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):


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?


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 $|z|<|w|$.

  • $\begingroup$ Thanks a lot Qmechanic, I haven't done all the steps of you are pointing out in detail yet, but I clearly that this is just an extension of the traditional Wick theorem in QFT, for 2-points function: $T(\phi_{x_1}\phi_{x_2}) = :\phi_{x_1}\phi_{x_2}: + [\phi_{x_1},\phi_{x_2}]$ I am sorry it wasn't obvious for me earlier, now I will redo the calculations in details, trying to familiarize with it better this time. Thanks again ;) $\endgroup$
    – toot
    Mar 25 '12 at 14:26
  • $\begingroup$ Yes, the method is similar to, e.g., this answer. $\endgroup$
    – Qmechanic
    Mar 25 '12 at 14:54

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