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 $\exp(i\varphi(z,\bar{z}))$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.