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I am very lost in this topic. I understand that there are $3\times 3$ possible combinations of a quark and an anti-quark, but why should one decide arbitrarily (that's how it appears to me) that one of these combination is a singlet, and so the rest is a octet? Comparing with coupling of two $\frac{1}{2}$ spins, I understand that a singlet is the "group" of state which satisfy a certain constraint (for example $S=0$ in the $\frac{1}{2}$ spin coupling case). Here with mesons what's the constraint?

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up vote 3 down vote accepted

The constraint on the singlet state $\eta' = \frac{1}{\sqrt{3}}(u\bar u+d\bar d+s\bar s)$ is that it be flavorless, just as the constraint on the singlet state for SU(2) is that it have no angular momentum.

To decompose $3\times 3^* = 8 + 1$ you first need to know what are the irreps. You can build these with raising and lowering operators as is done with SU(2).

If you are familiar with the graphical way of building and decomposing reps for SU(2) you will find there is an analogous method for SU(3). Basically, the reps look like triangles and hexagons and there is a way to multiply them graphically. (Of course, you can decompose them with raising and lowering operators as well, but you will find this tedious for SU(3).) This method allows you to calculate with pictures and very quickly that, for example, $3\times 3^* = 8 + 1$ and $3^*\times 3^* = 6^* + 3$. Of course, at some point you will probably want to learn Young-tableau.

I will avoid writing further details and drawing pictures and instead provide you with a reference that will make this all clear at the level you are looking for. See Ch 4, section 2.

Ta-Pei Cheng and Ling-Fong Li. Gauge Theory of Elementary Particle Physics

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1) The Lie group $G$ behind the meson octet is the $SU(3)_{\rm Flavor}$ Lie group over the three lightest quark flavors $u$, $d$, and $s$. More precisely, the fundamental $SU(3)$ representation $$V~=~\text{Fund} ~=~{\bf 3}$$ is a linear span of the $u$, $d$, and $s$ quarks. The three complex coefficients are collected in a $3\times 1$ column vector $v\in V$.

2) How does the $G=SU(3)_{\rm Flavor}$ act on $V$? It acts from left by multiplying $v\in V$ with a $3\times3$ special unitary group matrix $g\in G$, which results in a new column vector $v'=gv\in V$. Similarly, $G$ acts on the complex conjugated representation $\bar{V}$ as $\bar{v}'=\bar{g}\,\bar{v}$. In particular, if we write $\bar{v}$ as a $1\times 3$ row vector $v^{\dagger}$, the group acts as $v^{\dagger \prime}=v^{\dagger}g^{\dagger}=v^{\dagger}g^{-1}$.

3) Lie group representation theory explains how the meson nonet decomposes in irreps,

$$ {\bf 3} \otimes \overline{\bf 3}~=~V \otimes \bar{V}~=~\text{Fund} \otimes \overline{\text{Fund}}~\cong~ \text{Adj}\oplus \text{Sing}~=~ {\bf 8}\oplus {\bf 1}. $$

4) We may identify

$$V \otimes \bar{V}~\cong~V \otimes V^\dagger ~\cong~{\rm Mat}_{3 \times 3}(\mathbb{C}).\qquad\qquad (*)$$

Notice that $G$ acts on both spaces $V \otimes V^\dagger$ and ${\rm Mat}_{3 \times 3}(\mathbb{C})$ by similarity transformations.

5) The $SU(3)$ singlet is the eta prime meson

$$\eta'~=~\frac{u\otimes \bar{u}+ d\otimes \bar{d}+ s\otimes \bar{s}}{\sqrt{3}}.$$

Under the identification (*), the $\eta'$ corresponds to a $3 \times 3$ matrix proportional to the ${\bf1}_{3 \times 3}$ unit matrix. How do we know this is the unique flavor singlet? On one hand, the singlet is characterized by being invariant under the group action, i.e. similarity transformations. On the other hand, we know from Schur Lemma, that the only matrices that are invariant under all similarity transformations are the ones proportional to the unit matrix. Equivalently, in terms of the corresponding Lie algebra $su(3)$, the only matrices that commute with all the Lie algebra generators are the ones proportional to the unit matrix. This latter condition may be viewed as the constraint that OP is requesting.

6) Finally let us mention, that the $SU(3)_{\rm Flavor}$ symmetry is only an approximate symmetry in the standard model, as is evident from the different meson masses. The $SU(3)_{\rm Flavor}$ symmetry can be decomposed via branching rules in (strong) $SU(2)$ isospin symmetry, see e.g. chapter 10 of 't Hooft's lecture notes. The pdf file is available here.

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It is not arbitrary, but a result of the representation theory of $SU(3)$. Quark colors form a vector space $C^3$, and a quark antiquark-pair gives a tensor product $C^3 \otimes (C^3)^* \sim C^{3\times 3}$, hence is represented by 3 by 3 matrices, on which $SU(3)$ acts by conjugation. This is a 9-dimensional vector space which, as a representation space of $SU(3)$, is reducible.

Indeed, the space of 3 by 3 matrices is a direct sum of the 1-dimensional space of multiples of the identity, on which $SU(3)$ acts trivially, and the 8-dimensional space of matrices of trace zero, on which $SU(3)$ acts (as conjugation preserves tracelessness). It is not very difficult to show that this representation is in fact irreducibly.

Thus the decomposition $3 \otimes 3^* = 1 + 8$ into a singlet and an octet is not arbitrary, but determined by the properties of $SU(3)$.

It is a complete analogue of the decomposition $2 \otimes 2^* = 1 + 3$ for $SU(2)$ (i.e., spin).

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Both the replies defer the real answer to Lie group representation, fact which I already knew. Isn't there an immediate calculation as in the case of coupling two $\frac{1}{2}$ spins (the constraint thing)? That would help me to understand at least intuitively the problem, as it's not my aim to have a deep insight of the Lie group theory. By the way, the only one to blame here is my professor of nuclear physics, whose explanation of this subject (as well as all the others in the course) is... disarming. – Lorenzo Pistone Mar 13 '12 at 19:40
@LorenzoPistone have a look at the history of how these symmetries were observed: . At the time there was a plethora of particles all over the place and the organization into representations of a group seemed marvelous. When the omega minus, predicted by the symmetries, was discovered, symmetries took off. It is the strangeness quantum number that gives the y axis and the isospin the x in these plots. – anna v Mar 14 '12 at 4:53
@LorenzoPistone: I had given everything there is to calculate related to this question. There is nothing deep in the concepts I have been using. You said that 9=8+1 seemed arbitrary, and the above is the simple (and the only) answer to why it is not so. The Lie group terminology I used is very elementary, and is the very least you need to know if you want to understand anything about quarks beyond popular literature. – Arnold Neumaier Mar 16 '12 at 21:24

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