In Peskin's QFT P. 27, there is an integral $$-\frac{i}{2 (2 \pi )^2 r}\int_{-\infty}^{\infty} \mathrm{d}p\frac{p\ e^{ipr}}{\sqrt{p^2+m^2}}.\tag{2.51a}$$

He said that in order to push the contour up to wrap around the upper branch cut. After some manipulation, it gives the following integral $$\frac{1}{4\pi^2 r}\int_m^\infty d\rho \frac{\rho e^{-\rho r}}{\sqrt{\rho^2-m^2}}.\tag{2.52}$$

  1. I don't understand the phrase "push the contour up to wrap around the upper branch cut". The integral is on the real line and hence there is no singular point along the line.

  2. If we define $\rho =-i p$, I still can't get $\frac{1}{4\pi^2 r}\int_m^\infty d\rho \frac{\rho e^{-\rho r}}{\sqrt{\rho^2-m^2}}$.

  • $\begingroup$ 1. Peskin is talking about the branch cuts on the positive and negative imaginary axes. (Look for where the quantity in the square root becomes negative.) 2. You should show some of your work, because I can't know what you missed. $\endgroup$ – knzhou Oct 9 '16 at 2:39

In those paragraphs, P&S want to obtain the behaviour of the amplitude $D(x-y)$ in the limit $r \to \infty$. Starting from the integral $$ \tag{1} \label{int} \frac{-i}{2(2\pi)^2 r} \int_{-\infty}^\infty \text{d}p \frac{p e^{ipr}}{\sqrt{p^2 +m^2}}, $$ you can see that a limit for $r \to \infty$ is not well defined, since the exponential has an oscillating behaviour. Also, the final form of the integral can be recognized as a modified Bessel function of the second kind. See this Wikipedia page for the asymptotic expansion of this function. The modification of the integration contour is a way to obtain a more straightforward integral. I hope this answers your question 1.

About point 2., in order to evaluate this integral, P&S apply the Cauchy's theorem defining the integrand as a function of the complex variable $p$. The square root in the denominator causes the branch cut in Figure 2.3. When rewriting the contour integral as $$ \tag{2} \label{1} \int_{i \infty}^{i m} \dots + \int_{i m}^{i \infty} \dots $$ we have to be careful in writing the integrands, since the branch cut generates a discontinuity between the two sides of the cut. In fact the square root in the complex plane is a multi-valued function, and the argument of $p^2+m^2$ shifts of $2\pi$ when passing from the right of the branch cut to the left. Thus $$ \sqrt{p^2 + m^2} = \cases{|p^2+m^2|^{1/2} \exp\left(i\frac{\arg(p^2+m^2)}{2}\right), \\ |p^2+m^2|^{1/2} \exp\left(i\frac{\arg(p^2+m^2)+ 2\pi}{2}\right) = - |p^2+m^2|^{1/2} \exp\left(i\frac{\arg(p^2+m^2)}{2}\right),} $$ where the first value is at the right and the second at the left of the branch cut. You see that the difference is a minus sign.

Therefore, making the substitution $p = i \rho$, \eqref{1} becomes $$ i\int_\infty^m \text{d} \rho \frac{\rho e^{-r \rho}}{- \sqrt{\rho^2 - m^2}} + i\int_m^\infty \text{d} \rho \frac{\rho e^{-r \rho}}{\sqrt{\rho^2 - m^2}}, $$ which finally gives $$ 2i \int_m^\infty \text{d} \rho \frac{\rho e^{-r \rho}}{\sqrt{\rho^2 - m^2}}. $$ Substituting this in \eqref{int} you find your result. Note that if you had chosen a different contour, for example the analogous of the "pushed" contour of Figure 2.3 but with the U-turn at $p = 0 + i 0$, you would not have the phase difference in the integrands on the lines $(im, 0)$ and $(0, im)$, so these two pieces would have cancelled each other (see \eqref{1}) and you would have obtained the same result with $m$ as lower integration limit.

  • $\begingroup$ Your last phrase, I think if I were to choose U-turn at $p = 0 + i 0$, I would have zero due to the continuity but not the phase difference in the integrands. Otherwise, I would have something like $\int _{\text{im}}^0\fbox{$[-\text{INTEGRAND}]$}+\int _0^{\text{im}}\fbox{$[\text{INTEGRAND}]$}=\int _0^{\text{im}}\fbox{$[\text{INTEGRAND}]$}+\int _0^{\text{im}}\fbox{$[\text{INTEGRAND}]$}=2\int _0^{\text{im}}\fbox{$[\text{INTEGRAND}]$}$ $\endgroup$ – ZHANG Juenjie Oct 11 '16 at 3:08
  • $\begingroup$ Yes, that's what I said: "you would not have the phase difference in the integrands on the lines $(im,0)$ and $(0,im)$", and so you don't have that minus sign in the first term. Thus when inverting the integration limits you have zero, unlike when integrating around the branch cut. $\endgroup$ – ric_n Oct 11 '16 at 7:10
  • $\begingroup$ The argument of $p$ certainly shifts by $2\pi$ from one side to the other of the cut, but is the fact that the argument of $p^2 + m^2$ also shift by the same amount trivial? $\endgroup$ – Andrea Feb 20 '18 at 8:46

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