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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 ...


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^{\...


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

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} = \...


1

\begin{equation} \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} \end{equation} By definition, the bare coupling does not depend on the renormalization scale $\mu$. Hence \begin{equation} \...



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