What is a singlet under SU(3)?

[jinbaw]

what is a singlet under SU(3)?

 

[jinbaw]

and what is SU(3)xSU(3)?
What do we mean by 3 x 3(bar) = 8 + 1?

 

[fzero]

A singlet of any algebra is the state that has zero eigenvalue under the Casimir operators. These are the generators of the center of the algebra and are simultaneously diagonalizable. The singlet state is a one-dimensional representation. For instance, for SU(2), the singlet state satisfies

\vec{S}^2 |0\rangle = 0,

i.e., it is the s=0 state. It satisfies

S_{\pm} |0\rangle = 0

as well, so there are no other states in this representation.

SU(3)\times SU(3) refers to a direct product of two SU(3) groups. States or particles will simultaneously be in some representation of both groups.

\mathbf{3} \otimes \bar{\mathbf{3}} = \mathbf{1} \oplus \mathbf{8} refers to a direct product of the \mathbf{3} representation with the \bar{\mathbf{3}}. The resulting representation is what's known as reducible and can be written as a direct sum of the singlet \mathbf{1} and octet \mathbf{8} representations.

This relationship is analogous to the way we decompose two-particle spin states in QM. There, for SU(2), the decomposition of specific product states into states of irreducible representations is summed up by the Clebsch-Gordan coefficients.

 

[jinbaw]

Okay, that explains a singlet. Thanks.
But what I still can't get is how we know that it is 1 + 8. For example i need to find what 3 x 3 x 3 is. How can I do that?

 

[fzero]

Okay, that explains a singlet. Thanks.
But what I still can't get is how we know that it is 1 + 8. For example i need to find what 3 x 3 x 3 is. How can I do that?

You might want to look up Young tableaux, which are very useful for computing products of irreps for unitary groups. They are a pictorial way of keeping track of the symmetries of irreducible tensor representations. Alternatively, there are various places to find tables of products.

In your case, we should first compute \mathbf{3}\otimes\mathbf{3}. Thinking of the fundamental irrep \mathbf{3} like a vector, this product is a rank 2 tensor. There is an antisymmetric product which is the \bar{\mathbf{3}} and the symmetric product which is the \mathbf{6}. So we have

\mathbf{3}\otimes\mathbf{3} = \bar{\mathbf{3}}\oplus \mathbf{6}.

Now we already know that \mathbf{3} \otimes \bar{\mathbf{3}} = \mathbf{1} \oplus \mathbf{8}, so it remains to compute \mathbf{3} \otimes \mathbf{6}. We can either antisymmetrize the new index with the indices on the symmetric tensor, which gives us the \mathbf{8}, or we can totally symmetrize the rank 3 tensor, which gives us the \mathbf{10}. Therefore

\mathbf{3} \otimes \mathbf{6} = \mathbf{8}\oplus \mathbf{10}.

Putting it all together, we have

\mathbf{3}\otimes\mathbf{3}\otimes\mathbf{3} = \mathbf{1} \oplus \mathbf{8}\oplus\mathbf{8}\oplus \mathbf{10}.

As a sanity check, you can compare the total dimensions of the representations on both sides.