local commutator given by (\ref{local1}). The correspondence
principle (\ref{diffn}) readily yields any commutators of such a type,
for example,
\begin{align}
[t(z),\,j(w)] &= \frac{Q}{2}\,\delta''(z-w)-\delta'(z-w)\,j(w)
+ \delta(z-w)\,j'(w)\,, \\
\label{jzjw}
[j(z),\,j(w)] &= -\nu\,\delta'(z-w)\,.
\end{align}
Of course, these (and similar) formulas can also be derived from
the corresponding commutators of modes,
\begin{align}
[t_n,b_m] &= (\lambda n-m-n)b_{m+n} \label{tnbm} \\
[t_n,c_m] &= -(\lambda n+m)c_{m+n} \label{tncm} \\
[t_n,j_m] &= -m\,j_{m+n}
 + \frac{Q}{2}\,n(n+1)\,\delta_{n,-m}   \label{tnjm} \\
[t_n,t_m] &= (n-m)t_{m+n}
 - \frac{\nu}{6}\,(6\lambda^2-6\lambda+1)\,n(n^2-1)\,\delta_{n,-m}
  \label{tntm} \\
[j_n,b_m] &= b_{m+n} \label{jnbm} \\
[j_n,c_m] &= -c_{m+n} \label{jncm} \\
[j_n,j_m] &= \nu n\,\delta_{n,-m} \label{jnjm}
\end{align}
which are straightforwardly deduced from explicit expressions for
$j_n$ and $t_n$:
\begin{align}
-j_n &= \nu\sum_{k<\,0}c_k b_{n-k}
-\sum_{k\geqslant\,0}b_{n-k} c_k
+ \nu \lambda\,\delta_{n,0}\,\doteq\,\nu\sum_k:\!c_k b_{n-k}\!:
 + \nu\lambda\,\delta_{n,0}\,,\label{jn}\\
t_n &= \sum_k(\lambda n-k)\,:\!b_k c_{n-k}\!:
- \frac{\nu}{2}\,\lambda(\lambda-1)\,\delta_{n,0}\,. \label{tn}
\end{align}
Here we present a sample evaluation of $j_n$:
\begin{multline}
\label{evaljn}
-j_n = -\frac{1}{2\pi i}\oint dz\,z^n\,j(z)
= \frac{\nu}{2\pi i}\oint dz\,z^n\,:\!cb\!:(z)
= \frac{\nu}{(2\pi i)^2}\oint dz\,z^n\oint_z\frac{dw}{w-z}\,c(w)\,b(z) \\
= \frac{\nu}{(2\pi i)^2}\oint dz \oint dw\ \frac{z^n}{w-z}\,c(w)\,b(z)
\left(\bigl|_{|z|<|w|}\quad - \quad\bigr|_{|z|>|w|}\right) \\
= \frac{\nu}{(2\pi i)^2}\oint dz\oint dw\sum_{k\geqslant 0}\,
[\,z^{n+k}\,w^{-k-1}c(w)\,b(z) - \nu z^{n-k-1}\,w^k b(z)\,c(w)\,] \\
= \frac{\nu}{(2\pi i)^2}\oint dz\oint dw\sum_{k\geqslant 0}\sum_p\sum_q\,
[\,z^{n+k-q-\lambda}\,w^{-k-p-2+\lambda}\,c_p\,b_q
 - \nu z^{n-k-q-1-\lambda}\,w^{k-p-1+\lambda}\,b_q\,c_p\,] \\
= \sum_{k\geqslant 0}\sum_p\sum_q\,
[\,\nu\delta_{q,n+k+1-\lambda}\delta_{p,-k-1+\lambda}\,c_p\,b_q
 - \delta_{q,n-k-\lambda}\delta_{p,k+\lambda}\,b_q\,c_p\,] \\
= \sum_{k\geqslant 0}\,[\,\nu\,c_{-k-1+\lambda}\,b_{n+k+1-\lambda}
 - b_{n-k-\lambda}\,c_{k+\lambda}\,] \\
= \nu\sum_{k<0}c_k\,b_{n-k} - \sum_{k\geqslant 0}b_{n-k}\,c_k
 + \nu\lambda\,\delta_{n,0}
\doteq \nu\sum_k\,:\!c_k b_{n-k}\!: + \nu\lambda\,\delta_{n,0}\,.
\end{multline}
During this calculation (the same is true for $t_n$) we assumed $\lambda$
to be integer to avoid non-integer indexed modes. However, the final
result could be obtained without this reservation, by changing
the definition of normal product as follows:
\begin{equation}
\label{}
-j(z) = \nu\,:\!z^{-\lambda}c(z)\,z^{\lambda}b(z)\!:\
=\ \frac{\nu}{2\pi i}\oint_z\frac{dx}{x-z}x^{-\lambda}c(x)\,z^{\lambda}b(z)
\end{equation}
Generally, such modification shifts the result by a finite constant, which
can be accounted for by the uncertainty in the definition of normal product,
and therefore can be adjusted by hand to obtain a desired result (\ref{tnjm}).

