# Wick Order and Radial Ordering in CFT

+ 4 like - 0 dislike
288 views

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?

This post imported from StackExchange Physics at 2014-05-01 11:55 (UCT), posted by SE-user toot
retagged May 1, 2014

+ 3 like - 0 dislike

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
answered Mar 25, 2012 by (3,110 points)
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 ;)

This post imported from StackExchange Physics at 2014-05-01 11:55 (UCT), posted by SE-user toot
Yes, the method is similar to, e.g., this answer.

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

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOv$\varnothing$rflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.