In QFT we calculate quantities that ultimately depend on the way we normalize spinors; but because in perturbation theories spinor propagators are free then their normalization can be chosen arbitrarily: so in perturbative QFT computations give results that depend on arbitrary normalization. It is custom to choose the spinor square scalar equal to 2m and this gives the correct results; but for instance in condensed state physics the normalization would be to choose it equal to the number of particles, which in QFT would result in choosing it equal to 1 and this would be grossly out of scale. So given that 2m seems correct but not the only possibility, is there a way in which such a normalization could be justified a priori?
... ok, but still, in an expression involving products of \overline{\psi} and \psi if I change the normalization of \psi I would change the expression, right? For example, in Peskin & Schroeder, chapter 3, when they talk about plane-waves solutions of the Dirac equation, they explicitly say that the spinor will be multiplied by a factor \sqrt{m} "for future convenience", so I am asking, is this convenience somehow justifiable? And if yes, why in condensed states they employ a normalization that is totally different?
It doesn't matter what normalization is used. The results are independent of the choice. What does matter is that the reference state (here defined by the spinor) is normalizable at all. The mass isn't that of the field-which doesn't make sense-but of the one-particle excitation, that's created by the field. And the mass can be zero, nonetheless the corresponding state is normalizable, just the convention changes.
oh, nice, thanks... on the other hand, how can the results be independent on the choice? The Lagrangian density is proportional to the spinor squared and to get the results we integrate over the volume but not divide by the spinor squared: so the spinor squared must still be present and that is normalization-dependent, right? Can you elaborate on this, please?