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First I would recommend you to check this video: https://www.youtube.com/watch?v=hHTWBc14-mk The video mentions Renormalization in the context of one-loop Feynman Diagrams. In slide 14 (check slide 12 for a picture) of http://www.physics.indiana.edu/~dermisek/QFT_08/qft-II-11-1p.pdf you can check an example of an infinity (a divergent integral), and also ...

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Q1: The mass splitting he is referring to is between the components of a vector-like quark (VLQ) doublet, $Q=\begin{pmatrix} U \\D \end{pmatrix}$, i.e. between U and D. There is such a splitting because they have different charges, so they get different loop corrections to the tree- level mass term $M_Q \bar{Q}{Q}$. What Sher is saying is simply that we pass ...

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Without the factor $1/2$ for a complex field all observables constructed out of the Lagrangian in the standard way vie Noether theorem, like the energy $H:= \int T_{00} dx$ or the momentum $P_i = \int T_{i0} dx$, turn out automatically to be the ones of a system of identical particles of two types, {\em proper particles} and {\em anti particles}. E.g., $$H ... 2 A mass term in the photon self-energy tensor would look like Ag^{\mu\nu}, where A approaches a constant as q^{2}\rightarrow0. What is important is not the degree of divergence of A, but the Lorentz structure of the term. In particular, a term of the form Bq^{2}g^{\mu\nu} is not a mass term; when contracted with the external fields A^{\mu} and ... 3 Let's look at the analog of the massive photon. The propagator for this is$$ \frac{ g^{\mu\nu} - \frac{p^\mu p^\nu}{m^2} }{ p^2 + m^2 } $$At large energies, this scales like \frac{p^\mu p^\nu }{m^2 p^2} \sim \frac{1}{m^2}. In the same way, the massive graviton propagator takes the general form$$ \frac{g^{\mu\alpha} g^{\nu\beta} + g^{\mu\beta} ...

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To answer my own question after thinking about it for a while: it's because the external lines themselves must follow the same summation procedure - imagine adding the same quantum fluctuations to the external lines. Thus the propagator is written as $P = P^2(1/P)$ where $1/P = A + \lambda B + \lambda^2 C + \cdots$. Clearly to zeroth order in $\lambda$ the ...

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