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Okay, I think I have an answer but I would interested to hear some feedback on this. I initially thought that by writing the left and right handed fields in 2 component notation ($q_L $ and $q_R^c$), they were 'equivalent' objects and can freely rotated between one another. However, I realized now this is not the case. The left and right handed fields are ...


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From your comments, it sounds like you're looking for Deconfinement. At higher energies the quarks and antiquarks are no longer bound, they are asymptotically free, and thus chiral symmetry is restored.


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\begin{equation} \boldsymbol{3}\boldsymbol{\otimes}\boldsymbol{3}\boldsymbol{\otimes}\boldsymbol{3}= \boldsymbol{1}\boldsymbol{\oplus}\boldsymbol{10}\boldsymbol{\oplus} \boldsymbol{8}^{\boldsymbol{\prime}}\boldsymbol{\oplus}\boldsymbol{8} \end{equation} We talk about this because it explains the structure of a number of baryons in Particle Physics made ...


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I don't think this assumption is legit: Assume the you initially have two quarks (up and anti-up for instance) that are placed far outside of causal contact with each other for a real experimental setup. Indeed, what would have been the previous history of those two scorrelated (outside of each other's light cone) quarks, to be produced isolated? ...


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First Question You are correct that we have "lost" two degrees of freedom in defining $\epsilon^+$ and $\epsilon^-$ as above. This because they are just choices of basis vectors. In QFT we usually just work with some simple basis of polarization vectors, Indeed they'll be summed/averaged over anyway when calculating the cross-section. The two "lost" ...


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These are some nice answers! Wanted to add that, as you know, how strongly the quarks couple to each other (or interact with each other) is momentum-dependent. So within the nucleons (protons and neutrons) quark coupling is very strong (which is why the quarks are confined in the nucleons). Because the interquark interaction is so strong at these ...


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Great question! My answer would be that in order to get a bound state, we need to have a potential that is deeper than the kinetic energy the two particles have. We have a better chance of getting a potential of the right depth to bind two nucleons if: (1) They don't repel charge-wise. Compared to nucleon interaction the Coulomb force isn't that strong, ...



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