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Let me start by explaining my particle physics background is very patchy, so this question may not be as coherent as I would like it to be.

In general terms, what is the difference between a resonance found during a particle experiment and a particle?

From reading Wikipedia it seems to be mostly based on timescales. If a heavy particle decays quickly, how can we distinguish it from a resonance?

I would not be posting this question except for the fact that the Wikipedia page says "This page has some issues".

Wiki Page: Baryon Resonance

and I would be interested in knowing more about the subject.

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  • $\begingroup$ You could always read the cited papers. Wikipedia should not be treated as a primary source anyway... $\endgroup$ – Lightness Races with Monica Jun 7 '15 at 1:48
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A resonance (in the particle physics or related physics sense) and an unstable particle is exactly the same thing. The object has some complex mass and the imaginary part determines the decay width (and decay rate). But these two terms describe different aspects of the same thing.

"A particle" refers to the object, the particle species (in your URL's case, it's composite particles i.e. bound states, often excited states), and all conceivable properties it may have and processes it may undergo.

On the other hand, a "resonance" only describes one particular aspect of the object (particle), and the corresponding method how it may be discovered, namely its ability to produce a local peak ("bump") in a graph of a cross section as a function of energy. It's usually a cross section of a process with the particle in the initial state and a two-particle or multi-particle state in the final state, or vice versa.

The cross section goes up when the "energy is right" to produce (or come from) a particle of the particular mass. The local peak has the same mathematical reason as the resonances anywhere in physics – e.g. when a radio is amplifying the signal at a given frequency. When the frequency (or energy, and $E=hf$) is right, plus minus the width, the strength (or, in quantum mechanics, the probability) of a process is much higher.

When we see such a "bump", we may discover a new particle. That's how the Higgs boson was discovered in 2012 – and many other particles before the Higgs, too. The actual unstable particle, e.g. the Higgs boson, may also enter many other processes that can't be described as a simple resonance. It may be produced together with the Z-boson and/or other particles, for example, and in these more complicated processes, the Higgs boson is no longer a "resonance".

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  • $\begingroup$ Thanks for your time and quick answer, so my question about distinguishing them is redundant if they are the same thing, same decay pathway? $\endgroup$ – user81619 Jun 6 '15 at 16:07
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    $\begingroup$ Dear @AcidJazz - to talk about a resonance, you have to decide about a particular decay/fusion channel. But the particle that is seen as a resonance in this decay channel may also decay in other channels, or enter other processes. You can't distinguish a/the unstable particle and a resonance as an object because they may be the same object; but you should still distinguish the words "unstable particle" and "resonance" because they express different ideas. It's like asking how to distinguish Barack Obama from the U.S. president. Well, it's the same person, now, but it doesn't have to always be. $\endgroup$ – Luboš Motl Jun 6 '15 at 16:16
  • $\begingroup$ Yep, sort of getting the idea as I read through your answer, I will read up some more myself and ask as specific a question as I can, I appreciate your time and your blog has answered my question today, partially, re the lhc upgrade. Thanks for the two answers! $\endgroup$ – user81619 Jun 6 '15 at 16:21
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This is to be read in conjunction with the answer by Lubos

The particle data group has compiled a lot of cross sections in this paper, whence I have copied a particular plot, fig 49.5.

e+e-crossection $$\sqrt{s}~~ (\mathrm{GeV})$$

The blue part are the resonances that were found during the sixties, and are typical of other resonances in scattering crossections. This is the scattering of $e^+e^-$, and also after the decay products were identified, in the invariant mass spectrum of the decay products. The terminoloty "resonance" is carried over from classical physics resonances, where acoustic amplitudes become high at resonance, for example. The cross sections became large at that specific center of mass energy. Note the width, in contrast with the red peaks. When the $J/\psi$ was found the hoi polloi physicists were expecting widths comparable to the $\phi$ at best, and the sharpness of the resonance came as a surprise and fully confirmed the quark model. It was not made up of run of the mill quarks like the blue resonances, but of a new quark, the charm. The same is true with $\Upsilon$ and the beauty quark that completed the quark model.

If you look at the elementary particle table none of the resonances from the $\Upsilon$ and below exist there, because these resonances are composites of quarks. Continuing to higher energies though, we hit on the Z boson, which is an elementary particle and resonates in the $e^+e^-$ scattering experiment but also decays in a lot of other elementary particles.

Thus in the same experiment, $e^+e^-$ scattering with increasing center of mass energy (or invariant mass for the LHC plots) we have resonances that are composites of quarks and antiquarks on the one hand, and elementary particles on the other.

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  • $\begingroup$ Thank you very much Anna, I have a habit of copying your answers into my notes at this stage,it forces me to read them as carefully as possible. Much appreciated regards. $\endgroup$ – user81619 Jun 6 '15 at 18:59
  • $\begingroup$ Looking at the plot, it is amusing how different the experimental (pdg listing) status of $\omega$ is, being as it is as clear as Z. $\endgroup$ – arivero Sep 22 '15 at 13:50
  • $\begingroup$ @arivero Yes, it is the accumulation of data that confirmed the Z as the gauge boson of the model $\endgroup$ – anna v Sep 22 '15 at 13:55
  • $\begingroup$ Pity we can not do a neutrino positron collider, to see W as clear. $\endgroup$ – arivero Sep 22 '15 at 14:36

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