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I would like to stress the difference between 1) Perturbative Renormalization 2) Non-perturbative Renormalization By Perturbative Renormalization I mean removing infinities from the computation of a correlator/amplitude, order by order. This is done by introducing counterterms, i.e. re-writing the bare parameters of the lagrangian as $\lambda_{Bare} = \... 0 The emergence can't be captured, the emergence is just a phenomenon in our mind (which is a phenomenon too). You can just capture the transition stages. Capture them is finding some non-concern phenomena that do not show anything. 0 I found a paper that helped me a bit in understanding how things work. So here is what I understood. Given a random variable we define a formal power series and a formal derivation such that: $$\frac{\partial}{\partial U}\left(\sum_{n=0}^\infty a_nU^n\right)=\sum_{n=0}^\infty (n+1)a_{n+1}U^n$$ The "usual" Wick product$:\ :$is ... 0 To understand why renormalization may work you might first consider simpler situations in Classical/Quantum Mechanics. In this case there are explicitly solvable toy models where one can see exactly what happens and why. See my paper "A Toy Model of Renormalization and Reformulation" on arXiv. About how to cope with growing terms, see my short note here. 8 Renormalization is always needed when the Hamiltonian is singular. Singular means that the formal expression for the Hamiltonian resulting from the interaction specified is not a self-adjoint operator in a dense domain. Then the dynamics is formally ill-defined and must be renormalized by taking care to represent everything properly as a limit that makes ... 2 The trick is in the introduction of a renormalization scale. Once the perturbation theory has been regularized, one obtains a momentum (and cut-off) dependent interaction of the (schematic) form in 4D $$\lambda(p)=\lambda_0+\alpha\lambda_0^2 \ln(\Lambda^2/p^2),$$ where$\lambda_0$is the bare interaction, and$\alpha$some numerical factors. What one ... 2 Before I try to answer your question, one thing: Does Ryder really calculate the$\mathcal{O}(\lambda^2)$to the propagator as the first contribution in perturbation theory, because there is actually a$\mathcal{O}(\lambda^1)$to the propagator and the$\mathcal{O}(\lambda^2)$-loop is as far as I am concerned a two-loop diagram, i.e. having two loop momenta,... 1 $$\begin{split} \frac{d(g_b\mu^{\epsilon})}{d\log\mu^2}&=\frac{\mu}{2}\frac{d(g_b\mu^{\epsilon})}{d\mu}\\ &=\frac{\mu}{2}\left[\mu^{\epsilon}\frac{dg_b}{d\mu}+g_b\frac{d\mu^{\epsilon}}{d\mu}\right] \end{split}$$ By definition, the bare coupling does not depend on the renormalization scale$\mu$. Hence \... 3 The infinities of set theory aren't the same kind of infinities addressed with regulatisation or renormalisation. For example, if$\kappa$is cardinal (be it finite or transfinite, which for some reason is the name used rather than infinite) then$2^\kappa>\kappa>0$, but the "infinities" we regularise or renomalise are$\pm\infty$, satisfying$2^{\...