# Thinking about propagators

• vertices
In summary, the authors of the excerpts say that in the case of the neutral pion decay, there is no propagator involved, which would produce a factor of dimensionless decay rate. However, this is not the end of the story, as spinor factors coming from the outgoing fermions must also be taken into account.

#### vertices

i was thinking about propagators, and I've got bit of a silly question.

The amplitude for any given decay takes the form:

$$f(q^{2})=\frac{g^{2}}{q^{2}+M_{X}^{2}c^{2}}$$.

for EMAG decays, the Mx=0 (as we're dealing with photons), so the amplitude is just:

$$f(q^{2})=\frac{g^{2}}{q^{2}}$$,

Why do we not consider the propagator: $$\frac{1}{q^{2}}$$ when determining decay rates (we just say its proportional to the fine structure constant to power of the number of vertices)?

Hello.
In the case of EM interactions,
g IS the fine structure constant.. where's your problem ?

Atakor said:
Hello.
In the case of EM interactions,
g IS the fine structure constant.. where's your problem ?

thanks Atakor.

I understand that. However, the gauge boson (the carrier of the electromagnetic force), in this case a massless photon, transfers a momenmtum q. Why is the propagor 1/q^2, ignored when calculating the transition amplitude?

Hello vertices!
In feynman language:

to each vertex we associate the couplig constant : -ig
for each _internal_ line we associate the function you dubbed f(q)

but f(q) is not the amplitude of the transition..
we have to integrate over all momentum space (q) (with conservation of energy-momentum)
after taking into account the functions of the particles (spin 1/2 ->spinors ...)

vertices said:
i was thinking about propagators, and I've got bit of a silly question.

The amplitude for any given decay takes the form:

$$f(q^{2})=\frac{g^{2}}{q^{2}+M_{X}^{2}c^{2}}$$.

for EMAG decays, the Mx=0 (as we're dealing with photons), so the amplitude is just:

$$f(q^{2})=\frac{g^{2}}{q^{2}}$$,

Why do we not consider the propagator: $$\frac{1}{q^{2}}$$ when determining decay rates (we just say its proportional to the fine structure constant to power of the number of vertices)?

We do take that factor into account! Do you have any specific example to discuss? That factor will be there with q having the value of some energy scale in the problem (of order the mass of the decaying particle).

nrqed said:
We do take that factor into account! Do you have any specific example to discuss? That factor will be there with q having the value of some energy scale in the problem (of order the mass of the decaying particle).

Yes, as it happens I do. Please look at the attached files which consider the decays of the pi and pi- mesons. The bit I have highlighted bit in the file "Propagator 2", basically says that, in the case of the neutral pion, which decays electromagnetically, "there is no propagator". I don't get this!

#### Attachments

• Propagator1.pdf
60.9 KB · Views: 194
• Propagator2.pdf
60 KB · Views: 169
A propagator is used when a virtual particle propagates from one vertex to another. In the decay of the pi0, there are no virtual photons, so there are no photon propagators. In decays where you do have a virtual photon, e.g. $$J/\psi \rightarrow \mu^+ \mu^-$$, you have a photon propagator.

A propagator is used when a virtual particle propagates from one vertex to another. In the decay of the pi0, there are no virtual photons, so there are no photon propagators. In decays where you do have a virtual photon, e.g. $$J/\psi \rightarrow \mu^+ \mu^-$$, you have a photon propagator.

vertices said:
Yes, as it happens I do. Please look at the attached files which consider the decays of the pi and pi- mesons. The bit I have highlighted bit in the file "Propagator 2", basically says that, in the case of the neutral pion, which decays electromagnetically, "there is no propagator". I don't get this!

If you look at the Feynman diagram, you will see that there is no gauge boson propagator (no photon propagator or W or Z boson propagator) involved in the neutral pion decay.

However, this is not the end of the story. And the excerpts you provided don't make much sense. You see right away that what they say does not make sense because a decay rate must always have the same dimensions no matter what process is computed. They seem to say that in one case the decay rate goes like 1/M^4 and in the other case it is dimensionless. This is clearly nonsense.

two things are missing to the discussion. First, in the charged pion decay there will be spinor factors coming from the outgoing fermions taht are being produced. These will provide additional dimensionful factors that must be taken into account. These will produce an additional factor with dimension of energy.

Also, in the neutral pion decay there is a propagator in the diagram: it's the propagator of the virtual quark between the two photon vertices! This must be taken into account as well. This wil produce a factor of dimension $$E^{-1}$$. This way the two diagrams will have teh same dimensions (I am setting aside the factors coming from the initial quarks which are presumably of the same order in both diagrams so if we are taking a ratio of the two diagrams we don't have to think about those).

What is the source of those excerpts you provided. I hope it's not the book by D McMahon!

nrqed said:
However, this is not the end of the story. And the excerpts you provided don't make much sense. You see right away that what they say does not make sense because a decay rate must always have the same dimensions no matter what process is computed. They seem to say that in one case the decay rate goes like 1/M^4 and in the other case it is dimensionless. This is clearly nonsense.

Yes indeed it is, thanks for pointing that out!

Also, in the neutral pion decay there is a propagator in the diagram: it's the propagator of the virtual quark between the two photon vertices! This must be taken into account as well. This wil produce a factor of dimension $$E^{-1}$$. This way the two diagrams will have teh same dimensions (I am setting aside the factors coming from the initial quarks which are presumably of the same order in both diagrams so if we are taking a ratio of the two diagrams we don't have to think about those).

aahhh. didn't think of that. here comes a really stupid question - a virtual quark isn't a gauge boson, so how does it/why can it mediate the process in question?

What is the source of those excerpts you provided. I hope it's not the book by D McMahon!

Its "Nuclear and Particle Physics" by Brian Martin. I can overlook its complacency/lack of rigour because it does explain concepts quite well:)