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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 reguarization 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. $\endgroup$ – David Vercauteren May 19 '15 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 $\Lambda \to \infty$. $\endgroup$ – Prahar Mitra Nov 12 '15 at 22:01
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The first thing that you need to do is put the integral variable $p$ in a more clear way. You can doo 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. $\endgroup$ – Yoni Apr 22 '18 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. $\endgroup$ – Nogueira Apr 25 '18 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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