Are resonances formed in scattering between a baryon and meson?

Question

Consider the following reaction of strong interaction (in a scattering process) $$n+\pi^+\to \Lambda_0+K^+\tag{1}$$

Then the particle $\Lambda_0$ formed decays with weak interaction

$$\Lambda_0\to \pi^- +p\tag{2}$$

For each decay process I measure the four momenta of $K^+$, $\pi^+$ and $p$ in the final state, I calculate the center-of-mass energy $\sqrt{s}$ of reaction $(1)$. Then I plot the cross section vs $\sqrt{s}$.

Do I get a Breit Wigner resonance curve with central value equal to the sum of masses of $\Lambda_0$ and $K^+$ and width equal to $$\Gamma=\hbar/\tau$$ Where $\tau$ is $\sim 10^{-23}s$, i.e. the characteristic time of strong interaction?

I'm not sure about this because reaction $(1)$ is not a "decay" (while reaction $(2)$ is a decay) and I wonder if the resonances in the cross section are seen also in reactions that are not really a decay.

I suppose that in reaction $(1)$ a kind of "intermediate excited state" is formed and then it decays to the final state, but I'm quite confused about this.

Answer 1

Both resonant and non-resonant processes will be at work in $n\pi \to \Lambda K$. Resonant processes will manifest as one or more Breit-Wigner bumps at values of $\sqrt{s}>1115+494$ MeV, with kinetic energy to spare. In the pion-nucleon system, resonances with masses up to 2700 MeV and spins up to 13/2 have been observed. An unlimited number of heavier resonances with even higher spins are presumed to exist, but they are so broad that they blend into the relatively smooth non-resonant tail of the curve.

You can think of a resonance either as an excited intermediate state or as an unstable particle. (Six of one, half-a-dozen of the other.) Most resonances have multiple decay modes (also called channels). The total width is the sum of partial widths for its decay modes.

One common way for a resonance to form involves a quasi-bound state in a potential well with low (hence leaky) walls, e.g., $V(r<a)={{V}_{well}},$ $V(a<r<b)={{V}_{wall}}>{{V}_{well}},$ $V(b<r)=0$. This may be a poor model for the pion-nucleon system because the known resonances are $qqq$ rather than $qqq+\bar{q}q$states. The existence of resonances may rather be a consequence of an infinitely deep color-confining potential.

At a mathematical level, s-channel poles seen as resonances come from summing ladder diagrams, but what of non-resonant processes? Single-particle exchanges in the t-channel do not make for s-channel poles. In your process, exchanging an $s\bar{d}$meson should do the trick, but note that pristine CP-invariance would forbid a triple-pseudoscalar vertex, so simple kaon exchange would be suppressed.

Answer 2

Please read my answer here, to the question "What is the relationship between excitation and resonance? "

Your question is relevant to the subject. Particles that decay by the weak interaction are bound in a strong force potential. They do not form as stable a particle as the proton because the weak force and the energy balances allows them to decay. Only the proton is stable from the baryon octet.

The original scatter may form a resonance when plotting/scanning over the energy of the interaction, but that will be a different "particle" as its invariant mass would have to be at least the summed mass of the decay products. The crucial word here is "may".

If one looks at scattering crossections , for example e+ e- which is much easier to experiment with, the curves are smooth except where resonances occur.