# Operator product expansion energy momentum tensor

We have the following equation from Polchinski (2.4.6) $$T(z)X^{\mu}(0) \sim \frac{1}{z}\partial X^{\mu}(0) , \tag{2.4.6}$$ where $T(z)$ is defined as $T(z) = -\frac{1}{\alpha'} :\partial X^{\mu} \partial X_{\mu}:\tag{2.4.4}$ and : : is the normal ordering defined by $$:X^{\mu}(z,\bar{z}): = X^{\mu}(z,\bar{z})\tag{2.1.21a}$$ and $$:X^{\mu}(z_{1},\bar{z_{1}})X^{\nu}(z_{2},\bar{z_{2}}): = X^{\mu}(z_{1},\bar{z_{1}})X^{\nu}(z_{2},\bar{z_{2}}) + \frac{\alpha'}{2}\eta^{\mu \nu}\ln|z_{12}|^{2}.\qquad \tag{2.1.21b}$$

How exactly do we arrive at equation 2.4.6 from these definitions? I understand the previous assertions in the chapter where they simply taylor expanded in the normal ordering, but I can't see how the above is derived.

In particular, from http://arxiv.org/abs/0812.4408 (exercise 2.7), how is it concluded that $$T(z)\partial X^{\mu}(0) \sim \frac{1}{z^{2}}\partial X^{\mu}(z) \tag{18}$$

Edit: Just one more question: We have the expansion $$:F::G: = exp(-\frac{\alpha'}{2}\int d^{2}z_{1}d^{2}z_{2}ln|z_{12}|^{2}\frac{\delta}{\delta X^{\mu}_{F}(z_{1},\bar{z_{1}}}\frac{\delta}{\delta X_{G\mu}(z_{2},\bar{z_{2}})}):FG:, \tag{2.2.10}$$

given in Polchinski.

Is there any relation between this and the Ward Identity given by Polchinski (2.3.11) $$Res_{z \to z_{0}}j(z)A(z_{0},\bar{z_{0}}) + \bar{Res}_{\bar{z}\to \bar{z_{0}}}\tilde{j}(\bar{z})A(z_{0},\bar{z_{0}}) =\frac{1}{i\epsilon}\delta A(z_{0},\bar{z_{0}}), \tag{2.3.11}$$

Do these give two different ways to compute the weight of a given operator?

The reason for this question is when I attempt to compute the above following the answer given here Identity of Operator Product Expansion (OPE) , I can't seem to find how equation (18) would follow. It seems as if the solutions manual somehow concludes rhs of equation (18) and then taylor expands which gives $\frac{1}{z^{2}} \partial X^{\mu}(0) + \frac{1}{z} \partial^{2}X^{\mu}(0)$. If one were to follow the computation given in the above link, wouldn't one automatically arrive at this?

Thanks!

Equation (2.4.6): $T(z)X^\mu(0)\sim \frac{1}{z}\partial X^\mu(0)$ means that the RHS is the most singular term of the LHS. $T(z) = -\frac{1}{\alpha'} :\partial X^{\mu} \partial X_{\mu}:\tag{2.4.4}$ So \begin{align*} T(z)X^{\mu}(0) & =-\frac{1}{\alpha'}:\partial X^{\nu}(z)\partial X_{\nu}(z):X^{\mu}(0)\\ & =-\frac{2:\partial X^{\nu}(z):}{\alpha'}\left\langle \partial X_{\nu}(z)X^{\mu}(0)\right\rangle \\ & \sim-\frac{2\partial X^{\nu}(z)}{\alpha'}\partial\left(-\eta_{\nu}^{\ \mu}\frac{\alpha'}{2}ln\left|z\right|^{2}\right)\\ & \sim\partial X^{\mu}(z)\partial\left(lnz+ln\bar{z}\right)\\ & \sim\frac{\partial X^{\mu}\left(z\right)}{z}\\ & \sim\frac{1}{z}\partial X^{\mu}(0) \end{align*}