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Given the Operator Product Expansion (OPE) of a product of the energy momentum tensors $$T(z)T(0) = \frac{ \eta^{\mu}_{\mu} }{2z^4} - \frac{2}{\alpha' z^2} :\partial X^{\mu} \partial X_{\mu}(0): + :T(z) T(0): $$

$$ \sim \frac{D}{2z^4} + \frac{2}{z^2} T(0) +\frac{1}{z} \partial T(0) (2.4.22) $$

It is said Eq. (2.4.22) implies $$ \epsilon^{-1} \delta T = - \frac{D}{12} \partial_z^3 v(z) - 2 \partial_z v(z) T(z) - v(z) \partial_z T(z) (2.4.23) $$

Although I may start from the definition $$T(z) = - \frac{1}{\alpha'} :\partial X^{\mu} \partial X_{\mu}: (2.4.4)$$ and $$ \delta X^{\mu} = -\epsilon v(z) \partial X^{\mu} - \epsilon v(z)^* \bar{\partial} X^{\mu} (2.4.7)$$ I don't see how to use Eq. (2.4.22) to obtain (2.4.23). Would you at least provide me any hint about derivation?

Thank you very much in advance

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Use a Taylor serie for v, use j= ivT, and use the OPE in the Ward identity (keep only the $1/z$ term for residue calculus). – Trimok Jul 18 '13 at 19:55
> and use the OPE in the Ward identity Sorry don't get your point. Do you mean put (2.4.22) in the Ward identity? – user26143 Jul 18 '13 at 20:02
up vote 2 down vote accepted

You cannot derive $\delta T$ from $\delta X^\mu$, since there is a normal ordering and there might be contributions from contact terms (If you are careful, one could do it). But the easier way to do this is to use the Ward Identity for the current $j(z) = i v(z) T(z)$

\begin{equation} \begin{split} \delta T(w) &= i \epsilon \text{Res}_{z \to w} i v(z) T(z) T(w) \\ &= - \epsilon \text{Res}_{z \to w} v(z) \left[ \frac{D}{2(z-w)^4} + \frac{2}{(z-w)^2} T(w) + \frac{1}{(z-w)} \partial T(w)\right] \\ &= - \epsilon \ \left[ \frac{D}{2} \frac{\partial_w^3 v(w)}{6} + 2 \partial_w v(w) T(w) + v(w) T(w) \right] \end{split} \end{equation} Thus \begin{equation} \begin{split} \boxed{ \epsilon^{-1}\delta T(w) = - \frac{D}{12} \partial_w^3 v(w) - 2 \partial_w v(w) T(w) - v(w) T(w) } \end{split} \end{equation} where I used $$ \text{Res}_{z \to w} \frac{v(z)}{(z-w)^n} = \frac{1}{(n-1)!}\partial_w^{n-1} v(w) $$

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Thank you so much for your thoroughly expansions! I really appreciate your help! – user26143 Jul 18 '13 at 21:15
You're welcome. No problem. – Prahar Jul 18 '13 at 21:36

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