$\phi^4$ theory is not perturbatively renormalizable in 5 dimensions.

I have come across literature where renormalizability is discussed w.r.t $N$, for fields obeying $O(N)$ symmetry. But it is not clear to me if renormalizability w.r.t $N$ implies renormalizability overall? If I think of it naively, one can expand in terms of both the coupling constant $\lambda$, and the parameter, $N$. So, even if one proves renormalizability w.r.t $N$, the final observables may still depend on the UV cutoff for some powers of $\lambda$. Alternatively, one has to assume that the theory is renormalizable for all powers of $\lambda$ and $N^1$. And then show the theory is renormalizable for all values of $N$. Is the initial assumption valid? Does my argument(s) even make sense?

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    $\begingroup$ Which literature? Which page? $\endgroup$
    – Qmechanic
    Oct 31, 2021 at 6:53
  • $\begingroup$ In large $N$ theory for the $\phi^4$ interaction, $\lambda$ itself is taken to be of order $1/N$ in order to have a meaningful saddle point expansion. The Hubbard-Stratonovich transformation doesn't change this. $\endgroup$
    – octonion
    Oct 31, 2021 at 7:42
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    $\begingroup$ What is "renormalizability w.r.t. $N$"? $\endgroup$ Oct 31, 2021 at 9:35

1 Answer 1


Indeed, when we say a theory is renormalizable, it is tacitly assumed that there is only one perturbative expansion of the theory that we know how to do. When this is not the case, renormalizability is really a property of the parameter we are perturbing in. To say that the $\frac{1}{N}$ expansion is renormalizable means that a finite number of counter-terms added to the action will be enough to absorb all divergences. The $\lambda$ expansion being non-renormalizable, on the other hand, means that as we go to arbitrarily high powers of $\lambda$, the number of counter-terms one must add will grow without bound.


The trick that makes it easy to see why the two approaches can be so different is called the Hubbard-Stratonovich transformation. This introduces an auxiliary field in order to get a more explicit dependence on $N$. Namely a $\frac{1}{\sqrt{N}} \sigma \phi_i \phi_i$ vertex. Once you use this to build Feynman diagrams, it is no longer necessarily true that loop diagrams are higher order than tree diagrams. A $\phi$ loop in the $\sigma$ propagator, for instance, will involve $\delta^i_j \delta^j_i = \delta^i_i = N$ which cancels the extra two vertices and gives a diagram which is just as important as the one without the $\phi$ loop.

  • $\begingroup$ would you be aware of any introductory material on this? i.e. application of Hubbard-Stratonovich transformation to $\phi^4$ theory and subsequent renormalizability? $\endgroup$
    – Angela
    Oct 31, 2021 at 15:08
  • $\begingroup$ The best papers I know of are arxiv.org/abs/1404.1094 and arxiv.org/abs/1601.01310 which are consistent with the reviews by Zinn-Justin but presented in a more modern way. $\endgroup$ Oct 31, 2021 at 17:50
  • $\begingroup$ Thank you for the pointers. $\endgroup$
    – Angela
    Oct 31, 2021 at 22:35

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