Neutrinoless double beta decay: annihilation of neutrinos or nucleon absorption? I've a question regarding the interpretation of the neutrinos in neutrinoless double beta decay: 

The basic issue that I have is with papers that regard the decay as a neutrino-neutrino annihilation (as the neutrino is a Majorana particle in this case). The reason is that it can't be classical annihilation because we'd get no end-product (and therefore a loss of energy) like in electron-positron annihilation. 
Now you could say "they are virtual particles so they don't lie on their mass shell", but wouldn't it then be better to avoid this annihilation interpretation completely? Another interpretation could be nucleon-neutrino absorption and isn't that a better one? This process could at least in principal (neutrino-up -> W- -> electron-down) also happen with a "lone", real neutrino.
So what I want to say is that the annihilation interpretation is at least controversial, right (apart from the perturbative character of virtual particles)?
 A: From the particle data group it is shown that no reliable neutrinoless double beta decays have been measured. In any case this would happen if neutrinos were Majorana neutrinos, which is not the standard model hypothesis.

Revised August 2013 by P. Vogel (Caltech) and A. Piepke
  (University of Alabama).
  Neutrinoless double-beta (0νββ) decay would signal violation of total lepton number conservation. The process can be mediated by an exchange of a light Majorana neutrino, or by an exchange of other particles. However, the existence of 0νββ decay requires Majorana neutrino mass, no matter what
  the actual mechanism is. 

There are only limits.
Lepton number conservation is well tested in much cleaner interactions.
In any case in the diagram you give I do not see any annihilation, it would need an extra vertex. I see a W- decaying into a neutrino and an electron , and the Majorana neutrino (which is its own antiparticle) going into W+ ( reading W-the diagram line backwards)  + electron. The probability would be very small because in the nucleus the Ws are very much off shell, and the propagator comes in twice. 
A: Neutrinoless double beta "decay" is not actually a decay but a multi-body scattering ($2\to 4$) where two neutrons scatter into two protons and two electrons. Lepton number is clearly violated by two units. In the example you give, the underlying process where the violation takes place is in the 2-body scattering: $W^- W^- \to e^- e^-$. This could be mediated by a $t$-channel Majorana neutrino or something else, e.g. by a hypothetical doubly-charged $\phi^{--}$ scalar in the $s$-channel (or you can even have two BSM gauge bosons emitted by each neutron scatter into $e^- e^-$). The actual mechanism of how the violation takes place cannot be probed in the experiment, just lepton number violation.
