Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

At one point, I decided to make friends with the low-lying spectrum of QCD. By this I do not mean the symmetry numbers (the "quark content"), but the actual dynamics, some insight.

The pions are the sloshing of the up-down condensate, and the other pseudoscalars by extending to strangeness. Their couplings are by soft-particle theorems. The eta-prime is their frustrated friend, weighed down by the instanton fluid. The rho and omega are the gauge fields for flavor SU(2), and A1(1260) gauges the axial SU(2), and they have KaluzaKlein-like echoes at higher energies, these can decay into the appropriate "charged" hadrons with couplings that depend on the flavor symmetry multiplet. The proton and the neutron are the topological defects. That accounts for everything up to 1300 but a few scalars and the b1.

There are scalars starting at around 1300 MeV which are probably some combination of glue-condensate sloshing around and quark-condensate sloshing around, some kind of sound in the vacuum glue. Their mass is large, their lifetime is not that big, they have sharp decay properties.

On the other hand, there is nothing in AdS/QCD which should correspond to the sigma/f0(600), or (what seems to be) its strange counterpart f0(980). While looking around, I found this discussion: The literature that it pointed to suggests that the sigma is a very unstable bound state of pions (or, if you like, tetraquarks).

This paper gives strong evidence for an actual pole; another gives a more cursory review. The location of the pole is far away from the real axis, the width is larger than the mass by 20% or so, and the mass is about 400MeV. The authors though are confident that it is real because they tell me that the interpolation the interactions of pions is safe in this region because their goldstone properties dominate the interactions. I want to believe it, but how can you be sure?

I know this particle was controversial. I want to understand what kind of picture this is giving. The dispersion subtraction process is hard for me to visualize in terms of effective fields, and the result is saying that there is an unstable bound state.

Is there a physical picture of the sigma which is more field theoretical, perhaps even just an effective potential for pions? Did anyone who convinced himself of the reality of the sigma have a way of understanding the bound state properties? Is there an analog unstable bound state for other goldstone bosons? Any insight would be welcome.

share|cite|improve this question
I think my brain just exploded – Timtam Aug 13 '11 at 12:25
and also i don't think this particle exists, they probably made it up so they can get more government grant money. – Timtam Aug 13 '11 at 12:56
@Timatam: you're not the only person to have this idea. This particle was delisted for a long time, meaning people concluded that it was just a made up particle, just a bump in the cross section due to pion interactions. But the experimentalists shouted that it was really there, and people who do extrapolations of experimental data to extract particles now say that have a pole. I also used to think it was made up, this is why I am confused. – Ron Maimon Aug 13 '11 at 17:05

The literature that it pointed to suggests that the sigma is a very unstable bound state of pions (or, if you like, tetraquarks).

This has reminded me of a really interesting paper by Shifman and Vainshtein - - which talks about an exact symmetry between pions and diquarks in two-color QCD. They speculate that this symmetry should have an analogue in three-color QCD. I'm wondering if the sigma might be a sort of diquark-diquark correlation enhanced by the Shifman-Vainshtein symmetry.

I'd also look at Hilmar Forkel's work for AdS/QCD insight.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.