 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/iii/paper-50/2/b/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

Write

t = x^{0} - y^{0}

and

r = x - y

Δ_{F} (x - y) = θ (t) \int \frac{d ^{3} p}{( 2 π ) ^{3} 2 E _{p}} e^{- i E_{p} t + i p \cdot r} + θ (- t) \int \frac{d ^{3} p}{( 2 π ) ^{3} 2 E _{p}} e^{i E_{p} t - i p \cdot r} .

Reverse

p

in the second integral. At fixed spatial momentum consider

I (t) = \int_{- \infty}^{\infty} \frac{d p ^{0}}{2 π} \frac{i e ^{- i p^{0} t}}{( p ^{0} ) ^{2} - E _{p}^{2} + i 0} .

+ E_{p} - i 0

and

- E_{p} + i 0

. For

t > 0

, close in the lower half-plane, clockwise. The positive-energy residue and the orientation give

e^{- i E_{p} t} / (2 E_{p})

. For

t < 0

, close counterclockwise in the upper half-plane; the negative pole has derivative denominator

- 2 E_{p}

, and its residue gives

e^{i E_{p} t} / (2 E_{p})

. Thus

I (t) = e^{- i E_{p} ∣ t ∣} / (2 E_{p})

, reproducing the time-ordered mode expression. We obtain

Δ_{F} (x - y) = \int \frac{d ^{4} p}{( 2 π ) ^{4}} \frac{i e ^{- i p \cdot (x - y)}}{p ^{2} - μ ^{2} + i 0} .

This is a distributional boundary value, not an absolutely convergent Fourier integral. As an independent normalization check,

I^{'} (0^{+}) - I^{'} (0^{-}) = - i

, so

(□ + μ^{2}) Δ_{F} = - i δ^{(4)}

; the numerator is

i

with the question's definition of the propagator.

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