The massless two-body kinematics gives

q_{1} \cdot q_{2} = M^{2} /2

. The scalar scattering amplitude therefore has constant magnitude

∣ A_{aa} ∣ = ∣ λ_{a} ∣ M^{2} /2

. Use

\int d Ω = 4 π

and the identical-particle factor from part d:

Γ_{aa} = \frac{1}{2 !} \frac{4 π}{64 π ^{2} M} \frac{∣ λ _{a} ∣ ^{2} M ^{4}}{4} = \frac{∣ λ _{a} ∣ ^{2} M ^{3}}{128 π} .

(1)

For the photons, the two-photon vertex of a scalar field-strength interaction following from

Φ F_{μν} F^{μν}

is, up to an irrelevant overall phase,

A_{γγ} = λ_{γ} [(q_{1} \cdot q_{2}) (ϵ_{1} \cdot ϵ_{2}) - (q_{1} \cdot ϵ_{2}) (q_{2} \cdot ϵ_{1})] .

(2)

Vertex qualification: the source prints same-index polarization contractions in the second product, whereas the contractions must be crossed as above. The printed expression fails the Ward identity. For a concrete counterexample take

q_{1} = (E, 0, 0, E)

q_{2} = (E, 0, 0, - E)

and

ϵ_{2} = (1, 1, 0, - 1)

. This is a transverse unit spacelike photon polarization vector, since

q_{2} \cdot ϵ_{2} = 0

and

ϵ_{2}^{2} = - 1

. Replacing

ϵ_{1}

q_{1}

in the printed expression leaves

4 λ_{γ} E^{3}

instead of zero. The corrected crossed expression vanishes under either such replacement.

Choose physical spatial transverse photon polarizations along

x

and

y

in the parent rest frame. Then both crossed products vanish; the two equal-polarization combinations have

∣ ϵ_{1} \cdot ϵ_{2} ∣^{2} = 1

and the two orthogonal combinations give zero. There is no initial-spin average for a spin-zero parent. Hence

\sum_{pol} ∣ A_{γγ} ∣^{2} = 2∣ λ_{γ} ∣^{2} (\frac{M ^{2}}{2})^{2} = \frac{∣ λ _{γ} ∣ ^{2} M ^{4}}{2}, Γ_{γγ} = \frac{∣ λ _{γ} ∣ ^{2} M ^{3}}{64 π} .

(3)

The calculation in this spatial transverse gauge also yields the intended result from the misprinted expression, but only the corrected vertex makes its gauge invariance explicit. For equal-magnitude couplings, the photon decay width is twice the two-scalar decay width because of its two nonzero polarization combinations.