Dimensions of a probability amplitude (matrix element)?

[fliptomato]

Greetings. I'm a little bit confused about the dimensions of a probability amplitude in a QFT calculation. My understanding a the squared, spin averaged\summed Feynman diagram should be dimensionless. However, if we consider a decay process, say pair creation from a photon or the decay of the Higgs to two fermions, |M|^2 is dimensionful.

Quick check: the coupling constant of such a decay is dimensionless. There is no dimensionful factor for the initial state (scalar Higgs or photon). The matrix element is proportional to the external fermion factors. When these get spin averaged/summed we get traces of two quantities with dimensions of mass. Hence the matrix element squared has dimensions of mass squared.

Am I losing my mind?

 

[Rach3]

What about the integral over all of space (or momentum space)? Factors of dx^4 or dp^4 are not dimensionless.

 

[fliptomato]

At tree level there is no momenum integral because there are no propagators. Heuristically, we have a particle at rest (say, a Higgs) with a given 4 momentum, that decays into a particle and antiparticle. By symmetry the decay products have equal masses/energies and opposite 3-momenta, hence all momenta in the system are constrained. the amplitude is just proportional to a dimensionless coupling times the external fermion factors. When this is summed for spins, the fermion factors become a trace of something with dimensions of mass squared.

 

[nrqed]

Greetings. I'm a little bit confused about the dimensions of a probability amplitude in a QFT calculation. My understanding a the squared, spin averaged\summed Feynman diagram should be dimensionless. However, if we consider a decay process, say pair creation from a photon or the decay of the Higgs to two fermions, |M|^2 is dimensionful.

Quick check: the coupling constant of such a decay is dimensionless. There is no dimensionful factor for the initial state (scalar Higgs or photon). The matrix element is proportional to the external fermion factors. When these get spin averaged/summed we get traces of two quantities with dimensions of mass. Hence the matrix element squared has dimensions of mass squared.

Am I losing my mind?

Even if you are at tree level, you need to integrate over the phase space to get the total cross section or the total decay rate. The phase space factors are different in the two cases (see eqs 4.79 and 4.86 in Peskin and Schroeder for example).

In addition, a decay rate has different units than a cross section. A cross-section has dimensions of 1/(energy)^2 (in natural units, [itex] c= \hbar=1[\itex]). A decay rate is in $s^{-1}$ in conventional units, so it must have the dimensions of (energy)^1 in natural units.

So it should not be surprising that the matrix element squared has different dimensions!

As you can see from the eqs in P&S, the phase space for the decay rate has the dimension of energy times the phase space of the cross section
(that comes fom the difference in phase space factors. For the differential cross section there is a factor ${1 \over (2 E_A E_B (v_A-v_B))}$ where A and B refer to the two initial particles and the v_A-v_B is the relative speed whereas in the expression for the differential decay rate one has only ${1 \over 2E_A} = {1\over 2 m_A}$ in the rest frame of the decaying particle).

Therefore, since the squared of the matrix element is dimensionless for the cross section but has dim of energy squared for the decay rate, as you pointed out, one finds
$$dim(d\Gamma)= E^3 \,\,\, dim(d\sigma)$$
Since a cross section has dimensions of 1/(energy squared), this means that the decay rate will have dimensions of energy. Putting back the factors of hbar, this will turn into an inverse time, i.e $s^{-1}$ which is of course what is required for a decay rate.

Hope this helps.

Patrick