# Why Are Even and Odd Regge Trajectories Degenerate?

+ 7 like - 0 dislike
936 views

The Gribov-Froissart projection treats even angular momentum differently from odd angular momentum. But in QCD, I believe that the odd trajectories interpolate the even trajectories--- 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.

recategorized Aug 28, 2014
You might also want to take a look at the closely related concept of parity doubling in hadron spectrum, see e.g. this review: http://arxiv.org/abs/arXiv:0704.1639

This post has been migrated from (A51.SE)

+ 7 like - 0 dislike

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.

Edit

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.

This post imported from StackExchange Physics at 2015-02-16 11:18 (UTC), posted by SE-user QuantumDot
answered Aug 20, 2012 by (195 points)
That is nothing to be sorry about! I am glad that somebody remembers the question.

This post imported from StackExchange Physics at 2015-02-16 11:18 (UTC), posted by SE-user Ron Maimon
@RonMaimon I have updated my answer based on my conversations with Frautschi.

This post imported from StackExchange Physics at 2015-02-16 11:18 (UTC), posted by SE-user QuantumDot
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.

This post imported from StackExchange Physics at 2015-02-16 11:18 (UTC), posted by SE-user Ron Maimon

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar\varnothing$sicsOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.