At last, we show how the commutators of modes are derived. For example,
\begin{multline}
\label{}
[j_n\,,j_m] = -\nu\sum_{k<0}([j_n,c_k]b_{m-k} + c_k[j_n,b_{m-k}])
 + \sum_{k\geqslant0}([j_n,b_{m-k}]c_k + b_{m-k}[j_n,c_k]) \\
= \nu\sum_{k<0}(c_{n+k}b_{m-k} - c_k b_{n+m-k})
 - \sum_{k\geqslant0}(-b_{n+m-k}c_k + b_{m-k}c_{n+k}) \\
= \sum_{k=0}^{n-1}(\nu c_k b_{n+m-k} + b_{n+m-k}c_k)
= \nu n\,\delta_{n,-m}\,.
\end{multline}

\newpage

\begin{center}
\large\textbf{Sugawara construction}
\end{center}

Here we prove the Sugawara formula which represents the energy-momentum
tensor $T(z)$ as a normally ordered product of two currents,
\begin{equation}
\label{sug}
T(z) = \frac{\bar{\varkappa}_{ab}}{1+2K}:\!J^a(z)J^b(z)\!:\
\doteq \frac{\bar{\varkappa}_{ab}}{1+2K}\ \frac{1}{2\pi i}
\oint_z\frac{dw}{w-z}J^a(w)J^b(z)
\end{equation}
\begin{equation}
\label{km-vir}
L_n = \frac{\bar{\varkappa}_{ab}}{1+2K}:\!\sum_k J^a_k\,J^b_{n-k}:\
\doteq \frac{\bar{\varkappa}_{ab}}{1+2K}\,
(\sum_{k<0} J^a_k\,J^b_{n-k}+\sum_{k\geqslant 0}J^b_{n-k} J^a_k)
\end{equation}
One uses the Wick formula to check OPE (\ref{1})
for $T$ and obtain, in agreement with (\ref{2}),
\begin{equation}
\label{}
T(z)\,J^a(w) = \frac{J^a(w)}{(z-w)^2} + \frac{\partial J^a(w)}{z-w}\,.
\end{equation}
The calculations on the mode level are explicitly shown below.

Let's fix the notation and remind some useful formulas involving
the group structure constants. Let
\begin{equation}
\label{Killing}
\varkappa^{ab} = f_{j}^{ai}f_{i}^{bj}\,,\qquad
\varkappa^{ab}=\varkappa^{ba}\,,\qquad
\bar{\varkappa}_{ac}\varkappa^{cb} = \delta^b_a\,,\qquad
\bar{\varkappa}_{ab} = \bar{\varkappa}_{ba}
\end{equation}
be a (non-degenerate) Killing form and its inverse. Tensor
$\varkappa^{ai}f_{a}^{jk}$ is known to be fully antisymmetric
in $\{ijk\}$ (which is equivalent to the Ad-invariance of the
Killing form), whence it is easily shown that
\begin{equation}
\label{}
\bar{\varkappa}_{ia}f_{j}^{ak} = -\bar{\varkappa}_{ja}f_{i}^{ak}\,,
\qquad \bar{\varkappa}_{jk}f_{i}^{aj}f_b^{ik} = \delta^a_b
\end{equation}

