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I am considering the perturbative renormalization of a simple non-phenomenological QFT with Lagrangian ${\cal L}$ (for scalar fields with multiple generations). I understand that I can renormalize it, if I can, by choosing a (finite) number of counterterms ${\cal L}_{ct}$ whose divergent parts are chosen such that they cancel the UV divergencies in all graphs at the considered order of perturbation theory. Usually one is interested primarily in the cancellation of the divergencies and one would choose a renormalization prescription like MS-bar or on-shell scheme to fix the finite parts of the counterterms.

Here I am merely interested in the consequences of the freedom to choose the counterterms and the relation with fundamental (symmetry) properties of the Lagrangian (beyond the existence of the renormalization group), like hermiticity, C, P, T symmetries and the freedom to change the flavour basis for the different generations. It seems to be obvious to me that I can choose to shift finite parts to my liking between basic Lagrangian ${\cal L}_{B}$ and counterterms as long as ${\cal L}={\cal L}_B+{\cal L}_{ct}$ stays the same (because it will obviously have the same symmetries) if I add to my set of Feynman rules the rules implied by the counterterms. In particular, it seems, I can choose ${\cal L}_{B}$ and ${\cal L}_{ct}$ such that each of them separately violates fundamental properties of the total Lagrangian.

But after renormalization one usually ignores ${\cal L}_{ct}$ part and works with ${\cal L}_{B}$, while using a modified propagator with renormalized self-energy and renormalized expressions for the vertices. My question is to which extent the resulting perturbation theory will yield Greens-functions which exhibit the properties implied by the 'symmetries' of the bare Lagrangian. Will they, for instance, satisfy unitarity relations even if ${\cal L}_{B}$ is not hermitian.

A slightly related question: Is there a generalized procedure to construct counterterms along the logic of $\phi_0 = Z^{1/2}\phi$, $m_0=Z^{-1/2}(m+\delta m)$, $g_0=Z^{-3/2}g$ (in particular in view of multiple generations where such transformations are matrices) and what are the implications of choosing these?

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  • $\begingroup$ It's hard to see why you think that one "ignores" the counterterm part of the Lagrangian after renormalization. After renormalization, one works with the full Lagrangian and ignores the split into the two parts - this splitting was just an auxiliary tool to do some intermediate calculations. The bare Lagrangian and the total one typically look structurally the same but it's the total action, not its part, that governs the symmetries of physics and all other physically meaningful properties of physics. $\endgroup$ – Luboš Motl Jul 28 '13 at 8:56
  • $\begingroup$ Yes, I understand this. What I meant here is that the set of basic Feynman graphs (propagators and vertices) is taken to be the same as for the bare theory. This is because, as you say, the structure of the Lagrangian stays the same. Obviously the rules (analytic expressions) for the Feynman graphs are different, such that perturbative results computed based on the new rules are the same (possibly up to higher orders), as if I would take the rules for the basic Lagrangian plus that for the counterterms. $\endgroup$ – highsciguy Jul 28 '13 at 10:57
  • $\begingroup$ My problem is probably to see how 'symmetries' are respected in arbitrary Greens functions if I 'absorb' the counterterm rules in that for the basic Lagrangian. Surely my questions are very basic, but I might be a bit irritated because these concerns are largely irrelevant for the simple models treated in the literature. $\endgroup$ – highsciguy Jul 28 '13 at 10:59

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