Stupid question.

Consider a global SU(N) theory spontaneously broken. I want to write the EFT of the Goldstone bosons in terms of the field

$$ \Pi = e^{i\pi^a T^a} $$

where $T^a$ are the SU(N) generators normalized such that $\text{Tr}\left[T^a T^b\right]=1/2\delta^{ab}$. At two-derivatives order in the EFT expansion, the following term is for sure allowed

$$ \mathcal{L}_\pi = -\frac{f_\pi^4}{2}\text{Tr}\left[\partial_\mu \Pi\partial^\mu \Pi^\dagger\right] $$ This term gives the kinetic term plus pion self-interactions.

However, I can build another invariant term which does not contribute to the kinetic term but just give corrections to the self-interactions $$ \text{Tr}\left[\Pi^\dagger\partial_\mu\Pi\right]\text{Tr}\left[\partial_\mu\Pi^\dagger \Pi \right] $$

Notice that this term is of two-order in derivatives and four-order in field insertions.

It seems to me that this term is not considered in literature. Why? Is it zero? Is it a redundant operator?


For simplicity, I will denote $\hat{\pi} = \pi^a T^a$.

The invariant trace term is zero. Indeed

$$ \partial_\mu \Pi \cdot \Pi^\dagger = \left(i\partial_\mu \hat {\pi}\right)\Pi\cdot \Pi^\dagger = \left(i\partial_\mu \hat {\pi}\right) $$

Then you get $\text{Tr}\left[\partial_\mu \Pi \cdot \Pi^\dagger\right] = i\,\text{Tr}\left[\partial_\mu \hat {\pi}\right]=0 $ because the generators are traceless.


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