Let
$$
V=\epsilon_{\mu\nu}:\!\partial X^\mu\bar\partial X^\nu e^{ik\cdot X}\!:.
$$
The exponential is a <primary operator> with <conformal weights> $(\alpha'k^2/4,\alpha'k^2/4)$. In the $T(z)V(w,\bar w)$ OPE, contracting one derivative in $T$ with $\partial X^\mu$ and the other with the exponential produces a third-order pole proportional to $k^\mu\epsilon_{\mu\nu}$. Its antiholomorphic counterpart is proportional to $\epsilon_{\mu\nu}k^\nu$. Therefore $V$ is primary exactly when its <polarization tensor> is transverse in both indices,
$$
\boxed{k^\mu\epsilon_{\mu\nu}=0,
\qquad \epsilon_{\mu\nu}k^\nu=0}.
$$
The remaining second-order poles give
$$
\boxed{(h,\widetilde h)=
\left(1+\frac{\alpha'k^2}{4},
1+\frac{\alpha'k^2}{4}\right)}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 306 / 3 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-306/3/b/solution