Fast and slow modes, and the vanishing of certain diagrams during re-normalization

Question

In the middle of pg. 452 of Atland and Simonss Condensed Matter Field Theory, they state the following:

Terms of $\mathcal{O}(\phi _{\text{s}}^3\phi _{\text{f}})$ do not arise because the addition of a fast momentum and three slow momenta is incompatible with momentum conservation.

My question is simply: why? Naively, it seems that three slow momenta could add to give one fast momentum.

I'll now try to give more details so that you do not need to check the book yourself. The context is $\phi ^4$ theory: $$ S(\phi ):=\int \mathrm{d}x\, \left[ \tfrac{1}{2}|\nabla \phi |^2+\tfrac{r}{2}|\phi |^2+\tfrac{\lambda}{4!}|\phi |^4\right] . $$ We have chosen a re-normalization scaling factor $b\geq 1$ and a momentum cut-off $\Lambda$ sufficiently large so that $\int _{|p|>\Lambda}\hat{\phi}(p)\exp (\mathrm{i}px)$ is negligible. We then define $$ \phi _{\text{s}}(x):=\frac{1}{(2\pi )^{d/2}}\int _{|p|\leq \Lambda /b}\hat{\phi}(p)\exp (\mathrm{i}px)\text{ and }\frac{1}{(2\pi )^{d/2}}\phi _{\text{f}}:=\int _{\Lambda /b<p\leq \Lambda}\hat{\phi}(p)\exp (\mathrm{i}px), $$ where $d$ if of course the dimension of space, so that $\phi =\phi _{\text{s}}+\phi _{\text{f}}$. Then, we have that $$ S(\phi )=S(\phi _<)+S_0(\phi _>)+\frac{\lambda}{4!}\int \mathrm{d}x\, \left[ \phi _{\text{f}}^4+4\phi _{\text{f}}^3\phi _{\text{s}}+6\phi _{\text{f}}^2\phi _{\text{s}}^2+4\phi _{\text{f}}\phi _{\text{s}}^3\right] , $$ where $$ S_0(\phi ):=\int \mathrm{d}x\, \left[ \tfrac{1}{2}|\nabla \phi |^2+\tfrac{r}{2}|\phi |^2\right] . $$ We then compute $$ -\ln \left( \int \mathrm{d}\phi _{\text{f}}\exp \left( -S(\phi )\right) \right) , $$ which is a sum over connected vacuum diagrams in the $\phi _{\text{f}}$ theory.

An equivalent claim is that then every diagram which involves a vertex arising from the term $\phi _{\text{f}}\phi _{\text{s}}^3$ vanishes. Not only do I not understand his heuristics of this being in-compatible with momentum conservation, but I also do not see how these diagrams vanish when I carefully compute what their value should be using the Feynman rules. An ideal answer should be able to explain this vanishing diagrammatically.

Answer 1

Consider the partition function $$Z = \int D\phi ~ e^{-S_0 - S_I},$$ where $S_0$ is the Gaussian/free part and $S_I$ is the interaction part of the action. Within a perturbative framework we may aim to systematically include the contributions of fast modes to the (effective) action for slow modes. For this we expand in the interaction strength as $$Z = \int D\phi ~ e^{-S_0}\left[1 + \sum_{n=1}^{\infty} \frac{(-1)^n}{n!} (S_I)^n \right] = \int D\phi ~ e^{-S_0}\left[1 + \sum_{n=1}^{\infty} \frac{(-1)^n}{n!} \langle S_I \rangle_f^n \right],$$ where the average is done over the fast modes (hence the subscript $f$). The $\phi_s^3$ (or the $\phi_s$) vertex does not arise in the above for two reasons.

1. For $n = 1$: $\langle \phi_s^3 \phi_f \rangle_f = \phi_s^3 \langle \phi_f \rangle_f = 0$ because $\langle \phi_f \rangle_f = 0$, since we're not in a symmetry broken phase where the field can have a non-zero expectation value.

2. For $n>1$: Here we can contract fast modes originating from different vertices. However, $\phi_s^3$ vertex is not generated because there is no process in a $\phi^4$ theory that can produce a $\phi^3$ vertex. Consider a general diagram made out of $V$ vertices, containing $I$ internal lines or propagators, and $E$ external lines. These numbers are constrained as $$ 4V - 2I = E,$$ due to "conservation of number of legs" at each vertex. Clearly, $E$ cannot be odd since $V$ and $I$ are positive integers. So, $E=3$ diagrams (which contribute to $\phi_s^3$ vertex) are not generated, or equivalently are identically zero.