I'm reading some notes that say the reason that the strong interaction conserve isospin is because the up and down quarks have the same color... but I'm not very convinced. Is this a universal truth? Or does it just apply to e.g. baryons? Or is it that on average, there will be equal amounts of ups and downs with red, green and blue? Or is my understanding of color wrong altogether...?

  • $\begingroup$ Yes, up-quarks and down-quarks have the same color charges: they are what is called 'triplets of color' - each quark comes in 3 colors -. And when one says, e.g., the iso-spin of the up-quark is 1/2, it applies to all its colors. $\endgroup$
    – xi45
    May 12 '16 at 20:16
  • $\begingroup$ I get that, but doesn't that just say that "up and down quarks can have one of the three same colors", rather than "they have the same color"? Or am I just being too pedantic? $\endgroup$
    – hsnee
    May 12 '16 at 20:38
  • $\begingroup$ I deleted my answer as it became obvious I was a bit sleepy and couldn't devote enough time to ensure quality. The colour flow through the vertex handled by QCD rules as and @xi45 mentions will sort your isospin conservation as if coupled to the strong force there can not be a change in flavour. $\endgroup$ May 12 '16 at 23:16

The strong interaction works with colored gluons. The up and down quarks have the same type of color couplings--nothing about their color couplings can recognize whether they are up or down, they are R, G, or B, colorwise.

So you would expect the strong interactions to be indifferent to their flavor difference, up to their "slightly" differing masses (*). (They of course have different charges, so electromagnetic and weak interactions break strong isospin.)

So, yes, isospin symmetry is a "universal approximate truth" --- an "almost conserved quantum number in the strong interactions. The same type of symmetry, except less good (more approximate) holds for the strange quark, the celebrated origin of the SU(3) flavor symmetry of the strong interactions, which bind all hadrons together: baryons and mesons.

Now for the subtlety (*): The isospin symmetry would be perfect for the masses of the u and d quarks being equal, but they are evidently not: in fact, one of them is twice as heavy as the other! $$ m_d\sim 4.3–5.2 MeV, \qquad m_u \sim 1.8–2.8 MeV. $$

However, because both of these masses are much smaller than $v_{QCD}\sim 250 MeV$, the characteristic (condensation) energy scale of the strong interaction vacuum related to chiral symmetry breaking, they are comparably negligible w.r.t. that scale in this process: gluons can largely effectively regard either type of quark as massless.

As a consequence, they fetch up as almost degenerate constituent quarks, $$ M_d\sim 340 MeV, \qquad M_u\sim 336, $$ precisely because the nonperturbative QCD interactions largely ignored their small mass differences--and charge.

These are the additive building blocks of hadrons.

When you look at the isomultiplets in hadronic physics, 4MeV is the rough difference between the d and the u quark, so then, small potatoes. (A further factor affecting the symmetry of wavefunctions in baryon spectroscopy is also distinguishability, but this is not part of your question.)


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