Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What are Baryon Resonances?

  1. Apr 16, 2014 #1
    i've looked online and the closest i've found to a definition is this:

    what does this mean? i understand that an unstable baryon will decay into proton / neutron and a pion via the strong force. but what actually is this resonance? is it a resonance in the traditional sense? is it referring to the excited state of the baryon? does it have modes of vibration?

    if someone could shed a little light on this it would be appreciated as there appears to be very little information on the internet that explains what a baryon resonance actually is or to what property the term 'resonance' actually refers.
  2. jcsd
  3. Apr 16, 2014 #2
    A resonance is, in general, a very short living particle. In particular, for such particles it's pretty hard to define an actual value of their mass and this is basically a consequence of the uncertainty principle. If the particle is very short living then:
    \Delta t\Delta E\sim\hbar\Rightarrow \Delta E\sim\frac{\hbar}{\Delta t},
    and the spread on its energy (rest mass) is very large.

    This particles are called resonances since they are so short living that they basically only appear as virtual states. Consider for example the scattering between two particles, say A and B: [itex]A+B\to A+B[/itex]. In QFT the cross section (i.e. more or less the probability) for this scattering to occur is the sum of all the possible Feynman diagrams that give you the same final state.
    Now, if the energy of the two particles is reasonably close to the mass of this unstable particle, say X, you have one addition channel for your scattering: [itex]A+B\to X\to A+B[/itex]. Of course this is possible only if the quantum numbers are the right ones.

    Then it turns out that right near the mass of this particle the scattering cross section as a function of the energy drastically increases (basically because of the presence of one additional channel) showing the typical resonant behavior.
  4. Apr 16, 2014 #3


    User Avatar
    Gold Member

    I guess the name resonance came in use because of how they appear in experiments....
    They appear as peaks of the cross section with respect to energy....
    There's also a more satisfying explanation, in case you are familiar with classical resonances... For example if you check the wikipedia for resonances (http://en.wikipedia.org/wiki/Resonance) and take a look at the "intensity of oscillations I" formula, you will see a familiar formula (Breit Wigner cross section which describes resonances).
    If you also read below the formula, it says that "The intensity is defined as the square of the amplitude of the oscillations"
    And how is the differential cross section given? It's the square of the scattering amplitude...

    You don't have any mechanical oscillation though... since the amplitudes of oscillations represent some mechanical motion, but the scattering amplitudes come from the interaction of the wavefunction with a local potential...
    [itex] \psi_{scatt}= e^{ikx} + \frac{e^{ikr}}{r}f(k,k')[/itex]
    Last edited: Apr 16, 2014
  5. Apr 17, 2014 #4
    Excellent replies as always!! Thanks!!!
  6. May 3, 2014 #5


    User Avatar
    Science Advisor

    It's not just the peak in the scattering amplitude that matters. Even more important is its phase.

    Here's a recent paper from the LCHb collaboration concerning their observations of a meson state, the Z1-(4430). To address the essential question "is it a peak, or really a particle" they say,

  7. May 3, 2014 #6


    User Avatar
    Gold Member

    @Bill K
    what else could a peak be but a particle? I'm asking because as far as I know:
    peaks denote bound states... or semibound... depending on their branches...
    Bound states appear in general as resonances...
  8. May 4, 2014 #7


    User Avatar
    2017 Award

    Staff: Mentor

    Resonances are so short-living that you often see interference effects - and then you are sensitive to the phases. For long-living intermediate particles, interference effects (in the decays via those particles) are negligible.
  9. May 4, 2014 #8


    User Avatar
    Science Advisor

    This question has been extensively discussed in the context of the Z(4430) state, which thanks to the latest LCHb work now appears to be a genuine tetraquark, but several other proposals had previously been entertained.

    One possibility was a weakly bound "molecular" combination of two mesons.

    Another was a threshold effect, since its mass lies so close to the DD* threshold. See this paper, which says,

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook