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I know that resonances in particle physics are the creation of metastable bound states (i.e. particles). I also know that near such a resonance the decay width can be described by a Breit-Wigner curve. But I am confused about why we actually, physically get an increase in the cross section. Consider for example the reaction $e^{-}+e^{+}\rightarrow c \bar c\rightarrow q+\bar q$ and $e^{-}+e^{+}\rightarrow q+\bar q$ Feynman diagrams for which are shown below: enter image description here Naively I would expect second one to have the larger cross section due to the fewer number of interactions. But this is not the case, as can be seen from this pdg plot. My question is therefore:

Why do particle resonances cause an increase in the cross section of a reaction?

It must be pointed out that I am unsure if the reaction from $c$ to $d$ is via $\gamma$ or $Z$ (please let me know) but this should not change my argument.

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  • $\begingroup$ Can the formation of the bound state be described perturbativly by Feynman diagrams? $\endgroup$ – innisfree Mar 13 '17 at 10:49
  • $\begingroup$ Why can't the second vector in first diagram be a gluon? or n-gluons? Anyway, supposing it can be described by pQCD, why shouldn't a narrow resonance results in a spike in cross section as per usual? $\endgroup$ – innisfree Mar 13 '17 at 10:56
  • $\begingroup$ @innisfree Sorry, I was thinking gluons when I wrote $Z$, it could occur via any of them. I will have to think about your second question. $\endgroup$ – Quantum spaghettification Mar 13 '17 at 11:01
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Your higher order diagram is confusing, let us take this first order diagram of e+e- scattering .

e+e- Z

The large crossection at the mass of the Z is inevitable due to the integral which contains the propagator , when the momentum transfer squared equals the mass squared the contribution under the integral becomes huge.

Experimentally this is what is measured versus energy

enter image description here

As I understood in this lecture(page 6) each resonance on this crossection plot is represented by a sum of t channel diagrams that add into the resonance. I was as bemused as you trying to write an s channel feynman diagram for e+e- to e+e- resonating on the rho. So each resonance does not mean a new particle as happens with the Z.

Historical context, when the feynman diagrams used for the e+e- crossection assumed,before QCD was proposed and accepted as part of the standard model, that there was the vector meson dominance model

In particular, the hadronic components of the physical photon consist of the lightest vector mesons, ρ , ω and ϕ . Therefore, interactions between photons and hadronic matter occur by the exchange of a hadron between the dressed photon and the hadronic target.

Similar to your fig 1.

Thus this would give vector mesons of the experimental plot a probability of participating in a propagator in a first order Feynman diagram, and thus display the simple resonance behavior .

Now that we know QCD is playing the strong role, one needs additions of diagrams where the resonant behavior increases the probability in sums of many diagrams, as described in the link I gave.

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  • $\begingroup$ There's no integtral over a propagator in the tree-level ee -> z -> ee diagram. There's just a propagator. $\endgroup$ – innisfree Mar 19 '17 at 8:51
  • $\begingroup$ @innisfree Feynman diagrams are iconal representations of integrals that have to be carried out in order to get a probability crossection or decay time. The propagator is integrated over. If you mean in the vector meson dominance old model, they just assumed that a "dressed" propagator of gamma had a vector meson component, which would give the resonance when the energy was appropriate $\endgroup$ – anna v Mar 19 '17 at 9:11
  • $\begingroup$ Don't understand. If you calculate the amplitude from the Feynman diagram in your answer, there's no e.g., loop momenta to integrate over the propagator. The energy is just $\sqrt{s}$ flowing through the propagator. $\endgroup$ – innisfree Mar 19 '17 at 10:42
  • $\begingroup$ @innisfree Yes, this is the standard model with quarks and all. The VMD dominance model was before quarks entered the game, as was the Regge pole model. Theorists were trying to fit data. To get the equivalent of the VMD one has to add up all the ladder diagrams of the type shown by the OP as fig 1). as outlined in the link I gave. The scanning over s will give the plot of the resonance seen above, as s nears the Z pole the crossection is high and falls past it. This is due to the propagator in the matrix element. In VMD they introduced a "dressed" propagator that gave a VM contribution. $\endgroup$ – anna v Mar 19 '17 at 11:38

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