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  Non translation invariant correlator in CFT

+ 3 like - 0 dislike

I'm doing an exercise on vertex operators in the CFT book by Di Francesco & al.; exercise 9.2 p.329 :

Using mode expansion show that: $$\langle\tilde{\phi(z)}\tilde{\phi(w)}\rangle= - \text{ln} \;(1-\frac{w}{z})$$

I managed to complete the calculation but I'm bothered by the fact that the right hand side is not apparently translation invariant.

It seems that we have no charges (at infinity or wherever) and this should be a function of $(z-w)$. Am I missing out on something?

The same way one line below we have:

$$\langle V'_{\alpha_1}(z_1)\ldots V'_{\alpha_n}(z_n)\rangle = \Pi_{i<j} (z_i-z_j)^{2\alpha_i \alpha_j}z_i^{-2 \alpha_i \alpha_j}$$

which again doesn't seem translation invariant because of the last term.

This post imported from StackExchange Physics at 2014-08-24 09:13 (UCT), posted by SE-user Learning is a mess
asked Aug 23, 2014 in Theoretical Physics by Learning is a mess (75 points) [ no revision ]
I suspect that this could be in relation with the fact that $\partial \phi$ is no more a primary field, see page $300$ (while $\partial \phi$ correlators are still translation-invariant)

This post imported from StackExchange Physics at 2014-08-24 09:13 (UCT), posted by SE-user Trimok
@Trimok: the commutator you are referring to is $\langle\tilde{\phi(z)}\tilde{\phi(w)}\rangle= ln(z-w)$ which is translation invariant but different from the one given on this page (addictive factor $- ln(z)$)

This post imported from StackExchange Physics at 2014-08-24 09:13 (UCT), posted by SE-user Learning is a mess

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