Graduate Why do different regularization schemes give different answers?

[QFT1995]

My question is why is it okay that two different regularizations of a one loop contribution to the full propagator give two different answers? Are the finite parts for all regularization schemes the same? If that's the case, do the divergent parts only contain information about high energy momenta so the fact that the divergent parts are different just represents the fact that our theory isn't well defined for the high energy modes.

 

[Reggid]

My question is why is it okay that two different regularizations of a one loop contribution to the full propagator give two different answers?

Simply because a propagator is not a physical observable.

It depends on many things, like for example your gauge choice, etc... . So why would you care whether different regularization schemes give different answers?

 

[QFT1995]

Simply because a propagator is not a physical observable.

It depends on many things, like for example your gauge choice, etc... . So why would you care whether different regularization schemes give different answers?

But neither is the wavefunction in QM? I'm sure you would care if you had two different answers for the wavefunction when solving Schrodinger's equation even though it's not physically observable.

Also, you can tell the physical mass of the particle from the propagator so surely it matters if you get two different answers for the propagator.

 

[Reggid]

You cannot compare a propagator in quantum field theory with the wavefunction in non-releativistic quantum mechanics. The wavefunction is "physical" in the sense that you can calculate physical observables directly from it. Different wavefunction ---> different prediction for somehting you can measure in experiment. (But surely you would for example not care if somebody uses a different convention on the phase of the wavefuncion)

But what is a propagtor? It is one expression that appears in the calculation of an amplitude in a perturbative expansion. It is NOT one term in this expansion, it is only part of a calculation that you have not finished yet. You have to connect it to the external states, you have to calculate the full Feynman diagram, you maybe have to include other diagrams in order to get the amplitude at a given order in perturbation theory,... . That's why it is absolutely no problem that the propagator dependends for exmaple on the gauge fixing. One propagator is not a "thing", it is the Green's function of the equations of motion of the free theory. But when you are talking about loop-corrections, you clearly do not have a free theory any more.

Only when you finish your calculation you have to get something that is gauge invariant and independent on unphysical things like a regulator. When your regulator is regulating IR divergences, it will cancel out once you combine real and virtual contributions. But that really means calculating the full cross-section, which invovles much more than just a propagator. When you are talking about UV divergences, then you still have to do the renormalization procedure. Of course your counterterms will look very different depending on the choice of your regulator, such that in the end all divergences cancel.

Here you also have the freedom that you can choose different renormalization schemes. Again things like a propagator can look different depending on which scheme you choose. The result of your full calculation will even look different, but since you defined your renormalized parameters like couplings and masses in a different way, they will have different numerical values, precisely in such a way that the physical prediction does not change.

So yes, the form of a propagator is somewhat arbitrary, depending on choices you do on renormalization schemes and gauge fixing, but as long as you do your whole calculation consistently in that scheme, your physical predictions will not change.

Also, you can tell the physical mass of the particle from the propagator so surely it matters if you get two different answers for the propagator.

Actually, it's the other way round: The knowledge about what the physical mass is (something we only get from experiment) tells you what the propagator has to look like, given that you chose to use the pole mass renormalization scheme for the mass.

But there are also other mass schemes. In QCD for expample there quark propagator has no pole due to non-perturbative effects, so the pole mass is no more "physical" than any other mass scheme.

 

[vanhees71]

My question is why is it okay that two different regularizations of a one loop contribution to the full propagator give two different answers? Are the finite parts for all regularization schemes the same? If that's the case, do the divergent parts only contain information about high energy momenta so the fact that the divergent parts are different just represents the fact that our theory isn't well defined for the high energy modes.

Proper vertex functions are defined up to a finite renormalization prescription. They are not observable objects and thus this doesn't play any role in calculating observable quantities like S-matrix elements to get measurable cross sections. The independence of cross sections on the renormalization scheme is described by renormalization-group equations.

Regularizations are just intermediate calculational steps to evaluate the divergent integrals depicted by Feynman diagrams with loops. They just let you organize the renormalization of the corresponding proper vertex functions. You can as well use renormalization techniques without any regularization at all, e.g., the BPHZ description, where you directly subtract the diverging parts directly in the integrands of the loop integrals.

For details, see my QFT manuscript:

https://itp.uni-frankfurt.de/~hees/publ/lect.pdf

 

[Tendex]

BPHZ renormalization on the other hand is not BRST-invariant, it doesn't bother about gauge fixing and therefore is not concerned with regularization. Not very useful if one is limited to the perturbative domain though.