Suppose we have the Lagrangian in 3 dimensions: $$ \mathcal{L} = \frac{1}{2}\partial_\mu\phi\partial^\mu\phi-\frac{1}{2}m^2\phi^2-\frac{g_1}{4!}\phi^4-\frac{g_2}{6!}\phi^6 $$

The superficial degree of divergence could be found as $\omega = 3-(1/2)n-n_4$, where $n$ is the number of external lines, and $n_4$ is the number of 4-point vertices.

I'm not quite sure how to analyze the divergence of these 2-point functions: enter image description here

At high momenta, diagram (a) goes like $$ \int^\Lambda\frac{d^3k}{k^2}\sim \Lambda $$

diagram (b) goes like $$ \int^\Lambda\frac{d^6k}{k^6} \sim \log\Lambda $$

diagram (c) goes like $$ \int^\Lambda\frac{d^{12}k}{k^{10}} \sim \Lambda^2 $$

I actually don't quite understand how those approximation work, I was just told that the divergence of each diagram sometimes could be found by finding the superficial degree of divergence $\omega$, and then the approximation works like $\Lambda^\omega$ for $\omega>0$, and $\log\Lambda$ for $\omega = 0$. How can I justify the approximations above? For $(b)$, I remember I saw some reference that shows me

$$ \int^\Lambda\frac{d^6k}{k^6}\sim \frac{|k|^5d|k|d\Omega}{k^6} \sim \log\Lambda $$

But I don't know how the $d^6k$ is separated, and how to evaluate the integral of this form in general.

Thanks for the help!

  • 2
    $\begingroup$ What exactly do you mean by "I don't know how it works"? The issue is that you don't know how to evaluate the integrals $$\int_{|k|<\Lambda}\dfrac{d^nk}{k^{2m}}$$ in general or you don't know why these integrals capture the behavior of the diagram at large $k$? $\endgroup$
    – Gold
    Jan 2, 2023 at 17:08
  • $\begingroup$ @Gold Thanks for the comment! I don't know how to evaluate this integral in general. $\endgroup$
    – IGY
    Jan 2, 2023 at 17:13
  • $\begingroup$ Thanks for clarifying! I posted one answer on how these integrals can be evaluated, let me know if something is still not clear. $\endgroup$
    – Gold
    Jan 2, 2023 at 17:19

1 Answer 1


In this post we are going to consider all the integrals already in Euclidean signature, so that Wick rotation has already been performed.

We want to evaluate integrals of the form $$\int_{|k|<\Lambda} \dfrac{d^nk}{k^{2m}}$$

Since we are in an $n$-dimensional Euclidean space we can introduce hyperspherical coordinates. For our purposes, all we need to know is that the volume element is of the form

$$d^n k = |k|^{n-1} d|k| d^{n-1}\Omega,$$

where $|k|$ is the radial coordinate in $n$-dimensional $k$-space and $d^{n-1}\Omega$ is the volume element on $S^{n-1}$. In that case the integral becomes $$\int_{|k|<\Lambda}\dfrac{d^nk}{k^{2m}}=\int_0^\Lambda \dfrac{|k|^{n-1}}{|k|^{2m}}d|k|\int_{S^{n-1}}d^{n-1}\Omega={\rm Vol}(S^{n-1})\int_0^\Lambda |k|^{n-1-2m}d|k|$$

Now there are two cases: $n=2m$ in which case we get $\log \Lambda$ and $n\neq 2m$ in which case we get $\Lambda^{n-2m}$. The complete result is $$\int_{|k|<\Lambda}\dfrac{d^nk}{k^{2m}}=\begin{cases}{\rm Vol}(S^{n-1})\log \Lambda,& n=2m,\\{\rm Vol}(S^{n-1})\dfrac{\Lambda^{n-2m}}{n-2m},& n\neq 2m.\end{cases}$$

  • $\begingroup$ Many thanks for the answer! Is $|k|$ the magnitude of momentum? $\endgroup$
    – IGY
    Jan 2, 2023 at 17:57
  • 1
    $\begingroup$ You're welcome! Yes, after we Wick rotate the vector $k$ is in an Euclidean space and we can evaluate its magnitude $|k|$, which works as a radial coordinate in $k$-space then. $\endgroup$
    – Gold
    Jan 2, 2023 at 17:58
  • $\begingroup$ Thanks!! If we have another diagram that combines (a) and (b) above (so it contains a $\phi^4$ vertex and $\phi^6$ vertex), we can find the superficial degree of divergence as $\omega = 1$, then we have $m=4, n=9$. Using the equation above, we can find the integral is proportional to $\Lambda$. However, the composite diagram suggests it's $\Lambda\log\Lambda$, so can we conclude in this case Vol$(S^{n-1}) \sim \log\Lambda$? $\endgroup$
    – IGY
    Jan 2, 2023 at 18:10
  • 1
    $\begingroup$ @IGY I'm not sure what's giving rise to these two different results, I'll think about what is going on, but it's certainly not the case that ${\rm Vol}(S^{n-1})\sim \log \Lambda$. This is just the volume of a unit $(n-1)$-sphere and so it contains absolutely no dependence in $\Lambda$. In fact, we have an expression for it: $${\rm Vol}(S^{n-1}) = \dfrac{2\pi^{(n-1)/2}}{\Gamma\left(\frac{n-1}{2}\right)}$$ $\endgroup$
    – Gold
    Jan 2, 2023 at 18:32
  • $\begingroup$ Thanks so much!! $\endgroup$
    – IGY
    Jan 2, 2023 at 18:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.