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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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    $\begingroup$ 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 $\endgroup$ – Dilaton Jul 22 '13 at 8:23
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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 $$ c = 2 \left( -6\lambda^2 + 6 \lambda - 1 \right) = - 3 (2 \lambda - 1 )^2 + 1 $$

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  • $\begingroup$ 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? $\endgroup$ – user26143 Jul 22 '13 at 2:24
  • $\begingroup$ Ya, it should. $: \partial b c(z):$ means $: (\partial_z b(z) )c(z):$. I don't understand the rest of your question though. $\endgroup$ – Prahar Jul 22 '13 at 3:43
  • $\begingroup$ > I don't understand the rest of your question though. I think I got it. The normal ordering anyway clean the singular term. It's not a problem at all. Thank you for your answer! $\endgroup$ – user26143 Jul 30 '13 at 14:52
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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}

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