Let now
\begin{equation}
\label{}
\bar{\varkappa}_{ab}:\!J^a(z)J^b(z)\!: \ \
= \sum_n z^{-n-2}\mathcal{L}_n
\end{equation}
Then we proceed in analogy with (\ref{evaljn})\,:
\begin{multline}
\label{ln}
\mathcal{L}_n = \frac{\bar{\varkappa}_{ab}}{2\pi i}
\oint dz\,z^{n+1}:\!J^a(z)J^b(z)\!:
= \frac{\bar{\varkappa}_{ab}}{(2\pi i)^2}\oint dz\,z^{n+1}
\oint_z\frac{dw}{w-z}J^a(w)J^b(z) \\
= \frac{\bar{\varkappa}_{ab}}{(2\pi i)^2}\oint dz \oint dw\
\sum_{k\geqslant 0}(\,z^{k+n+1}w^{-k-1}J^a(w)J^b(z)
+ z^{n-k}w^k J^b(z)J^a(w)\,) \\
= \bar{\varkappa}_{ab}\sum_{k\geqslant 0}
(J^a_{-k-1}J^b_{k+n+1} + J^b_{n-k}J^a_k)
= \bar{\varkappa}_{ab}:\!\sum_k J^a_k\,J^b_{n-k}:
\end{multline}

To compare the normalizations of $\mathcal{L}_n$ and $L_n$,
one calculates
\begin{multline}
[\mathcal{L}_n,\,J^c_m] = \\
 = \bar{\varkappa}_{ab}\left[\sum_{k<0}\left([J^a_k,\,J^c_m]\,J^b_{n-k}
+ J^a_k\,[J^b_{n-k},\,J^c_m]\right) +
\sum_{k\geqslant 0}\left([J^b_{n-k},\,J^c_m]\,J^a_k
+ J^b_{n-k}\,[J^a_k,\,J^c_m]\right)\right] \\
= \bar{\varkappa}_{ab}\left[K\sum_k
\left(\varkappa^{ac}k\delta_{k,-m}J^b_{n-k}
+ \varkappa^{bc}(n-k)\delta_{n-k,-m}J^a_k\right)\right. \\
\left. + f_{d}^{ac}\left(\sum_{k<0}J^d_{k+m}J^b_{n-k}
+ \sum_{k\geqslant 0}J^b_{n-k}J^d_{k+m}\right)
+ f_{d}^{bc}\left(\sum_{k<0}J^a_k J^d_{m+n-k}
+ \sum_{k\geqslant 0}J^d_{m+n-k}J^a_k\right)\right] \\
= -2K\,m\,J^c_{n+m} + \bar{\varkappa}_{ab}f_d^{ac}
\left(\sum_{k<0}\left(J^d_{k+m}J^b_{n-k}-J^d_k J^b_{m+n-k}\right)\right.\\
+ \left.\sum_{k\geqslant 0}\left(J^b_{n-k}J^d_{k+m}
-J^b_{n+m-k}J^d_k\right)\right)
= -2K\,m\,J^c_{n+m} + \bar{\varkappa}_{ab}f_d^{ac}
\sum_{k=0}^{m-1}\,[J^d_k,\,J^b_{n+m-k}] \\
