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When computing loop integrals in QFT, one often encounters integrals of the form $$\int_{-\infty}^\infty\frac{dp^4}{(2\pi)^4}\frac{-i}{p^2+m^2-i\epsilon},$$ where we are in Minkowski space with metric $(-,+,+,+)$. One can Wick rotate to Euclidean space via the change of variable $p_0 \rightarrow ip_{0} $, and the integral becomes $$\int_{-\infty}^\infty\frac{dp^4}{(2\pi)^4}\frac{1}{p^2+m^2} = \frac{1}{(2\pi)^4}\int d\Omega_4 \int_0^{\Lambda}dp\frac{p^3}{p^2+m^2}= \frac{1}{8\pi^2} \frac{1}{2}\left(m^2\log\left(\frac{m^2}{m^2+\Lambda^2}\right)+\Lambda^2\right),$$ where we imposed a momentum cutoff $\Lambda$ to regulate UV divergence.

Alternatively, one can use the Sokhotski-Plemelj Formula to evaluate the integral. We first note that
$$\frac{1}{-p_0^2 + p_i^2 + m^2 -i\epsilon} = \mathcal{P}\left(\frac{1}{p_i^2 + m^2 - p_0^2} \right) + i\pi \delta(p_i^2 + m^2 - p_0^2).$$
Computing the $p_0$ integral first, the principle value will vanish since it is symmetric about $p_0$, and only the delta function term contributes. We have $$\int \frac{d^3p_i}{(2\pi)^3}\int\frac{dp_0}{2\pi} \pi \delta(p^2_i + m^2 - p_0^2) \\= \int \frac{d^3p_i}{(2\pi)^3}\int\frac{dp_0}{2} \frac{\delta \left(p_0-\sqrt{p_i^2+m^2}\right) + \delta\left(p_0+\sqrt{p_i^2+m^2}\right)}{|2p_0|_{p_0=\pm\sqrt{p_i^2+m^2}}} \\= \int \frac{d^3p_i}{(2\pi)^3} \frac{1}{2\sqrt{p_i^2+m^2}} \\ = \frac{1}{2(2\pi)^3}\int d\Omega_3 \int_0^\Lambda dp\frac{p^2}{\sqrt{p^2+m^2}} \\= \frac{1}{4\pi^2}\frac{1}{2}\left(\Lambda \sqrt{m^2+\Lambda^2} + m^2 \log\left(\frac{\sqrt{m^2+\Lambda^2} - \Lambda}{m} \right) \right), $$ which disagrees with the result calculated from Wick rotation.

I should expect both methods to produce the same result. Can anyone see why they disagree?

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  • $\begingroup$ Also, in the second line it seems that there is a typo: the factor in front of log should be $m^2$ instead of $\Lambda^2$ $\endgroup$ Nov 7, 2023 at 18:35
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    $\begingroup$ related? physics.stackexchange.com/q/192643 $\endgroup$
    – Diego
    Nov 7, 2023 at 19:14
  • $\begingroup$ No, you still have the typo. In the first method you should have $\Lambda^2+m^2\ln(m^2/(m^2+\Lambda^2))$. Note that here everywhere $\Lambda\gg m$, so both answers are identical in this sense. $\endgroup$ Nov 7, 2023 at 19:19
  • $\begingroup$ @Diego If I also impose $\Lambda$ cutoff to the $dp_0$ integral, it wont't change the $dp_0$ integral result since $\Lambda > \sqrt{p_i^2 + m^2}$ and the integral over the delta function will give the same result. $\endgroup$
    – Sean
    Nov 7, 2023 at 19:25

