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  Meaning of defining a quantum field theory

+ 2 like - 0 dislike
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I often heard from people saying (but not often found in textbooks or literature) in perturbative renormalization, we define quantum field theory perturbatively; in lattice field theory, we obtain a non-perturbative definition of quantum field theory.

I wonder what does it mean by saying defining a quantum field theory. Does it mean, in qft, we use some experimental data to predict other experimental data (since we are not in a stage to determine everything from theory), renormalization links to experimental data (physical mass etc), hence renormalization becomes a part of definition of a qft?

Any reference is appreciate

asked Aug 7 in Theoretical Physics by stupidquestions (10 points) [ revision history ]
recategorized Aug 21 by Dilaton

All constants are known from classical experiments. While formulating a QFT (which is a formulation for evoluitions of occupation numbers - that's its definition), we expected more precise calculation results in terms of already known constants. Unfortunately, perturbative corrections give some addenda to the fundamental constants (due to "self-action"), and renormalization is kind of discarding these addenda ("cheating" according to R. Feynman or "redefinition" according to renormalizators). You may read R. Feynman's chapter about electromagnetic mass in his lectures (Chapter 28, if I remember it right).

See the discussions on PhysicsForums starting here and  here.


thanks for the links. I am more interested in less rigorous physical aspects than constructive approaches though.

But it is the same. Lattice QCD and renormalized perturbative QCD with resummation based on the renormalization group are both approximate nonperturbative definitions of QCD. The former would become exact in the limit of zero lattice spacing, while the latter would become exact under full resummation of the perturbative series.

@ArnoldNeumaier is there a proof that the full resummation and the zero lattice spacing would converge to the "right" non-perturbative QCD? Or could we have a situation similar to a functions that can not be represented by their Taylor expansions even when an infinit number of terms is included?

A proof would constitute a rigorous construction - this is a widely open problem. Thus one only has opinions based on partial results that allow different interpretations but are suggestive depending on one"s prior expectations. For QCD there seems to be a consensus to expect convergence. For QED one expects triviality of the lattice limit but convergence (after resummation) of the causal perturbation series - though some believe that a rigorous QED cannot exists and any limit would be trivial.

@ArnoldNeumaier and @Dilaton: Let's distinguish practical calculations, for example, a cross section of Compton scattering in QED $d\sigma({\bf{q}},\text{e},\text{m})/d\Omega$ and "ill defined things" like a connection between a "bare" charge, a real charge, and a cutoff parameter. There is no doubt that the Compton cross section exists and all realistic calculations (summation of soft diagrams) only bring an additional dependence from the detector resolution $\text{E}$: $d\sigma({\bf{q}},\text{e},\text{m},\text{E})/d\Omega$.

As to "ill defined connections", they are out of physical and practical meaning.

1 Answer

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If we are too strict with our definition, defining a QFT is defining a set of n-point functions that satisfy wightman axioms (maybe cluster descomposition isn't necessary; I assume that the QFT is microlocal and defined in a minkowski space-time). I know that my answer isn't valid in perturbative QFT, nor in non-perturbative non-convergent. In four dimensions, we only have free theories and Grosse-whulkenhaar model. I asume that QFTs used in real life are well-defined, even if we don't know what's their definitions.
answered Aug 26 by Iliody [ no revision ]
reshown Sep 11

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