As I understand it, the Wilsonian picture of renormalisation says irrelevant interactions in the bare Lagrangian should not affect scattering amplitudes at energies far below than the fundamental cutoff $\Lambda_0$. That is, diagrams involving these vertices contribute negligibly when the external momenta are much less than $\Lambda_0$. However, I can't see how this plays out in a simple example:

Let $\phi$ be a scalar field in $d=4$, with bare action given by: $$S_{\Lambda_0}[\phi] = \int d^4x \left(\frac{1}{2}(\partial \phi)^2+\frac{1}{2}m^2\phi^2 - \frac{g}{6!}\phi^6\right).$$

The $\phi^6$ interaction is irrelevant, since $[g]=-2$. Therefore, I expect that it should not affect the low energy physics. Indeed, since there are no relevant interactions in $S_{\Lambda_0}[\phi]$, I expect the low energy physics to look like a free field theory. However, we have nontrivial 3 to 3 scattering:

$$\langle k_1, k_2 ,k_3 |S|k_4,k_4,k_6\rangle = g \times(2\pi)^4\delta(k_1+...+k_6).$$

I may have missed some constant factors, but the point is that this amplitude is nonzero, and does not depend on the incoming/outgoing momenta $k_i$. So, even at low energies, there is 3 to 3 scattering at $O(g)$.

My first thought is that contributions from loop diagrams should somehow cancel the amplitude above, so that the overall 3 to 3 amplitude ends up being very close to $0$ when external momenta are $\ll \Lambda_0$. But any loop diagrams with 6 external legs must involve at least 2 vertices, so they enter at $O(g^2)$. Therefore I don't see how they can cancel the $O(g)$ tree-level contribution.

What am I missing?

  • $\begingroup$ $\phi^4$ is much more relevant than $\phi^6$. Of course if you forget to include $\phi^4$ then $\phi^6$ becomes the most relevant interaction. If you write all the terms as instructed, you will observe that indeed $\phi^6$ is negligible: it is much smaller than the leading correction. $\endgroup$ Apr 27, 2021 at 18:33
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    $\begingroup$ Irrelevant couplings have negative mass dimension so they are of the form $g = c \Lambda^{-a}$ where $c$ is a dimensionless number and $a> 0$. $\Lambda$ is the UV cut-off of the QFT. As we now flow to the infrared by taking $\Lambda \to \infty$, we find that $g \to 0$ so this coupling does not survive in the IR. $\endgroup$
    – Prahar
    Apr 27, 2021 at 18:41
  • $\begingroup$ @AccidentalFourierTransform If I included a $\lambda\phi^4$ term in the bare Lagrangian, there would be a 1-loop correction to the 6-point vertex which enters at $O(\lambda^3)$. Is your point that this correction will dominate the tree-level $O(g)$ diagram? If so, why? $\endgroup$ Apr 27, 2021 at 18:52
  • $\begingroup$ There are six-legged tree level diagrams... $\endgroup$ Apr 27, 2021 at 23:08
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    $\begingroup$ because -- they come with factors of $1/p^2$ which blow up at low energies, unlike the irrelevant interaction which stays $\mathcal O(1)$. This is precisely the point of Wilson's classification: operators with scaling dimension $\Delta$ have a coupling with dimension $-\Delta$, and whose contribution to a scattering amplitude must come with a factor of $E^\Delta$, by dimensional analysis. As $E\to0$, the lower the $\Delta$ the higher the contribution, so more relevant the operator. $\endgroup$ Apr 28, 2021 at 11:16

1 Answer 1


The Lagrangian you have written down is only the $O(\hbar^0)$ piece. If you want to set up a proper theory you of course have to add counterterms which have higher orders of $\hbar$ and which diverge as you take $\Lambda_0$ to infinity. This will also generate terms proportional $\phi⁴$ (with a prefactor that should start like $\hbar g²$).

To fix the coefficients in front of the counter terms you need renormalisation conditions, which typically are a set of equations for correlation functions with momenta approximally equal to a new non zero scale $\mu$. Anyway when you just do a perturbative calculation without the use of the RG, powers of the coupling constant $g$ will be accompanied by (powers of) logarithms $\log(p²/\mu²)$, where $p²$ is the typical energy in this process. This means if you calculate something very far away from $\mu$ you can not trust your perturbation theory.

The solution to that is to use the RG, which first evolves the couplings down to the scale where your process happens. Then you can use perturbation theory with this new coupling $g(p)$, which produces reliable results. Then you should see that the coupling $g(p)$ goes like $p²$, meaning it vanishes at low energies. In terms of the old coupling $g$, the new coupling $g(p)$ can be interpreted as summing infinitely many loop diagrams.

This means for example that a tree diagram where the coupling $g(p)$ is used as an interaction vertex is equal to an infinite number of diagrams where $g$ is used for the interactions. So indeed the loop contributions cancel the tree contributions as you said in your post. You just have to go to all loops in some sense to see that. It is not valid to stop at some loop order since one needs to resum all those logarithms to see the true behavior.


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