Consider the following Lagrangian describing the interaction between a massless field $\phi$ and a massive field $\psi$: $$ {\scr L} = \frac12(\partial\phi)^2 + \frac12 (\partial\psi)^2(1 + f(\phi/M)) - \frac12m^2\psi^2.\tag{1} $$ The interaction between these two fields will generate an effective potential for $\phi$, which we can compute using the background field method. First we shift $$\phi\to\phi_b + \delta\phi\tag{2},$$ with $\phi_b$ constant $$ {\scr L} = \frac12(\partial\delta\phi)^2 + \frac12(\partial\psi)^2(1 + f[(\phi_b+\delta\phi)/M]) - \frac12 m^2\psi^2.\tag{3} $$ We can canonically normalize $\psi$ by rescaling: $$\psi_C = \sqrt{1 + f(\phi_b/M)}\psi\tag{4},$$ yielding $$ \frac12(\partial\delta\phi)^2 + \frac12(\partial\psi_C)^2\left(1 + \sum_{n = 1}^\infty \frac{f^{(n)}(\phi_b/M)}{(1 + f(\phi_b/M))n!} \left(\frac{\delta\phi}{M}\right)^n\right) - \frac12 m^2[\phi_b]\psi_C^2,\tag{5} $$ where $$ m^2[\phi_b] = \frac{m^2}{1 + f(\phi_b/M)}.\tag{6} $$ We can compute the path integral over $\psi$ in the gaussian approximation by dropping the 3 and higher-point interactions multiplying $(\partial\psi_C)^2$. The result is standard (see, e.g. Schwartz page 745 and surrounding discussion) $$ V_{\rm eff}^{1\,\rm loop} = \frac{-i}{2}\int\frac{{\rm d}^4 k}{(2\pi)^4}\log\left(1 - \frac{m^2[\phi_b]}{k^2}\right).\tag{7} $$ Wick rotating, we find $$ V_{\rm eff}^{1\,\rm loop} = \frac{1}{64 \pi ^2} \left(\Lambda ^2 m\left(\phi _b\right){}^2-m\left(\phi _b\right){}^4 \log \left(\frac{\Lambda ^2}{m\left(\phi _b\right){}^2}+1\right)+\Lambda ^4 \log \left(\frac{m\left(\phi _b\right){}^2}{\Lambda ^2}+1\right)\right),\tag{8} $$ where $\Lambda$ is a Euclidean cutoff on the loop momentum. At leading order, this result is quadratically divergent.

We can cross-check this answer by comparing $V_{\rm eff}''(\phi_b)$ to the self-energy diagram evaluated at zero external momentum. At one loop, there are two such diagrams, coming from the $(\partial\psi)^2\delta\phi$ and $(\partial\psi)^2\delta\phi^2$ interactions. Now comes the confusion: one can read off from these interactions that the resulting self-energy diagram contains a quartic divergence as opposed to a quadratic divergence. One may hope that these divergences cancel precisely, but this is not the case for general $f(\phi/M)$: $$ m_{\phi,{\rm eff}}^2 = \frac{\Lambda^4}{64\pi^2 M^2}\left(\frac{f''(\phi_b/M)}{1 + f(\phi_b/M)} - \frac{{f'(\phi_b/M)}^2}{(1 + f(\phi_b/M))^2}\right)+ V_{\rm eff}''(\phi_b).\tag{9} $$ I find this result extremely strange, since I was under the impression that it was a theorem that the effective potential generated all 1PI diagrams. Further, the fact that the results agree precisely up to the quartic divergence is further evidence to me that something pathological is going on.

One could resolve this tension by taking $f(\phi/M) = \exp(\lambda\phi/M) - 1$, but I hardly find this convincing. Another possible resolution is to canonically normalize $\psi$ before shifting to the background field, thus eliminating all vertices involving $\psi$-momentum. However, now the self-energy of $\psi$ contains a quartic divergence.

I will finally note that this problem is not present in dimensional regularization, since dim reg automatically drops polynomial divergences. However, if one is concerned about the UV sensitivity of the model, the cutoff scale $\Lambda$ is physical, and the quadratic vs quartic divergence dramatically changes the character of the effective potential.

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    $\begingroup$ This is all heavily dependent on the renormalization scheme, a.k.a you are asking meaningless questions. You can only compare RG-scheme invariant objects, anything else is ambiguous and has no intrinsic meaning. $\endgroup$ Aug 13, 2023 at 14:26
  • $\begingroup$ The bare mass of $\psi$ is a physical observable and is controlled by the VEV of $\phi$. New cutoff dependent terms in the effective potential can change the VEV and are therefore physical. Consider the case where $m_\psi$ is zero: in this case, the one loop mass renormalization is non-zero (no VEV), but the effective potential is non-zero, which appears to be a contradiction. $\endgroup$
    – Guy
    Aug 13, 2023 at 15:06
  • $\begingroup$ @AccidentalFourierTransform the discrepancy I find cannot be scheme dependent as I'm using the same scheme in both cases. $\endgroup$
    – Guy
    Aug 13, 2023 at 22:28
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    $\begingroup$ How do you know that you are using the same scheme in both cases? Recall that regularization is not the same as renormalization, the scheme dependence refers to the latter, not the former... $\endgroup$ Aug 13, 2023 at 23:34
  • $\begingroup$ Unless there is a subtlety i'm unaware of, i'm using a euclidean cutoff in both cases - the difference is the diagram. Whether $\phi$ has a mass is physical and should be scheme independent. $\endgroup$
    – Guy
    Aug 13, 2023 at 23:49

1 Answer 1


Comments to the post (v14):

  1. The way OP has normalized the 2 real scalar quantum fields $\widetilde{\phi}\equiv\delta\phi$ and $\widetilde{\psi}\equiv\psi_C$, the quadratic Lagrangian is indeed diagonal, but there are derivative coupling/interactions terms of the form $\widetilde{\phi}^n(\partial\widetilde{\psi})^2$.

  2. From a Wilsonian perspective, we would expect a massterm for $\widetilde{\phi}$ and other interaction terms as well, such as e.g. $\widetilde{\phi}^m$, as long as they are not forbidden by a symmetry. This makes OP's model unnatural/fine-tuned. (We can rule out $\widetilde{\phi}^m\widetilde{\psi}^n$ interaction terms by imposing $\widetilde{\psi}\to \widetilde{\psi}+\epsilon$ shift symmetry.)

    More generally, one would then have to work with the full $2\times 2$ Hessian matrix whereas OP's model gets away with a $1\times 1$ submatrix.

  3. Assuming that the UV cut-off $\Lambda\gg m(\phi_b)$, when Taylor expanding the last logarithm in eq. (8), the quartic $\Lambda^4$ term turns into a quadratic $\Lambda^2$ term (plus regular terms).

  4. The model should be renormalized, cf. Ref. 1.


  1. M.D. Schwartz, QFT & the standard model, 2014; subsection 34.2.1, eq. (34.49).

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