Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

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

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?

share|cite|improve this question
up vote 2 down vote accepted

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}

share|cite|improve this answer
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? – A friendly helper Apr 2 '13 at 16:33
But the log is to the power of two. If I repeat your computation I get: $\partial'\partial {ln|z-z'|^2}=\partial'\partial 2ln|z-z'|=2*\partial'\frac{1}{z-z'}=-2\frac{1}{(z-z')^2}$ – A friendly helper Apr 2 '13 at 16:41
@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 $|z-z'|^2 = (z-z')(\bar{z}-\bar{z}')$. – Olof Apr 2 '13 at 16:42
Ah, ok thanks. I secretly treated the z's as real variables :) My bad. That's where i got the factor of four. Thanks! – A friendly helper Apr 2 '13 at 16:43
One more question: How do I know that the first term in the answer is regular? By definition of the normal ordering? – A friendly helper Apr 2 '13 at 16:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.