This is an old classic which I don't think ever got a clear answer. The Gribov-Froissart projection that gives the relativistic version of Regge trajectories treats even angular momentum differently from odd angular momentum. The trajectory functions in general are separate for even and odd angular momentum.

But in QCD, the odd trajectories interpolate the even trajectories in every case--- the two trajectories are degenerate. Is there a way of understanding why the even-odd trajectories are degenerate? Is is a symmetry argument? Can you find a natural system where they are not degenerate?

A quick google search revealed this reference, which doesn't answer the question, but gives an experimental signature for the degeneracy: http://www.slac.stanford.edu/cgi-wrap/getdoc/slac-pub-0576.pdf , There is probably excellent data by now on this.


Once upon a time, I asked an experienced phenomenologist who worked on particle physics in the 60s why even and odd signatured trajectories lie on top of each other.

He said the phenomenon was called 'exchange degeneracy' and that so far no one has an explanation.

I'm looking back at my notes on Dual Resonance Models, and it looks like by introducing isospin a la Chan and Paton, the isospin multiplicity appears to be correlated with signature, and exhibit exchange degeneracy.

Sorry about replying to a 1-year old topic.


Earlier this summer while visiting the Caltech campus, I had the wonderful opportunity to speak with S. Frautschi (an early pioneer of the application of Regge theory to particle physics). At some point, I asked him about this phenomenon. He kindly reminded me that the origin of signature is due to exchange forces in relativistic scattering that generate the left-hand cut in the complex $s$-plane, absent in potential scattering. He told me that his interpretation of the phenomenon of 'exchange degeneracy' is that the exchange forces are 'especially weak'. For example, in $\pi^+ \pi^-$ elastic scattering, the exchange channel is exotic, and hence subdominant. To date, this is the best answer I have heard as it provides a dynamical bases for the phenomenon.

  • $\begingroup$ That is nothing to be sorry about! I am glad that somebody remembers the question. $\endgroup$ – Ron Maimon Aug 20 '12 at 2:45
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    $\begingroup$ @RonMaimon I have updated my answer based on my conversations with Frautschi. $\endgroup$ – QuantumDot Jul 26 '13 at 22:45
  • $\begingroup$ Oh that's interesting! This might be because the glue is larger than the quark, so it's a large N thing. But I don't have intuition for the left-hand cut arising from exchange force, I'll need to think about it before accepting. Already upvoted, I'll get back to it at some point. Thank you. $\endgroup$ – Ron Maimon Jul 27 '13 at 0:17

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