 Solution

The quadratic kinetic operator of the beta-gamma system couples $\beta$ only to $\gamma$. Its Green-function equation isin the stated normalization. Since $\bar{\partial }(1/(z-w))=2\pi {\delta }^{(2)}(z-w)$, the singular part isThe inverse kinetic matrix has no $\beta \beta$ or $\gamma \gamma$ entry, so those two operator product expansions are nonsingular. For commuting fields, reversing the order gives $\gamma (z)\beta (w)\sim 1/(z-w)$ after expanding about $w$.