# An integral in Peskin & Schroeder's Quantum Field Theory p. 27

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

They 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 $$\frac{1}{4\pi^2 r}\int_m^\infty d\rho \frac{\rho e^{-\rho r}}{\sqrt{\rho^2-m^2}}$$.

• 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. Commented Oct 9, 2016 at 2:39
• Thanks for posting this... its the year 2024 and I dare say physics stackexchange is even better than asking the author himself if you're learning something and are confused :D Commented Jul 9 at 4:34

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.

• 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}]$}$ Commented Oct 11, 2016 at 3:08
• 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. Commented Oct 11, 2016 at 7:10
• 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? Commented Feb 20, 2018 at 8:46

I'll just add a little more information about pushing the contour upward, since this was very confusing to me at first. We may push the contour of integration up, as long as we keep the endpoints fixed and don't pass through any singularities. This is because Cauchy's theorem says that the integral around any closed loop that doesn't enclose any singularities is zero. Therefore, any two contours with the same endpoints and no singularities between them must give the same integral so that they cancel each other when made into a loop contour.

We may create a U shaped contour wrapping around the fixed point $$p=im$$ and going up to infinity, as long as the endpoints are fixed at $$\pm\infty$$ on the real axis. This means the full contour is the U contour shown in Peskin and Schroesder plus two giant arcs at infinity, connecting the two top ends of the U contour to $$\pm\infty$$. However, in the limit as $$r\xrightarrow{}\infty$$, the imaginary part of $$p$$ along these arcs will combine with $$ir$$ in the exponential to make the integrand approach zero everywhere along these arcs everywhere except for a small region near the real axis. However, the portion near the real axis where the integral in non-negligible becomes vanishingly small as $$r\xrightarrow{}\infty$$. Therefore, the contribution from the arcs is zero in this limit, and we are left with the U contour.

• Hi, can you elaborate a bit more on how the two arcs on infinity have zero contribution? Commented Mar 17, 2021 at 17:15
• I 100% agree that there are two big arcs we need to worry about. (I have been worrying about them independently and that's why I found this post.) However I am not convinced that the answer given above is correct. I think it mixes up two different r's. The r in P&S's formula is the spatial separation between x and y in a frame in which tx=ty. In contrast the large arcs are at a different radius R which actually controls the magniture of mod(p) and it is for R (not r!) going to infinity that we must show the big-arcs vanish. I have not mananged to show this as Jordan's lemma fails for R->inf. Commented Apr 6, 2023 at 23:32
• OK - after a bit more thought I will qualify/amend some of what I said in my last comment. I now see that the "full" argument implicit (needed but not stated) in P&S is that at fixed r and R the missing big arcs do NOT vanish, and so should not be neglected, but can be shown (by Jordan's lemma) to contribute an amount to P&S's D(x-y) which is bounded in magnitude by pi/r^2, and so since the RHS of P&S's (2.52) looks only at the limit r->infinity, we can drop the contributions from the big arcs. Quite probably possibly this is what @Yachsut had in mind all along, but I ... Commented Apr 7, 2023 at 9:27
• ... but I (mis?)interpreted his/her explanation as a suggestion that the big arcs were negligible in the R->infinity limit. Commented Apr 7, 2023 at 9:28
• For what it's worth, I have created a specific question to address the issue that I've touched upon above but can't be described well in comments alone. The new question is here: physics.stackexchange.com/questions/758661/… Commented Apr 8, 2023 at 1:16

Peskin and Shroeder's argument is just wrong.

What is not correct is their argument that one can just evaluate the integral on a contour which has been pushed up on either side of the branch cut to establish that the propagator goes like $$e^{-mr}$$ at large $$r$$.

It is a case of "right answer wrong method" as the propagator in question does indeed go like $$e^{-mr}$$ at large $$r$$ as is nicely explained in an answer to a question of my own here: What is the missing part of the argument needed to justify the claim of (2.52) in Peskin and Schroeder's QFT textbook? The only problem is that Peskin and Shroeder's argument is an invalid way of trying to justify the result.

I applaud user @Yachsut for his/her answer elsehwere in the present question (the answer beginning

"I'll just add a little more information about pushing the contour upward, since this was very confusing to me at first .... "

) since he/she was right: (a) to have found Peskin and Schroeder's argument confusing, and (b) to have identified that the arcs that Peskin and Schroeder don't talk about are (or should be) in important part of the puzzle. But what @Yachsut and commenters on @Yachsut answer and on other answers elsewhere on the site appear not to have realised, though, is that those arcs can't be put back to keep Cauchy's Theormem happy without causing other (bigger) problems with Peskin and Schroeder's argument. The reason is because the relevant integral along those arcs is badly behaved.

In short: don't try to understand Peskin and Schroeder's argument because it's bogus, although you can believe the answer they give for the reason explained in the answer I accepted by @Siam to What is the missing part of the argument needed to justify the claim of (2.52) in Peskin and Schroeder's QFT textbook?