= -2K\,m\,J^c_{n+m} - m \bar{\varkappa}_{ab}f_d^{ca}f_e^{db}J^e_{m+n}
= -(1+2K)\,m\,J^c_{n+m}
\end{multline}
Now it is prompting to identify $\mathcal{L}_n$ with $(1+2K)L_n$
and check the Virasoro algebra:
\begin{multline}
[L_n,\,L_m] = \frac{\bar{\varkappa}_{ab}}{1+2K}
\left[L_n,\,:\!\sum_k J^a_k\,J^b_{m-k}:\,\right] \\
 =\frac{\bar{\varkappa}_{ab}}{1+2K}\left[\sum_{k<0}\left(
[L_n,\,J^a_k]\,J^b_{m-k} + J^a_k\,[L_n,\,J^b_{m-k}]\right) +
\sum_{k\geqslant 0}\left([L_n,\,J^b_{m-k}]\,J^a_k
+ J^b_{m-k}\,[L_n,\,J^a_k]\right)\right] \\
= -\frac{\bar{\varkappa}_{ab}}{1+2K}\left[\sum_{k<0}\left(
k\,J^a_{n+k}\,J^b_{m-k} + (m-k)\,J^a_k\,J^b_{n+m-k}\right)\right. \\
\left. + \sum_{k\geqslant 0}\left((m-k)\,J^b_{n+m-k}\,J^a_k
+ k\,J^b_{m-k}\,J^a_{n+k}\right)\right]
= -\frac{\bar{\varkappa}_{ab}}{1+2K}\left[\sum_{k<0}
k\,J^a_{n+k}\,J^b_{m-k}\right. \\
\left. + \sum_{k<-n}(m-n-k)\,J^a_{n+k}\,J^b_{m-k}+
\sum_{k\geqslant -n}(m-n-k)\,J^b_{m-k}\,J^a_{n+k}
+\sum_{k\geqslant 0} k\,J^b_{m-k}\,J^a_{n+k}\right] \\
= \frac{\bar{\varkappa}_{ab}}{1+2K}\left[(n-m)\left(
\sum_{k<-n}J^a_{n+k}\,J^b_{m-k}
+\sum_{k\geqslant -n}J^b_{m-k}\,J^a_{n+k}\right)
-\sum_{k=-n}^{-1}k\,[J^a_{n+k},\,J^b_{m-k}]\right] \\
= \frac{\bar{\varkappa}_{ab}}{1+2K}\left[
(n-m):\!\sum_k J^a_k\,J^b_{n+m-k}:
- \varkappa^{ab}K\delta_{n,-m}\sum_{k=-n}^{-1}k(n+k)\right] \\
= (n-m)L_{n+m} + \frac{\delta^a_a K}{1+2K}\,\delta_{n,-m}
\sum_{k=1}^n k(n-k)
= (n-m)L_{n+m} + \frac{c}{12}\,n(n^2-1)\,\delta_{n,-m}
\end{multline}
where $c$ is now explicitly given in terms of $K$ and $d=\delta^a_a$\,:
\begin{equation}
\label{}
c = \frac{2Kd}{1+2K}
\end{equation}

