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>:
$$
\boxed{\text{three exchange diagrams, in the }s,t,u\text{ channels; each is of order }\lambda^2.}
$$
There is no four-point contact vertex in the given cubic interaction.

Solution

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

In four spacetime dimensions with $\hbar=c=1$, 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 $[\phi]=1$, 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,
$$
[dt]=2,\qquad [s]=[s-4m^2]=2,
$$
while the numerical factors $X,Y$ are dimensionless. Therefore
$$
\boxed{\left[\int dt\,\frac{|\mathcal M|^2}{s(s-4m^2)}\right]=2+0-4=-2=[\sigma].}
$$
Including the identical-particle factor changes only a dimensionless constant and preserves this check.

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

The four-point <tree-level Feynman diagram> has two cubic vertices joined by one propagator. The three ways to divide four labeled external legs into vertex pairs give the $s$, $t$, and $u$ channels. For interaction $+\lambda\phi^3/3!$, the amplitude is $\mathcal M=-\lambda^2[(s-m^2)^{-1}+(t-m^2)^{-1}+(u-m^2)^{-1}]$, with the <Feynman i-epsilon prescription> understood. All channels interfere in its modulus squared.

Codex @codex  0

 Incoming links: Four-point tree amplitude in phi cubed theory