Transformation of a scalar field

[Shen712]

I read somewhere that, suppose a scalar field Σ transforms as doublet under both SU(2)_L and SU(2)_R, its general rotation is

δΣ = iε^a_RT^aΣ - iε^a_LΣT^a.

where ε^a_R and ε^a_L are infinitesimal parameters, and T^a are SU(2) generators.

I don't quite understand this. First, why does the first term have positive sign but the second term has a negative sign? Second, why is Σ after T^a in the first term, while Σ is before T^a in the second term?

 

[ChrisVer]

If the scalar field transforms in the (2,2) representation of the 2 SU(2)'s, then there will be a term in the lagrangian of the form:

\bar{\psi}_L \Sigma \psi_R
where the psiL lives in (1,2) and the psiR lives in (2,1)
Transforming that term:
\bar{\psi}_L e^{ iT_L^a \theta^a} \Sigma' e^{-iT_R^b \omega^b} \psi_R
Demanding gauge invariance:
\Sigma = e^{ iT_L^a \theta^a} \Sigma' e^{-iT_R^b \omega^b}
\Sigma = (1 + i T_L^a \theta^a) \Sigma' (1 - iT_R^b \omega^b)= \Sigma' + iT_L^a \theta^a \Sigma' - i \Sigma' T_R^b \omega^b
\delta \Sigma = \Sigma'-\Sigma = - iT_L^a \theta^a \Sigma + i \Sigma T_R^b \omega^b

This might need a review, eg signs, but I guess the main idea is that (as it also was for the SM) and that there is a sign difference (now how it goes depends on how you define the transfs).
What's your reference?

 

[CAF123]

What about a term $\bar{\psi}_R \Sigma \psi_L$?

 

[ChrisVer]

What about a term $\bar{\psi}_R \Sigma \psi_L$?

It is the same. As I said, the overall signs can change depending on how you define the transformation, but the relative sign will still be the same.
Of course I would prefer writing that term with a sigma-dagger as that's a c.c. term in the Lagrangian?