Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 337 / 1 / c

Solution

Z = \int D θ D δ ρ exp (i \int d t d^{d} x L_{2}) .

Introduce the positive spatial operator

A = 2 λ - \frac{\nabla ^{2}}{4 m ρ ˉ} .

δ ρ

yields

L_{eff} = \frac{1}{2} \dot{θ} A^{- 1} \dot{θ} - \frac{ρ ˉ}{2 m} (\nabla θ)^{2} .

A^{- 1} = (2 λ)^{- 1} + O (\nabla^{2})

therefore gives

L_{G} = \frac{1}{4 λ} [\dot{θ}^{2} - v^{2} (\nabla θ)^{2}], v^{2} = \frac{2 λ ρ ˉ}{m} = - \frac{r}{m} .

A

gives

ω^{2} = v^{2} k^{2} + k^{4} / (4 m^{2})

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