Under $SU(3)$, $3\times \bar{3}=8+1$ i.e., the tensor product of the fundamental and the conjugate (to the fundamental) representation is reducible: 1 represents the singlet under $SU(3)$, and 8 represents the octet. Assuming the quarks belong to $\psi^i\in 3$ and anti-quarks $\phi_i\in \bar{3}$, there will be 9 combinations, in the tensor product, of the form $\psi^i\phi_j$. I understand that, eight of the nine states transform into linear combinations of each other and one state remains unaltered.

What is the physical significance of this decomposition? In particular, I want to understand whether the singlet state is really different from the states in the octet in terms of its physical properties or interactions. I want to understand what is/are the physical conclusions that can be drawn mathematically from this decomposition.

  • $\begingroup$ Color singlets can be directly observable hadron states, in QCD, while color octets must be confined ones, such as gluons. That is, you will not se a color octet propagate farther than a fermi. $\endgroup$ Jan 17 '17 at 13:40
  • $\begingroup$ The comment is about color SU(3), the only way to properly interpret the question. Veering off to flavor SU(3) and the topological features of the η' is spectacularly inviting danger... $\endgroup$ Jan 17 '17 at 15:46
  • $\begingroup$ Related: physics.stackexchange.com/q/22321/2451 and links therein. $\endgroup$
    – Qmechanic
    Apr 15 '18 at 14:15

Ok, this is an interpretation question, so I can try to say something, but take it with a grain of salt.

The representation decomposition produces a singlet, which is the trace of the bilinear you wrote, i.e. $\psi^i \phi_i$. The physical interpretation of this singlet depends on the meaning you give to your fields $\psi$ and $\phi$. If you believe that all your quantities must transform as irreducible representation of your local symmetry group and the interaction must be invariant under this group (gauge principle), then it means that in your theory there is a composite state of two quarks that is a scalar at "low energy". For low energy I mean a scale energy where you don't see quarks, but only their composite states.

In particular, for a bilinear state you will have an adjoint field and a scalar field. And, according to the gauge principle, you can construct an effective lagrangian with those ingredients inserting all the interactions terms allowed by gauge symmetry (e.g. Yukawa couplings, mass terms, etc.).

  • 1
    $\begingroup$ Quarks aren't seen at low energy, not high... $\endgroup$ Jan 17 '17 at 12:37
  • $\begingroup$ This answer could use a mention of confinement for the non-singlets, since OP is evidently thinking about QCD. $\endgroup$
    – ACuriousMind
    Jan 17 '17 at 13:50
  • $\begingroup$ @ACuriousMind I'm thinking about both $SU(3)_f$ and $SU(3)_c$. $\endgroup$
    – SRS
    Jan 17 '17 at 16:55

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