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

Table of contents
- a 4
  - Solution a
- b 4
  - Solution b

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)} .

The Feynman i-epsilon prescription places the positive-energy pole

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.

b

Solution

The free complex field has the global unitary group

U (1)

Φ (x) \mapsto e^{i α} Φ (x) .

Gauging it permits

α = α (x)

and introduces a gauge field with

D_{μ} = \partial_{μ} + i e A_{μ}, A_{μ} \mapsto A_{μ} - \frac{1}{e} \partial_{μ} α .

The scalar quantum electrodynamics Lagrangian is

L = (D_{μ} Φ)^{†} D^{μ} Φ - M^{2} Φ^{†} Φ - \frac{1}{4} F_{μν} F^{μν} .

Its interaction terms are

L_{int} = i e A_{μ} {Φ^{†} \partial^{μ} Φ - (\partial^{μ} Φ^{†}) Φ} + e^{2} A_{μ} A^{μ} Φ^{†} Φ.

With momentum flowing along the scalar arrow, the momentum-space Feynman rules are:

an internal scalar line of momentum $p$ contributes $i / (p^{2} - M^{2} + i ϵ)$ ;
an internal photon in Feynman gauge contributes $- i η_{μν} / (k^{2} + i ϵ)$ ;
a scalar-scalar-photon vertex contributes $i e (p + p^{'})^{μ}$ , where $p$ enters and $p^{'}$ leaves along the scalar line;
a scalar-scalar-two-photon vertex contributes $2 i e^{2} η^{μν}$ ;
every vertex carries momentum conservation, and every undetermined internal momentum is integrated with $d^{4} q / (2 π)^{4}$ .

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