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:

enter image description here

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.


*: Here is the formula: $Im M(A->A) = 2E_{CM}|\vec{p_i}|\Sigma_x \sigma(A->X)$


1 Answer 1


It turns out that I was calculating the cross section incorrectly. $\Sigma_x \sigma(A->X)$ should come out to be $(2\pi) \frac{\lambda^2}{64\pi^2s}$ because I have to divide the phase space by 2 ( output particles are identical ).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.