Differential cross section of $\Phi^1\Phi^2\rightarrow \Phi^1\Phi^2$ process in linear sigma model: contribution due to forward scattering

Question

In Peskin and Schroeder exc. 4.3 we are asked to compute the differential cross section of $\Phi^1\Phi^2\rightarrow \Phi^1\Phi^2$ process for the linear sigma model: Compute, to leading order in $\lambda$, the differential cross section, in the center of mass frame, for the scattering process $\Phi^1\Phi^2 \rightarrow \Phi^1\Phi^2$.

Since all the particles involved have identical masses and we have two incoming and two outgoing particles, we can use (4.85): $$(d\sigma/d\Omega)_{CM} = \frac{|M|^2}{64\pi^2E_{CM}}.\tag{4.85}$$

Up to leading order in $\lambda$ we find $M=-2i\lambda$, since the Feynman rule for the vertex with particle types $i,j,k,l$ is $$-2i\lambda(\delta^{ij}\delta^{kl}+\delta^{il}\delta^{jk}+\delta^{ik}\delta^{jl}).$$ This would give a differential cross section of $$(d\sigma/d\Omega)_{CM} =\frac{\lambda^2}{16\pi^2E_{CM}}.$$ This is the answer that can be found in several sources (e.g. Zhong-Zhi Xianyu's solutions).

However, this only seems to take into account the interactions. As 4.85 is derived while ignoring the "trivial case" of forward scattering where no interaction takes place, I suspect a contribution due to forward scattering with no interaction is missing here. What is this contribution?

Answer 1

When you compute cross sections you trivially ignore the case of no interactions

You want to compute $\langle p_{1}p_{2}|S|k_{1}k_{2}\rangle$. If there are no interactions then $S = I$, the identity operator. For a theory with interactions we want to isolate the piece that is due to interactions, so we define $S = 1+iT$.

This is where the definition of the matrix element, $\mathcal{M}$ comes from:

$\langle p_{1}p_{2}|iT|k_{1}k_{2}\rangle = (2\pi)^{4}\delta^{(4)}(k_{1} + k_{2} - p_1 - p_2)\cdot i\mathcal{M}$.

Note we typically refer to forward scattering as in the the scattering angle of some incoming particle being $\theta \approx 0$, which is different than no scattering taking place at all.