Problem with OPE (from Polchinski) [closed]

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?

closed as off-topic by David Z♦Jul 27 '15 at 10:47

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – David Z
If this question can be reworded to fit the rules in the help center, please edit the question.

• Your first equation is not right. It should be $1/2z$ times the normal ordered exponents. See Vol. 1 page 40. – Haz Jul 27 '15 at 11:41
• Hi MaPo. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. – Qmechanic Jul 27 '15 at 13:18
• Thank you @Haz, for your answer. Now it seems all ok. But I found an apparent paradox: $$e^{iH(z)}e^{-iH(0)} = \frac{1}{z}:e^{iH(z)}e^{-iH(0)}: = :\frac{1}{z}e^{iH(z)}e^{-iH(0)}:$$ so that $$:e^{iH(z)}e^{-iH(0)}: = ::\frac{1}{z}e^{iH(z)}e^{-iH(0)}:: = \frac{1}{z}:e^{iH(z)}e^{-iH(0)}:$$ which seems to be a contraddiction... – MaPo Jul 28 '15 at 21:36