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Consider a 2d CFT in radial quantization. Let $A(z)$ be some primary field and $Q$ a charge that can be written as $$Q = \oint dz \, z^{(h-1)}J(z)\tag{1}$$ for some holomorphic current $J(z)$. I will denote Wick contraction by an overline, as I don't know how to typeset it in mathjax. The question is: is it true that inside expectation values (which implicitly include radial ordering) \begin{equation} \langle A(z) \,:e^Q: \rangle \stackrel{?}{=} \langle \overline{A(z) Q}\, :e^Q:\rangle\tag{2} \end{equation} where $$ \overline{A(z)\, Q} = \oint dw\, w^{(h-1)}\overline{ A(z) J(w)},\tag{3}$$ and $\overline{ A(z) J(w)}$ is the singular part of the OPE between $A$ and $J$.

Here are some preliminary thoughts:

Suppose $h=1$ for simplicity. Expand the (normal-ordered) exponential $$ :e^Q:\, \equiv \sum_{n=0}^\infty \frac{1}{n!} \oint_{\gamma_1} dz_1 \cdots \oint_{\gamma_n} dz_1 :J(z_1)\cdots J(z_n):\tag{4} $$ The contours $\gamma_i$ are non-intersecting and radially ordered. Define $$A_\varepsilon = \oint_\Gamma dz \,\varepsilon(z) A(z)\tag{5}$$ with $\Gamma$ larger than all $\gamma_i$. Then do the Wick contractions $$ \langle A_\varepsilon :e^Q: \rangle = \sum_{n=0}^\infty \frac{1}{n!} \sum_{i=1}^n \Big\langle \oint_{\gamma_1} dz_1 J(z_1) \cdots \oint_{\gamma_i} dz_i \oint_{z_i} dz\, \varepsilon(z) \overline{A(z) J(z_i)} \cdots \oint_{\gamma_n} dz_n J(z_n)\Big\rangle\tag{6} $$

Suppose $\langle A(z)\rangle =0$, so the sum starts at $n=1$. The singular part of the OPE in general is written as $$ \overline{A(z) J(z_i)} = \sum_k \frac{(AJ)_k(z_i)}{(z-z_i)^k}\tag{7} $$ Plugging this into the above formula, we see that the $z$-integral only has a pole in $z = z_i$. Does that imply we can deform the $z$-contour back to $\Gamma$? If so, can we also take out the contraction $ \oint_{\Gamma} dz\,\varepsilon(z)\oint_{\gamma_i} dz_i \overline{A(z) J(z_i)}$ as a common factor? My confusion lies in whether radial ordering puts $\oint_{\gamma_i} dz_i \overline{A(z) J(z_i)} \equiv B(z)$ to the left of all other $J(z_k)$ insertions or not.

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1 Answer 1

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Eq. (2) holds if the contraction (3) commutes with $Q$. See e.g. this related Phys.SE post.

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