I am trying to use optical theorem* ( given in box 24.2 in Quantum Field Theory and the Standard Model, M. Schwartz ). I am trying to calculate the imaginary part of this diagram for the scalar field:
But I am getting extra factor of 2 in the calculation. I have calculated the imaginary part using complex analysis, and also using the Cutkosky cutting rules. I have also checked some papers. I am quite sure that I am getting an extra 2 factor when I calculate using the optical theorem. I suspect that I might have misunderstood the formula (24.2 ).
Can anyone please verify my understanding of the formula?
Here it goes
( m is the mass of both the particles, s is a Mandelstam variable, $\lambda$ is the coupling constant ) :
$E_{CM}$ stands for the sum of energies of the two incident particles in the Center of Momentum frame. It should be equal to $\sqrt{s}$.
$|\vec{p_i}|$ stands for the magnitude of the four-momentum of either one of the incident particles. It comes out to be $\sqrt{\frac{s}{4} -m^2}$.
$\Sigma_x \sigma(A->X)$ is the sum of scattering cross section of all the possible diagrams at order $\lambda$. In this case there is just one term to be summed over. And it is $(4\pi) \frac{\lambda^2}{64\pi^2s}$ . Here $4\pi$ comes from the integration of total solid angle. Remaning is just the differential cross section $\frac{d\sigma}{d\Omega}$ for scattering at order $\lambda$ .
The total answer that I get is $\frac{\lambda^2}{8\pi} \sqrt{\frac{1}{4} - \frac{m^2}{s}}$.
I have spent quite some time on this but still don't understand what am I doing wrong here. By every other method, I am getting an answer half this value. I can show more calculations and more arguments if anyone asks for them. Thanks.
Update:
*: Here is the formula: $Im M(A->A) = 2E_{CM}|\vec{p_i}|\Sigma_x \sigma(A->X)$
