Wick Order and Radial Ordering in CFT I am not so much familiar with the computations tools of conformal field theory, and I just run into an exercise asking to demonstrate the following formula (related to the bosonic field case):
$$\cal{R}j(z_1)j(z_2)~=~\frac{1}{(z_1-z_2)^2}~+~:j(z_1)j(z_2):$$ 
with $j$ defined as 
$$j(z)~=~\sum_k \alpha_k z^{-k-1}.$$
My question is should I start the calculation form the Wick ordered term and make the two others appear, because starting from the left side, I don't see how could I develop some calculus?
 A: Here we will outline a strategy to prove the sought-for operator identity $(4)$ from the following definitions of what the commutator and the normal order of two mode operators $\alpha_m$ and $\alpha_n$ mean:
$$ [\alpha_m, \alpha_n]~=~ \hbar m~\delta_{m+n}^0, \tag{1}$$
$$\begin{align} :\alpha_m \alpha_n:~=~&\Theta(n-m) \alpha_m \alpha_n \cr~+~& \Theta(m-n) \alpha_n \alpha_m,\end{align}\tag{2} $$
where $\Theta$ denote the Heaviside step function.

*

*Note that the current $j(z)~=~j_{-}(z) + j_{+}(z)$ is a sum of a creation part $j_{-}(z)$ and an annihilation part $j_{+}(z)$.


*Recall that the radial order ${\cal R}$ is defined as
$$\begin{align}{\cal R}(j(z)j(w)) 
~=~&\Theta(|z|-|w|) j(z)j(w)\cr
~+~& \Theta(|w|-|z|)j(w)j(z).\end{align}\tag{3}$$


*Rewrite the sought-for operator identity as
$$\begin{align}\cal{R}(j(z)j(w))~-~&:j(z)j(w):\cr
~=~&\frac{\hbar}{(z-w)^2}.\end{align}\tag{4}$$


*Notice that each of the three terms in eq. $(4)$ are invariant under $z\leftrightarrow w$ symmetry. So we may assume from now on that $|z|<|w|$.


*Show that
$$\begin{align}j(w)j(z)~-~&:j(z)j(w):\cr
~=~&[j_{+}(w),j_{-}(z)].\end{align}\tag{5}$$


*Show (under the assumption $|z|<|w|$) that
$$\begin{align}j(w)j(z)~-~~&R(j(z)j(w))\cr
~\stackrel{|z|<|w|}{=}&0.\end{align}\tag{6}$$


*Subtract eq. (6) from eq. (5):
$$\begin{align}\cal{R}(j(z)j(w))~-~~&:j(z)j(w):\cr
~\stackrel{|z|<|w|}{=}&[j_{+}(w),j_{-}(z)].\end{align}
\tag{7}$$


*Evaluate rhs. of eq. (7):

$$\begin{align} [j_{+}(w),j_{-}(z)]~=~&\ldots\cr
~=~&\frac{\hbar}{w^2} \sum_{n=1}^{\infty}n \left(\frac{z}{w}\right)^{n-1}\cr
~=~&\ldots 
~=~  \frac{\hbar}{(z-w)^2}.\end{align} \tag{8} $$
In the last step we will use that the sum is convergent under the assumption $|z|<|w|$. $\Box$
