# Non translation invariant correlator in CFT

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.

• 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) – Trimok Aug 23 '14 at 13:00
• @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 $- ln(z)$) – Learning is a mess Aug 23 '14 at 20:03
• In my previous comment, I only noticed that, while $\langle\partial \tilde{\phi(z)}\partial \tilde{\phi(w)}\rangle$ is translation-invariant, $\partial \phi$ is not a primary field, and I suggest that this could be in relation with the fact $\langle\tilde{\phi(z)}\tilde{\phi(w)}\rangle$ is not translation-invariant. I have not a definitive argument about that, unfortunately. – Trimok Aug 25 '14 at 8:33
• @Trimok: Indeed taking two derivates (one with respect to $w$ and one with respect to $z$) gives the translation invariant power law. (The additive factor I was talking about gives zero when derived twice). Thus it works for the first correlation function but the second one is still obscure. Many thanks for the suggestion ;) – Learning is a mess Aug 25 '14 at 11:25