1

I am studying a couple of papers on the subject of neutrinoless double beta decay problem, somewhere in one of these papers, they emphasize that $ΔL=2$ operators have odd dimensions. What is the reason behind this?

https://arxiv.org/abs/2003.08493

Qmechanic
  • 201,751
Ymir Saber
  • 13

1 Answers1

1

It is a just-so fact. The proof of your statement is in Kobach 2016, based on Hypercharge and Lorentz invariance. If you really wish to know, you can sink time into the general proof, but it would distract you from things you must know.

For normal consumers in the street, the ΔΒ=0, ΔL=2 amps in the SM are all underlain by the Weinberg dimension 5 SM-singlet operator, $(LH)^2$; which has all the Lorentz and SM gauge symmetries, but is, ipso facto, unrenormalizable. When you build upon it by the appendage of $H^* H$ singlets, which have dimension 2, you end up with dimensions 7, 9, etc... You can convince yourself even the addition of new freak particles would not modify this rule, provided the gauge symmetries of the SM and Lorentz invariance are respected...

A more explicit, less abstract/elegant illustration of the point is the Wyler-Buchmueller 1985 paper which suffices to make the point by force of exhaustive exemplification... Do you now see how appendage of fermion bilinear singlets (dimension 3) can't work as a counterexample?

Cosmas Zachos
  • 62,595