3
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.