Effective action: Cancellation of odd powers (Altland & Simons)

Question

Consider a system of interacting electrons. Using the path integral formalism, we introduce the Hubbard Stratonovich transformation to decouple the interaction in the density channel. Then, we integrate out the fermionic degrees of freedom and extremize the action. The new effective action involves a term of the form $$ \ln\left(-\hat G^{-1}\right) + \ln\left(1 - i\hat G\hat\phi\right)\,, $$ where $\hat G$ is the diagonal bare propagator for the electrons and $\hat\phi$ is the auxiliary field. The first term just gives the partition function for the non-interacting system. The trace of the second term can be expanded as $$ \mathrm{tr} \sum_n \frac{\left(-i\hat G\hat\phi\right)^n}{n}\,. $$ The first term is $$ \mathrm{tr}\left(\hat G\hat\phi\right) = \phi_0\sum_n G_n\,, $$ where $G_n$ is the diagonal element of the propagator matrix.

The second term is $$ \sum_q \phi_q \phi_{-q}\left(\sum_p G_p G_{p+q}\right)\,. $$ The stuff inside the parentheses is the RPA polarization bubble. So far so good.

The third term becomes $$ \sum_{kp}\phi_k\phi_p\phi_{-k-p}\left(\sum_qG_qG_{k+q}G_{q-p}\right)\,. $$ This is a triangular loop diagram. Intuitively, it seems that it should cancel. Odd powers just seem, well, odd. In fact, consulting Altland and Simons book (second edition, after Eq.(6.6)), it does say that the odd powers cancel by symmetry. However, I don't see it. Am I missing something obvious?

Thank you

Answer 1

Since I can't comment (not enough reputation), I'll attempt a quick answer.

Altland and Simons make several comments about symmetry cancellation in a variety of sections. Let us proceed by deduction, if nothing else. In the present example, either the auxiliary field or the Green's function components must be antisymmetric with respect to momentum; $G_{-k} = -G_k$ or $\phi_{-k} = -\phi_k$.

From what little I know of Green functions, they are positive definite, which is true in any basis (momentum or otherwise). Therefore the auxiliary field must be antisymmetric with respect to the momentum variable; perhaps one can derive this property from conservation of momentum.

This answer may be way off; if so, please correct me!

Answer 2

I think the reason is rooted in the interpretation of the fluctuation around mean-field.

Back to the Hubbard-Stratonovich transformation, if we substitute $\phi_q$ with $-\phi_q$, the physics should not change. From another perspective of view, we can say the positive flucuation around the mean-field $\hat \phi=0$ is equivalent to its negative counterpart.

From above we can say odd powers of $\phi$ (ignoring $\phi$'s indices) must vanish.