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

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- Solution a

Solution

After integration by parts, define the Euclidean quadratic operator

K [λ] = - p a r t ia l_{τ}^{2} - v^{2} \nabla^{2} + iλ .

The action can then be written as

S = \frac{1}{2 vg} \sum_{a = 1}^{N} ⟨ n_{a}, K [λ] n_{a} ⟩ - \frac{i N}{2 vg} \int d^{2} x d τ λ .

Each component of

n

gives the same bosonic Gaussian functional integral, so

\int D n e^{- \frac{1}{2 vg} \sum_{a} ⟨ n_{a}, K n_{a} ⟩} \propto (det K)^{- N /2} .

Discarding a

λ

-independent normalization, the remaining functional determinant gives

Z = \int D λ e^{- S [λ]}, S [λ] = \frac{N}{2} Tr lo g K [λ] - \frac{i N}{2 vg} \int d^{2} x d τ λ .

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