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?


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  • $\begingroup$ Your first equation is not right. It should be $1/2z$ times the normal ordered exponents. See Vol. 1 page 40. $\endgroup$ – Haz Jul 27 '15 at 11:41
  • $\begingroup$ 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. $\endgroup$ – Qmechanic Jul 27 '15 at 13:18
  • $\begingroup$ 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... $\endgroup$ – MaPo Jul 28 '15 at 21:36