Past exam of the particle physics course of the University of Oxford / 2022 / 3 / b

For quarks in an $N_c$-dimensional internal space $V$, complete antisymmetrization of three labels corresponds to a Young diagram with one column of three boxes, the partition $(1,1,1)$. Its representation is the exterior power ${\bigwedge }^{3}V$ and has dimensionFor $N_c=2$, the column is too tall: ${\bigwedge }^3{\mathbb{C}}^2=0$. With only two labels at least two of the three coincide, and the Young tableau column antisymmetrizer annihilates the state. For $N_c=3$, the filled column $r,g,b$ spans a one-dimensional determinant representation. It is a colour singlet of SU(3) because every group element has determinant one.

Equivalently, the tensor product of group representations decomposes asThe two mixed-symmetry terms account for $16$ dimensions, so the total is $10+8+8+1=27$. Antisymmetry alone establishes at least three internal states. Exactly three follows in the usual special unitary group model when the observed three-quark baryon must be a colour singlet: for $N_c>3$ the antisymmetric third exterior power has dimension greater than one and is not a singlet. This extra assumption should be made explicit.