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

After <integration by parts>, define the Euclidean quadratic operator
$$
K[\lambda]=-partial_\tau^2-v^2\nabla^2+i\lambda.
$$
The action can then be written as
$$
S=\frac1{2vg}\sum_{a=1}^N
\langle n_a,K[\lambda]n_a\rangle
-\frac{iN}{2vg}\int d^2x\,d\tau\,\lambda.
$$
Each component of $\mathbf n$ gives the same bosonic <Gaussian functional integral>, so
$$
\int\mathcal D\mathbf n\,
e^{-\frac1{2vg}\sum_a\langle n_a,Kn_a\rangle}
\propto(\det K)^{-N/2}.
$$
Discarding a $\lambda$-independent normalization, the remaining <functional determinant> gives
$$
\boxed{
Z=\int\mathcal D\lambda\,e^{-\widetilde S[\lambda]},
\qquad
\widetilde S[\lambda]
=\frac N2\operatorname{Tr}\log K[\lambda]
-\frac{iN}{2vg}\int d^2x\,d\tau\,\lambda.}
$$

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 337 / 1 / a / Solution

{wiki}

A Gaussian path integral over a bosonic field with quadratic operator $A$ produces a factor proportional to $(\det A)^{-1/2}$. When $A$ depends on background fields, this functional determinant contributes to their quantum effective action.


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 Incoming links: Functional determinant