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Using Wick's theorem, I want to do this OPE (Polchinski 2.4.17) $$ T(z):e^{iK.X(0.0)}: = \frac{(\alpha)'K^2}{4z^2} :e^{iK.X(0.0)}: + \frac{1}{z}:\partial e^{iK.X(0.0)}: \tag{2.4.17} $$ but my problem is with the second term.

My attempt: $$ =\frac{\eta_{\mu \nu}}{\alpha^{'}} :\partial X^{\mu}(z)\partial X^{\nu}(z)::e^{iK.X(0.0)}: $$$$ =\frac{\eta_{\mu \nu}}{\alpha^{'}}(<\partial X^{\mu}(z)e^{iK.X(0.0)}>\partial X^{\nu}(z)+<\partial X^{\nu}(z)e^{iK.X(0.0)}>\partial X^{\mu}(z) $$$$ =\frac{\eta_{\mu \nu}}{\alpha^{'}}(\frac{-i \alpha ^{'}K^{\nu}}{2z}:e^{iK.X(0.0)}: \partial X^{\mu}(z)+\frac{-i \alpha ^{'}K^{\mu}}{2z}:e^{iK.X(0.0)}: \partial X^{\nu}(z)) $$ The problem occurs here, I do not know how to continue this to reach the final expression. Can anyone help?

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Things are more clear if (in analogy with $[AB, C] = A[B, C] + [A, C]B$) you write \begin{align} \frac{\eta_{\mu\nu}}{\alpha^\prime} \left [ \frac{-i\alpha^\prime K^\nu}{2z} \partial X^\mu(z) :e^{iK.X(0,0)}: + \frac{-i\alpha^\prime K^\mu}{2z} :e^{iK.X(0,0)}: \partial X^\nu(z) \right ]. \end{align} This shows that there is then a further OPE of $\partial X^\mu(z) :e^{iK.X(0,0)}:$ to take. But since there is a $\frac{1}{z}$ multiplying it, you need the first regular term. \begin{equation} \partial X^\mu(z) :e^{iK.X(0,0)}: = \frac{-i\alpha^\prime K^\mu}{2z} :e^{iK.X(0,0)}: + :e^{iK.X(0,0)} \partial X^\mu(0): + \dots \end{equation} Plugging this into the above, the $:e^{iK.X(0,0)}: \partial X^\nu(z)$ can be Taylor expanded, leading to \begin{align} \frac{\eta_{\mu\nu}}{\alpha^\prime} \left [ \frac{-(\alpha^\prime)^2 K^\mu K^\nu}{4z^2} :e^{iK.X}: - \frac{i\alpha^\prime K^\nu}{2z} :e^{iK.X} \partial X^\mu: - \frac{i\alpha^\prime K^\mu}{2z} :e^{iK.X} \partial X^\nu: \right ](0). \end{align} We can notice that the $O(1/z)$ term is exactly $-\partial :e^{iK.X(0,0)}:$. Combined with the fact that \begin{equation} T(z) = -\frac{\eta_{\mu\nu}}{\alpha^\prime} :\partial X^\mu \partial X^\nu:(z), \end{equation} the sign cancels to give the result you advertised.

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  • $\begingroup$ Great job! The point was how to regular the terms! Thanks a lot! $\endgroup$
    – ma p
    Jul 1, 2021 at 15:18

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