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

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 $$c = 2 \left( -6\lambda^2 + 6 \lambda - 1 \right) = - 3 (2 \lambda - 1 )^2 + 1$$

I know that a lot of time has passed, but maybe I can give my own answer to your question. Start by slightly modifying your relation in the following way \begin{align} :\mathcal{F}::\mathcal{G}:= exp\left(\int d^2z_1d^2z_2\frac{1}{z_{12}}\left[\overrightarrow{\frac{\delta}{\delta b_{\mathcal{F}}(z_1)}}\overleftarrow{\frac{\delta}{\delta c_{\mathcal{G}}(z_2)}}+\overleftarrow{\frac{\delta}{\delta c_{\mathcal{F}}(z_1)}}\overrightarrow{\frac{\delta}{\delta b_{\mathcal{G}}(z_2)}}\right]\right):\mathcal{FG}:. \end{align} Before checking that this formula actually gives back the correct results, notice that it has to be symmetric in b and c, since they are Grassmann variables. The arrows over the functional derivatives are necessary in order to obtain the correct result and they mean that you have to act on the left/right of the corresponding operator. Let us check the following trivial case \begin{align} :b(z)::c(w):&=exp\left(\int d^2z_1d^2z_2\frac{1}{z_{12}}\left[\overrightarrow{\frac{\delta}{\delta b_{\mathcal{F}}(z_1)}}\overleftarrow{\frac{\delta}{\delta c_{\mathcal{G}}(z_2)}}+\overleftarrow{\frac{\delta}{\delta c_{\mathcal{F}}(z_1)}}\overrightarrow{\frac{\delta}{\delta b_{\mathcal{G}}(z_2)}}\right]\right):b(z)c(w): \\ &=:b(z)c(w):+\int d^2z_1d^2z_2\frac{1}{z_{12}}\delta^2(z_1,z)\delta^2(z_2,w)\\ &=:b(z)c(w):+\frac{1}{z-w}, \end{align} which is exactly (2.5.7) in Polchinski's book. Here the role of the arrows was useless since $$\mathcal{F}$$ and $$\mathcal{G}$$ were made of a single operator. Let us move on with a more complicated (and useful) case \begin{align} :bc(z)::bc(w):&=:bc(z)bc(w):+\frac{1}{z-w}:c(z)b(w):+\frac{1}{z-w}:b(z)c(w):+\frac{1}{2}\left(\frac{1}{(z-w)^2}+\frac{1}{(z-w)^2}\right)\\ &=:bc(z)bc(w):+\frac{1}{z-w}:c(z)b(w):+\frac{1}{z-w}:b(z)c(w):+\frac{1}{(z-w)^2}. \end{align} Here you see that the highest term corresponds with what @Prahar wrote in his answer. If you do the whole computation of the OPE between two energy momentum tensors (which I did and which is quite lenghty), you end up with the following expression \begin{align} T(z)T(w)=\frac{-12\lambda^2+12\lambda-2}{2(z-w)^4}+\frac{2T(w)}{(z-w)^2}+\frac{\partial T(w)}{z-w}, \end{align} from which you read the central charge \begin{align} c=-12\lambda^2+12\lambda-2. \end{align}