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I'm trying to understand eq. (2.2.11) in Polchinski's first book.

He's computing

$$:\partial X^\mu(z)\partial X_\mu(z): :\partial' X^\nu(z')\partial' X_\nu(z'):$$

Now, I understand why this expression can be written as

$$\text{expression above}~=~:\partial X^\mu(z)\partial X_\mu(z)\partial' X^\nu(z')\partial' X_\nu(z'):\quad - 4\alpha'/2 (\partial\partial' \ln|z-z'|^2):\partial X^\mu(z)\partial'X_\mu(z'): + 2\eta_\mu^\mu(-\alpha'/2 \partial\partial'\ln|z-z'|^2)^2.\tag{2.2.11}$$

However, he then states to do a Taylor expansion inside the normal ordering to get the OPE in standard form, i.e.

$$\sim~ \frac{D\alpha'^2}{2(z-z')^4}-\frac{2\alpha'}{(z-z')^2}:\partial'X^\mu(z')\partial'X_\mu(z'): - \frac{2\alpha'}{z-z'}:\partial'^2X^\mu(z')\partial' X_\mu(z'): + \text{non-singular terms.}$$

I don't understand the last step. How exactly does he insert the Taylor expansion? Could someone please illuminate? For instance, I don't see where the first term goes? Does that disappear when he Taylor-expands?

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1 Answer 1

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The first term in your second equation does not contain any singularities and is hence part of the "non-singular terms" at the end of the last expression. To find the final form you just need to perform the derivatives of the logarithm terms and Taylor expand the term $:\!\partial X^\mu(z)\partial'X_\mu(z')\!:$ around $z=z'$. The singular contributions from the various terms are then given by \begin{align} :\partial X^\mu(z)\partial X_\mu(z)\partial' X^\nu(z')\partial' X_\nu(z')\!: \,\,&\sim\, 0 \\ -4\frac{\alpha'}{2} (\partial\partial' \ln|z-z'|^2) \,:\!\partial X^\mu(z)\partial'X_\mu(z')\!: \,\,\,&\sim -\frac{2\alpha'}{(z-z')^2} \,:\!\partial X^\mu(z)\partial'X_\mu(z')\!:\\ &\qquad-\frac{2\alpha'}{z-z'} \,:\!\partial' \partial' X^\mu(z')\partial'X_\mu(z')\!: \\ 2\eta^\mu_{\,\,\mu}(-\frac{\alpha'}{2} \partial\partial'\ln|z-z'|^2)^2 &\sim \frac{D(\alpha')^2}{2(z-z')^4} \end{align}

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  • $\begingroup$ Thanks for your answer. Still some parts are not clear to me: E.g. The last equation. I get from it $2D(-\alpha'/2 \partial'\partial ln|z-z'|^2)^2=2D \alpha'^2/4 \cdot 4\cdot 1/(z-z')^4=2D\alpha'^2\frac{1}{(z-z')^4}$, i.e. I cannot reproduce the 1/2 prefactor. Instead I get a prefactor of 2.Where does that enter here? $\endgroup$ Apr 2, 2013 at 16:33
  • $\begingroup$ But the log is to the power of two. If I repeat your computation I get: $\partial'\partial {ln|z-z'|^2}=\partial'\partial 2ln|z-z'|=2*\partial'\frac{1}{z-z'}=-2\frac{1}{(z-z')^2}$ $\endgroup$ Apr 2, 2013 at 16:41
  • $\begingroup$ @Afriendlyhelper: I'm not sure where your factors of $4$ come from, but the derivatives should give $\partial \partial' \ln|z-z'|^2 = 1/(z-z')^2$. Note that $|z-z'|^2 = (z-z')(\bar{z}-\bar{z}')$. $\endgroup$
    – Olof
    Apr 2, 2013 at 16:42
  • $\begingroup$ Ah, ok thanks. I secretly treated the z's as real variables :) My bad. That's where i got the factor of four. Thanks! $\endgroup$ Apr 2, 2013 at 16:43
  • $\begingroup$ One more question: How do I know that the first term in the answer is regular? By definition of the normal ordering? $\endgroup$ Apr 2, 2013 at 16:44

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