# Conformal transformation of a vertex operator before normal ordering

Let us consider a free scalar boson $$\varphi(z,\bar{z})$$ on the complex plane and assume the following two-point correlation function $$\begin{eqnarray} \langle\varphi(z,\bar{z})\varphi(w,\bar{w})\rangle&=&-\left[\ln\frac{z-w}{2L}+\ln\frac{\bar{z}-\bar{w}}{2L}\right]\nonumber\\ &=&-\ln\frac{|z-w|^2+a^2}{(2L)^2}, \end{eqnarray}$$ in which $$a$$ is the short-distance ultraviolet (UV) cut-off and $$L$$ is the infrared (IR) cut-off. Then, we have the following relation of vertex operator and its un-normal-ordered form (which seems scaling dependent) $$\begin{eqnarray} \exp(ik\varphi(z,\bar{z}))=\left(\frac{a}{2L}\right)^{k^2}:\exp(ik\varphi(z,\bar{z})):. \end{eqnarray}$$ I meet a contradiction when determining the conformal transformation rule of such an un-normal-ordered operator $$\exp(ik\varphi(z,\bar{z}))$$, or its conformal weight. From the relation above, $$\exp(ik\varphi(z,\bar{z}))$$ seems to have exactly the same conformal weight as $$:\exp(ik\varphi(z,\bar{z})):$$ since, when we do a dilatation transformation $$w=\lambda z$$, the second term of right-hand side (RHS) gains a Jacobian $$\lambda^{k^2}$$ while the first term of RHS is invariant. However, if we consider $$\exp(ik\varphi(z,\bar{z}))$$ as a polynomial of $$\varphi(z,\bar{z})$$, it seems to be invariant under conformal transformations because $$\varphi(z,\bar{z})$$ is a conformal invariant. How to solve this contradiction? Is there any other way to determine the conformal weight of $$\exp(i\varphi(z,\bar{z}))$$.