Problem with OPE (from Polchinski)

Question

I was reading Polchinski, Vol. 2 pag 12, while I found (10.3.12a):

$$e^{iH(z)}e^{-iH(z)}=\frac{1}{2z} + i\partial H(0) + 2zT^H_B(0) + O(z^2).\tag{10.3.12a}$$

Now I tried to do the OPE, what I get is

$$\begin{split} e^{iH(z)}e^{-iH(z)} &= e^{-\log(2z)} + :e^{iH(z)}e^{-iH(-z)}:\\ &= \frac{1}{2z} + :e^{i(H(0)+z\partial H(0)}e^{-i(H(0)+z\partial H(0)}: + O(z^2)\\ &= \frac{1}{2z} + :e^{iH(0)}i(1+z\partial H(0))e^{-iH(0)}(-i)(1-z\partial H(0)):\\ &= \frac{1}{2z} + :1+2z\partial H(0): + O(z^2). \end{split}$$

Where is the mistake? How can I get Polchinski formula?

This post imported from StackExchange Physics at 2015-07-27 20:42 (UTC), posted by SE-user MaPo

Comments

Your first equation is not right. It should be $1/2z$ times the normal ordered exponents. See Vol. 1 page 40.

This post imported from StackExchange Physics at 2015-07-27 20:42 (UTC), posted by SE-user Haz