# Wick Rotation vs Sokhotski-Plemeli Method to compute internal loop of Feynman correlators

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?

• Also, in the second line it seems that there is a typo: the factor in front of log should be $m^2$ instead of $\Lambda^2$ Commented Nov 7, 2023 at 18:35
• Commented Nov 7, 2023 at 19:14
• 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. Commented Nov 7, 2023 at 19:19
• @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.
– Sean
Commented Nov 7, 2023 at 19:25

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.
• 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?
• @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 Commented Nov 7, 2023 at 19:38