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Given a holomorphic field $H(z)$ with OPE: $$H(z)H(0)\sim -\ln z$$ What is the most smart way to calculate the OPE's of the exponential operators $e^{\pm iH(z)}$, given as follows? $$e^{iH(z)}e^{-iH(0)} \sim \frac{1}{z},$$ $$e^{iH(z)}e^{iH(0)} \sim 0,$$ $$e^{-iH(z)}e^{-iH(0)} \sim 0.$$ Should I expand $\exp$ and do term by term? Or is there a smarter way to do it?

Qmechanic
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BVquantization
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  • Expanding and rearranging terms certainly works, which is how this is computed in ie tong. Not sure what is the most “clever” approach or even whether that is well defined. – Prof. Legolasov Aug 30 '20 at 08:04
  • Hi @BVquantization. Is this from Polchinski section 10.3? – Qmechanic Aug 30 '20 at 10:18
  • Hint: Just use $$e^{iA H(z)} e^{iB H(w)}= (z-w)^{AB}e^{(iA H(z)+iB H(w))} $$ This gives you the leading behaviour you are looking for. It's a well known formula (apparently) and a proof can be found in e.g. Difrancesco CFT, Chap. 6. – user44895 Aug 30 '20 at 16:35

1 Answers1

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Hints:

  1. Note that there are implicitly written radial order ${\cal R}$ and normal order $::$ at various places in OP's equations.

  2. The starting point is the 2-point relation $$\begin{align}{\cal R}(A(z)B(w)) ~-~:A(z)B(w): ~=~& C(z,w)~{\bf 1}, \cr C(z,w)~\equiv~&\langle \Omega | {\cal R}(A(z)B(w))|\Omega\rangle,\end{align} \tag{1} $$ cf. this Phys.SE post.

  3. The relevant Wick's theorem is a nested Wick's theorem $$ \begin{align} {\cal R}(:e^{A(z)}::e^{B(w)}:)~=~&\exp\left( C(z,w)\frac{\partial}{\partial A(z)}\frac{\partial}{\partial B(w)}\right): e^{A(z)+B(w)}:\cr ~=~&\ldots~=~e^{C(z,w)}: e^{A(z)+B(w)}:\cr ~=~&e^{C(z,w)}\left(: e^{A(w)+B(w)}:~+~{\cal O}(z\!-\!w)\right) ,\end{align} \tag{2}$$ cf. my Phys.SE answer here.

Qmechanic
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