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Consider the $t$-channel diagram of $\phi^4$ one-loop diagrams. Evaluated it is, with loop momenta $p$,

$$\frac{\lambda^2}{2}\displaystyle\int\frac{d^4p}{(2\pi)^4}\frac{1}{(p+q)^2-m^2}\frac{1}{p^2-m^2}.$$

If I want to regularize this using Pauli-Villars regularization, which is the correct method? The procedure is to make the replacement $$\frac{1}{p^2-m^2}\rightarrow \frac{1}{p^2-m^2}-\frac{1}{p^2-\Lambda^2}. $$

My question is do I apply the regularization to one term in the integral or both terms?

I've seen variations where the propagators become $$\frac{1}{p^2-m^2}\frac{1}{(p+q)^2-m^2}\rightarrow \frac{1}{p^2-m^2}\frac{1}{(p+q)^2-m^2}-\frac{1}{p^2-\Lambda^2}\frac{1}{(p+q)^2-\Lambda^2}$$ and also where we have $$\frac{1}{p^2-m^2}\frac{1}{(p+q)^2-m^2}\rightarrow (\frac{1}{p^2-m^2}-\frac{1}{p^2-\Lambda^2})(\frac{1}{(p+q)^2-m^2}-\frac{1}{(p+q)^2-\Lambda^2}).$$

In the latter case one ends up with four terms and each term is then evaluated using a Feynman parameter and integrating over Wick-rotated momenta, obtaining a logarithmic expression.

I'm pretty sure I've also seen where it was only applied to one of the terms.

Which is correct? (or are they equivalent?)

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  • $\begingroup$ Where did you see that first variation? If you have to make that particular replacement you mention, then you have to do that with each and every propagator. If then you can come up with a clever argument as to why the cross terms cancel, you can drop them, but not before. $\endgroup$ Commented May 19, 2015 at 8:09
  • $\begingroup$ In Paulli-Villars, a spin-statistics violating particle of mass $\Lambda$ (which is large) is added to the theory. Any Feynman diagram must include all contributions from this particle. At the end of the diagram, we then take $\Lambda \to \infty$. $\endgroup$ Commented Nov 12, 2015 at 22:01

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The first thing that you need to do is put the integral variable $p$ more clearly. You can do this with Feynman Parameters. $$ \frac{1}{AB}=\int_{0}^{1} dx \frac{1}{[A+(B-A)x]^2} $$ I will give you an exercise: find the $A$ and $B$, as well as $\Delta$ and $C$, such that: $$ \int\frac{d^4p}{(2\pi)^4}\frac{1}{(p+q)^2-m^2}\frac{1}{p^2-m^2}=\int_{0}^{1}dx\,C(x,q,m)\int\frac{d^4k}{(2\pi)^4}\frac{1}{[k^2-\Delta(x,q,m)]^2} $$ You see that this is a loop of mass $\sqrt{\Delta}$. Applying the Pauli-Villars regularization to this loop integral is simple now: $$ \frac{1}{[k^2-\Delta]^2}\rightarrow\frac{1}{[k^2-\Delta]^2}-\frac{1}{[k^2-\Lambda^2]^2} $$ Using the Wick rotation we have: $$ \int\frac{d^4k}{(2\pi)^4}\left[\frac{1}{(k^2-\Delta)^2}-\frac{1}{(k^2-\Lambda^2)^2}\right]=\frac{i}{8\pi^2}\int_{0}^{\infty}dk_E\left[\frac{k_E^3}{(k_E^2-\Delta)^2}-\frac{k_E^3}{(k_E^2-\Lambda^2)^2}\right] $$ $$ =\frac{-i}{16\pi^2}\ln\left(\frac{\Delta}{\Lambda^2}\right) $$ Then, the result is: $$ \int\frac{d^4p}{(2\pi)^4}\frac{1}{(p+q)^2-m^2}\frac{1}{p^2-m^2}=\int_{0}^{1}dx\,\frac{-iC(x,q,m)}{16\pi^2}\ln\left(\frac{\Delta(x,q,m)}{\Lambda^2}\right) $$

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  • $\begingroup$ Raising the dead here, but shouldn't you treat $\frac{1}{(k^2-\Delta)^2}$ as a multiplication of two propagators and perform regularization to each of them separately? This yields a different expression, e.g. a $2m^2\Lambda^2$ term when in common-denominator form, which doesn't exist in your expression. I'm sure yours is correct, but I'm puzzled by why this is the case, more so due to Lê Dũng's answer. $\endgroup$ Commented Apr 22, 2018 at 15:32
  • $\begingroup$ @Yoni, we have some freedom in chose the regularization here. Also, different regularizations will give different expressions. The only thing that holds here is that when we renormalizes our expressions we should get the same answer given an experimental input. $\endgroup$ Commented Apr 25, 2018 at 20:40
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Actually, in Pauli - Villars regularization, a divergence arising from a loop integral is modulated by a spectrum of auxiliary particles added to the propagator. In a loop, we have two propagators, so we need to modulate two of them.

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I found this in a lecture note, hopefully it will add a suitable reference.

enter image description here from link.

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    $\begingroup$ Sorry for the late reply but this is important for posterity: Your answer is extremely misleading. Yes, the quotation from the link you have indeed says that, but even in that pdf, on page 13 Section 22.4 near Equation (22.55) the text literally treated a case where the insertion was of two replacements rather than one. You have to insert enough to suppress the divergence. $\endgroup$ Commented Apr 9 at 5:21

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