# Second baryon octet

Let's temporarily ignore spin. If 3 denotes the standard representation of SU(3), 1 the trivial rep, 8 the adjoint rep and 10 the symmetric cube then it's well-known that

3 x 3 x 3 = 1 + 8 + 8 + 10

Interpreting the 3's as the space of up/down/strange flavour states for a quark, the tensor cube is interpreted as the space of baryon states that can be obtained by combining three light quarks. Obviously spin matters, but at least this should give a classification of baryons modulo spin into SU(3)-multiplets.

There are two octets here, but in the literature I have only seen one of them described. What is the second octet?

I appreciate that the answer may be "it's more complicated than that".

-

## 3 Answers

I think I have worked out my confusion so I thought I should post it as an answer. The original question was not well-posed; hopefully this will help anyone else who has similar misunderstandings.

Keeping spin in the picture, the space of states for an individual quark is the tensor product of the three-dimensional flavour space with the two-dimensional spin representation of SU(2). However, this is not considered as a representation of SU(3) x SU(2), rather as a representation of SU(6) containing this product as a subgroup. In other words you can rotate flavours into spins and vice versa. As a representation of SU(6), the tensor cube of this standard 6-dimensional representation decomposes into pieces, one of which is the symmetric cube. This is an irreducible 56-dimensional representation. This decomposes under the subgroup SU(3) x SU(2) into a direct sum of two pieces: one is the SU(3)-decuplet tensored with the (4-dimensional) spin-3/2 representation of SU(2), one is the adjoint representation of SU(3) (the octet) tensored with the spin-1/2 representation of SU(2) (and indeed 10 x 4 + 8 x 2 = 56).

When you forget about spin, the spin-1/2 octet that appears here is actually a mixture of terms from the two SU(3)-octets in the decomposition of the tensor cube from the original question. In other words, the actual wavefunction of a proton is a sum of two terms, one involving terms from one SU(3)-octet and one involving terms from the other (both tensored with suitable spin wavefunctions).

I found these notes of Jiří Chýla very helpful in sorting out my misunderstanding:

http://www-hep2.fzu.cz/Theory/notes/text.pdf

-

I thought that, perhaps, the solution of the problem would lie in the fact that the two octet representations in which the tensor product 3x3x3 of SU(3) splits are two equivalent representations. That's right what Chyla says in his notes. So one representation can be obtained from the other performing a rotation in flavour space, and the two octets are physically indistinguishable

-

I think that the answer is that there is a flavor symmetric octet representation and a flavor antisymmetric octet representatio, while the decuplet is totally symmetric. Therefore, when you consider the spin and flavor wavefunction of a baryon for an octet baryon you have: $\chi(spin)\cdot\phi(flavor)=\frac{1}{\sqrt{2}}(\chi^{1/2}_s\cdot 8_s+\chi^{1/2}_a\cdot 8_a)$. where "s" denotes symmetric and "a" antisymmetric. 1/2 superscript denotes a 1/2 spin particle.

-