In Peskin and Schroeder, I have a problem with a claim in equation (2.54), which I will rewrite more concisely here. He claims that we have the following equality :

$$ \frac{1}{2E_p}e^{-iE_p(x_0)}-\frac{1}{2E_p}e^{iE_p(x_0)} = \int\frac{dp_0}{2\pi i}\frac{-1}{(p_0-E_p)(p_0+E_p)}e^{-ip_0x_0}.\tag{2.54} $$

Where $x_0 > 0, E_p>0$ and the complex integral should be carried along a contour that avoids with semi-circles "from above" the two singularities at $E_p$, and $-E_p$, and closing it from below.

Now, I have two problems with that. First of all, it seems in his book he takes into account only the contributions of the residues, which give exactly the equation above. However, when doing carefully the integration along the small semicircles around the poles, we find that they both contribute, so that we obtain the right expression only up to a factor of 2, i.e.:

$$\int\frac{dp_0}{2\pi i}\frac{-1}{(p_0-E_p)(p_0+E_p)}e^{-ip_0x_0} = -2\pi i(\frac{1}{2E_p*2\pi i}e^{-iE_p(x_0)}-\frac{1}{2E_p*2\pi i}e^{iE_p(x_0)})+\pi i(\frac{1}{2E_p*2\pi i}e^{-iE_p(x_0)}-\frac{1}{2E_p*2\pi i}e^{iE_p(x_0)}) = \frac{1}{4E_p}e^{-iE_p(x_0)}-\frac{1}{4E_p}e^{iE_p(x_0)}.$$

And this is done with the integration contour that he states, only it seems that he doesn't take into account the contribution of the small semicircles.

I think the problems stems from the fact that this integral is actually divergent, since we have first order poles on the real axis, so we need a prescription to regularise it. The one I did here, and the one he seems to be doing in his book, is using the principal value.

What I think he actually should have said in his book is that the prescription is the following:

$$\int\frac{dp_0}{2\pi i}\frac{-1}{(p_0-E_p+i\epsilon)(p_0+E_p+i\epsilon)}e^{-ip_0x_0}.$$

So that the poles are move from the real axis and then we don't have any contribution from the small semicircles.

Is this assumption right? Or have I somehow misunderstood/miscalculated Peskin's prescription?

P.S.: Since it was added in the comments, I detail here the semi-circle contribution, in case it is useful (which is, as stated in the answer, convergent ONLY if we take the principal value of the integral)

Let us do the integration along the semi-circle around $E_p$. The parametrisation is $z = E_p+\epsilon e^{i\theta}$ with $\theta \in [\pi,0]$(Note that in this parametrisation we implicitly took the principal value since the integration is symmetric around the pole). Writing the integral we obtain:

$$\int \frac{i \epsilon e^{i\theta}d\theta}{2\pi i} \frac{-1}{\epsilon e^{i\theta}(2E_p+\epsilon e^{i\theta})}\exp{\left(-i(E_p+\epsilon e^{i\theta})\right)}\\ = i\int_\pi^0 \frac{d\theta}{2\pi i} \frac{-1}{(2E_p+\epsilon e^{i\theta})}\exp{\left(-i(E_p+\epsilon e^{i\theta})\right)}.$$

Doing the series expansion in $\epsilon$ we see that all the terms that are of the form $e^{in\theta}$ vanish in the integration, so we can consider only the constant term which yields:

$$i\int_\pi^0 \frac{d\theta}{2\pi i} \frac{-1}{(2E_p)}e^{-i(E_p)} = \frac{1}{4E_p}e^{-iE_p}.$$

  • $\begingroup$ Looks good. Could you describe the careful cslculation of the semicircle contribution. Shdnt this part diverge if the whole integral diverges? Apart from that i would support your conclusion $\endgroup$
    – lalala
    Commented Apr 27, 2019 at 17:45
  • $\begingroup$ Indeed, even closing the semicircles in the $p_0 < 0$ region yields the same result. I believe that happens because we are actually calculating the principal value, but I don't know how to prove it. The prescription of shifting the poles seems to work though. $\endgroup$
    – ErickShock
    Commented Apr 28, 2019 at 0:20
  • 2
    $\begingroup$ The prescription is what they say, deform the contour by moving it off in complex plane, away from the poles. After this is done, the integral is convergent. There is no contribution from small semicircles. $\endgroup$ Commented Apr 28, 2019 at 5:04
  • 1
    $\begingroup$ It's not that he didn't include the semicircle contributions. The sum of residues is equal to the integral over the contour. Removing the half circle at infinity since it vanishes, the remaining contour is the sum of residues. In particular, the principal value of the integral should not be the sum of the residues; "including" the semicircle contributions is what gives you the right answer. In other words, you shouldn't be calculating the semicircles separately (or at all). This then is equivalent to deforming the contour or shifting the poles. $\endgroup$
    – Aaron
    Commented Apr 28, 2019 at 13:22

1 Answer 1

  1. P&S don't write explicitly the integration contour $$\begin{align}& -\int_{\color{red}{\gamma}}\frac{dp^0}{2\pi i}\frac{e^{-ip^0x^0}}{(p^0-E_{\bf p})(p^0+E_{\bf p})}\cr ~=~&-\frac{1}{2E_{\bf p}}\int_{\color{red}{\gamma}}\frac{dp^0}{2\pi i}e^{-ip^0x^0} \left(\frac{1}{p^0-E_{\bf p}}-\frac{1}{p^0+E_{\bf p}}\right)\end{align}$$ on the rhs. of eq. (2.54), but they mention in words and pictures that it is the retarded integration contour above the real axis $$\begin{align}& -\int_{\color{red}{\mathbb{R}+i\epsilon}}\frac{dp^0}{2\pi i}\frac{e^{-ip^0x^0}}{(p^0-E_{\bf p})(p^0+E_{\bf p})}\cr ~=~&-\int_{\mathbb{R}}\frac{dp^0}{2\pi i}\frac{e^{-ip^0x^0}}{(p^0-E_{\bf p}+\color{red}{i\epsilon})(p^0+E_{\bf p}+\color{red}{i\epsilon})},\end{align}$$ which by the residue theorem is equal to $$ \frac{1}{2E_{\bf p}}e^{-iE_{\bf p}x^0}-\frac{1}{2E_{\bf p}}e^{iE_{\bf p}x^0}$$ for $x^0>0$ (by closing the integration contour in the lower half-plane, clockwise).
  2. In particular, the integral on the rhs. of eq. (2.54) is not the principal value $$-\color{red}{P}\int_{\mathbb{R}}\frac{dp^0}{2\pi i}\frac{e^{-ip^0x^0}}{(p^0-E_{\bf p})(p^0+E_{\bf p})},$$ which would lead to half the correct value (still assuming $x^0>0$).

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