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By definition of P-Sh $\Sigma_2(p)$ is not a complete NLO the electron propagator. Instead it is the contribution of the one-loop diagram 7.15. In the naive perturbation theory this really is the only contribution. But in the renormalized perturbation theory in this order there is also the contribution of the counterterm, i.e. the full NLO is $$i(\gamma^\mu ...


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To understand the essence of perturbative renormalization you don't need any quantum field theory nor any quantum mechanics. A simple toy model suffices. Suppose your theory makes a prediction for two distinct observables $F$ and $G$ in terms of a perturbative parameter $g$: $$F = g + g^2 (S+1) + g^3 (S+1)^2 + g^4 (S+1)^3 + ...$$ $$G = g + g^2 (S-1) + g^3 ...


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Normally renormalization is necessary when the phenomenological constants acquire unnecessary perturbative corrections, not obligatorily divergent. But there may be other cases. Read, for example, http://www.physics.umd.edu/courses/Phys851/Luty/notes/renorm.pdf for QM.


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if my paper is correct then for every divergent integral $ \int_{0}^{\infty}x^{m}dx $ will depend on the value $ \zeta (-m) $ http://terrytao.wordpress.com/2010/04/10/the-euler-maclaurin-formula-bernoulli-numbers-the-zeta-function-and-real-variable-analytic-continuation/ and http://vixra.org/abs/1305.0171 for a recurrence realting divergent series and ...


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I find the whole notation here a bit confusing since we are talking about momenta modes while using real space notation (maybe I'm the only one...). To clarify what is going on we can switch to momentum space instead. Consider the quadratic term in the exponential: \begin{align} \int d^4x \phi ^2 & = \int d^4x \int d^4k d^4k' \phi _k \phi ^\ast _{ k ' ...


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I'm not being precise, but morally: Imagine you were integrating out all modes above a frequency $b\Lambda$. Consider $\omega < b\Lambda < 3 \omega$. A mode $\phi$ with frequency $\omega$ when cubed, will have some part of it as mode of frequency $3 \omega$, since: $\sin (3x) = 3 \sin (x) - 4 \sin^3 (x)$. (Easier to see that ${(e^{i \omega t})}^3 = ...



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