Let be $G$ a Lie Group and $\textbf{N}$ its complex representation.

It is known that any state $|\ ab\ \rangle\in \textbf{N}\otimes\textbf{N} = \oplus_I\textbf{r}_I$ may be decomposed through the Clebsch-Gordan decomposition, to wit $$ |ab\rangle = \sum_{I,i} C^{ab}_{Ii}|I,i\rangle \tag1 $$ where $I$ is a collective index for each irrep, and I am assuming there are no degenerate invariant subspaces in the decomposition.

I can also use a tensor notation instead of the bra-ket one. So I denote the single state $| a \rangle$ transforming under $\textbf{N}$ as $\pi^a$.

Can I write $$ \pi^a\pi^b = \sum_{I,i}\sum_{\phi} C^{ab}_{Ii}\phi^{Ii} \tag2 $$ where $\phi^{Ii}$ is a tensor which transforms under $\textbf{r}_I$? In this case, how can I identify the Clebsch-Gordan coefficients?

For example for SU(3), $\textbf{3}\otimes\textbf{3}=\textbf{6}+\bar{\textbf{3}}$ and the tensor decomposition reads $$ \pi^a\pi^b = \frac{1}{2}\left(\pi^a\pi^b + \pi^b\pi^a\right) + \frac{1}{2}\left( \pi^a\pi^b-\pi^b\pi^a \right) $$ How can I write this in the form indicated in $(2)$?

  • $\begingroup$ Somewhat related to 147243. $\endgroup$ Commented May 3, 2016 at 12:31

1 Answer 1


Well, you might have spared yourself confusion and grief by checking your peculiar language in SO(3), which any undergraduate is familiar with. Let me illustrate this for SO(3), before moving on to the much messier SU(3).

  • For SO(3), Kronecker-composing two vectors (spin 1, so 3 s) yields a spin 2 quintet (call it φ, so 5), a triplet (π) and a singlet (s), $$ \pi^a\pi^b = \frac{1}{2}\left(\pi^a\pi^b + \pi^b\pi^a+\frac{2(-1)^{ab}}{3}\delta^{a,-b} (-\pi^0 \pi^0+\pi^1\pi^{-1}+\pi^{-1}\pi^1) \right) + \frac{1}{2}\left( \pi^a\pi^b-\pi^b\pi^a \right) +\frac{(-1)^{ab}}{3} \delta^{a,-b} (\pi^0 \pi^0-\pi^1\pi^{-1}-\pi^{-1}\pi^1). $$ If your indices a and b represent $m_1$ and $m_2$ labels in the constituent vectors, so spherical instead of Cartesian tensors, you wish to express the r.h.side in terms of states labelled by the total J and M, as in your expression, with coefficients $C^{ab}_{JM}$. For example, $$ \pi^1 \pi^{-1}=\phi^0/\sqrt{6}+\pi^0/\sqrt{2}+s/\sqrt{3}= C_{20}^{1,-1}\phi^0 +C_{10}^{1,-1}\pi^0+C_{00}^{1,-1} s , $$ $$ \pi^0 \pi^{1}=\phi^1/\sqrt{2}-\pi^1/\sqrt{2}= C_{21}^{01}\phi^1 +C_{11}^{01}\pi^1, ~ ... $$ You know how to carry this out in angular momentum...

  • For SU(3), the decomposition of two triplets you wrote down, the SU(3) Clebsches, tabulated admirably in deSwart, require more labels for the complete set of commuting operators including the quadratic and cubic Casimir and isospin and hypercharge and 3rd component of isospin for each constituent quark, typically involving the total quadratic and cubic Casimir, isospin, 3rd-component of isospin and hypercharge. The tasteful classification, as you see in the WP article, opts for the total quantities and a further operator. So, in your correct unnumbered decomposition following your (2), the indices a and b have to have doublet values t,y, each, for the corresponding I3 and Y eigenvalues, as per the WP article cited.

To assist you, the quadratic Casimirs for the 3 and 6 are 4/3 and 10/3 , respectively; the cubic Casimirs are 10/9 and 35/9, respectively---and -10/9 for the antiquark, inevitably! The rest is drudgery, but it is your problem, after all.


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