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
$$
\boxed{
Z=\int\mathcal D\lambda\,e^{-\widetilde S[\lambda]},
\qquad
\widetilde S[\lambda]
=\frac N2\operatorname{Tr}\log K[\lambda]
-\frac{iN}{2vg}\int d^2x\,d\tau\,\lambda.}
$$

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-337/1/a/solution