# Week 6, Monday

We simplify the superficial degree of divergence to $$\begin{split} D =&\; dL – 2I \\ =&\; d – \frac{d-2}{2}E + V(d-4) \\ =&\; \begin{cases} 4-E & \text{in 4d} \\ 2-2V & \text{in 2d}. \end{cases} \end{split}$$ A superficial divergence (by power counting) does not necessarily translate into an actual divergence, for example the tree level interaction (E=4, V=1) has superficial degree of divergence $D=0$. But, since there is no loop integral, it obviously does not diverge logarithmically. Neither does a superficial convergence $D<0$ guarantee convergence, for example attach a tadpole to any convergent diagram. The tadpole integral remains divergent. But in that case the divergence just comes from a sub-diagram, it is not inherently due to the larger diagram. This suggests that we should only look at diagrams without superficially divergent sub-diagrams; If there is a divergent sub-diagram then you really only have to worry about that sub-diagram. This suggests the following:

Theorem:[Dyson-Weinberg convergence theorem] A superficially convergent diagram such that all of its sub-diagrams are also superficially convergent is actually convergent.

The theorem might seem obvious but is actually rather tricky due to the issue of overlapping divergences. We will explain what this in a second, but first let us classify the primitively divergent diagrams in $\phi^4$-theory. That is, look for the divergent diagrams that do not have a divergent sub-diagram. These are the real troublemakers, and we really only have to deal with their divergences.

• In the two-dimensional $\phi^4$-theory there is only a single primitively divergent diagram, namely the tadpole. It has superficial degree of divergence $D=0$.

• In the four-dimensional $\phi^4$-theory there are two primitive divergences at $O(\lambda)$, namely the tadpole $D=2$ and the “fish”, the only (up to permutations of the external vertices) connected one-loop diagram contributing to the 4-point function. There are further primitive divergences at all higher powers of $\lambda$, but only with two or four external legs.

• In the four-dimensional $\phi^k$-theory with $k\geq 5$ there are primitive divergences with any number of external legs. This is what makes the theory non-renormalizable, as we will see.

## Overlapping Divergences

There is one subtlety when dealing with potentially divergent loop integrals: Sometimes you can’t split the integrals into an outer integral times an inner integral and analyze the convergence separately. This happens when an internal line is part of two separate loops, giving rise to a propagator $\frac{1}{(\ell+p)^2+m^2}$ that depends on both the $\ell$ and $p$-loop momentum. More properly, these should be called overlapping integrals.

The simplest overlapping loop diagram would be in $\phi^3$-theory. So, just for this section, let us consider tri-valent interaction vertices and the diagram leading to the loop integral $$\begin{split} I =&\; \int d^4p \; d^4\ell \; \frac{1}{p^2+m^2-i\epsilon} \frac{1}{\ell^2+m^2-i\epsilon} \\ &\; \qquad \times \frac{1}{(p+\ell)^2+m^2-i\epsilon} \frac{1}{(p-k_1)^2+m^2-i\epsilon} \frac{1}{(\ell-k_2)^2+m^2-i\epsilon} \end{split}$$ where we left out constants and the external propagators, anything that does not depend on the loop momenta. We now

• Perform Wick rotation: drop the $-i\epsilon$.
• Set $m^2=0$ and $k_1=0=k_2$ since the UV divergence is at $k^2 \gg m^2, k_1^2, k_2^2$.

• Remove a region of small momenta by hand to avoid the IR divergence.

Hence the asymptotic behavior of the loop integral is that of $$I \sim \int \frac{d^4p \; d^4\ell}{p^4 (p+\ell)^2 \ell^4}.$$ This is an example of an overlapping divergence: by power counting, the diagram superficial degree of divergence $D=2\times 4-5\times 2 =-2$ and no divergent sub-diagram, so we expect convergence. But the actual integral is not so obviously convergent, for example there is a region where $p+\ell$ is constant. In that region, just the $p$-integral seems to have superficial degree of divergence $D=0$, which would indicate a logarithmic divergence. This is actually not true, the whole integral still converges.

To show that the integral does converge, we have to split up the domain according to which momentum factor is the smallest. For example, consider the region where $\ell^2$ is the smallest, $$U_p = \big\{ p \big| p^2 \geq \ell^2,~ (p+\ell)^2 \geq \ell^2 \big\},$$ and combine it with the necessary IR cutoff $$U_\ell = \big\{ \ell \big| \ell^2 \geq 1 \big\}.$$ Splitting the 4-momentum $\ell = \bar\ell e_\ell$ into its absolute value $\bar\ell\in\mathbb{R}$ and unit 4-vector $e_\ell$, the loop integral becomes $$I \sim \int_{U_p} \int_{U_\ell} \frac{d^4p \; d^4\ell}{p^4 (p+\ell)^2 \ell^4} = \int_0^\infty d\bar\ell \bar\ell^3 \frac{1}{\bar\ell^6} \int_{S^3} de_\ell \int_{U_{p’}} \frac{d^4 p’}{(p’)^4 (p’+e_\ell)^2}$$ where we rescaled $p’ = p \bar\ell$ and integrate over the rescaled region $$U_{p’} = \big\{ p’ \big| (p’)^2 \geq 1,~ (p’+\ell)^2 \geq 1 \big\}.$$ The $\bar\ell$-integration is, indeed, of superficial degree of divergence $D=-2$. The overlapping diagram does converge as expected by power counting, the potential overlapping divergence does not cause an actual divergence.