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

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- Solution b

Solution

The

p

-dependent part of the first-order action can be completed to a square:

π α^{'} V^{- 1} p^{α} p_{α} + i p^{α} \partial_{α} Z = π α^{'} V^{- 1} (p^{α} + \frac{iV}{2 π α ^{'}} \partial^{α} Z) (p_{α} + \frac{iV}{2 π α ^{'}} \partial_{α} Z) + \frac{V}{4 π α ^{'}} \partial^{α} Z \partial_{α} Z .

The Gaussian functional integral over

p

therefore leaves

S_{eff} [Y, Z] = \frac{1}{4 π α ^{'}} \int d^{2} σ (G_{ij} (Y) \partial^{α} Y^{i} \partial_{α} Y^{j} + V (Y) \partial^{α} Z \partial_{α} Z),

after the stipulated omission of its functional determinant. Identifying

X^{μ} = (Y^{i}, Z)

gives exactly

S_{1} [X]

, establishing the classical equivalence.

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