# How to reconcile infinite cross section of resonances with cross section formula from quantum mechanics?

If we consider $s$-wave scattering for two scalar fields $\phi$ and $\chi$ with an interaction $\frac{g}{2}\phi^2\chi$, then the Lorentz-invariant scattering amplitude to second order is:

$\mathcal{M}_{fi} = \frac{-ig^2}{s - m_\chi^2}$, where $s$ is the Mandelstam-s ($E_{cm}^2$).

The cross section, which goes as $\left| \mathcal{M}_{fi}^2 \right |$, is clearly singular at the $\chi$ resonance. But I don't know how to reconcile that with the scattering formulat given in Sakurai 7.6.17:

The cross section by $\left| f(\theta)\right|^2$, where $f(\theta)$ is given by the partial wave decomposition:

$f(\theta) = \frac{1}{k}\sum_l (2l+1)e^{i\delta_l}\sin\delta_l P_l(\cos\theta)$

For $s$-channel scattering, only $l=0$ contributes, and I do not see how we can have a singularity in $f(\theta)$ anywhere, as I would expect for resonance scattering.

• Anytime a physics calculation yields infinity, there is something wrong – Lewis Miller Feb 15 '18 at 16:24
• Usually, the fault lies in an approximation that may be accurate within some region but fails on other regions. Trust the finite resuly – Lewis Miller Feb 15 '18 at 16:26

The issue is that the $\mathcal{M}_{fi}$ in the question is only calculated to second order. When you include loop corrections and "dress" the propagator, it takes on a different structure and the width of the peak becomes finite.