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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!

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    $\begingroup$ 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

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

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  • $\begingroup$ Many thanks for the answer! Is $|k|$ the magnitude of momentum? $\endgroup$
    – IGY
    Jan 2, 2023 at 17:57
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    $\begingroup$ 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
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    $\begingroup$ @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

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