# Compute the central charge of $bc$ conformal field theory

I have a s****d question, how to calculate the central charge of $bc$ conformal-field theory in Polchinski's string theory, Eq. (2.5.12)? For a $bc$ CFT given by

$$S=\frac{1}{2\pi } \int d^2 z \,\,b \bar{\partial} c$$

where $b$ and $c$ are anticommuting fields, define normal ordering as $$:b(z_1) c(z_2): = b(z_1) c(z_2) - \frac{1}{z_{12}}. \tag{2.5.7}$$

Given the energy-momentum tensors $$T(z) = : (\partial b) c: - \lambda \partial ( : bc : ), \tilde{T}(\bar{z})=0$$ The $TT$ Operator Product Expansion (OPE) $$T(z) T(0) \sim \frac{c}{2 z^4} + \frac{2}{z^2} T(0) + \frac{1}{z} \partial T(0)$$ has central charges, $c=-3 (2 \lambda -1)^2+1$ and $$\tilde{c}=0. \tag{2.5.12}$$

For my understanding I should compute the cross-contraction to find the central charges. First I construct the relation $$:\mathcal{F}::\mathcal{G}: =\exp\left(\int d^2 z_1 d^2 z_2 \frac{1}{z_{12}} \frac{ \delta }{\delta b(z_1)} \frac{\delta }{\delta c(z_2)} \right) :\mathcal{FG}:$$ then apply it to $T(z) T(0)$.

My question starts from the very beginning, about $\partial ( : bc : )$ in $T(z)$, does it stand for $\partial_{z_1} ( : b(z_1) c(z_2):)$ or ? if it means $\partial_{z} ( : b(z) c(z):)$ , the right hand side of (2.5.7) is singular..

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Dear user26143, please do not think you or your questions are stupid, you are just learning that's all ;-). Your technical questions are among the better things that come in to the site these days and they, together with the nice answers people like Prahar for example gives, are very helpful for people who want to learn how to calculate in CFT etc. Please be patient with yourself ;-). Cheers – Dilaton Jul 22 '13 at 8:23
Thank you! I will try! – user26143 Jul 22 '13 at 19:01

You don't need to use that. You can simply do the cross-contractions by hand. Let's do that. Note that I only care about the the $\frac{1}{z^4}$ term to evaluate the central charge. We have
$$\begin{split} T(z) T(w) & =\left( : \partial_z b c(z): - \lambda \partial_z : b c (z): \right) \left( : \partial_w b c(w): - \lambda \partial_w : b c (w): \right) \\ & = : (\partial_z b) c(z): : (\partial_w b) c(w): - \lambda \partial_z : b c (z): : (\partial_w b) c(w): \\ &~~~~~~~~~~~~~~~~~~~~~~- \lambda : (\partial_z b) c(z):\partial_w : b c (w): + \lambda^2 \partial_z : b c (z):\partial_w : b c (w): \end{split}$$ Now at each step, we only keep the full contractions to extract the central charge. We then find $$\begin{split} T(z) T(w) &\sim \partial_z \frac{1}{z-w}\partial_w \frac{1}{z-w} - \lambda \partial_z \left( \frac{1}{z-w} \partial_w \frac{1}{z-w} \right)\\ &~~~~~ - \lambda \partial_w \left( \frac{1}{z-w} \partial_z \frac{1}{z-w} \right) + \lambda^2 \partial_z \partial_w \frac{1}{(z-w)^2} \\ &= \frac{-6\lambda^2 + 6 \lambda - 1 }{(z-w)^4} + \cdots \end{split}$$ We can then read off $$c = 2 \left( -6\lambda^2 + 6 \lambda - 1 \right) = - 3 (2 \lambda - 1 )^2 + 1$$
Thank you very much! In the first line of your equations, should $:\partial_z bc(z):$ be $:(\partial_z b) c(z):$? Also, does $:\partial bc(z):$ mean $:\partial_z b(z) c(z):$? In this case, will the normal ordering give $\frac{1}{0}$ singular term? – user26143 Jul 22 '13 at 2:24
Ya, it should. $: \partial b c(z):$ means $: (\partial_z b(z) )c(z):$. I don't understand the rest of your question though. – Prahar Jul 22 '13 at 3:43