# Degeneracy in mass of $8$ and $27$ reps of $SU(3)$ in Coleman's Aspects of Symmetry

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In Coleman's Aspect of symmetry he proposes an amusing problem in the first chapter. It asks us to consider a set of eight pseudo-scalar fields transforming in the adjoint representation of $SU(3)$. We are asked to write down interactions at quartic order (ignore cubic terms) and show that:

1. The interaction Lagrangian is controlled by only one term. This is easy to do. ${\rm Tr}\left(\Phi^2\right)^2$ and ${\rm Tr}\left(\Phi^4\right)$ are the only things we can write down and they are equal to one another by the tracelessness of $\Phi$ due to its belonging to the adjoint rep.

2. Show only $27$, $8$, and $1$ are possible representations. This was easy too. The only other thing that comes out of $8\otimes 8= 27\oplus\overline{10}\oplus10\oplus8\oplus8\oplus1$ is $10$ which is anti-symmetric in its lower indices which is incompatible with the Bose statistics of the mesons if they are forming a bound state.

3. Show that $8$ and $27$ are necessarily degenerate in mass. This one has me stumped. Any help would be great I will post if I solve it.

This post imported from StackExchange Physics at 2016-11-24 23:29 (UTC), posted by SE-user Ryan Plestid
retagged Nov 24, 2016

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Perhaps the lack of an answer is due to your skipping the actual problem stated, formation of scalar bound states out of the eight adjoint pseudoscalars Φ furnished. The problem is less profound than you gave it credit for. Coleman simply wants to ensure the the power of the machinery is appreciated, instead of serving as mere abstract sudoku mumbo-jumbo.

It pays to review standard SU(3) group technology, e.g. in the classic paper by Macfarlane et al, CMP 11 (1968) 77-90 , and, in particular, the symmetric coefficient of the Gell-mann matrices, $d_{ijk}$, which is traceless, $d_{idk}=0$, and satisfies $$d_{ijk} d_{lak}= \frac{5}{3} \delta_{il}.$$

• The cubic term is absent to preserve parity.

• 1) Indeed, for $\Phi=\phi_i \lambda_i$, so, then, traceless, its characteristic polynomial in the Cayley-Hamilton theorem relation, multiplied by Φ and traced readily yields TrΦ⁴= (TrΦ²)²/2, so only one interaction term is present; let us write it as the trace of the square of an 8x8 matrix in the adjoint indices, now, $2(\phi_i \phi_j)^2$.

• 2) The only real, symmetric reps in 88 are, indeed, 1, 8, and 27, so, then, 8(8+1)/2=36=1+8+27.

• 3) Now break up this symmetric matrix into its irreducible components transforming separately, which will serve as the interpolating fields of the scalar mesons, so the square of the matrix represents their mass terms. The traceful part is the singlet, $$s_{ij}=\frac{\delta_{ij}}{8} \phi_k\phi_k,$$ the traceless symmetric octet is $$\sigma_{ij}= \frac{3}{5} d_{ijk} d_{klm}\phi_l \phi_m,$$ and the traceless 27-plet $$\rho_{ij}= \phi_i\phi_j -\sigma_{ij} -s_{ij}.$$

Now, these three symmetric matrices are trace-orthogonal to each other, $$s_{ij}s_{ji}= \phi\cdot\phi/8, \qquad s_{ij} \sigma_{ji}=0, \qquad s_{ij} \rho_{ji}=0 ;$$ $$\sigma_{ij} \rho_{ji}=0, \qquad \sigma_{ij} \sigma_{ji}= (3/5) (d_{ijk}\phi_j \phi_k)(d_{ilm}\phi_l \phi_m) ;$$ so the quartic invariant reduces to $$(\phi_i\phi_j)^2= \mathrm{Tr}(\rho + \sigma+s)^2= \mathrm{Tr}(\rho^2+\sigma^2+s^2),$$ so the masses are all the same, coming from a unique, restricted parent interaction.

The theme of that flavor SU(3) world was to find symmetry restrictions correlating masses and couplings, in the absence of dynamics, so as to establish regularities to be explained; and the exhilarating successes of that game led the way to respect for Lie groups and application to more triumphant ambits...

This post imported from StackExchange Physics at 2016-11-24 23:29 (UTC), posted by SE-user Cosmas Zachos
answered Nov 24, 2016 by (80 points)

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