I tried to adopt the cut-off regulator to calculate a simple one-loop Feynman diagram in $\phi^4$-theory with two different math tricks. But in the end, I got two different results and was wondering if there is a reasonable explanation.

The integral I'm considering is the following $$ I=\int^\Lambda\frac{d^4 k}{(2\pi)^4}\frac{i}{k^2-m^2+i\epsilon} \qquad\text{where}\qquad \eta_{\mu\nu}=\text{diag}(-1,1,1,1) $$ $\Lambda$ is the cu-off energy scale and $\epsilon>0$. Then I do the calculations.

Method #1 - Residue Theorem:

Since $$ I=i\int\frac{d^3\vec{k}}{(2\pi)^3}\int_{-\infty}^{+\infty}\frac{dk^0}{2\pi}\left[\frac{(2k^0)^{-1}}{k^0+z_0}+\frac{(2k^0)^{-1}}{k^0-z_0}\right] \qquad\text{where}\qquad z_0=\sqrt{|\vec{k}|^2+m^2}-i\epsilon $$ choosing the upper contour in $k^0$-complex plane which encloses the pole, $-z_0$, we have $$ \begin{align} I &= \int\frac{d^3\vec{k}}{(2\pi)^3}\frac{1}{2\pi i}\oint dk^0\frac{(-2k^0)^{-1}}{k^0+z_0}\\ &= \frac{1}{2}\int\frac{d^3\vec{k}}{(2\pi)^3}\frac{1}{\sqrt{k^2+m^2}}\\ &= \frac{1}{4\pi^2}\int_0^\Lambda \frac{k^2dk}{\sqrt{k^2+m^2}}\\ &= \frac{1}{8\pi^2}\left[\Lambda^2\sqrt{1+\frac{m^2}{\Lambda^2}}-m^2\ln\left(\frac{\Lambda}{m}\bigg)-m^2\ln\bigg(1+\sqrt{1+\frac{m^2}{\Lambda^2}}\right)\right]\\ & \approx \frac{1}{8\pi^2}\left[\Lambda^2-m^2\ln\left(\frac{\Lambda}{m}\right)-m^2\ln2\right] \end{align} $$

Method #2 - Wick Rotation:

Drawing the poles, $-z_0, z_0$, one finds the integration contour can be rotated anticlockwise so that, $$ \begin{align} I &= i\int\frac{d^3\vec{k}}{(2\pi)^3}\int_{-i\infty}^{+i\infty}\frac{dk^0}{2\pi}\frac{1}{k^2-m^2+i\epsilon}\\ &= -i\int\frac{d^3\vec{k}}{(2\pi)^3}\int_{-\infty}^{+\infty}\frac{idk_4}{2\pi}\frac{1}{k^2_E+m^2} \end{align} $$ where $k_4=-ik^0$ and $k_E^2=-k^2$, which are $4d$ Euclidean variables. So we have $$ \begin{align} I&=\int\frac{d^4k_E}{(2\pi)^4}\frac{1}{k^2_E+m^2}\\ &=\frac{1}{16\pi^2}\int_0^{\Lambda^2}\frac{k_E^2 d(k_E^2)}{k^2_E+m^2}\\ &=\frac{1}{8\pi^2}\left[\frac{\Lambda^2}{2}-m^2\ln\left(\frac{\Lambda}{m}\right)-\frac{m^2}{2}\ln\left(1+\frac{m^2}{\Lambda^2}\right)\right] \end{align} $$ Comparing the results obtained from the above two methods, we will find only the $\ln\Lambda$ dependent parts are the same; Other two parts ($\Lambda^2$-dependence and finite piece) are different. Since I use the same regulator, it's a bit wired to me how could the math tricks affect the results.


I don't think it is exactly the same regulator: In the first method, you integrate $\int_{-\infty}^\infty dk^0 \int^\Lambda d^3k$, but in the second calculation you integrate $\int^\Lambda d^4 k_E$.

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  • $\begingroup$ Thanks for your reply. I agree the two methods don't share exactly the same integration regime, and may give different answers. But if this was the case, I expect the two results coincide in the infinite-$\Lambda$ limit. However the ratio between them is approaching to 2, say, $I_{residue}/I_{wick}\rightarrow 2\vert_{\Lambda\rightarrow\infty}$. So this still puzzles me. $\endgroup$ – Di Liu Jul 5 '15 at 13:58
  • 2
    $\begingroup$ Now I understand, perhaps the underlying reason could be concluded as: The first method adopt an axial symmetry ($S^2\times R$), while the second method adopt the spherical symmetry ($S^3$); The difference between the two results arise from the nonidentical volumes of 4-cylinder and 4-ball. $\endgroup$ – Di Liu Jul 5 '15 at 17:31

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