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..
Best Answer
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{equation} \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} \end{equation} Now at each step, we only keep the full contractions to extract the central charge. We then find \begin{equation} \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} \end{equation} We can then read off $$ c = 2 \left( -6\lambda^2 + 6 \lambda - 1 \right) = - 3 (2 \lambda - 1 )^2 + 1 $$