TITLE: Confusion on the renormalizability and the dimension of coupling constant (From Srednicki's book)
QUESTION [0 upvotes]: I am trying to understand the renormalizability and the dimension of couplings from section 18 of Srednicki's QFT book.
In section 18, it mentions if we can use finitely many new terms (counter-term) to absorb infinities, then we say the theory is renormalizable, otherwise is non-renormalizable. Now consider a diagram ($d$-dimensional scalar filed $\phi^n$-theory) with $E$ external legs, $V_n$ vertices of $n$ legs, the so-called superficial degree $D$ is then
$$D=[g_E]-\sum V_n[g_n],$$
where $[X]$ is the dimension of $X$, $g_n$ is the coupling of $g_n \phi^n$.
It is saying that if any of $[g_n]<0$, then the theory is non-renormalizable. I don't quite understand this point.
How can I see that when any of $[g_n]<0$, we need infinitely many new terms to absorb infinities? (Let's just say when $D\geq 0$ the diagram diverges for the time being ).
My understanding is like this：When you have a divergent diagram of $E$ external legs, you need a new vertex (counter-term) like $Z_Eg_E\phi^E$. Now $[g_E]=d-E(d-2)/2$, and one of $[g_n]<0$, so we can adjust corresponding $V_n$ such that there are infinitely many $E$ satisfying $D\geq0$. Therefore we need infinitely many new terms to absorb infinities. Is this logic right?

REPLY [1 votes]: You understanding is essentially correct.
The whole truth is quite complicated, but essentially well-understood (see for example John Collins' book on renormalization).
But let me give you some other rough considerations.
Take your formula
$$
D = d - E(d-2)/2 - \sum V_n [g_n].
$$
If $[g_n]$ is negative, since for any amplitude the amount of $n$-type vertices can be arbitrarily large, $D$ can be arbitrarily large.
This naively implies that there are UV divergent graphs with any number of external legs $E$, hence one needs terms proportional to $\phi^E$, for arbitrarily large $E$, in order to cancel the UV divergence in all graphs.
But this is maybe unsatisfying, since the UV divergence may also come only from divergent subgraphs, which can be cancelled also by terms $\phi^m$ with $m < E$.
The nested subdivergences are a reason that the superficial degree of divergence is not the actual degree of divergence for multiloop diagrams.
I will sketch another argument, very roughly, which is maybe more satisfying.
Notice that deriving a graph $\Gamma$ by some external momentum $p$ reduces its superficial degree of divergence by $1$
$$
\frac{\partial}{\partial p^{\mu}} \frac{1}{(k+p)^2} \sim - \frac{2k_{\mu}}{(k+p)^4}.
$$
Hence, naively, deriving $\Gamma$ with respect to external momentum sufficient amount of times, say $N+1$ (assume $N$ even), makes it finite.
$$
\frac{\partial^N \Gamma}{\partial p^{\mu_1} ... \partial p^{\mu_N}} = \text{finite} =: \frac{\partial^N \Gamma_{\text{fin}}}{\partial p^{\mu_1} ... \partial p^{\mu_N}}
$$
Integrating this $N$ (assume that $N$ is even) times gives
$$
\Gamma_{\text{fin}} = \Gamma + A_1 + A_2 p^2 + ... + A_{N/2} (p^2)^{N/2}.
$$
The terms $\sum_j A_j (p^2)^j$ cancel the divergences of $\Gamma$, i.e. they are the counterterms. In particular, we need $N > D$ and $D$ can be arbitrarily large, so we need counterterms $A_{N/2} (p^2)^{N/2}$ for arbitrarly large $N$. But such counterterms can only come from terms in the Lagrangian that involve arbitrarily high derivatives, $\Box^N$ with arbitrarily large $N$, i.e. an infinite amount of terms.
{"set_name": "stack_exchange", "score": 0, "question_id": 665512}