How can we expect the divergence of feynman diagram?

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:

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!

• 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 $k$?
– Gold
Commented Jan 2, 2023 at 17:08
• @Gold Thanks for the comment! I don't know how to evaluate this integral in general.
– IGY
Commented Jan 2, 2023 at 17:13
• Thanks for clarifying! I posted one answer on how these integrals can be evaluated, let me know if something is still not clear.
– Gold
Commented Jan 2, 2023 at 17:19

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 $$\int_{|k|<\Lambda}\dfrac{d^nk}{k^{2m}}=\begin{cases}{\rm Vol}(S^{n-1})\log \Lambda,& n=2m,\\{\rm Vol}(S^{n-1})\dfrac{\Lambda^{n-2m}}{n-2m},& n\neq 2m.\end{cases}$$

• Many thanks for the answer! Is $|k|$ the magnitude of momentum?
– IGY
Commented Jan 2, 2023 at 17:57
• 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.
– Gold
Commented Jan 2, 2023 at 17:58
• 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$(S^{n-1}) \sim \log\Lambda$?
– IGY
Commented Jan 2, 2023 at 18:10
• @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: $${\rm Vol}(S^{n-1}) = \dfrac{2\pi^{(n-1)/2}}{\Gamma\left(\frac{n-1}{2}\right)}$$
– Gold
Commented Jan 2, 2023 at 18:32
• Thanks so much!!
– IGY
Commented Jan 2, 2023 at 18:35