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 Incoming links: External-leg factorial cancellation at a quartic scalar vertex

π = \dot{ϕ}

H = \int d^{3} x [\frac{1}{2} π^{2} + \frac{1}{2} (\nabla ϕ)^{2} + \frac{1}{2} m^{2} ϕ^{2} + \frac{λ}{4 !} ϕ^{4}] .

The first three terms constitute

H_{0}

, the free massive Hamiltonian operator. The remaining term is

H_{I} = λ \int ϕ^{4} d^{3} x /4! = - \int L_{I} d^{3} x

. This sign is a consequence of having no time derivatives in the interaction: the canonical momentum is unchanged, while the interaction enters the Hamiltonian density with the opposite sign to the Lagrangian density.

In the interaction picture, operators evolve with

H_{0}

and states evolve with

H_{I} (t)

constructed from the free interaction-picture field. Their evolution operator satisfies

i \partial_{t} U_{I} (t, t_{0}) = H_{I} (t) U_{I} (t, t_{0}), U_{I} (t_{0}, t_{0}) = I .

Integrating once gives

U_{I} = I - i \int_{t_{0}}^{t} H_{I} (t_{1}) U_{I} (t_{1}, t_{0}) d t_{1}

. Repeated substitution orders the integrations as

t \geq t_{1} \geq \dots \geq t_{n} \geq t_{0}

. Replacing the ordered simplex by the full integration cube introduces

1/ n!

U_{I} (t, t_{0}) = \sum_{n = 0}^{\infty} \frac{( - i ) ^{n}}{n !} \int_{t_{0}}^{t} d t_{1} \dots d t_{n} T {H_{I} (t_{1}) \dots H_{I} (t_{n})} .

Taking the asymptotic limits with the usual adiabatic or regulated scattering prescription yields

S = T exp [- i \int_{- \infty}^{\infty} H_{I} (t) d t] = T exp [i \int d^{4} x L_{I} (ϕ_{I} (x))] .

The field in this expression is the free field evolving in the interaction picture, not an independently prescribed classical field. For quantum fields the expansion is a regulated perturbation series; the time-ordering symbol cannot be dropped because the operators at different times need not commute.

Use relativistically normalized two-particle states

∣ p_{1}, p_{2} ⟩ = a^{†} (p_{1}) a^{†} (p_{2}) ∣0 ⟩

, with

⟨ p ∣ q ⟩ = (2 π)^{3} 2 E_{p} δ^{3} (p - q)

. The connected first-order scattering contribution is

⟨ p_{3}, p_{4} ∣ (S - I) ∣ p_{1}, p_{2} ⟩_{conn} = - \frac{iλ}{4 !} \int d^{4} x ⟨ p_{3}, p_{4} ∣ : ϕ_{I} (x)^{4} : ∣ p_{1}, p_{2} ⟩ .

Here the connected S-matrix uses a normalized vacuum and physical external one-particle states. At this order, normal ordering the vertex separates the four-external-leg contribution from vacuum and tadpole pieces; those disconnected or external-mass contributions are not the two-to-two scattering amplitude.

Two fields annihilate the incoming particles and two create the outgoing ones. There are

(2 4)

choices of the annihilating fields, two assignments of the incoming momenta and two assignments of the outgoing momenta. Hence the total factor is

(2 4) 2! 2! = 4!

⟨ p_{3}, p_{4} ∣ : ϕ_{I} (x)^{4} : ∣ p_{1}, p_{2} ⟩_{conn} = 4! e^{i (p_{3} + p_{4} - p_{1} - p_{2}) \cdot x} .

⟨ p_{3}, p_{4} ∣ (S - I) ∣ p_{1}, p_{2} ⟩_{conn} = i (2 π)^{4} δ^{4} (p_{3} + p_{4} - p_{1} - p_{2}) T, T = - λ + O (λ^{2}) .

In particular, the tree scattering amplitude is independent of angle.

s = 2 E_{*}

p_{*}^{2} = E_{*}^{2} - m^{2}

. The spatial delta function sets the two final momenta opposite, and the energy delta function is

δ (s - 2 p^{2} + m^{2})

. Its radial Jacobian is

E_{*} / (2 p_{*})

. Thus

d Φ_{2} = \frac{1}{16 π ^{2}} \frac{p ^{2} d p}{E _{p}^{2}} δ (s - 2 E_{p}) d Ω = \frac{p _{*}}{16 π ^{2} s} d Ω.

F = 4 p_{*} s

for equal initial masses. Using

d σ = ∣ T ∣^{2} d Φ_{2} / F

, the labeled angular density is

λ^{2} / (64 π^{2} s)

. The outgoing quanta belong to one real scalar field, so they are identical. A full-sphere integration counts each event twice and requires the identical final-state symmetry factor

1/2!

. Consequently

σ = \frac{1}{2} \int_{S^{2}} \frac{λ ^{2}}{64 π ^{2} s} d Ω = \frac{λ ^{2}}{32 π s} + O (λ^{3}), s > 4 m^{2} .

Equivalently one may integrate the labeled density over a hemisphere representing each event once. The final-state factor is separate from the vertex factorial: applying it at the vertex instead would give the wrong elastic scattering from a quartic scalar contact interaction cross-section.