Background The fundamental representation of ${\rm SU(N)}$ is denoted by ${\bf N}$ and the conjugate of the fundamental is denoted by ${\bar{\bf N}}$. If the representation ${\bf N}$ is related to ${\bar{\bf N}}$ via a similarity transformation (i.e., equivalent), ${\bf N}$ is called a real representation. For example, the ${\bf 2}$ of ${\rm SU(2)}$ is a real representation while the ${\bf 3}$ of ${\rm SU(3)}$ is not.

Sub-context $1$ In the Standard Model, the left-handed lepton and quark fields belong to the ${\bf 2}$ of ${\rm SU(2)_L}$ and their antiparticle fields belong to ${\bar{\bf 2}}$.

Question $1$ What does the reality of ${\bf 2}$ tell us about the weak interaction?

Sub-context $2$ In the theory of strong interactions, quantum chromodynamics, the quark of a given flavor but three different colors belong to the ${\bf 3}$ of ${\rm SU(3)}$ and their antiparticle field belong to $\bar{{\bf 3}}$ which is not equivalent to ${\bf 3}$.

Question $2$ Also, how does the fact ${\bf 3}$ of ${\rm SU(3)}$, not being a real representation, affect the strong interaction of quarks?

  • $\begingroup$ Is your "real" and "conjugate " meaning connected with real and complex numbers? $\endgroup$ – anna v Sep 12 at 3:23
  • $\begingroup$ @annav en.wikipedia.org/wiki/… $\endgroup$ – probably_someone Sep 12 at 3:23
  • $\begingroup$ @probably_someone thanks $\endgroup$ – anna v Sep 12 at 3:26
  • A1 : It's not the quarks, it's the Higgs doublet (actually, you could invert everything and keep the action at the quarks, and not the Higgs, but this is too messy to teach to dyslexic audiences with no appeals to hermiticity).

Specifically, $(\phi_1, \phi_2)$ is a doublet, and so is its strictly equivalent conjugate, $(\phi_2^*, -\phi_1^*)$. As a result, if you pick your vacuum to be $\langle (\phi_1, \phi_2)\rangle =(0,v)$, then $\langle(\phi_2^*, -\phi_1^*)\rangle=(v,0)$. So you may dot either the Higgs doublet or its conjugate to a fermion weak doublet in your (independent!) $SU(2)_L$-invariant Yukawa couplings; and thereby give (independent) masses to both the lower and upper components of your fermion doublet, so both the d and u quarks! (A very good thing.)

An ancillary advantage of the equivalence is that the symmetric d - coefficient in the anticommutator of two generators vanishes for SU(2), so the anomaly coefficient of the doublet and the equivalent antidoublet amount to the same thing. ∴ There are no "unmixed" SU(2)$^3$ chiral anomalies in the SM, which would otherwise invalidate this gauge symmetry.

  • Parenthetically, an easy way to see reality for N=2 is to note that the one box Young tableau is the same as the one column of height N-1 one, uniquely for this value of N.

  • A2 : For SU(3), by contrast, there are chiral anomalies, but QCD is a vectorlike theory (non-chiral couplings), consisting of equal mixtures of left and right fermions, with opposite anomaly coefficients, so they cancel. ∴ QCD is anomaly free as well--phew!

I don't know of any other advantages or disadvantages associated with the inequivalence of conjugate representations. Of course, hadron spectroscopy is dominated by the difference in these two components, and is very different than a putative spectroscopy of e.g., SU(2) or SU(7) of color. The questions are a bit too broad to address...

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