1 Answer 1

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For the Wick rotation method I find $$\frac{1}{2}\left(\Lambda^2+m^2\ln\frac{m^2}{\Lambda^2}\right),$$ because for me there is no reason to preserve $m^2+\Lambda^2$ in the denominator due to $\Lambda\gg m$. For the second integral I find $$\frac{1}{2}\left(\Lambda\sqrt{\Lambda^2+m^2}-m^2\text{arctanh}\,\left\{\frac{\Lambda}{\sqrt{m^2+\Lambda^2}}\right\}\right).$$ In this expression I know that $m\ll\Lambda$, so $$\Lambda\sqrt{\Lambda^2+m^2}=\Lambda^2.$$ But for $\text{arctanh}$ I will be more gently. I know the following relation, $$\text{arctanh}\,z=\frac{1}{2}\ln(1+z)-\frac{1}{2}\ln(1-z)=-\frac{1}{2}\ln\frac{1-z}{1+z}.$$ Next, $$1-\frac{\Lambda}{\sqrt{m^2+\Lambda^2}}=\frac{\sqrt{\Lambda^2+m^2}-\Lambda}{\sqrt{m^2+\Lambda^2}}\approx \frac{m^2}{2\Lambda^2},$$ $$1+\frac{\Lambda}{\sqrt{m^2+\Lambda^2}}\approx 2,$$ therefore $$\frac{1}{2}\left(\Lambda\sqrt{\Lambda^2+m^2}-m^2\text{arctanh}\,\left\{\frac{\Lambda}{\sqrt{m^2+\Lambda^2}}\right\}\right)=\frac{1}{2}\Lambda^2+\frac{1}{2}\cdot\frac{1}{2}m^2\ln\frac{m^2}{\Lambda^2}.$$ Now I collect "angular" terms. For the Wick rotation method, I write $$\frac{1}{(2\pi)^4}\cdot(2\pi^2)\cdot\frac{1}{2}\left(\Lambda^2+m^2\ln\frac{m^2}{\Lambda^2}\right)=\frac{1}{8\pi^2}\cdot\frac{1}{2}\left(\Lambda^2+m^2\ln\frac{m^2}{\Lambda^2}\right).$$ For the S-P method, I find $$\frac{1}{(2\pi)^3}\cdot(4\pi)\cdot\left(\frac{1}{2}\Lambda^2+\frac{1}{2}\cdot\frac{1}{2}m^2\ln\frac{m^2}{\Lambda^2}\right)=\frac{1}{8\pi^2}\cdot\left(\Lambda^2+m^2\ln\frac{m^2}{\Lambda^2}\right),$$ where I do not distinguish $2\Lambda^2$ and $\Lambda^2$. Both answers are similar in sense of $\Lambda\gg m$ behavior, but the second is two times smaller. It seems that it happens due to taking into account both zeros in the delta-function when the author performs integation over $p_0$. It seems that in oder to have causality, we have to deal with time-like vectors, which implies $p_0>0$ and we should take into account only positive zero of delta-function, $+\sqrt{p_i^2+m^2}$, which adds the factor $1/2$ into the final answer. This situation realizes when one computes vacuum polarization diagram.

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  • $\begingroup$ Thanks! For the comment regarding regularizations, if I also impose $\Lambda$ cutoff to the $dp_0$ integral, it wont't change the $dp_0$ integral result since $\Lambda > \sqrt{p_i^2 + m^2}$ and the integral over the delta function will give the same result. In that sense, the regularizations should be the same in both cases? $\endgroup$
    – Sean
    Nov 7, 2023 at 19:29
  • $\begingroup$ @Mr.Chen , I feel that something happens when you take both zeros of delta-function into account. For instance, when one computes vacuum polarization diagram by Sokh-Plem formula, it happens that the integration domain corresponds to time-like 4-vectors $p^{\mu}$, which implies $p_0>0$ and you have take into accoun only positive zero, $+\sqrt{p_i^2+m^2}$. It gives additional $1/2$ factor into the final answer $\endgroup$ Nov 7, 2023 at 19:38
  • $\begingroup$ Interesting. Can you please explain more about some intuition behind using only one zero? Mathematically, we should include both, right? $\endgroup$
    – Sean
    Nov 7, 2023 at 19:49
  • $\begingroup$ @Mr.Chen , you can check Landau-Lifshitz vol. 4, chapter XII, paragraph 115. No, we take only one zero in case of vacuum polarization. $\endgroup$ Nov 7, 2023 at 19:51
  • $\begingroup$ @Mr.Chen , also check the point about the integration domains in the comment to answer for physics.stackexchange.com/q/192643 $\endgroup$ Nov 7, 2023 at 19:52

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