 Solution

ID: past-exam-of-the-advanced-quantum-field-theory-for-particle-physics-course-of-the-university-of-oxford/2020/3/b/iii/solution

For fixed

A

σ (x) = F [A] (x)

. Therefore its weighted functional integral is

\int D σ δ (F [A] - σ) exp [- \frac{i}{2 ξ} \int d^{4} x σ^{2}] = exp [- \frac{i}{2 ξ} \int d^{4} x F [A]^{2}] .

Since the action enters as

e^{i S}

, the gauge fixing contribution is

L_{gf} = - \frac{1}{2 ξ} (\partial_{μ} A^{μ} + λ A_{μ} A^{μ})^{2} .

σ

and remains in the path integral. The limit

ξ \to 0

enforces the original condition sharply; for nonzero

ξ

this is a weighted average of the gauge slices.

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