How to proof that $\phi^6$ theory is non-renormalizable? I'm studying QFT at beginner level. In the lecture, we never showed in detail that the $\phi^4$ theory is renormalizable and we are asked to show that $\phi^6$ theory is non-renormalizable.
I guess this should be easier as to show something is renormalizable, but I need some input how I can proof this rigorously?
 A: Let's call:
V = number of vertices
I = number of internal lines, E = number of external lines.
For the general connected Feynman diagrams of $\phi^6$-theory, each vertex has 6 half-lines, each internal line connects 2 half-lines, so the number of external lines is:
$$E = 6V-2I$$
The number of loops is:
$$L = 2V+1-\frac{E}{2}$$
(you can convince yourself)
The superficial degree of divergence is defined by:
$$D = \textrm{number of factors of internal momentum in the numerator} −
\textrm{number of factors of internal momentum in the denominator}$$
For the scalar theory, $D = 4L - 2I$ (convince yourself!). So, the superficial degree of divergence for $\phi^6$-theory:
$$D = 2V-E+4$$
Let's look back at the superficial degree of divergence for $\phi^4$-theory:
$$D = 4-E$$.


*

*D of $\phi^4$-theory does not depend on the number of vertices, or the order of pertubation theory. For a diagrams to be divergent, $D\ge 0$, the number of external lines is finite. This means there is a limit number of divergent diagrams => renormalizable.

*However that is not the case for the $\phi^6$-theory whose D depends on the number of vertices. So $\phi^6$-theory is non-renormalizable.
A: Let us assume we have a field theory with $V(\phi)~=~a\phi^6$ The Lagrangian density is then 
$$
{\cal L}~=~\frac{1}{2}\partial_\mu\phi\partial^\mu\phi~-~ a\phi^6.
$$
The action $S~=~\int d^4x{\cal L}$ is dimensionless in natural units. This means in units $[\phi]~=~1/\ell$. Given this potential we then must have $[a]~=~\ell^2$  
We have for a standard harmonic oscillator free field $V(\phi)~=~\frac{m^2}{2}\phi^2$ the straightforwards evaluation of the two point function 
$$
\langle|\phi(x)\phi(y)|\rangle~=~\frac{1}{4\pi}\frac{1}{|x~-~y|^2},
$$
where we think of $|x~-~y|~=~\delta$ as the cut off in the scalar potential $V(|\phi|)$. This then gives the renormalization correction to the mass
$$
\frac{m^2}{2}~=~\frac{m_b^2}{2}~+~k\frac{m_b^2 }{\delta^2},
$$
for $k$ a constant and $m_b$ the bare mass. This type of analysis can be done for any orders on $\phi$
Now let us look at $\phi^6$. The most elementary diagram involves $6$ lines at a vertex. For each line a $\Delta$ evaluated from $\delta$ to $\infty$ the amplitude will scale as
$$
a\int_\delta^\infty\frac{d^4\Delta}{\Delta^6}~=~\frac{a}{\delta^2}.
$$
This is rather inconvenient, for we do not normally work with 6 lines at a vertex. Instead real processes are going to be graphs similar to these below:

This means that we have an integration of the form
$$
a\int_\delta^\infty d^4\Delta\int_\delta^\infty d^4\Delta\frac{1}{\Delta^6}~=~\delta^2~-~\lim_{\Delta\rightarrow\infty}\Delta^2.
$$
That is really bad. This is terribly divergent for large distances. For this reason you never see field operators in standard QFT with powers larger than $4$.
