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:

$$ [\alpha_m, \alpha_n]~=~ \hbar m~\delta_{m+n}^0, \qquad\qquad(1)$$
$$ :\alpha_m \alpha_n:~=~\Theta(n-m) \alpha_m \alpha_n ~+~ \Theta(m-n) \alpha_n \alpha_m,\qquad\qquad(2) $$

where $\Theta$ denote the Heaviside step function.

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

2) Recall that the radial order ${\cal R}$ is defined as
$${\cal R}(j(z)j(w))
~=~\Theta(|z|-|w|) j(z)j(w)~+~ \Theta(|w|-|z|) j(w)j(z).\qquad\qquad(3)$$

3) Rewrite the sought-for operator identity as
$$\cal{R}(j(z)j(w))~-~:j(z)j(w):~=~\frac{\hbar}{(z-w)^2}.\qquad\qquad(4)$$

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 $|z|<|w|$.

5) Show that
$$j(w)j(z)~-~:j(z)j(w):~=~[j_{+}(w),j_{-}(z)].\qquad\qquad(5)$$

6) Show (under the assumption $|z|<|w|$) that
$$j(w)j(z)~-~R(j(z)j(w))~\stackrel{|z|<|w|}{=}~0.\qquad\qquad(6)$$

7) Subtract eq. (6) from eq. (5):

$$\cal{R}(j(z)j(w))~-~:j(z)j(w):~\stackrel{|z|<|w|}{=}~[j_{+}(w),j_{-}(z)].
\qquad\qquad(7)$$

8) Evaluate rhs. of eq. (7):

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

This post imported from StackExchange Physics at 2014-05-01 11:55 (UCT), posted by SE-user Qmechanic