When we compute scattering amplitude [itex]\mathcal{M}[/itex], using a coupling constant [itex]\lambda[/itex], and a cut-off energy [itex]\Lambda[/itex], it turns out that if [itex]\lambda[/itex] is constant, then [itex]\mathcal{M}\to\infty[/itex] when [itex]\Lambda\to\infty[/itex].(adsbygoogle = window.adsbygoogle || []).push({});

The idea of renormalization seems to be, that we relate some physical coupling constant [itex]\lambda_p[/itex] to the original constant [itex]\lambda[/itex] in such way, that if we demand [itex]\lambda_p[/itex] to be a constant, then [itex]\lambda\to 0[/itex] when [itex]\Lambda\to\infty[/itex], and at the same time we get some finite values for [itex]\mathcal{M}[/itex].

Would it make sense to think, that in fact the [itex]\lambda[/itex] was some kind of infinitesimal constant in the first place, and it cancels the divergences coming from some integrals in the scattering amplitude?

This doesn't make sense fully to me, because in the first order approximation, [itex]\lambda[/itex] must be left finite, but still..... well its value depends on the order of approximation used? And its infinitesimal when we use higher order approximations?

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# Renormalization, infinitesimal charges?

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