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In the book Group Theory in a Nutshell for Physicists by A. Zee p.339, the derivation of Gell-Mann-Okubo mass formula for meson is written.

It is described that in $SU(3)$, the equation

$$8 \otimes 8 = 27 \oplus 10 \oplus 10^{\ast} \oplus 8 \oplus 8 \oplus 1\tag{V.2.10}$$

holds, and the symmetric part and the antisymmetric part are given by

$$(8 \otimes 8)_{s} = 27 \oplus 8 \oplus 1\tag{V.4.3}$$

and

$$(8 \otimes 8)_{a} = 10 \oplus 10^{\ast} \oplus 8\tag{V.4.4}$$

respectively. Zee says that the effective Lagrangian is

$$\mathcal{L} = \frac{1}{2}tr\left((\partial_\mu \Phi)^2 - m_0^2\Phi^2 \right)\tag{V.4.2}$$

and since we are multiplying $\Phi$ by itself, the antisymmetric part $(8 \otimes 8)_{a}$ drops out. Then, it is concluded that there are two possible $SU(3)$-breaking terms, one is 27 and the other is 8.

My questions are following.

  1. Why is the symmetric part $(8 \otimes 8)_{s} = 27 \oplus 8 \oplus 1$? And, why is the antisymmetric part $(8 \otimes 8)_{a} = 10 \oplus 10^{\ast} \oplus 8$?

  2. Why does the antisymmetric part $(8 \otimes 8)_{a}$ drop out?

  3. Why isn't $1$ in the symmetric part $(8 \otimes 8)_{s} = 27 \oplus 8 \oplus 1$ in $SU(3)$-breaking terms?

Qmechanic
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konyonyo
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1 Answers1

3

  1. One of the shortest arguments is as follows:

    • $({\bf 8}\otimes {\bf 8})_s={\bf 36}$ and $({\bf 8}\otimes {\bf 8})_a={\bf 28}$.

    • Using the tensor method trick on p.216, we see that $\sum_{ij}\psi^i_j\chi^j_i$ [which corresponds to the irrep ${\bf 1}$] is symmetric under $\psi\leftrightarrow\chi$, so it belongs to the symmetric part $({\bf 8}\otimes {\bf 8})_s$.

    • Using the tensor method trick on p.216, we see that $T^{[ik]}_{\{jl\}}$ and $T^{\{ik\}}_{[jl]}$ [which correspond to the irreps $\overline{\bf 10}$ and ${\bf 10}$, respectively] are antisymmetric under $(i,j)\leftrightarrow (k,l)$, so they belong to the antisymmetric part $({\bf 8}\otimes {\bf 8})_a$.

    • Next counting dimensions, there is only one way to fit the rest of the irreps from eq. (V.2.10) into the symmetric and antisymmetric parts$^1$.

  2. On p.339 Zee merely assumes for simplicity that the $SU(3)_F$ symmetry-breaking term is made from $(\Phi\otimes\Phi)_s$, not $(\Phi\otimes\Phi)_a$, where the meson field $\Phi$ transforms in the adjoint irrep ${\bf 8}$ of $SU(3)_F$.

  3. The trivial irrep ${\bf 1}$ is an $SU(3)_F$ invariant, and can hence not serve as an $SU(3)_F$ symmetry-breaking term, cf. first paragraph on p. 340.

References:

  1. A. Zee, Group Theory in a Nutshell for Physicists, 2016; section V.2 p.316 + section V.4 p.339-340.

$^1$ The fusion rule (V.2.10) itself can be derived using a Wick-type theorem, explained in my Phys.SE answer here. The Wick-type calculus yields $$\begin{align} {\bf 8}\otimes {\bf 8}~=~& (1,1)\otimes (1,1)\cr\cr ~=~& :(1,1)\otimes (1,1):~\oplus~:(\overline{1,1)\otimes (1,1}):\cr\cr ~\oplus~& :(1,\overline{1)\otimes (1},1):~\oplus~ :(\overline{1,\overline{1)\otimes (1},1}):\cr\cr ~\oplus~& :(\underline{1,1)\otimes (1},1):~\oplus~ :(1,\underline{1)\otimes (1,1}):\cr\cr ~=~& (2,2)\oplus (1,1)\oplus (1,1)\oplus (0,0)\oplus (0,3)\oplus (3,0)\cr\cr ~=~& {\bf 27}\oplus {\bf 8}\oplus {\bf 8}\oplus {\bf 1}\oplus \overline{\bf 10}\oplus {\bf 10}, \end{align} \tag{V.2.10}$$ where a double antisymmetrization

$$\begin{matrix} :(1,&1)\otimes (1,&1):\cr |&---| &\cr &|---&|\end{matrix} \quad~=~\quad (1,1)~=~{\bf 8}$$

has been excluded.

Qmechanic
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