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I need to write the most general renormalized lagrangian under those conditions:

  1. The symmetry is global $SU(3)_G$

  2. There is 3 triplets of scalar field $\phi(3)_i$ , $i=1,2,3$

  3. There are no fermion fields.

I use Einstein`s notation $\mathcal L=(\partial_{\mu}\phi_i)^2-m_i^2\phi^{\dagger}_{i}\phi_i-\eta(some- combination -of -three -fields)-\lambda_i( \phi^{\dagger}_{i}\phi_i)^2$

So as you can see in the lagrangian I don`t know how to build the trilinear term - I know it must be a singlet, but how do I create singlet out of 3 triplets? The second question is does there is more quadratic and quartic elements exist in the lagrangian then of how I wrote this?

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    $\begingroup$ Presumably in 4D? $\endgroup$
    – Qmechanic
    Aug 7, 2019 at 10:13
  • $\begingroup$ Of course due to Lorentz invariance and Renormalizability $\endgroup$ Aug 7, 2019 at 10:16
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    $\begingroup$ Is there any symmetry acting on the $i$ index? Do you consider redefinitions of the $\phi_i$? What about $\phi_1^\dagger \phi_2$? $\endgroup$
    – Toffomat
    Aug 7, 2019 at 10:30
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    $\begingroup$ That's just it: $3\times 3 \times 3$ contains a singlet, but it is antisymmetric, so it's good for fermions but not scalars (unless you give those a flavor permitting antisymmetrization...). $\endgroup$ Aug 7, 2019 at 14:12
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    $\begingroup$ @CosmasZachos I don't see the ambiguity here, doesn't the parenthesis in $\phi (3)$ imply that the scalar fields sits in the triplet representation and he says there are 3 triplets which means the system he is referring to has, as you mention $3\times 3\times 2$ dof. $\endgroup$
    – Sanjay M
    Aug 8, 2019 at 19:45

1 Answer 1

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Yeah , it is a bit tricky to find the third order term. Essentially, it is pretty clear that we need use the epsilon tensor. Once you figure that out , it might seem that the antisymmetric tensor will give a zero term as mentioned by Cosmos Zachos , but we need to note that we have three different flavours in the model. Hence, we could have a term of the form , $$\epsilon^{abc}\phi_{1a}\phi_{2b}\phi_{3c}\sim \det[\Phi]$$ where the indices $a,b,c$ are the $SU(3)$ indices.

Note that this could only be done since we have at least 3 flavours. Also, if you want to give the most general renormalizable Lagrangian, you should in general allow varied coupling constants instead of a single $\lambda$ and also varied masses for the three flavours

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  • $\begingroup$ What is $\Phi$ ? $\endgroup$ Aug 9, 2019 at 8:52
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    $\begingroup$ It's the matrix with the three columns (or rows) being $\phi_1,\phi_2,\phi_3$ $\endgroup$
    – Sanjay M
    Aug 9, 2019 at 12:25

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