= Solution
The <odd symmetries of a zero-dimensional polynomial model> preserve both the action and the flat integration measure: their coefficients have zero superdivergence. The confining polynomial weight removes boundary terms at infinity. Thus the integral of an odd-symmetry derivative is zero, a <supersymmetric Ward identity> for this finite-dimensional integral in this <zero-dimensional supersymmetric field theory>.
Since $g(z)$ is a <holomorphic function>, the barred symmetry obeys $\overline V_1(\overline\theta_2g(z))=-P(z)g(z)$. Applying the <supersymmetric Ward identity> gives \b[the requested unnormalized insertion]:
$$
\boxed{\langle W'(z)g(z)\rangle=\langle P(z)g(z)\rangle=0.}
$$
One can also verify the result after <Berezin integration>, without invoking the symmetry terminology:
$$
\int_{\mathbb C}d^2z\,|P'|^2P g\,e^{-|P|^2}=-\int_{\mathbb C}d^2z\,g P'\,\partial_{\bar z}e^{-|P|^2}=0.
$$
The final <integration by parts> uses $\partial_{\bar z}(gP')=0$ and rapid decay. The derivative on $W'$ in the insertion is essential.
Back to article page
