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

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.

## 1 Answer

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.

• "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." ---- I feel as if this argument is circular . . . Part of what we would like to show is that, after a RG transformation, we do not generate any new terms in the action (at least to the order we are working). It is conceivable that we could generate a $\phi ^3$ term after re-normalization, in which case the theory obviously could produce a $\phi ^3$ vertex. Am I seriously mis-understanding something? – Jonathan Gleason Nov 30 '14 at 9:22
• "Clearly, $E$ cannot be odd since $V$ and $I$ are positive integers." ---- Sure, but what about, e.g., $E=6$. For example, consider a diagram with two $\phi _{\text{s}}^3\phi _{\text{f}}$ vertices with the two $\phi _{\text{f}}$ legs contracted. This yields a diagram with $6$ external slow legs, no? In fact, it is this specific diagram which I have been struggling to show vanishes for the past couple of days now. – Jonathan Gleason Nov 30 '14 at 9:24
• @JonathanGleason first comment: In a perturbative renormalization scheme the resultant vertices after integrating out high energy modes can be viewed as those that are generated by contracting the fast-mode legs of the tree-level vertices. In the $\phi^4$ theory there 4 vertices with fast-mode legs (ref: the body of question). Contracting these with each other cannot generate vertices with odd number of slow mode legs. A proof would follow from the constraint I quoted in my answer above. – vik Nov 30 '14 at 21:50
• 2nd comment: Higher order vertices such as $\phi_s^{2n}$ with $n\geq 3$ are always generated by the integrating out procedure. We can ignore these, because they do not affect the beta-functions of the flowing parameters in the theory. In other words, given the interaction vertex in the tree-level action is marginal, all higher order vertices are irrelevant. This general expectation should hold unless we're integrating out some zero-energy modes in our Wilsonian RG scheme. – vik Nov 30 '14 at 21:54
• How do we know a priori that vertices of the form $\phi ^{2n}$ with $n\geq 3$ will be irrelevant. Once again, don't we actually need to complete the re-normalization procedure to verify this precisely? Sure, one can do dimensional analysis and obtain that the 'engineering degree' suggests that they will be irrelevant, but to actually prove this expectation is correct, do we not have to do the full re-normalization procedure? – Jonathan Gleason Nov 30 '14 at 22:03