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In calculating transition amplitude for Klein-Gordon real-scalar field, I encountered the integral,

$$ \frac{-i}{2(2\pi)^2\Delta x} \int^{\infty}_{-\infty} \,dk \frac{ke^{ik\Delta x}}{\sqrt{k^2+m^2}} $$

I can see here the integrand has branch cuts at $ k= \pm im $ However, later they do a change of variables $ z= -ik $ and then the integral becomes,

$$ \frac{1}{2(2\pi)^2\Delta x} \int^{\infty}_{m} \,dz \frac{ze^{-z\Delta x}}{\sqrt{z^2-m^2}} $$

And it is said that they can wrap the contour around the upper branch cut for $ \Delta x > 0 $

enter image description here

I am not able to see how this transformation happens and how the contour can be wrapped around the upper branch. Thanks for your inputs.

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Here is a rough explanation of what they do. You start with an integral over a complex variable $K=k+iz$ along a path $\mathcal C_0$ (in the complex plan) that follow the real axis (that is a path characterized by $z=0$ and $k$ between $-\infty$ and $+\infty$). Calling $I$ the integral without the prefactors, we thus start with $$ I = \int_{\mathcal C_0} dK \frac{Ke^{iK\Delta x}}{\sqrt{K^2+m^2}}.$$ Now the integrand is analytic everywhere but on the cuts, so you can change the contour the way you want as long as it does not go through the cut, and the integral still converges.

The usual way to do this kind of integrals is to change the contour such that you just have to integrate around poles or along the cuts. But if you want to do that, you need that the contour has a contribution from $z$ at $\pm \infty$, depending if you go along the cut in the upper/lower half plan. Now you see the role of $\Delta x$: if $\Delta x>0$, $e^{iK\Delta x}$ converges/diverges if $z\to\pm\infty$, so you can't integrate along the lower cut (the integral is divergent), but you can for the upper cut. That's how you get the contour in your graph, let's call it $\mathcal C_1$.

To do the integral, you now have to parametrize $\mathcal C_1$ by two integrals : first the part along the left of the cut: $k=-\epsilon$ and $z$ going from $\infty$ to $m$; and second the part along the right of the cut: $k=\epsilon$ and $z$ going from $m$ to $\infty$. $\epsilon$ is an infinitesimal number that you have to send to zero at some point.

With all that, you should be able to recover the result you are looking for.

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  • $\begingroup$ Thanks for your answer (even though I did not ask the question). I got two questions: Are you (implicitly) using Jordan's lemma in the second parapraph? And are you talking about "principal value" in the third paragraph? $\endgroup$ – Hunter Jan 17 '14 at 15:34
  • $\begingroup$ @Hunter:I guess one can use a variation of Jordan's lemma (for each quarter of circle, since there's the cut) to show that. I'm not using the principal value. $\epsilon$ here insure that the integral is not on the cut, but slightly on the right or on the left. $\endgroup$ – Adam Jan 17 '14 at 16:20
  • $\begingroup$ @Adam : Thanks a lot for your answer. It clarifies a lot of things here. I will try to do the math in more in detail and get back to you. $\endgroup$ – user35952 Jan 17 '14 at 16:30
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    $\begingroup$ @Trimok: Good question. It's because $\sqrt{(\pm \epsilon+iz)^2+m^2}$ have opposite sign (that's why there's a cut in the complex plan !), which compensate for the opposite direction of the two parts of the integral. So all in all, the two parts give the same contribution (I'm wondering if a factor of two is not missing in the OP's second integral). $\endgroup$ – Adam Jan 17 '14 at 18:40
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    $\begingroup$ @YulOtani: I don't see your problem. The exponential makes the integral convergent in the limit $z\to \infty$, so we don't care that $z/\sqrt{z^2-m^2}\to 1$ is this limit... $\endgroup$ – Adam Jan 20 '14 at 15:36

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