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When evaluating the Klein-Gordon propagator, in the book by P&S, p. 31, I see that, it is customary to shift the poles and add $i\epsilon$ in the denominator. I don't understand, why this is necessary. Why can't we just use complex analysis? What is wrong in the following steps?

\begin{align} \int \frac{e^{ibz}}{z^2-a^2}\, dz &= (2\pi i) \left[\lim_{z\rightarrow a} (z-a) \frac{e^{ibz}}{z^2-a^2} + \lim_{z\rightarrow -a} (z+a) \frac{e^{ibz}}{z^2-a^2}\right] [\mathrm{Residue~theorem}]\nonumber\\ % &= (2\pi i) \left[\lim_{z\rightarrow a} \frac{e^{ibz}}{z+a} + \lim_{z\rightarrow -a} \frac{e^{ibz}}{z-a}\right]\nonumber\\ % &= (2\pi i) \left[ \frac{e^{iba}}{2\,a} - \frac{e^{-iba}}{2\,a}\right]\nonumber\\ % &= \frac{i\pi}{a} \left[ e^{iba} - e^{-iba}\right]\nonumber\\ % &= - \frac{2\, \pi\, \sin{ba}}{a} \end{align}

What goes wrong in proceeding this way? Can't we just do the integration $p^0$ as is done for the $z$-variable? Obviously, $a$ will be function of $\vec{p}$ and $m$.

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Note that the original integral you are trying to compute is over the real line, not over a closed contour, so the Cauchy theorem does not apply until you find a suitable way to close the contour. Due to the presence of the exponential factor $e^{ibz}$, as you have written it, one can close the contour in the upper half plane if $\mathrm{Re}\, b>0$. Let's assume that's the case. Now your two poles are actually on the real line, so we also need to specify which way to pass around them. Since you are closing the contour above, and you are picking up both of the residues, you are implying that you are passing below these two poles. If you passed above them, they would be outside your contour and would not contribute. Since you are passing below your two poles, we could equivalently describe what you did by saying that the two poles are shifted upwards on the complex plane by an infinitesimal amount $+i\epsilon$. This would guarantee that you pass below them as you integrate along the real axis. So you see that you also actually have included some $\epsilon$s in your calculation too, although you didn't acknowledge it.

For calculations in QFT, there is a correct physical prescription for which way to go around the poles, which is called the Feynman prescription, and differs from what you did above. This is covered well in P&S.

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  • $\begingroup$ Thanks for the clear explanation. $\endgroup$ Dec 27 '20 at 13:46
  • $\begingroup$ take a look at he following link: math.stackexchange.com/questions/564952/… $\endgroup$ Dec 27 '20 at 15:46
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    $\begingroup$ @SamapanBhadury yes that math post seems to have the same moral value as ours. Also, I thought I should note that your calculation corresponds to P&S Eq. 2.54, which indeed is just a certain choice of propagator, i.e., a certain choice of boundary conditions. It's not wrong, but it's just not the one that's appropriate for QFT, and my point is that any time one writes down an expression for a propagator there is implicitly some choice of boundary conditions, and those can be encoded in some choice of $i\epsilon$ prescription. $\endgroup$
    – kaylimekay
    Dec 27 '20 at 16:02
  • $\begingroup$ Right. After looking around the web on this topic, I am getting a sense that due to the possible boundary conditions, this $i\epsilon$ method is chosen. Now I am wondering whether at some point the QFT formulation and the mathematical one should overlap or not. If possible, I would like to know how different choices of the boundary condition affect the $i\epsilon$ choice. $\endgroup$ Dec 27 '20 at 16:09
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    $\begingroup$ @SamapanBhadury you are right, there definitely is a physical reason for it. It's more than we can talk about in comments. Also, I don't know how far along you are in studying these things. But P&S discuss the significance of these boundary conditions more in Chapter 4. So if you haven't got there yet, check it out, and please come back to Physics SE if you have more questions then. $\endgroup$
    – kaylimekay
    Dec 27 '20 at 16:23

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