Massive versus Massless $\phi^4$ Sunset Diagram - does $\frac{1}{\epsilon^2}$ term vanish for $m=0$?

Question

In a real scalar massive $\phi^4$-interacting theory consider the amputated sunset diagram. This is the integral out of Kleinert and Schulte-Frohlinde Critical Properties of $\phi^4$-Theories:

The above two-loop integral is in Euclidean space, and $\mathbf{q}$ is the incoming momentum.

Using dimensional regularization, where $\epsilon = \frac{4-D}{2}$ and $\mu$ is the reference scale, the above integral takes on the following form when regularized:

Now consider that we take $m \to 0^{+}$ and consider the diagram in the massless case. Then the above simplifies to: $$ - \frac{g^2}{(4\pi)^4} \frac{\mathbf{q}^2}{2\epsilon} + \mathcal{O}(\epsilon^{0}) $$

So we see in the massless case, the $\frac{1}{\epsilon^2}$ term vanishes.

Why does this happen? My understanding is that a $\frac{1}{\epsilon^2}$ signifies a quadratic UV-divergence, while a $\frac{1}{\epsilon}$ signifies a logarithmic UV divergence (maybe this is wrong?).

From a naive power-counting point of view if we look at the original loop integral, we have an integral roughly like $\frac{d^{8}\mathbf{p}}{\mathbf{p}^6}$ for large momenta, which looks like a quadratic divergence - independent of whether $m$ has vanished or not.

Why does the power-counting argument fail when $m=0$? Or is it false that $\frac{1}{\epsilon^2}$ signifies a quadratic UV divergence?

Answer 1

The order of the $\varepsilon$ pole is not related to the power of the divergence. Indeed the poles are only sensitive to $\log$-like divergences. Power-like divergences are killed by dimensional regularization. By this I mean that if you were to regularize with a cutoff, terms with $\Lambda^k$ will disappear and terms with $\log\Lambda$ will become poles in $\varepsilon$. This is related to the fact that in dim-reg the following class of integrals vanish: $$ \int d^Dp\, (p^2)^\alpha = 0\,.\tag{1}\label{int} $$ There is however a way to determine the order of the poles in $\varepsilon$. For an integral with $L$ loops it's bounded by $L$, but it might be less. A theorem shows that the maximal order of the pole of an $L$ loop diagram with $V$ vertices in the massless theory is given by $\min(V-1,L)$. This is in agreement with what you found as $m\to 0$ because $V=2$.

This theorem can also be used in general, upon considering massive propagators are vertices with two legs (with coupling $m^2$).

The intuitive explanation of this particular case is as follows: the "overall" divergences (i.e. those that show up when all loop momenta are sent to infinity at the same speed) only produce simple poles in $\varepsilon$, while the higher poles are given by subdivergences. Subdivergences can be interpreted as subdiagrams of the original diagram that are themselves divergent. In this case the only subdiagrams are the two single loop tadpoles, which are known to vanish when $m^2 = 0$.

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I can suggest a book that shows both that the integral $\eqref{int}$ vanishes (around p. 63) and proves the theorem for the maximal order of the poles (Sec. 2.8).

Answer 2

The power counting of momentum $p$ has nothing to do with the divergence power like $1/\epsilon^a$, actually what we are doing here for calculating loop diagrams is translating(mapping) divergence into the language of Gamma function. I think we can call it Gamma function regularization. I will show you examples as the following,

as we know the $\beta$-function is defined to be \begin{align} B(\alpha, \beta ) \equiv \int_0^1 d x x^{\alpha-1}(1-x)^{\beta-1}=\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}. \end{align} All the divergence like $1/\epsilon$ comes from $\Gamma(\epsilon)$, for example, \begin{align} B(\epsilon, 1 ) \equiv \int_0^1 d x x^{-1+\epsilon}=\frac{\Gamma(\epsilon)\Gamma(1)}{\Gamma(1+\epsilon)}=\frac{1}{\epsilon}. \end{align} However, the power counting of $x$ tell us the integral is approximately $x^{\epsilon}$. Also, we can see \begin{align} B(-1+\epsilon, 1 ) \equiv \int_0^1 d x x^{-2+\epsilon}=\frac{\Gamma(-1+\epsilon)\Gamma(1)}{\Gamma(\epsilon)}=-1-\epsilon+O(\epsilon^2). \end{align} Thus, we can conclude \begin{align} \int_0^1 d x \frac{1}{x^2}=-1. \end{align} It looks like regularization always gives us weird results, for example, Riemann Zeta regularization can give us $1+2+3+4+\cdots =-1/12$. Anyway, here, the power of momentum $p$ has nothing to do with the the power of $\epsilon$, at least, they are not simply related with the same power.