Virasoro TT OPE (2.2.11) in Polchinski's book

Question

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?

Answer 1

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}