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Here is my tentative answer. First of all, this answer is all based on the conventional QED, where we know the electron's self-energy is like $\Sigma=\Sigma({\not}{p})$. My original question deals with nonconventional QED where $\Sigma$ may also have momentum dependence on $p^\mu$ (which could be contracted with a background field of some kind, for instance, ...
As Trimok said, the probability of scattering of some nicely focused packets will still go like the cross section and like $|{\mathcal M}|^2$. For bosons, there are no energy-dependent extra factors, so $|{\mathcal M}|^2$ itself has to be smaller than a number of order one for the probability to stay smaller than one. This is related to $T^\dagger ... 0 Thanks to @Trimok , the reference he provided gives some detailed calculation. To see one of the missing Feynman diagrams (you can draw others once you understand that) and more detailed calculation of the leading order (not one loop), you can also check http://www.itp.phys.ethz.ch/research/qftstrings/archive/13FSProseminar/LEEre_Guns which gives more useful ... 6 The correct option is really option 3. Most of the time when a physicist says a theory is renormalizable, they mean that the theory is a relevant deformation of some conformal field theory. This is a non-perturbative definition. It contains the physically meaningful content that the other more technical definitions about counterterms in perturbation ... 2 Usually one refers to option 2) when talking about the renormalizability of a theory. Often power counting is used to determine at a glance whether a theory has a chance of being renormalizable or not. 0 I've found the answer to this question, and it does basically follow from holomorphicity. The holomorphicity property is essentially the statement that the operator is BPS:$\Phi$is annihilated by a supercharge$Q$. This is also the reason the property does not hold for vector superfields, which are not BPS. The reason that the dimension of the operator is ... 1 I will give some hints: It is anomalous scaling dimension. scaling dimension is defined as $$x \rightarrow \lambda x,\\ \phi(x) \rightarrow \lambda^\Delta \phi(\lambda x)$$ From the foemula (3.45) (maybe it is better to be$\phi(x) \rightarrow \Lambda^\frac{d-2}{2}\phi(\Lambda^{-1 }x)$), we know that the classical dimension of$\phi$is$\Delta\$. Due to ...