\newpage

\begin{center}
\large\textbf{Free scalar field}
\end{center}

On the classical level, free massless scalar theory in 2-dimensional
Euclidean space is conformally invariant and characterized by the following
Lagrangian, equations of motion, and stress tensor:
\begin{gather}
\mathcal{L} = \frac{1}{2}\,d^2x\,\partial_\mu\varphi\,\partial_\mu\varphi
 = dzd\bar{z}\,\partial_z\varphi\,\partial_{\bar z}\varphi\,, \\
\partial_z\partial_{\bar z}\varphi(z,\bar{z}) = 0 \ \ \ \ \
\Rightarrow \ \ \ \varphi(z,\bar{z})=\varphi(z)+\bar\varphi(\bar{z})\,,
\ \ \ \partial_{\bar z}\varphi(z) = 0\,, \ \ \
\partial_z\bar\varphi(\bar{z}) = 0\,,  \\
T_{\mu\nu} = \partial_\mu\varphi\,\partial_\nu\varphi
-\frac{1}{2}\,\delta_{\mu\nu}\partial_\lambda\varphi\,\partial_\lambda\varphi
\,, \ \ \ T_{z\bar z} = 0\,, \ \ \
T_{zz}=\partial_z\varphi\,\partial_z\varphi\,, \ \ \
\partial_{\bar z}T_{zz}=0\,.
\end{gather}
As a quantum field, $\varphi$ formally displays logarithmic behaviour
of the two-point correlation function,
\begin{equation}
\label{}
<\varphi(z,\bar{z})\,\varphi(w,\bar{w})> \ = \
(-4\partial_z\partial_{\bar z})^{-1} \ = \ \frac{-\ln\!|z-w|^2}{4\pi}\,,
\end{equation}
which violates positivity. However, its derivative $i\,\partial\varphi$ is
a valid quantum field due to
\begin{equation}
\label{}
<\partial_z\varphi(z)\,\partial_w\varphi(w)> \ = \ \frac{-1}{4\pi(z-w)^2}\,.
\end{equation}
This suggests the following quantum conformal theory
of a free massless field
\begin{equation}
\label{a}
a(z) = \sum_n\frac{a_n}{z^{n+1}} = 2i\sqrt\pi\partial_z\varphi
\end{equation}
with $T$ multiplied by \,$-2\pi$ (usual practice in Euclidean CFT${}_2$)\,:
\begin{equation}
\label{tfree}
a(z)\,a(w) = \frac{1}{(z-w)^2}\,, \ \ \ \ \ \ \
T(z) = \frac{1}{2}:\!a\,a\!:\!(z)\,.
\end{equation}
Using the Wick formula, one readily cheks (\ref{1}),\,(\ref{2}) and
establishes that $a(z)$ is a primary field with $\Delta=1$ and $T(z)$
a genuine stress tensor with $c=1$\,. Quite analogously to (\ref{jzjw}),
(\ref{jnjm}) and (\ref{ln}) we deduce
\begin{equation}
\label{}
[a(z),\,a(w)] = -\delta'(z-w)\,, \ \ \ \ \
[a_n,\,a_m] = n\,\delta_{m,-n}\,, \ \ \ \ \
T_n = \frac{1}{2}\sum_{k}\,:\!a_k a_{n-k}\!:
\end{equation}
and verify eq.\,(\ref{lphi})\,:\ \ $[T_n,\,a_m] = -m\,a_{n+m}$\,.

It's interesting to observe that if we define, instead of (\ref{tfree}),
\begin{equation}
\label{tlam}
t(z) = \frac{1}{2}:\!a\,a\!:\!(z) \,+\, (\frac{1}{2}-\lambda)\,\partial a(z)
\end{equation}
we also obtain a stress tensor whose OPE properties exactly coincide
with those of (\ref{t}), whereas $a(z)$ plays the role of the current
$j$ (\ref{j}), and $\nu$ is set to $1$\,. This fact allows one to treat
the free-field conformal theory just described as a bosonized version
of the generalized ghost model. Note that for $\lambda\neq \frac{1}{2}$\,,
both $a(z)$ and $j(z)$ (\ref{j}) are not primary fields due to (\ref{tj})\,.

Let us now study the properties of so-called \emph{vertex operators}
\begin{equation}
\label{Vz}
V^\alpha(z) = \ :\!e^{2i\alpha\sqrt{\pi}\varphi(z)}\!:
\ = Y^\alpha \exp(-\alpha\sum_{n<0}\frac{a_n}{nz^n})
\ z^{\alpha a_0}\, \exp(-\alpha\sum_{n>0}\frac{a_n}{nz^n})\,,
\end{equation}
where $\varphi$ is written in the form prompted by (\ref{a})\,,
\begin{equation}
\label{iphi}
2i\sqrt{\pi}\varphi(z) = \ln Y + a_0\ln z - \sum_{n\neq0}\frac{a_n}{nz^n}\,,
\end{equation}
and $Y$ is an invertible operator specified by
\begin{equation}
\label{Ja}
a_n Y = Y a_n \ \ (n\neq0)\,, \ \ \ [a_0,\,\ln Y] = 1 \ \ \ \ \
\Longrightarrow \ \ \ \ [a_0,Y] = Y,
\end{equation}
so it is natural to treat $\ln Y$ as a creation operator. Using
\begin{equation}
\label{}
[a(z),\,\frac{a_n}{nw^n}] = -\frac{z^{n-1}}{w^n}\,, \ \ \ \ \ \
[a(z),\,\ln Y] = \frac{1}{z}\,,
\end{equation}
one obtains
\begin{equation}
\label{[aV]}
[a(z),\,V^\alpha(w)] = \alpha V^\alpha(w)\left(\frac{1}{z}
 + \sum_{n<0}\frac{z^{n-1}}{w^n} + \sum_{n>0}\frac{z^{n-1}}{w^n}\right)
 = \alpha V^\alpha(w)\,\delta(z-w)\,,
\end{equation}
which implies the following OPE\,:
\begin{equation}
\label{aV}
a(z)\,V^\alpha(w) = \frac{\alpha V^\alpha(w)}{z-w}\,, \ \ \ \ \ \
V^\alpha(z)\,a(w) = -\frac{\alpha V^\alpha(w)}{z-w}\,.
\end{equation}

