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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...

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  • $\begingroup$ Comment to the question (v2): It would be good if OP (or somebody else?) could make the question formulation self-contained and understandable for readers without Polchinski's book. $\endgroup$ – Qmechanic Jul 17 '13 at 17:02
  • $\begingroup$ I will try somehow... $\endgroup$ – user26143 Jul 17 '13 at 17:03
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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}

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  • $\begingroup$ Thank you very much! I still have two questions (1) In the last three line, $\partial \log |z-w|^2 = \partial \left( \log (z - w) + \log (\bar{z} - \bar{w}) \right) $. $\log (\bar{z} - \bar{w}) $ is not holomorphic at $\bar{z}=\bar{w}$, can we conclude its derivative of $z$ is zero? like en.wikipedia.org/wiki/Holomorphic_function "equivalently, the Wirtinger derivative of f with respect to the complex conjugate of z is zero"; (2) About the last step, how to make sure $:\frac{1}{\alpha'}\partial X^{\mu}(z) e^{ik \cdot X(w,\bar{w})}:$ does not contribute to $\sim$? $\endgroup$ – user26143 Jul 17 '13 at 16:45
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    $\begingroup$ (1) The OPE is an expression for the product of two operators at $z$ and $w$ which is valid only when $|z-w|$ is small but not zero. More explicitly, we are interested in the limiting case when $z \to w$. For all such cases, the holomorphic derivative of $\log {\bar z}$ vanishes. $\endgroup$ – Prahar Jul 17 '13 at 16:55
  • $\begingroup$ (2) The operator $: {\cal O} :$ is defined so as to not have any poles anywhere. Therefore, $:{\cal O}:$ admits a Taylor series expansion. We can then write $:j^\mu(z) e^{i k \cdot X(w,{\bar w})}: = \sum\limits_{i=0}^\infty (z-w)^n :\partial_w^n j^\mu(w) e^{i k \cdot X(w,{\bar w})}:$, which is completely regular $z \to w$ and does not contribute to $\sim$. $\endgroup$ – Prahar Jul 17 '13 at 17:00
  • $\begingroup$ Excuse me, where do we know $:\mathcal{O}:$ is defined to be without pole.... $\endgroup$ – user26143 Jul 17 '13 at 17:02
  • $\begingroup$ Well, thats the definition of normal ordering right. One defines $:{\cal O}:$ by taking the operator ${\cal O}$ and subtracting out all the poles so that $:{\cal O}:$ is well defined at all points. $\endgroup$ – Prahar Jul 17 '13 at 17:05
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An other way is to begin with (2.2.1.4)

$$:e^{i k_1.X(z,\bar z)}:~ :e^{i k_2.X(0,0)}: ~\sim ~|z|^{\alpha' k_1.k_2} ~:e^{i (k_1 + k_2).X(0,0)}:$$

Now, derive this expression relatively to $k_1^\mu$, then doing $k_1=0$, we get :

$$:i X_\mu(z, \bar z):~:e^{i k_2.X(0,0)}: ~\sim(\alpha' (k_2)_{\mu} \ln|z| + i :X_\mu(0,0):):e^{i k_2.X(0,0)}:$$

Now, we derive relatively to $z$ and divide by $\alpha'$, and we get :

$$: \frac{i}{\alpha'} \partial _zX_\mu(z, \bar z):~:e^{i k_2.X(0,0)}: ~\sim \frac{k_2^\mu}{2z} :e^{i k_2.X(0,0)}:$$

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  • $\begingroup$ (2.2.1.4) formula is not a shortcut. It is the basic formula (2.2.10) (used by Prahar), applied to $F=e^{i k_1.X(z,\bar z)}$ and $G = e^{i k_2.X(0,0)}$. So, you should be able to recover the formula (2.2.1.4) from (2.2.10) $\endgroup$ – Trimok Jul 17 '13 at 17:09
  • $\begingroup$ I mean your solution comparing Prahar's approach is a shortcut... $\endgroup$ – user26143 Jul 17 '13 at 17:11

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