Suppression of $W$ boson propagator by its mass

Question

In my experimental particle physics introductory class it was often said that quantum electrodynamics (QED) is very predictive for sufficiently small center of mass energys since the $W^\pm$-propagators are suppressed by the mass of the boson.

I am a beginner in quantum field theory (QFT) so if you spot a misconception feel free to correct me in the following.

At tree-level processes the above statement makes sense to me but at next to leading order we integrate over all possible loop momenta. So the $W$-boson propagator should no longer be "that" suppressed by the non vanishing rest mass of the $W$-boson.

Why is it that QED is still predictive in this case?

Answer 1

Good question! Their is lot to unpack to so I will do my best here. I think I can give you a global answer (or try to anyway), but a true answer to your question would involve knowning what process you are looking at.
I am going to assume that you are thinking about $W_\pm/Z$-boson mediated interaction and not a more complicated process with $W$ boson loops such as the higgs decay $H \rightarrow \gamma \gamma$.

For simplicity, let us consider the electrically neutral interaction of $q \overline{q} \rightarrow l \overline{l}$, for a quark/anti-quark pair and lepton/anti-lepton pair. I am thinking here of a photon and $Z$ boson interaction, which is called the Drell-Yan process.

First of all, the observed mass of the mediated boson is the renormalised mass that includes all loop corrections to the propagator. Sure we start with a bare mass, but upon integrating all loop momentum in a propagator, the observed mass is the one used. Nature knows all orders of perturabation theory.

Second, the probabilty of a massive particle being mediated follows the Breit-Wigner Distribution, while a massless particule contributes roughly as $\frac{1}{s}$ for centre of mass energy $s$.
Their is a lot of physics here already but, roughly speaking, one interpretation is that the center of mass-energy of the system determines which process is going to contribute dominantly to the probability amplitude. In the case considered, a photon has $0$ width (does not decay) whereas the $Z$ boson has quite a big width ($ \Gamma\sim 3GeV$), meaning that the peak of the distribution will be around the mass of the $Z$ boson and be litteraly of size $\sim \Gamma$

Here is a graph of the cross-section for the mass of the out state. We see clearly that the probability distribution changes drastically depending on the invariant mass, and we identify 2 parts to the curve: the $Z$ peak and the background $\gamma$.

Finally, I think one should not forget that photons and $W/Z$ bosons are remnants of a $U(1) \times SU(2)$ symmetry. At high enough energies (above the weak scale), these concepts are unified, but I'm don't think these energies are attained yet, because this would mean the Higgs itself could act as a propagator.

Thank you for your question, it's been a while since I have thought about these concepts, so hopefully I have not made any glaring mistakes.

EDIT: For a much more comprehensive and better phrased answer about virtual particles and unstable particles, see this post : Are W & Z bosons virtual or not?

Answer 2

It is not clear what do you mean in statement "QED is very predictive...": QED does not contain any particles except electrons & photons. QED describes only electron-photon proccesses and when you include additional particles (for instance, muon) you obtain a different theory (strictly speaking).

In Standard Model (SM) there is the following fact:

On tree-level we can neglect momenum of intermediate $W$-boson in its propagator and study effective 4-fermion theory (which is called Fermi theory)

This statement can be sketched as $$\frac{1}{k^2+m_W^2}\rightarrow \frac{1}{m_W^2}$$

Of course, this approch falls due to several facts (I do not want to discuss them). In next order (=1-loop level) if you want to study radiative corrections to coupling constants or masses, you should consider 'honest' propagator (as you have noticed). However, in most of these calculations there is virtually always a small parameter (for instance, electron mass) which makes derivation a little bit easier. For instance, if you want to consider correction to decay $Z\rightarrow e^{+}e^{-}$, you should evaluate all of these diagrams,

and use 'honest' propagator when loop contains $W$-boson (for instance).

As was mentioned by Guillaume Trojani, one of the key points of tree-level calculations is using of renormalized couplings constants. For me, it seems useful to sketch derivation of $gg\rightarrow H$ procces on tree-level,

which is dominantly caused by $t$-quarks (due to restriction on $m_H$ mass). Honest and comprehensive derivation is complicated. However, we can easy consider the limit $m_H/(2m_t)\ll 1$. Next step is to set $\alpha_s=\alpha_s(m_t)$ and then notice that $t$-quark obtains his mass by Higgs mechanism, $m_t\rightarrow m_t(1+H/\eta)$. Finally, one should extract the leading order. Resulting expression for amplitude $tt\rightarrow H$ is very precise for $m_H<200$ GeV.

I try to demonstrate than in SM you have a lot of parameters and in many cases complicated proccesses are calculating approximately and sometimes even in 'loop'-processes approximation with point-like $W$-boson will be enough good. May be my last words are wrong, but for me it seems true.

Finally: You are absolutely right that when you integrate over all possilbe momenta of intermediate boson you should use 'honest' propagator. When you consider proper QED you do not know about any other particles except electrons and photons, so with these particles QED is predictive. When you go to higher energies ($>m_W$), you should add other particles and not it is not a QED. After adding particles, you can compare results (for instance $(g-2)$ correction to electron moment) with and without other particles (for instance, muonic contribution to electron $(g-2)$. In most cases due to interplay between particles masses/other parameters results from QED is quite good.