For two triplet angular momenta states, say $J=1$ and $I=1$, if we wanna look at it in the coupled basis $F=I+J$, we use the regular Angular Momentum rules:

$$|I-J|\leq F\leq I+J,$$

and from that you get

$$|0\rangle,|1\rangle,|2\rangle,$$

which are $1\oplus 3\oplus 5$, because if we project this on $m$ (magnetic number) space

$$|0\rangle \rightarrow \{|0,0\rangle\},$$ $$|1\rangle \rightarrow\{|1,-1\rangle\,|1,0\rangle\,|1,1\rangle\},$$ $$|2\rangle \rightarrow\{|2,-2\rangle,|2,-1\rangle,|2,0\rangle,|2,1\rangle,|2,2\rangle\}.$$

So we can write

$$3\otimes3=1\oplus 3\oplus 5.$$

However, what we see in QCD when talking about quark color-mixing is $$3\otimes3=1\oplus 8.$$

Why is this the case in QCD? Why is it different? Don't we just use the same coupling? I thought people still use the same Clebsch-Gordan coefficients with exactly the same procedure.

Thanks for any efforts.

For two triplet angular momenta states, say $J=1$ and $I=1$, if we wanna look at it in the coupled basis $F=I+J$, we use the regular Angular Momentum rules:

$$|I-J|\leq F\leq I+J,$$

and from that you get

$$|0\rangle,|1\rangle,|2\rangle,$$

which are $1\oplus 3\oplus 5$, because if we project this on $m$ (magnetic number) space

$$|0\rangle \rightarrow \{|0,0\rangle\},$$ $$|1\rangle \rightarrow\{|1,-1\rangle\,|1,0\rangle\,|1,1\rangle\},$$ $$|2\rangle \rightarrow\{|2,-2\rangle,|2,-1\rangle,|2,0\rangle,|2,1\rangle,|2,2\rangle\}.$$

So we can write

$$3\otimes3=1\oplus 3\oplus 5.$$

However, what we see in QCD when talking about quark color-mixing is $$3\otimes3=1\oplus 8.$$

Why is this the case in QCD? Why is it different? Don't we just use the same coupling? I thought people still use the same Clebsch-Gordan coefficients with exactly the same procedure.

Thanks for any efforts.

EDIT: Michio Kaku's book (Modern intro to QFT) says that there's no equivalence between SU(2) and SU(3)... so the question is still open, how can you use Clebsch-Gordan coefficients there?