Renormalized Mass: What is Physically Going On?

[Malamala]

Hello! I am reading Schwarz QFT and I reached the mass renormalization part. So he introduces, after renormalization, a physical mass, defined as the pole of the renormalized propagator, and a renormalized mass which can be the physical mass, but can also take other values, depending on the subtraction schemes used. Are these masses, other than the physical one, observable in any way experimentally, or are they just ways to do the math easier (using minimal subtraction instead of on-shell scheme, for example)? Also, in the case of charge renormalization, the explanation was that due to the vacuum polarization, the closer you are to the charge, the more you see of it, so the charge of the particle increases with the momentum you are testing the particle with. However, I am not sure I understand, from a physical point of view, why do you need to renormalize the mass. Is this physical mass (defined as the pole of the renormalized propagator) the same no matter how close you get, or it also changes with energy? And if it changes, what is shielding it at big distances?

 

[Orodruin]

Anything that can vary depending on the renormalisation scheme (or scale) cannot be physical (i.e., you cannot measure it directly).

However, I am not sure I understand, from a physical point of view, why do you need to renormalize the mass.

In the free theory, the propagator is $i/(p^2 - m^2)$. This has a rather nice and easy functional dependence on $p$. However, in the interacting theory the two-point Green's function will have a different (and more complicated) functional form. The field strength and mass counter-terms are therefore typically chosen in a way such that the two-point function looks as close to the free propagator as possible at the scale you are renormalising.

However, I am not sure I understand, from a physical point of view, why do you need to renormalize the mass.

Because you cannot make predictions with your theory if you do not. The two-point function has a degree of superficial divergence of 1 or 2 (depending on whether you are looking at a fermion or boson field) and you therefore end up with a lot of infinities that you need to deal with.

Is this physical mass (defined as the pole of the renormalized propagator) the same no matter how close you get, or it also changes with energy?

Sorry, but this question does not make sense. The energy scale of the two-point function is given by the external momentum squared. Its pole is what it is (and it is a physical measurable quantity, since it is a feature of the two-point function). If you look at a higher energy (higher $p^2$), then you are not at the pole. It is like asking if the pole position of $1/x$ changes when $x$ is different from zero. Given an external momentum, the two-point function either has a pole there or not.

I think it is also worth pointing out an issue regarding running coupling constants that I always struggled with as a student: What the ****** does it mean? The entire point of renormalisation is to get rid of the infinities that otherwise will plague your observables (but appear alongside formally infinite non-measurable quantities such as the bare parameters). In order to do so, you are free to choose renormalisation conditions to your heart's content as long as you do it properly using physical observables (typically based on the n-point functions of the theory). Now, the n-point functions are physical and should not depend on whatever renormalisation conditions you chose, so regardless of whether you chose to put your renormalisation conditions at 1 GeV or 1 PeV, they should be the same. However, the renormalised parameters that you get out of the renormalisation procedure may (and in general will not) be the same simply because your counter terms will be different, but will typically differ by a (finite) amount. Since the renormalisation conditions are typically chosen in such a way that they are easily relatable to some cross section (through the appropriate n-point function) at the particular energy scale, the cross section at that scale sets the renormalised coupling parameter.

However, the renormalisation scale is not unique and just as you picked one renormalisation scale you could just as well have picked another, but you might have ended up with a different renormalised coupling. Now here is the main point of running coupling constants: Depending on your renormalisation scale, you will have different renormalised couplings and different counter terms. However, this should not affect your physical observables. You should still be describing the same theory. What does this mean? It means that all the predictions of physical measurables should be independent of the renormalisation conditions you chose, or in other words, your n-point functions should not depend on your renormalisation conditions. This ultimately leads you to the Callan-Symanzik equations and beta functions, which tell you how the renormalised parameters vary as you vary the renormalisation conditions. But the renormalisation conditions were chosen such that they correspond to observables at different scales! Hence, you can use the beta functions to relate these physically observable quantities at different scales.

This is a huge deal! If you want to compute a cross section at an energy scale $M_1$, but renormalised your theory at $M_2$, you now have two choices. You can write down all of the diagrams that are needed to compute the appropriate n-point function at the different scale (including diagrams with counter terms) and go through the entire diagrammatic machinery. Since your theory is renormalised you will end up with a finite result. The alternative is to say "Would it not be much more convenient if I had renormalised my theory at $M_1$? I wonder what the coupling constant would have been then because it is just what I need to compute my cross section." Well, lucky you. The beta function tells you exactly how the renormalised coupling constant changes with the renormalisation scale so you just use that to compute the coupling constant at $M_2$, which is directly related to your desired cross section.

 

[Malamala]

Thank you so much for this, it is really illuminating. Just one small thing that still bothers me. I understood the explanation as to why you need to renormalize the mass, how is the propagator changing and why you pick the pole of the propagator as a definition for a mass. I was hoping, however, for a more physical explanation. What I mean is, in the case of charge renormalization, the explanation is due to vacuum polarization (particle-antiparticle pairs appearing on the photon propagator and acting similar to an electric field in a dielectric). The existence of an interaction between fermions and photons is clear, and it is also clear why the charge needs to be renormalized. In the case of the mass, I am not sure I can see the reason for renormalization in the same way. I understand you need to do it, to get finite results, and it makes sense to be different in the interacting theory compared to the free one, but I am not sure I see the actual mechanism by which it is produced i.e. electron-positron pair on the photon line screen the original charge of the particle that produced the photon, as in a dielectric, but why does a photon appearing on the electron propagator changes the mass of the electron?