The $p$-dependent part of the first-order action can be completed to a square:
$$
\pi\alpha'V^{-1}p^\alpha p_\alpha+i p^\alpha\partial_\alpha Z
=\pi\alpha'V^{-1}
\left(p^\alpha+\frac{iV}{2\pi\alpha'}\partial^\alpha Z\right)
\left(p_\alpha+\frac{iV}{2\pi\alpha'}\partial_\alpha Z\right)
+\frac{V}{4\pi\alpha'}\partial^\alpha Z\partial_\alpha Z.
$$
The <Gaussian functional integral> over $p$ therefore leaves
$$
S_{\rm eff}[Y,Z]=\frac1{4\pi\alpha'}\int d^2\sigma
\left(G_{ij}(Y)\partial^\alpha Y^i\partial_\alpha Y^j
+V(Y)\partial^\alpha Z\partial_\alpha Z\right),
$$
after the stipulated omission of its <functional determinant>. Identifying $X^\mu=(Y^i,Z)$ gives exactly $S_1[X]$, establishing the classical equivalence.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 306 / 4 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-306/4/b/solution