# Where does $\vert\mathcal{M}(k)\vert^2$ come from in the last equation, p235 in Peskin&Schroeder's book?

In the Optical Theorem section, in the middle paragraph of p235, it is said

"What is left over in expression (7.52) is just the factor $\lambda^2$, the square of the leading-order scattering amplitude, and the symmetry factor (1/2)".

However in (7.52) (which is the invariant amplitude of the 1-loop order diagram in $\phi^4$ theory) there is no $|\mathcal{M}(k)|^2$. Therefore, where does $|\mathcal{M}(k)|^2$ come from in the last equation, p235 which is

$$\operatorname{Disc}\mathcal{M}=2i\operatorname{Im}\mathcal{M}=\frac{i}{2}\int\frac{d^3p_1}{(2\pi)^3}\frac{1}{2E_1}\frac{d^3p_2}{(2\pi)^3}\frac{1}{2E_2}\vert\mathcal{M}(k)\vert^2(2\pi)^4\delta^{(4)}(p_1+p_2-k)?$$

$$i\delta\mathcal{M}=\frac{\lambda^2}{2}\int\frac{d^4q}{(2\pi)^4}\frac{1}{(k/2-q)^2-m^2+i\epsilon}\frac{1}{(k/2+q)^2-m^2+i\epsilon}\tag{7.52}$$

• P&S are showing two ways to derive the imaginary part of the 1 loop diagram in phi^4 scattering. The first way is the systematic replacement of propagators with a delta function c.f Cutkosky cutting rules. The second way is via the optical theorem. P&S show of course that both ways are equivalent - the last equation on p.235 is a statement of the optical theorem.
– CAF
Commented Aug 18, 2018 at 11:06
• Why are they saying that what is left over in (7.52) is the square of the leading-order scattering amplitude? I cannot see it in (7.52). Commented Aug 18, 2018 at 12:45
• ‘The square of the leading order scattering amplitude’ in the middle paragraph of p.235 is simply a parentheses. That is to say, $|\mathcal M(k)|^2 = \lambda^2$.
– CAF
Commented Aug 18, 2018 at 13:55
P&S are showing two ways to derive the imaginary part of the $1$ loop diagram in $\phi^4$ scattering. The first way is the systematic replacement of propagators with an on shell constraining delta function c.f Cutkosky cutting rules. This originates in $$\text{Im} \frac{1}{k^2-m^2+ \mathrm{i}\epsilon} = -\pi \delta(k^2-m^2).$$
The issue in question may actually just be a confusion in the english - ‘The square of the leading order scattering amplitude’ in the cited paragraph above is simply a parentheses to $\lambda^2$, that is to say, $$|\mathcal M(k) |^2 = \lambda^2,$$ no $k$ dependence at tree level on the rhs because we are in the scalar field theory.