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  1. I am having a bit of trouble following a simple integral from the book on QFT by Mark Srednicki - free draft can be accessed at http://web.physics.ucsb.edu/~mark/qft.html - and I was hoping you could help me fill some gaps. The part of interest is chapter 14, pages 111-113, calculating loop contributions to the propagator in $\phi^3$. After he writes the expression for $I(k^2)$ (14.18): \begin{equation} I(k^2) = \int_{0}^{1}dx \int\frac{d^d\overline{q}}{(2\pi)^d}\frac{1}{(\overline{q}^2+D)^2}, \tag{14.18} \end{equation} he then differentiates twice w.r.t. $k^2$, to arrive at (14.20): \begin{equation} I''(k^2) = \int_{0}^{1}dx ~6x^2(1-x)^2 \int\frac{d^d\overline{q}}{(2\pi)^d}\frac{1}{(\overline{q}^2+D)^4}, \tag{14.20} \end{equation} which I get naively only if I treat the $\overline{q}^2$ in the denominator of the integrand as independent of $k^2$. Does the integral measure affect the computation? $D$ and $\overline{q}^2 = q^2$ are given by: \begin{equation} q=l+xk \tag{14.13} \end{equation} \begin{equation} D=x(1-x)k^2+m^2,\tag{14.14} \end{equation} where $x$ is a Feynman parameter, $k$ is the momentum of the particle and $l$ is momentum in the loop.

  2. On a slightly related note, below eq. (14.14), when he performs the Wick rotation, he shows anti-clockwise rotation and since poles are at $q_0=\omega-i\epsilon$ and $-\omega+i\epsilon$, the integral value is unchanged if we change the range to $-i\infty$ and $i\infty$. What happens if we close contours over the poles (say we rotate clockwise, or poles were at $q_0=\omega+i\epsilon$ and $-\omega-i\epsilon$)?

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1) No. You use translation invariance of the measure to neutralize the explicit $xk$ with a shift of integration variable (which it appears you've already done--note $\bar{q}$ is just a dummy variable in that integral). Therefore you just need to differentiate the $k^2$ dependence in $D$.

2) I'm not exactly sure what you're asking here, but I'll try to provide useful information. You can't rotate clockwise, you'll hit the poles. The point of the Wick rotation is that it takes into account the $i\epsilon$ needed for causality while allowing simple evaluation of the integral (just closing the contour would spoil the four dimensional symmetry, leading to ugly equations). If your question is why the rotation is performed counterclockwise instead of clockwise (that is, why we use $\epsilon$ and $-\epsilon$), I think it's easier to go back to the path integral derivation of the $i\epsilon$ trick. Recall the $i\epsilon$ is inserted so that the path integral projects in the far past and future onto the ground state, so that we produce the appropriate correlation functions for the LSZ formula. Flipping the sign on $\epsilon$ would fail to damp out the excited states, defeating the point of doing the trick in the first place. Alternatively, the $i\epsilon$ in the action acts as a Gaussian convergence factor in the functional integral (it looks like $e^{-\epsilon \int d^d x \phi^2(x)}$) so the sign is essential.

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  • $\begingroup$ Thank you for your answers! 1) I am not sure I follow. If I hadn't made the change $d^d l = d^d \overline{q}$, would I then have to worry about differentiating $\overline{q}^2 + D$ instead of only $D$? 2) In the text, it says : "In making this Wick rotation, the contour does not pass over any poles. (The $i\epsilon$’s are needed to make this statement unambiguous.). Thus the value of the integral is unchanged." How would choosing to go over the poles (rotate the other way around) alter the value of the integral? $\endgroup$ – Vangi Feb 12 '19 at 17:33
  • $\begingroup$ For 1) I haven’t looked at the calculation in the book in too much detail, but in these sorts of things you always complete the square after combining denominators. After completing the square you shift the integration variable to knock off the external momentum. For 2) it’s a basic fact of complex analysis that you can only deform integration contours in a domain in which the function is holomorphic (no poles). Going the other way will cause you to run into poles, so you can’t conclude that the deformation is allowed without changing the value of the integral. $\endgroup$ – Spencer Tamagni Feb 13 '19 at 19:05

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