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In Polchinski's String Theory Vol. 2, equations 10.3.12 are

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

$$\psi(z)\bar\psi(-z)~=~\frac{1}{2z}+\psi\bar\psi(0)+2zT_B^\psi(0)+O(z^2)\tag{10.3.12b}$$

How are these two OPEs calculated, especially the second and third terms?

Qmechanic
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Han Yan
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1 Answers1

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I've got the answer by myself.

Simply do Taylor expansion of the left hand side. Expand both the exponential, and the field around $H(0)$ or $\psi(0)$, then the right hand follows naturally after plugging in definitions of $T_B$.

Han Yan
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