# Explicit construction of states from $3\times \bar{3}=8+1$

Under $SU(3)$, $3\times \bar{3}=8+1$ i.e., the tensor product of the fundamental and the conjugate (to the fundamental) representation is reducible: 1 represents the singlet under $SU(3)$, and 8 represents the octet. Assuming the quarks belong to $\psi^i\in 3$ and anti-quarks $\phi_i\in \bar{3}$, there will be 9 combinations, in the tensor product, of the form $\psi^i\phi_j$. 8 of them transform into linear combinations of each other.

If one takes, $\Psi=\begin{pmatrix}u\\d\\s\end{pmatrix}$, then one can show that the anti-quarks transform as $\Psi^c=\begin{pmatrix}u^c\\d^c\\s^c\end{pmatrix}$. With this, one obtains 9 meson states.

How can I obtain the wavefunctions representing these meson states in terms of the elements of $\Psi$ and $\Psi^c$ and taking proper symmetrization into account?

• The singlet subspace is the trace in $\mathbf{3}\otimes\bar{\mathbf{3}}$. I'm not sure what more you might be searching for, that already gives an explicit recipe to split an element in that space into the singlet and octet parts. Commented Jan 17, 2017 at 13:52

Some about the $SU(3)$ algebra

First, note that $SU(3)$ group algebra has two Casimir operators in covering group, namely, $$C_{1}= \sum_{i}t_{i}t_{i}, \quad C_{2} = \sum_{i,j,k}d_{ijk}t_{i}t_{j}t_{k},$$ where $t_{i}, i = 1,...,8$ are Gell-Mann matrices and $d_{ijk}$ is defined as $$\{t_{i},t_{j}\} = \frac{1}{3}\delta_{ij}+d_{ijk}t_{k}$$ This means that each irreducible representation of the $SU(3)$ group is characterized by two numbers.

Let's define these numbers. Precisely, let's introduce new generators: $$\tag 0 I_{\pm}=t_{1}\pm it_{2}, \quad I_{3} = t_{3}, \quad V_{\pm} =t_4 \pm it_5, \quad U_{\pm} = t_6 \pm it_7, Y =\frac{2}{\sqrt{3}}t_8$$ Two generators $Y, I_{3}$ commute and thus have the same set of eigenstates (other combinations will be used below).

Let's write these eigenstates as $|m,y\rangle$ (Note that here and everywhere below I neglect the physical spin part of the state). We have $$I_{3}|m,y\rangle = m |m,y\rangle , \quad Y|m,y\rangle = y|m,y\rangle$$ Note that there is a linear relation between $I_{3}, Y$ called Gell-Mann-Nishijima formula: $$\tag 1 Q = I_{3} +\frac{Y}{2},$$ where $Q$ is electric charge, and also $$\tag 2 Y = B+S,$$

where $B$ is the baryon number and $S$ is the strangeness.

Constructing the quarks states

Let's now construct the quark states in the space of the generators $t$. Each quark has the baryon number $B=\frac{1}{3}$; u-quark has electric charge $Q=\frac{2}{3}$, while $s-,d-$quarks have $Q=-\frac{1}{3}$; finally, the strangeness of $s$-quark is $S=-1$, while the strangeness of $u-,d$-quarks are zero. By using $(1),(2)$ we have that $u$-quark has $Y = \frac{1}{3}, I_{3} = \frac{1}{2}$, $d$-quark has $Y=\frac{1}{3}, I_{3}=-\frac{1}{2}$, while $s$-quark has $Y=-\frac{2}{3}, I_{3}=0$ (note that in fact the quantum number of quarks like charge and the baryon number were historically determined by knowing the quantum numbers of mesons and other bounded states, see the answer).

By using explicit form of Gell-Mann matrices, You can construct corresponding states: $$u = \begin{pmatrix} 1\\ 0 \\ 0\end{pmatrix}, \quad d = \begin{pmatrix} 0\\ 1 \\ 0\end{pmatrix}, \quad u = \begin{pmatrix} 0\\ 0 \\ 1\end{pmatrix}$$ Completely analogically You can define the antiquarks states. Corresponding generators have the form $\bar{t}_{i}= -t_{i}^{*}$, so the numbers $m, y$ for them are with different sign.

Mesons states

Next, suppose the product $3\times \bar{3} = 8\oplus 1$. This product contains the states $u\bar{s}, s\bar{u}, s\bar{d}, d\bar{s},u\bar{d}, d\bar{u}$ with non-zero numbers $y,m$ and three linearly independent combinations $u\bar{u}, d\bar{d}, s\bar{s}$ with zero $m,y$. One of these three states is the $SU(3)$ singlet. Note that all of these nine states have zero baryon number. By calculating the quantum numbers (strangeness, charge, isospin) of first six states You can identify them as $K^{\pm}, K^{0}, \bar{K}^{0}, \pi^{\pm}$ mesons. The last three states are $\eta, \pi^{0},\eta'$.

Let's construct the states $\eta, \pi^{0},\eta'$. What do we know about them? For example, the facts that $\pi^{0}$ form the isospin triplet (together with $\pi^{\pm}$) and must have $I_{3} = 0$ (for the fixed value of $y$), while $\eta'$ is the $SU(3)$ singlet. This is enough for constructing of all states.

We know that $\pi^{-}$ is identified with the state $|I_{3} = -1, Y = 1\rangle$. By using algebra of Gell-Mann matrices we know that $I_{+}|m,y\rangle = \sqrt{2}|m+1,y\rangle$ (for the given representation $3$). Therefore the $\pi^{0}$ is can be identified with $\frac{1}{\sqrt{2}}I_{3}\pi^{-}$: $$\tag 3 \pi^{0} = \frac{1}{\sqrt{2}}I_{+}\pi^{-} \equiv \frac{1}{\sqrt{2}}(I_{+}^{3} + I_{+}^{\bar{3}})d\bar{u} = \frac{1}{\sqrt{2}}(u\bar{u} - d\bar{d})$$ As for $\eta'$, let's define it as $$\eta' = \alpha u\bar{u} + \beta d\bar{d} + \gamma s\bar{s}$$ It is the singlet, so action of all of lowering-uppering operators $U_{\pm}, V_{\pm}, I_{\pm}$ from $(1)$ on it must give zero. We thus obtain $$\tag 4 \eta' = \frac{1}{\sqrt{3}}(u\bar{u} + d\bar{d} + s\bar{s})$$ Finally, let's find $\eta^{0}$ by requiring its state to be orthogonal to $(3),(4)$. By using $\langle q_{i}| q_{j}\rangle = \langle \bar{q}_{i}|\bar{q}_{j}\rangle = \delta_{ij}$, one obtains $$\eta^{0} = \frac{1}{\sqrt{6}}(u\bar{u} + d\bar{d} - 2s\bar{s})$$