homework and exercises - Identity of Operator Product Expansion (OPE)

I have one more s****d question in Polchinski's string theory book, Eqs. (2.3.14a)

$$j^{\mu}(z) :e^{ik \cdot X(0,0)}:~ \sim~ \frac{k^{\mu}}{2 z} :e^{ik \cdot X(0,0)}:,$$

where $j^{\mu}_a =\frac{i}{\alpha'} \partial_a X^{\mu}$, $::$ is normal ordered, defined as

$$:X^{\mu}(z,\bar{z}): = X^{\mu} (z,\bar{z})$$ $$:X^{\mu}(z_1,\bar{z}_1) X^{\nu}(z_2,\bar{z}_2): = X^{\mu}(z_1,\bar{z}_1) X^{\nu}(z_2,\bar{z}_2) + \frac{\alpha'}{2} \eta^{\mu \nu} \ln |z_{12}|^2$$ . $\sim$ means equal up to nonsingular terms.

I thought I have derived it in analogy of integration by part, but it turns out that not as I thought. Actually how to derive Eq. (2.3.14a)? Eq.(2.3.14b) will be expected in analogous...

Answer

We can use

$$\begin{equation} \begin{split} : F : : G: = \exp \left( - \frac{\alpha'}{2} \int d^2 z_1 d^2 z_2 \log|z_{12}|^2\frac{\delta }{\delta X_F^\mu(z_1, {\bar z}_1)} \frac{\delta }{\delta X_{G\mu}(z_2, {\bar z}_2)} \right) :F G: \end{split} \end{equation}$$ This gives $$\begin{equation} \begin{split} : \frac{i}{\alpha'} \partial X^\mu(z) : : e^{i k \cdot X(w,{\bar w})}: &= \exp \left( - \frac{\alpha'}{2} \int d^2 z_1 d^2 z_2 \log|z_{12}|^2\frac{\delta }{\delta X_F^\mu(z_1, {\bar z}_1)} \frac{\delta }{\delta X_{G\mu}(z_2, {\bar z}_2)} \right) \\ &~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~: \frac{i}{\alpha'} \partial X^\mu(z) e^{i k \cdot X(w,{\bar w})}: \\ &= : \frac{i}{\alpha'} \partial X^\mu(z) e^{i k \cdot X(w,{\bar w})}: \\ &~~~~~~~~~~~~~~~~ - \frac{i}{2} : \int d^2 z_1 d^2 z_2\log|z_{12}|^2 \frac{\delta( \partial X^\mu(z) ) }{\delta X_F^\mu(z_1, {\bar z}_1)} \frac{\delta ( e^{i k \cdot X(w,{\bar w})}) }{\delta X_{G\mu}(z_2, {\bar z}_2)}: \\ &= : \frac{i}{\alpha'} \partial X^\mu(z) e^{i k \cdot X(w,{\bar w})}: \\ &~~~~~ - \frac{i}{2} : \int d^2 z_1 d^2 z_2 \log|z_{12}|^2\partial \left( \delta^\mu{}_\alpha \delta^2(z_1, z) \right) i k^\alpha \delta^2(z_2, w) e^{i k \cdot X(w,{\bar w})} \\ &= : \frac{i}{\alpha'} \partial X^\mu(z) e^{i k \cdot X(w,{\bar w})}: \\ &~~~~~ +\frac{k^\mu}{2} : \partial \left( \int d^2 z_1 d^2 z_2 \log|z_{12}|^2 \delta^2(z_1, z) \delta^2(z_2, w) e^{i k \cdot X(w,{\bar w})}\right) : \\ &= : \frac{i}{\alpha'} \partial X^\mu(z) e^{i k \cdot X(w,{\bar w})}: +\frac{k^\mu}{2} : \partial \left( \log|z-w|^2 e^{i k \cdot X(w,{\bar w})}\right) : \\ &= : \frac{i}{\alpha'} \partial X^\mu(z) e^{i k \cdot X(w,{\bar w})}: + \frac{k^\mu}{2(z-w)} : e^{i k \cdot X(w,{\bar w})} : \\ \end{split} \end{equation}$$ Therefore $$\begin{equation} \begin{split} : j^\mu(z) : : e^{i k \cdot X(w,{\bar w})}: \sim \frac{k^\mu}{2(z-w)} : e^{i k \cdot X(w,{\bar w})} : \end{split} \end{equation}$$