= Solution
The quadratic kinetic operator of the <beta-gamma system> couples $\beta$ only to $\gamma$. Its Green-function equation is
$$
\bar\partial_z\langle\beta(z)\gamma(w)\rangle
=-2\pi\delta^{(2)}(z-w)
$$
in the stated normalization. Since $\bar\partial(1/(z-w))=2\pi\delta^{(2)}(z-w)$, the singular part is
$$
\boxed{\beta(z)\gamma(w)\sim-\frac1{z-w}}.
$$
The 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 $\gamma(z)\beta(w)\sim1/(z-w)$ after expanding about $w$.
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