5
$\begingroup$

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?

|cite|improve this question
$\endgroup$
  • $\begingroup$ 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). $\endgroup$ – Qmechanic♦ Nov 4 '12 at 19:31
4
$\begingroup$

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.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Heidar Nov 4 '12 at 1:12
  • $\begingroup$ 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) $\endgroup$ – Heidar Nov 4 '12 at 1:19
  • $\begingroup$ 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 $h_1 = h_2$. $\endgroup$ – Heidar Nov 4 '12 at 1:23
  • 2
    $\begingroup$ @Heidar: The resolution to your example seems to be that the decedent field $\partial\psi(z)$ is not a quasi-primary field. $\endgroup$ – Qmechanic♦ Nov 4 '12 at 17:19
  • 1
    $\begingroup$ @Heidar: Try to work directly from the definition of a quasi-primary field, see e.g. Di Franscesco et. al. eqs. (5.22-23). I believe that the decedent field $\partial\psi(z)$ fails to behave as a quasi-primary field under special conformal transformations. $\endgroup$ – Qmechanic♦ Nov 4 '12 at 22:12
2
$\begingroup$

The answer seems to be that, technically at least, the two point function $\langle b(z) c(w) \rangle$ does vanish on the sphere. In the context of the standard $bc$ ghost system that shows up in bosonic string theory, the simplest nonzero correlation function on the sphere that involves both $b$ and $c$ is $$ \langle c(z_1) c(z_2) c(z_3) c(z_4) b(w) \rangle. $$ This correlation function is a special case of (6.3.5) in Polchinski volume 1 and evaluates to $$ \frac{(z_1-z_2)(z_1-z_3)(z_2-z_3)(z_1-z_4)(z_2-z_4)(z_3-z_4)}{(z_1-w)(z_2-w)(z_3-w)(z_4-w)}. $$

The Kronecker delta function $\delta_{h_i h_j}$ that shows up in the result for $\langle \phi_i(z) \phi_j(w) \rangle$ described in OP's first formula can be argued for based on the transformations of the fields under inversion $z \to 1/z$. While $\langle b(z) c(w) \rangle$ cannot transform properly, as OP has noted, it would be reassuring to see that this five-point function does transform in the right way. Indeed, the conformal factor $$ \frac{z_1^2 z_2^2 z_3^2 z_4^2}{w^4} $$ takes precisely the right form to convert this correlation function into one involving only $1/z_i$ and $1/w$.

(Note I am using the $SL(2, R)$ invariant vacuum state to compute these correlation functions. There is also the option of using so-called Q-vacua, some of which are obtained by acting with $c$ operators at the poles of the sphere. In this case, translation invariance is lost, and the correlation function $\langle \phi_i(z) \phi_j(w) \rangle$ can depend on a further dimensionless ratio $z/w$.)

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.