# How to write the Clebsch-Gordan decomposition in tensor notation

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 $(2)$?

• Somewhat related to 147243. Commented May 3, 2016 at 12:31

• 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...