Why do we need two representations of SU(3)

[Jelly-bean]

TL;DR

if we use up, down and staring quarks and their own antiparticle we can create the Eightfold way and understand mesons by the hyper charge and isospin projections.

Summary: if we use up, down and staring quarks and their own antiparticle we can create the Eightfold way and understand mesons by the hyper charge and isospin projections.

I don't understand how the conjugate representation of SU(3) allows us to create a vector space of dimension 3, while SU(3) by itself cannot.

 

[Vanadium 50]

You have three quarks, each in a triplet representation. 3 x 3 x 3 = 3 x (6 x 3bar) = (3 x 6) + (3 x 3bar) = 10 + 8 + 8 + 1. One of the octets is the eightfold way.

 

[strangerep]

staring quarks

Now there's a caption begging for a cartoon.

 

[Jelly-bean]

You have three quarks, each in a triplet representation. 3 x 3 x 3 = 3 x (6 x 3bar) = (3 x 6) + (3 x 3bar) = 10 + 8 + 8 + 1. One of the octets is the eightfold way.

sorry I don't think I get the second step. Where did the 6 come from?

 

[Vanadium 50]

Where did the 6 come from?

The algebra of SU(3). 3x3 has 9 elements, and they fall into the sextet and 3bar representations.

 

[samalkhaiat]

q^{a} \left( = u , d , s \right) \in \{ 3 \}, a 3-vector in the fundamental (or defining) representation space \{ 3 \} of \mbox{SU}(3). The 9-component tensor q^{a}q^{b} \in \{ 3 \} \otimes \{ 3 \} can be decomposed as follows q^{a}q^{b} = S^{ab} + A^{ab} , where S^{ab} = S^{ba} = \frac{1}{2} \left( q^{a}q^{b} + q^{b}q^{a}\right) , is symmetric (therefore 6-component) tensor, and A^{ab} = - A^{ba} = \frac{1}{2} \left( q^{a}q^{b} - q^{b}q^{a}\right), is anti-symmetric (i.e., 3-component) tensor. Since S^{ab} and A^{ab} don’t mix under \mbox{SU}(3) transformation, they must belong to different representation spaces: S^{ab} \in D^{6} \equiv \{ 6 \}, and A^{ab} \in D^{3} \equiv \{ \bar{3}\}. In order to complete the proof of \{ 3 \} \otimes \{ 3 \} = \{ 6 \} \oplus \{ \bar{3} \}, you need to know the answers to the following 2 questions: The tensor A^{ab} has only 3 independent components, i.e., it belongs to a 3-dimensional vector space, which we denoted by D^{3}. So, 1) why did we identify D^{3} with the conjugate representation space \{ \bar{3} \} and not with the space \{ 3 \}? In other words, where did the bar on the 3 come from? And 2) how do we know that D^{6} \equiv \{ 6 \} is an irreducible space?

 

[vanhees71]

From a physics point of view we need it, e.g., for QCD's color-charge space, i.e., we want to have antiquarks carrying the "anticolors" of the quarks' colors.

 

[DarMM]

Try reading the Appendix on group theory in A. Zee's "Quantum Field Theory in a Nutshell" @Jelly-bean . I know a good few people who found it helpful for clearing up things like why $3\otimes 3 = 6 \oplus \bar{3}$. It's basically about two things:

1. Symmetric and Antisymmetric tensors
2. The Levi-Cevita symbol converting between upper and lower indicies

 

[vanhees71]

Another good source is the old classic by Lipkin: "Lie groups for pedestrians".