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Let's discuss about $SU(3)$. I understand that the most important representations (relevant to physics) are the defining and the adjoint. In the defining representation of $SU(3)$; namely $\mathbf{3}$, the Gell-Mann matrices are used to represent the generators $$ \left[T^{A}\right]_{ij} = \dfrac{1}{2}\lambda^{A}, $$ where $T^A$ are the generators and $\lambda^A$ the Gell-Mann matrices. In adjoint representation, on the other hand, an $\mathbf{8}$, the generators are represented by matrices according to $$ \left[ T_{i} \right]_{jk} = -if_{ijk}, $$ where $f_{ijk}$ are the structure constants.

My question is this, how can one represent the generators in the $\mathbf{10}$ of $SU(3)$, which corresponds to a symmetric tensor with 3 upper or lower indices (or for that matter how to represent the $\mathbf{6}$ with two symmetric indices). What is the general procedure to represent the generators in an arbitrary representation?

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  • $\begingroup$ The simplest and most direct way that I know of uses the canonical technique of highest weight vectors, and root operators (the generalization of ladder operators beyond SO(3)). I'm a little short on time at the moment, but I would suggest giving answer based on this approach. Like SO(3), the matrix realizations obtained this way may not be the most physically natural choice, but they are certainly the simplest (analogous to using helicity instead of coordinate components of a polarization). $\endgroup$ – TotallyRhombus Mar 4 '16 at 17:20
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The generic, abstract answer to your question is in Coleman's excellent book. However, I have not seen these 10×10 explicit matrices around, as they are not exactly low-lying fruit. I have not stumbled upon them, but it may not be impossible they are in somebody's thesis.

In principle, you might Kronecker-multiply 3 triplets, and shed off the two octets and the singlet in the Clebsch-Gordan decomposition, ${\bf 3}\otimes{\bf 3}\otimes{\bf 3}={\bf 10} \oplus {\bf 8} \oplus {\bf 8} \oplus{\bf 1}$, having worked out the Casimirs of all of those, as in the WP article which has a somewhat better basis than Gell-Mann's. The 10 is the D(3,0) in that notation, and both its quadratic and cubic Casimirs are equal to 6. In practice, it might take you about 1-2 days to work them out.

Alternatively, you might slug through the WP article and tease them out in a baryon decuplet basis, where the 10-vector they act on is the obvious $(\Delta^{++},\Delta^{+},\Delta^{0},\Delta^{-}, \Sigma^{* +},\Sigma^{* 0},\Sigma^{* -}, \Xi^{* 0},\Xi^{* -}, \Omega^- )$. The normalized Cartan subalgebra generators for $I_3$ and hypercharge (B+S), then, are the predictable diagonal ones, $F_3$=diag(3/2,1/2,-1/2,-3/2, 1,0,-1, 1/2,-1/2, 0)/$\sqrt{15}$ and $F_8$=diag(1,1,1,1, 0,0,0, -1,-1, -2)/$\sqrt{20}$, etc. Try $U_+= F_6+iF_7$ first!

Edit: Indeed, Richard Shurtleff arXiv:0908.3864 has produced a mathematica notebook which produces these generators and checks their algebra.

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