# Breit-Wigner curve derivation of energy dependence?

The Breit-Wigner cross section for resonant particle decay takes the form: $$\sigma_{fi}=\frac{\pi\hbar^2}{q^2} \frac{2j+1}{(2S_1+1)(2S_2+1)} \frac{\Gamma_i \Gamma_f}{(E-Mc^2)^2+\Gamma^2/4}$$ I am looking to (non-relativistically) explain the origin of each of these terms. This question is concerned however with just the: $$\frac{1}{(E-Mc^2)^2+\Gamma^2/4}$$ factor. Here is a rough overview of it's origin:

The resonant state is of finite life time so we can write it's wave function as: $$\psi(t)=\psi(0) e^{iEt/\hbar}e^{-\Gamma t/2}$$ taking the Fourier transform of this we get: $$\chi(E)\propto \frac{1}{(E-Mc^2)+i\Gamma/2}$$ the cross section is then proportional to $\chi \chi^*$ meaning: $$\sigma \propto \frac{1}{(E-Mc^2)^2+\Gamma^2/4}$$

I am ok with this explanation apart from the part in bold. How do we know that $\sigma \propto \chi \chi^*$? should we not be integrating over the matrix element of this with the initial (final?) state?

• you have taken the initial (final) states in your xx* and it is a probability density and the crossection is proportional to the probability density. If you integrate over E you will find the total production crossection for that resonance. – anna v Mar 21 '17 at 16:49
• @annav Sorry what do you mean by 'you have taken the initial (final) states in your xx* '? That I have already taken the matrix elements or something else? – Quantum spaghettification Mar 21 '17 at 16:54
• yes, what you have written is the matrix element squared which is the probabilty density. look at page6 of this cpp.edu/~pbsiegel/phy40413/lectures/lecture10.pdf – anna v Mar 21 '17 at 17:16
• To further expound on @anna v's (virtual) answer, the E in your wave function should actually be M. You wish to compute the decay amplitude $\psi(t=0)^* ~\psi(t)$. This is what you are re-expressing as a function of E instead of t in your Fourier transform, so of course the square of the amplitude χ(E) yields the cross section. – Cosmas Zachos Sep 7 '17 at 16:01