Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 301 / 3

In phi cubed theory, each vertex has three legs. A connected four-point tree-level Feynman diagram therefore needs two cubic vertices and one internal propagator. There are exactly three ways to partition the labeled external legs between these vertices:

       s channel               t channel              u channel

 p \          / p'         p ---o--- p'            p ---o--- q'
    o--------o                  |                       |
 q /          \ q'         q ---o--- q'            q ---o--- p'

 internal: p+q             internal: p-p'          internal: p-q'

Solid lines denote the real scalar field propagator. The labels specify which external pair meets at each vertex, so the last two graphs are different even though their unlabeled shapes agree. This is the four-point tree amplitude in phi cubed theory:There is no four-point contact vertex in the given cubic interaction.

For the interaction sign in this paper, the cubic Feynman rule is $+i\lambda$, while the scalar Feynman propagator is $i/(k^2-m^2+iϵ)$. Each exchange graph therefore contributes $-i{\lambda }^2/(k^2-m^2+iϵ)$ to $i\mathcal{M}$. The three graphs giveThe equal-mass Mandelstam variables satisfy $s+t+u=4m^2$, so $u-m^2=3m^2-s-t$. In the physical elastic region with $m>0$, none of these denominators vanishes, and the tree amplitude is real up to its irrelevant overall convention. ThusThe square includes interference between all three channels; summing their individual squared contributions would omit physical cross terms. If a denominator approaches a pole, retain the Feynman i-epsilon prescription in the modulus before taking a limit.

Use the Lorentz-invariant phase-space measure from the original PDF, including the $1/(2E_{p^{\prime }})$ and $1/(2E_{q^{\prime }})$ factors missing from the local TeX. In the center of mass frame put $E=\sqrt{s}/2$ and $k=\sqrt{s-4m^2}/2$, assuming $s>4m^2$. The spatial Dirac delta distribution sets ${\mathbf{q}}^{\prime }=-{\mathbf{p}}^{\prime }$, leavingSince $d(2E_{k^{\prime }})/d{k}^{\prime }=2k^{\prime }/E_{k^{\prime }}$, the radial integral givesThe invariant flux factor is $2\sqrt{\kappa (s,m^2,m^2)}=2\sqrt{s(s-4m^2)}$, using the Källén function. Dividing the phase space by this flux yields $d\sigma /d\Omega =|\mathcal{M}{|}^2/(64{\pi }^{2}s)$, with the final-state labels retained as in the printed formula.

Let $x=\cos \theta$. Of the Mandelstam variables, $t$ isThe azimuthal integral contributes $2\pi$, and the angular endpoints are thereforeChanging variables gives the requested expression from the formula supplied in the paper:There is a normalization qualification: the displayed starting formula integrates over labeled final momenta and contains no $1/2!$. For the physical cross-section of two indistinguishable outgoing quanta of this real scalar field, the identical-particle factor in a final-state phase-space integral divides the full integral by two. With that convention $X=1/(32\pi )$; equivalently, integrate only one representative of each exchanged pair. The boxed value $1/(16\pi )$ follows the given formula exactly.

In four spacetime dimensions with $\hslash =c=1$, a relativistic scattering cross-section has mass dimension $-2$, since it is an area. The action is dimensionless, so the Lagrangian density has dimension four. The scalar kinetic term then gives $[\varphi ]=1$, and the cubic term gives $[\lambda ]=1$.

Each term of the four-point tree amplitude in phi cubed theory has dimension $[{\lambda }^2]-[s-m^2]=2-2=0$. Hence $[|\mathcal{M}{|}^2]=0$. In the final integral,while the numerical factors $X,Y$ are dimensionless. ThereforeIncluding the identical-particle factor changes only a dimensionless constant and preserves this check.