More than one 15 dimensional irrep of $mathfrak{su}(3)$

Question

The irreps of (complexified) $mathfrak{su}(n)$ are labelled by highest weight Dynkin labels $(a_1, ldots, a_{n-1})$, and are often referred to simply by their dimension, e.g. $mathbf{3}$ to label $(1, 0)$ of $mathfrak{su}(3)$. Now consider the $(2, 1)$ and $(4, 0)$ irreps of $mathfrak{su}(3)$: both have dimension 15, but they are not dual (the dual of $(2, 1)$ is $(1, 2)$). Is there a standard way in the physicists notation to distinguish such irreps? If not, then is there ever any (physical) need to do so?

Cosmas Zachos · Accepted Answer

? Of course physicists use the standard Dynkin indices, as you confirm by using your Slansky 1981, Table 23, p 92.
Even though your two irreps have the same
$$
d(p,q)= tfrac{1}{2} (p+1)(q+1)(p+q+2)  ~~~~leadsto ~~~ 15,
$$
Their quadratic Casimirs entering in the β function of QCD are different, eigenvalues
$$
(p^2+q^2+3p+3q+pq)/3  ~~~~leadsto ~~~  28/3  leftrightarrow  16/3,
$$
respectively; and the cubic Casimir eigenvalues, rescaled to the respective anomaly coefficients, are
$$
(p-q)(3+p+2q)(3+q+2p)/18  ~~~~leadsto ~~~  154/9   leftrightarrow 28/9 ,
$$
so they'd contribute differently to anomalies.

ZeroTheHero · Answer

The irreps are really different, as illustrated by branching rules to subgroup and other relevant quantum numbers.
For instance the $(4,0)$ contains angular momenta $L=4,2,0$ but the $(2,1)$ contains $L=3,2,1$.  The branch to $mathfrak{su}(2)$ irreps is also different.
The irrep $(4,0)$ does not have weight multiplicities but the irrep $(2,1)$ has three weights occurring twice.  Thus the physical contents are certainly different.

More than one 15 dimensional irrep of $mathfrak{su}(3)$

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