Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 301 / 4 / a

Solution

In Feynman gauge, use the mode expansion

A^{μ} (x) = \int \frac{d ^{3} k}{( 2 π ) ^{3} 2∣ k ∣} \sum_{λ = 0}^{3} [ϵ_{λ}^{μ} (k) a_{λ} (k) e^{- ik x} + ϵ_{λ}^{μ *} (k) a_{λ}^{†} (k) e^{ik x}],

with

k^{0} = ∣ k ∣

and the covariant polarization completeness relation. For

x^{0} > y^{0}

, time ordering retains the annihilation-creation contraction and gives the positive-frequency term; for

x^{0} < y^{0}

, it gives the negative-frequency term. Combining them by a

k^{0}

contour integral gives

⟨ 0∣ T A^{μ} (x) A^{ν} (y) ∣0 ⟩ = \int_{C_{F}} \frac{d ^{4} k}{( 2 π ) ^{4}} \frac{- i η ^{μν}}{k ^{2}} e^{- ik \cdot (x - y)} .

k^{0} = ∣ k ∣ - i ϵ

below the real axis and the negative-energy pole

k^{0} = - ∣ k ∣ + i ϵ

above it. Equivalently the denominator is

k^{2} + i ϵ

with the contour along the real axis.

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