# Going from width to cross section

Given the decay width of a process, $\Gamma(A\to B+C)$, is it possible to turn this around to find the production cross section, $\sigma(B+C\to A)$?

Edit: In particular I have been thinking of something like a Breit-Wigner resonance type example assuming that $\sigma(B+C)$ is dominated by $\sigma(B+C\to A)$ at relevant energies.

## 2 Answers

It is not possible to get the cross section from the decay width. The reason is that when calculating the decay width, one squares the amplitude and then sums over the momenta of B and C. While the amplitude squared part is the same for cross section, summing over final momenta means you cannot use it for specific momenta of B and C in the cross section.

There is optical theorem in quantum field theory and it enables you to study with the decay rates to find cross section. However there are some difficulties to calculate the cross section of explicit decay channel.
$$Im(\Sigma(m^2))=m\Gamma_{tot}$$ So the propagator becomes $$iG(p^2)=\frac{i}{p^2-m^2+im\Gamma_{tot}}$$ You can jump up to cross section then considering s channel

$$\sigma = g^4{\mid\frac{i}{p^2-m^2+im\Gamma_{tot}} \mid}^2=g^4\frac{1}{(p^2-m^2)^2+(m\Gamma_{tot})^2}\sim g^4\frac{\pi}{m\Gamma_{tot}}\delta(p^2-m^2)$$ Now the resonant become on-shell particle. But the thing is you need to calculate all the decay channels such as $A\rightarrow B+C$ and $A\rightarrow B+X$ and $A\rightarrow Y+C$ etc... For a specific decay channel one should use branching ratio as well $$\sigma(B+C\rightarrow A \rightarrow f )= \sigma(B+C\rightarrow A )\frac{\Gamma_{f}}{\Gamma_{tot}}$$ As you see it is very tedious to calculate each decay process and their width(decay rate) but I believe one can succeed.