To deduce similar relations for the stress tensor $T(z)$\,, we need
some normal products. First of all, differentiating (\ref{Vz}) gives
\begin{equation}
\label{:aV:}
:\!a(z)V^\alpha(z)\!: \ = \ \frac{1}{\alpha}\,\partial_z V^\alpha(z)\,.
\end{equation}
Then
\begin{equation}
\label{:Va:}
:\!V^\alpha(z)\,a(z)\!: \
 = \ (\frac{1}{\alpha}-\alpha)\,\partial_z V^\alpha(z)\,,
\end{equation}
as is shown by the following calculation:
\begin{multline*}
:\!V^\alpha(z)a(z)\!: \ =
\frac{1}{2\pi i}\oint_z\frac{dx}{x-z}V^\alpha(x)\,a(z)
 = \frac{1}{2\pi i}\oint_z\frac{dx}{x-z}\,a(z)V^\alpha(x) \\
= \frac{1}{2\pi i}\oint_z\frac{dx}{x-z}\left(\frac{\alpha V^\alpha(x)}{z-x}
 + :\!aV^\alpha\!:\!(x) + \ldots\right) \\
= \frac{1}{2\pi i}\oint_z\frac{dx}{x-z}\left[\frac{\alpha}{z-x}
  (V^\alpha(z) + (x-z)\partial V^\alpha(z))
 + :\!aV^\alpha\!:\!(z)\right]
 = \ :\!aV^\alpha\!:\!(z) - \alpha\partial V^\alpha(z)\,. \\
\end{multline*}
Now we can compute
\begin{multline}
\label{VT}
V^\alpha(z)\,T(w) = \frac{1}{4\pi i}\oint_w\frac{dx}{x-w}
 (\underbrace{V^\alpha(z)\,a(x})a(w) + a(x)\underbrace{V^\alpha(z)a(w})) \\
= -\frac{\alpha}{4\pi i}\oint_w\frac{dx}{x-w}
 \left(\frac{V^\alpha(x)a(w)}{z-x} + \frac{a(x)V^\alpha(w)}{z-w}\right) \\
= -\frac{\alpha}{4\pi i}\oint_w\frac{dx}{x-w}
 \left(\frac{-\alpha V^\alpha(w)}{(z-x)(x-w)}
 + (\frac{1}{\alpha} - \alpha)\frac{\partial V^\alpha(w)}{z-x}
 + \frac{\alpha V^\alpha(w)}{(z-w)(x-w)}
 + \frac{1}{\alpha} \frac{\partial V^\alpha(w)}{z-w}\right) \\
= \frac{\alpha^2}{2}\frac{V^\alpha(w)}{(z-w)^2}
 + (\frac{\alpha^2}{2} - 1)\frac{\partial V^\alpha(w)}{z-w}\,,
\end{multline}
whence
\begin{equation}
\label{TV}
T(z)\,V^\alpha(w) = \frac{\alpha^2}{2}\frac{V^\alpha(w)}{(z-w)^2}
 + \frac{\partial V^\alpha(w)}{z-w}\,.
\end{equation}
We see that vertex operator $V^\alpha(z)$ is a primary field of conformal
dimension $\Delta=\alpha^2/2$\,. Note that using a generalized
stress tensor (\ref{tlam}) results in
$\Delta=\alpha^2/2+\alpha(\lambda-1/2)$\,. This is evident from
\begin{equation}
\label{}
[\partial a(z),\,V^\alpha(w)] = \alpha V^\alpha(w)\,\delta'(z-w)\,, \ \ \ \ \
\partial a(z)\,V^\alpha(w) = -\frac{\alpha V^\alpha(w)}{(z-w)^2}\,.
\end{equation}

