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In page 319 in Peskin's An Introduction to Quantum Field Theory, renormalization of QED is discussed, and it's shown that there are only four divergent quantities involved. But this conclusion is based on the assumption that potentially divergent amplitude is an analytic function of the external momenta, for example for the electron self energy $i\Sigma(\displaystyle{\not p})=A_0+A_1\displaystyle{\not p}+A_2 \displaystyle{\not p}^2+\ldots$. Higher order coefficients in the Taylor series are more and more convergent by simple power counting as is argued in the book, hence we always get a finite number of divergent quantities in an amplitude. However, I can't see an obvious reason why the amplitude should be an analytic function of the external momenta. In my simple mind free from math rigor now, I think this would be true if the internal propagators can't go on-shell, but I can't see why this must be true either.

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  • $\begingroup$ See 302405 and 211499. $\endgroup$ Sep 30, 2017 at 21:38
  • $\begingroup$ just differentiate enough times the amplitude with respect to the external momentum and you will get at some point a convergent integral. Integrate back the same amount of times in the momentum, and you will get a polynomial in the external momenta with unknown integration constants. Still is a polynomial hence analytic. $\endgroup$
    – TwoBs
    Oct 2, 2017 at 20:46
  • $\begingroup$ @TwoBs Thanks for the interest but I am not sure I understand what you mean. I think you mean a formal series not a polynomial, right? By simple power counting the coefficients are eventually finite, but I think internal propagator going on-shell can ruin this argument. $\endgroup$ Oct 2, 2017 at 21:36
  • $\begingroup$ I really mean a polynomial of finite order for renormalizable theories, a series for non-renormalizable ones. No pole is touched in the last integral, even when particles go on-shell (unelss there are IR divergence for massless particles), which are absolutely convergent in the euclidean space that you can get to by Wick rotating, thanks to the $+i\epsilon$ prescription that tell you how to analytically continue. For an example look at e.g. how Weinberg pass from eq. 16.2.14 to 16.2.15 in QFT-vol.II $\endgroup$
    – TwoBs
    Oct 2, 2017 at 21:42
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    $\begingroup$ don't get me wrong, the amplitude is not an analytic function of the external momenta in general. Particles are associated to poles, particle production is associated to branch cuts, resonances are pole in the second Riemann sheets and so on. In fact, most of the actual physical beef is in the non-analytic terms. The correct claim is that the divergent parts are analytic functions of the external momenta, which is the very reason why they can be reabsorbed by local counterterms in the lagrangian. $\endgroup$
    – TwoBs
    Oct 2, 2017 at 21:49

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