Wigner teaches us that the mass of a particle is the Casimir invariant of the corresponding representation of the Poincaré group. I don't understand how the bare and renormalised masses in QFT are related to representation theory.

(EDIT1) The question is probably too vaguely stated, but I would appreciate any hint.

Do I correctly understand that the Hilbert space of the interacting theory contains the Hilbert space of the free theory (asymptotic states), thus the free theory is a sub-representation of the interacting theory?

(EDIT2) Recently I observed a comment by Arnold Neumaier (https://physics.stackexchange.com/users/7924/arnold-neumaier) in the discussion of the question related to renormalisation (https://physics.stackexchange.com/q/22524) that states: "...(Renormalisation) gives a family of representations of the field algebra depending on an energy scale. All these representations are equivalent, due to the renormalization group...".

I am used to the fact that the renomalisation is introduced as a tool for making divergent perturbation series terms finite and the renormalisation scale appearing in the calculations being unphysical parameter. Nevertheless, the renormalisation invariance seems to be an intrinsic property of QFT independent of the approach used to solve the theory.

I don't know much about it, but I guess, in the models that can be solved exactly one can show that masses and couplings do depend on the scale, although one doesn't have to resort to divergent integrals argument. In particular, there must be a sort of renomalisation in free QFTs, although it is not apparent to me. I guess, this has to be related to the fact that one can split the mass term into "true" mass and "perturbation" pieces arbitrarily.

Now, following Arnold's comment, I wonder if there a way to see from onset that QFTs have a one-parameter family of representations resulting in identical predictions for observable quantities. I guess this is shown in other approach to renormalisation, namely BPHZ/Epstein-Glaser, where the renormalization is a way of defining products of distributions at coinciding space-time points. I suppose there is a sort of arbitrariness in such a definition that results in the renormalisation group, eventually. I must admit, I have rather surface knowledge in this area and would very much appreciate a bird's eye view explanation.


1 Answer 1


In causal perturbation theory (i.e., the Epstein-Glaser approach), the distribution splitting procedure leaves naturally some freedom forming a low-dimensional manifold of possibilities. These possibilities can be parameterized by specifying appropriate renormalization conditions matching masses, response coefficients, etc. These renormalization conditions are not uniquely specified but are usually chosen by convenience.

The (exact) renormalization group is discussed, e.g., in Section 4.8 of the 2nd edition of Scharf's book on QED. The main body discusses the renormalized construction with conditions imposed at zero mass, one section shows how one gets an extra mass parameter by modifying the renormalization conditions. This is just a reparamerization of the manifold of renormalized coupling constants, hence has no effect on the theory. The exact renormalization group equation is just the specification of the infinitesimal reparemeterizing transformation.


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