# 2D Ghost CFT and two-point functions

For some reason I am suddenly confused over something which should be quit elementary.

In two-dimensional CFT's the two-point functions of quasi-primary fields are fixed by global $SL(2,\mathbb C)/\mathbb Z_2$ invariance to have the form

$$\langle \phi_i(z)\phi_j(w)\rangle = \frac{d_{ij}}{(z-w)^{2h_i}}\delta_{h_i,h_j}.$$ So a necessary requirement for a non-vanishing two-point function is $h_i = h_j$. Now consider the Ghost System which contains the two primary fields $b(z)$ and $c(z)$ with the OPE's

$$T(z)b(w)\sim \frac{\lambda}{(z-w)^2}b(w) + \frac 1{z-w}\partial b(w),$$ $$T(z)c(w)\sim \frac{1-\lambda}{(z-w)^2}c(w) + \frac 1{z-w}\partial c(w).$$ These primary fields clearly don't have the same conformal weight for generic $\lambda$, $h_b\neq h_c$. However their two-point function is

$$\langle c(z)b(w)\rangle = \frac 1{z-w}.$$

Why isn't this forced to be zero? Am I missing something very trivial, or are there any subtleties here?

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Comment to the question(v1): OP's last formula is indeed eqs. (5.107) and (5.109) in the textbook Di Franscesco et. al., CFT (up to an inessential normalization factor). –  Qmechanic Nov 4 '12 at 19:31

1) Everything OP writes(v1) above his last equation is correct. The $bc$ OPE reads

$${\cal R}c(z)b(w) ~\sim~ \frac 1{z-w} ,$$

where ${\cal R}$ denotes radial ordering.

2) To calculate the two-point function

$$\langle c(z)b(w)\rangle$$

(which as OP writes must vanish if the conformal dimensions for $b$ and $c$ are different) is more subtle due to the presence of the ghost number anomaly, i.e. the vacuum should be prepared with certain modes of the $bc$ system, see e.g. Polchinski, String Theory, Vol. 1, Sections 2.5-2.7.

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I am very sorry for this late response. For some reason I completely missed this answer, thought nobody had replied. I am not sure I can follow, the discussion in Polchinski is based on applying the ghost system in superstring theory and this is why the vacuum needs to be prepared with some $bc$ modes right? I am more concerned with this CFT by itself (using it for something different than string theory). In di Francesco section 5.3.3 $\langle c(z)b(w)\rangle$ is derived directly from the action, without preparing the vacuum with any modes. –  Heidar Nov 4 '12 at 1:12
Let me give another example which (might be) related to the concern I have, but here there is for sure no ghost number anomaly. Take the usual free fermion CFT (section 5.3.2 di Francesco) where $\psi(z)$ is a (chiral) primary with $h=\frac 12$. The two-point function is given by $\langle\psi(z)\psi(w)\rangle\propto \frac 1{z-w}$. The field $\partial\psi(z)$ is a decedent field with conformal weight $h=\frac 32$. From the basic theorem cited in the question one would expect $\langle\partial\psi(z)\psi(w)\rangle$ to vanish since they don't have equal conformal weights. (continued) –  Heidar Nov 4 '12 at 1:19
But as is typically written in CFT books (for example di Francesco section 5.3.2), this is just given by differentiation and is equation to $\langle\partial\psi(z)\psi(w)\rangle \propto \frac 1{(z-w)^2}$. This two-point function cannot respect special conformal transformations since they usually demand that $h_1 = h_2$. –  Heidar Nov 4 '12 at 1:23
@Heidar: The resolution to your example seems to be that the decedent field $\partial\psi(z)$ is not a quasi-primary field. –  Qmechanic Nov 4 '12 at 17:19
(I just deleted an earlier comment I think is wrong, I try again). But isn't $\partial\psi(z)$ quasi-primary? From $T(z)\psi(w)$ one can calculate the OPE $T(z)\partial\psi(w)\sim \frac{\psi(w)}{(z-w)^3} + \frac{3/2\,\partial\psi(w)}{(z-w)^2} + \frac{\partial^2\psi(w)}{z-w}$, which I guess implies that it transforms correctly under global conformal transformations and so is quasi-primary. But its not primary because of the extra $\frac 1{(z-w)^3}$ pole. –  Heidar Nov 4 '12 at 21:00