- #1
Sunset
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Hi!
As far as I understood Renormalisation you try to find renormalized quantities R1,R2,... which are related to the bare quantities B1,B2,... in the following way:
- Ri is finite when you send Bi to infinity
- you can write every lorentz-invariant-amplitude of the theory in terms of the Ri instead of the Bi
I have one concern about this: is your choice of Ri unique (up to a additive constant)? I mean, is the relation between Ri and Bi unambigously given? What we want to have are physically measurable quantities, which can all derived from lorentz-invariant-amplitudes for processes. If there's only one possible choice for your renormalized quantity, Renormalisation is a method that makes really sense, otherwise it would depend on arbitrariness.
LI-amplitudes are constructed from vertex-function and propagators, so you have to make sure you can write all vertex-functions and propagators in dependence of unambigously defined Ri .
Does anybody know the textbooks of Griffiths or Ryder? So we could discus it explicitly...
Best regards Martin
As far as I understood Renormalisation you try to find renormalized quantities R1,R2,... which are related to the bare quantities B1,B2,... in the following way:
- Ri is finite when you send Bi to infinity
- you can write every lorentz-invariant-amplitude of the theory in terms of the Ri instead of the Bi
I have one concern about this: is your choice of Ri unique (up to a additive constant)? I mean, is the relation between Ri and Bi unambigously given? What we want to have are physically measurable quantities, which can all derived from lorentz-invariant-amplitudes for processes. If there's only one possible choice for your renormalized quantity, Renormalisation is a method that makes really sense, otherwise it would depend on arbitrariness.
LI-amplitudes are constructed from vertex-function and propagators, so you have to make sure you can write all vertex-functions and propagators in dependence of unambigously defined Ri .
Does anybody know the textbooks of Griffiths or Ryder? So we could discus it explicitly...
Best regards Martin