Renormalization, infinitesimal charges?

[jostpuur]

When we compute scattering amplitude \mathcal{M}, using a coupling constant \lambda, and a cut-off energy \Lambda, it turns out that if \lambda is constant, then \mathcal{M}\to\infty when \Lambda\to\infty.

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

Would it make sense to think, that in fact the \lambda 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, \lambda 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?

 

[kdv]

When we compute scattering amplitude \mathcal{M}, using a coupling constant \lambda, and a cut-off energy \Lambda, it turns out that if \lambda is constant, then \mathcal{M}\to\infty when \Lambda\to\infty.

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

Would it make sense to think, that in fact the \lambda 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, \lambda 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?

I am not sure exactly what your question is but as you point out, the bare coupling constant is not a well defined quantity as the cutoff goes to infinity.

In traditional renormalization, the bare coupling constant has no physical meaning at all and it must be related to some measurable before any comparison to experiment can be done. And then one gets that the bare constant is ill defined in the limit Lambda goes to infinity. This does not bother people since the measurable quantities are finite in that limit.

In the context of effective field theories, Lambda is never taken to infinity anyway so there are no infinities.

I am sure you know all that but I wrote just to put down some thoughts about the topic.

 

[kdv]

When we compute scattering amplitude \mathcal{M}, using a coupling constant \lambda, and a cut-off energy \Lambda, it turns out that if \lambda is constant, then \mathcal{M}\to\infty when \Lambda\to\infty.

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

Would it make sense to think, that in fact the \lambda 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, \lambda 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?

Yes, the relationship between the renormalized coupling constant and the bare coupling constant depends on the order (in number fo loops) at which we are working.

You are right that at zeroth order (tree level), the bare coupling constant is finite. Then, at one loop, it becomes essentially zero as Lambda goes to infinity! So it's kind of crazy. What happens is that a certain assumption is always implicit in those calculations. In the intermediate steps of relating the bare and renormalized coupling constants it is always assumed that Lambda is kept small enough that higher order loops are negligible. For example, in QED, one would assume that \alpha \log (\Lambda) \ll 1. Only after renormalization has been carried out and Lambda has disappeared does one truly take Lambda to infinity.