I'm interested in studying the theory $$S=\int d^dx\left(\frac{1}{2}(\partial\phi)^2-\frac{g}{(2k)!}\phi^{2k}\right)$$ in $d=2, 3, 4$ and for $k=2, 3, 4\dots$ and I'm having trouble with the the 1-loop renormalization of the said theory. The questions are boldfaced.

I know that for the theory to be renormalizable the mass dimension of the coupling constant must be $\left[g\right]\geq 0$, and therefore for a said dimension only some values of $k$ are worth studying:

$d=4\quad\Rightarrow\quad k=2$
$d=3\quad\Rightarrow\quad k=2,3$
$d=2\quad\Rightarrow\quad k=2,3,4\dots$

I'm also pretty confident that the Feynman rules are $k$-independent: $\frac{i}{p^2}$ for the propagator and $-ig$ for the vertex (with the appropriate number of external legs, which is obviously $k$-dependent).
The doubts come with the loops: my idea was to draw all the possible 1-loop diagrams starting with 1 vertex, and then two, and so on, with the appropriate number of external legs coming out of the vertices depending on $k$. The only diagrams that I could draw are the "tadpoles", the "spiders" and the various polygons:enter image description here
First question: are these the only 1-loop diagrams there are, or am I missing something?
The divergence of the loop integrals are obviously $d$-dependent, and in $d\leq4$ only the tadpoles and the spiders can be divergent (by simple power-counting), so let's focus on these two.
I know that for a massless scalar theory with $k=2$ in $d=4$ the tadpoles diagrams can't be regularized and are therefore just considered as an infinite constant shift of the theory. Can one do this with general $k$ and $d$?
Assuming this is correct, we're left with the spiders. Is there a general formula to find the number of diagrams with the same number of vertices, and external lines? With this I don't mean the symmetry factor (that should be $\frac{1}{2}$ for all these spider diagrams), I mean for example the three diagrams in the $k=2$ case:enter image description here
I couldn't find a general formula so I decided to move on and leave this problem for later. Now for the real question: these spider diagrams, what 0-loop diagrams are they "correcting" for $k\geq3$? I mean, in the $k=2$ case, the diagram is the quantum correction to the vertex, which has 4 external legs. But for the $k=3$ case, for example, there isn't a 0-loop diagram with 8 external legs! The first correction to the 6-legged vertex is a 2-loop diagram, so what is this loop correcting?
My attempt to answer this is that for $k>2$ some diagrams are just not there and there is no problem with this, but it seems a bit odd: enter image description here
(the diagrams in brackets are some 2-loop diagrams that I'm not interested in).

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    $\begingroup$ Every spider has degree of divergence $d - 4$. So the $k \geq 3$ ones converge and therefore don't need a 0-loop vertex to correct. $\endgroup$ Commented Nov 20, 2021 at 23:11
  • $\begingroup$ @ConnorBehan so among all the 1-loop diagrams in a general scalar theory, with general (even) coupling and general dimension, the only one that actually matters is the $k=2$ spider? $\endgroup$ Commented Nov 21, 2021 at 0:15
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    $\begingroup$ Assuming finite diagrams don't matter for what you're doing, yes. $\endgroup$ Commented Nov 21, 2021 at 1:00
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    $\begingroup$ Suggestion: Consider to replace the word tadpole with self-loop for clarity. $\endgroup$
    – Qmechanic
    Commented Nov 21, 2021 at 11:50

1 Answer 1


OP asks many questions. If we jump to OP's last question (which appears to be OP's real question), then the main point is that one cannot artificially exclude vertex interaction terms with less legs (such as, e.g. mass terms) unless they are prohibited by some symmetry. Even if they are absent in the UV, they get generated during the RG flow of integrating out modes. This fact modifies OP's tables.


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