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

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?

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.