To evaluate (ordinary) products of vertex operators, we use the relations
\begin{gather*}
z^{\alpha a_0} Y^\beta = z^{\alpha\beta} Y^\beta z^{\alpha a_0}\,, \ \ \ \ \ \
\left[\sum_{n>0}\frac{a_n}{nz^n}\,, \ \sum_{m<0}\frac{a_m}{mw^m}\right]
 = -\sum_{n>0}\frac{w^n}{nz^n} = \ln(1-\frac{w}{z})\,, \\
\exp(-\alpha\sum_{n>0}\frac{a_n}{nz^n})\exp(-\beta\sum_{m<0}\frac{a_m}{mw^m})
 = \left(\frac{z-w}{z}\right)^{\alpha\beta}\!
\exp(-\beta\sum_{m<0}\frac{a_m}{mw^m})\exp(-\alpha\sum_{n>0}\frac{a_n}{nz^n})
\end{gather*}
and come to general formula
\begin{equation}
\label{VV}
V^\alpha(z)V^\beta(w) = (z-w)^{\alpha\beta}:\!V^\alpha(z)V^\beta(w)\!:
\end{equation}
which has very interesting particular cases. For instance, operators
$\psi^{\pm}(z)\doteq V^{\pm 1}(z)$ behave like fermionic fields:
\begin{equation}
\label{V1}
[\psi^+(z),\psi^+(w)]_+ = [\psi^-(z),\psi^-(w)]_+ = 0\,, \ \ \ \ \
[\psi^+(z),\psi^-(w)]_+ = \delta(z-w)
\end{equation}
(eq.\,(\ref{diff}) is useful in deriving the last relation)\,. They have
conformal weight $1/2$ (however, $\Delta(\psi^+)=\lambda, \
\Delta(\psi^-)=1-\lambda$ \ for the general case $\lambda\neq \frac{1}{2}$)\,,
and develop the following OPE\,:
\begin{equation}
\label{psipsi}
\psi^+(z)\,\psi^+(w) = \psi^-(z)\,\psi^-(w) = 0\,, \ \ \ \ \ \
\psi^+(z)\,\psi^-(w) = \psi^-(z)\,\psi^+(w) = \frac{1}{z-w}\,.
\end{equation}
Thus, vertex operators $V^{\pm 1}(z)$ \,(\ref{Vz}) provide an explicit
\emph{bosonization} of the (conformal) theory of free fermions.

Another remarkable example is produced by a triple of fields (with $\Delta=1$)
\begin{equation}
\label{three}
J^0(z) \doteq \frac{a(z)}{\sqrt{2}}\,, \ \ \ \ \ \
J^{\pm}(z) \doteq V^{\pm\sqrt{2}}(z)\,.
\end{equation}
Using eq.\,(\ref{diffn}) for $n=1$ and expanding in $z\!-\!w$ inside
$:\!J^+(z)J^-(w)\!:$ and $:\!J^-(w)J^+(z)\!:$ leads to
\begin{equation}
\label{[JJ]}
[J^+(z),\,J^-(w)] = -\delta'(z-w) + 2J^0(w)\delta(z-w)\,.
\end{equation}
Together with (\ref{aV})\,, this corresponds to the (centrally extended)
\,$\text{sl}_2$ OPE algebra:
\begin{equation}
\label{sl2}
J^+(z)\,J^-(w) = \frac{1}{(z-w)^2} + \frac{2J^0(w)}{z-w}\,, \ \ \ \ \
J^0(z)\,J^{\pm}(w) = \frac{\pm J^{\pm}(w)}{z-w}\,.
\end{equation}

\end{document}